# Fourier Series and the Fourier Transform

• Boris Makarov
• Anatolii Podkorytov
Chapter
Part of the Universitext book series (UTX)

## Abstract

Sections 10.1 and 10.2 are devoted to the general theory of orthogonal systems. Besides basic results (Bessel’s inequality, Riesz–Fischer theorem, etc.) we consider various examples of orthogonal systems (trigonometric system, Rademacher functions, Legendre polynomials, etc.). At the end of Sect. 10.2, we consider orthogonal series of independent functions, which play an important role in probability theory.

In Sects. 10.3 and 10.4 we discuss facts related to trigonometric Fourier series. For such series, we establish convergence conditions, the possibility of termwise integration, and the uniqueness theorem (for both summable functions and measures). We consider summation methods for Fourier series (including the classical methods of Fejér and Abel–Poisson) and the corresponding approximate identities. At the end of Sect. 10.4, we discuss comparatively recent results showing that the results concerning trigonometric series for functions of one variable cannot be carried over to Fourier series of functions of several variables. The rest of Sect. 10.4 is devoted to multiple Fourier series. Along with the counterparts of certain statements in the one-dimensional case, we discuss some facts (failure of the localization principle, etc.) showing that certain classical results cannot be carried over to the multi-dimensional case.

Section 10.5 is devoted to the Fourier transform, which is one of the most important concepts in harmonic analysis. We consider both $$\mathcal{L}^{1}$$ and $$\mathcal{L}^{2}$$- theory of the Fourier transform, and, in particular, prove the inversion formula, Plancherel’s theorem, and the uncertainty principle. We also study the Fourier transforms of finite Borel measures. Using the Fourier transform, we prove that the $$\mathcal{L}^{1}$$-norms of the Dirichlet kernels for balls (such kernels arise when studying the multiple Fourier series) have power rate of growth.

In Sect. 10.6, we discuss the Poisson summation formula and its applications. In particular, we show how this formula can be used to estimate the number of integer points in a ball (Gauss’s problem).

## Keywords

Multiple Fourier Series Arbitrary Orthogonal System Trigonometric System Poisson Summation Formula Approximate Identity
These keywords were added by machine and not by the authors. This process is experimental and the keywords may be updated as the learning algorithm improves.

## 10.1 Orthogonal Systems in the Space Open image in new window In the present section, we consider only the norm in the space . For brevity, we denote it by ∥⋅∥ without index.

### 10.1.1

The norm in the space has an important specific feature: just like a norm in a finite dimensional Euclidean space, it is generated by a scalar product. The scalar product of functions f and g belonging to the (in general, complex) space is defined by the formula
$$\langle f,g\rangle=\int_Xf\overline{g}\,d\mu$$
(the product $$f\,\overline{g}$$ is summable since $$2|f\,\overline{g}|\leqslant|f|^{2}+|g|^{2}$$).
Obviously, $$\langle g,f\rangle=\overline{\langle f,g\rangle}$$ and 〈f,f〉=∥f2. Moreover, by the Cauchy–Bunyakovsky inequality, we have |〈f,g〉|⩽∥f∥ ∥g∥, which implies the continuity of the scalar product with respect to convergence in norm. Indeed, if $$f_{n}\underset{n\to\infty}{\longrightarrow} f$$ and $$g_{n}\underset{n\to\infty}{\longrightarrow} g$$, then
\begin{aligned} \big|\langle f_n,g_n\rangle-\langle f,g\rangle\big|\leqslant& \big| \langle f_n-f,g_n\rangle\big|+\big|\langle f,g_n-g \rangle\big|\\\leqslant& \big\|f_n-f\big\|\|g_n\|+\|f\| \|g_n-g\|\underset{n\to\infty}{\longrightarrow} 0. \end{aligned}

From the continuity of the scalar product, it follows that the scalar multiplication of a series convergent in norm by a function can be carried out termwise, $$\langle\sum_{n=1}^{\infty} f_{n},g\rangle=\sum_{n=1}^{\infty}\langle f_{n},g\rangle$$. To verify this, it is sufficient to pass to the limit in the equation $$\langle\sum_{n=1}^{k} f_{n},g\rangle=\sum_{n=1}^{k}\langle f_{n},g\rangle$$ (the limit on the left-hand side of the equation exists since the series converges and the scalar product is continuous).

We point out one more property of the norm in , the so-called parallelogram identity
which is connected with the fact that the norm is generated by a scalar product.

The reader can easily verify that if a measure is non-degenerate (more precisely, if there exist two disjoint sets of positive finite measure), then in each space with p≠2 the parallelogram identity is violated.

### 10.1.2

In the presence of a scalar product, as in a finite-dimensional Euclidean space, we can introduce the notion of the angle between vectors. We are not going to do this in the general setting, instead restricting ourselves to the most important case where the angle is π/2. We introduce the following definition.

### Definition

We remark that if 〈f,g〉=0, then also $$\langle g,f\rangle=\overline{\langle g,f\rangle}=0$$, and so the orthogonality relation is symmetric. We denote it by fg. A function that is zero almost everywhere is orthogonal to every function in and, obviously, the converse is also true. For orthogonal functions the Pythagorean 1 theorem is valid: if fg, then ∥f+g2=∥f2+∥g2. This result remains valid for an arbitrary number of pairwise orthogonal summands: if f j f k for jk (j,k=1,…,n), then
$$\|f_1+\cdots+f_n\|^2=\|f_1 \|^2+\cdots+\|f_n\|^2.$$
(1)
Indeed, since 〈f j ,f k 〉=0 for jk, we have
$$\|f_1+\cdots+f_n\|^2=\langle f_1+\cdots+f_n,f_1+\cdots+f_n \rangle= \sum_{j,k=1}^n\langle f_j,f_k\rangle=\sum_{k=1}^n \|f_k\|^2.$$
The Pythagorean theorem is also valid for an “infinite number of summands”. If functions f 1,f 2,… are pairwise orthogonal and the series $$\sum_{k=1}^{\infty} f_{k}$$ converges, then
$$\Bigg\|\sum_{k=1}^{\infty} f_k \Bigg\|^2=\sum_{k=1}^{\infty} \|f_k\|^2.$$
(1′)
For the proof, it remains only to pass to the limit in Eq. (1).

Due to the scalar product, every n-dimensional space L contained in is isomorphic (as a Euclidean space) to $$\mathbb{R}^{n}$$ or $$\mathbb{C}^{n}$$ (depending on the field of scalars under consideration). Therefore, we can speak of the orthogonal projection of a function f onto a subspace L. In particular, the projection of f onto the one-dimensional subspace generated by the unit vector e, is 〈f,ee.

In the space , the families of pairwise orthogonal functions play a role similar to that of the orthogonal bases in finite dimensional Euclidean spaces.

### Definition

A family of functions {e α } αA is called an orthogonal system (briefly, OS) if e α e α for αα′ and ∥e α ∥≠0 for every αA. An orthogonal system is called orthonormal if ∥e α ∥=1 for every αA.

It follows immediately from the Pythagorean theorem (1) that the functions from an OS are linearly independent. Obviously, dividing each element of an orthogonal system by its norm, we obtain an orthonormal system.

Let the functions e 1,…,e n form an OS, and let L be the subspace generated by e 1,…,e n (i.e., the set of all linear combinations of these functions). It is important to know how to find the best approximation to a given function f by elements of L. The following theorem gives a solution of this extremal problem.

### Theorem

The minimum value of the norm $$\|f-\sum_{k=1}^{n}a_{k}e_{k}\|$$ is attained if and only if a k =c k (f), where
$$c_k(f)=\frac{\langle f,e_k\rangle}{\|e_k\|^2}\quad(k=1,\ldots,n).$$
(2)
The function $$f-\sum_{k=1}^{n}c_{k}(f)e_{k}$$ is orthogonal to every element of L.

Thus, the function $$\sum_{k=1}^{n} c_{k}(f)e_{k}$$ is the best approximation for f in the set L. The above-stated theorem can be regarded as a generalization of the following well-known fact of school geometry: “the perpendicular dropped from a point f to L”, i.e., the difference $$f-\sum_{k=1}^{n} c_{k}(f)e_{k}$$, is shorter than any “slant” $$f-\sum_{k=1}^{n}a_{k}e_{k}$$.

### Proof

We begin with the second assertion of the theorem. We put $$S_{n}= \sum_{k=1}^{n} c_{k}(f)e_{k}$$ and verify that $$f-S_{n}\perp\sum_{k=1}^{n} a_{k}e_{k}$$. It is sufficient to prove that fS n e m for all m=1,…,n. Indeed,
\begin{aligned} \langle f-S_n,e_m\rangle=& \langle f,e_m \rangle-\langle S_n,e_m\rangle= \langle f,e_m \rangle-\sum_{k=1}^nc_k(f) \langle e_k,e_m\rangle\\=& \langle f,e_m \rangle-c_m(f)\|e_m\|^2 =0. \end{aligned}
The last equality holds by the definition of c m (f).
Now, the extremal property of the sum S n follows from the Pythagorean theorem. Indeed, if $$g=\sum_{k=1}^{n}a_{k}e_{k}$$ is an arbitrary function L, then S n gL, and, consequently, fS n S n g. Therefore, by the Pythagorean theorem, we obtain
\begin{aligned} \|f-g\|^2=&\big\|(f-S_n)+(S_n-g) \big\|^2=\|f-S_n\|^2+\|S_n-g\|^2 \\=&\|f-S_n\|^2+\sum_{k=1}^n\big|a_k-c_k(f)\big|^2 \|e_k\|^2. \end{aligned}
(3)
From this it follows that
$$\Bigg\|f-\sum_{k=1}^na_ke_k \Bigg\|^2\geqslant \Bigg\|f-\sum_{k=1}^nc_k(f)e_k \Bigg\|^2,$$
and the equality holds only in the case where a k =c k (f) for all k. □
For g=0 Eq. (3) takes the form
$$\|f\|^2= \Bigg\|f-\sum_{k=1}^nc_k(f)e_k \Bigg\|^2+ \sum_{k=1}^n\big|c_k(f)\big|^2 \|e_k\|^2,$$
and, therefore, the Bessel 2 inequality
$$\sum_{k=1}^n\big|c_k(f)\big|^2 \|e_k\|^2\leqslant\|f\|^2$$
(4)
holds.

### 10.1.3

Let $$\{e_{n}\}_{n\in\mathbb{N}}$$ be an OS in the space . Obviously, there are functions in that cannot be represented as linear combinations of functions e n . Therefore, the question naturally arises, what are the conditions under which a function is the sum of a series of the form $$\sum_{n=1}^{\infty} a_{n}e_{n}$$. From the theorem proved above it follows that such a series can converge to f only if it coincides with the series $$\sum_{n=1}^{\infty} c_{n}(f)e_{n}$$ whose coefficients are calculated by formula (2). Indeed, Eq. (3) shows that if a m c m (f) and nm, then
$$\Bigg\|f-\sum_{k=1}^na_ke_k \Bigg\|\geqslant\big|a_m-c_m(f)\big|\|e_m \|^2>0,$$
and, therefore, the series $$\sum_{n=1}^{\infty} a_{k}e_{k}$$ cannot converge to f.

The series with coefficients calculated by formula (2) play an important role, which justifies the following definition.

### Definition

Let $$\{e_{n}\}_{n\in\mathbb{N}}$$ be an orthogonal system, and let . The numbers c n (f) obtained by formula (2) are called the Fourier 3 coefficients, and the series $$\sum_{n=1}^{\infty} c_{n}(f)e_{n}$$ is called the Fourier series of f with respect to the given OS.

As we will see, the Fourier series of an arbitrary function converges in the norm ∥⋅∥ (but not necessarily to f).

In the case of an orthonormal system, formula (2) becomes simpler and takes the form c n (f)=〈f,e n 〉. If an orthogonal system $$\{e_{n}\}_{n\in \mathbb{N}}$$ is not orthonormal, then we can pass to the system $$\widetilde{e}_{n}=e_{n}/ \|e_{n}\|$$ (to “normalize” the given system). The Fourier coefficients, obviously, can change, but the terms of the Fourier series do not change as the following relation shows:
$$c_n(f)e_n= \biggl\langle f,\frac{e_n}{\|e_n\|} \biggr\rangle\frac{e_n}{\| e_n\|} =\langle f,\widetilde{e}_n\rangle\widetilde{e}_n.$$
Thus, the terms of the Fourier series of a function f are simply the projections of f onto the lines generated by the elements of the orthogonal system.
Passing to the limit in Bessel inequality (4) as n→∞, we obtain the estimate
$$\sum_{k=1}^{\infty}\big|c_k(f)\big|^2 \|e_k\|^2\leqslant\|f\|^2$$
(4′)
also called Bessel’s inequality. As follows from (1′), inequality (4′) becomes an equality if $$f=\sum_{n=1}^{\infty} c_{n}(f)e_{n}$$.

### 10.1.4

We do not yet know whether a Fourier series converges or, in the case of convergence, what its sum is. The following important theorem establishes that the sum of a Fourier series always exists. As a preliminary, we prove the following lemma.

### Lemma

Let $$\{e_{n}\}_{n\in\mathbb{N}}$$ be an orthogonal system. A series
$$\sum_{n=1}^{\infty} a_ne_n$$
(5)
converges in norm if and only if
$$\sum_{n=1}^{\infty}|a_n|^2 \|e_n\|^2<+\infty.$$
(5′)
In the case of convergence, series (5) is the Fourier series of its sum.

### Proof

Let S n and T n be the partial sums of series (5) and (5′), respectively. Then, for all $$n,p\in\mathbb{N}$$, we have
$$\|S_{n+p}-S_n\|^2= \Bigg\|\sum_{k=n+1}^{n+p}a_ke_k\Bigg\|^2= \sum_{k=n+1}^{n+p}|a_k|^2 \|e_k\|^2=T_{n+p}-T_n.$$
It follows that the partial sums of series (5) and (5′) are fundamental simultaneously. Since the space is complete (see Theorem 9.1.3), we obtain the first assertion of the lemma. The concluding assertion follows from the fact that scalar multiplication of a convergent series by a function can be performed termwise, i.e., if S is the sum of series (5), then the relation
$$\langle S,e_m\rangle=\sum_{n=1}^{\infty} a_n\langle e_n,e_m\rangle= a_m \|e_m\|^2$$
is valid for every $$m\in\mathbb{N}$$. Thus, a m =c m (S) for all m, i.e., series (5) is the Fourier series of its sum. □

### Theorem

(Riesz–Fischer4)

For every orthogonal system $$\{e_{n}\}_{n\in\mathbb{N}}$$, the Fourier series of a function converges in norm and
$$f=\sum_{n=1}^{\infty} c_n(f)e_n+h, \quad\text{where} \ h\perp e_n \ \text{for all}\ n\in\mathbb{N}.$$
(6)

### Proof

By Bessel’s inequality, we obtain $$\sum_{n=1}^{\infty}|c_{n}(f)|^{2}\|e_{n}\|^{2}\leqslant\|f\|^{2}<+\infty$$, and so the series $$\sum_{n=1}^{\infty} c_{n}(f)e_{n}$$ converges by the lemma. Let S be its sum. By the second assertion of the lemma, we have c n (f)≡c n (S). Therefore, the Fourier coefficients of the difference h=fS are zero, i.e., he n for all n. □

### 10.1.5

Obviously, the sum of the Fourier series may not coincide with the function generating this series. For example, if we replace an OS e 1,e 2,… by the system e 2,e 3,… obtained by deleting the first vector, then the Fourier coefficients of the function e 1 with respect to the new system are zeros, and e 1 is not equal to the sum of its Fourier series (with respect to the new system).

### Definition

An orthogonal system $$\{e_{n}\}_{n\in\mathbb{N}}$$ is called a basis if every function in coincides with the sum of its Fourier series almost everywhere.

If $$\{e_{n}\}_{n\in\mathbb{N}}$$ is a basis, then, by (1′), the relation $$f=\sum_{n=1}^{\infty} c_{n}(f)e_{n}$$ implies that $$\|f\|^{2}=\sum_{n=1}^{\infty}|c_{n}(f)|^{2}\|e_{n}\|^{2}$$. Thus, for a basis, the Bessel inequality becomes an equality. We will prove that this property characterizes a basis.

We remark that if $$\{e_{n}\}_{n\in\mathbb{N}}$$ is a basis, then the scalar product of two functions can be calculated by their Fourier coefficients since
$$\langle f,g\rangle= \Biggl\langle\sum_{n=1}^{\infty} c_n(f)e_n,g \Biggr\rangle= \sum _{n=1}^{\infty} c_n(f)\langle e_n,g\rangle= \sum_{n=1}^{\infty} c_n(f)\overline{c_n(g)}\|e_n\|^2.$$
This relation (as well as the special case where g=f) is called Parseval’s 5 identity.

We introduce one more important property which, like Parseval’s identity, is characteristic for a basis.

### Definition

A family of functions {f α } αA in is called complete if the condition
implies that f=0 almost everywhere, i.e., ∥f∥=0.

### Lemma

A family {f α } αA is complete if the set of all linear combinations of functions contained in this family is everywhere dense, i.e., if, for every function and every ε>0, there exists a linear combination $$g=\sum_{k=1}^{n} c_{k} f_{\alpha _{k}}$$ such thatfg∥<ε.

### Proof

Let ff α for each α. If ∥f∥≠0, then there is a function $$g=\sum_{k=1}^{n} c_{k} f_{\alpha _{k}}$$ such that ∥fg∥<∥f∥. Since fg, we obtain a contradiction:
$$\|f\|^2>\|f-g\|^2=\|f\|^2+\|g \|^2\geqslant\|f\|^2.$$
□

### Theorem

(On the characterization of bases)

Let $$\{e_{n}\} _{n\in \mathbb{N}}$$ be an orthogonal system. The following conditions are equivalent:
1. (1)

the system $$\{e_{n}\}_{n\in\mathbb{N}}$$ is a basis;

2. (2)

for every function , Parseval’s identity $$\sum_{n=1}^{\infty}|c_{n}(f)|^{2}\|e_{n}\|^{2}=\|f\|^{2}$$ holds;

3. (3)

the system $$\{e_{n}\}_{n\in\mathbb{N}}$$ is complete.

### Proof

We prove the chain of implications (1)⇒(2)⇒(3)⇒(1).

(1)⇒(2) This implication was proved just after the definition of a basis.

(2)⇒(3) Assume that fe n , i.e., c n (f)=0 for all n=1,2,…. By hypothesis, $$\|f\|^{2}=\sum_{n=1}^{\infty}|c_{n}(f)|^{2}\| e_{n}\|^{2}=0$$, which means that the system $$\{e_{n}\}_{n\in\mathbb{N}}$$ is complete.

(3)⇒(1) Let . By the Riesz–Fischer theorem, f=g+h, where $$g=\sum_{n=1}^{\infty} c_{n}(f)e_{n}$$ and he n for all n. Since the system is complete, we obtain that h=0 almost everywhere. Taking account of the arbitrariness of f, we obtain that the OS in question is a basis. □

Comparing the theorem with the preceding lemma, we see that the following statement is valid.

### Corollary

An orthogonal system $$\{e_{n}\}_{n\in\mathbb{N}}$$ is complete if and only if the set of all linear combinations of the functions contained in this system is everywhere dense.

### 10.1.6

We will see in the next section (see also Sect. 10.2) that it is often convenient to label naturally arising orthogonal systems not by positive integers but by some other indices. Therefore, it is useful to generalize the definition of the Fourier series and coefficients. Let {e α } αA be an arbitrary OS in the space , and let . As above, the numbers $$c_{\alpha }(f)=\frac{\langle f,e_{\alpha }\rangle}{\|e_{\alpha }\|^{2}}$$ will be called the Fourier coefficients of the function f with respect to the given OS. Since Bessel’s inequality $$\sum_{k=1}^{n}|c_{\alpha _{k}}(f)|^{2}\|e_{\alpha _{k}}\|^{2}\leqslant\|f\|^{2}$$ is valid for every finite set of indices α 1,…,α n , the family {|c α (f)|2e α 2} αA is summable (see Sect. 1.2.2). Therefore, the set A f of indices of the non-zero coefficients c α (f) is at most countable (see Sect. 1.2.2), which, after enumeration, can be written in the form {α 1,α 2,…}. By the Riesz–Fischer theorem, the series $$\sum_{k=1}^{\infty} c_{\alpha _{k}}(f)e_{\alpha _{k}}$$ converges, and its sum will also be called the sum of the Fourier series of f with respect to {e α } αA . To verify that the sum is well-defined, we must prove that different enumerations of the set A f give the same sum. A change of enumeration of the set A f results in a series obtained by rearranging the terms of the series $$\sum_{k=1}^{\infty} c_{\alpha _{k}}(f)e_{\alpha _{k}}$$. Therefore, it is sufficient to prove the following auxiliary statement.

### Lemma

Let $$\{e_{n}\}_{n\in\mathbb{N}}$$ be an orthogonal system and $$\omega:\,\mathbb{N}\to\mathbb{N}$$ be a bijection. Then the series
\begin{aligned} &(\mathrm{a}) \quad \sum_{n=1}^{\infty} a_ne_n \quad \textit{and} \\&(\mathrm{b}) \quad \sum_{k=1}^{\infty} a_{\omega(k)}e_{\omega(k)} \end{aligned}
converge simultaneously and, in the case of convergence, their sums are equal.

### Proof

As established in Lemma 10.1.4, series (a) and (b) converge simultaneously with the series $$\sum_{n=1}^{\infty}|a_{n}|^{2}\|e_{n}\|^{2}$$ and $$\sum_{k=1}^{\infty}|a_{\omega(k)}|^{2}\|e_{\omega(k)}\|^{2}$$, respectively. The last two series converge simultaneously because the sum of a positive series is independent of any rearrangement of the terms. This proves that series (a) and (b) converge simultaneously. Now, let series (a) and (b) converge and S n be a partial sum of (a). By the Pythagorean theorem (see Eq. (1′)), we obtain
$$\Bigg\|\sum_{k=1}^{\infty} a_{\omega(k)}e_{\omega(k)}-S_n \Bigg\|^2= \sum_{\omega(k)>n}^{\infty}|a_{\omega(k)}|^2 \|e_{\omega(k)}\|^2= \sum_{j=n+1}^{\infty}|a_{j}|^2 \|e_{j}\|^2 \underset{n\to\infty}{\longrightarrow}0,$$
which implies that the sums of series (a) and (b) coincide. □

As in the case of sequences, a family {e α } αA is called a basis if every function is the sum of its Fourier series. It can easily be seen that the theorem on the characterization of bases and its corollary remain valid in the more general setting in question.

### 10.1.7

Let $$\{e_{k}\}_{k\in\mathbb{N}}$$ and $$\{g_{n}\} _{n\in\mathbb{N}}$$ be orthogonal systems in the spaces and , respectively. We use these systems to construct an OS $$\{h_{k,n}\}_{k,n\in\mathbb{N}}$$ in the space by putting
$$h_{k,n}(x,y)=e_k(x)g_n(y)\quad(x\in X, y\in Y).$$
Using Fubini’s theorem, we can easily verify that the functions h k,n are square-summable and pairwise orthogonal. We will prove that the above construction preserves completeness.

### Theorem

If orthogonal systems $$\{e_{k}\}_{k\in\mathbb{N}}$$ and $$\{g_{n}\}_{n\in\mathbb{N}}$$ are complete, then the system $$\{h_{k,n}\}_{k,n\in\mathbb{N}}$$ is also complete.

### Proof

Let fh k,n for all $$k,n\in\mathbb{N}$$. This means that
\begin{aligned} &\int_{X\times Y}f(x,y)\overline{e_k(x)}\, \overline{g_n(y)}\,d(\mu\times \nu) (x,y) \\ &\quad = \int_X \biggl(\int_Yf(x,y)\overline{g_n(y)}\,d\nu(y) \biggr)\overline{e_k(x)}\, d\mu(x)=0 \end{aligned}
(7)
for all $$k,n\in\mathbb{N}$$. We fix an arbitrary n and consider the function
$$x\mapsto \varphi _n(x)=\int_Yf(x,y) \overline{g_n(y)}\,d\nu(y).$$
This function is measurable by Corollary 2 to Tonelli’s theorem. Moreover, since
$$\big|\varphi _n(x)\big|\leqslant \biggl(\int_Y\big|f(x,y)\big|^2 \,d\nu(y) \biggr)^{1/2}\|g_n\|,$$
and, therefore,
$$\int_X\big|\varphi _n(x)\big|^2\,d\mu(x) \leqslant\int_X \biggl(\int_Y\big|f(x,y)\big|^2 \,d\nu (y) \biggr)\, d\mu(x)\|g_n\|^2<+\infty.$$
Equation (7) means that the Fourier coefficients of φ n with respect to the system $$\{e_{k}\}_{k\in\mathbb{N}}$$ are zero. Since the system is complete, we have φ n (x)=0 almost everywhere. Since this is true for all indices n, we have
$$\sum_{n=1}^{\infty}\big|\varphi _n(x)\big|^2=0 \quad\text{almost everywhere on}\ X.$$
(8)
Since ∫ X Y |f(x,y)|2(y) (x)<+∞, Fubini’s theorem implies that ∫ Y |f(x,y)|2(y)<+∞ almost everywhere. In other words, the function yf x (y)=f(x,y) is square-summable for almost all x. The numbers φ n (x) are simply the Fourier coefficients of this function with respect to the system $$\{g_{n}\}_{n\in\mathbb{N}}$$. Since the system $$\{g_{n}\}_{n\in \mathbb{N}}$$ is complete, Eq. (8) means that
$$\int_Y\big|f(x,y)\big|^2\,d\nu(y)=\|f_x \|^2=\sum_{n=1}^{\infty}\big| \varphi _n(x)\big|^2=0\quad \text{almost everywhere on}\ X.$$
Integrating the above equation over X, we obtain
$$0=\int_X\int_Y\big|f(x,y)\big|^2 \,d\nu(y)\,d\mu(x)=\|f\|^2.$$
Consequently, f=0 almost everywhere, which proves that the system $$\{ h_{k,n}\}_{k,n\in\mathbb{N}}$$ is complete. □

By induction, the statement just proved can obviously be carried over to the case of more than two orthogonal systems.

### 10.1.8

Lemma 10.1.4 shows that, for a given orthonormal system, an arbitrary sequence {a n } n⩾1 satisfying the condition $$\sum_{n=1}^{\infty}|a_{n}|^{2}<+\infty$$ can serve as the sequence of Fourier coefficients of a square-summable function. It is natural to assume that the smaller the class of functions in question, the greater, in general, the rate of decrease of the Fourier coefficients. In Sect. 10.3, we will find more evidence for this conjecture. However, if, instead of square-summable functions, we consider arbitrary bounded functions (assuming, naturally, that they belong to , i.e., that the measure μ is finite) then our conjecture is false: the Fourier coefficients of bounded functions tend to zero “no faster” than the Fourier coefficients of arbitrary functions from . A more precise formulation of this result of F.L. Nazarov6 [Na] is as follows.

### Theorem

Let $$\{e_{n}\}_{n\in\mathbb{N}}$$ be an orthonormal system in , μ(X)<+∞, such that X |e n | β>0, where β does not depend on n. Then, for every series $$\sum_{n=1}^{\infty} a_{n}^{2}=1$$ (a n >0), there exists a measurable function F a such that |F a |⩽1 and |c n (F a )|⩾θa n for all n (the coefficient θ>0 depends only on μ(X) and β).

We note that the condition ∫ X |e n | β>0 is certainly fulfilled if the orthogonal system consists of uniformly bounded functions since 1=∫ X |e n |2⩽∥e n X |e n | .

### Proof

We consider only the real case, leaving the complex case to the reader (see Exercises 6 and 7).

For an arbitrary sequence of signs ε={ε n }, where ε n =±1, we construct the sum
$$f_{\varepsilon }=\sum_{n=1}^{\infty} \varepsilon _na_ne_n$$
(the series on the right-hand side converges by Lemma 10.1.4). Let A be the set formed by all functions f ε . This set is compact as a continuous image of the Cantor set (the reader can verify independently the continuity of the mapping that takes a number $$\sum_{n=1}^{\infty} t_{n}3^{-n}$$ (t n =0 or 2) from the Cantor set to the point $$\sum_{n=1}^{\infty}(t_{n}-1)a_{n}e_{n}$$ of the set A).
Now, we consider the function Φ of class $$C^{2}(\mathbb{R})$$ such that |Φ′|,|Φ″|⩽1 (the choice of Φ will be specified later). Since |Φ(u)|⩽|Φ(0)|+|u| and the measure is finite, the integral I(f)=∫ X Φ(f)  is finite for every function . Obviously, the integral continuously depends on f and so, by the Weierstrass extreme value theorem, it assumes its maximum value on A: there exists a sequence of signs ε={ε n } such that I(f ε )⩾I(f) for every function f in A. We show that the required function has the form F a =Φ′(f ε ) for an appropriate choice of Φ. Since |F a |⩽sup|Φ′|⩽1, it only remains for us to estimate the Fourier coefficients c n (F a ). To this end, we use the fact that the replacement of ε n by −ε n leaves a function in the class A and, therefore, does not increase the integral I,
$$\int_X \bigl(\Phi (f_{\varepsilon } )-\Phi (f_{\varepsilon }-2\varepsilon _na_ne_n ) \bigr)\, d \mu\geqslant0.$$
The application of the Taylor formula to the integrand leads to the inequality
$$\int_X \biggl(2\varepsilon _na_n e_n\Phi'(f_{\varepsilon })- \frac{1}{2}(2 \varepsilon _na_ne_n)^2 \Phi''(g_n) \biggr)d\mu\geqslant0,$$
(9)
where g n is a function whose values lie between f ε and f ε −2ε n a n e n .
Dividing both sides of inequality (9) by 2a n , we obtain the required estimate for the Fourier coefficients of the function F a =Φ′(f ε ),
$$\big|c_n(F_a)\big|\geqslant \varepsilon _n\int _X e_n\Phi'(f_{\varepsilon }) \,d\mu\geqslant a_n\int_Xe_n^2 \Phi''(g_n)\,d\mu.$$
Now, it is necessary to choose Φ so that the integrals $$J_{n}=\int_{X} e_{n}^{2}\Phi''(g_{n})\,d\mu$$ be separated from zero. If we take an antiderivative of $$\frac{2}{\pi}\,\text{arctan}\,u$$ as Φ, then
$$J_n=\frac{2}{\pi}\int_X\frac{e_n^2}{1+g_n^2} \,d\mu.$$
To estimate this integral, we use the Cauchy–Bunyakovsky inequality,
$$\beta \leqslant\int_X|e_n|\,d\mu=\int _X\frac{|e_n|}{\sqrt{1+g_n^2}}\cdot \sqrt{1+g_n^2} \,d\mu\leqslant\sqrt{\frac{\pi}{2}\,J_n}\cdot\sqrt{\int _X\bigl(1+g_n^2\bigr)\,d \mu}.$$
Since |g n |⩽|f ε |+|f ε −2ε n a n e n |, we obtain $$\int_{X} g_{n}^{2}\,d\mu\leqslant2(\|f_{\varepsilon }\|^{2}+\|f_{\varepsilon }-2\varepsilon _{n}a_{n}e_{n}\|^{2})=4$$. Therefore, $$J_{n}\geqslant\theta=\frac{2\beta ^{2}}{\pi(\mu(X)+4)}$$ and |c n (F a )|⩾θa n for all n. □

### EXERCISES

1.

Supplement Lemma 10.1.4 by the following statement: if a system (in general, nonorthogonal) $$\{e_{n}\}_{n\in\mathbb{N}}$$ is such that the inequality ∥a 1 e 1+⋯+a n e n 2⩽|a 1|2+⋯+|a n |2 is valid for all n and all scalars a 1,…,a n , then the series $$\sum_{n=1}^{\infty} a_{n}e_{n}$$ converges as soon as $$\sum_{n=1}^{\infty}|a_{n}|^{2}<+\infty$$.

2.

Let an orthonormal system $$\{e_{n}\}_{n\in\mathbb{N}}$$ in be uniformly bounded. Prove that $$\int_{X} f\overline{e}_{n}\,d\mu\underset{n\to\infty}{\longrightarrow}0$$ for every function f not only from but also from .

3.

Let $$\{e_{n}\}_{n\in\mathbb{N}}$$ be an orthonormal basis in , and let EX be such that 0<μ(E)<+∞. Prove that $$\sum_{n=1}^{\infty}\int_{E}|e_{n}|^{2}\,d\mu\geqslant1$$.

4.

Supplement the previous exercise by proving that $$\sum_{n=1}^{\infty}|e_{n}(x)|^{2}=+\infty$$ is valid almost everywhere if the σ-finite measure μ is such that every set of positive measure can be partitioned into two sets of positive measure. Can this additional condition be dropped?

5.

Let {φ n } be an orthonormal basis. Prove that the system of functions {ψ n } is complete if ∑ n φ n ψ n 2<1. If, in addition, we know that {ψ n } is an orthonormal system, then it is complete if ∑ n φ n ψ n 2<2. Hint. Assuming that a function f=∑c n φ n is orthogonal to all functions ψ n , estimate the norm of the difference f−∑ n c n ψ n from above and from below.

6.

Verify that Theorem 10.1.8 remains valid in the real case if the orthonormality condition is replaced by the condition from Exercise 1 (the quantities 〈F a ,e n 〉 are estimated instead of the Fourier coefficients).

7.

Generalize the result of the previous exercise to complex systems.

## 10.2 ⋆Examples of Orthogonal Systems

Throughout this section, we consider the convergence of Fourier series only with respect to the -norm, which is denoted by ∥⋅∥. Instead of , where $$X\subset\mathbb{R}^{m}$$, we will write briefly , omitting the indication of a measure.

### 10.2.1

Trigonometric Systems. The most important orthogonal systems are the following real and complex trigonometric systems in the space :
$$1,\quad \cos\frac{\pi x}{\ell},\quad \sin\frac{\pi x}{\ell},\quad \ldots, \quad \cos \frac{\pi nx}{\ell},\quad \sin\frac{\pi nx}{\ell},\quad \ldots \quad\text{and} \quad \bigl\{e^{\frac{\pi inx}{\ell}}\bigr\}_{n\in\mathbb{Z}}.$$
We leave the simple verification of the orthogonality to the reader. The Fourier series with respect to these systems have, respectively, the form
$$A(f)+\sum_{n=1}^{\infty} \biggl(a_n(f)\cos\frac{\pi nx}{\ell}+ b_n(f)\sin \frac{\pi nx}{\ell}\biggr)\quad\text{and}\quad \sum _{n=-\infty}^{\infty} c_n(f) e^{\frac{\pi inx}{\ell}},$$
where the Fourier coefficients are calculated by the formulas
\begin{aligned} A(f)=&\frac{1}{2\ell}\int_a^{a+2\ell}f(x)\,dx, \\a_n(f)=&\frac{1}{\ell}\int_a^{a+2\ell}f(x) \cos\frac{\pi nx}{\ell}\, dx,\\b_n(f)=&\frac{1}{\ell}\int _a^{a+2\ell}f(x)\sin\frac{\pi nx}{\ell}\,dx \quad(n\in\mathbb{N});\\c_n(f)=&\frac{1}{2\ell}\int_a^{a+2\ell}f(x) e^{-\frac{\pi inx}{\ell}}\,dx\quad(n\in\mathbb{Z}). \end{aligned}

In the study of Fourier series, we may assume that the functions are defined on the intervals of the form (0,2), because the general case can be reduced to the case a=0 by a translation. It is often convenient to use a symmetric interval (−,).

The study of Fourier series with respect to a trigonometric system with some period can be reduced to the study of Fourier series with a different period. Following tradition, we will consider (with rare exceptions) only the Fourier series
$$A(f)+\sum_{n=1}^{\infty} \bigl(a_n(f)\cos nx+b_n(f)\sin nx \bigr)\quad \text{and}\quad\sum_{n=-\infty}^{\infty} c_n(f) e^{inx}$$
with respect to more natural and convenient 2π-periodic systems
$$1,\quad \cos x,\quad \sin x,\quad \ldots,\quad \cos nx, \quad \sin nx,\quad \ldots \quad\text{and} \quad\bigl\{e^{inx}\bigr\}_{n\in\mathbb{Z}}.$$
(T)
In this case, the Fourier coefficient c n (f) will also be denoted by the symbol $$\widehat{f}(n)$$. Thus,
$$\widehat{f}(n)=\frac{1}{2\pi}\int_0^{2\pi}f(x) e^{-inx}\,dx\quad(n\in \mathbb{Z}).$$
The transition from the expansion in one system to the expansion in a different system proceeds as follows. For a function , we define a function g by putting $$g(y)=f(\frac{\ell}{\pi}y)$$, where y∈(0,2π). It is clear that . There is an obvious relation connecting the Fourier coefficients of these functions (with respect to the corresponding systems):
$$c_k(f)=\frac{1}{2\ell}\int_0^{2\ell}f(x) e^{-\frac{\pi ikx}{\ell}}\,dx= \frac{1}{2\pi}\int_0^{2\pi}f \biggl(\frac{\ell}{\pi}y \biggr) e^{-iky}\,dy= \widehat{g}(k)$$
for each $$k\in\mathbb{Z}$$. Consequently,
$$\sum_{|k|\leqslant n}c_k(f) e^{\frac{\pi ikx}{\ell}}= \sum _{|k|\leqslant n}\widehat{g}(k) e^{\frac{\pi ikx}{\ell}}= \sum _{|k|\leqslant n}\widehat{g}(k) e^{iky},$$
i.e., the partial sums of the Fourier series of the functions f and g at the corresponding points coincide. From this, it follows, in particular, that both series converge simultaneously and their sums coincide (or do not coincide) simultaneously with the values of the functions f and g. Thus, the transition from f to g makes it possible to reduce the study of a Fourier series in a system with an arbitrary period to the study of a Fourier series in the 2π-periodic system.
By Euler’s formula, the systems (T) are tightly connected with each other: their linear spans coincide (the functions from these spans are called trigonometric polynomials), and the Fourier coefficients in one system are expressed in terms of the Fourier coefficients in the other one by the following formula:
$$\widehat{f}(\pm n)=\frac{1}{2\pi}\int_0^{2\pi} f(x) (\cos nx\mp i\sin nx)\,dx= \frac{a_n(f)\mp ib_n(f)}{2}\quad(n\in\mathbb{N})$$
and
$$a_n(f)=\widehat{f}(n)+\widehat{f}(-n)\quad\text{and}\quad b_n(f)=i\bigl(\widehat{f}(n)- \widehat{f}(-n)\bigr)\quad (n\in \mathbb{N}).$$
It follows that the Fourier series in systems (T) essentially coincide. More precisely, the relation
$$A(f)+\sum_{k=1}^n \bigl(a_k(f) \cos kx+b_k(f)\sin kx \bigr)= \sum_{k=-n}^n \widehat{f}(k) e^{ikx} ,$$
showing that the partial sums of the Fourier series in the real system (T) coincides with symmetric partial sums of the Fourier series in the complex system, is valid for each n.

In the following theorem, we establish one of the most important properties of the systems (T).

### Theorem

The real and complex trigonometric systems form bases in .

### Proof

The assertion of the theorem follows immediately from Corollary 10.1.5 the assumptions of which are fulfilled by Theorem 4 of Sect. 9.3.7. □

Since each of the systems (T) is a basis, it satisfies Parseval’s identity: if , then
\begin{aligned} \frac{1}{2\pi} \int_0^{2\pi} f(x) \overline{g(x)}\,dx =& A(f)\overline{A(g)}+\frac{1}{2} \sum _{n=1}^{\infty} \bigl(a_n(f) \overline{a_n(g)}+b_n(f)\overline{b_n(g)} \,\bigr)\\=& \sum_{n=-\infty}^{\infty}\widehat{f}(n) \overline{\widehat{g}(n)} \end{aligned}
in particular, every function f in satisfies the equation
$$\frac{1}{2\pi}\int_0^{2\pi} \big|f(x)\big|^2\,dx=\big|A(f)\big|^2+ \frac{1}{2}\sum _{n=1}^{\infty} \bigl(\big|a_n(f)\big|^2+\big|b_n(f)\big|^2 \bigr)= \sum_{n=-\infty}^{\infty}\big| \widehat{f}(n)\big|^2,$$
which is often called the closeness relation. As we have already noted, in these formulas and in the theorem, the interval (0,2π) can be replaced by an arbitrary interval of length 2π, in particular, by (−π,π).

We will now give several examples that illustrate the importance of this formula.

### Example 1

Let f(x)=x for x∈(−π,π). The Fourier series of this function has the form $$\sum_{n=1}^{\infty}(-1)^{n-1}\frac{2}{n}\sin nx$$. By Parseval’s identity, we have
$$\frac{1}{\pi}\int_{-\pi}^{\pi}x^2 \,dx=\sum_{n=1}^{\infty}\big|b_n(f)\big|^2= 4\sum_{n=1}^{\infty}\frac{1}{n^2}.$$
Thereby we have arrived at the following result first obtained by Euler: $$\sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{\pi^{2}}{6}$$. The same reasoning applied to the function f(x)=x 2 (|x|⩽π) gives another result of Euler’s: $$\sum_{n=1}^{\infty}\frac{1}{n^{4}}=\frac {\pi^{4}}{90}$$.

### Example 2

As we have seen (see Corollary 9.2.4), 2π-periodic functions in , i.e., square integrable functions on (−π,π) are continuous in mean. By the closeness equation, we can obtain an exact value for the deviation of a function from its translation.

We will assume that a function is extended by periodicity from [−π,π] to $$\mathbb{R}$$. Let $$h\in\mathbb{R}$$, and let f h be the corresponding translation of f, i.e., f h (x)=f(xh) for $$x\in\mathbb{R}$$. It can easily be verified that $$\widehat {f_{h}}(k)=e^{-ikh}\widehat{f}(k)$$. Therefore, by Parseval’s identity, we obtain
$$\|f_h-f\|^2=2\pi\sum_{k=-\infty}^{\infty}\big| \widehat{f}(k)\big|^2 \big|e^{-ikh}-1 \big|^2= 8\pi\sum _{k=-\infty}^{\infty}\big|\widehat{f}(k)\big|^2 \sin^2\frac{kh}{2}.$$

From this formula, the continuity in the mean, $$f_{h}\underset{h\to0}{\longrightarrow} f$$, follows directly.

### Example 3

We apply Parseval’s identity to prove an elegant inequality (see [EF]), which, in some cases, makes it possible to estimate from above the mean value of a function on an interval by its mean value on a smaller interval.

Let the Fourier coefficients of a function φ in be non-negative. Then the inequality
$$\frac{1}{2\pi}\int_{-\pi}^{\pi}\big| \varphi (t)\big|^2\,dt\leqslant \frac{3}{2\alpha }\int_{-\alpha }^{\alpha }\big| \varphi (t)\big|^2\,dt$$
is valid for every α∈(0,π).
Since the function $$h(t)= (1-\frac{|t|}{\alpha })_{+}$$ does not exceed 1, it is sufficient for us to estimate the integral $$I=\int_{-\pi}^{\pi}|\varphi (t)h(t)|^{2}\,dt$$ from below. The product f=φh, obviously, belongs to . We calculate its Fourier coefficients (in what follows, e n (t)=e int ),
\begin{aligned} \widehat{f}(n)=&\frac{1}{2\pi}\langle \varphi h,e_n\rangle= \frac{1}{2\pi}\langle \varphi ,he_n\rangle=\frac{1}{2\pi} \sum _{k=-\infty}^{\infty}\widehat{\varphi }(k)\langle e_k,he_n\rangle\\=& \sum_{k=-\infty}^{\infty} \widehat{\varphi }(k)\,\widehat{h}(n-k) =\sum_{k+j=n}\widehat{\varphi }(k)\,\widehat{h}(j). \end{aligned}
Now, by Parseval’s identity we obtain
$$I=2\pi\sum_{n=-\infty}^{\infty}\big| \widehat{f}(n)\big|^2=2\pi\sum_{n=-\infty}^{\infty} \bigg|\sum_{k+j=n}\widehat{\varphi }(k)\,\widehat{h}(j)\bigg|^2.$$
A direct calculation shows that $$\widehat{h}(j)\geqslant0$$ for all $$j\in\mathbb{Z}$$ (this also follows from the result of Example 2 of Sect. 4.6.6 since the function h is convex on (0,π)). Therefore, replacing the square of the sum by the sum of squares (here we use the inequalities $$\widehat{\varphi }(k)\geqslant0$$), we obtain
\begin{aligned} I\geqslant&2\pi\sum_{n=-\infty}^{\infty} \sum _{k+j=n}\widehat{\varphi }^2(k)\, \widehat{h}^2(j)= 2\pi\sum_{k=-\infty}^{\infty} \widehat{\varphi }^2(k)\sum_{j=-\infty}^{\infty} \widehat{h}^2(j) \\=&\frac{1}{2\pi}\int_{-\pi}^{\pi}\big| \varphi (t)\big|^2\,dt\int_{-\pi}^{\pi} h^2(t)\,dt= \frac{\alpha }{3\pi}\int_{-\pi}^{\pi}\big| \varphi (t)\big|^2\,dt. \end{aligned}
Thus,
$$\int_{-\alpha }^{\alpha }\big|\varphi (t)\big|^2dt\geqslant I \geqslant \frac{\alpha }{3\pi}\int_{-\pi}^{\pi}\big| \varphi (t)\big|^2\,dt.$$

### Example 4

Hurwitz7 found an unexpected application of trigonometric series. It turns out that they can be used to obtain a very simple proof of the classical isoperimetric inequality connected with the problem of determining a closed plane curve that has a given circumference L and bounds a figure of the largest area. This inequality has the form
$$4\pi S\leqslant L^2,$$
where S is the area of the figure. The equality is attained only in the case where the curve is a circle (the multi-dimensional case of the isoperimetric inequality is considered in Sects. 2.8.2 and 13.4.7).

The proof given by Hurwitz is analytic. It uses only the closeness equation and the formula for the area in terms of a curvilinear integral.

Let $$K\subset\mathbb{R}^{2}$$ be a compact set whose boundary is a closed smooth curve. Without loss of generality, we may assume that the length of the curve is 2π. Let z(t)=(x(t),y(t)), 0⩽t⩽2π be the natural parametrization (see Sect. 8.2.3) of the curve ∂K. Then z(0)=z(2π) because the curve ∂K is closed and |z′(t)|≡1 because the parametrization is natural.

Using the closeness equation and the identity |z′(t)|≡1, we can represent the relation L=2π in the form
$$L^2=2\pi\int_0^{2\pi}\big|z'(t)\big|^2 \,dt=4\pi^2\sum_{n\in\mathbb{Z}}\big|\widehat {z}'(n)\big|^2.$$
(1)
To calculate the area S=λ 2(K), we apply the relation
$$S=\frac{1}{2}\int_{\partial^+K}(-y\,dx+x\,dy)=\frac{1}{2} \int_0^{2\pi} \bigl(x(t)y'(t)-y(t)x'(t) \bigr)\,dt,$$
which follows from Green’s formula with P(x,y)=−y and Q(x,y)=x (see Sect. 8.6.7). Since $$x(t)y'(t)-y(t)x'(t)=\mathcal{I} m(z'(t)\overline{z(t)})$$ and $$\int_{0}^{2\pi}\mathcal{R}e(z'(t)z(t))\,dt= \int_{0}^{2\pi} (x^{2}(t)+y^{2}(t) )'\,dt=0$$, we have
$$S=\frac{1}{2i}\int_0^{2\pi} z'(t)\overline{z(t)}\,dt.$$
Transforming the integral by Parseval’s identity, we obtain
$$S=-\pi i\sum_{n\in\mathbb{Z}}\widehat{z}'(n) \overline{\widehat{z}(n)}.$$
(2)
Now, we eliminate the Fourier coefficients of the derivative from Eqs. (1) and (2), expressing them in terms of the Fourier coefficients of the function z. Integrating by parts and taking into account that z(0)=z(2π), we have
$$\widehat{z}'(n)=\frac{1}{2\pi}\int_0^{2\pi}z'(t)e^{-int} \,dt= \frac{1}{2\pi}z(t)e^{-int}\Big\vert_{t=0}^{2\pi}+ \frac{in}{2\pi} \int_0^{2\pi}z(t)e^{-int} \,dt=in\widehat{z}(n).$$
Substituting the resulting expressions for $$\widehat{z}'(n)$$ in (1) and (2), we obtain
$$L^2=4\pi^2\sum_{n\in\mathbb{Z}}n^2\big| \widehat{z}(n)\big|^2\quad\text{and}\quad S=\pi\sum _{n\in\mathbb{Z}}n\big|\widehat{z}(n)\big|^2.$$
Consequently,
$$L^2-4\pi S=4\pi^2\sum_{n\in\mathbb{Z}} \bigl(n^2-n\bigr)\big|\widehat{z}(n)\big|^2\geqslant0,$$
which proves the isoperimetric inequality. Moreover, the last formula implies that the equality holds only if $$\widehat{z}(n)=0$$ for n≠0,1, i.e., only if $$z(t)=\widehat{z}(0)+\widehat{z}(1)e^{it}$$. We have $$|\widehat{z}(1)|=1$$, since |z′(t)|≡1. Thus, the curve of length 2π for which the isoperimetric inequality becomes an equality is the unit circle $$|z-\widehat{z}(0)|=1$$.

### 10.2.2

Considering the product of m copies of the complex trigonometric system (see Sect. 10.1.7), we obtain its multi-dimensional analog in the space , where Q=(−π,π) m (a multi-dimensional version of the real trigonometric system is quite cumbersome and we do not consider it). The new system consists of the complex exponential functions e n numbered by multi-indices n=(n 1,…,n m ):
$$e_n(x)=e^{i\langle n,x\rangle},\quad \text{where}\ x\in Q,\ n\in \mathbb{Z}^m.$$
The Fourier coefficients of a function in this system are calculated by the formulas
$$\widehat{f}(n)=\frac{\langle f,e_n\rangle}{\|e_n\|^2}=\frac{1}{(2\pi )^m}\int_Qf(x) e^{-i\langle n,x\rangle}\,dx \quad\bigl(n\in\mathbb{Z}^m\bigr).$$
From Theorem 10.1.7, it follows that the system $$\{e^{i\langle n,x\rangle }\}_{n\in\mathbb{Z}^{m}}$$ is complete, which implies Parseval’s identity
Of course, the cube Q=(−π,π) m in the two last formulas can be replaced by a shifted cube.

### Example

Let 0<ρπ. We consider the function that is equal to 1/∥x∥ for ∥x∥<ρ and vanishes on (−π,π)3B(0,ρ). Its norm is easily calculated in spherical coordinates,
$$\|f\|^2=\int_{B(0,\rho)}\frac{1}{\|x\|^2}\,dx= 4\pi \int_0^{\rho}\frac{1}{r^2}\,r^2 \,dr=4\pi\rho.$$
To calculate the Fourier coefficients, we use the formula obtained in the example of Sect. 6.2.5 with f 0(r)=1/r on (0,ρ), f 0(r)=0 for rρ and y=n/2π:
\begin{aligned} \widehat{f}(n)=&\frac{1}{(2\pi)^3}\int_{B(0,\rho)} \frac{1}{\|x\|} e^{-i\langle n,x\rangle}\,dx=\frac{1}{2\pi^2\|n\|} \int _0^{\rho}\frac{1}{r}r\sin\bigl(\|n\|r\bigr)\,dr\\=& \biggl(\frac{\sin\frac{\rho}{2}\|n\|}{\pi\|n\|} \biggr)^2 \end{aligned}
if n≠0 and $$\widehat{f}(0)=\rho^{2}/4\pi^{2}$$. By Parseval’s identity for the function f, we obtain
$$4\pi\rho=(2\pi)^3\sum_{n\in\mathbb{Z}^3} \biggl( \frac{\sin\frac{\rho}{2}\|n\|}{\pi\|n\|} \biggr)^4.$$
Thus, the identity
$$\frac{\pi^2}{t^3}=\sum_{n\in\mathbb{Z}^3} \biggl( \frac{\sin\|n\|t}{\|n\| t} \biggr)^4$$
is valid for $$t=\frac{\rho}{2}\in(0,\frac{\pi}{2}]$$ (the summand for n=0 is equal to 1).

### 10.2.3

The trigonometric system is closely connected with the orthogonal system $$\{z^{n}\}_{n\in\mathbb{Z}}$$ in the space , where $$S^{1}=\{z\in\mathbb{C}\,|\,|z|=1\}$$ is the unit circle and σ is the arc length. Knowing that the trigonometric system is complete in , we use the change of variable z=e ix (−π<x<π) and easily verify that the system $$\{z^{n}\}_{n\in\mathbb{Z}}$$ is complete in . Therefore, every function f in this space is the sum of the series $$\sum_{n\in\mathbb{Z}} c_{n} z^{n}$$, where $$c_{n}=\frac{1}{2\pi}\int_{S^{1}}f(z)\overline{z}^{n}\,d\sigma (z)$$. The reader familiar with the theory of holomorphic functions will see that this formula coincides with the formula for the nth coefficient of the Laurent expansion of f in the annulus r<|z|<R, where r<1<R. Therefore, the Fourier series in the system $$\{z^{n}\}_{n\in\mathbb{Z}}$$ can be regarded as the limit form of the Laurent series, when the annulus degenerates to a circle.

We consider an example connected with the system $$\{z^{n}\}_{n\in\mathbb{Z}}$$. Let T:S 1S 1 be a rotation of the circle, i.e., the map zT(z)=ζz, where ζS 1 is a fixed number. We now address the question of how much the points of the circle “mix” under the iterations of T. Does there exist an invariant subset of the circle, that is, a set which retains all of its points after rotation? More precisely, a set ES 1 is called invariant if it differs from its image only on a set of measure zero, i.e., if χ E =χ T(E) almost everywhere. Of course, such sets exist: the circle S 1 and the set $$\{\zeta^{n}\}_{n\in\mathbb{Z}}$$ are examples. It is easy to construct more examples of invariant sets of measure 2π or zero. Therefore, we are interested in the question of whether there are non-trivial invariant sets, i.e., sets satisfying the condition 0<σ(E)<2π. If ζ m =1 for some m, then the map T is repeated after m iterations (T m+1=T), and a non-trivial invariant subspace can easily be constructed. We leave this construction to the reader. However, if ζ is not a root of unity, then the map T has no non-trivial invariant sets (such maps are called ergodic). Let us prove this.

Let ES 1 be an invariant set. Then χ E =χ T(E) almost everywhere, and therefore, c n (χ T(E))=c n (χ E ). At the same time, by a change of variable (Corollary 6.1.1), we obtain
$$c_n(\chi_{T(E)})=\frac{1}{2\pi}\int _{T(E)}\overline{z}^n\,d\sigma (z)= \frac{1}{2\pi} \int_E\overline{(\zeta z)}^n\,d\sigma (z)= \zeta^{-n}c_n(\chi_E).$$
Thus, c n (χ E )(1−ζ n )=0 for all $$n\in\mathbb{Z}$$. Since 1−ζ n ≠0 for n≠0, it follows that all Fourier coefficients of χ E , except, possibly, c 0(χ E ), are zero. Since the system $$\{z^{n}\}_{n\in\mathbb{Z}}$$ is complete, the function χ E coincides with the sum of its Fourier series almost everywhere. Therefore, χ E is a constant almost everywhere. Consequently, either χ E (x)=0 almost everywhere (the invariant set has measure zero) or χ E (x)=1 almost everywhere (the invariant set is a set of full measure).

### 10.2.4

We will now give other examples of orthogonal systems. Let P n (x)=((x 2−1) n )(n), n=0,1,…. The polynomials P n are called the Legendre polynomials. Obviously, deg P n =n, and so every polynomial is a linear combination of Legendre polynomials, which form an orthogonal system in the space . Indeed, for m<n, we have
\begin{aligned} \langle P_m,P_n\rangle=&\int_{-1}^1P_m(x) \bigl(\bigl(x^2-1\bigr)^n \bigr)^{(n)}\,dx\\=& P_m(x) \bigl(\bigl(x^2-1\bigr)^n \bigr)^{(n-1)}\bigg\vert _{-1}^1 -\int_{-1}^1P'_m(x) \bigl(\bigl(x^2-1\bigr)^n \bigr)^{(n-1)}\,dx\\=&-\int_{-1}^1P'_m(x) \bigl( \bigl(x^2-1\bigr)^n \bigr)^{(n-1)}\,dx. \end{aligned}
Integrating by parts n times, we arrive at the equation
$$\langle P_m,P_n\rangle=(-1)^n\int_{-1}^1P^{(n)}_m(x) \bigl(x^2-1\bigr)^n\,dx,$$
where $$P^{(n)}_{m}(x)\equiv0$$, since deg P m <n. Thus, 〈P m ,P n 〉=0 for mn.

### Theorem

The Legendre polynomials form a basis in the space .

### Proof

As in the proof of Theorem 10.2.1, we use Corollary 10.1.5. We must verify that every function in can be approximated arbitrarily closely (in the -norm) by linear combinations of polynomials P n , i.e., by arbitrary algebraic polynomials. This, however, has already been established in Corollary 9.2.3. □

We mention one more useful orthogonal system. In the space , we consider the Hermite 8 functions
$$h_n(x)=e^{x^2/2}\bigl(e^{-x^2}\bigr)^{(n)},\quad n=0,1,\ldots.$$
It is easy to verify that $$h_{n}(x)=H_{n}(x)e^{-x^{2}/2}$$, where H n is an nth degree polynomial called a Hermite polynomial. The orthogonality of the Hermite functions can be established by integrating by parts the equation
$$\langle h_m,h_n\rangle=\int_{-\infty}^{\infty} H_m(x) \bigl(e^{-x^2}\bigr)^{(n)}\,dx$$
in the same way as in the proof of the orthogonality of the Legendre polynomials. It is obvious that the orthogonality of the Hermite functions in implies the orthogonality of the Hermite polynomials in with measure $$d\mu(x)=e^{-x^{2}}dx$$.

Later on (see the corollary in Sect. 10.5.6) we prove that the system of functions h n is complete in or, equivalently, the system of polynomials H n is complete in .

### 10.2.5

In the applications of probability theory in analysis, the sequence of Rademacher functions r n defined in Sect. 6.4.5 plays an important role. As has already been proved, these functions are independent in the sense of Definition 4.4.4. Since, in addition, $$\int_{0}^{1}r_{n}(x)\,dx=0$$, we see that the relation
$$\int_0^1r_{n_1}(x)r_{n_2}(x) \cdots r_{n_m}(x)\,dx =\prod_{k=1}^m \int_0^1r_{n_k}(x)\,dx=0$$
(1)
holds for 1⩽n 1<n 2<⋯<n m .

In particular, the Rademacher functions form an orthonormal system in the space . Of course, this system is not complete: for example, the pairwise products r j r k are orthogonal to all Rademacher functions. To obtain a complete system containing the Rademacher functions, we proceed as follows. For every non-empty finite set $$A\subset\mathbb{N}$$, we consider the function w A =∏ nA r n . Furthermore, we will assume, by definition, that w ≡1. The functions w A are called the Walsh 9 functions. The Rademacher functions are the Walsh functions corresponding to the one-element sets. By Eq. (1), the functions w A are pairwise orthogonal. The system of Walsh functions is complete in . To prove this, we need the following lemma.

### Lemma

Let $$n\in\mathbb{N}$$. The set of linear combinations of the functions w A such that A⊂{1,2,3,…,2 n } coincides with the set of linear combinations of the characteristic functions of the intervals Δ n,k =(k2n ,(k+1)2n ) for k=0,1,…,2 n −1.

### Proof

Let L 1 and L 2 be the linear spans of the first and second systems, respectively. Since the functions r 1,…,r n are constant on the intervals Δ n,k , the Walsh functions in question are also constant on these intervals. Therefore, L 1L 2. At the same time, the dimensions of L 1 and L 2 are, obviously, equal (to 2 n ). Hence it follows that L 1=L 2. □

### Theorem

The system of Walsh functions is complete in the space .

### Proof

We use Corollary to Theorem 10.1.5 on the characterization of bases. We will prove that every function f in can be approximated arbitrarily closely in norm by linear combinations of Walsh functions. If f is the characteristic function of an interval (p,q)⊂(0,1), then, for a given ε, we can find a large n such that p and q can be approximated by the points j/2 n and k/2 n within ε. Then ∥fχ Δ2<2ε, where χ Δ is the characteristic function of the interval (j/2 n ,k/2 n ), which almost everywhere coincides with the sum $$\sum_{s=j}^{k-1}\chi_{\Delta_{n,s}}$$ equal, by the lemma, to a certain linear combination of Walsh functions. Being able to approximate the characteristic functions of the intervals, we can also approximate their linear combinations, i.e., the step functions. Now, we consider the general case. By Theorem 9.2.2, for each ε, we can find a step function g such that ∥fg∥<ε. Approximating g within ε by a linear combination h of Walsh functions, we obtain ∥fh∥⩽∥fg∥+∥gh∥<2ε. Since ε was arbitrary, this completes the proof. □

### 10.2.6

From the viewpoint of probability theory, the Rademacher functions give an example of a sequence of independent trials with two equiprobable outcomes (the simplest “Bernoulli scheme”). Here, a “simple” random event is a roll of a number x∈(0,1), and the probability that a point will fall in the interval (p,q) is the length of the interval. A “trial” consists of the calculation of the values of the Rademacher functions: the first trial is the calculation of r 1(x), the second trial is the calculation of r 2(x), etc. Taking into account the connection between the values of a Rademacher function at a given point and the binary digits of the point, we can replace r n (x) by ε n (x) (the binary digits of x).

One of the first results of probability theory is Bernoulli’s law of large numbers, which says that in the scheme described above the frequency of occurrence of 0 or 1 becomes close to 1/2 with probability arbitrarily close to 1. In the language of measure theory, this result means that, on the interval (0,1), the arithmetic mean $$\frac{1}{n}(\varepsilon _{1}(x)+\cdots+\varepsilon _{n}(x))$$ (the frequency of occurrence of the digit 1 in the binary expansion of a point x) tends to 1/2 in measure. Returning to the Rademacher functions, we can say that
$$\frac{r_1(x)+\cdots+r_n(x)}{n}\underset{n\to\infty}{\longrightarrow} 0\quad \text{in measure.}$$
This assertion follows from the fact that
$$\frac{1}{n}\|r_1+\cdots+r_n\|=\frac{1}{\sqrt{n}} \underset{n\to\infty}{\longrightarrow}0,$$
and the convergence in norm implies the convergence in measure.

Two centuries after Bernoulli, Borel proved a stronger statement.

### Theorem

(Strong law of large numbers)

$$\frac{r_1(x)+\cdots+r_n(x)}{n}\underset{n\to\infty}{\longrightarrow} 0\quad \textit{almost everywhere on}\ (0,1).$$

### Proof

We put S n (x)=r 1(x)+⋯+r n (x) and estimate the integral $$\int_{0}^{1}S_{n}^{4}(x)\,dx$$. Obviously,
$$S_n^2(x)=\sum_{k=1}^nr_k^2(x)+2 \sum_{1\leqslant j<k\leqslant n}r_j(x)r_k(x) = n+2\sum_{1\leqslant j<k\leqslant n}w_{\{j,k\}}(x).$$
Since the Walsh functions w {j,k} form an orthonormal system, the Pythagorean theorem implies
$$\int_0^1S_n^4(x)\,dx= \bigg\|nw_{\varnothing}+2\sum_{1\leqslant j<k\leqslant n}w_{\{j,k\}}\bigg\|^2= n^2+4\sum_{1\leqslant j<k\leqslant n}1<3n^2.$$
Consequently, $$\sum_{n=1}^{\infty}\int_{0}^{1} (\frac{1}{n}\,S_{n}(x) )^{4} < \sum_{n=1}^{\infty}\frac{3}{n^{2}}<+\infty$$, and, therefore, the series $$\sum_{n=1}^{\infty} (\frac{1}{n}\,S_{n}(x) )^{4}$$ converges almost everywhere (see Corollary 2 of Sect. 4.8.2). This implies the assertion of the theorem since the terms of a convergent series tend to zero. □

### 10.2.7

The theorem just proved admits various generalizations also called the strong laws of large numbers. The statements concerning sequences of independent functions with zero mean values (obviously, these functions form an orthogonal system) are of most interest. Before passing to this question, we consider an inequality playing a decisive role in the study of series of such functions.

Throughout this section, we consider real functions in the space , assuming that the measure μ is normalized (μ(X)=1).

### Theorem

(Kolmogorov’s10 inequality)

Let f 1,…,f n in be independent and have zero means, ∫ X f 1=⋯=∫ X f n =0. Then the inequality
$$\mu \Bigl( \Bigl\{x\in X\, |\,\max_{1\leqslant k\leqslant n} \big|f_1(x)+ \cdots+f_k(x)\big| \geqslant t \Bigr\} \Bigr)\leqslant \frac{1}{t^2} \sum_{k=1}^n\int_X f_k^2\,d\mu$$
holds for every t>0.

### Proof

We put S k =f 1+⋯+f k , $$S_{k}^{*}=\max_{1\leqslant j\leqslant k}|S_{j}|$$ and R k =S n S k . We need to estimate the measure of the set $$E= \{x\in X\,|\,S_{n}^{*}(x)\geqslant t \}$$. To this end, we divide the set into disjoint parts $$E_{k}= \{x\in X\,|\,S_{k-1}^{*}(x)<t\leqslant S_{k}^{*}(x) \}$$ (we assume that $$S_{0}^{*}\equiv0$$). Then
\begin{aligned} \sum_{k=1}^n\int_X f_k^2\,d\mu=&\int_XS_n^2 \,d\mu\geqslant \int_ES_n^2\,d\mu= \sum_{k=1}^n\int_{E_k}(S_k+R_k)^2 \,d\mu \\= &\sum_{k=1}^n \biggl(\int _{E_k}S_k^2\,d\mu+2\int _{E_k}S_kR_k\,d\mu+ \int _{E_k}R_k^2\,d\mu \biggr)\\\geqslant&\sum _{k=1}^n\int_{E_k}S_k^2 \,d\mu+ 2\sum_{k=1}^n\int _{E_k}S_kR_k\,d\mu. \end{aligned}
By the corollary to Lemma 6.4.4, the functions $$S_{k}\chi_{E_{k}}$$ and R k are independent. Therefore,
$$\int_{E_k}S_kR_k\,d\mu=\int _XS_k\chi_{E_k}R_k\,d \mu= \int_XS_k\chi_{E_k}\,d\mu \cdot\int_X R_k\,d\mu=0.$$
Since $$|S_{k}|=S_{k}^{*}\geqslant t$$ on the set E k , we obtain the required inequality,
$$\sum_{k=1}^n\int_Xf_k^2 \,d\mu\geqslant \sum_{k=1}^n\int _{E_k}S_k^2\,d\mu\geqslant\sum _{k=1}^nt^2 \mu(E_k)=t^2 \mu(E).$$
□

We supplement the theorem (preserving the notation) and verify that, for a sequence of independent functions f n satisfying the assumptions of the theorem, the following statement is true.

### Corollary

If $$A^{2}=\sum_{n=1}^{\infty}\int_{X}f_{n}^{2}\,d\mu<+\infty$$, then the function $$S^{*}=\sup_{k\geqslant1}|S_{k}|=\sup_{k\geqslant1}S_{k}^{*}$$ is summable and X S ⩽2A.

### Proof

For every t>0, the set X(S t) is exhausted by the expanding sequence of sets $$X(S_{k}^{*}\geqslant t)$$. By the theorem, the measure of each of these sets does not exceed A 2/t 2. Consequently, μ(X(S t))⩽A 2/t 2. Thus, F(t)⩽A 2/t 2, where F is the decreasing distribution function for S . Using the formula of Proposition 6.4.3 with p=1, we see that
\begin{aligned} \int_X S^*\,d\mu=&\int_0^{\infty} F(t)\,dt=\int_0^A\cdots+\int _A^{\infty}\cdots\\\leqslant& AF(0)+ \int _A^{\infty}\frac{A^2}{t^2}\,dt\leqslant A+A=2A. \end{aligned}
□

### 10.2.8

The estimate for the integral of the function S established in the previous corollary leads to an important result concerning the behavior of the series $$\sum_{n=1}^{\infty} f_{n}$$, which, in turn, implies a generalization of Borel’s theorem.

### Theorem 1

Let $$\{f_{n}\}_{n=1}^{\infty}$$ be a sequence of independent functions with zero means. If $$\sum_{n=1}^{\infty}\int_{X}f_{n}^{2}\,d\mu<+\infty$$, then the series $$\sum_{n=1}^{\infty} f_{n}$$ converges almost everywhere.

### Proof

We put
$$S_n=f_1+\cdots+f_n\quad\text{and}\quad R_n=\sup_{p\geqslant1}|S_{n+p}-S_n|.$$
Since |S n+p S n |⩽2R m for nm and all m and p, we must verify only that inf n R n =0 almost everywhere. For this, it is sufficient to verify the relation $$\int_{X}R_{n}\,d\mu\underset{n\to\infty}{\longrightarrow}0$$, which follows immediately from Corollary 10.2.7,
$$\int_XR_n\,d\mu\leqslant2 \Biggl(\sum _{k=n+1}^{\infty}\int_Xf_k^2 \,d\mu \Biggr) ^{1/2} \underset{n\to\infty}{\longrightarrow}0.$$
□

### Corollary

Let $$\{f_{n}\}_{n=1}^{\infty}$$ be a sequence of independent functions with zero means. If $$\sum_{n=1}^{\infty}\frac{1}{n^{2}}\int_{X} f_{n}^{2}\,d\mu<+\infty$$, then $$\sigma _{n}=\frac{1}{n}\sum_{k=1}^{n}f_{k}\underset{n\to\infty}{\longrightarrow}0$$ almost everywhere.

### Proof

By the theorem, the sums $$T_{n}=\sum_{k=1}^{n}\frac{1}{k}f_{k}$$ have a finite limit almost everywhere. The quantities $$\theta_{n}=\frac{1}{n+1}(T_{1}+\cdots+T_{n})$$ have the same limit. Therefore, the difference T n θ n tends to zero almost everywhere. At the same time, it is easy to verify that $$T_{n}-\theta_{n}=\frac{1}{n+1}(f_{1}+\cdots+f_{n})$$, which completes the proof. □

A similar statement can be obtained for an arbitrary orthogonal system if we drop the independence requirement and strengthen the restriction on the quantities ∥f n ∥ (see Exercise 11).

If we impose quite natural additional restrictions on the independent functions f n , then the condition $$\sum_{n=1}^{\infty}\int_{X}f_{n}^{2}\,d\mu<+\infty$$ will turn out to be not only sufficient but also necessary for the convergence of the series $$\sum_{n=1}^{\infty} f_{n}$$ almost everywhere (or, equivalently by the zero-one law, on a set of positive measure).

### Theorem 2

Let $$\{f_{k}\}_{k=1}^{\infty}$$ be a sequence of independent bounded functions with zero means. If the series $$\sum_{k=1}^{\infty} f_{k}$$ converges almost everywhere, then $$\sum_{k=1}^{\infty}\int_{X}f_{k}^{2}\,d\mu<+\infty$$.

### Proof

We put $$S=\sum_{k=1}^{\infty} f_{k}$$ and $$S_{n}=\sum_{k=1}^{n} f_{k}$$ (n=1,2,…). Since the sum S is finite almost everywhere, the sequences {S n (x)} n are bounded for almost all x. They are uniformly bounded on some set of positive measure. Therefore, for a sufficiently large t, the intersection $$E=\bigcap_{n=1}^{\infty} E_{n}$$ of the sets E n ={xX | |S k (x)|⩽t for k=1,…,n} has a positive measure. We find a recurrence estimate for the integrals
$$I_n=\int_{E_n}S_n^2\,d \mu.$$
For this, we use the independence of the functions f n+1 and $$S_{n}\chi_{E_{n}}$$ (see the corollary of Lemma 6.4.4). This gives us the relations
$$\int_{E_n}S_nf_{n+1}\,d\mu= \int _X\chi_{E_n}S_n\,d\mu\cdot\int _Xf_{n+1}\,d\mu=0$$
and
$$\int_{E_n}f^2_{n+1}\,d\mu=\int _X\chi_{E_n}f^2_{n+1}\,d \mu= \mu(E_n)\int_Xf^2_{n+1} \,d\mu\geqslant\mu(E)\int_Xf^2_{n+1} \,d\mu.$$
Therefore, putting F n =E n E n+1, we arrive at the inequality
$$I_{n+1}=\int_{E_n}(S_n+f_{n+1})^2 \,d\mu-\int_{F_n}S_{n+1}^2\,d\mu \geqslant I_n+\mu(E)\int_Xf^2_{n+1} \,d\mu-\int_{F_n}S^2_{n+1}\,d\mu.$$
By assumption, there is a number c such that, for all n, the inequality |f n |⩽c holds almost everywhere. Then
$$\big|S_{n+1}(x)\big|\leqslant\big|S_n(x)\big|+\big|f_{n+1}(x)\big| \leqslant t+c\quad\text{for almost all}\ x\ \text{in}\ E_n.$$
Thus,
$$I_{n+1}-I_n+(t+c)^2\mu(F_n) \geqslant\mu(E)\int_Xf_{n+1}^2\,d\mu.$$
Since $$\sum_{k=1}^{n}(I_{k+1}-I_{k})\leqslant I_{n+1}\leqslant t^{2}$$ and $$\sum_{k= 1}^{n}\mu(F_{k})\leqslant1$$, it follows that the series $$\sum_{k}\mu(E)\int_{X}f^{2}_{k+1}\,d\mu$$ converges, which is equivalent to the assertion of the theorem since μ(E)>0. □

### EXERCISES

1.

Prove that the systems 1,cos x,cos 2x,… and sin x,sin 2x,… are complete in the space .

2.

Let μ be a measure on the interval (−1,1) having density $$\frac{1}{\sqrt{1-x^{2}}}$$ with respect to Lebesgue measure. Prove that the functions T n (x)=cos(n arccos x) (n=0,1,2,…) form an orthogonal basis in the space . Verify that T n is an algebraic polynomial of degree n (a Chebyshev polynomial).

3.

Prove that the functions xe x/2(x n e x )(n) (n=0,1,…), called the Laguerre 11 functions, form an orthogonal system in the space .

4.
Let m be one of the digits 0,1,…,9, and let c m (x)=1 if the decimal expansion of the fractional part of x has the form 0.m… and c m (x)=0 otherwise. Verify that
$$\frac{1}{n}\sum_{0\leqslant k<n}c_m \bigl(10^kx\bigr)\underset{n\to\infty}{\longrightarrow} \frac{1}{10}\quad\text{almost everywhere on}\ (0,1).$$
5.

Generalize the result of the previous exercise by proving that almost all numbers x∈(0,1) are normal, i.e., for each $$p\in\mathbb{N}$$, the p-ary expansion of x contains all digits (the numbers 0,1,…,p−1) “equally often”.

In Exercises 6–9, by r 1,r 2,…,r n … we denote the Rademacher functions.

6.
Use the Khintchine inequality to supplement Theorem 10.2.6 by proving that
$$\frac{r_1(x)+\cdots+r_n(x)}{n^p} \underset{n\to\infty}{\longrightarrow} 0\quad \text{almost everywhere on}\ (0,1)$$
for p>1/2.
7.

Verify that the result of the previous exercise is false for p=1/2. Hint. Find the limits of the integrals $$\int_{0}^{1}e^{i\sigma _{n}(x)}dx$$, where $$\sigma _{n}(x)=\frac{r_{1}(x)+\cdots+r_{n}(x)}{\sqrt{n}}$$.

8.

Show that if the sum of the series $$\sum_{n=1}^{\infty} a_{n}r_{n}$$ is bounded almost everywhere on some non-degenerate interval, then $$\sum_{n=1}^{\infty}|a_{n}|<+\infty$$.

9.

Let $$f_{n}(x,y)=\sum_{j=1}^{n}r_{j}(x)r_{j}(y)$$. Show that |∬ A×B f n (x,y) dxdy|⩽1 for any measurable sets A,B⊂(0,1), but nevertheless $$\iint_{(0,1)^{2}} |f_{n}(x,y)|\,dx\,dy\to+\infty$$. Hint. Use Bessel’s inequality and the inequality from Exercise 7 of Sect.  with p=1.

10.

Refine the assertion of Corollary 10.2.8 by proving that the functions σ n are dominated by a summable function.

11.

Let $$\{f_{n}\}_{n=1}^{\infty}$$ be an orthogonal system in such that $$\sum_{n=1}^{\infty}\frac{\|f_{n}\|^{2}}{n^{3/2}}<+\infty$$. Prove that $$\frac{1}{n}(f_{1}+\cdots+f_{n})\,\underset{n\to\infty}{\longrightarrow}0$$ almost everywhere on X and the function $$\sup_{n}\frac{1}{n}|f_{1}+\cdots+f_{n}|$$ belongs to .

12.

Verify that the assumptions of Theorem 2 of Sect. 10.2.8 can be weakened by replacing the convergence of the series $$\sum_{k=1}^{\infty} f_{k}$$ almost everywhere by the boundedness of its partial sums at the points of a set of positive measure.

## 10.3 Trigonometric Fourier Series

The present and following sections are devoted to harmonic analysis. Without striving to expose this important and vast subject in its entirety, we restrict ourselves to the exposition of selected topics the choice of which is motivated only by the desire to demonstrate the methods developed above.

In Sect. 10.1, we established important properties of Fourier series in arbitrary orthogonal systems. Now, we consider the properties of Fourier series in trigonometric systems in more detail. This is historically the first example of an orthogonal system, and the problem of the representability of a function as the sum of a trigonometric series was one of the central problems in mathematics for nearly two hundred years.

Suffice to say that the lively discussion in the 18th century devoted to this problem provided an important impetus for the formulation of the modern concept of function. Riemann introduced his definition of an integral in connection with the study of trigonometric series, and Cantor, studying the uniqueness of the expansion of a function as a trigonometric series, came up with his foundation of set theory.

### 10.3.1

We recall that, according to the general definition 10.2.1, the Fourier series of a function in the systems
$$1,\quad \cos x,\quad \sin x,\quad \ldots,\quad \cos nx,\quad \sin nx,\quad \ldots,\quad\text{and} \quad \bigl\{e^{inx}\bigr\}_{n\in\mathbb{Z}}$$
have, respectively, the forms
$$A(f)+\sum_{n=1}^{\infty}\bigl(a_n(f) \cos nx+b_n(f)\sin nx\bigr)$$
(1)
and
$$\sum_{n=-\infty}^{\infty}\widehat{f}(n) e^{inx},$$
(1′)
where the Fourier coefficients are calculated by the formulas
\begin{aligned} A(f)=&\frac{1}{2\pi}\int_0^{2\pi}f(x)\,dx, \\a_n(f)=&\frac{1}{\pi}\int_0^{2\pi}f(x) \cos nx\,dx, \end{aligned}
(2)
\begin{aligned} b_n(f)=&\frac{1}{\pi}\int_0^{2\pi}f(x) \sin nx\,dx \quad(n\in\mathbb{N}); \\\widehat{f}(n)=&\frac{1}{2\pi}\int_0^{2\pi}f(x) e^{-inx}dx\quad(n\in \mathbb{Z}). \end{aligned}
(2′)
Unlike the previous section, where we considered only functions of class , here we will deal with arbitrary functions summable on (0,2π). It is obvious that, in this case, the integrands in formulas (2) and (2′) will also be summable. Therefore, we keep the terminology introduced above (a Fourier coefficient, a Fourier series) for the functions in . We are now interested not in convergence in the -norm, but in other types of convergence, and first of all, pointwise convergence. Here, by the sum of the series (1′), we always mean the limit of the symmetric partial sums
$$S_n(f,x)=\sum_{|k|\leqslant n}\widehat{f}(k) e^{ikx},$$
(3)
which are also called the Fourier sums of the function f. As noted in Sect. 10.2.1, the partial sums of series (1) and (1′) are equal. Thus, all results obtained for one of the series are valid for the other one. In the sequel, we will mainly consider series (1′) because this leads to some technical simplifications.

In conclusion, we touch on a question that may arise when solving the problem of the expansion of a function as a trigonometric series. Up to now, the choice of its coefficients have been dictated by geometric considerations presented in Sect. 10.1 and has led to formulas (2) and (2′). Can it happen that, for a different mode of convergence (e.g., pointwise or in measure) the coefficients of the trigonometric series must be chosen in a different way? It is easy to verify, however, that, under mild additional assumptions, there is essentially no freedom in the choice of the coefficients. Indeed, if, for example, a trigonometric series $$\sum_{k=-\infty}^{\infty} c_{k} e^{ikx}$$ converges to a function f almost everywhere or in measure and its partial sums S n (x)=∑|k|⩽n c k e ikx have a summable majorant, i.e., a function such that |S n (x)|⩽g(x) for all x∈(0,2π) and $$n\in\mathbb{N}$$, then the coefficients of the series coincide with the Fourier coefficients of the function f, $$c_{k}\equiv\widehat{f}(k)$$. Indeed, by Lebesgue’s theorem, the integral $$\widehat{f}(k)=\frac{1}{2\pi}\int_{0}^{2\pi}f(x) e^{-ikx}\,dx$$ is the limit (as n→∞) of the integrals $$\frac{1}{2\pi}\int_{0}^{2\pi}S_{n}(x) e^{-ikx}\,dx$$, each of which is equal to c k for n⩾|k|.

### 10.3.2

Instead of functions defined only on the interval (0,2π), it will be more convenient for us to deal with 2π-periodic functions. Since every function defined on (0,2π) can be extended to a periodic function, we will assume in what follows that all functions in question are periodic (in the sequel, periodicity means 2π-periodicity). Being summable on an interval of length 2π, such functions are summable on each finite interval. We will repeatedly use the fact that the integral $$\int_{a}^{a+2\pi}f(x)\,dx$$ does not depend on the parameter a (the reader is invited to prove this independently). Often, especially when dealing with odd and even functions, it is more convenient to integrate over the interval (−π,π) in formulas (2) and (2′). By $$\widetilde{C}$$ and $$\widetilde{C}^{r}$$ (1⩽r⩽+∞), we denote the classes of periodic functions that are continuous and, respectively, r times continuously differentiable on $$\mathbb{R}$$; by , we denote the class of periodic functions summable on (−π,π) with power p⩾1. For a function , by ∥f p we mean the -norm of its restriction to (−π,π).

We note the following elementary properties of the Fourier coefficients.
1. (a)

$$|\widehat{f}(n)|\leqslant\frac{1}{2\pi}\|f\|_{1}$$ (see formula (2′)).

2. (b)

$$\widehat{f}(n)\underset{|n|\to+\infty}{\longrightarrow}0$$ (see the Riemann–Lebesgue theorem).

This qualitative result can be supplemented by an estimate connected with the continuity in the mean (see Exercise 1).
The properties connecting Fourier coefficients with translation, differentiation, and convolution play an important role. We recall that the translation f h of a function corresponding to a number h is defined by the formula f h (x)=f(xh). Making the change of variable xhx in the integral $$\int_{0}^{2\pi}f(x-h) e^{-inx}\,dx$$, we arrive at the formula
(c)

$$\widehat{f_{h}}(n)=e^{-inh}\widehat{f}(n)$$.

(d)
If a periodic function f is absolutely continuous on $$\mathbb{R}$$ (in particular, if it is piecewise differentiable), then
$$\widehat{f'}(n)=in\widehat{f}(n)\quad(n\in\mathbb{Z})$$
(for the proof, it is sufficient to integrate by parts). In particular, $$\widehat{f}(n)=o(1/n)$$. We note a weak version of this estimate for a function of bounded variation.
(d′)
If f is a function of bounded variation on the interval [0,2π], then $$\widehat{f}(n)=O(1/n)$$. Indeed, integrating by parts (see Sect. 4.11.4), we obtain
\begin{aligned} 2\pi\widehat{f}(n)=&\int_0^{2\pi}f(x) e^{-inx}\,dx= f(x)\frac{e^{-inx}}{-in}\bigg\vert _0^{2\pi}+ \frac{1}{in}\int_0^{2\pi}e^{-inx} \,df(x)\\ =&O \biggl(\frac{1}{n} \biggr). \end{aligned}
(e)
Let . Then
$$\widehat{f*g}(n)=2\pi\widehat{f}(n)\cdot\widehat{g}(n)\quad\text{for all}\ n \in\mathbb{Z}$$
(for the definition of the convolution of periodic functions, see Sect. 7.5.5). The proof is obtained by direct calculation using the change of the order of integration,
\begin{aligned} \widehat{f*g}(n)=&\frac{1}{2\pi}\int_{-\pi}^{\pi}(f*g) (x) e^{-inx}\,dx\\=& \frac{1}{2\pi}\int_{-\pi}^{\pi} \biggl(\int_{-\pi}^{\pi} f(x-t)g(t)\, dt \biggr)e^{-inx}\,dx \\=&\frac{1}{2\pi}\int_{-\pi}^{\pi} g(t) e^{-int} \biggl(\int_{-\pi}^{\pi} f(x-t) e^{-in(x-t)}dx \biggr)dt \\=&\frac{1}{2\pi}\int_{-\pi}^{\pi} g(t) e^{-int} \biggl(\int_{-\pi}^{\pi} f(u) e^{-inu}du \biggr)dt=2\pi\widehat {g}(n)\cdot\widehat{f} (n). \end{aligned}

### 10.3.3

The problem of the Fourier series expansion of a function is rather complicated and has a long history. The famous work “The analytical theory of heat” by Fourier, in which the series that were later named after him were first studied and used systematically, did not contain an explicit formulation of a condition providing the expandability of a function as a Fourier series. Such criteria arose later. Still later it became clear that the Fourier series of a continuous function can diverge at some points, and, as Kolmogorov proved, the Fourier series of a summable function can diverge everywhere.

So far, even knowing that a Fourier series of a differentiable function converges at a point, we cannot be sure that its sum coincides with the value of the function.

At the moment, we know (see Sect. 10.2.1) that if f is a square-summable function, then series (1′) converges in the -norm and its sum is equal to f. If a function f is only assumed to be summable, the question of the convergence of a Fourier series (pointwise, in an -norm, or in some other sense) remains open for the time being.

We begin the investigation of a Fourier series’ convergence with the derivation of an important formula for its partial sums discovered by Dirichlet. Relying on formula (2′), we transform Eq. (3) as follows:
$$S_n(f,x)=\sum_{|k|\leqslant n} \biggl( \frac{1}{2\pi}\int_{-\pi}^{\pi} f(t) e^{-ikt}\,dt \biggr) e^{ikx}=\frac{1}{2\pi}\int _{-\pi}^{\pi} f(t)\sum_{|k|\leqslant n}e^{ik(x-t)} \,dt.$$
The function
$$D_n(u)=\frac{1}{2\pi}\sum_{|k|\leqslant n}e^{iku}= \frac{1}{2\pi}+ \frac{1}{\pi}\sum_{k=1}^n \cos ku$$
(4)
is called the nth Dirichlet kernel. Obviously, the Dirichlet kernel is even and periodic. Summing the geometric sequence ∑|k|⩽n e iku , we obtain
$$D_n(u)=\frac{\sin (n+\frac{1}{2} )u}{2\pi\sin\frac{u}{2}}\quad \text{for }\ u\notin2\pi\mathbb{Z}.$$
(4′)
From this, we see that the function D n is strongly oscillating for large n, and, in a neighborhood of zero, it takes extreme values with alternating signs and absolute values comparable with $$\max D_{n}=D_{n}(0)=\frac{1}{\pi} (n+\frac{1}{2} )$$ (see Fig. 10.1). Fig. 10.1 Graph of the Dirichlet kernel
It follows directly from the definition that the sum of the Fourier series is the convolution of the function and the Dirichlet kernel,
$$S_n(f,x)=\int_{-\pi}^{\pi} f(t)D_n(x-t)\,dt=(f*D_n) (x).$$
Since the integrands are periodic, we can also represent the above equation in the form
$$S_n(f,x)=\int_{-\pi}^{\pi} f(x-u)D_n(u)\,du.$$
(5)
Considering periodic approximate identities, we have encountered similar formulas (see Sect. 7.6.5). The Dirichlet kernels satisfy conditions (b) and (c) of the definition of a periodic approximate identity; it immediately follows from Eq. (4) that
$$\int_{-\pi}^{\pi} D_n(u)\,du=1.$$
Moreover, we have
$$\int_{\delta <|u|<\pi} D_n(u)\,du=\int_{\delta <|u|<\pi} \frac{\sin (n+\frac{1}{2} )u}{2\pi\sin\frac{u}{2}}\,du \underset{n\to\infty}{\longrightarrow} 0$$
for each δ∈(0,π) (the passage to the limit can be justified by integration by parts or by referring to the Riemann–Lebesgue theorem).
However, D n does not satisfy the most important property of an approximate identity, namely, the positivity. Moreover, the Dirichlet kernels do not satisfy the periodic analog of condition (a′) of Sect. 7.6.1, i.e., they have unbounded -norms. Indeed,
\begin{aligned} \int_{-\pi}^{\pi}\big|D_n(u)\big|\,du=& \int _0^{\pi}\frac{|\!\sin(n+\frac{1}{2})u|}{\pi\sin\frac{u}{2}}\,du\geqslant \frac{2}{\pi}\int_0^{\pi}\frac{|\!\sin(n+\frac{1}{2})u|}{u} \,du \\=&\frac{2}{\pi}\int_0^{\pi(n+\frac{1}{2})} \frac{|\!\sin v|}{v}\,dv\geqslant \frac{2}{\pi}\sum _{k=1}^n\int_{(k-1)\pi}^{k\pi} \frac{|\!\sin v|}{k\pi}\,dv =\frac{4}{\pi^2}\sum_{k=1}^n \frac{1}{k}. \end{aligned}
Since $$\sum_{k=1}^{n}\frac{1}{k}\geqslant\int_{1}^{n}\frac{1}{x}\,dx=\ln n$$, we have $$\|D_{n}\|_{1}\geqslant\frac{4}{\pi^{2}}\ln n$$ (see also Exercise 11).

Thus, the general theorems connected with the use of approximate identities cannot be applied here. This is the cause of considerable difficulties in the study of the convergence of Fourier series. Here, we meet not just technical questions, but those of a fundamental nature. We will see later that the proofs of Theorems 2 and 3 of Sect. 9.3.7 cannot be carried over to convolutions with Dirichlet kernels.

At the same time, in many problems, it is essential that the norms ∥D n 1 increase quite slowly. Indeed, the estimate from above for ∥D n 1 just obtained is exact in order,
$$\|D_n\|_1=\int_0^{\pi} \frac{|\!\sin (n+\frac{1}{2} )u|}{\pi\sin\frac {u}{2}}\,du \leqslant\int_0^{\pi} \frac{|\!\sin (n+\frac{1}{2} )u|}{u}\,du= \int_0^{\pi(n+\frac{1}{2})} \frac{|\!\sin v|}{v}\,dv.$$
Consequently, $$\|D_{n}\|_{1}\leqslant1+\int_{1}^{\pi(n+\frac{1}{2})}\frac{dv}{v}$$, and, therefore, ∥D n 1⩽2lnn for n⩾10. Since S n (f)=fD n , we obtain the following estimate for the Fourier sums of a bounded function (n⩾10):
$$\big\|S_n(f)\big\|_{\infty}\leqslant\|f\|_{\infty} \|D_n\|_1\leqslant2\|f\|_{\infty} \ln n.$$
(6)

The partial sums of the Fourier series are calculated by formula (5), and so depend on the values of the function on an interval of length 2π. It is all the more surprising that, as we will now verify, the convergence of the Fourier series at a point x and the value of its sum are local properties of the function, i.e., they are preserved under an arbitrary change of the function outside an arbitrarily small neighborhood of the point. More formally, we have the following.

### Theorem

(Riemann’s localization principle)

If functions coincide in a neighborhood of a point x, then their Fourier series have the same behavior at x, S n (f 1,x)−S n (f 2,x)→0 as n→∞.

### Proof

From the assumptions it follows that the function $$\varphi _{x}(u)=\frac{f_{1}(x+u)-f_{2}(x-u)}{\sin(u/2)}$$ (equal to zero in a neighborhood of the point u=0) is summable on (−π,π). Since
$$S_n(f_1,x)-S_n(f_2,x)= \frac{1}{2\pi} \int_{-\pi}^{\pi} \varphi _x(u)\sin \biggl(n+\frac{1}{2} \biggr)u\,du$$
by Eq. (5), it remains to refer to the Riemann–Lebesgue theorem according to which the integral on the right-hand side of this equation tends to zero. □

### 10.3.4

Among a great variety of convergence tests for Fourier series, we mention only two of the most applicable ones, the Dini12 test and the Dirichlet–Jordan13 test. They supplement each other and can be applied to a wide range of cases.

First, we establish a useful property of the Dirichlet kernel.

### Lemma

Let $$n\in\mathbb{N}$$. Then:
$$(\mathrm{a})\quad D_n(u)=\frac{\sin nu}{\pi u}+\frac{1}{2\pi} \bigl(\cos nu+\Delta(u)\sin nu\bigr),$$
where Δ is a function independent of n and |Δ(u)|<1 for |u|⩽π;
$$(\mathrm{b})\quad \biggl \vert \int_0^x D_n(u)\,du\biggr \vert \leqslant 2\quad\text{\textit{for}}\ |x|\leqslant2 \pi.$$

### Proof

(a) It is clear that
$$D_n(u)=\frac{\sin nu}{2\pi\text{tan}\,\frac{u}{2}}+\frac{1}{2\pi}\cos nu= \frac{\sin nu}{\pi u}+\frac{1}{2\pi} \biggl(\cos nu+ \biggl(\frac{1}{\text{tan}\,\frac{u}{2}}- \frac{2}{u} \biggr)\sin nu \biggr).$$
It remains to observe that the difference $$\Delta(u)=\frac{1}{\text {tan}\,\frac{u}{2}}- \frac{2}{u}$$ (Δ(0)=0) decreases on [−π,π], and, therefore, $$|\Delta(u)|\leqslant|\Delta(\pi)|=\frac{2}{\pi}<1$$.
(b) It is sufficient to consider the case where x∈(0,2π). First let x∈(0,π]. Then assertion (a) proved above implies the inequality
$$\biggl \vert \int_0^x D_n(u) \,du-\int_0^x\frac{\sin nu}{\pi u}\,du\biggr \vert \leqslant \frac{1}{2\pi}\int_0^x2 \,du\leqslant1.$$
Now, we prove that the integral
$$J_n(x)=\int_0^x \frac{\sin nu}{\pi u}\,du= \int_0^{nx} \frac{\sin v}{\pi v}\,dv,$$
lies between 0 and 1. To verify this, we divide the interval of integration [0,nx] into parts on which sinv preserves its sign. Then the integral J n (x) splits into the alternating sum of terms whose absolute values decrease since $$\frac{1}{v}$$ decreases. Therefore,
$$0\leqslant J_n(x)\leqslant\int_0^{\pi} \frac{\sin v}{\pi v}\,dv\leqslant \int_0^{\pi} \frac{dv}{\pi}=1.$$
Thus, the integral $$\int_{0}^{x} D_{n}(u)\,du$$ lies between −1 and 2 provided 0<xπ.
For x∈(π,2π), we use the easily verifiable relation
$$\int_0^xD_n(u)\,du=1-\int _0^{2\pi-x} D_n(u)\,du$$
from which it follows that the inequality $$-1\leqslant\int_{0}^{x}D_{n}(u)\,du \leqslant2$$ also holds in this case. □
Using the first assertion of the lemma, we can represent Eq. (5) in the following form:
$$S_n(f,x)=\int_{-\pi}^{\pi} f(x-u) \frac{\sin nu}{\pi u}\,du+\varepsilon _n,$$
(5′)
where the quantity $$\varepsilon _{n}=\frac{1}{2\pi}\int_{-\pi}^{\pi} f(x-u)(\cos nu+\Delta(u)\sin nu)\,du$$ tends to zero by the Riemann–Lebesgue theorem.
In particular, if f≡1, then
$$1=\int_{-\pi}^{\pi}\frac{\sin nu}{\pi u}\,du+o(1).$$
(5″)
Making the change of variable nu=t and passing to the limit as n→∞, we once again obtain the equality $$\int_{0}^{\infty}\frac{\sin t}{t}\,dt=\frac{\pi}{2}$$ established in Sect. 7.1.6 by a different method.

### Theorem

(Dini test)

If a function satisfies the Dini condition
$$\int_0^{\pi}\biggl \vert \frac{f(x+u)+f(x-u)}{2}-C \biggr \vert \,\frac{du}{u}<+\infty$$
at a point $$x\in\mathbb{R}$$ for some $$C\in\mathbb{C}$$, then its Fourier series converges to C at the point x.
In particular, if f is differentiable at x, then the Dini condition is fulfilled with C=f(x), and so the sum of the Fourier series is equal to f(x). However, if only the one-sided limits f(x±0) exist and
$$\big|f(x\pm u)-f(x\pm0)\big|=O\bigl(u^{\alpha }\bigr)\quad\text{as}\ u\to+0$$
for some α>0, then the Fourier series of f at x converges to the average $$\frac{f(x-0)+f(x+0)}{2}$$.

### Proof

From (5′), it follows that
$$S_n(f,x)=\int_{-\pi}^{\pi} f(x-u) \frac{\sin nu}{\pi u}\,du+o(1)= \int_{-\pi}^{\pi} f(x+u) \frac{\sin nu}{\pi u}\,du+o(1)$$
as n→∞. Thus,
$$S_n(f,x)=\int_{-\pi}^{\pi} \frac{f(x-u)+f(x+u)}{2}\frac{\sin nu}{\pi u}\,du+o(1).$$
Subtracting Eq. (5″) multiplied by C from the above equation, we see that
\begin{aligned} S_n(f,x)-C=&\int_{-\pi}^{\pi} \biggl( \frac{f(x-u)+f(x+u)}{2}-C \biggr) \frac{\sin nu}{\pi u}\,du+o(1) \\ =&\frac{2}{\pi}\int_0^{\pi} g_x(u)\sin nu\,du+o(1), \end{aligned}
where $$g_{x}(u)=\frac{f(x-u)+f(x+u)-2C}{2u}$$. Since the function g x is summable on (0,π) by the assumptions of the theorem, the integral on the right-hand side of this equation tends to zero by the Riemann–Lebesgue theorem. □

### Theorem

(Dirichlet–Jordan test)

If a periodic function f has bounded variation on the interval [−π,π], then, for each $$x\in\mathbb{R}$$, the Fourier series of f converges to the average (f(x+0)+f(x−0))/2. Moreover, $$|S_{n}(f,x)|\leqslant\sup_{\mathbb{R}}|f|+2\mathbf{V}_{-\pi}^{\pi}(f)$$.

We remark that the convergence of a Fourier series at a point x is preserved by the localization principle if we assume that f has bounded variation only locally, in a neighborhood of this point.

### Proof

By Eq. (5′), we must find the limit of the integrals
$$I_n=\int_{-\pi}^{\pi} f(x-u) \frac{\sin nu}{\pi u}\,du= \int_0^{\pi} \varphi (u) \frac{\sin nu}{\pi u}\,du,$$
where φ(u)=f(xu)+f(x+u). This function has bounded variation on [0,π], and so can be represented as the difference of decreasing functions. Therefore, it is sufficient for us to find the limit of the integrals I n under the assumption that the function φ is non-negative and decreases on the interval [0,π]. To this end, we represent I n in the form
$$I_n=\int_0^{\infty}\Phi(u) \frac{\sin nu}{\pi u}\,du =\int_0^{\infty}\Phi \biggl( \frac{t}{n} \biggr)\frac{\sin t}{\pi t}\,dt,$$
where Φ(u)=φ(u)χ (0,π)(u). By Corollary 2 of Sect. 7.4.7, the integral on the right-hand side of the above equation tends to φ(+0)/2=(f(x−0)+f(x+0))/2.
To obtain a uniform estimate for the sums S n (f), we put $$H_{n}(u)=\int_{0}^{u}D_{n}(t)\,dt$$. Then
\begin{aligned} S_n(f,x)=&\int_{-\pi}^{\pi} f(x-u)D_n(u)\,du\\ =& H_n(u)\,f(x-u)\Big\vert _{u=-\pi}^{\pi}-\int_{-\pi}^{\pi} H_n(u)\,df(x-u). \end{aligned}
Since $$H_{n}(\pm\pi)=\pm\frac{1}{2}$$, the first summand is equal to (f(xπ)+f(x+π))/2. Furthermore, |H n (u)|⩽2 by the lemma, and so
$$\biggl \vert \int_{-\pi}^{\pi} H_n(u) \,df(x-u)\biggr \vert \leqslant 2\mathbf{V}_{x-\pi}^{x+\pi}(f)=2 \mathbf{V}_{-\pi}^{\pi}(f),$$
hence the required estimate for the sums S n (f,x) follows. □
In conclusion, we prove the Dini test in a different way, without using Dirichlet kernels (see [Ch]). The Dini condition means that the function
$$g(u)= \biggl(\frac{f(x+u)+f(x-u)}{2}-C \biggr)\frac{1}{e^{iu}-1}\quad (u\notin2\pi \mathbb{Z})$$
belongs to the class . Multiplying both sides of the equation
$$\frac{f(x+u)+f(x-u)}{2}-C= \bigl(e^{iu}-1 \bigr)g(u)$$
by $$\frac{1}{2\pi} e^{-iku}$$ and then integrating with respect to u∈(−π,π), we obtain
$$\begin{array}{rcl@{\quad}l} \displaystyle\frac{1}{2} \bigl(\widehat{f}(k) e^{ikx}+\widehat{f}(-k) e^{-ikx} \bigr) &=&\widehat{g}(k-1)-\widehat{g}(k),&\text{if}\ k \ne0, \\[12pt] \widehat{f}(0)-C&=&\widehat{g}(-1)-\widehat{g}(0),& \text{if}\ k=0. \end{array}$$
It remains to sum all these equations for |k|⩽n,
$$S_n(f,x)-C=\sum_{k=-n}^n \widehat{f}(k) e^{ikx}-C= \widehat{g}(-n-1)-\widehat{g}(n)\underset{n\to+ \infty}{\longrightarrow}0.$$

If we sum them for k=0,1,…,n and for k=−1,…,−n separately, it becomes clear that the Dini condition implies the convergence of not only the symmetric sums $$\sum_{k=-n}^{n}\widehat{f}(k) e^{ikx}$$, but also the “one-sided” sums $$\sum_{k=0}^{n}\widehat{f}(k) e^{ikx}$$ and $$\sum_{k=-n}^{-1}\widehat{f}(k) e^{ikx}$$. In other words, the Dini condition ensures the convergence of each of the series $$\sum_{k=0}^{\infty}\widehat{f}(k) e^{ikx}$$ and $$\sum_{k=-\infty }^{-1}\widehat{f}(k) e^{ikx}$$. In particular, it ensures the convergence of the series $$\sum_{n\in\mathbb{Z}}\text{sign}(n)\widehat{f}(n) e^{inx}$$ called the conjugate to series (1′).

### 10.3.5

We give some examples of Fourier series expansions.

### Example 1

We supplement Example 1 of Sect. 10.2.1 as follows: since the periodic function equal x on (−π,π) is differentiable at all points distinct from (2k+1)π $$(k\in\mathbb{Z})$$, its Fourier series converges not only in the -norm, but also pointwise,
$$x=2\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n} \sin nx\quad \text{for}\ x\in(-\pi,\pi).$$
At the points (2k+1)π, the sum of the series is equal to the average of the one-sided limits of the function. At $$x=\frac{\pi}{2}$$, the Fourier series expansion yields the relations
$$\frac{\pi}{4}=\sum_{m=0}^{\infty}(-1)^m \frac{1}{2m+1}.$$

Considering the Fourier series expansion of the function equal to x 2 on [−π,π] at the point π, we again obtain the Euler identity $$\sum_{n=1}^{\infty}\frac{1}{n^{2}}=\frac{\pi^{2}}{6}$$ (see Example 1 of Sect. 10.2.1 or Example 2 of Sect. 4.6.2).

### Example 2

Let $$w\in\mathbb{C}\setminus\mathbb{Z}$$. We consider the periodic function equal to coswx on the interval [−π,π]. This function has finite one-sided derivatives everywhere on $$\mathbb{R}$$ and, therefore, can be expanded in a Fourier series. After elementary transformations, we obtain that the equation
$$\cos wx=\frac{\sin\pi w}{\pi w}+ \frac{2}{\pi}w\sin\pi w\sum _{n=1}^{\infty}\frac{(-1)^n}{w^2-n^2} \cos nx$$
holds for all |x|⩽π.
For x=π and x=0, the equation implies the following expansions of cotangent and cosecant as sums of partial fractions:
\begin{aligned} \text{cot}\,\pi w=&\frac{1}{\pi w}+\frac{2w}{\pi}\sum _{n=1}^{\infty}\frac {1}{w^2-n^2}= \frac{1}{\pi} \sum_{n=-\infty}^{\infty}\frac{1}{w-n}, \\\frac{1}{\sin\pi w}=&\frac{1}{\pi w}+\frac{2w}{\pi}\sum _{n=1}^{\infty} \frac{(-1)^n}{w^2-n^2}=\frac{1}{\pi} \sum_{n=-\infty}^{\infty}\frac{(-1)^n}{w-n}. \end{aligned}

### Example 3

Here, we verify the existence of a convergent non-zero numerical series $$\sum_{n=1}^{\infty} a_{n}$$ with the unusual property $$\sum_{m=1}^{\infty} a_{km}=0$$ for every k. In the construction, we follow F.L. Nazarov who suggested the use of Fourier series for this purpose. We consider periodic functions equal to zero in a neighborhood of each point of the form πt, $$t\in\mathbb{Q}$$. Among them, we can, obviously, find an even function f satisfying the conditions $$\widehat{f}(0)=\frac{1}{2\pi}\int_{-\pi}^{\pi} f(x)\,dx=0$$ and $$0<\int_{-\pi}^{\pi}|f(x)|^{2}\,dx<+\infty$$. We take the required series equal to $$\sum_{n=1}^{\infty}\widehat{f}(n)$$. This is a non-zero series since
$$0<\int_{-\pi}^{\pi}\big|f(x)\big|^2\,dx= \frac{1}{2\pi}\sum_{n=-\infty}^{\infty} \big|\widehat{f}(n)\big|^2=\frac{1}{\pi}\sum_{n=1}^{\infty}\big| \widehat{f}(n)\big|^2$$
(here we have used Parseval’s identity).
At each point $$x=2\pi\frac{j}{k}$$ ($$j\in\mathbb{Z}$$, $$k\in\mathbb{N}$$), the function f satisfies the Dini condition and, therefore,
$$\sum_{n=1}^{\infty}\widehat{f}(n)\cos \biggl(2\pi\frac{j}{k}n \biggr)=0.$$
Summing these equations for j=0,1,…,k−1, we obtain
$$\sum_{n=1}^{\infty}\widehat{f}(n)\sum _{j=0}^{k-1}\cos \biggl(2\pi\frac{j}{k}n \biggr)=0.$$
If the index n is divisible by k, then the inner sum is equal to k, since otherwise this sum is obviously zero. Consequently,
$$k\sum_{m=1}^{\infty}\widehat{f}(km)=0\quad \text{for all}\ k\in\mathbb{N}.$$

We remark that the series just constructed does not converge absolutely. It can be proved (see [PS], Part 1, Problem 129) that it is impossible to construct an absolutely convergent series with the property in question.

### 10.3.6

As we have already noted, the Fourier series of a summable, or even of a continuous, function may diverge (see also Sect. 10.3.9). However, such a series has the remarkable property that it can be integrated termwise over an arbitrary finite interval without worrying about convergence.

### Theorem 1

Let . Then the equation
$$\int_a^b f(x)\,dx=\sum _{n=-\infty}^{\infty}\widehat{f}(n)\int_a^b e^{inx}\,dx$$
(where the sum is regarded as the limit of the symmetric partial sums) is valid for all $$a,b\in \mathbb{R}$$.

### Proof

Taking into account the periodicity, we restrict ourselves, without loss of generality, to the case where −πa<bπ. Let χ be the characteristic function of the interval (a,b). Then a partial sum of the series on the right-hand side of the required equation can be represented in the form
\begin{aligned} \sum_{k=-n}^n \widehat{f}(k)\int _a^b e^{ikx}\,dx = &\sum _{k=-n}^n \biggl(\frac{1}{2\pi}\int _{-\pi}^{\pi} f(t) e^{-ikt}\,dt \biggr) 2\pi \widehat{\chi}(-k) \\= &\int_{-\pi}^{\pi} f(t)S_n(\chi,t)\,dt. \end{aligned}
(7)
By Dini’s test, we have $$S_{n}(\chi,t)\underset{n\to\infty}{\longrightarrow} \chi(t)$$ for t∈(−π,π) and ta,b. Moreover,
\begin{aligned} S_n(\chi,t)=&\int_a^b D_n(x-t)\,dx=\int_{a-t}^{b-t} D_n(u)\,du\\=& \int_0^{b-t} D_n(u)\,du-\int_0^{a-t}D_n(u) \,du. \end{aligned}
Therefore, Lemma 10.3.4 gives use the uniform estimate |S n (χ,t)|⩽4. By Lebesgue’s theorem, we can pass to the limit on the right-hand side of Eq. (7) and obtain
$$\sum_{k=-n}^n\widehat{f}(k)\int _a^b e^{ikx}dx=\int _{-\pi}^{\pi} f(t) S_n(\chi,t)\,dt \underset{n\to\infty}{\longrightarrow}\int_{-\pi}^{\pi} f(t)\chi(t)\,dt= \int_a^b f(t)\,dt.$$
□

Theorem 1 allows us to considerably strengthen the assertion on the completeness of the trigonometric system, according to which two functions of the class that have the same Fourier coefficients coincide almost everywhere. Now, we can extend this result to the class .

### Corollary 1

Functions having the same Fourier coefficients coincide almost everywhere on $$\mathbb{R}$$.

### Proof

By the theorem, the integrals of f and g are equal on every finite interval. Therefore, (see Corollary 4.5.4) f and g coincide almost everywhere. □

### Corollary 2

For every function , the series $$\sum_{n=1}^{\infty} b_{n}(f)/n$$ converges.

We recall that $$b_{n}(f)=\frac{1}{\pi}\int_{-\pi}^{\pi} f(x)\sin nx\,dx= i(\widehat{f}(n)-\widehat{f}(-n))$$ is the Fourier sine coefficient of f.

### Proof

As established in the theorem, the equation
$$\int_0^u f(x)\,dx=\sum _{n=-\infty}^{\infty}\widehat{f}(n)\int_0^u e^{inx}\,dx$$
holds for all u. From (7) and the estimate |S n (χ,t)|⩽4, it follows that the symmetric partial sums of this series are uniformly bounded for u∈[−π,π]. Therefore, we can integrate the series termwise,
$$\int_{-\pi}^{\pi} \biggl(\int_0^u f(x)\,dx \biggr)\,du= \sum_{n=-\infty}^{\infty} \widehat{f}(n)\int_{-\pi}^{\pi} \biggl(\int _0^u e^{inx}\,dx \biggr)\,du=-2\pi\sum _{n\ne0}\frac{\widehat{f}(n)}{in}.$$
The convergence of the symmetric partial sums of the series $$\sum_{n\ne 0}\frac{\widehat{f} (n)}{n}$$ is equivalent to the required statement. □

Corollary 2 gives a necessary condition for a trigonometric series $$\sum_{n=1}^{\infty}(a_{n}\cos nx+b_{n}\sin nx)$$ to be a Fourier series. The everywhere convergent series $$\sum_{n=2}^{\infty}(\sin nx)/\ln n$$ does not satisfy this condition and, therefore, cannot be the Fourier series of a summable function. It is interesting to note that, in contrast to the sine coefficients, the cosine coefficients can tend to zero arbitrarily slowly. For example, the series $$\sum_{n=2}^{\infty}(\cos nx)/\ln n$$ is the Fourier series of a summable function (see Theorem 10.4.2).

The relation obtained in Theorem 1 can be regarded as a new version of Parseval’s identity in which the assumption about one function is weakened (it belongs to but not to ) and the assumption concerning the other is strengthened considerably (it is the characteristic function of an interval). At the same time, the proof of the theorem uses the properties of the function χ only partially. This makes it possible to extend considerably the applicability conditions of Parseval’s identity.

### Theorem 2

Let , and let g be a bounded (measurable and periodic) function whose Fourier sums S n (g,x) are uniformly bounded (with respect to x and n). Then the following Parseval identity is valid:
$$\int_{-\pi}^{\pi} f(x)\overline{g}(x)\,dx=2\pi\sum _{n=-\infty}^{\infty} \widehat{f}(n)\overline{ \widehat{g}(n)}.$$

The class of functions with uniformly bounded partial sums of Fourier series is sufficiently wide. In particular, it contains all smooth functions on [−π,π]. As follows from the Dirichlet–Jordan test, this class also contains all functions with finite variation on [−π,π] (see also Exercises 9 and 10).

The assumption that the function g is bounded is superfluous (see Exercise 8 or Fejér’s theorem in Sect. 10.4).

### Proof

Since , the sums S n (g) converge to g in the -norm and, a fortiori, in measure. This implies, as one can easily verify, that
$$f(x)\overline{S_n}(g,x)\to f(x)\overline{g}(x)\quad \text{in measure.}$$
Therefore, we can use Lebesgue’s theorem and pass to the limit on the right-hand side of the equation
$$\int_{-\pi}^{\pi} f(x)\overline{S_n}(g,x) \,dx= 2\pi\sum_{|k|\leqslant n} \widehat{f}(k)\overline{\widehat{g}(k)},$$
as required. □

### 10.3.7

To obtain a further generalization of the uniqueness theorem for Fourier series, (see Corollary 1 of the previous section), we introduce the notion of Fourier coefficients and Fourier series for a measure.

### Definition

Let μ be a finite Borel measure on the interval [−π,π]. The Fourier coefficients of μ are defined by the formula
$$\widehat{\mu}(n)=\frac{1}{2\pi}\int_{[-\pi,\pi]}e^{-inx} \,d\mu(x)\quad (n\in\mathbb{Z}).$$
The series $$\sum_{n=-\infty}^{\infty}\widehat{\mu}(n) e^{inx}$$ is called the Fourier series of μ.
If a measure μ has density f with respect to Lebesgue measure, then $$\widehat{\mu}(n)=\widehat{f}(n)$$ for all $$n\in\mathbb{Z}$$ and, consequently, the Fourier series of the measure μ and of the function f coincide. As in the case of Fourier series, it follows directly from definition that the nth (symmetric) partial sum of the Fourier series of a measure, which will be denoted by S n (μ,x), is the convolution of this measure and a Dirichlet kernel,
$$S_n(\mu,x)=\int_{[-\pi,\pi]} D_n(x-t)\,d \mu(t)=(D_n*\mu) (x).$$
Extending Corollary 1 to measures, we must take into account the relation
$$\widehat{\mu}(n)=\frac{(-1)^n}{2\pi}\bigl(\mu\bigl(\{-\pi\}\bigr)+\mu\bigl(\{\pi\} \bigr)\bigr)+ \frac{1}{2\pi}\int_{(-\pi,\pi)} e^{-inx}\,d \mu(x).$$
Thus, the Fourier coefficients do not change under redistribution of the loads (preserving their sum) at the points ±π. This will be the case when we replace these loads by, for example, μ({−π})+μ({π}) (at the point −π) and by 0 (at the point π). Therefore, it makes sense to pose the question of whether a measure is uniquely determined by its Fourier coefficients only if we fix the load at one of the points ±π. For definiteness, we will consider only the measures that have zero load at the point π.

### Theorem

Let μ and ν be finite Borel measures on the interval [−π,π] satisfying the condition μ({π})=ν({π})=0. If the Fourier coefficients of these measures coincide, then the measures also coincide.

### Proof

First, we verify that the Fourier series of a measure, as well as the Fourier series of a function, can be integrated termwise, i.e., if μ({a})=μ({b})=0, then
$$\sum_{n=-\infty}^{\infty}\widehat{\mu}(n)\int _a^be^{inx}\,dx=\mu\bigl([a,b)\bigr)$$
(8)
for [a,b)⊂[−π,π). Indeed, let χ=χ [a,b). Then
\begin{aligned} \sum_{|k|\leqslant n}\widehat{\mu}(k)\int_a^be^{ikx} \,dx= &\sum_{|k|\leqslant n}\widehat{\chi}(-k)\int _{[-\pi,\pi]}e^{-ikx}d\mu(x) \\=& \int_{[-\pi,\pi]} S_n(\chi,x)\,d\mu(x). \end{aligned}
(9)
In the proof of Theorem 1 of Sect. 10.3.6, we have established that |S n (χ,t)|⩽4. Moreover, $$S_{n}(\chi,t)\underset{n\to\infty}{\longrightarrow}\chi(t)$$ for ta,b and, consequently, μ-almost everywhere. Therefore, we can use Lebesgue’s theorem and pass to the limit on the right-hand side of Eq. (9), which leads to Eq. (8). Thus, if measures μ and ν have the same Fourier coefficients, then μ([a,b))=ν([a,b)) for every interval [a,b)⊂[−π,π) satisfying the condition μ({a})=μ({b})=ν({a})=ν({b})=0. Since the set of points of non-zero measure is at most countable (see Sect. 1.2.2), this condition is fulfilled on a dense subset of (−π,π). Hence it follows (see Remark 1.1.7) that the measures μ and ν coincide on all Borel subsets of the interval (−π,π). At the same time, μ({π})=ν({π})=0 and $$\mu([-\pi,\pi])=\widehat{\mu}(0)=\widehat{\nu}(0)=\nu([-\pi,\pi])$$ by assumption. Consequently, the measures μ and ν have the same loads at the point −π, which completes the proof of the theorem. □

Generalizations of this theorem are given in Sects. 10.4.7, 11.1.9, and 12.3.3.

### 10.3.8

Considering Fourier series with coefficients that tend to zero sufficiently fast, we must take into account that if the Fourier series of a function f converges uniformly, then its sum coincides with f almost everywhere by the uniqueness theorem. Therefore, if the Fourier series of a continuous function converges uniformly, then its sum coincides with the function. Taking this into account, we consider only continuous functions in the theorems of this section. Lifting the assumption of continuity, we must replace the equality of a function and its Fourier series by their equivalence.

The Fourier coefficients of smooth functions tend to zero sufficiently fast. For example, if a function satisfies the Lipschitz condition of order α, then $$\widehat{f}(n) = O(|n|^{-\alpha })$$. Indeed, if $$h = \frac{\pi}{n}$$, then property (c) of Sect. 10.3.2 implies that $$\widehat{f}_{\frac{\pi}{n}}(n) = -\widehat{f}(n)$$. Consequently, $$2\widehat{f}(n) = \frac{1}{2\pi}\int_{0}^{2\pi}(f(x)-f(x-\frac{\pi}{n})) e^{-inx}\,dx$$, and, therefore,
$$2\big|\widehat{f}(n)\big|\leqslant\frac{1}{2\pi}\int_0^{2\pi} \bigg|f(x)-f \biggl(x-\frac{\pi}{n} \biggr) \bigg|\,dx\leqslant L\biggl \vert \frac{\pi}{n} \biggr \vert ^{\alpha },$$
where L is a Lipschitz constant for f.

The repeated application of the relation $$\widehat{f'}(n)=in\widehat {f}(n)$$ (see property (d) of Sect. 10.3.2) shows that the Fourier coefficients of a function f of class $$\widetilde{C}^{r}$$ satisfy the relation $$\widehat{f}(n)=o(|n|^{-r})$$ as |n|→+∞. The converse is “almost true”: if $$\widehat{f}(n)=O(|n|^{-r-2})$$ for some $$r\in\mathbb{N}$$, then the continuous function f coincides with a function of class $$\widetilde{C}^{r}$$. Indeed, the series $$\sum_{n}\widehat{f}(n) e^{inx}$$ converges uniformly, and, by the above remark, its sum coincides with f. Moreover, since the coefficients decrease fast, the Fourier series admits r-fold differentiation, which implies that $$f\in\widetilde{C}^{r}$$. For infinitely smooth functions, this gives a complete description.

### Theorem 1

In order that a function $$f\in\widetilde{C}$$ be infinitely differentiable it is necessary and sufficient that the limit relation $$n^{r}\widehat{f}(n)\to0$$ as |n|→+∞ be fulfilled for every $$r\in\mathbb{N}$$.

The smaller class of holomorphic periodic functions can also be well described in terms of the Fourier coefficients: these coefficients must tend to zero not slower that a geometric sequence. We note that a periodic function f is analytic at all points of the line $$\mathbb{R}$$ if and only if, on $$\mathbb{R}$$, the function f coincides with a function holomorphic in some horizontal strip $$\{z\in\mathbb{C}\,|\,|\mathcal{I} m\,z|<L\}$$. In the proof of the following theorem, we use some elementary properties of holomorphic functions (see, for example, [Ca]).

### Theorem 2

Let $$f\in\widetilde{C}$$. The following two statements are equivalent:
1. (a)

there is a function F holomorphic in a strip $$|\mathcal{I} m\,z|<L$$ and coinciding with f on the real axis;

2. (b)

the relation $$\widehat{f}(n) = O (e^{-a|n|} )$$ as |n|→+∞ holds for every a∈(0,L).

### Proof

(a)⇒(b). Assuming that n>0 and 0<a<L, we consider the integral ∫ C F(z)e inz dz, where C is the boundary of the rectangle P with vertices at the points ±π, ±πai lying in the strip $$|\mathcal{I} m\,z|<L$$. Since the function F is holomorphic in a neighborhood of P, this integral is equal to zero. Moreover, F has period 2π, and, therefore, the sum of the integrals over the vertical sides of P are equal to zero. Consequently,
$$\widehat{f}(n)=\frac{1}{2\pi}\int_{-\pi}^{\pi} f(x) e^{-inx}\,dx= \frac{1}{2\pi}\int_{-\pi-ai}^{\pi-ai}F(z) e^{-inz}\,dz.$$
Therefore,
$$\big|\widehat{f}(n)\big|\leqslant\max_{x\in\mathbb{R}}\big|F(x-ai)\big| \big|e^{-in(x-ai)} \big|= e^{-an}\max _{x\in\mathbb{R}}\big|F(x-ai)\big|=C_a e^{-an}.$$

The coefficients with negative indices can be estimated in the same way, only in this case the rectangle is replaced by a rectangle symmetric with respect to the real axis.

(b)⇒(a). The series $$\sum_{n=-\infty}^{\infty}\widehat{f}(n) e^{inz}$$ converges uniformly in the strip $$|\mathcal{I} m\,z|\leqslant a$$ if 0<a<L. By Weierstrass’s theorem the sum of the series is holomorphic in the strip $$|\mathcal{I} m\,z|<L$$ and coincides with the function f on the real axis. □

### 10.3.9

As we have already mentioned, the Fourier series of a periodic continuous function may diverge (compare this with the result of Exercise 5). There are several such examples. We give here a slight modification of an example suggested by Schwartz. We define an even function $$f\in\widetilde{C}$$ whose oscillation frequency increases rapidly when approaching zero. More precisely, we will assume that
$$f(0)=0 \quad \text{and}\quad f(t)=\frac{1}{\sqrt{k}}\sin n_kt \quad \text{for}\ t\in[t_k,t_{k-1}],\ k=2,3,\ldots,$$
where n k =2 k!, t k =2π/n k for $$k\in\mathbb{N}$$ (see Fig. 10.2). Fig. 10.2 Sketch of the graph of f
We prove that the sums S n (f,0) tend to infinity along the indices n k . Since
$$S_n(f,0)=\frac{2}{\pi}\int_0^{\pi} \frac{\sin nt}{t}f(t)\,dt+o(1),$$
by (5′), it is sufficient to prove that the integrals
$$I_k=\int_0^{\pi}\frac{\sin n_k t}{t} f(t)\,dt= \int_0^{t_k}\cdots+\int _{t_k}^{t_{k-1}}\cdots+\int_{t_{k-1}}^{\pi} \cdots =F_k+J_k+H_k$$
tend to infinity. We verify that the main contribution comes from the integral J k . Indeed, since |sinn k t|⩽n k t and $$|f(t)|<\frac{1}{\sqrt{k}}$$ on (0,t k ), we have
$$|F_k|= \bigg|\int_0^{t_k}\cdots \bigg| \leqslant\frac{n_k}{\sqrt{k}}\,t_k= \frac{2\pi}{\sqrt{k}}\to0.$$
Since the absolute value of the integrand does not exceed 1/t, we have
$$|H_k|\leqslant\int_{t_{k-1}}^{\pi}\frac{1}{t} \,dt=\ln\pi/t_{k-1}= \ln\frac{n_{k-1}}{2}<(k-1)!\ln2.$$
Now, we calculate the integral over the middle interval,
$$J_k=\int_{t_k}^{t_{k-1}}\frac{\sin n_kt}{t} f(t)\,dt= \frac{1}{\sqrt{k}}\int_{t_k}^{t_{k-1}} \frac{\sin^2 n_kt}{t}\,dt= \frac{1}{\sqrt{k}}\int_{2\pi}^{A_k} \frac{\sin^2u}{u}\,du,$$
where A k =n k t k−1=2πn k /n k−1. Consequently, for sufficiently large k, we have
$$J_k=\frac{1}{2\sqrt{k}}\int_{2\pi}^{A_k} \frac{1-\cos2u}{u}\,du= \frac{\ln A_k+O(1)}{2\sqrt{k}}>\frac{k!\ln2}{3\sqrt{k}}.$$
Thus,
$$I_k=F_k+J_k+H_k\geqslant \frac{k!\ln2}{3\sqrt{k}}-(k-1)!\ln2+o(1)\to +\infty,$$
and, therefore, $$S_{n_{k}}(f,0)=\frac{2}{\pi}\,I_{k}+o(1)\to+\infty$$.

In the above example, we could select a subsequence $$\{S_{n_{k}}(f,0)\}$$ that tends to +∞. As we will see in Sect. 10.4.1, it is impossible to construct a continuous function for which $$S_{n}(f,0)\underset{n\to\infty}{\longrightarrow}+\infty$$. We also remark that, in the above example, estimate (6) for the Fourier sums is almost attained (in order) on the sequence {n k } (see also Exercise 13).

By a slightly more complicated construction it is possible to give an example of a continuous function whose Fourier series diverges on a countable set. Must this series converge almost everywhere? This famous problem was open for more than half a century. It was answered in the affirmative only in 1966 by L. Carleson.14 It turned out that the Fourier series of an arbitrary function of class (for example a continuous function) converges to the function almost everywhere (see [C]). Since that time, several modifications and strengthenings of the original proof have been obtained, but all of them are quite difficult and lie far beyond the scope of this book.

### 10.3.10

Using the Riemann–Lebesgue theorem, we can obtain an important result concerning arbitrary trigonometric series, i.e., series of the form
$$A+\sum_{n=1}^{\infty}(a_n\cos nx+b_n\sin nx)\quad(A,a_n,b_n\in\mathbb{C}).$$
(10)
As we know, (see Sect. 10.3.6) even an everywhere convergent trigonometric series may not be a Fourier series. At the same time, the following statement holds:

### Theorem

(Denjoy15–Luzin)

If series (10) converges absolutely on a set of positive measure, then
$$\sum_{n=1}^{\infty}\bigl(|a_n|+|b_n|\bigr)<+ \infty.$$
(11)

In particular, if a trigonometric series converges absolutely on a set of positive measure, then it converges uniformly on $$\mathbb{R}$$, and, therefore, is the Fourier series of its sum.

### Proof

Without loss of generality, we may assume that the coefficients a n and b n are real. We put φ n (x)=|a n cosnx+b n sinnx|. Since the series $$\sum_{n=1}^{\infty} \varphi _{n}$$ converges on a set of positive measure, its sum is bounded on a smaller set X of positive measure,
$$\sum_{n=1}^{\infty} \varphi _n(x) \leqslant C\quad\text{for all}\ x\in X.$$
Consequently (in what follows, λ is the one-dimensional Lebesgue measure),
$$\sum_{n=1}^{\infty}\int_X \varphi _n(x)\,dx\leqslant C \lambda (X).$$
We represent the functions φ n in the form φ n (x)=c n |sin(nx+θ n )|, where $$c_{n}=\sqrt{a^{2}_{n}+b^{2}_{n}}$$ and $$\theta_{n}\in\mathbb{R}$$. Using the obvious inequality |sint|⩾sin2 t and the Lebesgue–Riemann theorem, we see that
$$\int_X\frac{1}{c_n}\varphi _n(x)\,dx \geqslant\int_X\sin^2(nx+\theta_n) \,dx= \int_X\frac{1-\cos2(nx+\theta_n)}{2}\,dx\underset{n\to\infty}{\longrightarrow} \frac{\lambda (X)}{2}.$$
Therefore,
$$0<\frac{\lambda (X)}{3}\leqslant\int_X\frac{1}{c_n} \varphi _n(x)\,dx\quad\text{for}\ n \geqslant N$$
for some N, and, consequently,
$$\sum_{n=N}^{\infty}\frac{\lambda (X)}{3}c_n \leqslant \sum_{n=N}^{\infty}\int _X\varphi _n(x)\,dx\leqslant C\lambda (X).$$
Thus, the following estimate is valid for the remainder of series (11):
$$\sum_{n=N}^{\infty}\bigl(|a_n|+|b_n|\bigr) \leqslant2\sum_{n=N}^{\infty}\sqrt {a^2_n+b_n^2} = 2\sum _{n=N}^{\infty} c_n\leqslant6C.$$
□

### EXERCISES

1.

Let . By the method used in the second proof of the Lebesgue–Riemann theorem, prove that $$|\widehat{f}(n)|\leqslant\frac{1}{2}\|f-f_{\tau}\|_{1}$$, where f τ is the translation of the function f by the vector τ=πn/∥n2.

2.

Prove that the Fourier sums of the function $$f(x)=\sum_{k=1}^{\infty} 2^{-k}\cos kx$$ provide almost the best uniform approximations for f, more precisely, ∥fS n (f)∥⩽3∥fT for every trigonometric polynomial T of order n.

3.

Show by example that an absolutely continuous function may not satisfy Dini’s condition.

4.

Verify by examples that neither of the Dini and Dirichlet tests implies the other.

5.
Prove (see [HR], Sect. 6.7) that
$$\frac{1}{2} \biggl(S_n \biggl(f,x+\frac{\pi}{2n} \biggr)+ S_n \biggl(f,x-\frac{\pi}{2n} \biggr) \biggr) \underset{n\to\infty}{ \longrightarrow} f(x)$$
if the function f in is continuous at x. Hint. Verify that the functions $$\frac{1}{2} (D_{n} (x+\frac{\pi}{2n} )+D_{n} (x-\frac{\pi}{2n} ) )$$ form a periodic approximate identity.
6.

Verify that belongs to the class if and only if the series $$\sum_{n=-\infty}^{\infty}|\widehat{f}(n)|^{2}$$ converges.

7.

Prove that a function $$f\in\widetilde{C}$$ is the restriction of an entire function to $$\mathbb{R}$$ if and only if $$\root{n}\of{|\widehat{f}(\pm n)|}\to0$$ as n→∞.

8.

Prove that if the sequence of partial sums of a Fourier series is uniformly bounded, then the function belongs to .

9.
Let . Prove that the Fourier sums S n (hg,x) of the product hg are uniformly bounded (with respect to n and x) if the function g has the same property and the function h satisfies the Dini condition uniformly:
$$\int_{-\pi}^{\pi} \bigg|\frac{h(u)-h(x)}{u-x} \bigg|\,du \leqslant \text{const}\quad\text{for all}\ x.$$
10.

Assume that a function f (possibly discontinuous) is such that the interval [−π,π] can be divided into a finite number of intervals inside each of which the function f satisfies the Lipschitz condition of order α>0. Prove that the Fourier sums S n (f,x) are uniformly bounded with respect to x and n.

11.

Prove that $$\int_{-\pi}^{\pi}|D_{n}(u)|\,du=\frac{4}{\pi^{2}}\ln n+O(1)$$.

12.

Supplement inequality (6) by proving that ∥fS n (f)∥=o(lnn) as n→∞ if $$f\in\widetilde{C}$$. Verify that, for bounded functions, this refinement is, generally, false. Hint. Modify the Schwartz example by putting f(t)=sinn k t on the interval [t k ,t k−1].

13.

Modifying the Schwartz example, verify that the result of the previous exercise is precise, i.e., that for every sequence ε n ↓0, there is a function $$f\in \widetilde{C}$$ such that S n (f,0)⩾ε n lnn along some sequence of indices n k →∞.

14.

Show that the convergence of a Fourier series of a function $$f\in \widetilde{C}$$ does not imply the convergence of the Fourier series of the function f 2. Hint. Modifying the Schwartz example, construct a non-negative even function $$F\in\widetilde{C}$$, F(0)=Fπ)=0, with Fourier series divergent at zero and consider an odd function f equal to $$\sqrt{F}$$ on [0,π].

15.

Prove that $$a_{n},b_{n}\underset{n\to\infty}{\longrightarrow}0$$ if the sums a n cosnx+b n sinnx converge to zero on a set of positive measure.

16.

Find a set E⊂(0,2π) of the cardinality of the continuum and a sequence n k →+∞ such that sinn k x⇉0 on the set E.

17.
Consider the series $$\sum_{n=1}^{\infty}\sin(n!\pi x)$$.
1. (a)

Prove that the series converges at the points x=sin1, x=cos1, $$x=\frac{2}{e}$$, and their multiples and converges at the points ke ($$k\in\mathbb{N}$$) only for odd k. Is the convergence absolute?

2. (b)

Prove that the given series diverges at the points x=sinh 1 and $$x=\frac{1}{2}\text{cosh}\,1$$.

3. (c)

Find a set of the cardinality of the continuum at all points of which the given series converges.

18.

Prove that $$\frac{1}{n}S'_{n}(f,x)\underset{n\to\infty}{\longrightarrow} \frac{1}{\pi}(f(x+0)-f(x-0))$$ if the periodic function f has a bounded variation on the interval [−π,π].

## 10.4 ⋆Trigonometric Fourier Series (Continued)

### 10.4.1

The fact that a Fourier series may diverge, even at points of continuity, suggests that we might obtain information on its behavior if we consider a weaker definition of convergence than the classical one. One of the possible approaches is to investigate the convergence of the arithmetic means of the partial sums rather than the partial sums themselves. The limit of a sequence {a n } in the sense of arithmetic means or in the sense of Cesaro16 is, by definition, the limit $$\lim_{n\to\infty}\frac{1}{n}(a_{0}+\cdots+a_{n-1})$$. It can exist even in the case where the sequence itself diverges, for example, if a n =(−1) n . At the same time, if $$a_{n}\underset{n\to\infty}{\longrightarrow} a$$, then also $$\frac{1}{n}(a_{0}+\cdots+a_{n-1})\underset{n\to\infty}{\longrightarrow} a$$ (the permanence of the method of arithmetic means). For numerical series, this approach leads to the concept of a generalized sum of a series. We say that a series Cesaro converges to a number C if the limit of the partial sums of the series in the sense of Cesaro is equal to C.

Based on these considerations, we put
$$\sigma _n(f, x)=\frac{1}{n} \bigl(S_0(f,x)+ \cdots+S_{n-1}(f,x) \bigr),$$
where S 0(f,x),…,S n−1(f,x) are partial sums of the Fourier series of f. The sums σ n are called the Fejér 17 sums. From Eq. (5) of Sect. 10.3.3, it follows that
$$\sigma _n(f, x)= \Biggl(f*\frac{1}{n}\sum _{j=0}^{n-1}D_j \Biggr) (x)= \frac{1}{2\pi n}\int_{-\pi}^{\pi} \frac{f(x-u)}{\sin\frac{u}{2}} \sum_{j=0}^{n-1}\sin \biggl(j+\frac{1}{2} \biggr)u\,du.$$
(1)
The trigonometric identity
$$\sin\frac{u}{2}+\sin\frac{3}{2}u+\cdots+\sin \biggl(n- \frac{1}{2} \biggr)u= \frac{1-\cos nu}{2\sin\frac{u}{2}} =\frac{\sin^2\frac{n}{2}u}{\sin\frac{u}{2}}\quad(u \notin2\pi\mathbb{Z}),$$
the verification of which is left to the reader, allows one to represent the right-hand side of Eq. (1) in the form
$$\sigma _n(f,x)=\frac{1}{2\pi n}\int_{-\pi}^{\pi} f(x-u) \biggl(\frac{\sin\frac{n}{2}u}{\sin\frac{u}{2}} \biggr)^2\,du.$$
Thus, a Fejér sum can be represented as the convolution of f and the function
$$\Phi_n(u)=\frac{D_0(u)+\cdots+D_{n-1}(u)}{n}= \frac{1}{2\pi n} \biggl( \frac{\sin\frac{n}{2}u}{\sin\frac{u}{2}} \biggr)^2,$$
(2)
which is called the nth Fejér kernel. We advise the reader to sketch the graph of Φ n and compare it with the graph of the Dirichlet kernel.
The Fejér kernel can be represented in the form
$$\Phi_n(u)=\frac{1}{n}\sum_{j=0}^{n-1}D_j(u)= \frac{1}{2\pi n}\sum_{j=0}^{n-1} \sum _{|k|\leqslant j}e^{iku}=\frac{1}{2\pi}\sum _{|k|<n} \biggl(1-\frac{|k|}{n} \biggr)e^{iku},$$
and, therefore,
$$\widehat{\Phi}_n(k)= \left\{ \begin{array}{l@{\quad}l} \frac{1}{2\pi}(1-\frac{|k|}{n}) & \text{for}\ |k|<n,\\[6pt] 0 & \text{for}\ |k|\geqslant n. \end{array} \right.$$
(3)
We verify that the sequence {Φ n } is an approximate identity. It follows from Eq. (2) that the Fejér kernels are non-negative and periodic. Since $$\int_{-\pi}^{\pi} D_{j}(u)\,du=1$$ for all j, we have
$$\int_{-\pi}^{\pi}\Phi_n(u)\,du= \frac{1}{n} \biggl(\int_{-\pi}^{\pi} D_0(u)\, du+\cdots+ \int_{-\pi}^{\pi} D_{n-1}(u)\,du \biggr)=1.$$
Finally, the Fejér kernels have a strong form of the localization property (see condition (c′)) of Sect. 7.6.5), namely, Eq. (2) implies that
$$\Phi_n(u)=\frac{1}{2\pi n} \biggl(\frac{\sin\frac{n}{2}u}{\sin\frac {u}{2}} \biggr)^2\leqslant \frac{1}{2\pi n\sin^2\frac{\delta }{2}}=\frac{C_{\delta }}{n}$$
for δ<|u|<π. Now, we are ready to state the main result of this section, which plays an important role in harmonic analysis.

### Theorem

(Fejér)

Let and $$x\in\mathbb{R}$$. Then:
1. (a)

if the limits L ±=lim tx±0 f(t) exist and are finite, then $$\sigma _{n}(f,x)\underset{n\to\infty}{\longrightarrow}\frac{L_{+}+L_{-}}{2}$$;

2. (b)

if $$f\in\widetilde{C}$$, then $$\sigma _{n}(f)\underset{n\to\infty}{\rightrightarrows} f$$ on $$\mathbb{R}$$;

3. (c)

if for some p∈[1,+∞), then $$\|\sigma _{n}(f)-f\|_{p}\underset{n\to\infty}{\longrightarrow}0$$;

4. (d)

$$\sigma _{n}(f)\underset{n\to\infty}{\longrightarrow} f$$ almost everywhere.

### Proof

In the case where the limit lim tx f(t) exists and is finite, both assertions (a) and (b) are special cases of Theorem 7.6.5 (in view of the remarks to it) for m=1, $$T=\mathbb{N}$$ and t 0=+∞. If the one-sided limits are distinct, then we must use the fact that the Fejér kernels are even and apply the result obtained to the function f 0(u)=(f(x+u)+f(xu))/2, which tends to (L ++L )/2 as u→0.

Assertion (c) is already known. It is a special case of Theorem 2 of Sect. 9.3.7.

To prove the last assertion, we estimate the “hump-shaped” majorant of a Fejér kernel ψ n (x)=sup|x|⩽yπ Φ n (y). Since $$\Phi_{n}(y)\leqslant\Phi_{n}(0)=\frac{n}{2\pi}$$ and $$\Phi_{n}(y)\leqslant \frac{1}{2\pi n \sin^{2}\frac{y}{2}}\leqslant \frac{\pi}{2ny^{2}}$$, we see that $$\psi_{n}(x)\leqslant\frac{1}{2\pi}\min \{n,\frac{\pi^{2}}{nx^{2}} \}$$. From this, it immediately follows that
$$\int_{-\pi}^{\pi}\psi_n(x)\,dx \leqslant2\quad\text{for all }\ n\in \mathbb{N}.$$
Thus, the assumptions of Theorem 3 of Sect. 9.3.7 are fulfilled and, therefore, $$\sigma _{n}(f)\underset{n\to\infty}{\longrightarrow} f$$ almost everywhere. □

By statement (a) of the theorem and the permanence of the method of arithmetic means, we are now able to answer the question we posed before taking up the investigation of the convergence of Fourier series (see Sect. 10.3.3): if the Fourier series of a summable function f converges at a point of continuity of f, then the sum of the Fourier series is necessarily equal to the value of f at this point.

Since the convolution f∗Φ n is a trigonometric polynomial, the Fejér theorem supplements the Weierstrass theorem (see Corollary 7.6.5) by providing specific approximating polynomials.

### Remark

The third and, obviously, the fourth statements of the theorem give new proofs of the uniqueness theorem. Indeed, if functions have the same Fourier coefficients, then σ n (f)=σ n (g) for all n. Therefore, statement (c) of the theorem for p=1 implies the relation
$$\|f-g\|_1=\lim_{n\to\infty}\big\|\sigma _n(f)- \sigma _n(g)\big\|_1=0.$$
Consequently, the functions f and g are equal almost everywhere.

As we have verified, the Fejér sums σ n (f) have the obvious advantage over the partial sums S n (f) of the Fourier series that they approximate an arbitrary summable function in the integral metric and converge uniformly to f if f is continuous. It should be mentioned, however, that there is a price to pay for the universality of the Fejér sums: they cannot converge to the function rapidly (see Exercise 2). Therefore, if a Fourier series converges rapidly, then the Fejér sums are a poorer approximation of the function than the Fourier sums (see Exercise 2, Sect. 10.3).

In some cases, Fejér sums allow one to obtain an additional information concerning the behavior of Fourier sums.

### Corollary 1

The Fourier series of an absolutely continuous periodic function f converges to f uniformly on  $$\mathbb{R}$$.

### Proof

By the second statement of the Fejér theorem, it is sufficient to verify that the difference S n (f)−σ n (f) converges uniformly to zero. It is clear that
$$\big|S_n(f,x)-\sigma _n(f,x) \big|= \Bigg|\sum _{k=-n}^n\frac{|k|}{n}\widehat{f}(k) e^{ikx} \Bigg|\leqslant\frac{1}{n}\sum_{k=-n}^n|k|\big| \widehat{f}(k)\big|.$$
By assumption, we have $$f(x)=f(0)+\int_{0}^{x} g(t)\,dt$$, where , and, therefore (see property (d) of Sect. 10.3.2) $$|k\widehat {f}(k)|=|\widehat{f}'(k)|$$. It remains to use the permanence of the method of arithmetic means: since $$\widehat{f}'(k)\underset{|k|\to+\infty}{\longrightarrow}0$$, we have
$$\max_{x\in\mathbb{R}} \big|S_n(f,x)-\sigma _n(f,x) \big| \leqslant\frac {1}{n}\sum_{k=-n}^n\big| \widehat{f'}(k)\big| \underset{n\to\infty}{\longrightarrow}0.$$
□

### Corollary 2

Let f be a periodic function satisfying the Lipschitz condition of order α, 0<α<1, i.e.,
$${\textit{there exists a number}\ L \ \textit{such that}}\ |f(x)-f(y)|\leqslant L|x-y|^{\alpha }\ \textit{for all}\ x,y\in\mathbb{R}.$$
Then the Fourier series of f converges uniformly on $$\mathbb{R}$$ and
$$\big|S_n(f,x)-f(x)\big|\leqslant C_{\alpha } L\,\frac{\ln n}{n^{\alpha }} \quad\textit{for all}\ x\in\mathbb{R}.$$

The coefficient C α depends only on α (for the case α=1, see Exercise 3).

### Proof

First, we estimate the deviation of the Fejér sums. Since
$$\sigma _n(f,x)-f(x)=\int_{-\pi}^{\pi} \bigl(f(x-t)-f(x)\bigr)\Phi_n(t)\,dt,$$
we obtain
\begin{aligned} \big|\sigma _n(f,x)-f(x)\big|\leqslant&\int_{-\pi}^{\pi}\big|f(x-t)-f(x)\big| \Phi_n(t)\, dt\leqslant \frac{L}{\pi n}\int_0^{\pi} t^{\alpha } \biggl(\frac{\sin\frac{n}{2}t}{\sin\frac{t}{2}} \biggr)^2\,dt \\\leqslant&\frac{\pi L}{n}\int_0^{\pi} \frac{\sin^2\frac{n}{2}t}{t^{2-\alpha }}\,dt= \frac{\pi L}{n} \biggl(\frac{n}{2} \biggr)^{1-\alpha } \int_0^{\pi n/2}\frac{\sin^2u}{u^{2-\alpha }}\,du. \end{aligned}
Thus, $$\|\sigma _{n}(f)-f\|_{\infty}\leqslant\widetilde{C}_{\alpha } L/n^{\alpha }$$, where $$\widetilde{C}_{\alpha }=\pi2^{\alpha -1}\int_{0}^{\infty}\frac{\sin^{2}u}{u^{2-\alpha }}\,du$$.
To estimate the deviation of the Fourier sums, we observe that S n (f)−f=S n (φ n )−φ n , where φ n =fσ n (f), since S n (σ n (f))=σ n (f). Therefore, inequality (6) of Sect. 10.3.3 implies
$$\big|S_n(f,x)-f(x)\big|\leqslant\big|S_n(\varphi _n,x)\big|+\big| \varphi _n(x)\big|\leqslant 3\|\varphi _n\|_{\infty}\ln n\leqslant3 \widetilde{C}_{\alpha } L\,\frac{\ln n}{n^{\alpha }}.$$
□

### 10.4.2

The non-negativity of the Fejér kernels yields interesting properties of the Fourier series. For example, if the coefficients of the Fourier series of a function are non-negative, then the Fourier series converges absolutely if the function is bounded in a neighborhood of zero. Indeed, let a function be such that $$\widehat{f}(k)\geqslant0$$ for all $$k\in\mathbb{Z}$$ and |f(x)|⩽C if |x|<δ. Since 0⩽Φ n (x)⩽C δ for δ⩽|x|⩽π, we obtain
\begin{aligned} \big|\sigma _n(f,0)\big|=& \bigg|\int_{-\pi}^{\pi} f(x) \Phi_n(x)\,dx \bigg|\leqslant \int_{-\delta }^{\delta } C\Phi_n(x)\,dx+\int_{\delta \leqslant|x|\leqslant\pi}C_{\delta }\big|f(x)\big| \, dx\\\leqslant& C+C_{\delta }\|f\|_ 1. \end{aligned}
Therefore,
$$\sum_{|k|<\frac{n}{2}}\widehat{f}(k)\leqslant2\sum _{|k|<n} \biggl(1-\frac {|k|}{n} \biggr) \widehat{f}(k) =2 \sigma _n(f,0)\leqslant2\bigl(C+C_{\delta }\|f\|_1\bigr).$$
Since this inequality is fulfilled for all n, the series $$\sum_{k=-\infty}^{\infty}\widehat{f}(k)$$ converges.
Now, we consider an arbitrary trigonometric series of the form
$$\frac{1}{2} c_0+\sum_{n=1}^{\infty} c_n\cos nx,$$
(4)
where the coefficients form a convex sequence tending to zero (the convexity of a sequence {c n } n⩾0 means that $$c_{n}\leqslant\frac{1}{2}(c_{n-1}+c_{n+1})$$ for all $$n\in\mathbb{N}$$). The fact that series (4) converges pointwise for $$x\notin2\pi\mathbb{Z}$$ can easily be verified by the Dirichlet test.

Before passing to the study of the sum of series (4), we prove a lemma on numerical series.

### Lemma

Let {c n } n⩾0 be a convex sequence tending to zero. Then:
1. (1)

c n−1c n ⩾0 for all $$n\in\mathbb{N}$$;

2. (2)

$$\sum_{k=1}^{\infty} k(c_{k-1}-2c_{k}+c_{k+1})=c_{0}$$.

### Proof

We put b n =c n−1c n . The convexity implies that b n b n+1, and, consequently, b n ⩾0 (because $$b_{n}\underset{n\to\infty}{\longrightarrow}0$$), which proves the first statement.

It is obvious that the series $$\sum_{k=1}^{\infty} b_{k}$$ converges and its sum is equal to c 0. Since
$$nb_n\leqslant2\sum_{n/2\leqslant k\leqslant n}b_k \leqslant2\sum_{k \geqslant n/2}b_k,$$
we see that $$nb_{n}\underset{n\to\infty}{\longrightarrow}0$$ together with the remainder of a convergent series. This relation and the equality c k−1−2c k +c k+1=b k b k+1 allow us to prove the second statement,
\begin{aligned} \sum_{k=1}^n k(c_{k-1}-2c_k+c_{k+1})=& \sum_{k=1}^nk(b_k-b_{k+1})= \sum_{k=1}^nkb_k-\sum _{k=2}^{n+1}(k-1)b_k\\=& \sum _{k=1}^nb_k-nb_{n+1} \underset{n\to\infty}{\longrightarrow} c_0. \end{aligned}
□

### Theorem

Let the coefficients of series (4) form a convex sequence tending to zero. Then the sum f of the series is non-negative and summable on (−π,π) and series (4) is the Fourier series of f.

### Proof

We transform a partial sum S n (x) of series (4). Using the relation coskx=π(D k (x)−D k−1(x)), we obtain
\begin{aligned} \frac{1}{\pi}S_n(x)=&c_0D_0(x)+ \sum_{k=1}^nc_k \bigl(D_k(x)-D_{k-1}(x)\bigr)\\=& c_n D_n(x)+\sum_{k=0}^{n-1}(c_k-c_{k+1})D_k(x). \end{aligned}
Since D k (x)=(k+1)Φ k+1(x)−kΦ k (x), after elementary transformations, we arrive at the relation
$$\frac{1}{\pi}\,S_n(x)=c_n\,D_n(x)+(c_{n-1}-c_n)n \Phi_n(x)+ \sum_{k=1}^{n-1}(c_{k-1}-2c_k+c_{k+1})k \Phi_k(x).$$
We observe that c k−1−2c k +c k+1⩾0 since the sequence {c k } k⩾0 is convex. Because the sequences {D n (x)} n⩾1 and {nΦ n (x)} n⩾1 are bounded for x≠0 (see formula (4′) of Sect. 10.3.3 and (2) above) and $$c_{n}\underset{n\to\infty}{\longrightarrow}0$$, the first two terms on the right-hand side of the last equation tend to zero. Therefore, passing to the limit in this equation as n→∞, we see that
$$f(x)=\pi\sum_{k=1}^{\infty}(c_{k-1}-2c_k+c_{k+1})k \Phi_k(x)= \frac{1}{2}\sum_{k=1}^{\infty}(c_{k-1}-2c_k+c_{k+1}) \biggl(\frac{\sin\frac{k}{2}x}{\sin\frac{x}{2}} \biggr)^2.$$
(5)
This proves that the function f is non-negative. Because non-negative series can be integrated termwise (see Corollary 1 of Sect. 4.8.2), we obtain
\begin{aligned} \int_{-\pi}^{\pi} f(x)\,dx=&\pi\sum _{k=1}^{\infty} (c_{k-1}-2c_k+c_{k+1})k \int_{-\pi}^{\pi}\Phi_k(x)\,dx\\=& \pi\sum_{k=1}^{\infty}(c_{k-1}-2c_k+c_{k+1})k= \pi c_0 \end{aligned}
(the last equation is valid by statement (2) of the lemma). Thus, $$\int_{-\pi}^{\pi} f(x)\,dx<+\infty$$.
Now, we verify that series (4) is the Fourier series of the function f. Obviously, the function f remains a majorant of the partial sums after multiplication of series (5) by cosjx. Therefore, the series obtained by multiplication can be integrated termwise, which leads to the following relation for the Fourier cosine coefficients
$$a_j(f)=\pi\sum_{k=1}^{\infty}(c_{k-1}-2c_k+c_{k+1})ka_j( \Phi_k).$$
Since $$a_{j}(\Phi_{k})=\widehat{\Phi}_{k}(j)+\widehat{\Phi}_{k}(-j)$$, we obtain by Eq. (3) that
$$a_j(f)=\sum_{k=j+1}^{\infty}(c_{k-1}-2c_k+c_{k+1}) (k-j)= \sum_{k=1}^{\infty}(c_{k+j-1}-2c_{k+j}+c_{k+j+1})k$$
for $$j\in\mathbb{N}$$. The numbers $$\widetilde{c}_{k}=c_{k+j}$$ (k=0,1,2…) obviously form a convex sequence. Applying statement (2) of the lemma to this sequence, we see that the sum of the last series is equal to $$\widetilde{c}_{0}$$. Therefore, $$a_{j}(f)=\widetilde{c}_{0}=c_{j}$$. The relation $$A(f)=\frac{1}{2\pi}\int_{-\pi}^{\pi} f(x)\,dx=\frac{1}{2}c_{0}$$ has already been obtained in the proof that f is summable. □

A sequence of coefficients satisfying the conditions of the theorem can tend to zero arbitrarily slow (see Exercise 4). For example, since the sequence {1/lnn} n⩾2 is convex, the theorem we have just proved implies that the sum of the series $$\sum_{n=2}^{\infty}\frac{\cos nx}{\ln n}$$ belongs to and the series itself is the Fourier series of its sum. In this connection, we recall that the everywhere convergent series $$\sum_{n=2}^{\infty}\frac{\sin nx}{\ln n}$$ is not a Fourier series, as established in Sect. 10.3.6.

### 10.4.3

The Fejér method is based on Cesaro’s generalization of the sum of a numerical series. Other generalizations of the concept of the sum of a series are also possible. One of them is based on the following well-known Abel theorem for numerical series: if a series $$\sum_{n=1}^{\infty} c_{n}$$ converges to the sum S, then the sum of the series $$\sum_{n=1}^{\infty} e^{-\varepsilon n}c_{n}$$ tends to S as ε→+0. This limit can exist also for a divergent series, and, therefore, can be regarded as a generalized sum of the series in question.

With a view to applications to Fourier series, it is natural to replace the summation over $$\mathbb{N}$$ by a summation over $$\mathbb{Z}$$ and use the symmetric partial sums S j =∑|k|⩽j c k . In this case, we should assume that, by Cesaro’s method, to each numerical series $$\sum_{n\in\mathbb{Z}}c_{n}$$ (convergent or not), we must assign the sequence of sums
$$\sigma _n=\frac{1}{n}(S_0+\cdots+S_{n-1})= \frac{1}{n}\sum_{j=0}^{n-1} \biggl( \sum_{|k|\leqslant j}c_k \biggr)=\sum _{|k|<n} \biggl(1-\frac {|k|}{n} \biggr)c_k$$
and, by Abel’s method, we must assign the function
$$A(\varepsilon )=\sum_{n\in\mathbb{Z}}e^{-\varepsilon |n|}c_n$$
(to simplify the exposition, we will consider only bounded sequences $$\{ c_{n}\}_{n\in\mathbb{Z}}$$; for Fourier series, this condition is always fulfilled). In the first case, we calculate the limit lim n→∞ σ n , and, in the second case, the limit lim ε→+0 A(ε). It can be proved that if the first limit exists (i.e., the series converges in the sense of Cesaro), then the second limit also exists (i.e., the series converges in the sense of Abel) and the limits are equal. In this sense, Abel’s method is stronger that the method of Cesaro. We will not dwell on the comparison of these methods. Instead, we show that both methods are special cases of the following general scheme.

Let M be a decreasing function summable on [0,+∞), and let M(0)=lim u→0 M(u)=1. It is clear that M⩾0 and the series $$\sum_{n\in\mathbb{Z}}M(\varepsilon |n|)$$ converges for every ε>0.

To an arbitrary numerical series $$\sum_{n\in\mathbb{Z}}c_{n}$$ with bounded terms, we assign the absolutely convergent series
$$S_M(\varepsilon )=\sum_{n=-\infty}^{\infty} M\bigl( \varepsilon |n|\bigr)c_n,$$
which converges uniformly with respect to ε>0, provided the initial series converges. Under this assumption $$\lim_{\varepsilon \to+0}S_{M}(\varepsilon )=\sum_{n\in\mathbb{Z}}c_{n}$$. The limit lim ε→+0 S M (ε), if it exists, is called the generalized sum of the given series. We have M(u)=(1−u)+ and M(u)=e u for the Cesaro and the Abel method, respectively. We obtain the usual sum of a series if we take $$M(u)=\chi_{{}_{[0,1]}}(u)$$.
Turning to Fourier series, we see that, to each function , we must assign the sums
$$S_{M,\varepsilon }(f,x)=\sum_{n=-\infty}^{\infty} M\bigl( \varepsilon |n|\bigr)\widehat{f}(n) e^{inx}\quad (x\in\mathbb{R}).$$
To find integral representations for them, we use property (e) of Sect. 10.3.2, $$\widehat{f*g}(n)=2\pi\widehat{f}(n)\cdot\widehat{g}(n)$$, which allows us to regard S M,ε (f) as the sum of Fourier series of the convolution fω ε , where
$$\omega_{\varepsilon }(x)=\frac{1}{2\pi}\sum_{n=-\infty}^{\infty} M\bigl(\varepsilon |n|\bigr) e^{inx}= \frac{1}{2\pi}+\frac{1}{\pi}\sum _{n=1}^{\infty} M(\varepsilon n)\cos nx \quad (x\in\mathbb{R}).$$
By the uniqueness theorem (Corollary 1 of Sect. 10.3.6), the functions fω ε and S M,ε (f) coincide almost everywhere. Since the functions are continuous, they coincide everywhere.

To study the behavior of the sums S M,ε (f) as ε→+0, we impose an additional constraint on the function M. We will assume that it is convex on [0,+∞). Then, the sequence {M(εn)} n⩾0 is convex and ω ε (x)⩾0 by Theorem 10.4.2.

### Lemma

Let M be a continuous function convex on [0,+∞), tending to zero at infinity, and let M(0)=1. Then the functions ω ε form a periodic approximate identity with the strong localization property as ε→+0. Moreover, there exist even functions Ω ε that decrease on [0,π], majorize ω ε , and satisfy the inequality $$\int_{-\pi}^{\pi}\Omega_{\varepsilon }(x)\,dx\leqslant2$$.

For the definition of the strong localization property, see Sect. 7.6.5.

### Proof

Obviously, $$\int_{-\pi}^{\pi}\omega_{\varepsilon }(x)\,dx=1$$ and, as mentioned above, the function ω ε is non-negative. We verify that it has the localization property. For this, we use Eq. (5) with f=πω ε and c n =M(εn), which implies that, for δ<|x|<π and C δ =1/(2πsin2 δ/2),
\begin{aligned} \omega_{\varepsilon }(x)=&\sum_{n=1}^{\infty} \bigl(M(n\varepsilon -\varepsilon )-2M(n\varepsilon )+M(n\varepsilon +\varepsilon ) \bigr)\,n \Phi_n(x) \\\leqslant& C_{\delta }\sum_{n=1}^{\infty} \bigl(M(n\varepsilon -\varepsilon )-2M(n\varepsilon )+M(n\varepsilon +\varepsilon ) \bigr)\\= &C_{\delta }\bigl(M(0)-M(\varepsilon )\bigr) \underset{\varepsilon \to+0}{\longrightarrow}0. \end{aligned}
To prove the last statement of the lemma, we use the previous equation one more time. Replacing the Fejér kernel Φ n in it by the majorant ψ n constructed in the proof of Theorem 10.4.1, we obtain
$$\omega_{\varepsilon }(x)\leqslant\sum_{n=1}^{\infty} \bigl(M(n\varepsilon -\varepsilon )-2M(n\varepsilon )+M(n\varepsilon +\varepsilon ) \bigr)n\psi_n(x) \equiv \Omega_{\varepsilon }(x).$$
As established in the same proof, $$\int_{-\pi}^{\pi}\psi_{n}(x)\, dx\leqslant2$$, and, therefore,
\begin{aligned} \int_{-\pi}^{\pi}\Omega_{\varepsilon }(x)\,dx=& \sum_{n=1}^{\infty} \bigl(M(n\varepsilon -\varepsilon )-2M(n\varepsilon )+M(n \varepsilon +\varepsilon ) \bigr)\,n \int_{-\pi}^{\pi} \psi_n(x)\,dx \\\leqslant &2\sum_{n=1}^{\infty} \bigl(M(n\varepsilon - \varepsilon )-2M(n\varepsilon )+M(n\varepsilon +\varepsilon ) \bigr)\,n=2M(0) \end{aligned}
(the last equality is valid by the second statement of Lemma 10.4.2). □

Now, we are able to study the behavior of the sums S M,ε (f,x).

### Theorem

Let , $$x\in\mathbb{R}$$, and let the function M be summable on [0,+∞) and satisfy the conditions of the lemma. Then:
1. (a)

if the limits L ±=lim tx±0 f(t) exist and are finite, then $$S_{M,\varepsilon }(f,x)\underset{\varepsilon \to0}{\longrightarrow}\frac{L_{+}+L_{-}}{2}$$;

2. (b)

if $$f\in\widetilde{C}$$, then $$S_{M,\varepsilon }(f)\underset{\varepsilon \to0}{\rightrightarrows} f$$ on $$\mathbb{R}$$;

3. (c)

if for some p∈[1,+∞), then $$\|S_{M,\varepsilon }(f)-f\|_{p}\underset{\varepsilon \to0}{\longrightarrow}0$$;

4. (d)

$$S_{M,\varepsilon }(f,x)\underset{\varepsilon \to0}{\longrightarrow} f(x)$$ almost everywhere.

### Proof

The proof can be obtained by repeating verbatim the proof of the Fejér theorem. Indeed, the proof of the Fejér theorem uses only the fact that the sum σ n (f) is the convolution of the function f and an even approximate identity satisfying the strong localization property and having a hump-shaped majorant, which, by the lemma, is also valid for the sum S M,ε (f). □

In conclusion, we note that, applying Abel’s method to a Fourier series, i.e., taking M(u)=e u , we obtain the sums
$$S_{\varepsilon }(f,x)=\sum_{n=-\infty}^{\infty} e^{-\varepsilon |n|}\widehat{f}(n) e^{inx}$$
called the Abel–Poisson sums. We have already encountered the corresponding approximate identity ω ε . Indeed, let r=e ε and z=re ix . Then
\begin{aligned} 2\pi\omega_{\varepsilon }(x)=&1+2\sum_{n=1}^{\infty} r^n\cos nx= \mathcal{R} e \Biggl(1+2\sum _{n=1}^{\infty} z^n \Biggr)\\=& \mathcal{R} e\, \frac{1+z}{1-z}=\frac{1-r^2}{1-2r\cos x+r^2}. \end{aligned}
Thus, in the case in question, the function ω ε is nothing but the Poisson kernel for the disc (for the definition, see Sect. 8.7.10).

### 10.4.4

The rest of the section is devoted to multiple trigonometric Fourier series, i.e., to series in the system $$\{e^{i\langle n,x\rangle}\}_{n\in\mathbb{Z}^{m}}$$. As was mentioned in Sect. 10.2.2, this system is an orthogonal basis in the space . Therefore, the -theory of multiple Fourier series is a special case of the general theory where the convergence of these series is in the -norm, and we will not touch on this here. The situation is completely different in regard to other forms of convergence. The problems arising here are connected with the fact that there is no preferred definition of the sum of a multiple series. It is possible to find the limit of the partial sums over unboundedly expanding balls, cubes, parallelepipeds, etc. It turns out that the answers to most problems depend essentially on the choice of the definition of a partial sum. We will not dwell on this topic, instead confining ourselves to partial sums over rectangles. Even in this case, for m>1, there are phenomena that did not occur in the one-dimensional situation.

We introduce the necessary notation. When speaking of periodic functions in $$\mathbb{R}^{m}$$, we will always mean 2π-periodicity with respect to each variable. By $$\widetilde{C}(\mathbb{R}^{m})$$ and $$\widetilde{C}^{r}(\mathbb{R}^{m})$$ ($$r\in\mathbb{N}$$), we denote the classes of periodic continuous and periodic smooth functions, respectively, defined on $$\mathbb{R}^{m}$$; by (1⩽p⩽+∞), we denote the class of periodic pth power summable functions on the cube Q=(−π,π) m . For a function f in , we denote the -norm of its restriction to Q by ∥f p . For a function and a multi-index $$n\in\mathbb{Z}^{m}$$,
$$\widehat{f}(n)=\frac{1}{(2\pi)^m}\int_Q f(x) e^{-i\langle n,x\rangle}\,dx$$
is the nth Fourier coefficient of f.

When solving the problem of expanding a periodic function in the Fourier series $$\sum_{n\in\mathbb{Z}^{m}}c_{n} e^{i\langle n,x \rangle}$$, as in the one-dimensional case, there is no freedom in the choice of coefficients. The reasoning used at the end of Sect. 10.3.1 also remains also in the case in question. More precisely, let $$S_{j}(x) =\sum_{n\in A_{j}}c_{n}e^{i\langle n,x\rangle}$$ be the partial sums of this series corresponding to expanding bounded sets $$A_{j}\subset \mathbb{R}^{m}$$ such that $$\bigcup_{j=1}^{\infty} A_{j}=\mathbb{R}^{m}$$. If $$S_{j}\underset{j\to\infty}{\longrightarrow} S$$ almost everywhere or in measure, then the function S can be called the sum of the series. If, in addition, the partial sums S j are dominated by some function g in , then the coefficients of the given trigonometric series are determined uniquely. Indeed, as follows from Lebesgue’s theorem, $$\widehat{S}(n)=\lim_{j\to\infty}\widehat{S}_{j}(n)=c_{n}$$. Therefore, under the present hypothesis, the expansion of a function in a multiple Fourier series is unique.

For absolutely convergent Fourier series, all definitions of partial sums give the same result because, in this case, the terms of the Fourier series form a summable family. Moreover, an absolutely convergent trigonometric series is, obviously, the Fourier series of its sum. Since the trigonometric system is complete (see Theorem 10.2.2) the sum of an absolutely convergent Fourier series coincides with the function almost everywhere, and if the function is continuous, it converges everywhere (in the one-dimensional case, this has been noted at the beginning of Sect. 10.3.8). As we will soon verify, the Fourier series of smooth functions converge absolutely.

In the multi-dimensional case, all basic properties of the coefficients of a Fourier series, as well as their proofs, are preserved (in what follows, $$n=(n_{1},\ldots,n_{m})\in\mathbb{Z}^{m}$$):
1. (a)

$$|\widehat{f}(n)|\leqslant\frac{1}{(2\pi)^{m}}\|f\|_{1}$$;

2. (b)

$$|\widehat{f}(n)|\to0$$ as ∥n∥→+∞;

3. (c)

the Fourier coefficients of a translation f h ($$h\in\mathbb {R}^{m}$$), i.e., of the function f h (x)=f(xh), are connected with the Fourier coefficients of f by the formulas $$\widehat{f}_{h}(n)=e^{-i\langle n,h\rangle}\widehat{f}(n)$$;

4. (d)

if $$f\in\widetilde{C}^{r}(\mathbb{R}^{m})$$ and $$g=\frac{\partial^{r} f}{\partial x_{j_{1}}\ldots\partial x_{j_{r}}}$$, then $$\widehat{g}(n)=i^{r}n_{j_{1}}\cdots n_{j_{r}}\,\widehat{g}(n)$$;

5. (e)

$$\widehat{f*g}(n)=(2\pi)^{m} \widehat{f}(n)\cdot\widehat{g}(n)$$ for all functions f and g in .

By property (d), it is easy to show that the functions of class $$\widetilde{C}^{(m+1)}(\mathbb{R}^{m})$$ have absolutely convergent Fourier series. The following theorem shows that the smoothness requirement can be weakened considerably.

### Theorem

If $$f\in\widetilde{C}^{r}(\mathbb{R}^{m})$$ and r>m/2, then $$\sum_{n\in\mathbb{Z}^{m}}|\widehat{f}(n)|<+\infty$$, and, therefore, $$f(x)=\sum_{n\in\mathbb{Z}^{m}}\widehat{f}(n) e^{i\langle n,x\rangle}$$ for all x.

### Proof

Indeed, by property (d), we have $$|n_{k}|^{r}|\widehat{f} (n)|=|\widehat{g}_{k}(n)|$$, where $$n= (n_{1},\ldots, n_{m})\in\mathbb{Z}^{m}_{+}$$ and $$g_{k}=\frac{\partial^{r} f}{\partial x_{k}^{r}},\ k=1,\ldots,m$$. We put
$$c_n=\big|\widehat{g}_1(n)\big|+\cdots+\big|\widehat{g}_m(n)\big|= \bigl(|n_1|^r+\cdots+|n_m|^r \bigr)\big|\widehat{f}(n)\big|.$$
Since ∥n r m r/2max{|n 1| r ,…,|n m | r }⩽m r/2(|n 1| r +⋯+|n m | r ), we obtain
$$\big|\widehat{f}(n)\big| =\frac{c_n}{|n_1|^r+\cdots+|n_m|^r}\leqslant\mathrm{const}\frac{c_n}{\|n \|^r}$$
for all n≠0. It is clear that
\begin{aligned} \sum_{n\in\mathbb{Z}^m}c_n^2=&\sum _{n\in\mathbb{Z}^m} \bigl(\big|\widehat{g}_1(n)\big|+ \cdots+\big| \widehat{g}_m(n)\big| \bigr)^2\\\leqslant& m\sum _{n\in\mathbb{Z}^m} \bigl(\big|\widehat{g}_1(n)\big|^2+ \cdots+\big|\widehat{g}_m(n)\big|^2 \bigr)<+\infty \end{aligned}
by Bessel’s inequality. Consequently,
$$\sum_{n\ne0}\big|\widehat{f}(n)\big|\leqslant\mathrm{const} \sum_{n\ne0}\frac {c_n}{\|n\|^r} \leqslant \frac{\mathrm{const}}{2}\sum_{n\ne0} \biggl( \frac{1}{\|n\| ^{2r}}+c_n^2 \biggr)<+\infty$$
(the series $$\sum_{n\ne0}\frac{1}{\|n\|^{2r}}$$ converges since 2r>m). □

### 10.4.5

Now, we turn to the problem of the Fourier series representation of functions from a wider class. As already mentioned, we confine ourselves to partial sums of multiple series over rectangles. Here, we will use the notation introduced in Sect. 1.1.6. For vectors a=(a 1,…,a m ) and b=(b 1,…,b m ), the inequalities ab and a<b mean that a 1b 1,…,a m b m and a 1<b 1,…,a m <b m , respectively. By |a|, we denote the vector (|a 1|,…,|a m |).

For a function and a multi-index $$n=(n_{1},\ldots,n_{m})\in\mathbb{Z}^{m}_{+}$$, we consider the partial sum
$$S_n(f,x)=\sum_{|k|\leqslant n}\widehat{f}(k) e^{i\langle k,x\rangle}$$
of the Fourier series over the corresponding “rectangle”. Reasoning as in Sect. 10.3.3, we can easily verify that this sum is the convolution of f and a multi-dimensional Dirichlet kernel equal to the product to the product of one-dimensional Dirichlet kernels, S n (f)=fD n , where
$$D_n(u)=D_{n_1}(u_1)\cdots D_{n_m}(u_m) \quad\bigl(u=(u_1,\ldots,u_m)\bigr).$$
As in the one-dimensional case, the kernel D n satisfies the equation (below, Q=(−π,π) m )
$$\int_Q D_n(u)\,du=1.$$
(6)
It was noted in Sect. 10.3.3 that the -norms of one-dimensional Dirichlet kernels have logarithmic rate of growth. The same is also true for the Dirichlet kernels corresponding to rectangles,
$$\|D_n\|_1=\int_Q\big|D_n(u)\big| \,du=\prod_{j=1}^m\int _{-\pi}^{\pi} \big|D_{n_j}(u_j)\big| \,du_j \asymp\prod_{j=1}^m \ln n_j.$$
Hence it follows that the following estimate is valid for periodic bounded (in particular, continuous) functions and n 1,…,n m >1:
$$\big\|S_n(f)\big\|_{\infty}\leqslant C_m \Biggl(\prod _{j=1}^m\ln n_j \Biggr)\|f\| _{\infty}.$$
(7)

The following theorem shows that the class of functions with uniformly convergent Fourier series is rather wide.

### Theorem

Assume that a periodic function f satisfies the Lipschitz condition of order α, 0<α⩽1, i.e., there is an L such that |f(x)−f(y)|⩽Lxy α for all $$x,y\in\mathbb{R}^{m}$$. Then the sums S n (f) converge uniformly to f as min{n 1,…,n m }→+∞.

### Proof

It is sufficient to consider the case where α<1. In addition, we will assume that m=2 because no new ideas are required for the proof in the general case. Subtracting Eq. (6) multiplied by f(x) from S n (f,x)=(fD n )(x)=∫ Q f(xu)D n (u) du, we obtain
$$\Delta=S_n(f,x)-f(x)=\int_Q\bigl(f(x-u)-f(x) \bigr)D_n(u)\,du.$$

To simplify the subsequent formulas, we change the notation as follows: n=(j,k), x=(a,b) and u=(s,t). Estimating the difference Δ, we may assume, without loss of generality, that the first coordinate of the vector n does not exceed its second coordinate, i.e., jk.

We represent the increment of the function f as the sum of the increments in each coordinate,
$$f(x-u)-f(x)=f(a-s,b-t)-f(a-s,b)+f(a-s,b)-f(a,b).$$
Therefore, the difference Δ splits into the sum of the integrals I and J, where
$$I=\int_Q\bigl(f(a-s,b-t)-f(a-s,b)\bigr)D_j(s)D_k(t) \,ds\,dt$$
and
\begin{aligned} J=&\int_Q\bigl(f(a-s,b)-f(a,b)\bigr)D_j(s)D_k(t) \,ds\,dt\\ =& \int_{-\pi}^{\pi}\bigl(f(a-s,b)-f(a,b) \bigr)D_j(s)\,ds. \end{aligned}
The integral J is small by Corollary 2 of Fejér’s theorem, |J|⩽C α L(lnj)/j α . The same corollary makes it possible to estimate the integral I,
\begin{aligned} |I| \leqslant&\int_{-\pi}^{\pi} \bigg|\int _{-\pi}^{\pi}\bigl(f(a-s,b-t)-f(a-s,b)\bigr) D_k(t)\,dt \,\bigg| \big|D_j(s) \big|\,ds\\\leqslant& C_{\alpha } \,L\,\frac{\ln k}{k^{\alpha }}\int_{-\pi}^{\pi} \big|D_j(s) \big|\,ds. \end{aligned}
Since $$\int_{-\pi}^{\pi} |D_{j}(s) |\,ds\leqslant2\ln j$$ for j⩾10 (see inequality (6) of Sect. 10.3.3), we obtain
$$|I|\leqslant2C_{\alpha }L\frac{\ln k}{k^{\alpha }}\ln j\leqslant2C_{\alpha } L\frac {\ln^2k}{k^{\alpha }}.$$
Thus, the inequality
$$\big|S_n(f,x)-f(x)\big|=|\Delta|\leqslant|I|+|J|\leqslant C_{\alpha } L \biggl(\frac {\ln j}{j^{\alpha }}+ 2\frac{\ln^2k}{k^{\alpha }} \biggr)$$
holds for all x. □

### 10.4.6

Here, we present two negative results illustrating some phenomena that may occur in the behavior of the double Fourier series and which are impossible in the one-dimensional case.

The first of them is connected with the Riemann localization principle (see Theorem 10.3.3). It turns out that this principle is not true for sums over rectangles: there is a function $$f\in\widetilde{C}$$ equal to zero in a neighborhood of the origin and such that the partial sums (over rectangles) of its Fourier series are unbounded at the origin. To verify this, consider a function f of the form f(s,t)=φ(s)ψ(t), where $$\varphi ,\psi\in\widetilde{C}$$. It is easy to find a function φ equal to zero in the vicinity of the origin and such that S j (φ,0)≠0 for an infinite number of indices j. We take a function ψ for which the sequence $$\{S_{k}(\psi,0)\}_{k\in\mathbb{N}}$$ is unbounded as the second factor (see the Schwartz example in Sect. 10.3.9). Then the product f(s,t)=φ(s)ψ(t) is equal to zero not only in the vicinity of the origin, but also in a vertical strip containing the second coordinate axis. Moreover, S j,k (f,(0,0))=S j (φ,0)S k (ψ,0). Taking an arbitrarily large j for which S j (φ,0)≠0, we can choose a k such that the sum S j,k (f,(0,0)) is larger than every preassigned number.

It can be proved that the localization principle is preserved if the usual neighborhoods of a point (a,b) are replaced by “cross neighborhoods”, i.e., by sets of the form {(s,t) |min(|sa|,|tb|)<δ}.

The second fact that we want to mention is connected with Carleson’s theorem (see the end of Sect. 10.3.9) and is considerably harder. C.L. Fefferman18 observed that this theorem cannot be carried over to functions of several variables: there is a periodic function of two variables that is uniformly bounded in the square (0,2π)2 and whose Fourier sums over rectangles are unbounded at every point of this square. The “divergence phenomenon” manifests itself on the functions f N equal to e iNst for 0<s,t<2π (N is a large parameter). It turns out that, despite the fact that |f N |=1, for every N>1 and every point (s,t), there are numbers j and k such that the quantity |S j,k (f N ,(s,t))| is comparable with lnN. We will not discuss the Fefferman’s example in detail, but refer the reader to Exercise 10.

### 10.4.7

As in the one-dimensional case, one can use the Fejér sums to approximate a function of several variables by trigonometric polynomials. In the multi-dimensional case, these sums, as well as their partial sums, can be defined in different ways. We define the m-dimensional Fejér sums by the equation (in what follows, $$n,j\in\mathbb{Z}^{m}_{+}$$ and $$k\in\mathbb{Z}^{m}$$)
$$\sigma _n(f,x)=\frac{1}{n_1\cdots n_m}\sum_{0\leqslant j<n}S_j(f,x)= \sum_{|k|<n} \biggl(1-\frac{|k_1|}{n_1} \biggr) \cdots \biggl(1-\frac{|k_m|}{n_m} \biggr)\widehat{f}(k) e^{i\langle k,x\rangle}.$$
Since S j (f,x)=(fD j )(x), we obtain σ n (f,x)=(f∗Φ n )(x), where
$$\Phi_n(y)=\frac{1}{n_1\cdots n_m}\sum_{0\leqslant j<n}D_j(y)= \Phi_{n_1}(y_1)\cdots\Phi_{n_m}(y_m).$$
It is natural to call the function Φ n the m-dimensional Fejér kernel. The properties of the one-dimensional Fejér kernel established in Sect. 10.4.1 imply immediately that:
1. (a)

Φ n (y)⩾0;

2. (b)

Q Φ n (y) dy=1;

3. (c)

$$\int_{Q\setminus B(\delta )}\Phi_{n}(y)\,dy\leqslant\frac{C_{\delta }}{\min\{n_{1}, \ldots,n_{m}\}}$$ for every δ∈(0,π).

Thus, we can regard the functions $$\{\Phi_{n}\}_{n\in\mathbb{Z}^{m}_{+}}$$ as an approximate identity with the proviso that it is now parametrized by an integer vector n and the localization property is valid as min{n 1,…,n m }→+∞. Therefore, the following analogs of statements (b) and (c) of Theorem 10.4.1 are valid for the sums σ n (f,x).

### Theorem

1. (1)

If $$f\in\widetilde{C}(\mathbb{R}^{m})$$, then σ n (f)⇉f on $$\mathbb{R}^{m}$$ as min{n 1,…,n m }→+∞.

2. (2)

If for some p∈[1,+∞), thenσ n (f)−f p →0 as min{n 1,…,n m }→+∞.

As in the one-dimensional case (see the Remark in Sect. 10.4.1), the convergence of the Fejér sums in mean implies, in particular, the uniqueness theorem for multiple Fourier series:

### Corollary

Functions in coincide almost everywhere if they have the same Fourier coefficients.

(For more general results, see Sects. 11.1.9 and 12.3.3.)

In the multi-dimensional case, however, the Fejér kernels do not satisfy the strong localization property. Therefore, an analog of statement (a) of Theorem 10.4.1 can be obtained for them only under the additional assumption that the function f is bounded (see Theorem 7.6.5). This restriction cannot be lifted as can be shown by modifying the example from the previous section. Indeed, we preserve the first factor φ in the example and change the second factor, rejecting the continuity (which now is not needed), as follows:
$$\psi(t)=\cos t+\frac{1}{2}\cos2t+\frac{1}{3}\cos3t+\cdots.$$
Since $$S_{k}(\psi,0) \underset{k\to\infty}{\longrightarrow}+\infty$$, the same is also valid for the Fejér sums, $$\sigma _{k}(\psi,0) \underset{k\to\infty}{\longrightarrow}+\infty$$. Furthermore, there are infinitely many non-zero sums σ j (φ,0) because the Fourier sums S j (φ,0) have the same property. Consequently, the sums σ j,k (f,(0,0))=σ j (φ,0)σ k (ψ,0) are unbounded, and so do not tend to zero, even though the function f is zero in a strip containing the second coordinate axis.

If f belongs to the class for some p>1, then the sums σ n (f) converge almost everywhere. Although this assumption can be weakened somewhat, it is impossible to drop it completely.

The difficulties arising in the study of the multi-dimensional Fejér method occur also in the “coordinatewise” generalizations of other methods, for example, of the Abel–Poisson method. We will not discuss this question in detail, instead referring the reader to the literature [Zy], vol. II, Chap. XVII.

### 10.4.8

In conclusion, we note that, for m>1, there are different natural ways of constructing partial sums of a multiple Fourier series and their averages. For example, instead of sums over rectangles, we could consider only sums over cubes centered at the origin and their arithmetic means. Although the corresponding kernels will not preserve sign, it can be proved that they define a generalized approximate identity satisfying the assumption (a′), less restrictive than assumption (a) (see Sect. 7.6.1).

Even more difficulties arise for another natural definition of the partial sums of a Fourier series, namely, when the summation is performed over balls. In this case, we form “ball” partial sums, putting
$$S_R(f,x)=\sum_{\|k\|<R}\widehat{f}(k) e^{i\langle k,x\rangle}$$
for an R>0. This partial sum can, of course, be represented as the convolution with the corresponding “ball Dirichlet kernel” D R (y)=(2π)m k∥<R e ik,y for which, unfortunately, no compact expression is known. The sums S R (f) have the important property that they do not satisfy a “logarithmic” estimate similar to inequality (7). This is due to the fast growth of the norms ∥D R 1. It turns out that, as R→+∞, the norms ∥D R 1 grow (in order) as R (m−1)/2 (see [AIN1, AIN2]). We will return to this unexpected result at the end of the next section.

In the two-dimensional case, the arithmetic means of the “disc” Fourier sums of a periodic continuous function f converge to f uniformly (and if for p<+∞, then in the -norm), but this is not the case for a larger number of variables.

### EXERCISES

1.

Let $$T(x)=\sum_{k=-n}^{n} c_{k} e^{ikx}$$ be a trigonometric series of order n, and let p∈[1,+∞]. Prove the Bernstein 19 inequalityT′∥ p ⩽2nT p . Hint. Verify that T′=−2nT∗Ψ n , where Ψ n (x)=Φ n (x) sinnx and Φ n is a Fejér kernel.

2.

Prove that the Fejér sums cannot converge rapidly: either there is a δ>0 such that $$\|f-\sigma _{n}(f)\|_{1}\geqslant\frac{\delta}{n}>0$$ for all $$n\in\mathbb{N}$$, or f≡const almost everywhere. Hint. Calculate the Fourier coefficients of the difference fσ n (f) and apply inequality (a) of Sect. 10.3.2.

3.

Supplement the result of Corollary 2 of Sect. 10.4.1 by proving that $$\|S_{n}(f)-f\|_{\infty}\leqslant CL\frac{\ln n}{n}$$ for α=1. Hint. To estimate the integral $$S_{n}(f,x) - f(x) = \int_{-\pi}^{\pi}(f(x-u) - f(x))D_{n}(u)\,du$$, replace the difference f(xu)−f(x) on each interval $$[\frac{2k-1}{n+1/2}\pi,\frac{2k+1}{n+1/2}\pi]$$ by its value at the midpoint of this interval and then verify that the integral of the Dirichlet kernel over the interval admits an estimate O(1/k 2).

4.

Prove that the Fourier cosine coefficients can tend to zero arbitrarily slowly, i.e., for every sequence {c n } n⩾1 decreasing to zero, there is a function such that c n a n (f) for all $$n\in\mathbb{N}$$. Hint. Dominate {c n } n⩾1 by a convex sequence and apply Theorem 10.4.2.

5.

Let the sequence of coefficients of a series $$\sum_{n=1}^{\infty} c_{n} \cos nx$$ be convex. Prove that the -norms of the partial sums are bounded if and only if c n =O(1/lnn); prove that the given series converges in if and only if c n =o(1/lnn) as n→∞.

6.

Let $$S(x)=\sum_{n=1}^{\infty} c_{n}\sin nx$$ where c n ↓0. Prove that the boundedness and the continuity of the function S is equivalent to the relation $$c_{n}=O (\frac{1}{n} )$$ and $$c_{n}=o (\frac{1}{n} )$$, respectively.

7.

Prove that the following version of Parseval’s identity is valid for functions and : the series $$\sum_{n\in\mathbb{Z}}\widehat{f}(n)\overline{\widehat {g}(n)}$$ Cesaro converges to $$\frac{1}{2\pi}\int_{-\pi}^{\pi} f(x)\overline{g(x)}\,dx$$.

8.
Prove that the Fourier series of every function f in can be integrated termwise over every rectangular parallelepiped P,
$$\int_Pf(x)\,dx=\sum_{n\in\mathbb{Z}^m} \widehat{f}(n) \int_Pe^{i\langle n,x\rangle}\,dx$$
(the sum of the series on the right-hand side of the equation is regarded as the limit of the partial sums over rectangles).
9.

Let a function be such that $$\widehat{f}(n)\geqslant0$$ for all $$n\in\mathbb{Z}^{m}$$. Prove that if f is bounded and continuous at the origin, then its Fourier series converges absolutely (therefore, f coincides with a function of class $$\widetilde{C}$$ almost everywhere). Hint. Use the fact that the sums σ n (f,0) are bounded.

10.
To construct a continuous function of two variables for which the Fourier series diverges everywhere in the square (0,2π)2 (see Sect. 10.4.6), prove that the Fourier sums S j,k (f N ) of the function f N (s,t)=e iNst (N⩾1, 0<s,t<2π) over the rectangles [−j,j]×[−k,k] satisfy the following inequalities at each point of the given square (the constants at the O-terms depend only on s and t):
1. (a)

|S j,k (f N ;s,t)|=O(lnj);

2. (b)

if k>2πN, then $$|S_{j,k}(f_{N};s,t)|= O (1+\frac{\ln j}{k-2\pi N} )$$;

3. (c)

$$|S_{j,k}(f_{N};s,t)|\geqslant\frac{1}{2\pi}\ln N+O(1)$$ for j=[Ns] and k=[Nt].

Conclude from this that, for a sufficiently small r>0 and N n =e n!, the Fourier series of the function $$F(s,t)=\sum_{n=1}^{\infty} r^{n} e^{iN_{n}st}$$ diverges unboundedly (the sums S j,k (F) are unbounded) at each point of the square (0,2π)2.

## 10.5 The Fourier Transform

### 10.5.1

We introduce one of the main concepts of this chapter.

### Definition

The Fourier transform $$\widehat{f}$$ of a function f in is defined by the formula
$$\widehat{f}(y)=\int_{\mathbb{R}^m}f(x) e^{-2\pi i\langle y,x\rangle}\,dx \quad \bigl(y\in\mathbb{R}^m\bigr)$$
(here, as before, 〈y,x〉 is the scalar product of vectors y and x).
Theorem 7.1.3 on the continuity of an integral depending on a parameter implies that the function $$\widehat{f}$$ is continuous. This function is bounded since
$$\big|\widehat{f}(y)\big|\leqslant\|f\|_1\quad\text{for all}\ y\in \mathbb{R}^m.$$
Moreover, by the Riemann–Lebesgue theorem, we have $$\widehat{f}(y)\to 0$$ as ∥y∥→+∞.
We recall that the translation f h of a function f by a fixed vector $$h\in\mathbb{R}^{m}$$ is defined by the equation f h (x)=f(xh). An easy calculation shows that $$\widehat{f}$$ and $$\widehat{f_{h}}$$ are related as follows (see also Exercise 1):
$$\widehat{f_h}(y)=\int_{\mathbb{R}^m}f(x-h) e^{-2\pi i\langle y,x\rangle}\,dx= \int_{\mathbb{R}^m}f(t) e^{-2\pi i\langle y,t+h\rangle}\,dt= e^{-2\pi i\langle y,h\rangle}\,\widehat{f}(y).$$
Another operation with the argument of a function, a contraction, is also connected with the Fourier transform: if $$a\in\mathbb {R}\setminus\{0\}$$ and g(x)=f(ax), then
$$\widehat{g}(y)=\int_{\mathbb{R}^m}f(ax) e^{-2\pi i\langle y,x\rangle}\,dx= \frac{1}{|a|^m}\int_{\mathbb{R}^m}f(t) e^{-2\pi i\frac{1}{a}\langle y,t\rangle}\,dt= \frac{1}{|a|^m}\widehat{f} \biggl(\frac{y}{a} \biggr).$$

An important property of the Fourier transform relates the operations of convolution and multiplication.

### Theorem

If , then $$\widehat{f*g}(y)=\widehat{f}(y)\,\widehat{g}(y)$$ $$(y\in\mathbb{R}^{m})$$. Moreover, $$\int_{\mathbb{R}^{m}}\widehat{f}(y)\,g(y)\,dy= \int_{\mathbb{R}^{m}}f(y)\,\widehat{g}(y)\,dy$$.

### Proof

The proof is an almost verbatim repetition of the corresponding reasoning for Fourier coefficients (see property (e)) of Sect. 10.3.2),
\begin{aligned} \widehat{f*g}(y)=&\int_{\mathbb{R}^m} \biggl(\int _{\mathbb {R}^m}f(u)g(x-u)\,du \biggr) e^{-2\pi i\langle y,x\rangle}\,dx \\ =&\int_{\mathbb{R}^m}f(u) e^{-2\pi i\langle y,u\rangle} \biggl(\int _{\mathbb{R}^m}g(x-u) e^{-2\pi i\langle y,x-u\rangle}\, dx \biggr)\,du \\ =&\int_{\mathbb{R}^m}f(u) e^{-2\pi i\langle y,u\rangle} \biggl(\int _{\mathbb{R}^m}g(v) e^{-2\pi i\langle y,v\rangle}\,dv \biggr)\,du=\widehat{f}(y) \widehat{g}(y). \end{aligned}

The second relation is proved similarly. □

We consider some examples.

### Example 1

The Fourier transform of the characteristic function χ of the interval (−1,1) is calculated very simply:
$$\widehat{\chi}(y)=\int_{-\infty}^{\infty}\chi(x) e^{-2\pi iyx}\,dx= \int_{-1}^1 e^{-2\pi i yx}\,dx=\frac{\sin2\pi y}{\pi y}.$$
We remark that (this is established in Example 1 of Sect. 4.6.6).

### Example 2

We consider the function $$f_{t}(x)=e^{-\pi t^{2}x^{2}}$$ $$(x\in\mathbb{R},\ t>0)$$. Its Fourier transform is actually calculated in Example 1 of Sect. 7.1.6:
$$\widehat{f}_t(y)=\int_{-\infty}^{\infty} e^{-\pi t^2 x^2} e^{-2\pi iyx}\,dx =2\int_0^{\infty} e^{-\pi t^2 x^2}\cos2\pi yx\,dx=\frac{1}{t}e^{-\frac{\pi}{t^2}y^2}.$$
(1)
It is interesting to note that $$\widehat{f}_{t}=\frac{1}{t} f_{\frac{1}{t}}$$ and, in particular, $$\widehat{f_{1}}=f_{1}$$.
From Eq. (1), we immediately obtain its multi-dimensional counterpart,
$$\int_{\mathbb{R}^m} e^{-\pi t^2 \|x\|^2} e^{-2\pi i\langle y,x\rangle}\,dx = \frac{1}{t^m} e^{-\frac{\pi}{t^2}\|y\|^2}.$$
(1′)

### Example 3

Let f(x)=e −|x| $$(x\in\mathbb{R})$$. Then
\begin{aligned} \widehat{f}(y)=&\int_{-\infty}^{\infty} e^{-|x|} e^{-2\pi iyx}\, dx\\ =&2\mathcal{R}e \biggl(\int_0^{\infty} e^{-(1+2\pi iy)x}\,dx \biggr)=\mathcal{R}e\frac{2}{1+2\pi iy}=\frac{2}{1+(2\pi y)^2}. \end{aligned}

### Example 4

It is considerably harder to obtain a multi-dimensional generalization of Example 3, i.e., to calculate the Fourier transform of the function f(x)=e −∥x ($$x\in\mathbb{R}^{m}$$) because, in this case, it is impossible to use separation of variables. The complication can be overcome by an artificial trick based on an integral representation of the function e −∥x. We need the formula
$$e^{-t}=\frac{2}{\sqrt{\pi}}\int_0^{\infty} e^{-u^2-\frac{t^2}{4u^2}}\,du\quad\text{for every}\ t>0.$$
To obtain it, we must represent the integral on the right-hand side in the form $$e^{-t}\int_{0}^{\infty} e^{-(u-\frac{t}{2u})^{2}}\,du$$. After the change of variables $$v=u-\frac{t}{2u}$$, this integral reduces to the Euler–Poisson integral $$\int_{-\infty}^{\infty} e^{-v^{2}}\,dv=\sqrt{\pi}$$.
Now, we use the relation established above to calculate $$\widehat{f}$$,
$$\widehat{f}(y)=\int_{\mathbb{R}^m}e^{-\|x\|}e^{-2\pi i\langle y,x\rangle}\,dx= \frac{2}{\sqrt{\pi}}\int_{\mathbb{R}^m} \biggl(\int_0^{\infty} e^{-u^2-\frac{\|x\|^2}{4u^2}}du \biggr)e^{-2\pi i\langle y,x\rangle}\,dx.$$
Changing the order of integration and applying Eq. (1′) with $$t=\frac{1}{2\sqrt{\pi}u}$$, we obtain
\begin{aligned} \widehat{f}(y)=&\frac{2}{\sqrt{\pi}}\int_0^{\infty} e^{-u^2} \biggl(\int_{\mathbb{R}^m} e^{-\frac{\|x\|^2}{4u^2}}e^{-2\pi i\langle y,x\rangle} \,dx \biggr)\,du \\=&\frac{2}{\sqrt{\pi}}\int_0^{\infty} e^{-u^2}(2\sqrt{\pi}u)^m e^{-4\pi^2u^2\|y\|^2} \,du\\=&2^{m+1}\pi^{\frac{m-1}{2}}\int_0^{\infty} u^me^{-(1+4\pi^2\|y\|^2)u^2}du. \end{aligned}
Now, the change of variables v=(1+4π 2y2)u 2 allows us to express the last integral in terms of the Gamma function, and we obtain the required result
$$\widehat{f}(y)=2^m\pi^{\frac{m-1}{2}}\, \frac{\Gamma (\frac{m+1}{2} )}{(1+4\pi^2\|y\|^2)^{\frac{m+1}{2}}}.$$

### Example 5

Let a,u>0, and let f(x)=x a−1 e ux for x>0 and f(x)=0 for x<0. Then $$\widehat{f}(y)=\frac{\Gamma(a)}{(u+2\pi iy)^{a}}$$ (we use the branch of the power function z a equal to 1 at z=1). This was established in Example 1 of Sect. 7.1.7.

Before passing to a more detailed study of the properties of the Fourier transform, we show the usefulness of this notion by one more example.

### Example 6

Let f be a function in equal to zero outside (−π,π), and let f 0 be its 2π-extension from this interval to $$\mathbb{R}$$ (). The Fourier coefficients of f 0 can easily be expressed in terms of the Fourier transform of f, namely, $$\widehat{f_{0}}(n)=\frac{1}{2\pi}\widehat{f} (\frac{n}{2\pi} )$$ for all $$n\in\mathbb{Z}$$. We also consider the function g(x)=e itx for x∈(−π,π), where t is a fixed number. Since the Fourier coefficients of g are equal to
$$\widehat{g}(n)=\frac{1}{2\pi}\int_{-\pi}^{\pi} e^{i(t-n)x}\,dx= \frac{\sin\pi(t-n)}{\pi(t-n)},$$
we obtain by Parseval’s generalized identity (see Theorem 2 of Sect. 10.3.6)
\begin{aligned} \widehat{f} \biggl(\frac{t}{2\pi} \biggr)=&\int_{-\pi}^{\pi} f_0(x)\overline{g}(x)\,dx= 2\pi\sum_{n=-\infty}^{\infty} \widehat{f_0}(n)\overline{\widehat{g}(n)}\\=& \sum _{n=-\infty}^{\infty}\widehat{f} \biggl(\frac{n}{2\pi} \biggr) \frac{\sin\pi(t-n)}{\pi(t-n)}. \end{aligned}
Thus, the following sampling formula is valid for the function $$F(t)=\widehat{f}(t/2\pi)$$:
$$F(t)=\sum_{n=-\infty}^{\infty} F(n) \frac{\sin\pi(t-n)}{\pi(t-n)}= \frac{\sin\pi t}{\pi}\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{t-n}F(n).$$
This formula allows one to recover the value of a function F at an arbitrary point t knowing the values of F on the integer lattice. This fact plays a fundamental role in optics and radio engineering because it is easier to deal with a discrete system of values than with a continuously varying signal. A multi-dimensional version of the sampling formula is given in Exercise 3.

### 10.5.2

We establish elementary relations between differentiation and the Fourier transform.

### Theorem

Let . Then:
1. (1)
if the partial derivative $$g=\frac{\partial f}{\partial x_{k}}$$ is summable and continuous for some k=1,…,m, then
$$\widehat{g}(y)=2\pi iy_k \widehat{f}(y)\quad\bigl(y\in \mathbb{R}^m\bigr);$$

2. (2)
if the productxf(x) is summable, then $$\widehat{f}\in C^{1}(\mathbb{R}^{m})$$ and the equation
$$\frac{\partial\widehat{f}(y)}{\partial y_k}=-2\pi i\widehat {f}_k(y),\quad {\textit{where}}\ f_k(x)=x_kf(x)\ \bigl(x\in\mathbb{R}^m \bigr),$$

holds for all $$y\in\mathbb{R}^{m}$$ and k=1,…,m

### Proof

(1) Without loss of generality, we will assume that k=m. We identify a point x=(x 1,…,x m−1,t) with the pair (u,t), where $$u=(x_{1},\ldots,x_{m-1})\in\mathbb{R}^{m-1}$$. First, we verify that f(u,t)→0 as t→±∞ for almost all points $$u\in\mathbb{R}^{m-1}$$. Indeed, since the derivative $$f'_{t}=g$$ is continuous, we have
$$f(u,t)-f(u,0)=\int_0^tg(u,s)\,ds.$$
From Fubini’s theorem, it follows that the function tg(u,t) is summable for almost all u, and, therefore,
$$f(u,t)-f(u,0)=\int_0^tg(u,s)\,ds\underset{t \to\pm\infty}{\longrightarrow} \int_0^{\pm\infty} g(u,s)\,ds.$$
Thus, the limits lim t→±∞ f(u,t) exist and are finite for almost all $$u \in\mathbb{R}^{m-1}$$. However, since (again, by Fubini’s theorem) the function tf(u,t) is summable for almost all u, we see that the limits are zero for such u and, therefore,
\begin{aligned} \int_{-\infty}^{\infty} g(u,t) e^{-2\pi iy_mt}\,dt=& f(u,t) e^{-2\pi iy_mt}\Big\vert _{-\infty}^{\infty}- (-2\pi iy_m)\int_{-\infty}^{\infty} f(u,t) e^{-2\pi iy_mt}\,dt \\=&2\pi iy_m\int_{-\infty}^{\infty} f(u,t) e^{-2\pi iy_mt}\,dt. \end{aligned}
To obtain the required result, it only remains to multiply both sides of this equation by $$e^{-2\pi i(y_{1}x_{1}+\cdots+y_{m-1}x_{m-1})}$$ and integrate with respect to u.

To obtain the equation of (2), we must apply the Leibnitz rule (see Sect. 7.1.5). The functions f 1,…,f m are summable by assumption. Therefore, their Fourier transforms and the first order partial derivatives of $$\widehat{f}$$ are continuous everywhere. Consequently, $$\widehat{f}\in C^{1}(\mathbb{R}^{m})$$. □

### Corollary

If is a compactly supported function, then $$\widehat{f}\in C^{\infty}(\mathbb{R}^{m})$$; if $$f\in C_{0}^{\infty}(\mathbb {R}^{m})$$, then the product $$\|y\|^{p}\widehat{f}(y)$$ is summable in $$\mathbb{R}^{m}$$ for every p>0.

### Proof

The fact that $$\widehat{f}$$ is infinitely differentiable follows directly from the second assertion of the theorem because the product ∥x n f(x) is summable for every $$n\in\mathbb{N}$$.

If $$f\in C_{0}^{\infty}(\mathbb{R}^{m})$$, then the derivatives of all orders of f are summable and the relation
$$\biggl(\frac{\partial^nf}{\partial y_k^n} \biggr)\ \widehat{\phantom{ )}}\ (y)=(2\pi i y_k)^n\widehat{f}(y)$$
is fulfilled for all k=1,…,m and $$n\in\mathbb{N}$$. Since the functions $$(\frac{\partial^{n}f}{\partial y_{k}^{n}} ) \ \widehat{}\ (y)$$ are bounded, we obtain the estimate
$$\big|\widehat{f}(y)\big|\leqslant\mathrm{const}\cdot\bigl(1+|y_1|^n+ \cdots+|y_m|^n\bigr)^{-1}$$
providing (if we take sufficiently large n) the summability of $$\|y\|^{p}\widehat{f}(y)$$. □

In many problems, it is important to know the rate of decrease of the Fourier transform at infinity. The theorem proved above shows that a fast decrease can be provided by the smoothness of the function in question. How accurate are these conditions? What can be expected if smoothness fails on a “small” set? The following examples are devoted to such results.

### Example 1

Supplementing Examples 2 and 3 of Sect. 10.5.1, we investigate the asymptotic behavior of the Fourier transform of the function $$f(x)=e^{-|x|^{p}}$$ at infinity for 0<p<2. After integrating by parts, we see that
$$\widehat{f}(y)=2\int_0^{\infty} e^{-x^p} \cos(2\pi xy)\,dx= \frac{p}{\pi y}\int_0^{\infty} e^{-x^p}x^{p-1}\sin(2\pi xy)\,dx,$$
(2)
which implies the crude estimate $$\widehat{f}(y)=o(1/y)$$ as y→+∞. We study the behavior of $$\widehat{f}(y)$$ for large y in detail. If 0<p<1, then the change 2πxy=u leads to the equation
$$\widehat{f}(y)=\frac{2p}{(2\pi y)^{p+1}} \int_0^{\infty} e^{- (\frac{u}{2\pi y} )^p}\frac{\sin u}{u^{1-p}}\,du.$$
The integral $$\int_{0}^{\infty}\frac{\sin u}{u^{1-p}}\,du$$ of the limit function (as y→+∞) converges, and Corollary 2 of Sect. 7.4.7 justifies the passage to the limit,
$$\int_0^{\infty} e^{- (\frac{u}{2\pi y} )^p} \frac{\sin u}{u^{1-p}}\,du \underset{y\to+\infty}{\longrightarrow} \int _0^{\infty}\frac{\sin u}{u^{1-p}}\,du=\Gamma(p)\,\sin \frac{\pi p}{2}$$
(the equality was established in Example 1 of Sect. 7.4.8). Consequently, the estimate
$$\widehat{f}(y)\sim\frac{C_p}{y^{p+1}}\quad \text{as}\ y\to+\infty$$
(3)
is valid for 0<p<1 with constant $$C_{p}=\frac{2\Gamma(p+1)}{(2\pi)^{p+1}}\sin\frac{\pi p}{2}$$. This, in particular, implies that the function $$\widehat{f}$$ is summable.
Now, let 1<p<2 (the case where p=1 was considered in Example 3 of Sect. 10.5.1). Integrating the right-hand side of (2) one more time, we arrive at the equation
\begin{aligned} \widehat{f}(y)=&\frac{p}{2(\pi y)^2} \biggl((p-1)\int _0^{\infty} x^{p-2}e^{-x^p} \cos(2\pi yx)\,dx\\&{}-p\int_0^{\infty} x^{2(p-1)}e^{-x^p}\cos(2\pi yx)\, dx \biggr). \end{aligned}
Here, the second integral admits the estimate o(1/y) as y→+∞, but the first integral tends to zero more slowly. Indeed, by an almost verbatim repetition of the reasoning given in the case where 0<p<1, we obtain
$$\int_0^{\infty} x^{p-2} e^{-x^p}\cos(2\pi yx)\,dx\underset{y\to+\infty}{\sim} \frac{1}{(2\pi y)^{p-1}} \Gamma(p-1)\sin\frac{\pi p}{2}.$$
Thus, we again come to relation (3), which is also valid for p=1; for p=2, the coefficient C p vanishes, and the asymptotic of $$\widehat{f}$$ changes completely (see Examples 2 and 3 of Sect. 10.5.1).

It can be proved that $$\widehat{f}(y)>0$$ for 0<p⩽2 (for 0<p⩽1, this follows from the result of Example 2 of Sect. 4.6.6 and the fact that the function $$e^{-x^{p}}$$ is convex).

### Example 2

Let us determine how fast the Fourier transform of the characteristic function of the unit ball
$$\widehat{\chi}_B(y)=\int_Be^{-2\pi i\langle y,x\rangle} \,dx= \int_B e^{-2\pi i\|y\|x_1}\,dx=\alpha _{m-1}\int _{-1}^1 \bigl(1-t^2 \bigr)^{\frac{m-1}{2}} e^{-2\pi i\|y\|t}\,dt$$
decreases at infinity (α m−1 is the volume of the unit ball in $$\mathbb{R}^{m-1}$$). For odd m the “integral can be calculated” and $$\widehat{\chi}_{B}$$ can be expressed explicitly in terms of ∥y∥. In particular, in the one-dimensional case, we have B=(−1,1) and $$\widehat{\chi}_{B}(y)=\frac{\sin2\pi y}{\pi y}$$. For m=3, we have $$\widehat{\chi}_{B}(y)=\frac{1}{\pi\|y\|^{2}} (\frac{\sin2\pi\|y\|}{2\pi \|y\|}- \cos2\pi\|y\| )$$ (this also follows from the result of the Example in Sect. 6.2.5 with f 0=χ (0,1)).
For even m, the situation is more complicated. In this case, χ B can be expressed in terms of the Bessel function. However, an exact formula for $$\widehat{\chi}_{B}(y)$$ is not our main concern here. We want to study the asymptotic behavior of this function as ∥y∥→+∞. To this end, we put r=2πy∥ and consider the integrals
$$I_m(r)=\int_{-1}^1 \bigl(1-t^2\bigr)^{\frac{m-1}{2}} e^{-irt}\,dt\quad (m=0,1,2,\ldots).$$
The larger m is, the more derivatives of the function $$(1-t^{2})^{\frac{m-1}{2}}$$ vanish at the endpoints of the interval of integration. Therefore, as m increases, the rate at which the integrals I m (r) tend to zero as r→+∞ also increases. To describe this in more detail, we use the recurrence formula
$$I_m(r)=\frac{m-1}{r^2} \bigl((m-2)I_{m-2}(r)-(m-3)I_{m-4}(r) \bigr) \quad(m\geqslant4),$$
which can easily be obtained by twofold integration by parts. From this relation, we see that, to obtain the asymptotic of the integral I m (r), it is sufficient to know only the asymptotics of the integrals I 0(r) and I 2(r) or of I 1(r) and I 3(r), depending on the parity of m. The integrals I 1(r) and I 3(r) can easily be calculated,
$$I_1(r)=2\frac{\sin r}{r},\qquad I_3(r)= \frac{4}{r^2} \biggl(\frac{\sin r}{r}- \cos r \biggr).$$
The integrals I 0(r) and I 2(r) coincide with the integrals C(r) and S(r), respectively, considered in the example of Sect. 9.2.5:
\begin{aligned} I_0(r)=&C(r)=\sqrt{\frac{\pi}{r}}(\sin r+\cos r)+O \biggl( \frac{1}{r} \biggr),\\ I_2(r)=&S(r)=\frac{\sqrt{\pi}}{r^{3/2}}(\sin r-\cos r)+O \biggl(\frac {1}{r^2} \biggr). \end{aligned}
The last four formulas can be written uniformly as follows:
$$I_m(r)=\frac{\gamma_m}{r^{\frac{m+1}{2}}}\cos(r-\varphi _m)+ O \biggl( \frac{1}{r^{\frac{m}{2}+1}} \biggr) \quad(r\to+\infty),$$
where m=0,1,2,3, $$\varphi _{m}=\frac{\pi}{4}(m+1)$$, and γ m is a positive coefficient depending only on m. The recurrence formula allows us to extend this relation to all positive integers m.
Returning to the Fourier transform of the function χ B , we see that
$$\widehat{\chi}_B(y)=\alpha _{m-1}I_m\bigl(2\pi\|y\|\bigr)= \frac{C_m}{\|y\|^{\frac{m+1}{2}}}\cos\bigl(2\pi\|y\|-\varphi _m\bigr)+ O \biggl(\frac{1}{\|y\|^{\frac{m}{2}+1}} \biggr).$$
It can be verified that C m =1/π for all m.
It is interesting to compare $$\widehat{\chi}_{B}$$ with the function $$\widehat{\chi}_{Q}$$, where Q=(−1,1) m . It is clear that
$$\widehat{\chi}_Q(y)=\prod_{j=1}^m \frac{\sin2\pi y_j}{\pi y_j}.$$
If the angles between the vector y and the coordinate axes are non-zero, then this function admits the estimate O(∥ym ). Thus, for most directions, the function decreases considerably faster than $$\widehat{\chi}_{B}$$. One possible sharpening of this assertion is as follows: the integrals $$L_{B}(R)=\int_{\|y\|<R}|\widehat{\chi}_{B}(y)|\,dy$$ grow considerably faster than the integrals $$L_{Q}(R)=\int_{\|y\|<R}|\widehat{\chi}_{Q}(y)|\,dy$$ as R→+∞. Indeed,
\begin{aligned} L_Q(R)\leqslant&\int_{[-R,R]^m}\big|\widehat{\chi}_Q(y)\big|\,dy= \prod_{j=1}^m \int_{-R}^R \biggl \vert \frac{\sin2\pi y_j}{\pi y_j} \biggr \vert \,dy_j\\=& \biggl(\frac{2}{\pi}\int _0^{2\pi R}\frac{|\!\sin t|}{t}\,dt \biggr)^m. \end{aligned}
It follows that L Q (R)=O(ln m R) as R→+∞. At the same time
\begin{aligned} L_B(R)=&\alpha _m\int_0^R \biggl \vert \frac{C_m}{r^{\frac{m+1}{2}}}\cos(2\pi r-\varphi _m)+ O \biggl( \frac{1}{r^{\frac{m}{2}+1}} \biggr)\biggr \vert \,r^{m-1}\,dr \\=&\alpha _mC_m\int_0^Rr^{\frac{m-3}{2}}\big| \cos(2\pi r-\varphi _m)\big|\,dr+O\bigl(R^{\frac{m}{2}-1}\bigr) \end{aligned}
for m>2 (for m=2, the remainder term has order O(lnR)). Therefore, $$L_{B}(R)=O(R^{\frac{m-1}{2}})$$, and the estimate is exact by order,
\begin{aligned} L_B(R)\geqslant&\alpha _mC_m\int _{R/2}^Rr^{\frac{m-3}{2}}\cos^2(2\pi r-\varphi _m)\,dr+ O\bigl(R^{\frac{m}{2}-1}\bigr) \\ \geqslant&\mathrm{const}\,R^{\frac{m-3}{2}}\int_{R/2}^R \bigl(1+\cos2(2\pi r-\varphi _m)\bigr)\,dr+ O\bigl(R^{\frac{m}{2}-1}\bigr) \\ =&\frac{\mathrm{const}}{2}R^{\frac{m-1}{2}}+O\bigl(R^{\frac{m}{2}-1}\bigr). \end{aligned}

### Example 3

It follows from the theorem that the condition f(x)=O(∥xp ) as ∥x∥→+∞ implies the smoothness of the Fourier transform for p>m+1. It turns out that this restriction cannot be weakened essentially. To verify this, we show that if f(x)∼∥xp as ∥x∥→+∞, then the differentiability of $$\widehat{f}$$ at zero implies the inequality p>m+1.

Without loss of generality, we may assume that f⩾0. Indeed, we know that f(x)⩾0 for large ∥x∥, but changing the function on an arbitrary ball (for example, putting f(x)=0 inside the ball), we change the Fourier transform of f by an infinity differentiable function.

Assuming that f⩾0, we study the mean value of the difference $$\widehat{f}(0)- \widehat{f}$$ in the vicinity of zero (in what follows, B is the unit ball centered at zero and v is the volume of B). We put
$$I(r)=\frac{1}{v}\int_B\bigl(\widehat{f}(0)-\widehat{f}(ry) \bigr)\,dy.$$
Since $$\widehat{f}$$ is differentiable at zero, we obtain that I(r)=o(r) as r→+0. Now, we estimate the integral I(r) from below. Since
$$\widehat{f}(0)-\widehat{f}(ry)=\int_{\mathbb{R}^m}f(x) \bigl(1-e^{-2\pi ir\langle y,x \rangle} \bigr)\,dx,$$
we obtain, by Fubini’s theorem, that
$$I(r)=\int_{\mathbb{R}^m}f(x) \biggl(1-\frac{1}{v}\int _B e^{-2\pi irs\langle y,x\rangle}\,dy \biggr)\,dx= \int _{\mathbb{R}^m}f(x) \biggl(1-\frac{1}{v}\widehat{\chi}(rx) \biggr)\,dx,$$
where χ is the characteristic function of B. Obviously, $$\widehat {\chi} (x)\in\mathbb{R}$$ and $$|\widehat{\chi}(x)|\leqslant v$$, and the Riemann–Lebesgue theorem implies that $$\widehat{\chi}(x)\to0$$ as ∥x∥→+∞. We take a sufficiently large radius R so that $$f(x)>\frac{1}{2\|x \|^{p}}$$ and $$|\widehat{\chi}(x)|<\frac{v}{2}$$ for ∥x∥>R. Then, since f⩾0, we have
\begin{aligned} I(r)\geqslant&\int_{\|x\|>R/r}f(x) \biggl(1-\frac{1}{v}\widehat{ \chi}(rx) \biggr)\,dx\\ \geqslant& \frac{1}{4}\int_{\|x\|>R/r} \frac{dx}{\|x\|^p}= \frac{\sigma (S^{m-1})}{4}\int_{R/r} \frac{dt}{t^{p-m+1}}. \end{aligned}
Thus, I(r)⩾const r pm . Since I(r)=o(r) for r→0, we obtain that p>m+1.

### 10.5.3

In the one-dimensional case, for a function f differentiable at a point x, there is an important formula allowing one to find the value of f(x) from $$\widehat{f}$$. This formula is called the inversion formula and has the following form:
$$f(x)=\int_{-\infty}^{\infty}\widehat{f}(y) e^{2\pi ixy}\,dy.$$
The integral on the right-hand side of this equation is called the Fourier integral of f. In general, this is an improper integral because the Fourier transform can be non-summable on $$\mathbb{R}$$ (see Sect. 10.5.1). We will say that the integral converges if there exists a limit of the partial integrals
$$I_A(f,x)=\int_{-A}^A \widehat{f}(y) e^{2\pi ixy}\,dy$$
as A→+∞.

There is an obvious analogy between the expansion of a periodic function in a Fourier series and the Fourier integral representation of a non-periodic function. The following theorem shows that these two problems share not only some superficial analogies but are connected in essence. To show this, we need the following easy lemma.

### Lemma

Let and $$x\in\mathbb{R}$$. Then the following holds for every A>0
$$I_A(f,x)=\int_{-A}^A \widehat{f}(y) e^{2\pi ixy}\,dy= \int_{-\infty}^{\infty} f(x-t)\frac{\sin2\pi At}{\pi t}\,dt.$$

### Proof

It is clear that
$$I_A(f,x)=\int_{-A}^A \biggl(\int _{-\infty}^{\infty} f(u) e^{2\pi i(x-u)y}\,du \biggr)\,dy.$$
Since the function (y,u)↦f(u)e 2πi(xu)y is summable in the strip $$(-A,A)\times\mathbb{R}$$, we may use Fubini’s theorem,
$$I_A(f,x)=\int_{-\infty}^{\infty} \biggl(\int _{-A}^A f(u) e^{2\pi i(x-u)y}\,dy \biggr)\,du=\int _{-\infty}^{\infty} f(u)\frac{\sin2\pi A(x-u)}{\pi(x-u)}\,du.$$
It remains to change the integration variable t=xu. □
By the Riemann–Lebesgue theorem, the integral $$\int_{|t|\geqslant \delta }f(x-t)\frac{\sin2\pi At}{\pi t}\,dt$$ tends to zero as A→+∞ for every δ>0. Therefore, the lemma implies the asymptotic relation
$$I_A(f,x)=\int_{-\delta }^{\delta } f(x-t) \frac{\sin2\pi At}{\pi t}\,dt+o(1) \quad\text{as}\ A\to+\infty$$
(4)
(we already know a similar result for the partial sums of Fourier series; see Eq. (5′) of Sect. 10.3.4). Thus, the behavior of the integrals I A (f,x) as A→+∞ is determined only by the values of f in the vicinity of x. In other words, we have the same localization principle for Fourier integrals as for Fourier series. Furthermore, it is easy to prove the equiconvergence of the expansions in the Fourier series and the Fourier integral. More precisely, the following statement holds.

### Theorem

If functions and coincide in a neighborhood of a point x, then the convergence of the Fourier integral of f at x is equivalent to the convergence of the Fourier series of f 0 at x, and, in the case of convergence, the following holds:
$$\int_{-\infty}^{\infty}\widehat{f}(y) e^{2\pi ixy}\,dy= \sum_{n=-\infty}^{\infty} \widehat{f}_0(n) e^{inx}.$$

From the theorem, it obviously follows that the convergence tests for Fourier series, obtained in Sect. 10.3.4, can be carried over to the Fourier integrals. In particular, the inversion formula is valid at a point x if Dini’s condition is fulfilled at x with C=f(x). We leave it to the reader to state an analog of the Dirichlet–Jordan test.

### Proof

We show that the following holds:
$$I_A(f,x)-S_{[2\pi A]}(f_0,x)\underset{A\to+ \infty}{\longrightarrow}0,$$
where [u], as usual, is the integer part of u.
Let f(xt)=f 0(xt) for |t|<δ, where 0<δ<π. By Eq. (4) and Eq. (5′) of Sect. 10.3.4, we have
\begin{aligned} I_A(f,x)=&\int_{-\delta }^{\delta } f(x-t) \frac{\sin2\pi At}{\pi t}\,dt+o(1)=\int_{-\delta }^{\delta } f_0(x-t)\frac{\sin2\pi At}{\pi t}\, dt+o(1), \\S_n(f_0,x)=&\int_{-\pi}^{\pi} f_0(x-t)\frac{\sin nt}{\pi t}\,dt+o(1)= \int_{-\delta }^{\delta } f_0(x-t)\frac{\sin nt}{\pi t}\,dt+o(1) \end{aligned}
as A,n→+∞. If $$2\pi A=n\in\mathbb{N}$$, then we immediately obtain the required relation. If 2πA is not integer, then we have n<2πA<n+1 for n=[2πA], and, therefore,
\begin{aligned} \big|I_A(f,x)-I_{n/2\pi}(f,x)\big|\leqslant&\int_{A-\frac{1}{2\pi}}^A\big| \widehat {f}(y)\big|\,dy+ \int_{-A}^{-A+\frac{1}{2\pi}}\big| \widehat{f}(y)\big|\,dy\\ \leqslant&2\max_{|y|\geqslant A-1}\big| \widehat{f}(y)\big| \underset{A\to+\infty}{\longrightarrow}0. \end{aligned}
Thus,
$$I_A(f,x)-S_n(f_0,x)= \bigl(I_A(f,x)-I_{n/2\pi}(f,x) \bigr)+ \bigl(I_{n/2\pi}(f,x)-S_n(f_0,x) \bigr),$$
where each of the two differences on the right-hand side tend to zero as A→+∞. □

Now, we once again turn to Examples 2 and 3 considered in Sect. 10.5.1.

### Example 1

From the theorem, it follows that the inversion formula is valid for the function $$f_{t}(x)=e^{-\pi t^{2}x^{2}}$$ $$(x\in\mathbb{R},\ t>0)$$. However, this already follows from the relation $$\widehat{f}_{t}=\frac{1}{t} f_{\frac{1}{t}}$$ established in Example 2 of Sect. 10.5.1. Indeed, since the function $$\widehat{f}_{t}$$ is even and summable, we have
\begin{aligned} \int_{-\infty}^{\infty}\widehat{f}_t(y) e^{2\pi ixy}\,dy=(\widehat{f}_t )\ {}^{\widehat{}}\ (x)= \frac{1}{t} (f_{\frac{1}{t}} )\ {}^{\widehat{}}\ (x)=f_t(x). \end{aligned}

### Example 2

The function $$f(x)=e^{-|x|}\ (x\in\mathbb{R})$$ satisfies Dini’s condition at every point (in particular, at zero). The Fourier transform of f was calculated in Example 3 of Sect. 10.5.1. By the inversion formula, we obtain
\begin{aligned} e^{-|x|}=& \int_{-\infty}^{\infty}\widehat{f}(y) e^{2\pi iyx}\,dy= \int_{-\infty}^{\infty} \frac{2 e^{2\pi iyx}}{1+4\pi^2y^2}\,dy\\=& \int_0^{\infty} \frac{4\cos2\pi yx}{1+4\pi^2y^2}\,dy= \frac{2}{\pi}\int_0^{\infty} \frac{\cos xt}{1+t^2}\,dt. \end{aligned}
Thus, we again obtain the value of the Laplace integral
$$\int_0^{\infty}\frac{\cos xt}{1+t^2}\,dt= \frac{\pi}{2} e^{-|x|},$$
which was calculated in a different way in Example 2 of Sect. 7.4.8.

### 10.5.4

Generalizing the inversion formula to functions of several variables, we confine ourselves to the most important case where the Fourier transform is summable. In this connection, we note that Dini’s condition providing the validity of the inversion formula in the one-dimensional case is a local property of a summable function, whereas the summability of the Fourier transform is a global property.

In contrast to the one-dimensional setting, now, when deriving the inversion formula, we cannot use the equiconvergence of the expansions in the Fourier series or Fourier integral since Theorem 10.5.3 cannot be carried over to the multi-dimensional case (see Exercise 6).

The transformation that assigns the function defined by the formula
to is called the inverse transform. Obviously, , and so the properties of the Fourier transform can easily be carried over to the inverse transform. Using the inverse transform, we can represent the inversion formula proved in the one-dimensional case in the following form: This justifies the choice of the term “inverse transform”.

### Theorem

We remark that the right-hand side of Eq. (5) continuously depends on x since $$\widehat{f}$$ is summable. Therefore, the condition of the theorem (the summability of $$\widehat{f}$$) can be fulfilled only if the function f is equivalent to a continuous function. Moreover, Eq. (5) is valid at all points where f is continuous because it is valid on a set of full measure. In particular, if f is continuous and its Fourier transform is summable, then for all $$x\in\mathbb{R}^{m}$$.

### Proof

We use the approximate identity W t , which played an important role in the proof of the Weierstrass theorem in Sect. 7.6.4. We recall that $$W_{t}(x)=\frac{1}{t^{m}} e^{-\frac{\pi}{t^{2}}\|x\|^{2}}$$ ($$x\in\mathbb{R}^{m}$$, t>0). Our proof of the theorem is based on inversion formula (1′) for this function,
$$W_t(x)=\int_{\mathbb{R}^m}e^{-\pi t^2\|y\|^2} e^{2\pi i\langle x,y\rangle}\,dy.$$
(6)
First, for the smoothened function fW t , we obtain an equation close to (5). Then, we obtain the statement of the theorem by passage to the limit.
Using Eq. (6) and changing the order of integration, we obtain, for every t>0 that
\begin{aligned} (f*W_t) (x)=&\int_{\mathbb{R}^m}f(y)W_t(x-y) \,dy\\= &\int_{\mathbb{R}^m}f(y) \biggl(\int_{\mathbb{R}^m}e^{-\pi t^2\|u\|^2} e^{2\pi i \langle x-y,u\rangle}\,du \biggr)\,dy \\=&\int_{\mathbb{R}^m}e^{-\pi t^2\|u\|^2} e^{2\pi i\langle x,u\rangle} \biggl(\int _{\mathbb{R}^m}f(y) e^{-2\pi i\langle y,u\rangle}\,dy \biggr)\,du. \end{aligned}
Thus, we have established the required relation,
$$(f*W_t) (x)=\int_{\mathbb{R}^m}e^{-\pi t^2\|u\|^2} e^{2\pi i\langle x,u\rangle} \widehat{f}(u)\,du.$$
(7)
Since the absolute value of the integrand in the last integral does not exceed $$|\widehat{f}|$$, we obtain, by Lebesgue’s theorem, that, for every x, this integrals tends to as t→+0.

Now, we can finish the proof, referring to Theorem 10.3.4, from which it follows that the limit on the left-hand side of Eq. (7) coincides with f(x) almost everywhere. However, it is possible to dispense with the use of the theorem based on the notion of a Lebesgue point and on Theorem 4.9.2 on differentiation of an integral with respect to a set. We show that the left-hand side of Eq. (7) tends to f(x) almost everywhere as t tends to zero along a sequence.

Indeed, let {t n } be a sequence such that $$t_{n}\underset{n\to\infty}{\longrightarrow}0$$. Theorem 9.3.3 implies that $$f*W_{t_{n}}\underset{n\to\infty}{\longrightarrow} f$$ in mean, and, consequently, in measure (see Theorem 9.1.2). By Riesz’s theorem (see Sect. 3.3.4), there is a subsequence $$\{t_{n_{k}}\}$$ of {t n } such that, almost everywhere $$f*W_{t_{n_{k}}}\to f$$ as k→∞. Replacing t by $$t_{n_{k}}$$ in Eq. (7) and passing to the limit, we obtain the required result. □

### Example

We give the inversion formula for the function f(x)=e −∥x ($$x\in\mathbb{R}^{m}$$) whose Fourier transform is calculated in Example 4 of Sect. 10.5.1,
\begin{aligned} e^{-\|x\|}=&2^m\pi^{\frac{m-1}{2}}\Gamma \biggl( \frac{m+1}{2} \biggr) \int_{\mathbb{R}^m}\frac{e^{2\pi i\langle x,y\rangle}\,dy}{ (1+4\pi^2\|y\|^2)^{\frac{m+1}{2}}}\\=& \frac{\Gamma(\frac{m+1}{2})}{\pi^{\frac{m+1}{2}}} \int_{\mathbb{R}^m}\frac{\cos\langle x,t\rangle\,dt}{(1+\|t\| ^2)^{\frac{m+1}{2}}}. \end{aligned}
In the one-dimensional case, this formula was obtained in Example 2 of Sect. 10.5.3.

The summability of the Fourier transform is important in many problems (see, for example, Sect. 10.6.4). The result of Exercise 7 shows that this condition is necessarily fulfilled if $$\widehat{f}\geqslant0$$ and the function f is continuous (or at least bounded in a neighborhood of zero). In this connection, we recall (see Example 2 of Sect. 4.6.6) that $$\widehat{f}\geqslant0$$ if f is an even function summable on $$\mathbb{R}$$ and convex on (0,+∞). Together with the inversion formula, this proves the following statement.

### Corollary

If an even continuous function f is summable on the real line and is convex on the positive semi-axis, then f is the Fourier transform of a non-negative summable function.

The fact just proved remains valid even if, instead of the summability of f, we assume only that $$f(x)\underset{x\to+\infty}{\longrightarrow}0$$, but, in this case, the proof invokes a subtler reasoning (see [Luk], Pólya’s theorem).

### 10.5.5

Here, we discuss one more important property of the Fourier transform, its injectivity on the entire set of summable functions. Of course, there is no injectivity in the literal sense because distinct equivalent (i.e., coinciding almost everywhere) functions have the same Fourier transform. However, Theorem 10.5.4 shows that the injectivity holds up to equivalence on the set of functions with summable Fourier transform. To strengthen this result, we generalize Definition 10.5.1 somewhat.

### Definition

Let μ be a finite Borel measure on $$\mathbb {R}^{m}$$. The function $$y\mapsto\widehat{\mu}(y)\equiv\int_{\mathbb{R}^{m}}e^{-2\pi i\langle y,x\rangle}\,d\mu (x)$$ is called the Fourier transform of μ.

If a measure μ has a density f with respect to Lebesgue measure, then $$\widehat{\mu}=\widehat{f}$$.

Now, we establish an important result connected with the injectivity of the Fourier transform of a measure.

### Theorem

If two finite Borel measures μ and ν have the same Fourier transform, then the measures coincide.

### Proof

Let H j (t)={(x 1,…,x m ) | x j =t} be a plane perpendicular to the jth coordinate axis, and let
$$E=\bigl\{t\in\mathbb{R}\,|\,\mu\bigl(H_j(t)\bigr)=\nu \bigl(H_j(t)\bigr)=0\ \text{ for every }\ j=1,\ldots,m\bigr\}.$$
The set E is everywhere dense because the set $$\{t\in\mathbb{R}\,|\, \mu(H_{j}(t))>0\}$$ is at most countable for each j (see Sect. 1.2.2). Therefore, the Borel hull of the semiring consisting of the cells whose vertices have coordinates belonging to E coincides with the σ-algebra of Borel subsets of the space $$\mathbb{R}^{m}$$ (see the remark after Theorem 1.1.6). We express the measure of the cell $$P=\prod_{j=1}^{m}\Delta_{j}$$ in terms of $$\widehat{\mu}$$, assuming that .
Obviously, $$\chi_{P}(x)=\prod_{j=1}^{m}\chi_{\Delta_{j}}(x_{j})$$, where x 1,…,x m are the coordinates of a vector x. By Fubini’s theorem, $$\widehat{\chi}_{P}(y)=\prod_{j=1}^{m}\widehat{\chi}_{\Delta_{j}}(y_{j})$$ for y=(y 1,…,y m ), and, therefore,
$$I_A(\chi_{\Delta},x)=\int_{(-A,A)^m}\widehat{\chi}_P(y) e^{2\pi i\langle x,y \rangle}\,dy =\prod _{j=1}^m\int_{-A}^A \widehat{\chi}_{\Delta_j}(y_j) e^{2\pi ix_jy_j} \,dy_j.$$
The characteristic function of an interval satisfies Dini’s condition everywhere except the endpoints of the interval. Therefore, we have
$$\int_{-A}^A\widehat{\chi}_{\Delta_j}(y_j) e^{2\pi ix_jy_j}\,dy_j \underset{A\to+\infty}{\longrightarrow} \chi_{\Delta_j}(x_j)$$
for all j=1,…,m, provided that x j is distinct from the endpoints of the interval Δ j . Since , we see that $$I_{A}(\chi_{\Delta},x) \underset{A\to+\infty}{\longrightarrow}\chi_{P}(x)$$ μ-almost everywhere. Moreover, putting Δ j =[a j ,b j ), we obtain (see Lemma 10.5.3) that
$$\int_{-A}^A\widehat{\chi}_{\Delta_j}(y_j) e^{2\pi ix_jy_j}\,dy_j= \int_{-\infty}^{\infty} \chi_{\Delta_j}(x_j-t)\frac{\sin2\pi At}{t}\,dt= \int _{A(x_j-b_j)}^{A(x_j-a_j)}\frac{\sin2\pi u}{u}\,du.$$
All these integrals are bounded (since the integral $$\int_{0}^{\infty}\frac{\sin2\pi u}{u}\,du$$ converges), and so, the integral I A (χ Δ,x) is also bounded (uniformly with respect to x and A). Therefore, we can use Lebesgue’s theorem on passing to the limit under the integral sign,
\begin{aligned} \mu(P)=&\int_{\mathbb{R}^m}\chi_P(x)\,d\mu(x)= \lim _{A\to+\infty}\int_{\mathbb{R}^m}I_A( \chi_{\Delta},x)\,d\mu(x) \\=&\lim_{A\to+\infty}\int_{\mathbb{R}^m} \biggl(\int _{(-A,A)^m}\widehat{\chi}_P(y) e^{2\pi i\langle x,y\rangle} \,dy \biggr)\,d\mu(x). \end{aligned}
Changing the order of integration, we obtain
\begin{aligned} \mu(P)=&\lim_{A\to+\infty}\int_{(-A,A)^m}\widehat{ \chi}_P(y) \biggl(\int_{\mathbb{R}^m}e^{2\pi i\langle x,y\rangle}\,d \mu(x) \biggr)\,dy\\=& \lim_{A\to+\infty}\int_{(-A,A)^m} \widehat{\chi}_P(y)\widehat{\mu}(-y)\,dy. \end{aligned}
This relation shows that the values of the measure on the cells belonging to can be expressed in terms of its Fourier transform. Since the measures μ and ν have the same Fourier transform, they coincide on the semiring , and, consequently, (by the uniqueness theorem) on all Borel sets. □

It follows from the above theorem that the Fourier transform is injective up to equivalence on the set of summable functions.

### Corollary 1

If two summable functions f and g have the same Fourier transform, they coincide almost everywhere.

### Proof

It is clear that the Fourier transform of the functions $$\overline{f}$$ and $$\overline{g}$$ also coincide. Consequently, the functions $$\mathcal{R}e\,f=(f+\overline{f})/2$$ and $$\mathcal{R}e\,g=(g+\overline{g})/2$$, as well as the imaginary parts of the functions f and g, have the same Fourier transform. Therefore, we may assume that the functions f and g are real.

If they are non-negative, the theorem just proved implies that the measures with the densities f and g coincide. It was proved in Sect. 4.5.4 that, in this case, the densities coincide almost everywhere.

In the general case, we represent f and g in the form f=f +f and g=g +g , where f ±,g ±⩾0. Then
$$\widehat{f}=\widehat{f}_+-\widehat{f}_-=\widehat{g}=\widehat {g}_+- \widehat{g}_-.$$
Consequently, the non-negative functions f ++g and f +g + have the same Fourier transform, and, therefore, they coincide almost everywhere, which is equivalent to the assertion of the corollary. □

### Corollary 2

If finite Borel measures μ and ν have the same values on all half-spaces (in $$\mathbb{R}^{m}$$), then they coincide.

### Proof

By the theorem, it is sufficient to verify that $$\widehat{\mu}(y)=\widehat{\nu}(y)$$ for all $$y\in\mathbb{R}^{m}$$. For y=0, the equality holds since $$\widehat{\mu} (0)=\mu(\mathbb{R}^{m})$$ and $$\widehat{\nu}(0)=\nu(\mathbb{R}^{m})$$, and if two measures coincide on half-spaces, they coincide on the entire space. For y≠0, we consider the half-spaces
$$H_t=\bigl\{x\in\mathbb{R}^m\,|\,\langle x,y\rangle<t \bigr\}\quad(t\in\mathbb{R})$$
and put g(t)=μ(H t )=ν(H t ) and Φ(x)=〈x,y〉. The function g increases, and the Stieltjes measure μ g is the Φ-image of the measures μ and ν since Φ−1((−∞,t))=H t . It remains to use Theorem 6.1.1 on integration with respect to a weighted image of a measure,
\begin{aligned} \widehat{\mu}(y)=&\int_{\mathbb{R}^m}e^{-2\pi i\langle x,y\rangle}\,d\mu(x)= \int _{\mathbb{R}}e^{-2\pi it}\,dg(t)\\ =&\int_{\mathbb{R}^m}e^{-2\pi i\langle x,y\rangle} \,d\nu (x) =\widehat{\nu}(y). \end{aligned}
□

### 10.5.6

Using the results of the previous section, we will prove here that the system of Hermite polynomials is complete. The method we use enables us to consider a more general situation and prove that the family of monomials in m variables, i.e., the products $$x^{n}=x_{1}^{n_{1}}\cdots x_{m}^{n_{m}}$$, where $$x=(x_{1},\ldots,x_{m})\in\mathbb{R}^{m}$$ and $$n=(n_{1},\ldots, n_{m})\in\mathbb{Z}^{m}_{+}$$, is complete in for a wider class of measures.

### Theorem

If a Borel measure μ on $$\mathbb{R}^{m}$$ satisfies the condition $$\int_{\mathbb{R}^{m}}e^{a\|x\|}\,d\mu(x)<+\infty$$ for some a>0, then the family of all monomials is complete in the space .

### Proof

Let a function f in be orthogonal to all monomials. Obviously, fP for every polynomial P in m variables. We put
$$F(y)=\int_{\mathbb{R}^m}f(x) e^{i\langle y,x\rangle}\,d\mu(x).$$
Since |e iy,x|≡1 and all polynomials are summable with respect to μ, the function F is infinitely differentiable and, for each y, the derivatives of F can be found by the Leibnitz rule.
We prove that F≡0. If ∥y∥<a/2, then expanding the exponential factor in a Taylor series and integrating termwise, we obtain that F(y)=0. The legitimacy of termwise integration follows from the fact that the partial sums of the series have a summable majorant, namely, |f(x)|e x∥ ∥y (this function is summable because the functions |f| and e x∥ ∥y belong to ). To prove that F≡0, we show that the interior G of the set where F(y)=0 coincides with $$\mathbb{R}^{m}$$. Since G≠∅ (because it contains a neighborhood of zero), it is sufficient to verify that the set G is closed, in which case the equality $$G=\mathbb{R}^{m}$$ will follow from the fact that the space $$\mathbb{R}^{m}$$ is connected. Let $$y\in \overline{G}$$. The function F and all its derivatives vanish at y by continuity. Calculating the derivatives by Leibnitz’s rule, we see that
$$0=F^{(n)}(y)=\int_{\mathbb{R}^m}f(x) (ix)^ne^{i\langle x,y\rangle} \,d\mu(x) \quad\bigl(n\in\mathbb{R}^m_+\bigr).$$
Thus, the function f 1(x)=f(x)e ix,y is orthogonal to all monomials. Replacing f by f 1, we may assert by what has just been proved that the function $$F_{1}(\eta)=\int_{\mathbb{R}^{m}}f_{1}(x) e^{i\langle x,\eta\rangle}\,d\mu(x)$$ assumes only zero values in a neighborhood of zero. However, F 1(η) is nothing but F(y+η). Therefore, F≡0 in a neighborhood of y, i.e., yG. Thus, $$G=\overline{G}=\mathbb{R}^{m}$$ and, consequently, F≡0. Now, we can easily complete the proof. Indeed, without loss of generality, we may assume that the function f is real. The identity F≡0 means that the measures f + and f have the same Fourier transform. Consequently, these measures coincide by Theorem 10.5.5, which implies (by Theorem 4.5.4) that the functions f + and f coincide almost everywhere with respect to μ. □

### Corollary

The Hermite polynomials are complete in with $$d\mu(x)=e^{-x^{2}}\,dx$$.

This is a special case of the theorem for m=1. We also remark that the theorem implies that the Laguerre functions are complete (for the definition, see Exercise 3 of Sect. 10.2).

The following example shows that the result obtained in the theorem is quite sharp.

### Example

We verify that the polynomials are not complete in the space with measure μ having density $$e^{-|x|^{p}}$$ (0<p<1) with respect to the one-dimensional Lebesgue measure (for p⩾1 this effect is ruled out by the theorem just proved).

We will need the following formula from Example 1 of Sect. 7.1.7: if a>0 and z=e , where $$\theta\in(0,\frac{\pi}{2})$$, then $$z^{-a}\Gamma(a)=\int_{0}^{\infty} t^{a-1}e^{-z t}\,dt$$. Comparing the imaginary parts and using the substitution t=x p /cosθ, we obtain
\begin{aligned} \Gamma(a)\sin a\theta=&\int_0^{\infty} t^{a-1}e^{-t\cos\theta}\sin(t\sin \theta)\,dt\\=& \frac{p}{ \cos^a \theta}\int_0^{\infty} x^{ap-1}e^{-x^p}\sin\bigl(x^p\text{tan}\, \theta \bigr)\,dx. \end{aligned}
Now, we use the freedom in the choice of the parameters a and θ. Putting $$\theta=\frac{\pi}{2}p$$ and $$a=\frac{2}{p}(n+1)$$, we obtain
$$\int_0^{\infty} x^{2n+1}e^{-x^p} \sin \biggl(x^p\text{tan}\,\frac{\pi }{2}p \biggr)\,dx=0 \quad \text{for}\ n=0,1,2\ldots.$$
This means that the odd function equal to $$\sin (x^{p}\text{tan}\,\frac{\pi}{2}p )$$ for x⩾0 is orthogonal to all polynomials in the space with measure $$d\mu(x)=e^{- |x|^{p}}\,dx$$.

### 10.5.7

The present and two following sections are devoted to an important theorem, due to Plancherel,20 and its corollaries. The traditional formulation of the theorem would require us to invoke some concepts from functional analysis and operator theory. To avoid this, we first establish an analytic fact constituting the core of the theorem.

(Plancherel)

### Proof

Let {ω t } t>0 be a Sobolev approximate identity in $$\mathbb{R}^{m}$$ (see Sect. 7.6.2) and f t =fω t .

First, we prove the assertion of the theorem for the smoothened function f t . By properties of convolution, we have . By Theorem 10.5.1, we obtain $$\widehat{f}_{t}=\widehat{f}\widehat{\omega}_{t}$$. This product is summable since the function $$\widehat{f}$$ is bounded and by Corollary 10.5.2. Using Fubini’s theorem and inversion formula (5), we obtain
\begin{aligned} \int_{\mathbb{R}^m}\widehat{f}_t(y)\overline{ \widehat{f}_t(y)}\,dy=& \int_{\mathbb{R}^m} \widehat{f}_t(y)\overline{\biggl(\int_{\mathbb{R}^m}f_t(x) e^{-2\pi i\langle y,x\rangle}\,dx \biggr)}\,dy \\ =&\int_{\mathbb{R}^m}\widehat{f}_t(y) \biggl(\int _{\mathbb{R}^m}\overline{f_t(x)} e^{2\pi i\langle y,x\rangle}\,dx \biggr)\,dy \\ =&\int_{\mathbb{R}^m}\overline{f_t(x)} \biggl(\int _{\mathbb{R}^m}\widehat{f}_t(y) e^{2\pi i\langle y,x\rangle}\,dy \biggr)\,dx =\int_{\mathbb{R}^m}\overline{f_t(x)} \,f_t(x)\,dx. \end{aligned}
Thus,
$$\|\widehat{f}_t\|^2_2=\|f_t\|^2_2.$$
(8)
It remains to verify that we can pass to the limit in this equation as t→0.
Since $$f_{t}\underset{t\to0}{\longrightarrow} f$$ in the -norm, the continuity of the norm implies $$\|f_{t}\|_{2}\underset{t\to0}{\longrightarrow}\|f\|_{2}$$. We verify that and $$\|\widehat{f}_{t}\|_{2}\underset{t\to0}{\longrightarrow}\|\widehat{f}\| _{2}$$. To this end, we write the left-hand side of Eq. (8) in more detail,
$$\|\widehat{f}_t\|_2^2=\int_{\mathbb{R}^m}\big|\widehat{f}_t(y)\big|^2\,dy= \int _{\mathbb{R}^m}\big|\widehat{f}(y)\big|^2\big|\widehat{\omega}_t(y)\big|^2\,dy.$$
(9)
Since $$\widehat{\omega}_{t}(y)\underset{t\to0}{\longrightarrow}1$$ (see Corollary 7.6.3 with t 0=0 and g(x)=e −2πiy,x), Fatou’s theorem and Eq. (8) imply
\begin{aligned} \int_{\mathbb{R}^m}\big|\widehat{f}(y)\big|^2\,dy\leqslant&\lim _{t\to0}\int_{\mathbb{R}^m}\big| \widehat{f}(y)\big|^2\big| \widehat{\omega}_t(y)\big|^2\,dy=\lim_{t\to0} \|\widehat{f}_t\|_2^2= \lim _{t\to0}\|f_t\|_2^2=\|f \|_2^2\\ <&+\infty. \end{aligned}
Thus, . Returning to Eq. (9), we see that the integrand in the integral on the right has a summable majorant, namely, $$|\widehat{f}|^{2}$$. Therefore, we can pass to the limit in this integral by Lebesgue’s theorem,
$$\int_{\mathbb{R}^m}\big|\widehat{f}(y)\big|^2 \big|\widehat{\omega}_t(y)\big|^2\,dy \underset{t\to0}{\longrightarrow} \int_{\mathbb{R}^m}\big| \widehat{f}(y)\big|^2\,dy.$$
Now, the passage to the limit in Eq. (8) leads to the required result. □

The concluding part of the proof can be somewhat shortened. Indeed, since $$|\widehat{\omega}_{t}|\leqslant\int_{\mathbb{R}^{m}} \omega(x)\,dx=1$$, we have $$|\widehat{f}_{t}|\leqslant|\widehat{f}|$$. Since $$\widehat{f}_{t}\underset{t\to0}{\longrightarrow}\widehat{f}$$, we can pass to the limit on the right-hand side of Eq. (8) by the generalization of B. Levi’s theorem given in Exercise 4 of Sect. .

### 10.5.8

We show how Plancherel’s theorem can be used to generalize the concept of the Fourier transform to functions in .

### Lemma

Let . If {f n } n⩾1 is a sequence of functions in convergent to f in the -norm, then the sequence $$\{\widehat{f}_{n}\}_{n\geqslant1}$$ also converges in the -norm. Its limit does not depend (up to equivalence) on the choice of the sequence {f n } n⩾1.

### Proof

From Plancherel’s theorem, it follows that the sequence $$\{ \widehat{f}_{n}\}_{n\geqslant1}$$ is fundamental,
$$\|\widehat{f}_n-\widehat{f}_k\|_2=\| \widehat{f_n-f_k}\|_2=\|f_n-f_k \| _2\underset{n,k\to\infty}{\longrightarrow}0.$$
The limit exists because the space is complete (see Theorem 9.1.3). If {g n } n⩾1 is another sequence of functions in convergent to f in the -norm, then the sequence f 1,g 1,f 2,g 2,… obtained by “shuffling” the sequences {f n } n⩾1 and {g n } n⩾1 converges to f. By what has just been proved, the sequence $$\widehat{f}_{1},\widehat{g}_{1},\widehat{f}_{2}, \widehat{g}_{2},\ldots$$ has a limit, which is unique up to equivalence and coincides with the limits of its subsequences. □

The lemma just proved allows us to extend the definition of the Fourier transform to the functions in .

### Definition

By the Fourier transform of a function , we mean the limit in the -norm of the functions $$\widehat{f}_{n}$$, where {f n } n⩾1 is an arbitrary sequence of functions in such that $$\|f_{n}-f\|_{2}\underset{n\to\infty}{\longrightarrow}0$$.

Thus, the Fourier transform of a function in is also square-summable. As before, we will denote the Fourier transform of f by $$\widehat{f}$$. However, one must keep in mind that now the Fourier transform is defined up to equivalence and the symbol $$\widehat{f}$$ refers to many functions. If f is summable, then, among these functions, is the Fourier transform defined in Sect. 10.5.1. For definiteness, the latter is sometimes called the classical Fourier transform. What has just been said also applies to the inverse transform, which, as before, is denoted by .

Elementary properties of the Fourier transform of square-summable functions can be obtained from the properties of the classical Fourier transform by a passage to the limit.

### Theorem

Let . Then:
1. (1)

$$\|\widehat{f}\|_{2}=\|f\|_{2}$$;

2. (2)

if and $$\|f_{n}-f\|_{2} \underset{n\to\infty}{\longrightarrow}0$$, then $$\|\widehat{f}_{n}-\widehat{f}\|_{2} \underset{n\to\infty}{\longrightarrow}0$$, and a similar statement holds for the inverse transform;

3. (3)

4. (4)

$$\langle\widehat{f},\widehat{g}\rangle=\langle f,g\rangle$$ for every function . In particular, the Fourier transform preserves orthogonality: if fg, then $$\widehat{f}\perp\widehat{g}$$.

### Proof

Let $$\{\varphi _{n}\}_{n\geqslant1}\in C_{0}^{\infty}(\mathbb{R}^{m})$$ be a sequence of functions converging to f in the -norm. It is obvious that these functions and their Fourier transforms belong to .

(1) It is clear that $$\|\widehat{\varphi }_{n}-\widehat{f}\|_{2}\underset{n\to\infty}{\longrightarrow}0$$ by the definition of $$\widehat{f}$$. By Plancherel’s theorem, we have $$\|\widehat{\varphi }_{n}\|_{2}=\|\varphi _{n}\|_{2}$$. Therefore, it is sufficient for us to use the continuity of the norm and pass to the limit in this equation.

(2) Obviously, $$\|\widehat{f}_{n}-\widehat{f}\|_{2}=\|\widehat{f_{n}-f}\|_{2}=\| f_{n}-f\|_{2} \underset{n\to\infty}{\longrightarrow}0$$.

(3) We will prove only the equality (the other one is proved similarly). Since $$\varphi _{n}\underset{n\to\infty}{\longrightarrow} f$$, we obtain by definition that $$\widehat{\varphi }_{n}\underset{n\to\infty}{\longrightarrow}\widehat{f}$$, and, by property 2) applied to the inverse transform, we have . At the same time, by Theorem 10.5.4. Thus, it only remains to pass to the limit (in the -norm) in the last equality.

(4) For the proof, we must use the identity $$4f\overline{g}=|f+g|^{2}+|f+ig|^{2}- |f-g|^{2}-|f-ig|^{2}$$ and apply relation (1) to the functions f±g and f±ig. □

### 10.5.9

Plancherel’s theorem implies an inequality known as the uncertainty principle. Without touching on its physical meaning (the impossibility of simultaneously determining the exact values of the coordinates and impulse of a quantum object), we mention only its consequence: if f≠0 only in the vicinity of the origin, then the quantity $$|\widehat{f}|$$ is not small at some remote points (the Fourier transform “blurs”). In the one-dimensional case, the reader can see this effect in the example of functions $$\frac{1}{2t}\chi_{(-t,t)}$$ forming an approximate identity.

In the precise formulation of the uncertainty principle, we confine ourselves to infinitely differentiable compactly supported functions of one variable (more general statements are given in Exercises 10 and 11).

### Theorem

If $$f\in C_{0}^{\infty}(\mathbb{R})$$ andf2=1, then
$$\int_{-\infty}^{\infty} x^2\big|f(x)\big|^2 \,dx\cdot \int_{-\infty}^{\infty} x^2\big| \widehat{f}(x)\big|^2\,dx\geqslant\frac{1}{16\pi^2}.$$

### Proof

Since
$$\int_{-\infty}^{\infty} x \bigl(\big|f(x)\big|^2 \bigr)'\,dx=x\big|f(x)\big|^2 \Big\vert _{-\infty}^{\infty}-\int_{-\infty}^{\infty}\big|f(x)\big|^2 \,dx=-1,$$
the Cauchy–Bunyakovsky inequality implies
$$1=\biggl \vert \int_{-\infty}^{\infty} x \bigl(\big|f(x)\big|^2 \bigr)'\,dx\biggr \vert \leqslant 2\int _{-\infty}^{\infty}\big|xf(x)\big|\cdot\big|f'(x)\big|\,dx \leqslant2\|g\|_2\|f'\|_2,$$
where g(x)=|xf(x)|. By Plancherel’s theorem, we have $$\|f'\|_{2}=\|\widehat{f'}\|_{2}$$, and, by Theorem 10.5.2, we obtain $$\widehat{f'}(y)=2\pi iy\widehat{f}(y)$$. Consequently,
\begin{aligned} 1\leqslant4\|g\|_2^2\big\|f' \big\|_2^2=4\|g\|_2^2\big\| \widehat{f}'\big\|_2^2 =4\int _{-\infty}^{\infty} x^2\big|f(x)\big|^2\,dx \cdot 4\pi^2 \int_{-\infty}^{\infty} y^2\big|\widehat{f}(y)\big|^2\,dy.\quad \end{aligned}
□

### 10.5.10

In the conclusion of this section, we apply the Fourier transform to estimate the Dirichlet kernels for a ball (see Sect. 10.4.8)
$$D_R(x)=\frac{1}{(2\pi)^m}\sum_{\|k\|<R}e^{-i\langle k,x\rangle} \quad \bigl(x\in\mathbb{R}^m\bigr)$$
(the summation is taken over the points k of the integer lattice $$\mathbb{Z}^{m}$$). We show that, in the case where m>1, their -norms (in contrast to the norms of the Dirichlet kernels for cubes (−R,R) m ) have not a logarithmic, but a power order of growth as R→+∞,
$$\|D_R\|_1=\frac{1}{(2\pi)^m}\int_{[-\pi,\pi]^m} \bigg| \sum_{\|k\|<R}e^{-i\langle k,x\rangle} \bigg|\,dx\asymp R^{\frac{m-1}{2}}.$$
Being unable to represent the kernel D R in a compact form, we obtain for D R an approximate integral representation, replacing the sum over the ball B(R) by an integral over a set close to B(R). For this, we use the fact that the mean value of the exponential function e iat on the interval (a−1/2,a+1/2) differs from the function itself only by a factor independent of a,
$$e^{-iat}=\frac{t/2}{\sin\,t/2}\int_{a-1/2}^{a+1/2}e^{-ist} \,ds.$$
Therefore, in the multiple integral for the shifted unit cube $$Q_{k}=k+[-\frac{1}{2},\frac{1}{2}]^{m}$$ at the point x=(x 1,…,x m ), we have
$$e^{-i\langle k,x\rangle}=\theta(x)\int_{Q_k}e^{-i\langle y,x\rangle}\,dy, \quad\text{where}\ \theta(x)=\prod_{j=1}^m \frac{x_j/2}{\sin x_j/2}.$$
Putting T(R)=⋃k∥<R Q k , we arrive at the equation
$$D_R(x)=\frac{\theta(x)}{(2\pi)^m}\int_{T(R)}e^{-i\langle y,x\rangle}\,dy.$$
Thus,
$$\|D_R\|_1=\frac{1}{(2\pi)^m}\int_{[-\pi,\pi]^m} \theta(x) \bigg|\int_{T(R)}e^{-i\langle y,x\rangle}\,dy \bigg|\,dx.$$
Since $$1\leqslant\frac{t}{\sin t}\leqslant\frac{\pi}{2}$$ for $$t\in [-\frac{\pi}{2},\frac{\pi}{2}]$$, we obtain $$1\leqslant\theta(x)\leqslant (\frac{\pi}{2} )^{m}$$ in this integral, and, therefore,
$$\|D_R\|_1\asymp\int_{[-\pi,\pi]^m} \bigg|\int _{T(R)}e^{-i\langle y,x\rangle}\,dy \bigg|\,dx =(2\pi)^m\int _{[-\frac{1}{2},\frac{1}{2}]^m}\big|\widehat{\chi}_{T(R)}(u)\big|\,du.$$
We show that, for m>1, the integral on the right-hand side of this relation grows as $$R^{\frac{m-1}{2}}$$. It is more convenient to deal with the integral over a ball rather than over a cube. Therefore, we consider the integral
$$I_R(\rho)=\int_{B(\rho)}\big|\widehat{\chi}_{T(R)}(u)\big|\,du.$$
Since
$$I_R \biggl(\frac{1}{2} \biggr)\leqslant\int _{[-\frac{1}{2},\frac {1}{2}]^m}\big|\widehat{\chi}_{T(R)} (u)\big|\,du\leqslant I_R \biggl(\frac{\sqrt{m}}{2} \biggr),$$
it is sufficient to verify that $$I_{R}(\rho)\asymp R^{\frac{m-1}{2}}$$ as R→+∞ for every fixed ρ>0.
For large R, the set T(R) is close to the ball B(R). Therefore, it is natural to replace $$\widehat{\chi}_{T(R)}$$ with $$\widehat{\chi }_{B(R)}$$ and compare the integral I R (ρ) with a “similar” integral
$$J_R(\rho)=\int_{B(\rho)}\big|\widehat{\chi}_{B(R)}(u)\big|\,du.$$
The rate of its growth was essentially found in Example 2 of Sect. 10.5.2. Indeed, since
$$\widehat{\chi}_{B(R)}(u)=\int_{\|x\|<R}e^{-2\pi i\langle u,x\rangle}\,dx= R^m\int_{\|x\|<1}e^{-2\pi iR\langle u,x\rangle}\,dx=R^m \widehat{\chi}_B(Ru),$$
the integral J R (ρ) can be reduced to the integral $$L_{B}(R)=\int_{\|y\|<R}|\widehat{\chi}_{B}(y)|\,dy$$ considered in this Example,
$$J_R(\rho)=\int_{B(\rho)}\bigl \vert R^m\widehat{\chi}_B(Ru)\bigr \vert \,du= \int _{B(\rho R)}\bigl \vert \widehat{\chi}_B(y)\bigr \vert \,dy=L_B(\rho R) \asymp(\rho R)^{\frac{m-1}{2}}.$$
Therefore,
$$0<C_m(\rho\,R)^{\frac{m-1}{2}}\leqslant J_R(\rho) \leqslant C_m'(\rho R)^{\frac{m-1}{2}}.$$
(10)
To estimate the difference I R (ρ)−J R (ρ), we introduce the function η R =χ B(R)χ T(R). It is clear that
$$\big|I_R(\rho)-J_R(\rho)\big|\leqslant\int _{B(\rho)}\big|\widehat{\eta}_R(u)\big|\, du\leqslant \sqrt{ \alpha _m\rho^m\int_{B(\rho)}\big|\widehat{\eta}_R(u)\big|^2\,du}\leqslant \sqrt{\alpha _m \rho^m}\|\widehat{\eta}_R\|_2.$$
The next step is possible due to Plancherel’s theorem allowing us to pass from the norm $$\widehat{\eta}_{R}$$ to the norm η R , which can easily be estimated (since |η R |⩽1 and the function η R differs from zero only in the spherical layer $$R-\sqrt{m}\leqslant\|x\|\leqslant R+\sqrt{m}$$),
$$\|\widehat{\eta}_R\|_2=\|\eta_R \|_2\leqslant \sqrt{\alpha _m \bigl((R+\sqrt{m}\,)^m-(R-\sqrt{m}\,)^m \bigr)}.$$
Therefore, we obtain for R>1
$$\big|I_R(\rho)-J_R(\rho)\big|\leqslant \alpha _m \rho^{\frac{m}{2}}\sqrt{2m^{\frac{3}{2}}} (R+\sqrt{m})^{\frac{m-1}{2}}\leqslant A_m\rho^{\frac{m}{2}}R^{\frac{m-1}{2}},$$
(11)
where A m is a coefficient depending only on the dimension m. Taking into account inequality (10), we obtain the following estimate from above: $$I_{R}(\rho)=O (R^{\frac {m-1}{2}} )$$ as R→+∞.
Because the integrals I R (ρ) grow with the growth of ρ, it is sufficient to establish an estimate from below for small ρ. For this, we again use inequalities (10) and (11),
\begin{aligned} I_R(\rho)\geqslant& J_R(\rho)-\big|I_R( \rho)-J_R(\rho)\big|\geqslant C_m(\rho R)^{\frac{m-1}{2}}-A_m \rho^{\frac{m}{2}}R^{\frac{m-1}{2}}\\=& (C_m-A_m\sqrt{\rho}) (\rho R)^{\frac{m-1}{2}}. \end{aligned}
We obtain the required result if we take, for example, $$\rho=C_{m}^{2}/(2A_{m})^{2}$$.

### EXERCISES

1.

Find the Fourier transform of the product e 2πih,x f(x), where and $$h\in\mathbb{R}^{m}$$.

2.

Let and $$f_{E}(x)=\frac{1}{\lambda _{m}(E)}\int_{E}f(x+t)\,dt$$, where $$E\subset\mathbb{R}^{m}$$ is a set of finite positive measure. Prove that $$|\widehat{f}_{E}|\leqslant|\widehat{f}|$$.

3.
Let a function f in vanish outside the cube (−π,π) m , and let $$F(t)=\widehat{f}(t/2\pi)$$. Prove that
$$F(t)=\sum_{n\in\mathbb{Z}^m}F(n)\prod _{j=1}^m\frac{\sin\pi(t_j-n_j)}{\pi (t_j-n_j)}$$
(the sum on the right-hand side of the equation is understood as the limit of the partial sums over rectangles).
4.
A function f defined on $$\mathbb{R}^{m}$$ is called positive definite if
$$\sum_{1\leqslant j,k\leqslant n}f(x_j-x_k)z_j \overline{z}_k\geqslant0$$
for all $$n\in\mathbb{N}$$, $$x_{j}\in\mathbb{R}^{m}$$ and $$z_{j}\in\mathbb{C}$$. Prove that the Fourier transform of a finite Borel measure μ is a positive definite function.
5.

Prove that if g is the generalized derivative of f with respect to the kth coordinate (see Exercise 5 of Sect. ), then the relation $$\widehat {g}(y)=2\pi iy_{k}\widehat{f}(y)$$ of statement (1) of Theorem 10.5.2 remains valid.

6.

Verify that, in general, the equiconvergence of the expansions in Fourier series and Fourier integrals does not take place in the multi-dimensional case. Hint. Use the same idea as in the first part of Sect. 10.4.6.

7.

Let a function be bounded in a neighborhood of zero. Prove that if $$\widehat{f} \geqslant0$$, then . Hint. By Eq. (7), prove that the integrals $$\int_{\mathbb{R}^{m}}e^{-\pi t^{2}\|u\|^{2}}\widehat{f}(u)\,du$$ are bounded for t>0 and apply Fatou’s theorem.

8.
Let a measure μ on $$\mathbb{R}^{m}$$ be such that $$\int_{\mathbb{R}^{m}}e^{a\|x\|}\,d\mu(x)<+\infty$$ for some a>0. Generalizing Theorem 10.5.6, prove that, for p>1, every function satisfying the condition
$$\int_{\mathbb{R}^m}f(x)x^n\,d\mu(x)=0\quad\text{for all}\ n\in \mathbb{Z}^m_+$$
is equal to zero μ-almost everywhere.
9.

Let φ 1,φ 2,… be functions in . Prove that the systems $$\{\varphi _{n}\}_{n\in\mathbb{N}}$$ and $$\{\widehat{\varphi }_{n}\} _{n\in\mathbb{N}}$$ are complete or not simultaneously.

10.
Let and ∥f2=1. Prove that the inequality
$$\int_{\mathbb{R}^m}\|x-a\|^2 \big|f(x)\big|^2\,dx \cdot \int_{\mathbb{R}^m}\|y-b\|^2\big|\widehat{f}(y)\big|^2 \,dy\geqslant\frac{m^2}{16\pi^2}$$
is valid for all $$a,b\in\mathbb{R}^{m}$$.
11.
Assume that the values of a function are small outside a ball B(a,r) in the sense that
$$\int_{\mathbb{R}^m\setminus B(a,r)}\|x-a\|^2\big|f(x)\big|^2\,dx< \frac{1}{2} \int_{\mathbb{R}^m}\|x-a\|^2\big|f(x)\big|^2\,dx,$$
and the values of $$\widehat{f}$$ are small (in the same sense) outside a ball B(b,R). Prove that $$rR>\frac{m}{8\pi}$$.
12.

Supplement the result of Example 3 of Sect. 10.5.2 by proving that if $$f(x)\underset{\|x\|\to+\infty}{\sim}\|x\|^{-p-m}$$, as ∥x∥→+∞ for some p∈(0,2) and the function f is even, then $$\widehat{f}(0)-\widehat{f}(y)\underset{y\to0}{\sim} C_{p}\|y\|^{p}$$; for p=2, this relation must be replaced by $$\widehat{f}(0)-\widehat{f}(y)\underset{y \to0}{\sim} C\|y\|^{2}\ln\frac{1}{\|y\|}$$. The coefficients C p and C depending on the dimension can be expressed in terms of the gamma function.

13.

Let P, Q be algebraic polynomials with deg Q>deg P. Show that the Fourier transform of the fraction $$f=\frac{P}{Q}$$ vanishes on the negative half-axis ($$\widehat{f}(y)=0$$ for y⩽0) if all roots of the denominator Q lie in the lower half-plane (i.e., their imaginary parts are negative).

## 10.6 ⋆The Poisson Summation Formula

In this section, by the periodicity of a function of several variables, we mean 1-periodicity with respect to each variable. Speaking of the Fourier series of a periodic function summable on the cube $$(-\frac{1}{2},\frac{1}{2} )^{m}$$, we mean a series in the system of exponential functions $$\{e^{2\pi i\langle n,x\rangle} \}_{n\in\mathbb{Z}^{m}}$$.

### 10.6.1

As we have already seen, the properties of the Fourier transform of a summable function f can be far from the properties of f. A smooth function can have a non-smooth Fourier transform, which can be non-summable, and the order of decrease of $$\widehat{f}$$ can be distinct from that of f, etc. However, remarkably it turns out that, under quite mild assumptions, there is a characteristic that does not change when passing from f to $$\widehat{f}$$. Confining ourselves to the functions of one variable and not touching now the problem of convergence of the series in question, we can say that the required characteristic is the sum of the values of a function at the integer points. In other words, we are talking about the relation
$$\sum_{n=-\infty}^{\infty} f(n)=\sum _{k=-\infty}^{\infty}\widehat{f}(k),$$
(1)
known as the Poisson summation formula. Generalizing Eq. (1) somewhat, we can represent it in the form
$$\sqrt{t}\sum_{n=-\infty}^{\infty} f(tn)=\sqrt{s} \sum _{k=-\infty}^{\infty}\widehat{f}(sk),$$
(1′)
where st=1 (s>0,t>0). Thus, the sum of the values of f at the equidistant lattice points tn (t>0) is proportional to a similar sum for $$\widehat{f}$$, provided that the values of $$\widehat{f}$$ are calculated at compatible lattice points. Equation (1′) can be obtained by applying Eq. (1) to the function g obtained from f by the similarity (g(x)=f(tx)).
Our goal is to justify Poisson’s formula and give examples of its application. As often happens, to solve a problem, one must generalize it. We will study not the sum $$\sum_{n=-\infty}^{\infty} f(n)$$ itself, but the function S defined by the formula
$$S(x)=\sum_{n=-\infty}^{\infty} f(x+n) \quad(x\in \mathbb{R}).$$
(2)
Thus, to study a non-periodic function, we assign to it a periodic function and study the properties of the latter, using, in particular, the machinery of Fourier series developed in Sects. 10.3 and 10.4.

We need the following statement.

### Lemma

Let a function f be summable on $$\mathbb{R}$$. Then:
1. (a)

series (2) converges absolutely for almost all x;

2. (b)

its sum S is a 1-periodic function; this function is summable on $$(-\frac{1}{2},\frac{1}{2})$$, and Eq. (2) can be integrated termwise;

3. (c)
the Fourier coefficients of the function S are equal to the values of $$\widehat{f}$$ at the integer points,
$$\int_{-\frac{1}{2}}^{\frac{1}{2}} S(x) e^{-2\pi inx}\,dx=\widehat {f}(n)\quad (n\in\mathbb{Z}).$$

### Proof

We consider the non-negative measurable function
$$F(x)=\sum_{n=-\infty}^{\infty}\big|f(x+n)\big|\quad(x\in \mathbb{R}).$$
Since a positive series can be integrated termwise, we have
$$\int_{-\frac{1}{2}}^{\frac{1}{2}}F(x)\,dx=\sum _{n=-\infty}^{\infty} \int_{-\frac{1}{2}}^{\frac{1}{2}}\big|f(x+n)\big| \,dx=\int_{\mathbb{R}}\big|f(x)\big|\,dx$$
(the second equation is valid since the integral is countably additive). Therefore, the function F is summable on $$(-\frac{1}{2},\frac{1}{2})$$ and, therefore, F is almost everywhere finite, which proves the fact that the series $$\sum_{n=-\infty}^{\infty} |f(x+n)|$$ converges almost everywhere and so statement (a) holds.

Since the 1-periodicity of the function S is obvious, statement (b) follows from the inequality |S|⩽F. Since F also dominates all partial sums of series (2), the series can be integrated termwise.

To complete the proof, we multiply Eq. (2) by e −2πikx and integrate termwise over $$(-\frac{1}{2},\frac{1}{2})$$,
\begin{aligned} \int_{-\frac{1}{2}}^{\frac{1}{2}} S(x) e^{-2\pi ikx}\,dx=& \sum _{n=-\infty}^{\infty}\int_{-\frac{1}{2}}^{\frac{1}{2}}f(x+n) e^{-2\pi ikx}\,dx\\=&\int_{\mathbb{R}} f(x) e^{-2\pi ikx} \,dx=\widehat{f}(k). \end{aligned}
The termwise integration here is allowed since F is also a summable majorant of the partial sums of the series $$\sum_{n=-\infty}^{\infty} f(x+n) e^{-2\pi ikx}$$. □
As established in the lemma, the series $$\sum_{k=-\infty}^{\infty}\widehat {f}(k) e^{2\pi ikx}$$ is the Fourier series of the function S. Therefore, the relation
$$\sum_{n=-\infty}^{\infty} f(x+n)=\sum _{k=-\infty}^{\infty} \widehat{f}(k) e^{2\pi ikx}$$
(3)
(also called the Poisson summation formula) means simply that, at a point x, the function S is the sum of its Fourier series. In particular, if S is continuous at x, then Eq. (3) holds only under the assumption that the series on the right-hand side converges.

### Example 1

Let $$f(x)=e^{-\pi(tx)^{2}}$$ (where t is a positive parameter). As established in Example 2 of Sect. 10.5.1, $$\widehat{f}(y)=\frac{1}{t} e^{-\pi(y/t)^{2}}$$. Obviously, the sum $$\sum_{n=-\infty}^{\infty} e^{-\pi t^{2}(x+n)^{2}}$$ is a smooth function. Therefore, formula (3) is valid for f everywhere,
$$\sum_{n=-\infty}^{\infty} e^{-\pi t^2(x+n)^2}= \frac{1}{t}\sum_{n=-\infty}^{\infty} e^{-\pi(n/t)^2} e^{2\pi inx}= \frac{1}{t} \Biggl(1+2\sum_{n=1}^{\infty} e^{-\pi(n/t)^2}\cos2\pi nx \Biggr).$$
For x=0, the left-hand side of this equation is the so-called θ-function
$$\theta(t)=\sum_{n=-\infty}^{\infty} e^{-\pi(tn)^2}.$$
From this equation, it follows that the θ-function satisfies the Jacobi identity $$\theta(t)=\frac{1}{t}\theta(\frac{1}{t})$$, which plays an important role in heat transfer theory and in the theory of elliptic functions.

### Example 2

Let g(x)=(1−|x|)+ for $$x\in\mathbb{R}$$. Obviously,
$$\widehat{g}(y)=\int_{-1}^1\bigl(1-|x|\bigr) e^{-2\pi ixy}\,dx= 2\int_0^1(1-x)\cos2\pi xy\,dx= \biggl(\frac{\sin\pi y}{\pi y} \biggr)^2.$$
We apply formula (3) to the function $$f=\widehat{g}$$. It can easily be verified that the sum $$S(x)=\sum_{n=-\infty}^{\infty} f(x+n)$$ is everywhere continuous and , and so, the inversion formula implies $$\widehat{f}=g$$. Therefore, by (3), we obtain
$$\sum_{n=-\infty}^{\infty} \biggl( \frac{\sin\pi(x+n)}{\pi(x+n)} \biggr)^2= \sum_{n=-\infty}^{\infty} f(x+n)=\sum_{k=-\infty}^{\infty} \widehat{f}(k) e^{2\pi ikx}=\widehat{f}(0)=1.$$
Since sin2 π(x+n)=sin2 πx for all $$n\in\mathbb{Z}$$, we obtain the following partial fraction expansion of $$\frac{1}{\sin^{2}\pi x}$$:
$$\frac{\pi^2}{\sin^2\pi x}=\sum_{n=-\infty}^{\infty} \frac{1}{(x+n)^2}\quad (x\in\mathbb{R}\setminus\mathbb{Z}).$$
For $$x=\frac{1}{2}$$, this leads to the equality $$\pi^{2}=4\sum_{n=-\infty}^{\infty}\frac{1}{(2n+1)^{2}}= 8\sum_{n=0}^{\infty}\frac{1}{(2n+1)^{2}}$$ from which the well-known result $$\frac{\pi^{2}}{6}=\sum_{n=1}^{\infty}\frac{1}{n^{2}}$$ follows easily. Differentiating the expansion obtained termwise an even number of times, one can calculate the sums of the series $$\sum_{n=1}^{\infty}\frac {1}{n^{2k}}$$ $$(k\in\mathbb{N})$$ first found by Euler.

### Example 3

Let a>1 and u>0. Let f(x)=x a−1 e ux for x>0 and f(x)=0 for x⩽0. This function is continuous everywhere and the series $$S(x)= \sum_{n=-\infty}^{\infty} f(x+n)$$ converges uniformly on every finite interval. Since $$\widehat{f}(y)=\frac{\Gamma(a)}{(u+2\pi iy)^{a}}$$ (see Example 5 of Sect. 10.5.1), we see that the Fourier series of S converges absolutely. Consequently, Eq. (3) is valid everywhere. In particular, it takes the following form for x=0:
$$\sum_{n=1}^{\infty} n^{a-1}e^{-nu}= \sum_{n=-\infty}^{\infty}\frac{\Gamma(a)}{(u+2\pi in)^a}.$$
Hence, we see that the sum on the right-hand side of the equation is exponentially small as u→+∞. For u=1, the sum on the left-hand side can be regarded as a discrete analog of the integral $$\Gamma(a)=\int_{0}^{\infty} t^{a-1}e^{-t}\,dt$$. The formula obtained yields the following interesting relation: $$\sum_{n=1}^{\infty} n^{a-1}e^{-n}=\Gamma(a)\sum_{n=-\infty}^{\infty} (1+2\pi in)^{- a}$$.

Without any assumptions on the function f (except the summability), Eq. (3) is valid in the following “weak” sense: after integrating both sides of the equation over an arbitrary interval, we obtain convergent series with equal sums. This follows immediately from our ability to integrate Fourier series termwise (Theorem 1 of Sect. 10.3.6).

We also remark that not only does (1) follow from (3), but also (3) can be regarded as Eq. (1) for the shift f x of the function f since $$\widehat{f}_{-x}(n)=\widehat{f}(n) e^{2\pi inx}$$ (see the beginning of Sect. 10.5.1).

To derive Eq. (1) from (3), we must be sure that the latter is valid for x=0. For this, it is not sufficient, for example, that series (3) converges almost everywhere. Therefore, of particular interest to us is to find conditions under which the function S has a Fourier series expansion everywhere. One such conditions is given in Exercise 2. Other versions of sufficient conditions for functions of several variables will be established in the next section.

### 10.6.2

Here, we discuss the following multi-dimensional version of the Poisson summation formula for a function f in :
$$\sum_{n\in\mathbb{Z}^m}f(n)=\sum_{k\in\mathbb{Z}^m} \widehat{f}(k),$$
(4)
or, in a more general form,
$$\sum_{n\in\mathbb{Z}^m}f(x+n)=\sum_{k\in\mathbb{Z}^m} \widehat{f}(k) e^{2\pi i\langle k,x\rangle}.$$
(5)
Their derivation is based on an obvious modification of the lemma of Sect. 10.6.1, in which the one-dimensional lattice $$\mathbb{Z}$$ is replaced by the multi-dimensional lattice and the interval of integration $$(-\frac{1}{2},\frac{1}{2})$$ is replaced by the cube $$(-\frac {1}{2},\frac{1}{2})^{m}$$. Thus, the series on the right-hand side of (5) is the Fourier series of the function $$S(x)=\sum_{n\in\mathbb{Z}^{m}}f(x+n)$$. Therefore, as in the one-dimensional case, Eq. (5) for the continuous function S means that S has a Fourier series expansion.
Formula (4) can be modified by a linear change of variables (cf. Eq. (1′)),
$$\sqrt{|\text{det}(T)|}\sum_{n\in\mathbb{Z}^m}f\bigl(T(n)\bigr)= \sqrt{|\text{det}(S)|} \sum_{n\in\mathbb{Z}^m}\widehat{f} \bigl(S(n)\bigr),$$
where T is an arbitrary non-degenerate linear transformation, S=(T )−1 (here T is the adjoint mapping).

We give two types of conditions under which Eq. (5) is valid.

### Theorem 1

Let f be a continuous function on $$\mathbb {R}^{m}$$ such that f(x)=O(∥xp ) and $$\widehat{f}(x)=O(\|x\|^{-p})$$ asx∥→+∞ for some p>m. Then Eq. (5) is valid for all $$x\in \mathbb{R}^{m}$$.

### Proof

First, we observe that since the summability on an arbitrary ball follows from the continuity of f and the summability outside the ball follows from the estimate f(x)=O(∥xp ).

Under our assumptions, the series on both sides of Eq. (5) converge absolutely and uniformly on every ball, and, since their terms are continuous, the sums of the series are also continuous. Thus, the right-hand side of (5) is a uniformly convergent Fourier series of the sum on the left-hand side of (5). □

It is not difficult to give an interpretation of the sum of a multiple series in the case of absolute convergence. Otherwise, it is necessary to clarify the definition of this sum. First of all, this concerns the series on the right-hand side of Eq. (5) (the series on the left-hand side of (5) converges absolutely for almost all x). The following theorem enables us to consider the situation in which the series on the right-hand side of (5) does not converge absolutely (see also Exercises 3 and 4).

### Theorem 2

Let a function f satisfy the Lipschitz condition with exponent α in $$\mathbb{R}^{m}$$, i.e., there is a positive L such that |f(x)−f(x′)|⩽Lxx′∥ α for all $$x,x'\in\mathbb{R}^{m}$$. We assume, in addition, that the function decreases rapidly at infinity, i.e., f(x)=O(∥xp ) asx∥→+∞ for some p>m. Then Eq. (5) is valid for all $$x\in \mathbb{R}^{m}$$ (the sum on the right-hand side of (5) is understood as the limit of rectangular partial sums).

### Proof

As in the previous theorem, . It is sufficient to verify that the function S satisfies the Lipschitz condition with an exponent β (in this case, the rectangular partial sums converge uniformly to S by Theorem 10.4.5). Estimating the difference S(x+h)−S(x), we will assume that x∈[0,1] m and ∥h∥⩽1. It is clear that
$$\big|S(x+h)-S(x)\big|\leqslant\sum_{n\in\mathbb{Z}^m}\big|f(x+h+n)-f(x+n)\big|.$$
(6)
Now, fixing a large parameter R (its choice will be specified later), we partition the terms of the series obtained into two sets depending on whether ∥n∥⩽R or ∥n∥>R. Using the Lipschitz condition |f(x+h+n)−f(x+n)|⩽Lh α , we estimate the terms of the first set (the number of them has order R m ). In the second case, we apply the following estimate for f at infinity:
$$\big|f(x+h+n)-f(x+n)\big|\leqslant\big|f(x+h+n)\big|+\big|f(x+n)\big|=O \bigl(\|n\|^{-p} \bigr).$$
Substituting these estimates into inequality (6), we obtain
$$\big|S(x+h)-S(x)\big|\leqslant\mathrm{const} \biggl(R^m\|h \|^{\alpha }+\sum_{\|n\| >R}\|n\|^{-p} \biggr) =O \bigl(R^m\|h\|^{\alpha }+R^{m-p} \bigr).$$
Now, we use the freedom in the choice of R and equate the terms R m h α and R mp and, for R=∥hα/p , we obtain that |S(x+h)−S(x)|=O(∥h β ) with β=α(1−m/p). □

### Corollary

If a function f having bounded first order derivatives satisfies the condition f(x)=O(∥xp ) asx∥→+∞ for some p>m, then Eq. (5) holds for every point x (the sum on the right-hand side of the equation is understood as the limit over rectangular partial sums).

### 10.6.3

The Poisson summation formula has proved to be an effective tool for solving various problems (see Sects. 10.6.4 and 10.6.5). However, before passing to these technically more involved applications, we use the summation formula to supplement the uncertainty principle established in Sect. 10.5.9, according to which the functions f and $$\widehat{f}$$ cannot be concentrated on “small sets” simultaneously (see also Exercises 10 and 11 of Sect. 10.5). The following statement is valid (see [B]).

### Theorem

Let a summable function f on $$\mathbb{R}^{m}$$ be such that the sets $$A = \{x \in\mathbb{R}^{m}|f(x)\ne0\}$$ and $$B=\{y\in\mathbb {R}^{m}|\,\widehat{f}(y)\ne0\}$$ have finite measures, then f(x)=0 almost everywhere.

### Proof

We will assume that λ m (A)<1 (this can be achieved by the change of variables xcx). Let $$Q=[-\frac{1}{2},\frac{1}{2})^{m}$$. Since
$$\sum_{k\in\mathbb{Z}^m}\int_Q \chi_B(y+k)\,dy= \int_{\mathbb{R}^m}\chi_B(y) \,dy=\lambda _m(B)<+\infty,$$
the series $$\sum_{k\in\mathbb{Z}^{m}}\chi_{B}(y+k)$$ converges for almost all y. Consequently, χ B (y+k)≠0, i.e., $$\widehat{f}(y+k)\ne0$$ only for a finite number of multi-indices k. Therefore, for almost all y, the function f y (x)=f(x)e −2πiy,x is such that only a finite number of the values $$\widehat{f_{y}}(k)$$ ($$k\in\mathbb{Z}^{m}$$) are distinct from zero.

Since the kth Fourier coefficient of the function S y is equal to $$\widehat{f_{y}} (k)$$ (see statement (c)) of Lemma 10.6.1), we obtain that S y coincides with a trigonometric polynomial almost everywhere. The set E y ={xQ | S y (x)≠0} is contained in the union ⋃ n (−n+A)∩Q, and, therefore, λ m (E y )⩽λ m (A)<1=λ m (Q). Since a non-zero trigonometric polynomial does not vanish almost everywhere, we obtain that S y (x)=0 almost everywhere. Consequently, $$0=\widehat{S_{y}}(0)=\widehat{f_{y}}(0)=\int_{\mathbb{R}^{m}}f_{y}(x)\, dx=\widehat{f}(y)$$ for almost all y. By the uniqueness theorem, f=0 almost everywhere. □

We remark that the proof of the theorem does not use Eq. (5). It is based only on statement (c) of Lemma 10.6.1 (more precisely, on its m-dimensional modification).

### 10.6.4

Generalizing the reasoning of Sect. 10.4.3 to multiple Fourier series, we consider a summation method generated by a function M (M(0)=1) continuous and summable on $$\mathbb{R}^{m}$$. The method is as follows: for each ε>0 and each 1-periodic function f summable on the cube $$Q=[-\frac{1}{2},\frac{1}{2}) ^{m}$$, we consider the sum
$$S_{M,\varepsilon }(f,x)=\sum_{n\in\mathbb{Z}^m}M(\varepsilon n) \widehat{f}(n) e^{2\pi i\langle n,x \rangle}$$
and study its limit as ε→0.
To simplify the exposition, we will assume that M tends to zero at infinity so fast that
$$\sum_{n\in\mathbb{Z}^m}\big|M(\varepsilon n)\big|<+\infty\quad\text{for every}\ \varepsilon >0$$
(this condition is necessarily fulfilled if M is a compactly supported function).
It is clear that S M,ε (f)=fω ε , where
$$\omega_{\varepsilon }(x)=\sum_{n\in\mathbb{Z}^m}M(\varepsilon n) e^{2\pi i\langle n,x\rangle}.$$
If it turns out that the functions ω ε form an approximate identity as ε→0, then our problem simplifies considerably: it will be possible to use general theorems 7.6.5 and 9.3.7 in the study of the sums S M,ε (f,x).

When is the family {ω ε } ε>0 an approximate identity as ε→0? In Sect. 10.4.3, we obtained a sufficient condition. Now, using the Poisson summation formula, we can a give much more complete answer to this question. It turns out that it is sufficient that the Fourier transform $$\widehat{M}$$ be summable (as follows from the result of Exercise 6, this condition is also necessary).

We verify the conditions characterizing a periodic approximate identity (see Sect. 7.6.5). The equality $$\int_{\mathbb{R}^{m}}\omega_{\varepsilon }(x)\, dx=1$$ is valid because the series defining the function ω ε converges absolutely, and we can integrate it termwise over the cube Q,
$$\int_Q\omega_{\varepsilon }(x)\,dx=\sum _{n\in\mathbb{Z}^m}M(\varepsilon n)\int_Q e^{2\pi i\langle n,x \rangle} =M(0)=1.$$
We will show below that the functions ω ε are non-negative if $$\widehat{M}\geqslant0$$. However, regardless of this condition, the integrals ∫ Q |ω ε (x)| dx are bounded. To verify this, we put $$N(x)=\widehat{M}(-x)$$ and N ε (x)=ε m N(x/ε). Then the inversion formula for the Fourier transform (see Theorem 10.5.4) implies
$$M(\varepsilon n)=\int_{\mathbb{R}^m}\widehat{M}(y) e^{2\pi i\langle \varepsilon n,y\rangle}\,dy= \varepsilon ^{-m}\int_{\mathbb{R}^m}N \biggl(-\frac{u}{\varepsilon }\biggr) e^{2\pi i\langle n,u\rangle}\,du =\widehat{N_{\varepsilon }}(n),$$
and, by the Poisson summation formula, we obtain
$$\omega_{\varepsilon }(x)=\sum_{n\in\mathbb{Z}^m} \widehat{N}_{\varepsilon }(n) e^{2\pi i\langle n,x\rangle} = \sum _{n\in\mathbb{Z}^m}N_{\varepsilon }(x+n).$$
(7)
Therefore, ω ε ⩾0 if N⩾0, i.e., $$\widehat{M}\geqslant0$$, and
$$\int_Q\big|\omega_{\varepsilon }(x)\big|\,dx\leqslant\int _Q\sum_{n\in\mathbb{Z}^m} \big| N_{\varepsilon }(x+n) \big|\,dx= \int_{\mathbb{R}^m}\big|N_{\varepsilon }(x)\big| \,dx=\|N\|_1=\|\widehat{M}\|_1$$
in the general case. Thus, the -norm (on the cube Q) of each function ω ε does not exceed the -norm (on $$\mathbb{R}^{m}$$) of the Fourier transform $$\widehat{M}$$.
By refining this argument a little, we can establish a localization property. Indeed, for every $$\delta \in (0,\frac{1}{2} )$$, we have
$$\int_{Q\setminus B(\delta )}\big|\omega_{\varepsilon }(x)\big|\,dx\leqslant\int _{\|x\|\geqslant \delta }\big|N_{\varepsilon } (x)\big|\,dx= \int_{\|y\|\geqslant \delta /\varepsilon }\big| \widehat{M}(y)\big|\,dy\underset{\varepsilon \to0}{\longrightarrow}0.$$
Thus, {ω ε } ε>0 is a periodic approximate identity, and, consequently, statements (a) and (b) of Theorem 7.6.5 are valid for it.

### Example

The function M(u)=e −∥u generates the “radial” Abel–Poisson summation method for multiple Fourier series, where, to each 1-periodic function f summable on the cube Q, we assign the sums
$$S_{\varepsilon }(f,x)=\sum_{n\in\mathbb{Z}^m}e^{-\varepsilon \|n\|} \widehat{f}(n) e^{2\pi i\langle n,x \rangle}.$$
The Fourier transform of M was calculated in Example 4 of Sect. 10.5.1, $$\widehat{M}(y)=C_{m}(1+4\pi^{2}\|y\|^{2})^{-\frac{m+1}{2}}$$. It, obviously, is summable. The corresponding kernel has the form (see formula (7))
$$\omega_{\varepsilon }(x)=\sum_{n\in\mathbb{Z}^m} e^{-\varepsilon \|n\|} e^{2\pi i\langle n,x\rangle} =\sum_{n\in\mathbb{Z}^m} \frac{C_m\varepsilon }{(\varepsilon ^2+4\pi^2\|n+x\|^2)^{\frac{m+1}{2}}}.$$
As follows from the result of Exercise 7, not only does this kernel have the strong localization property, but it is also dominated by a summable “hump-shaped” majorant (see Sect. 9.3.4). Therefore, the theorems of Sects. 7.6.5 and 9.3.7 can be applied to the sums S ε (f)=fω ε . Consequently, $$S_{\varepsilon }(f)\underset{\varepsilon \to0}{\rightrightarrows} f$$ if the 1-periodic function f is continuous, and $$\int_{Q}|S_{\varepsilon }(f,x)-f(x)|^{p} \,dx\underset{\varepsilon \to0}{\longrightarrow}0$$ if , where p⩾1. Moreover, $$S_{\varepsilon }(f,x)\underset{\varepsilon \to0}{\longrightarrow} L$$ if $$f(x+t)\underset{t\to0}{\longrightarrow} L$$, and $$S_{\varepsilon }(f,x)\underset{\varepsilon \to0}{\longrightarrow} f(x)$$ almost everywhere.

### 10.6.5

An interesting application of the Poisson identity is in one of the solutions to the Gauss problem on determining the number N m (R) of points of the integer lattice $$\mathbb{Z}^{m}$$ that lie in the closed ball of a large radius R. The number N m (R) is close to the volume of the ball, $$N_{m}(R)-\lambda _{m}(\overline{B}(R))=O(R^{m-1})$$ as R→+∞. This can easily be verified by considering the unit cubes centered at the lattice points, i.e., the cubes $$n+[-\frac{1}{2},\frac{1}{2}]^{m}$$, $$n \in\mathbb{Z}^{m}$$. Since, for $$n \in\overline{B}(R)$$, their union contains the ball $$B(R-\sqrt{m})$$ and is contained in the ball $$\overline{B}(R+\sqrt{m})$$, we see that the number N m (R), being equal to the volume of the union of these cubes, lies between $$\lambda _{m}(B(R-\sqrt{m}))$$ and $$\lambda _{m}(B(R+\sqrt{m}))$$. In other words, $$\alpha _{m}(R - \sqrt{m})^{m} \leqslant N_{m}(R) \leqslant \alpha _{m}(R + \sqrt{m})^{m}$$, where α m is the volume of the unit cube in $$\mathbb{R}^{m}$$.

Much greater efforts are needed to sharpen this elementary estimate. However, we first observe that the exponent θ in the relation N m (R)=α m R m +O(R θ ) cannot be less than m−2. Indeed, the function RN m (R) makes a jump at $$R^{2}\in\mathbb{N}$$, the value of which is equal to the number of lattice points on the sphere of radius R. Since the principal term α m R m of the asymptotic depends continuously on R, the exponent θ must be so large that the number of points on the sphere of radius R be dominated by a summand proportional to R θ . The number of lattice points in the spherical layer R<∥x∥<2R is of order R m . The number of spheres containing these points is at most 3R 2 (every such sphere is defined by the equation ∥x2=t with an integer parameter t lying between R 2 and (2R)2). At least one of the spheres contains at least const R m−2 points. Therefore, necessarily θm−2.

It is clear that (in what follows, $$n\in\mathbb{Z}^{m}$$)
$$N_m(R)=\sum_{\|n\|\leqslant R}1= \sum _{n\in\mathbb{Z}^m}\chi \biggl(\frac{n}{R} \biggr),$$
(8)
where χ is the characteristic function of the unit ball $$\overline{B}$$. To calculate the sum on the right-hand side of (8), we apply the Poisson summation formula. Unfortunately, this cannot be done directly because the function χ is discontinuous. Therefore, we smoothen it and estimate N m (R), applying the Poisson formula to smooth compactly supported functions χ + and χ approximating χ from above and from below. It is desired, of course, that the functions χ ± be as close as possible to χ.
To construct them, we take a non-negative function $$\psi\in C_{0}^{\infty} (\mathbb{R}^{m})$$ with the properties $$\operatorname {supp}(\psi)\subset\overline{B}$$ and $$\int_{\mathbb{R}^{m}}\psi(x)\,dx=1$$ and put ψ ε (x)=ε m ψ(x/ε), where ε is a small positive parameter the choice of which will be specified later. The infinitely differentiable function χ ε =χψ ε (see Corollary 7.5.4) is, obviously, equal to 1 in the ball $$\overline{B}(1-\varepsilon )$$ and to zero outside the ball $$\overline{B}(1+\varepsilon )$$. Consequently, the functions χ (x)=χ ε ((1+ε)x) and χ +(x)=χ ε ((1−ε)x) satisfy the inequality χ χχ +. Thus,
$$\sum_{n\in\mathbb{Z}^m}\chi_{\varepsilon } \biggl((1+\varepsilon )\frac{n}{R} \biggr)\leqslant N_m(R) \leqslant \sum _{n\in\mathbb{Z}^m}\chi_{\varepsilon } \biggl((1-\varepsilon )\frac{n}{R} \biggr),$$
i.e.,
$$S_{\varepsilon } \biggl(\frac{R}{1+\varepsilon } \biggr)\leqslant N_m(R) \leqslant S_{\varepsilon } \biggl(\frac{R}{1-\varepsilon } \biggr),$$
(9)
where $$S_{\varepsilon }(r)=\sum_{n\in\mathbb{Z}^{m}}\chi_{\varepsilon }(n/r)$$.
To apply the Poisson summation formula (4) to the function f(x)=χ ε (x/r), we observe that
\begin{aligned} &\widehat{f}(y)=r^m\widehat{\chi}_{\varepsilon }(ry)=r^m \widehat{\chi}(ry)\widehat{\psi}_{\varepsilon } (ry)= r^m \widehat{\chi}(ry)\widehat{\psi}(r\varepsilon y);\\&\quad\text{in particular,}\ \widehat{f}(0)= \alpha _mr^m. \end{aligned}
Since the function ψ belongs to the class $$C_{0}^{\infty}(\mathbb {R}^{m})$$, its Fourier transform (and, consequently, also $$\widehat{f}$$) rapidly decreases at infinity. Thus, the conditions of Theorem 1 of Sect. 10.6.2 are fulfilled, and we obtain
\begin{aligned} S_{\varepsilon }(r)=&\sum_{n\in\mathbb{Z}^m}\chi_{\varepsilon } \biggl(\frac{n}{r} \biggr)= \sum_{n\in\mathbb{Z}^m}f(n)=\sum _{n\in\mathbb{Z}^m}\widehat {f}(n)=r^m\sum _{n\in\mathbb{Z}^m}\widehat{\chi}(rn) \widehat{\psi}(r\varepsilon n) \\=&r^m\alpha _m+r^m\sum _{n\ne0}\widehat{\chi }(rn)\widehat{\psi} (r\varepsilon n). \end{aligned}
Therefore,
$$\big|S_{\varepsilon }(r)-\alpha _mr^m\big|\leqslant r^m \sum_{n\ne0}\big|\widehat{\chi}(rn)\big|\cdot\big| \widehat{\psi}(r \varepsilon n)\big|.$$
Using the estimates $$\widehat{\psi}(y)=O((1+\|y\|)^{-m})$$ and $$\widehat{\chi}(y)=O(\|y\|^{-\frac{m+1}{2}})$$ (see Example 2 of Sect. 10.5.2), we obtain
\begin{aligned} \big|S_{\varepsilon }(r)-\alpha _mr^m\big|\leqslant&\sum _{n\ne0}\frac{Cr^m}{\|rn\|^{\frac {m+1}{2}}(1+\|r\varepsilon n\|)^m}\\=&C\varepsilon ^{-\frac{m-1}{2}}\sum _{n\ne0}\frac{(r\varepsilon )^m}{\|r\varepsilon n\|^{\frac{m+1}{2}} (1+\|r\varepsilon n\|)^m}. \end{aligned}
Since the estimate $$\|x\|\leqslant\|r\varepsilon n\|+\frac{\sqrt{m}}{2}r\varepsilon \leqslant(1+\frac{\sqrt{m}}{2})\|r\varepsilon n\|$$ is valid for each point x in the cube of edge length and center at rεn, n≠0, we see that the last sum is dominated (with a coefficient depending only on the dimension) by the integral
$$\int_{\mathbb{R}^m}\frac{dx}{\|x\|^{\frac{m+1}{2}}(1+\|x\|)^m} =\alpha _m\int _0^{\infty}\frac{t^{\frac{m-3}{2}}\,dt}{(1+t)^m}<+\infty.$$
Therefore,
$$\big|S_{\varepsilon }(r)-\alpha _mr^m\big|=O\bigl(\varepsilon ^{-\frac{m-1}{2}} \bigr).$$
For $$r=\frac{R}{1\pm \varepsilon }$$ and $$0<\varepsilon <\frac{1}{2}$$, we have r m =R m (1+O(ε)), so $$|S_{\varepsilon } (\frac{R}{1\pm \varepsilon } )-\alpha _{m}R^{m}|=O(\varepsilon R^{m}+\varepsilon ^{-\frac{m-1}{2}})$$. Taking into account (9), we see that
$$\big|N_m(R)-\alpha _mR^m\big|=O\bigl(\varepsilon R^m+ \varepsilon ^{-\frac{m-1}{2}}\bigr).$$
The sum $$\varepsilon R^{m}+\varepsilon ^{-\frac{m-1}{2}}$$ has a minimal order of growth if $$\varepsilon R^{m}=\varepsilon ^{-\frac{m-1}{2}}$$, i.e., if $$\varepsilon =R^{-\frac{2m}{m+1}}$$. Choosing this value of ε, we arrive at the relation
$$N_m(R)=\alpha _mR^m+O\bigl(R^{\theta} \bigr)\quad\text{as}\ R\to+\infty$$
(10)
with $$\theta=m\frac{m-1}{m+1}<m-1$$.

### 10.6.6

What can be said about the exactness of formula (10)? Since $$\theta= m - 2+\frac{2}{m+1}$$, we see that, for large m, its error is close to the minimum possible value O(R m−2). As we know, for m>4, the best estimate is achieved, namely, relation (10) is valid with θ=m−2 (for m=4, this relation is valid for every θ>2). For m=3 the minimum value of the exponent θ is still unknown. As we have verified in the previous section, it does not exceed $$\frac{3}{2}$$ and is not less than 1. There is a conjecture stating that the exponent θ can be taken arbitrarily close to 1, but it has only been proved that θ⩽29/22. For the history of the problem, see the paper [CI] or the book [LK].

We consider the two-dimensional case in more detail. We have proved that, for m=2, formula (10) is valid with θ=2/3. In the study of the Gauss problem, this is the first non-trivial result, which was obtained by Sierpiński in 1906 and has been sharpened several times since then. More sophisticated methods made it possible to decrease the exponent θ to $$\frac{131}{208}$$, but it is still unknown whether the value of θ can be taken arbitrarily close to $$\frac{1}{2}$$. This bound cannot be lowered. As Hardy and Landau21 proved independently in 1915, the relation N 2(R)=πR 2+O(R θ ) holds only for $$\theta\geqslant\frac{1}{2}$$. We give the proof of this result, based on the paper [EF] (see also Exercise 9).

Let $$N(R)=N_{2}(R)=\text{card}\{n\in\mathbb{Z}^{2}\,|\,\|n\|\leqslant R\}$$ and Δ(R)=N(R)−πR 2. We will need the function $$f(z)=\sum_{k=-\infty}^{\infty} z^{k^{2}}$$ (|z|<1) tightly connected with the quantities N(R) and Δ(R). Indeed,
$$f^2(z)=\sum_{k=-\infty}^{\infty} z^{k^2}\sum_{j=-\infty}^{\infty} z^{j^2}=\sum_{(k,j)\in\mathbb{Z}^2} z^{k^2+j^2}=\sum _{k=0}^{\infty}\nu (m) z^m,$$
where ν(m) is equal to the number of points (k,j) lying on the circle of radius $$\sqrt{m}$$. Since ν(0)=1 and $$\nu(m)=N(\sqrt{m})-N(\sqrt{m-1})$$ for m⩾1, we obtain
$$f^2(z)=\sum_{m=0}^{\infty} N( \sqrt{m})z^m-\sum_{m=1}^{\infty} N( \sqrt{m-1})z^m=(1-z)\sum_{m=0}^{\infty} N(\sqrt{m})z^m.$$
Since $$N(\sqrt{m})=\pi m+\Delta(\sqrt{m})$$, we see that
$$f^2(z)=\frac{\pi z}{1-z}+(1-z)\sum_{m=0}^{\infty} \Delta(\sqrt{m})z^m.$$
(11)
The required inequality $$\theta\geqslant\frac{1}{2}$$ can be obtained by comparing estimates from above and from below for the integrals
$$I(r)=\int_{-\pi}^{\pi} \big|f^2 \bigl(re^{it}\bigr) \big|\,dt \quad \biggl(\frac{1}{2}<r<1 \biggr).$$
Let us estimate the integral from below. Since $$f(re^{it})=1+2\sum_{k=1}^{\infty} r^{k^{2}}e^{ik^{2}t}$$, Parseval’s identity implies
\begin{aligned} I(r) =& 2\pi \Biggl(1+4\sum_{k=1}^{\infty} r^{2k^2} \Biggr) \geqslant2\pi \sum_{k=0}^{\infty} r^{2k^2}\geqslant2\pi\sum_{k=0}^{\infty} \int_k^{k+1}r^{2t^2}\,dt\\ =& 2\pi\int _0^{\infty} r^{2t^2}\,dt= \frac{\pi^{\frac{3}{2}}}{\sqrt{2\ln\frac{1}{r}}}. \end{aligned}
Therefore,
$$I(r)\geqslant\frac{C_1}{\sqrt{1-r}}$$
(here and below, C 1,C 2,… are positive coefficients independent of r).
Now, we obtain an estimate from above. Applying the result established in Example 3 of Sect. 10.2.1 to the function φ(t)=f(re it ), we obtain that the inequality
$$I(r)\leqslant\frac{3\pi}{ \alpha }\int_{-\alpha }^{\alpha }\big|f^2 \bigl(re^{it}\bigr)\big|\,dt$$
is valid for α∈(0,π) (the choice of this parameter will be specified later). Denoting the sum on the right-hand side of Eq. (11) by S(z), we obtain
$$I(r)\leqslant\frac{3\pi}{ \alpha }\biggl(\int_{-\alpha }^{\alpha } \frac{\pi dt}{|1-re^{it}|}+ \int_{-\alpha }^{\alpha }\big|1-re^{it}\big| \big|S\bigl(re^{it}\bigr)\big|\,dt \biggr)=\frac{3\pi}{ \alpha }(I_1+I_2).$$
By the inequality
$$\big|1-re^{it}\big|=\sqrt{(1-r)^2+4r\sin^2\frac{t}{2}} \geqslant \frac{(1-r)+|\!\sin\frac{t}{2}|}{2}\geqslant\frac{(1-r)+|t|}{2\pi},$$
it is easy to verify that
$$I_1\leqslant C_2\big|\!\ln(1-r)\big|.$$
Since |1−re it |⩽(1−r)+r|t|⩽(1−r)+α, we obtain the following inequality for α⩾1−r:
\begin{aligned} I_2\leqslant&2\alpha \int_{-\alpha }^{\alpha } \big|S \bigl(re^{it}\bigr) \big|\,dt\leqslant 2\alpha \sqrt{2\alpha \int _{-\pi}^{\pi} \big|S\bigl(re^{it}\bigr) \big|^2\,dt}\\=& (2\alpha )^{\frac{3}{2}}\sqrt{2\pi\sum _{m=0}^{\infty}\Delta^2(\sqrt{m})r^{2m}}. \end{aligned}
Therefore, if Δ(R)=O(R θ ) as R→+∞ for some θ>0, then
\begin{aligned} I_2\leqslant &C_3\alpha ^{\frac{3}{2}}\sqrt{1+\sum _{m=1}^{\infty} m^{\theta} r^{2m}} \leqslant C_3\alpha ^{\frac{3}{2}}\sqrt{1+ \sum_{m=1}^{\infty} \int_m^{m+1}t^{\theta} r^{2(t-1)}dt} \\\leqslant&2C_3\alpha ^{\frac{3}{2}}\sqrt{1+\int _0^{\infty} t^{\theta} r^{2t}dt}= 2C_3\alpha ^{\frac{3}{2}}\sqrt{1+\frac{\Gamma(1+\theta)}{\ln^{1+\theta}\frac {1}{r^2}}} \leqslant C_4\frac{\alpha ^{\frac{3}{2}}}{(1-r)^{\frac{1+\theta}{2}}}. \end{aligned}
Thus, for α>1−r>0, we obtain the double inequality
$$\frac{C_1}{\sqrt{1-r}}\leqslant I(r)\leqslant\frac{C_5}{\alpha }\biggl(\big|\! \ln(1-r)\big|+ \frac{\alpha ^{\frac{3}{2}}}{(1-r)^{\frac{1+\theta}{2}}} \biggr),$$
and, consequently,
$$0<C_6\leqslant\frac{1}{\alpha }\sqrt{1-r}\big|\!\ln(1-r)\big|+ \frac{\sqrt{\alpha }}{(1-r)^{\frac{\theta}{2}}}.$$
Now, using the freedom in the choice of the parameter α, we decrease the right-hand side, which is minimal (in order) if the summands are equal, i.e., if $$\alpha = (1 - r)^{\frac{1+\theta}{3}}|\!\ln(1 - r)|^{\frac{2}{3}}$$. Taking this value of α, we obtain that, for all $$r\in(\frac{1}{2},1)$$, the inequality
$$0<C_6\leqslant\frac{2}{\alpha }\sqrt{1-r}\big| \ln(1-r)\big|=2(1-r)^{\frac{1-2\theta}{6}} \big|\!\ln(1-r)\big|^{\frac{1}{3}}$$
is valid, which is possible only if $$\theta\geqslant\frac{1}{2}$$.

### 10.6.7

It is clear that the number of points of the integer lattice $$\mathbb{Z}^{m}$$ lying in a shifted ball $$\overline{B}(t,R)$$ of a large radius R is asymptotically equal to the volume of the ball. As we have already verified, it is not easy to obtain a good estimate for the difference
$$\Delta_R(t)=\operatorname {card}\bigl\{n\in\mathbb{Z}^m\, |\, \|n-t\| \leqslant R \bigr\}-\alpha_mR^m$$
as R→+∞ at a fixed point t. However, it is considerably easier to estimate the mean value (with respect to t) of the error Δ R (t). As established in [Ke],
$$\int_{[0,1]^m} \big|\Delta_R(t) \big|^2\,dt \leqslant C_mR^{m-1}.$$
(12)
To verify this, we find the Fourier coefficients of the function
$$f(t)=\Delta_R(t)+\alpha_mR^m= \text{card} \bigl\{n\in\mathbb{Z}^m\, |\,\|n-t\|\leqslant R \bigr\}$$
(obviously, this function has period 1 with respect to each variable). Let χ be the characteristic function of the closed unit ball centered at zero. Then $$f(t)=\sum_{n\in\mathbb{Z}^{m}}\chi (\frac{n-t}{R} )$$. For Q=[0,1) m and each $$k\in\mathbb{Z}^{m}$$, we obtain
\begin{aligned} \widehat{f}(k)=&\int_Q f(t) e^{-2\pi i\langle k,t\rangle}\,dt =\sum _{n\in\mathbb{Z}^m}\int_Q\,\chi \biggl( \frac{n-t}{R} \biggr) e^{-2\pi i\langle k,t\rangle}\,dt \\=& \sum_{n\in\mathbb{Z}^m}\int_{n+Q} \chi \biggl(\frac{t}{R} \biggr) e^{-2\pi i\langle k,t\rangle}\,dt = \int_{\mathbb{R}^m} \chi \biggl(\frac{t}{R} \biggr) e^{-2\pi i\langle k,t\rangle}\,dt =R^m\widehat{ \chi}(Rk). \end{aligned}
Therefore, $$\widehat{\Delta}_{R}(k) = \widehat{f}(k) = R^{m}\widehat{\chi}_{B}(Rk)$$ for k≠0 and $$\widehat{\Delta}_{R}(0) = \widehat{f}(0) - \alpha_{m} R^{m} = (\widehat{\chi}_{B}(0) - \alpha_{m})R^{m} =0$$. By Parseval’s identity, we have
$$\int_{[0,1]^m} \big|\Delta_R(t) \big|^2\,dt=\sum _{k\in\mathbb{Z}^m} \big|\widehat{\Delta}_R(k) \big|^2=R^{2m}\sum_{k\ne0} \big| \widehat{\chi}_B(Rk) \big|^2.$$
Now, inequality (12) follows from the estimate $$\widehat{\chi}_{B}(y)=O (\|y\|^{-\frac{m+1}{2}} )$$ that follows from the asymptotic formula for $$\widehat{\chi}_{B}(y)$$ (see Example 2 of Sect. 10.5.2).
It is interesting to note that, for m≠1 (mod 4), the above-mentioned asymptotic formula for $$\widehat{\chi}_{B}(y)$$ enables us to obtain the inequality
$$\int_{[0,1]^m} \big|\Delta_R(t) \big|^2\,dt \geqslant\widetilde{C}_mR^{m-1}>0.$$
(12′)
In particular, for m=2, we obtain that, in the problem in question, the typical error for the shifted discs $$\overline{B}(t,R)$$ has order of growth $$\sqrt{R}$$.

However, if m=4l+1, then the superior limit of the quotient $$\frac{1}{R^{m-1}}\int_{[0,1]^{m}} |\Delta_{R}(t) |^{2}\,dt$$ as R→+∞ is positive and the inferior limit is zero.

### EXERCISES

1.

Supplement the statement of Lemma 10.6.1 by proving that the series $$\sum_{n\in\mathbb{Z}}f (x+n)$$ converges to S(x) not only almost everywhere, but also in mean.

2.

Let an absolutely continuous function f and its derivative be summable on $$\mathbb{R}$$. Prove that Eq. (3) is valid for all $$x\in\mathbb{R}$$.

3.

Verify that, in Theorem 2 of Sect. 10.6.2, the Lipschitz condition can be weakened to the assumption that the inequality |f(x+h)−f(x)|⩽R q h α , where ∥h∥⩽1 and ∥x∥⩽R (here, q is a fixed non-negative number), holds for every R>1.

4.

Using the result of the previous exercise, show that, in Corollary 10.6.2, the assumption that the derivatives are bounded can be replaced by the requirement that they are dominated by a polynomial.

5.

Let and $$\widehat {f}\geqslant0$$ everywhere. Prove that Eq. (5) is valid at every point $$x\in\mathbb{R}^{m}$$ (the series on the right-hand side of (5) converges absolutely).

6.

Prove that the condition is not only sufficient but also necessary for the boundedness of the -norms of the sums ω ε (see formula (7)).

7.

Prove that the function ω ε in the example of Sect. 10.6.4 admits the estimate $$\omega_{\varepsilon }(x)=O(\varepsilon ) (1 + (\varepsilon ^{2} +4\pi^{2}\|x\|^{2})^{-\frac{m+1}{2}} )$$ (the constant in the O-term depends only on the dimension) and, therefore, ω ε is dominated by a summable “hump-shaped” majorant.

8.

Prove that, as ε→0, the functions $$\omega_{\varepsilon }(x) = \sum_{n\in\mathbb{Z}^{m}} e^{-\varepsilon \|n\|^{2}}e^{2\pi i\langle n,x \rangle}$$ form an approximate identity having the strong localization property and a “hump-shaped” majorant.

9.

Verify that the reasoning of Sect. 10.6.6 can be used to obtain the following stronger result (see [EF]): the fraction $$\Delta(R) /\sqrt{\frac{R}{\ln R}}$$ does not tend to zero as R→+∞.

## Footnotes

1. 1.

Pythagoras (Πυϑαγóρας) (circa 570–500 BC)—Greek philosopher and mathematician.

2. 2.

Friedrich Wilhelm Bessel (1784–1846)—German mathematician.

3. 3.

Jean Baptiste Joseph Fourier (1768–1830)—French mathematician.

4. 4.

Ernest Sigismund Fisher (1875–1954)—German mathematician.

5. 5.

Marc-Antoine Parseval (1755–1836)—French mathematician.

6. 6.

Fedor L’vovich Nazarov (born 1967)—Russian mathematician.

7. 7.

8. 8.

Charles Hermite (1822–1901)—French mathematician.

9. 9.

Joseph Leonard Walsh (1895–1973)—American mathematician.

10. 10.

Andrei Nikolaevich Kolmogorov (1903–1987)—Russian mathematician.

11. 11.

Edmond Nicolas Laguerre (1834–1886)—French mathematician.

12. 12.

Ulisse Dini (1845–1918)—Italian mathematician.

13. 13.

Marie Ennemond Camille Jordan (1838–1922)—French mathematician.

14. 14.

Lennart Axel Edvard Carleson (born 1928)—Swedish mathematician.

15. 15.

Arnaud Denjoy (1884–1974)—French mathematician.

16. 16.

Ernesto Cesaro (1859–1906)—Italian mathematician.

17. 17.

Lipót Fejér (1880–1959)—Hungarian mathematician.

18. 18.

Charles Louis Fefferman (born 1949)—American mathematician.

19. 19.

Sergei Natanovich Bernstein (1880–1968)—Russian mathematician.

20. 20.

Michel Plancherel (1885–1967)—Swiss mathematician.

21. 21.

Edmund Georg Hermann Landau (1877–1938)—German mathematician.

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