The $L^p$ Spaces

Abstract

Let $\{X,\mathcal {A},\mu \}$ be a measure space and let $E\in \mathcal {A}$.

Appendices

Problems and Complements

1c Functions in $L^p(E)$ and Their Norm

1.1 1.1c The Spaces $L^p$ for $0<p<1$

A measurable function $f:E\rightarrow \mathbb {R}^*$ is in $L^p(E)$ for $0<p<1$ if $|f|^p\in L^1(E)$. A norm-like function $f\rightarrow \Vert f\Vert _p$ might be defined as in (1.1). The collection $L^p(E)$ for $0<p<1$ is a linear space. This is a consequence of the following lemma.

Lemma 1.1c

Let $0<p<1$. Then for nonnegative x, y

$$\begin{aligned} (x+y)^p\le x^p+y^p. \end{aligned}$$

Proof

If either x or y is zero, the inequality is trivial. Otherwise, letting $t=x/y$, the inequality is equivalent to

$$\begin{aligned} f(t)=(1+t)^p-(1+t^p)\le 0\quad \text { for }\> t\ge 0\quad \text { and }\> 0<p<1 \end{aligned}$$

$\blacksquare$

1.2 1.2c The Spaces $L^q$ for $q<0$

If $0<p<1$, its Hölder conjugate q is negative. A measurable function $f:E\rightarrow \mathbb {R}^*$ is in $L^q(E)$ for $q<0$ if

$$\begin{aligned} 0<\int _E|f|^qd\mu = \int _E\frac{1}{|f|^{\frac{p}{1-p}}} d\mu <\infty ,\quad \qquad \frac{1}{p}+\frac{1}{q}=1. \end{aligned}$$

A norm-like function $f\rightarrow \Vert f\Vert _q$ might be defined as in (1.1). If $f\in L^q(E)$ for $q<0$, then $f\ne 0$ a.e. in E and $|f|\not \equiv \infty$. If $q<0$ the set $L^q(E)$ is not a linear space.

1.3 1.3c The Spaces $\ell _p$ for $1\le p\le \infty$

Let $\mathbf {a}=\{a_n\}$ be a sequence of real numbers and set

$$\begin{aligned} \begin{array}{ll} \Vert \mathbf {a}\Vert _p=\left( \mathop {\textstyle {\sum }}\limits |a_n|^p\right) ^{1/p} \qquad &{}\text { if }\>1\le p<\infty \\ \Vert \mathbf {a}\Vert _\infty =\sup |a_n|\>&{}\text { if }\>p=\infty . \end{array} \end{aligned}$$

(1.1c)

Denote by $\ell _p$ the set of all sequences $\mathbf {a}$ such that $\Vert \mathbf {a}\Vert _p<\infty$. One verifies that $\ell _p$ is a linear space for all $1\le p\le \infty$ and that (1.1c) is a norm. Moreover $\ell _\infty$ satisfies analogues of (1.2) and (1.4).

2c The Inequalities of Hölder and Minkowski

Corollary 2.1c

Let $p,q\in (1,\infty )$ be conjugate. For $a,b\in \mathbb {R}$ and $\varepsilon >0$

$$\begin{aligned} |ab|\le \frac{\varepsilon ^p}{p}|a|^p +\frac{1}{\varepsilon ^qq}|b|^q. \end{aligned}$$

Corollary 2.2c

Let $\alpha _i\in (0,1)$ for $i=1,\dots ,n$, and $\sum _{i=1}^n\alpha _i=1$. Then for any n-tuple of real numbers $\xi _1,\dots ,\xi _n$

$$\begin{aligned} \mathop {\textstyle {\prod }}\limits _{i=1}^n|\xi _i|^{\alpha _i}\le \mathop {\textstyle {\sum }}\limits _{i=1}^n \alpha _i|\xi _i|. \end{aligned}$$

State and prove a variant of these corollaries when p is permitted to be one, or some of the $\alpha _i$ is permitted to be one.

1.1 2.1c Variants of the Hölder and Minkowski Inequalities

Corollary 2.3c

Let $f_i\in L^{p_i}(E)$ , for $1\le p_i\le \infty$ and $i=1,\dots ,n$. Then

$$\begin{aligned} \int _E\mathop {\textstyle {\prod }}\limits _{i=1}^n|f_i|d\mu \le \mathop {\textstyle {\prod }}\limits _{i=1}^n\Vert f_i\Vert _{p_i} \quad \text { whenever }\quad \mathop {\textstyle {\sum }}\limits _{i=1}^n\frac{1}{p_i}=1. \end{aligned}$$

Prove the following convolution inequality.

Corollary 2.4c

Let $f\in L^p(\mathbb {R}^N)$, $g\in L^q(\mathbb {R}^N)$ and $h\in L^r(\mathbb {R}^N)$, where $p,q,r\ge 1$ satisfy

$$\begin{aligned} \frac{1}{p}+\frac{1}{q}+\frac{1}{r}=2. \end{aligned}$$

Then

$$\begin{aligned} \int _{\mathbb {R}^N}f(g*h) dx\le \Vert f\Vert _p\Vert g\Vert _q\Vert h\Vert _r. \end{aligned}$$

(2.1c)

Proof

Write

$$\begin{aligned} f(x)g(x-y)h(y)=[f^p(x) g^q(x-y)]^{\frac{1}{r^\prime }} [g^q(x-y) h^r(y)]^{\frac{1}{p^\prime }} [f^p(x) h^r(y)]^{\frac{1}{q^\prime }}. \end{aligned}$$

where $p^\prime$, $q^\prime$ and $r^\prime$ are the Hölder conjugate of p, q and r.$\blacksquare$

Corollary 2.5c

Let $1\le p,q\le \infty$ be conjugate. Then for $\mathbf {a}\in \ell _p$ and $\mathbf {b}\in \ell _q$

$$\begin{aligned} |\mathbf {a}\cdot \mathbf {b}|=\mathop {\textstyle {\sum }}\limits |a_ib_i|\le \Vert \mathbf {a}\Vert _p\Vert \mathbf {b}\Vert _q. \end{aligned}$$

For $\mathbf {a},\,\mathbf {b}\in \ell _p$ and $1\le p\le \infty$ [109]

$$\begin{aligned} \Vert \mathbf {a}+\mathbf {b}\Vert _p\le \Vert \mathbf {a}\Vert _p+\Vert \mathbf {b}\Vert _p. \end{aligned}$$

1.2 2.2c Some Auxiliary Inequalities

Lemma 2.1c

Let x and y be any two positive numbers. Then for $1\le p<\infty$

$$\begin{aligned} |x-y|^p\le |x^p-y^p|; \end{aligned}$$

(2.2c)

$$\begin{aligned} (x+y)^p\ge x^p+y^p; \end{aligned}$$

$$\begin{aligned} (x+y)^p\le 2^{p-1}(x^p+y^p). \end{aligned}$$

(2.3c)

Proof

Assuming $x\ge y$ and $p>1$

$$\begin{aligned} x^p-y^p&=\int _0^1\frac{d}{ds}[sx+(1-s) y]^pds =p(x-y)\int _0^1[sx+(1-s)y]^{p-1}ds\\&=p(x-y)\int _0^1[s(x-y)+y]^{p-1}ds\\&\ge p(x-y)\int _0^1s^{p-1}(x-y)^{p-1}ds =(x-y)^p. \end{aligned}$$

This proves the first of (2.2c). The second follows from the first, since

$$\begin{aligned} (x+y)^p-y^p\ge (x+y-y)^p. \end{aligned}$$

To prove (2.3c) we may assume that both x and y are nonzero and that $p>1$. Setting $t=x/y$, the inequality is equivalent to

$$\begin{aligned} f(t)=\frac{(1+t)^p}{1+t^p}\le 2^{p-1}\quad \text { for all }\> t>0. \end{aligned}$$

$\blacksquare$

1.3 2.3c An Application to Convolution Integrals

Proposition 2.1c

Let $f\in L^p(\mathbb {R}^N)$ and $g\in L^q(\mathbb {R}^N)$ for some $1\le p\le \infty$, with p and q Hölder conjugate. Then

$$\begin{aligned} f*g\in L^\infty (\mathbb {R}^N)\quad \text { and }\quad \Vert f*g\Vert _\infty \le \Vert f\Vert _p\Vert g\Vert _q. \end{aligned}$$

Proposition 2.2c

Let $f\in L^1(\mathbb {R}^N)$ and $g\in L^p(\mathbb {R}^N)$ for some $1\le p\le \infty$. Then

$$\begin{aligned} f*g\in L^p(\mathbb {R}^N)\quad \text { and }\quad \Vert f*g\Vert _p\le \Vert f\Vert _1\Vert g\Vert _p. \end{aligned}$$

Hint: Assume $\Vert f\Vert _1=1$ and observe that $|t|^p$ is a convex function of t in $\mathbb {R}$.

1.4 2.4c The Reverse Hölder and Minkowski Inequalities

Proposition 2.3c

(Reverse Hölder Inequality) Let $p\in (0,1)$ and $q<0$ satisfy (2.1). Then for every $f\in L^p(E)$ and $g\in L^q(E)$

$$\begin{aligned} \int _E|fg|d\mu \ge \Vert f\Vert _p\Vert g\Vert _q,\quad \qquad \frac{1}{p}+\frac{1}{q}=1. \end{aligned}$$

Proof

We may assume that $fg\in L^1(E)$. Apply Hölder’s inequality with

$\blacksquare$

Proposition 2.4c

(Reverse Minkowski Inequality) Let $p\in (0,1)$. Then for all $f,\,g\in L^p(E)$

$$\begin{aligned} \Vert \,|f|+|g|\,\Vert _p\ge \Vert f\Vert _p+\Vert g\Vert _p. \end{aligned}$$

1.5 2.5c $L^p(E)$-Norms and Their Reciprocals

Proposition 2.5c

Let $E\subset \mathbb {R}^N$ be measurable with $\mu (E)=1$. Then for every nonnegative measurable function f defined in E, and all $p>0$

$$\begin{aligned} \Vert f\Vert _1\,\Big \Vert \frac{1}{f}\Big \Vert _p\ge 1. \end{aligned}$$

(2.4c)

Proof

Apply inequality (14.4c) of Chap. 5.$\blacksquare$

3c More on the Spaces $L^p$ and Their Norm

3.1.:

Let $\{f_n\}$ be a sequence in $L^p(E)$ for some $p\ge 1$. Then

$$\begin{aligned}&\Big (\int _E\left| \mathop {\textstyle {\sum }}\limits f_n\right| ^pd\mu \Big )^{\frac{1}{p}}\le \mathop {\textstyle {\sum }}\limits \,\Vert f_n\Vert _p.\\&\Big (\mathop {\textstyle {\sum }}\limits \big |\int _Ef_nd\mu \big |^p\Big )^{\frac{1}{p}}\le \int _E\left( \mathop {\textstyle {\sum }}\limits |f_n|^p\right) ^{\frac{1}{p}} d\mu . \end{aligned}$$

3.2.:

The spaces $\ell _p$ satisfy analogues of Propositions 3.1 and 3.2.

3.3.:

Let $f\in L^p(E)$ for all $1\le p<\infty$. Assume that $\Vert f\Vert _p\le K$ for all $1\le p<\infty$, for some constant K, and that $\mu (E)<\infty$. Then $f\in L^\infty (E)$ and $\Vert f\Vert _\infty \le K$.

3.4.:

Give an example of $f\in L^p(E)$ for all $p\in [1,\infty )$, and $f\notin L^\infty (E)$.

3.5.:

Let $E\subset$ be open set. Give an example of $f\in L^p(E)$ for all $p\in [1,\infty )$, and $f\notin L^\infty (E^\prime )$ for any open subset $E^\prime \subset E$. Hint: Properly modify the function in 17.9. of the Complements of Chap. 4.

1.1 3.4c A Metric Topology for $L^p(E)$ when $0<p<1$

If $0<p<1$, the norm-like function $f\rightarrow \Vert f\Vert _p$ defined as in (1.1), is not a norm. Indeed (4.3) is violated in view of the reverse Minkowski inequality. By the same token $(f,g)\rightarrow \Vert f-g\Vert _p$ is not a metric in $L^p(E)$. A topology could be generated by the balls

$$\begin{aligned} B_\rho (g)=\left\{ f\in L^p(E)\bigm | \Vert f-g\Vert _p<\rho \right\} ,\qquad p\in (0,1). \end{aligned}$$

where $g\in L^p(E)$ is fixed. By the reverse Minkowski inequality these balls are not convex. A distance in $L^p(E)$ for $0<p<1$ is introduced by setting

$$\begin{aligned} d(f,g)=\Vert f-g\Vert _p^p=\int _E|f-g|^pd\mu . \end{aligned}$$

(3.1c)

One verifies that $d(\cdot ,\cdot )$ satisfies the requirements (i)–(iii) of § 13 of Chap. 2, to be a metric. The triangle inequality (iv) follows from Lemma 1.1c.

1.2 3.5c Open Convex Subsets of $L^p(E)$ for $0<p<1$

Let $\{X,\mathcal {A},\mu \}$ be $\mathbb {R}^N$ with the Lebesgue measure.

Proposition 3.1c

(Day [31]) Let $L^p(E)$ for $0<p<1$ be equipped with the topology generated by the metric in (3.1c). Then the only open, convex subsets of $L^p(E)$ are $\emptyset$ and $L^p(E)$ itself.

Proof

Let $\mathcal {O}$ be a nonempty, open, convex neighborhood of the origin of $L^p(E)$ and let f be an arbitrary element of $L^p(E)$. Since $\mathcal {O}$ is open it contains some ball $B_\rho$ centered at the origin. Let n be a positive integer such that

$$\begin{aligned} \frac{1}{n^{1-p}}\int _E|f|^p dx\le \rho \qquad \text { i.e., }\quad nf\in B_{n\rho }. \end{aligned}$$

Partition E into exactly n disjoint, measurable subsets $\{E_1,\dots , E_n\}$ such that

$$\begin{aligned} \int _{E_j}|f|^pdx=\frac{1}{n}\int _E|f|^pdx\qquad j=1,\dots ,n. \end{aligned}$$

Such a partition can be carried out in view of the absolute continuity of the Lebesgue integral. Set

$$\begin{aligned} h_j=\left\{ \begin{array}{ll} nf\>&{}\text { in }\> E_j\\ 0\>&{}\text { in }\> E-E_j \end{array} \right. \end{aligned}$$

and compute

$$\begin{aligned} \int _E|h_j|^pdx=n^p\int _{E_j}|f|^pdx\le \rho . \end{aligned}$$

Thus $h_j\in B_\rho \subset \mathcal {O}$ for all $j=1,\dots ,n$. Since $\mathcal {O}$ is convex

$$\begin{aligned} f=\frac{h_1+h_2+\cdots +h_n}{n}\in \mathcal {O}. \end{aligned}$$

Since $f\in L^p(E)$ is arbitrary $\mathcal {O}\equiv L^p(E)$.$\blacksquare$

Corollary 3.1c

The topology generated by the metric (3.1c) in $L^p(E)$ for $0<p<1$ is not locally convex.

5c Convergence in $L^p(E)$ and Completeness

5.1.:

A sequence $\{f_n\}\subset L^\infty (E)$ converges in $L^\infty (E)$ to some f, if and only if there exists a set $\mathcal {E}\subset E$ of measure zero, such that $\{f_n\}\rightarrow f$ uniformly in $E-\mathcal {E}$.

5.2.:

A sequence $\{f_n\}\subset L^p(E)$ converges to some f in $L^p(E)$ if and only if every subsequence $\{f_{n^\prime }\}\subset \{f_n\}$, contains in turn a subsequence $\{f_{n^{\prime \prime }}\}$ converging to f in $L^p(E)$.

5.3.:

$\ell _p$ is complete for all $1\le p\le \infty$.

5.4.:

Let $\{f_n\}$ be a sequence of functions in $L^p(E)$ for $p\in (1,\infty )$, converging a.e. in E to a function $f\in L^p(E)$. Then $\{f_n\}$ converges to f in $L^p(E)$ if and only if $\>\lim \Vert f_n\Vert _p =\Vert f\Vert _p$.

5.5.:

Let $L^p(E)$ for $0<p<1$ be endowed with the metric topology generated by the metric in (3.1c). With respect to such a topology $L^p(E)$ is a complete metric space. In particular $L^p(E)$ is of second category.

1.1 5.1c The Measure Space $\{X,\mathcal {A},\mu \}$ and the Metric Space $\{\mathcal {A};d\}$

Let $\{X,\mathcal {A},\mu \}$ be a measure space and let d(E; F) be the distance of any two measurable sets E and F, as introduced in 3.7 of the Complements of Chap. 3. One verifies that

$$\begin{aligned} d(E;F)=\int _X\big |\chi _E-\chi _F\big |d\mu . \end{aligned}$$

Two measurable sets E and $E^\prime$ are equivalent if $d(E;E^\prime )=0$. This identifies equivalence classes in $\mathcal {A}$. Continue to denote by $\mathcal {A}$ the collection of such equivalence classes. By the Riesz-Fisher theorem, $L^1(X)$ is complete (Theorem 5.1). Therefore $\{\mathcal {A};d\}$ is a complete metric space and, as such, is of second category.

1.1.1 5.1.1c Continuous Functions on $\{\mathcal {A};d\}$

Lemma 5.1c

Let $\{X,\mathcal {A},\mu \}$ be a measure space, and let $\lambda$ be a signed measure on $\mathcal {A}$, absolutely continuous with respect to $\mu$, and of finite total variation $|\lambda |$. Then the function $\lambda :\{\mathcal {A};d\}\rightarrow \mathbb {R}$, is continuous with respect to the metric topology of $\{\mathcal {A};d\}$.

Proof

Let $\{E_n\}$ be a sequence of equivalence classes in $\mathcal {A}$, converging to $E\in \mathcal {A}$, in the sense that $d(E;E_n)\rightarrow 0$. Equivalently,

$$\begin{aligned} \lim \mu (E-E\cap E_n)=\lim \mu (E_n-E\cap E_n)=0. \end{aligned}$$

Since $|\lambda |$ is finite and $\lambda \ll \mu$, by (c) and (d) of Chap. 3, this implies

$$\begin{aligned} \lim |\lambda |(E-E\cap E_n)=\lim |\lambda |(E_n-E\cap E_n)=0. \end{aligned}$$

Hence $\lim \lambda (E_n)=\lambda (E)$.$\blacksquare$

6c Separating $L^p(E)$ by Simple Functions

For a measure space $\{X,\mathcal {A},\mu \}$ and $E\in \mathcal {A}$, Proposition 6.1 asserts that the simple functions, is dense in $L^p(E)$ for all $1\le p\le \infty$.

When $E\subset \mathbb {R}^N$ is Lebesgue measurable, the simple functions dense in $L^p(E)$ can be given specific forms, provided $1\le p<\infty$. Let $\mathcal {S}_o$ denote the collection of simple functions of the form

$$\begin{aligned} \varphi _o= \mathop {\textstyle {\sum }}\limits _{j=1}^n \varphi _j\chi _{E_{j,o}} \end{aligned}$$

(6.1c)

where $\varphi _j\in \mathbb {R}$ and $E_{j,o}$ are open for $j=1,\dots ,n$.

Proposition 6.1c

Let $E\subset \mathbb {R}^N$ be Lebesgue measurable. Then $\mathcal {S}_o$ is dense in $L^p(E)$ for $1\le p<\infty$.

Proof

Fix $f\in L^p(E)$ and $\varepsilon >0$. There exists a simple function

$$\begin{aligned} \varphi =\mathop {\textstyle {\sum }}\limits _{j=1}^n \varphi _j\chi _{E_j} \quad \text { such that }\quad \Vert f-\varphi \Vert _{p,E}<{\textstyle \frac{1}{2}}\varepsilon . \end{aligned}$$

Each $E_j$ is of finite measure. Therefore, by Proposition 12.2 of Chap. 3, there exist open sets $E_{j,o}\supset E_j$ such that

$\blacksquare$

If $N=1$, the sets $E_{j,o}$ can be taken to be open intervals $(a_j,b_j)$.

Corollary 6.1c

Let $E\subset \mathbb {R}$ be Lebesgue measurable. Then the collection of simple functions of the form

$$\begin{aligned} \varphi = \mathop {\textstyle {\sum }}\limits _{j=1}^n \varphi _j\chi _{(a_j,b_j)} \end{aligned}$$

(6.2c)

is dense in $L^p(E)$ for $1\le p<\infty$.

Proof

Fix $f\in L^p(E)$ and $\varepsilon >0$, and let $\varphi _o$ of the form (6.1c) be such that

$$\begin{aligned} \Vert f-\varphi \Vert _p<\frac{1}{2}\varepsilon . \end{aligned}$$

Each $E_{j,o}$ is the countable union of disjoint open intervals $\{(a_{j,i},b_{j,i})\}_{i\in \mathbb {N}}$. Since each $E_{j,o}$ is of finite measure, for each $j\in \{1,\dots ,n\}$, there exists an index $m_j$, such that

$\blacksquare$

7c Weak Convergence in $L^p(E)$

Throughout this section $\{X,\mathcal {A},\mu \}$ is $\mathbb {R}^N$ with the Lebesgue measure and $E\subset \mathbb {R}^N$ is Lebesgue measurable and of finite measure.

7.1.:

Let $\{f_n\}$ be a sequence of functions in $L^p(E)$ converging weakly in $L^p(E)$ to some $f\in L^p(E)$, and converging a.e. in E to some g. Then $f=g$, a.e. in E.

7.2.:

Let $\{f_n\}\rightarrow f$ weakly in $L^p(E)$ and $\{f_n\}\rightarrow g$ in measure. Then $f=g$, a.e. in E.

1.1 7.3c Comparing the Various Notions of Convergence

Denote by $\{\text {a.e.}\}$ the set of all sequences $\{f_n\}$ of functions from E into $\mathbb {R}^*$, convergent a.e. in E. Analogously denote by $\{\text {meas}\},\,\{L^p(E)\}$ and {w-$L^p(E)$} the set of all sequences $\{f_n\}$ of measurable functions defined in E and convergent respectively, in measure, strongly and weakly in $L^p(E)$.

By the remarks in § 7 and the counterexample of § 7.1

$$\begin{aligned} \{L^p(E)\}\subset \{\text {w- }L^p(E)\} \quad \text { and }\quad \{\text {w- }L^p(E)\}\not \subset \{L^p(E)\}. \end{aligned}$$

By Proposition 4.2 of Chap. 4 and the counterexample (4.1)

$$\begin{aligned} \{\text {a.e.}\}\subset \{\text {meas}\}\qquad \text { and }\qquad \{\text {meas}\}\not \subset \{\text {a.e.}\}. \end{aligned}$$

Weak convergence in $L^p(E)$ does not imply convergence in measure, nor does convergence in measure imply weak convergence in $L^p(E)$, i.e.,

$$\begin{aligned} \{\text {w- } L^p(E)\}\not \subset \{{\text {meas }}\}\quad \text { and }\quad \{{\text {meas }}\}\not \subset \{\text {w- } L^p(E)\}. \end{aligned}$$

The sequence $\{\cos {nx}\}$ for $x\in [0,2\pi ]$, converges weakly to zero, but not in measure. Indeed

$$\begin{aligned} \mu \big (x\in [0,2\pi ]\,|\,|\cos {nx}| \ge {\textstyle \frac{1}{2}}\big )={\textstyle \frac{4}{3}}\pi . \end{aligned}$$

This proves the first statement. For the second consider the sequence

$$\begin{aligned} f_n(x)=\left\{ \begin{array}{ll} n\quad &{}\text { for }\> x\in [0,\frac{1}{n}]\\ 0\quad &{}\text { for }\> x\in (\frac{1}{n},1]. \end{array} \right. \end{aligned}$$

Such a sequence converges to zero in measure and not weakly in $L^p[0,1]$. Almost everywhere convergence does not imply weak convergence in $L^p(E)$. Strong convergence in $L^p(E)$ does not imply a.e. convergence, i.e.,

$$\begin{aligned} \{\text {a.e.}\}\not \subset \{\text {w- } L^p(E)\} \quad \text { and }\quad \{L^p(E)\}\not \subset \{\text {a.e.}\}. \end{aligned}$$

Indeed the sequence $\{f_n\}$ above, converges a.e. to zero and does not converge to zero weakly in $L^p(E)$. For the second statement consider the sequence $\{\varphi _{nm}\}$ introduced in (4.1) of Chap. 4.

The relationships between these notions of convergence can be organized in the diagram below in form of inclusion of sets where each square represents convergent sequences (Fig. 1c).⁶

7.4.:

Exhibit a sequence $\{f_n\}$ of measurable functions in [0, 1] convergent to zero in measure and weakly in $L^2[0,1]$, but not convergent almost everywhere in [0, 1].

Fig. 1

Various notions of convergence

1.2 7.5c Weak Convergence in $\ell _p$

Let $1\le p,q\le \infty$ be conjugate. Let $\{\mathbf {a}_n\}$ be a sequence of elements in $\ell _p$ and let $\mathbf {b}\in \ell _q$. The sequence $\{\mathbf {a}_n\}$ converges weakly in $\ell _p$ to some $\mathbf {a}\in \ell _p$ if

$$\begin{aligned} \lim \mathop {\textstyle {\sum }}\limits \,a_{j,n}b_j=\mathop {\textstyle {\sum }}\limits \,a_j b_j\qquad \text { for all }\>\mathbf {b}\in \ell _q. \end{aligned}$$

Strong convergence implies weak convergence. For $1<p\le \infty$, the converse is false. For example, let

$$\begin{aligned} \begin{aligned} a_{j,n}&={\left\{ \begin{array}{ll} {\displaystyle n^{-\frac{1}{p}}}\quad &{}\text {for }\> 1\le j\le n\\ 0&{}\text {for }\> j>n \end{array}\right. } \quad&\text { for }\> 1<p<\infty ;\\ b_{j,n}&={\left\{ \begin{array}{ll} 0\qquad &{}\text {for }\> 1\le j\le n\\ 1&{}\text {for }\> j>n \end{array}\right. } \quad&\text { for }\> p=\infty . \end{aligned} \end{aligned}$$

Then $\Vert \mathbf {a}_n\Vert _p=1$ for all n and $\{\mathbf {a}_n\}\rightarrow 0$ weakly in $\ell _p$. Likewise $\Vert \mathbf {b}_n\Vert _\infty =1$ for all n and $\{\mathbf {b}_n\}\rightarrow 0$ weakly in $\ell _\infty$. Discuss and compare the various notions of convergence in $\ell _p$, for $1<p\le \infty$, and construct examples. For $p=1$ weak and strong convergence are equivalent (Corollary 22.1c and § 23c).

9c Weak Convergence and Norm Convergence

When $p=2$, the proof of Proposition 9.1 is particularly simple and elegant. Indeed

$$\begin{aligned} \Vert f_n-f\Vert _2^2 =\Vert f_n\Vert _2^2+ \Vert f\Vert _2^2-2\int _Ef_nfd\mu . \end{aligned}$$

1.1 9.1c Proof of Lemmas 9.1 and 9.2

For $t\ne 0$ and $x=t^{-1}$, consider the function

$$\begin{aligned} f(t)=\frac{|1+t|^p-1-pt}{|t|^p} =|1+x|^p-|x|^p- p|x|^{p-2}x=\varphi (x). \end{aligned}$$

It suffices to prove that $\varphi (x)\ge c$ for all $x\in \mathbb {R}$ for some $c>0$. If $x\in (-1,0]$ we estimate directly

$$\begin{aligned} \varphi (x)\ge (1-|x|)^p+(p-1)|x|^{p-1}\ge \frac{1}{2^p} \min \{1;(p-1)\}. \end{aligned}$$

If $x\in (0,1]$

$$\begin{aligned} (1+x)^p-x^p&=\int _0^1\frac{d}{ds}(s+x)^pds =p\int _0^1(s+x)^{p-1}ds\\&=-p\int _0^1(s+x)^{p-1}\frac{d}{ds}(1-s)ds\\&=p x^{p-1}+p(p-1)\int _0^1 (1-s)(s+x)^{p-2}ds\\&\ge px^{p-1}+ \min \big \{1;\frac{p(p-1)}{4}\big \}. \end{aligned}$$

Therefore for $|x|\le 1$

$$\begin{aligned} \varphi (x)\ge c_p (p-1)\quad \text { for some positive constant } c_p. \end{aligned}$$

The Case $p\ge 2$ and $|x|>1$: By direct calculation

$$\begin{aligned} |1+x|^p-|x|^p=\int _0^1 \frac{d}{ds}|s+x|^pds =p\int _0^1|s+x|^{p-1}\text {sign }(s+x)ds. \end{aligned}$$

From this, for $x\ge 1$ and $p\ge 2$, making use of the second of (2.2c)

$$\begin{aligned} |1+x|^p-|x|^p\ge p\int _0^1\left( x^{p-1}+s^{p-1}\right) ds \ge p |x|^{p-2}x +1. \end{aligned}$$

For $x\le -1$ and $p\ge 2$, making use of the first (2.2c)

$$\begin{aligned} |1+x|^p-|x|^p&=-p\int _0^1(|x|-s)^{p-1}ds\\&\ge -p\int _0^1\left( |x|^{p-1}-s^{p-1}\right) ds =p|x|^{p-2}x+1. \end{aligned}$$

The Case $1<p<2$ and $|x|>1$: Assume first $t\in (0,1)$. Then by repeated integration by parts

$$\begin{aligned} (1+t)^p-1&=-\int _0^t\frac{d}{ds}[1+(t-s)]^pds =p\int _0^t[1+(t-s)]^{p-1}ds\\&=pt+p(p-1)\int _0^t s[1+(t-s)]^{p-2}ds\\&\ge pt+\frac{p(p-1)}{4} t^2. \end{aligned}$$

Therefore

$$\begin{aligned} \frac{(1+t)^p-1-pt}{t^2}\ge \frac{p(p-1)}{2} \qquad \text { for all } \>t\in (0,1). \end{aligned}$$

A similar calculation holds for $t\in (-1,0)$ with the same bound below.

11c The Riesz Representation Theorem

1.1 11.1c Weakly Cauchy Sequences in $L^p(X)$ for $1< p\le \infty$

Let $\{X,\mathcal {A},\mu \}$ be a measure space. A sequence of functions $\{f_n\}:E\rightarrow \mathbb {R}^*$ is weakly Cauchy in $L^p(X)$ if it is bounded in $L^p(X)$, and if, for all measurable subsets E, of finite measure, the sequence

$$\begin{aligned} \Big \{\int _Ef_n d\mu \Big \}\quad \text { is a Cauchy sequence in } \mathbb {R}\end{aligned}$$

(11.1c)

In such a case, for all such sets, there exists the limit

$$\begin{aligned} \nu (E)=\lim \int _Ef_n d\mu . \end{aligned}$$

(11.2c)

Prove that for $1<p<\infty$ a sequence $\{f_n\}\subset L^p(X)$ is weakly convergent to some $f\in L^p(E)$, if and only if it is weakly Cauchy in $L^p(X)$. Thus for $1<p<\infty$, the space $L^p(X)$ is weakly complete. Prove that the same conclusion holds for $p=\infty$ if $\{X,\mathcal {A},\mu \}$ is $\sigma$-finite.

1.2 11.2c Weakly Cauchy Sequences in $L^p(X)$ for $p=1$

If $p=1$, the notion of weakly Cauchy sequence is modified by requiring that (11.1c) holds for all measurable sets E. Prove that if $\{X,\mathcal {A},\mu \}$, is $\sigma$-finite, then $L^1(X)$ is weakly complete. Hint: Use the indicated modified notion of weakly Cauchy sequence and the Radon–Nikodým theorem.

1.3 11.3c The Riesz Representation Theorem in $\ell _p$

Let $\mathbf {a}\in \ell _p$ and $\mathbf {b}\in \ell _q$, where $1\le p,q\le \infty$ are conjugate. Every element $\mathbf {b}\in \ell _q$ induces a bounded linear functional on $\ell _p$ by the formula

$$\begin{aligned} T(\mathbf {a})=\mathbf {a}\cdot \mathbf {b}=\mathop {\textstyle {\sum }}\limits \, a_i b_i. \end{aligned}$$

Theorem 11.1c

Let $1\le p<\infty$. For every bounded, linear functional $\mathcal {F}$ in $\ell _p$, there exists a unique element $\mathbf {b}\in \ell _q$ such that $\mathcal {F}(\mathbf {a})=\mathbf {a}\cdot \mathbf {b}$ for all $\mathbf {a}\in \ell _p$.

14c The Riesz Representation Theorem By Uniform Convexity

1.1 14.1c Bounded Linear Functional in $L^p(E)$ for $0<p<1$

Let $\{X,\mathcal {A},\mu \}$ be $\mathbb {R}^N$ with the Lebesgue measure and let $E\subset \mathbb {R}^N$ be Lebesgue measurable. The topology generated in $L^p(E)$ for $0<p<1$ by the metric (3.1c) is not locally convex. A consequence is that there are no linear, bounded maps $T:L^p(E)\rightarrow \mathbb {R}$ except the identically zero map.

Proposition 14.1c

(Day [31]) Let E be a Lebesgue measurable subset of $\mathbb {R}^N$ and let $L^p(E)$ for $0<p<1$ be equipped with the topology generated by the metric (3.1c). Then the only bounded, linear functional on $L^p(E)$ is the identically zero functional.

Proof

Let T be a bounded linear functional in $L^p(E)$ for some $0<p<1$. Since T is continuous and linear, the pre-image of any open interval must be open and convex in $L^p(E)$. However by Proposition 3.1c, $L^p(E)$ for $0<p<1$, with the indicated topology, does not have any open convex sets except the empty set and $L^p(E)$ itself. Let $(-\alpha ,\alpha )$ for some $\alpha >0$ be an interval about the origin of $\mathbb {R}$. Then $T^{-1}(-\alpha ,\alpha )=L^p(E)$ for all $\alpha >0$. Thus $T\equiv 0$.$\blacksquare$

1.2 14.2c An Alternate Proof of Proposition 14.1c

The alternate proof below is independent of the lack of open, convex neighborhoods of the origin in the metric topology of $L^p(E)$.⁷

Let T be a continuous, linear functional in $L^p(E)$ and let $f\in L^p(E)$ be such that $T(f)\ne 0$. There exists $A\in E\cap \mathcal {A}$ such that

$$\begin{aligned} \int _E\chi _A|f|^pd\mu =\frac{1}{2}\Vert f\Vert _p^p. \end{aligned}$$

Set $f=f_o$ and $f_{o,1}=f_o\chi _A$ and $f_{o,2}=f_o\chi _{E-A}$. Therefore $f_o=f_{o,1}+f_{o,2}$, and

$$\begin{aligned} \Vert f_o\Vert _p^p=\Vert f_{o,1}\Vert _p^p+\Vert f_{o,2}\Vert _p^p,\qquad \Vert f_{o,1}\Vert _p^p=\Vert f_{o,2}\Vert _p^p={\textstyle \frac{1}{2}}\Vert f_o\Vert _p^p. \end{aligned}$$

Since T is linear, $T(f_o)=T(f_{o,1})+T(f_{o,2})$, and

$$\begin{aligned} |T(f_o)|\le |T(f_{o,1})|+|T(f_{o,2})|. \end{aligned}$$

Therefore either

$$\begin{aligned} |T(f_{o,1})|\ge {\textstyle \frac{1}{2}}|T(f_o)| \qquad \text { or }\qquad |T(f_{o,2})|\ge {\textstyle \frac{1}{2}}|T(f_o)|. \end{aligned}$$

Assume the first holds true and set $f_1=2f_{o,1}$. For this choice

$$\begin{aligned} |T(f_1)|\ge |T(f_o)|\quad \text { and }\quad \Vert f_1\Vert _p^p=2^p\Vert f_{o,1}\Vert _p^p=2^{p-1}\Vert f_o\Vert _p^p. \end{aligned}$$

Now repeat this construction with $f_o$ replaced by $f_1$ and generate a function $f_2$ such that

$$\begin{aligned} |T(f_2)|\ge |T(f_o)|\quad \text { and }\quad \Vert f_2\Vert _p^p=2^{p-1}\Vert f_1\Vert _p^p=2^{2(p-1)}\Vert f_o\Vert _p^p. \end{aligned}$$

By iteration generate a sequence of functions $\{f_n\}$ in $L^p(E)$ such that

$$\begin{aligned} |T(f_n)|\ge |T(f_o)|\qquad \text { and }\qquad \Vert f_n\Vert _p^p=2^{n(p-1)}\Vert f_o\Vert _p^p. \end{aligned}$$

Since $p\in (0,1)$ the sequence $\{f_n\}$ converges to zero in the metric topology of $L^p(E)$. However $T(f_n)$ does not converge to zero. $\blacksquare$

15c If $E\subset \mathbb {R}^N$ and $p\in [1,\infty )$, then $L^p(E)$ Is Separable

15.1.:

The spaces $\ell _p$ for $1\le p<\infty$ are separable, whereas $\ell _\infty$ is not separable.

15.2.:

BV[a, b] is not separable (§ 1.2c of the Complements of Chap. 5).

15.3.:

Let E be a measurable subset in $\mathbb {R}^N$ and let $L^p(E)$ for $0<p<1$ be endowed with the metric topology generated by the metric in (3.1c). With respect to such a topology $L^p(E)$ is separable. In particular it satisfies the second axiom of countability.

18c Approximating Functions in $L^p(E)$ with Functions in $C^\infty (E)$

A function $f\in L^p(\mathbb {R}^N)$ can approximated by smooth functions by forming the convolution with kernels other than the Friedrichs mollifying kernels $J_\varepsilon$.

We mention here two such kernels. Their advantage with respect to the Friedrichs kernels is that they satisfy specific Partial Differential Equations and therefore they are more suitable in applications related to such equations. Their disadvantage is that they are not compactly supported. Therefore even if f is of compact support in $\mathbb {R}^N$, its approximations will not be.

1.1 18.1c Caloric Extensions of Functions in $L^p(\mathbb {R}^N)$

For $x\in \mathbb {R}^N$ and $t>0$ set

$$\begin{aligned} \varGamma (x-y;t)=\frac{1}{(4\pi t)^{N/2}} e^{-\frac{|x-y|^2}{4t}}. \end{aligned}$$

Set formally

$$\begin{aligned} \varDelta _x=\text {div}\nabla _x=\mathop {\textstyle {\sum }}\limits _{j=1}^N\, \frac{\partial ^2}{\partial x_j^2} \end{aligned}$$

and define $\varDelta _y$ similarly. Verify by direct calculation that for all $x,y\in \mathbb {R}^N$ and $t>0$

$$\begin{aligned} \varGamma _t-\varDelta _x\varGamma =0\qquad \text { and }\qquad \varGamma _t-\varDelta _y\varGamma =0. \end{aligned}$$

This partial differential equation is called the heat equation. The variables x are referred to as the space variables and t is referred to as the time.

A function $(x,t)\rightarrow u(x,t)$ that satisfies the heat equation in a space-time open set $E\subset \mathbb {R}^N\times \mathbb {R}$, is said to be caloric in E. For example $(x,t)\rightarrow \varGamma (x-y;t)$ is caloric in $\mathbb {R}^N\times \mathbb {R}^+$ for all $y\in \mathbb {R}^N$.

Verify that

$$\begin{aligned} \int _{\mathbb {R}^N}\varGamma (x-y;t)dy=\int _{\mathbb {R}^N}\varGamma (x-y;t)dx=1 \end{aligned}$$

for all $x,y\in \mathbb {R}^N$ and all $t>0$. Hint: Introduce the change of variables

$$\begin{aligned} \frac{x-y}{2\sqrt{t}}=\eta \qquad \text { for a fixed } \quad y\in \mathbb {R}^N\quad \text { and }\quad t>0 \end{aligned}$$

and use 14.2 of the Complements of Chap. 4.

Let E be an open set in $\mathbb {R}^N$ and regard functions in $L^p(E)$ as functions in $L^p(\mathbb {R}^N)$ by extending them to be zero in $\mathbb {R}^N-E$. For $f\in L^p(E)$ and $t>0$ set

$$\begin{aligned} f_t(x)= (\varGamma *f)(x)=\int _{\mathbb {R}^N}\varGamma (x-y;t)f(y)dy. \end{aligned}$$

The function $(x,t)\rightarrow f_t(x)$ is caloric in $\mathbb {R}^N\times \mathbb {R}^+$ and is called the caloric extension of f in the upper half space $\mathbb {R}^N\times \mathbb {R}^+$. Such an extension is a mollification of f since $x\rightarrow f_t(x)\in C^\infty (\mathbb {R}^N)$.

Proposition 18.1c

Let $f\in L^p(E)$ for some $1\le p<\infty$. Then $\varGamma *f\in L^p(E)$ and $\Vert \varGamma *f\Vert _p\le \Vert f\Vert _p$. The mollifications $\varGamma *f$ approximate f in the sense

$$\begin{aligned} \lim _{t\rightarrow 0}\Vert f_t-f\Vert _p=0. \end{aligned}$$

Moreover if $f\in C(E)$ and f is bounded in $\mathbb {R}^N$ then for every compact subset $\mathcal {K}\subset E$

$$\begin{aligned} \lim _{t\rightarrow 0}f_t(x)=f(x)\quad \text { uniformly in }\>\mathcal {K}. \end{aligned}$$

Proof

The first two statements are proved as in Proposition 18.1. To prove the last, fix a compact set $\mathcal {K}\subset E$ and a positive number $\varepsilon _o$. If $\varepsilon _o$ is sufficiently small, there exists a compact set $\mathcal {K}_{\varepsilon _o}$ such that $\mathcal {K}\subset \mathcal {K}_{\varepsilon _o}\subset E$, and ${\text {dist }}\{\mathcal {K};\mathcal {K}_{\varepsilon _o}\}\ge \varepsilon _o$. For all $x\in \mathcal {K}$ and all $\varepsilon \in (0,\varepsilon _o)$, write

$$\begin{aligned} |f_t(x)-f(x)|&=\Big |\int _{\mathbb {R}^N}\varGamma (x-y;t)[f(y)-f(x)]dy\Big |\\&\le \Big |\int _{|x-y|\le \varepsilon }\varGamma (x-y;t) [f(y)-f(x)]dy\Big |\\&\quad +\Big |\int _{|x-y|>\varepsilon }\varGamma (x-y;t) [f(y)-f(x)]dy\Big |\\&\le \sup _{|x-y|\le \varepsilon ; x\in \mathcal {K}} |f(y)-f(x)|\int _{\mathbb {R}^N}\varGamma (x-y;t)dy\\&\quad +2\Vert f\Vert _\infty \int _{|x-y|>\varepsilon }\varGamma (x-y;t)dy\\&\le \omega (\varepsilon )+ 2\Vert f\Vert _\infty \int _{|x-y|>\varepsilon }\varGamma (x-y;t)dy \end{aligned}$$

where $\omega _o(\cdot )$ is the uniform modulus of continuity of f in $\mathcal {K}_{\varepsilon _o}$. The last integral is transformed into

$$\begin{aligned} \int _{|x-y|>\varepsilon }\varGamma (x-y;t)dy= \frac{1}{\pi ^{N/2}}\int _{|\eta |>\frac{\varepsilon }{2\sqrt{t}}} e^{-|\eta |^2}d\eta . \end{aligned}$$

From this, for $\varepsilon \in (0,\varepsilon _o)$ fixed

$$\begin{aligned} \lim _{t\rightarrow 0}\int _{|x-y|>\varepsilon }\varGamma (x-y;t)dy=0. \end{aligned}$$

Letting now $t\rightarrow 0$ in the previous inequality gives

$\blacksquare$

Remark 18.1c

The assumption that f be bounded in $\mathbb {R}^N$ can be removed. Indeed a similar approximation would hold if f grows as $|x|\rightarrow \infty$ not faster than $e^{\gamma |x|^2}$, for a positive constant $\gamma$ ([34] Chap. V).

1.2 18.2c Harmonic Extensions of Functions in $L^p(\mathbb {R}^N)$

For $x,y\in \mathbb {R}^N$ and $t>0$ set

$$\begin{aligned} H(x-y;t)=\frac{1}{\omega _{N+1}}\frac{2t}{[|x-y|^2 +t^2]^{\frac{N+1}{2}}}. \end{aligned}$$

Set formally

$$\begin{aligned} \varDelta _{(x,t)}=\varDelta _x+\frac{\partial ^2}{\partial t^2} \end{aligned}$$

and define $\varDelta _{(y,t)}$ similarly. Verify by direct calculation that for all $x,y\in \mathbb {R}^N$ and all $t>0$

$$\begin{aligned} \varDelta _{(x,t)} H=\varDelta _{(y,t)}H=0. \end{aligned}$$

This is the Laplace equation in the variables (x, t). A function that satisfies the Laplace equation in an open set $E\subset \mathbb {R}^{N+1}$ is called harmonic in E. As an example, $(x,t)\rightarrow H(x-y;t)$ is harmonic in $\mathbb {R}^N\times \mathbb {R}^+$ for all $y\in \mathbb {R}^N$.

Verify that for all $x,y\in \mathbb {R}^N$ and all $t>0$

$$\begin{aligned} \int _{\mathbb {R}^N}H(x-y;t)dy=\int _{\mathbb {R}^N}H(x-y;t)dx=1. \end{aligned}$$

Hint: The change of variables $(x-y)=t\eta$ transforms these integrals in

$$\begin{aligned} \frac{2}{\omega _{N+1}}\int _{\mathbb {R}^N}\frac{d\eta }{\left( 1 +|\eta |^2\right) ^{\frac{N+1}{2}}}= 2\frac{\omega _N}{\omega _{N+1}}\int _0^\infty \frac{\rho ^{N-1}d\rho }{\left( 1+\rho ^2\right) ^{\frac{N+1}{2}}}=1. \end{aligned}$$

Use also 14.1 of the Complements of Chap. 4.

Let E be an open set in $\mathbb {R}^N$ and regard functions in $L^p(E)$ as functions in $L^p(\mathbb {R}^N)$ by extending them to be zero in $\mathbb {R}^N-E$. For $f\in L^p(E)$ and $t>0$ set

$$\begin{aligned} f_t(x)= (H*f)(x)=\int _{\mathbb {R}^N}H(x-y;t)f(y)dy. \end{aligned}$$

The function $(x,t)\rightarrow f_t(x)$ is harmonic in $\mathbb {R}^N\times \mathbb {R}^+$ and is called the harmonic extension of f in the upper half space $\mathbb {R}^N\times \mathbb {R}^+$. Such and extension is a mollification of f since $x\rightarrow f_t(x)\in C^\infty (\mathbb {R}^N)$.

The integral defining $f_t(\cdot )$ is called the Poisson Integral of f ([34] Chap. II).

Proposition 18.2c

Let $f\in L^p(E)$ for some $1\le p<\infty$. Then $H*f\in L^p(E)$ and $\Vert H*f\Vert _p\le \Vert f\Vert _p$. The mollifications $H*f$ approximate f in the sense,

$$\begin{aligned} \lim _{t\rightarrow 0}\Vert f_t-f\Vert _p=0. \end{aligned}$$

Moreover if $f\in C(E)$ and f is bounded in $\mathbb {R}^N$ then for every compact subset $\mathcal {K}\subset E$

$$\begin{aligned} \lim _{t\rightarrow 0}f_t(x)=f(x)\quad \text { uniformly in }\>\mathcal {K}. \end{aligned}$$

If $p=\infty$ neither $C_o^\infty (E)$ nor $C(\bar{E})$ is dense in $L^\infty (E)$.

1.3 18.3c Characterizing Hölder Continuous Functions

Let $E\subset \mathbb {R}^N$ be open and let $C^\alpha (E)$ be the space of Hölder continuous functions in E, endowed with the topology generated by the distance $d(\cdot ,\cdot )$ introduced in (15.6) of Chap. 2. With respect to such a topology a function $u\in C^{\alpha }(E)$ cannot be approximated, in general, by smooth functions (15.5–15.7 of § 15.1c of the Complements of Chap. 2).

However u can be approximated in the topology of the uniform convergence by its mollifications $\{u_\varepsilon \}$. Indeed the rate of convergence of $\{u_\varepsilon \}$ to u in $L^\infty (E)$ characterizes $C^\alpha (E)$.

Proposition 18.3c

Let $u\in C^\alpha (E)$ for some $\alpha \in (0,1)$. Then

$$\begin{aligned} \Vert u-u_\varepsilon \Vert _\infty \le [u]_\alpha \varepsilon ^\alpha ,\quad \text { and }\quad \Vert u_{\varepsilon ,x_j}\Vert _\infty \le \Vert J_{x_j}\Vert _1 [u]_\alpha \varepsilon ^{\alpha -1} \end{aligned}$$

(18.1c)

for $j=1,\dots ,N$.

Proof

Without loss of generality we may assume $E=\mathbb {R}^N$. Indeed, since u is uniformly continuous in E, with concave modulus of continuity, it can be extended to a Hölder continuous function defined in $\mathbb {R}^N$ with the same upper and lower bounds and with the same Hölder exponent $\alpha$ (Theorem 15.1 of Chap. 5). Then

$$\begin{aligned} |u(x)-u_\varepsilon (x)|\le \int _{|x-y|<\varepsilon } J_\varepsilon (x-y)|u(x)-u(y)|dy\le [u]_\alpha \varepsilon ^\alpha . \end{aligned}$$

Also by the properties of $J_\varepsilon$

$\blacksquare$

This rate of convergence characterizes $C^\alpha (E)$ in the following sense.

Proposition 18.4c

Let u be a continuous function defined in $\mathbb {R}^N$ and assume that for some fixed $\alpha \in (0,1)$, for all $\varepsilon >0$ there exists $v_\varepsilon \in C^1(\mathbb {R}^N)$ such that

$$\begin{aligned} \Vert u-v_\varepsilon \Vert _\infty \le \gamma \varepsilon ^\alpha ,\quad \text { and }\quad \Vert v_{\varepsilon ,x_j}\Vert _\infty \le \gamma \varepsilon ^{\alpha -1} \end{aligned}$$

(18.2c)

for some fixed constant $\gamma >0$. Then $u\in C^{\alpha }(\mathbb {R}^N)$ and $[u]_{\alpha }\le 3\gamma$.

Proof

For any pair $x,y\in \mathbb {R}^N$ with $|x-y|<\varepsilon$

$\blacksquare$

19c Characterizing Pre-compact Sets in $L^p(E)$

19.1.:

A closed, bounded subset C of $\ell _p$ for $1\le p<\infty$ is compact if and only if for every $\varepsilon >0$, there exists an index $n_\varepsilon$ such that $\sum _{n>n_\varepsilon }|a_n|^p\le \varepsilon$ for all $\mathbf {a}\in C$.

19.2.:

The closed unit ball of $L^p(E)$ is not compact since is not sequentially compact. The same conclusion holds for the unit ball of $\ell _p$ and $C(\bar{E})$.

1.1 19.1c The Helly’s Selection Principle

When $E=(a,b)$ is an open interval and $\{f_n\}\subset BV[a,b]$ then $L^p(E)$-convergence of a subsequence can be replaced with everywhere pointwise convergence, provided $\{f_n\}$ is uniformly bounded. Continue to denote by $\mathcal {V}_f[a,b]$ the variation of f in [a, b].

Proposition 19.1c

(Helly [73]) Let (a, b) be an open interval of $\mathbb {R}$ and let $\{f_n\}$ be a sequence of real valued functions defined in [a, b], such that

$$\begin{aligned} \sup _{[a,b]} |f_n|\le M,\quad \text { and }\quad \mathcal {V}_{f_n}\le M \end{aligned}$$

(19.1c)

for some constant M independent of n. Then, there exists a function $f\in BV[a,b]$, with $\mathcal {V}_f[a,b]\le M$ and a subsequence $\{f_{n^\prime }\}\subset \{f_n\}$ such that $\{f_{n^\prime }\}\rightarrow f$ everywhere in (a, b).

Proof

By the Jordan’s decomposition we may assume that each of the $f_n$ are nondecreasing. Define $f_n$ in the whole $\mathbb {R}$ by extending them to be zero in $\mathbb {R}^N-[a,b]$. Then for $h>0$ however small, compute

$$\begin{aligned} \int _{\mathbb {R}} |f_n(x+h)-f_n(x)|dx&=\mathop {\textstyle {\sum }}\limits _{j\in \mathbb {Z}}\int _0^h |f_n(x+(j+1)h)-f_n(x+jh)|dx\\&=\int _0^h\mathop {\textstyle {\sum }}\limits _{j\in \mathbb {Z}}|f_n(x+(j+1)h)-f_n(x+jh)|dx\\&\le h\mathcal {V}_{f_n}[a,b]\le h\,M \quad \text { for all }\> n\in \mathbb {N}. \end{aligned}$$

Hence $\{f_n\}$ satisfy (19.1) uniformly in n. Therefore there exists $f\in L^1[a,b]$ and a subsequence $\{f_{n_1}\}\subset \{f_n\}$ such that $\{f_{n_1}\}\rightarrow f$ in $L^1[a,b]$. From this a further subsequence $\{f_{n_2}\}\subset \{f_{n_1}\}$ can be selected converging to f a.e. in [a, b]. Since $\{f_n\}$ is equibounded in [a, b] a further subsequence $\{f_{n_3}\}\subset \{f_{n_2}\}$ can be selected converging to f at all rationals of [a, b]. Here f is properly redefined on a set of measure zero. Since the $f_n$ are all nondecreasing, the function f is nondecreasing at the points of convergence. Also by properly redefining f on a set of measure zero we may assume that f is nondecreasing in [a, b]. One also verifies that $\{f_{n_3}\}\rightarrow f$ at all points of continuity of f. Thus $\{f_{n_3}\}$ might fail to converge to f only at the points of discontinuity of f. However f being nondecreasing, it has at most countably many points of discontinuity. Hence a further selection of $\{f_{n^\prime }\}\subset \{f_{n_3}\}$ can be effected such that $\{f_{n^\prime }\}\rightarrow f$ everywhere in [a, b]. $\blacksquare$

20c The Vitali-Saks-Hahn Theorem [59, 138, 170]

Theorem 20.1c

(Vitali-Saks-Hahn [59, 138, 170]) Let $\{X,\mathcal {A},\mu \}$ be a measure space and let $\{\lambda _n\}$ be a sequence of signed measures on $\mathcal {A}$, absolutely continuous with respect to $\mu$, each of finite variation $|\lambda _n|$, and such that

$$\begin{aligned} \lim _n\lambda _n(E)\quad \text { exists for all }\> E\in \mathcal {A}. \end{aligned}$$

Then

$$\begin{aligned} \lim _{\mu (E)\rightarrow 0}\lambda _n(E)=0\quad \text { uniformly in } n. \end{aligned}$$

Proof

Continue to denote by $\{\mathcal {A};d\}$ the collection of equivalence classes of measurable sets at zero mutual distance, endowed with the metric topology, generated by $d(\cdot ;\cdot )$. Having fixed $\varepsilon >0$, consider the collection of equivalence classes

$$\begin{aligned} E_{m,n}=\big \{E\in \mathcal {A}\bigm | |\lambda _n(E)-\lambda _m(E)|\le \varepsilon \big \},\quad \text { for }\> m,n=1,2\dots . \end{aligned}$$

By Lemma 5.1c, $\lambda _m$ and $\lambda _n$ are continuous functions in $\{\mathcal {A};d\}$. Therefore the sets $E_{m,n}$ are closed in the metric topology of $\{\mathcal {A};d\}$. Set

$$\begin{aligned} E_k={\mathop {\textstyle {\bigcap }}\limits }_{m,n\ge k} E_{m,n}. \end{aligned}$$

The assumption implies that every $E\in \mathcal {A}$ belongs to some $E_k$, and hence $\mathcal {A}=\mathop {\textstyle {\bigcup }}\limits E_k$. Since $\{\mathcal {A};d\}$ is a complete metric space, by the Baire category theorem (Theorem 16.1 of Chap. 2), at least one of the $E_k$ has nonempty interior. Thus there exists $k\in \mathbb {N}$, a positive number r, and an equivalence class $A\in E_k$ such that all equivalence classes F in the ball $B_r(A)$ of radius r centered at A

$$\begin{aligned} B_r(A)=\big \{F\in \mathcal {A}\bigm | d(A;F)=\mu (A\varDelta F)<r\big \} \end{aligned}$$

belong to $E_k$. Equivalently

$$\begin{aligned} |\lambda _n(F)-\lambda _n(F)|\le \varepsilon ,\>\text { for all }\> m,n\ge k,\quad \text { and for all }\>F\in B_r(A). \end{aligned}$$

Determine $0<\delta <r$ such that for any $E\in \mathcal {A}$ of $\mu$-measure $\mu (E)<\delta$, there holds

$$\begin{aligned} |\lambda _n(E)|\le \varepsilon ,\quad \text { for }\> n=1,\dots ,k. \end{aligned}$$

Such choice is possible since $\lambda _n$ are absolutely continuous with respect to $\mu$, and k is finite. For any such set E, one verifies that both $A\cup E$, and $A - E$, belong to the ball $B_r(A)$. Then compute

$$\begin{aligned} \lambda _n(E)&=\lambda _k(E)r+ \lambda _n(E)-\lambda _k(E)\\&=\lambda _k(E)+\big [\lambda _n(A\cup E)-\lambda _k(A\cup E)\big ]\\&\qquad \qquad - \big [\lambda _n(A-E)-\lambda _k(A-E)\big ] \end{aligned}$$

Hence $|\lambda _n(E)|\le 3\varepsilon$, for all $n\in \mathbb {N}$.$\blacksquare$

Remark 20.1c

A consequence of the Vitali-Saks-Hahn theorem is that there exists a Lebesgue measurable set $E\subset [0,1]$ such that

$$\begin{aligned} \lim \int _En\chi _{[0,\frac{1}{n}]} dx \quad \text { does not exist}. \end{aligned}$$

Likewise there exists a Lebesgue measurable set $E\subset \mathbb {R}^N$ such that

$$\begin{aligned} \lim \Big (\frac{n}{4\pi }\Big )^{\frac{N}{2}}\int _Ee^{-n\frac{|y|^2}{4}} dy \quad \text { does not exist}. \end{aligned}$$

Construct such sets explicitly.

Corollary 20.1c

Let $\{X,\mathcal {A},\mu \}$ be a finite measure space and let $\{\lambda _n\}$ be a sequence of finite measures on $\mathcal {A}$, absolutely continuous with respect to $\mu$, and such that

$$\begin{aligned} \lambda (E)\buildrel {{{def}}}\over {=}\lim _n\lambda _n(E)\quad \text { exists for all }\> E\in \mathcal {A}. \end{aligned}$$

Then $\lambda (\cdot )$ is a measure on $\mathcal {A}$.

Proof

There is only to prove that $\lambda$ is countably additive. By the Vitali-Saks-Hahn theorem, for all $\varepsilon >0$ there exists $\delta =\delta (\varepsilon )$ independent of n such that

$$\begin{aligned} \lambda _n(E)<\varepsilon \quad \text { for all }\> E\in \mathcal {A}\>\text { such that }\> \mu (E)<\delta , \quad \text { for all }\>n\in \mathbb {N}. \end{aligned}$$

Let $\{E_j\}$ be a countable collection of disjoint, measurable sets. Since $\mu$ is finite, for all $\delta >0$ there exists an index $m=m(\delta )$ such that

$$\begin{aligned} \mu \big (\mathop {\textstyle {\bigcup }}\limits _{j>m} E_j\big )<\delta . \end{aligned}$$

Fix $\varepsilon >0$, determine $\delta$ as claimed by the Vitali-Saks-Hahn theorem, and choose $m=m(\delta )$ accordingly. Then

$$\begin{aligned} \lambda \big (\mathop {\textstyle {\bigcup }}\limits E_j\big )&=\lim _n\lambda _n\big (\mathop {\textstyle {\bigcup }}\limits _{j=1}^m E_j\big ) + \lim _n\big (\mathop {\textstyle {\bigcup }}\limits _{j>m} E_j\big )\\&=\mathop {\textstyle {\sum }}\limits _{j=1}^m \lambda (E_j)+ \lim _n\big (\mathop {\textstyle {\bigcup }}\limits _{j>m} E_j\big ). \end{aligned}$$

Thus

$\blacksquare$

Corollary 20.2c

(Nikodým [117]) Let $\mathcal {A}$ be a $\sigma$-algebra on a set X, and let $\{\lambda _n\}$ be a sequence signed measures on $\mathcal {A}$, each with finite variation $|\lambda _n|$, and such that

$$\begin{aligned} \lambda (E)\buildrel {{{def}}}\over {=}\lim _n\lambda _n(E)\quad \text { exists for all }\> E\in \mathcal {A}. \end{aligned}$$

Then $\lambda (\cdot )$ is countably additive on $\mathcal {A}$.

Proof

For all $E\in \mathcal {A}$ and all $n\in \mathbb {N}$ set

$$\begin{aligned} \mu _n(E)=\frac{|\lambda _n|(E)}{|\lambda _n|(X)} \qquad \text { and }\qquad \mu (E)=\mathop {\textstyle {\sum }}\limits \frac{1}{2^n}\mu _n(E). \end{aligned}$$

One verifies that $\mu$ is a finite measure on $\mathcal {A}$ and $\lambda _n\ll \mu$ for all n. Thus the conclusion follows from the Vitali-Saks-Hahn theorem and Corollary 20.1c.$\blacksquare$

21c Uniformly Integrable Sequences of Functions

The next assertions are a direct consequence of the Vitali-Saks-Hahn Theorem 20.1c. As a consequence give conditions on a sequence of functions $\{f_n\}$ to be uniformly integrable in X, in the sense of § 11c of Chap. 4.

Proposition 21.1c

Let $\{X,\mathcal {A},\mu \}$ be a measure space and let $\{\lambda _n\}$ be a sequence of signed measures on $\mathcal {A}$, absolutely continuous with respect to $\mu$, each of finite variation $|\lambda _n|$, and such that

$$\begin{aligned} \lim _n\lambda _n(E)\quad \text { exists for all }\> E\in \mathcal {A}. \end{aligned}$$

Then for all $\varepsilon >0$ there exists $\delta$ such that

$$\begin{aligned} \mu (E)<\delta \quad \text { implies }\quad |\lambda _n|(E)<\varepsilon \quad \text { uniformly in} \ n. \end{aligned}$$

Proof

If the conclusion does not hold, there exists $\varepsilon >0$ and a sequence $\{E_m\}$ of measurable subsets of X, such that

$$\begin{aligned} \mu (E_m)<\frac{1}{m}\quad \text { and }\quad |\lambda _m(E_m)|>\varepsilon . \end{aligned}$$

For each m fixed

$$\begin{aligned} \varepsilon <|\lambda _m|(E_m)=\lambda _m^+(E_m)+\lambda _m^-(E_m). \end{aligned}$$

Therefore either

$$\begin{aligned} \lambda _m^+(E_m)> {\textstyle \frac{1}{2}}\varepsilon ,\quad \text { or }\quad \lambda _m^-(E_m){\textstyle \frac{1}{2}}\varepsilon . \end{aligned}$$

Let $X_m^+\cup X_m^-$ be the Hanh’s decomposition of X induced by $\lambda$. By replacing $E_m$ with either $E_m\cap X_m^+$ or $E_m\cap X_m^-$, the sets $\{E_m\}$ can be chosen so that

$$\begin{aligned} \mu (E_m)<\frac{1}{m}\qquad \text { and }\qquad \lambda _m(E_m)>{\textstyle \frac{1}{2}}\varepsilon . \end{aligned}$$

This contradicts the conclusion of the Vitali-Hahn-Saks theorem and establishes the proposition.$\blacksquare$

Corollary 21.1c

Let $\{X,\mathcal {A},\mu \}$ be a measure space and $\{f_n\}$ be a sequence of integrable functions in X such that

$$\begin{aligned} \Big \{\int _E f_n d\mu \Big \}\quad \text { is a Cauchy sequence in } \mathbb {R}\end{aligned}$$

for all $E\in \mathcal {A}$. Then $\{f_n\}$ is uniformly integrable in X , in the sense of § 11c of Chap. 4.

Proof

Since $f_n\in L^1(X)$, setting

$$\begin{aligned} \mathcal {A}\ni E\longrightarrow \lambda _n(E)=\int _Ef_nd\mu . \end{aligned}$$

defines signed measures $\lambda _n$ on $\mathcal {A}$, absolutely continuous with respect to $\mu$, and each of finite variation $|\lambda _n|$. By the assumptions $\{\lambda _n(E)\}$ has a limit for all $E\in \mathcal {A}$. Therefore by the Vitali-Saks-Hahn theorem

$$\begin{aligned} \lim _{\mu (E)\rightarrow 0}\Big |\int _Ef_n d\mu \Big |=0\quad \text { uniformly in } n. \end{aligned}$$

If $\{f_n\}$ is not uniformly integrable, there exists $\varepsilon >0$ and a sequence $\{E_m\}$ of measurable subsets of X, such that

$$\begin{aligned} \mu (E_m)<\frac{1}{m}\quad \text { and }\quad \int _{E_m}|f_m|d\mu >\varepsilon . \end{aligned}$$

For each m fixed

$$\begin{aligned} \varepsilon <\int _{E_m}f_m^+d\mu +\int _{E_m} f^-d\mu . \end{aligned}$$

Therefore either

$$\begin{aligned} \int _{E_m}f_m^+d\mu> {\textstyle \frac{1}{2}}\varepsilon ,\quad \text { or }\quad \int _{E_m} f_m^-d\mu >{\textstyle \frac{1}{2}}\varepsilon . \end{aligned}$$

Thus by replacing $E_m$ with either $E_m\cap [f_m>0]$ or $E_m\cap [f_m\le 0]$, the sets $\{E_m\}$ can be chosen so that

$\blacksquare$

Corollary 21.2c

Let $\{X,\mathcal {A},\mu \}$ be a measure space and $\{f_n\}$ be a sequence of integrable functions in X weakly convergent in $L^1(X)$ to some $f\in L^1(X)$. Then $\{f_n\}$ is uniformly integrable in X, in the sense of § 11c of Chap. 4.

22c Relating Weak and Strong Convergence and Convergence in Measure

The previous statements permit one to give necessary and sufficient conditions for weak convergence to imply strong convergence.

Proposition 22.1c

Let $\{X,\mathcal {A},\mu \}$ be a finite measure space and let $\{f_n\}$ be a sequence of integrable functions in X. Then $\{f_n\}$ converges strongly in $L^1(X)$ if and only if $\{f_n\}\rightarrow f$ weakly and in measure.

Proof

Strong convergence implies weak convergence and convergence in measure. To prove the converse, fix $\varepsilon >0$ and determine $\delta >0$ such that

$$\begin{aligned} \mu (E)<\delta \quad \text { implies }\quad \int _E|f_n-f|d\mu <\varepsilon \quad \text { uniformly in }\>n. \end{aligned}$$

This is possible, since by Corollary 21.2c, $\{f_n\}$ is uniformly integrable. Since $\{f_n\}\rightarrow f$ in measure, there exists $n_\varepsilon$ such that

$$\begin{aligned} \mu (|f_n-f|>\varepsilon ]\big )<\varepsilon \qquad \text { for all } \quad n>n_\varepsilon . \end{aligned}$$

Then since $\mu$ is finite

$\blacksquare$

The next proposition extends Proposition 22.1c to $\sigma$-finite measure spaces.

Proposition 22.2c

Let $\{X,\mathcal {A},\mu \}$ be a $\sigma$-finite measure space and let $\{f_n\}$ be a sequence of integrable functions in X converging weakly in $L^1(X)$ to some $f\in L^1(X)$. Then $\{f_n\}$ converges strongly in $L^1(X)$ if and only if $\{f_n\}\rightarrow f$ in measure, on every subset of X of finite measure.

Proof

By replacing $f_n$ with $f_n-f$, one may assume that $f=0$. For a measurable set $E\subset X$ and $n\in \mathbb {N}$ set

$$\begin{aligned} \lambda _n(E)=\int _Ef_n d\mu ;\quad \nu _n(E) =\frac{\int _E|f_n|d\mu }{\int _X|f_n|d\mu };\qquad \nu (E)=\mathop {\textstyle {\sum }}\limits \frac{1}{2^n}\nu _n(E). \end{aligned}$$

One verifies that $\nu$ is a finite measure on $\mathcal {A}$ and that $\lambda _n\ll \nu$ for all n. Moreover, since $\{f_n\}\rightarrow 0$ weakly in $L^1(X)$, the $\lim \lambda _n(E)$ exists for all $E\in \mathcal {A}$. Thus, by Proposition 22.1c, for all $\varepsilon >0$ there exists $\delta$ such that

$$\begin{aligned} \nu (E)<\delta \quad \text { implies }\quad \int _E|f_n| d\mu \Big |<\varepsilon \quad \text { uniformly in }\>n. \end{aligned}$$

Having fixed $\varepsilon$ and the corresponding $\delta$, there is $n=n_\delta$ such that

$$\begin{aligned} \mathop {\textstyle {\sum }}\limits _{j>n_\delta }\frac{1}{2^j}\nu _j(E)<\frac{1}{2}\delta \qquad \text { for all }\> E\in \mathcal {A}. \end{aligned}$$

Since $\{X,\mathcal {A},\mu \}$ is $\sigma$-finite, there exists a countable collection of measurable sets $\{E_m\}$ of finite $\mu$-measure, such that $E_m\subset E_{m+1}$, and $X=\mathop {\textstyle {\bigcup }}\limits E_m$. Since $f_j\in L^1(X)$, there exists $m=m(n_\delta )$ such that

$$\begin{aligned} \int _{X-E_{m(n_\delta )}}|f_j|d\mu <\frac{1}{2}\delta \qquad \text { for }\>j=1,\dots , n_\delta . \end{aligned}$$

Therefore, by the definition of $\nu$, the set $X-E_{m(n_\delta )}$ satisfies

$$\begin{aligned} \nu (X-E_{m(n_\delta )})<\delta \end{aligned}$$

and hence,

$$\begin{aligned} \int _{X-E_{m(n_\delta )}} |f_n| d\mu <\varepsilon \qquad \text { uniformly in }\>n. \end{aligned}$$

Then

$$\begin{aligned} \lim \int _X|f_n|d\mu&=\lim \int _{X-E_{m(n_\delta )}} |f_n|d\mu +\lim \int _{E_{m(n_\delta )}} |f_n|d\mu \\&<\varepsilon + \lim \int _{E_{m(n_\delta )}} |f_n|d\mu . \end{aligned}$$

Since $E_{m(n_\delta )}$ is of finite $\mu$-measure, the last limit is zero (Proposition 22.1c).$\blacksquare$

Corollary 22.1c

In $\ell _1$ weak and strong convergence coincide.

23c An Independent Proof of Corollary 22.1c

Proposition 23.1c

A sequence $\{\mathbf {x}_n\}\subset \ell _1$ converges weakly to some $\mathbf {x}\in \ell _1$ if and only if $\Vert \mathbf {x}_n-\mathbf {x}\Vert _1\rightarrow 0$ as $n\rightarrow \infty$.

Proof

Strong convergence implies weak convergence. To show the converse assume $\mathbf {x}=\mathbf {0}$. Thus the assumption is

$$\begin{aligned} \lim \langle \mathbf {x}_n,\mathbf {y}\rangle =\lim \mathop {\textstyle {\sum }}\limits x_{j,n}y_j=0\quad \text { for all }\> \mathbf {y}\in \ell _{\infty }. \end{aligned}$$

(23.1c)

Choosing $\mathbf {y}$ as the base elements of $\ell _\infty$, gives

$$\begin{aligned} \lim _n x_{j,n}=0\quad \text { for all }\> j\in \mathbb {N}. \end{aligned}$$

(23.2c)

If $\limsup \Vert \mathbf {x}_n\Vert _1>0$, there exists $\varepsilon >0$ and a subsequence $\{\mathbf {x}_{n^\prime }\}\subset \{\mathbf {x_n}\}$ such that

$$\begin{aligned} \Vert \mathbf {x}_{n^\prime }\Vert _1>\varepsilon \quad \text { and }\quad \lim \langle \mathbf {x}_{n^{\prime \prime }},\mathbf {y}\rangle =0 \end{aligned}$$

(23.3c)

for all subsequences $\{\mathbf {x}_{n^{\prime \prime }}\}\subset \{\mathbf {x}_{n^\prime }\}$ and all $\mathbf {y}\in \ell _\infty$. The proof consists of extracting a subsequence $\{\mathbf {x}_{n^{\prime \prime }}\}\subset \{\mathbf {x}_{n^\prime }\}$ and an element $\mathbf {y}\in \ell _\infty$ fow which that last statement fails to hold.

Fix the index $m_1=n^\prime _1$ and consider the sequence $\mathbf {x}_{m_1}=\{x_{j,m_1}\}$. Since $\mathbf {x}_{m_1}\in \ell _1$, there exists an index $j_{m_1}$ such that

$$\begin{aligned} \mathop {\textstyle {\sum }}\limits _{j>j_{m_1}} |x_{j,m_1}|<{\textstyle \frac{1}{4}}\varepsilon . \end{aligned}$$

Then choose

$$\begin{aligned} y_j=\text {sign }\{x_{j,m_1}\},\quad \text { for }\> j=1,\dots ,j_{m_1}. \end{aligned}$$

For such a choice, and in view of the first of (23.3c),

$$\begin{aligned} \Big |\mathop {\textstyle {\sum }}\limits _{j=1}^{j_{m_1}}y_j x_{j,m_1}\Big |>\Vert \mathbf {x}_{m_1}\Vert _1- \mathop {\textstyle {\sum }}\limits _{j>j_{m_1}} |x_{j,m_1}| >{\textstyle \frac{3}{4}}\varepsilon . \end{aligned}$$

The index $j_{m_1}$ being fixed, by the pointwise convergence in (23.2c), there exists an integer $m_1<m_2\in \{n^\prime \}$ such that

$$\begin{aligned} \Big |\mathop {\textstyle {\sum }}\limits _{j=1}^{j_{m_1}} y_j x_{j,m_2}\Big |<{\textstyle \frac{1}{8}}\varepsilon . \end{aligned}$$

Since $\mathbf {x}_{m_2}\in \ell _1$, there is an index $j_{m_2}$, such that

$$\begin{aligned} \mathop {\textstyle {\sum }}\limits _{j>j_{m_2}} |x_{j,m_2}|<{\textstyle \frac{1}{8}}\varepsilon . \end{aligned}$$

Without loss of generality we may take $j_{m_2}>j_{m_1}+1$ and set

$$\begin{aligned} y_j=\text {sign }\{x_{j,m_2}\},\quad \text { for }\> j=j_{m_1}+1,\dots ,j_{m_2}. \end{aligned}$$

Taking into account the first of (23.3c) the element $\mathbf {x}_{m_2}$ satisfies

$$\begin{aligned} \Big |\mathop {\textstyle {\sum }}\limits _{j=1}^{j_{m_1}}y_j x_{j,m_2}\Big |<{\textstyle \frac{1}{8}}\varepsilon ;\quad \mathop {\textstyle {\sum }}\limits _{j_{m_1}+1}^{j_{m_2}}y_j x_{j,m_2}>{\textstyle \frac{3}{4}}\varepsilon ;\quad \mathop {\textstyle {\sum }}\limits _{j>j_{m_2}} |x_{j,m_2}|<{\textstyle \frac{1}{8}}\varepsilon . \end{aligned}$$

(23.4c)

Proceeding by induction, assume that for a positive integer $s\ge 2$, an element $\mathbf {x}_{m_s}\in \{\mathbf {x}_{n^\prime }\}$, an index $j_{m_s}$ and numbers $y_j$ for $j=1,\dots j_{m_s}$, have been selected satisfying

$$\begin{aligned} \Big |\mathop {\textstyle {\sum }}\limits _{j=1}^{j_{m_{s-1}}}y_j x_{j,m_s}\Big |<{\textstyle \frac{1}{8}}\varepsilon ;\quad \mathop {\textstyle {\sum }}\limits _{j_{m_{s-1}}+1}^{j_{m_s}}y_j x_{j,m_s}>{\textstyle \frac{3}{4}}\varepsilon ;\quad \mathop {\textstyle {\sum }}\limits _{j>j_{m_s}} |x_{j,m_s}|<{\textstyle \frac{1}{8}}\varepsilon . \qquad {(23.4\mathrm{c})_{s}} \end{aligned}$$

An element $\mathbf {x}_{m_{s+1}}\in \{\mathbf {x}_{n^\prime }\}$, an index $j_{m_{s+1}}$ and numbers $y_j$ for $j=j_{m_s}+1,\dots , j_{m_{s+1}}$ are constructed by first choosing $m_{s+1}\in \{n^\prime \}$ so that

$$\begin{aligned} \Big |\mathop {\textstyle {\sum }}\limits _{j=1}^{j_{m_s}}y_j x_{j,m_{s+1}}\Big |<{\textstyle \frac{1}{8}}\varepsilon . \end{aligned}$$

Such a choice is possible in view of the pointwise convergence in (23.2c). Then choose $j_{m_{s+1}}$ so that

$$\begin{aligned} \mathop {\textstyle {\sum }}\limits _{j>j_{m_{s+1}}} |x_{j,m_{s+1}}|<{\textstyle \frac{1}{8}}\varepsilon . \end{aligned}$$

Such a choice is possible since $\mathbf {x}_{m_{s+1}}\in \ell _1$. Then choose

$$\begin{aligned} y_j=\text {sign }\{x_{j,m_{s+1}}\},\quad \text { for }\> j=j_{m_s}+1, \dots ,j_{m_{s+1}}. \end{aligned}$$

Taking into account the first of (23.3c) one verifies that $\mathbf {x}_{m_{s+1}}$ satisfies (23.4c)$_{s+1}$. This procedure identifies a subsequence $\{\mathbf {x}_{m_s}\}\subset \{\mathbf {x}_{n^\prime }\}$, and an element $\mathbf {y}\in \ell _\infty$, of norm 1, such that

$\blacksquare$