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The Ramanujan Journal

, Volume 48, Issue 3, pp 685–690

# Correction to: Almost periodic functions in terms of Bohr’s equivalence relation

• J. M. Sepulcre
• T. Vidal
Correction
• 375 Downloads

## Correction to: Ramanujan J (2018) 46:245–267  https://doi.org/10.1007/s11139-017-9950-1

The authors wish to draw the attention to a mistake which appears in the proof of Proposition 3 of the above quoted paper [4]. Specifically, after label (17), the statement “we can also construct a vector $$\mathbf {x}_0\in \mathbb {R}^{\sharp G_{\varLambda }}$$ such that, by taking into account Remarks 1 and 2, verifies
\begin{aligned} a_{j}=a_{1,j}e^{i\langle \mathbf {r}_j,\mathbf {x}_0\rangle },\ j=1,2\ldots , \text{ with } \lambda _j\in \varLambda \text {''} \end{aligned}
is correct when it is possible to obtain an integral basis for the set of exponents or frequencies $$\varLambda$$, but it is not correct in the general case. We next define the concept of integral basis.
Let $$G_{\varLambda }=\{g_1, g_2,\ldots , g_k,\ldots \}$$ be a basis of the $$\mathbb {Q}$$-vector space generated by a set $$\varLambda =\{\lambda _1,\lambda _2,\ldots \}$$ of exponents (by abuse of notation, it is said that $$G_{\varLambda }$$ is a basis for $$\varLambda$$), which implies that $$G_{\varLambda }$$ is linearly independent over the rational numbers and each $$\lambda _j$$ is expressible as a finite linear combination of terms of $$G_{\varLambda }$$, say
\begin{aligned} \lambda _j=\sum _{k=1}^{i_j}r_{j,k}g_k\ \quad \text{ for } \text{ some } r_{j,k}\in \mathbb {Q},\ i_j\in \mathbb {N}. \end{aligned}
(2)
We will say that $$G_{\varLambda }$$ is an integral basis for $$\varLambda$$ when $$r_{j,k}\in \mathbb {Z}$$ for any jk.

By taking into account this situation, in order to maintain the validity of Proposition 3 (not only for the case when it is possible to obtain an integral basis for the set of exponents) and some other subsequent results of our paper, the equivalence relation which is inspired by that of Bohr is now revised to adapt correctly the situation in the general case. In this way, Definition 3 of the above quoted paper is now modified in the following terms.

### Definition 3′

Given $$\varLambda =\{\lambda _1,\lambda _2,\ldots ,\lambda _j,\ldots \}$$ a set of exponents, consider $$A_1(p)$$ and $$A_2(p)$$ two exponential sums in the class $$\mathcal {S}_{\varLambda }$$, say $$A_1(p)=\sum _{j\ge 1}a_je^{\lambda _jp}$$ and $$A_2(p)=\sum _{j\ge 1}b_je^{\lambda _jp}.$$ We will say that $$A_1$$ is equivalent to $$A_2$$ if for each integer value $$n\ge 1$$, with $$n\le \sharp \varLambda$$, it is satisfied $$a_{n}^*\sim b_{n}^*$$, where $$a_{n}^*,b_{n}^*:\{\lambda _1,\lambda _2,\ldots ,\lambda _{n}\}\rightarrow \mathbb {C}$$ are the functions given by $$a_{n}^*(\lambda _j):=a_j$$ y $$b_{n}^*(\lambda _j):=b_j$$, $$j=1,2,\ldots ,n$$ and $$\sim$$ is in Definition 1.

It is clear that the relation defined in the foregoing definition is an equivalence relation. That is, it is reflective, symmetric and transitive. By abuse of notation, we will use $$\sim$$ for both equivalence relations introduced in Definitions 1 and 3$$^\prime$$.

Analogously, Definition 5 must be rewritten in the following terms.

### Definition 5′

Given $$\varLambda =\{\lambda _1,\lambda _2,\ldots ,\lambda _j,\ldots \}$$ a set of exponents, let $$f_1$$ and $$f_2$$ denote two equivalence classes of $$B(\mathbb {R},\mathbb {C})/\simeq$$ (resp. $$B(U,\mathbb {C})/\simeq$$) whose associated Fourier series (resp. Dirichlet series) are given by
\begin{aligned}&\sum _{j\ge 1}a_je^{i\lambda _jt}\ \text{ and }\ \sum _{j\ge 1}b_je^{i\lambda _jt},\ a_j,b_j\in \mathbb {C},\ \lambda _j\in \varLambda .\\&\left( \text{ resp. } \sum _{j\ge 1}a_je^{\lambda _js}\ \text{ and }\ \sum _{j\ge 1}b_je^{\lambda _js},\ a_j,b_j\in \mathbb {C},\ \lambda _j\in \varLambda .\right) \end{aligned}
We will say that $$f_1$$ is equivalent to $$f_2$$ if for each integer value $$n\ge 1$$, with $$n\le \sharp \varLambda$$, it is satisfied $$a_n^*\sim b_n^*$$, where $$a_n^*,b_n^*:\{\lambda _1,\ldots ,\lambda _n\}\rightarrow \mathbb {C}$$ are the functions given by $$a_n^*(\lambda _j):=a_j$$ and $$b_n^*(\lambda _j):=b_j$$, $$j=1,2,\ldots ,n$$, and $$\sim$$ is in Definition 1.

Under the modification of Definition 3, Proposition 1 must now be rewritten in the following way.

### Proposition 1′

Given $$\varLambda =\{\lambda _1,\lambda _2,\ldots ,\lambda _j,\ldots \}$$ a set of exponents, consider $$A_1(p)$$ and $$A_2(p)$$ two exponential sums in the class $$\mathcal {S}_{\varLambda }$$, say $$A_1(p)=\sum _{j\ge 1}a_je^{\lambda _jp}$$ and $$A_2(p)=\sum _{j\ge 1}b_je^{\lambda _jp}.$$ Fixed a basis $$G_{\varLambda }$$ for $$\varLambda$$, for each $$j\ge 1$$ let $$\mathbf {r}_j\in \mathbb {R}^{\sharp G_{\varLambda }}$$ be the vector of rational components verifying (2). Then $$A_1\sim A_2$$ if and only if for any integer value $$n\ge 1$$, with $$n\le \sharp \varLambda$$, there exists $$\mathbf {x}_n=(x_{n,1},x_{n,2},\ldots ,x_{n,k},\ldots )\in \mathbb {R}^{\sharp G_{\varLambda }}$$ such that $$b_j=a_j e^{\langle \mathbf {r}_j,\mathbf {x}_n\rangle i}$$ for $$j=1,2,\ldots ,n$$. Furthermore, if $$G_{\varLambda }$$ is an integral basis for $$\varLambda$$ then $$A_1\sim A_2$$ if and only if there exists $$\mathbf {x}_0=(x_{0,1},x_{0,2},\ldots ,x_{0,k},\ldots )\in \mathbb {R}^{\sharp G_{\varLambda }}$$ such that $$b_j=a_j e^{\langle \mathbf {r}_j,\mathbf {x}_0\rangle i}$$ for every $$j\ge 1$$.

### Proof

For each integer value $$n\ge 1$$, let $$V_n$$ be the $$\mathbb {Q}$$-vector space generated by $$\{\lambda _1,\ldots ,\lambda _n\}$$, V the $$\mathbb {Q}$$-vector space generated by $$\varLambda$$, and $$G_{\varLambda }=\{g_1, g_2,\ldots , g_k,\ldots \}$$ a basis of V. If $$A_1\sim A_2$$, by Definition 3$$^\prime$$ for any integer value $$n\ge 1$$, with $$n\le \sharp \varLambda$$, there exists a $$\mathbb {Q}$$-linear map $$\psi _n:V_n\rightarrow \mathbb {R}$$ such that
\begin{aligned} b_j=a_je^{i\psi _n(\lambda _j)},\ j=1,2\ldots ,n. \end{aligned}
Hence $$b_j=a_je^{i\sum _{k=1}^{i_j}r_{j,k}\psi _n(g_k)},\quad \ j=1,2\ldots ,n$$ or, equivalently,
\begin{aligned} b_j=a_j e^{i\langle \mathbf {r}_j,\mathbf {x}_n\rangle },\quad \ j=1,2\ldots ,n, \end{aligned}
with $$\mathbf {x}_n:=(\psi _n(g_1),\psi _n(g_2),\ldots )$$. Conversely, suppose the existence, for any integer value $$n\ge 1$$ with $$n\le \sharp \varLambda$$, of a vector $$\mathbf {x}_n=(x_{n,1},x_{n,2},\ldots ,x_{n,k},\ldots )\in \mathbb {R}^{\sharp G_{\varLambda }}$$ such that $$b_j=a_j e^{\langle \mathbf {r}_j,\mathbf {x}_n\rangle i}$$, $$j=1,2\ldots ,n$$. Thus a $$\mathbb {Q}$$-linear map $$\psi _n:V_n\rightarrow \mathbb {R}$$ can be defined from $$\psi _n(g_k):=x_{n,k}$$, $$k\ge 1$$. Therefore
\begin{aligned} \psi _n(\lambda _j)=\sum _{k=1}^{i_j}r_{j,k}\psi (g_k)=\langle \mathbf {r}_j,\mathbf {x}_n\rangle ,\ j=1,2\ldots ,n, \end{aligned}
and the result follows.
Now, suppose that $$G_{\varLambda }$$ is an integral basis for $$\varLambda$$ and $$A_1\sim A_2$$. Thus, by above, for each fixed integer value $$n\ge 1$$, let $$\mathbf {x}_n=(x_{n,1},x_{n,2},\ldots )\in \mathbb {R}^{\sharp G_{\varLambda }}$$ be a vector such that
\begin{aligned} b_j=a_j e^{i\langle \mathbf {r}_j,\mathbf {x}_n\rangle },\ j=1,2\ldots ,n. \end{aligned}
Since each component of $$\mathbf {r}_j$$ is an integer number, without loss of generality, we can take $$\mathbf {x}_n\in [0,2\pi )^{\sharp G_{\varLambda }}$$ as the unique vector in $$[0,2\pi )^{\sharp G_{\varLambda }}$$ satisfying the above equalities, where we assume $$x_{n,k}=0$$ for any k such that $$r_{j,k}=0$$ for $$j=1,\ldots ,n$$. Therefore, under this assumption, if $$m>n$$ then $$x_{m,k}=x_{n,k}$$ for any k so that $$x_{n,k}\ne 0$$. In this way, we can construct a vector $$\mathbf {x}_0=(x_{0,1},x_{0,2},\ldots ,x_{0,k},\ldots )\in [0,2\pi )^{\sharp G_{\varLambda }}$$ such that $$b_j=a_j e^{\langle \mathbf {r}_j,\mathbf {x}_0\rangle i}$$ for every $$j\ge 1$$. Indeed, if $$r_{1,k}\ne 0$$ then the component $$x_{0,k}$$ is chosen as $$x_{1,k}$$, and if $$r_{1,k}=0$$ then each component $$x_{0,k}$$ is defined as $$x_{n+1,k}$$ where $$r_{j,k}=0$$ for $$j=1,\ldots ,n$$ and $$r_{n+1,k}\ne 0$$. Conversely, if there exists $$\mathbf {x}_0=(x_{0,1},x_{0,2},\ldots ,x_{0,k},\ldots )\in \mathbb {R}^{\sharp G_{\varLambda }}$$ such that $$b_j=a_j e^{\langle \mathbf {r}_j,\mathbf {x}_0\rangle i}$$ for every $$j\ge 1$$, then it is clear that $$A_1\sim A_2$$ under Definition 3$$^\prime$$. $$\square$$
From Proposition 1$$^\prime$$, it is worth noting that old Definition 3 and Definition 3$$^\prime$$ are equivalent in the case that it is possible to obtain an integral basis for the set of exponents $$\varLambda$$. Consequently, all the results of the quoted above paper which can me formulated in terms of an integral basis are also valid under old Definition 3$$^\prime$$ (in particular, the mistake above does not concern the results on the finite exponential sums in Section 3 and the Riemann zeta function in Section 5). Moreover, in this case, the set of all exponential sums A(p) in an equivalence class $$\mathcal {G}$$ in $$\mathcal {S}_{\varLambda }/\sim$$ can be determined by a function $$E_{\mathcal {G}}:\mathbb {R}^{\sharp G_{\varLambda }}\rightarrow \mathcal {S}_{\varLambda }$$ of the form
\begin{aligned} E_{\mathcal {G}}(\mathbf {x}):=\sum _{j\ge 1}a_je^{\langle \mathbf {r}_j,\mathbf {x}\rangle i}e^{\lambda _jp}\text {, } \mathbf {x}=(x_1,x_2,\ldots ,x_k,\ldots )\in \mathbb {R}^{\sharp G_{\varLambda }}, \end{aligned}
where $$a_1,a_2,\ldots ,a_j,\ldots$$ are the coefficients of an exponential sum in $$\mathcal {G}$$ and the $$\mathbf {r}_j$$’s are the vectors associated with a prefixed integral basis $$G_{\varLambda }$$ for $$\varLambda$$.
However, both Definitions 3 and 3$$^\prime$$ are not equivalent in the general case. For example, consider the set of exponents $$\varLambda _0=\{\lambda _1,\lambda _2,\ldots ,\lambda _j,\ldots \}$$ given by
\begin{aligned} \lambda _j=2j-1+\frac{1}{2(2j-1)}\in \mathbb {Q},\ j=1,2,\ldots , \end{aligned}
then $$G_{\varLambda _0}=\{1\}$$ is a basis for $$\varLambda _0$$ but it is not an integral basis for $$\varLambda _0$$. In fact, it is not possible to get an integral basis for $$\varLambda _0$$. Now, consider the exponential sums $$B_1(p) = \sum _{j\ge 1} e^{-\lambda _j p}$$ and $$B_2(p)=-B_1(p)=\sum _{j\ge 1} -e^{-\lambda _j p}$$. Then $$B_1(p)$$ and $$B_2(p)$$ are equivalent according to Definition 3$$^\prime$$. Indeed, take $$a_j=1$$ and $$b_j=-1$$ for each $$j=1,2,\ldots$$, $$G_{\varLambda _0}=\{1\}$$, $$r_j=\lambda _j$$ for each $$j=1,2,\ldots$$ and $$x_n=2\pi \prod _{k=1}^n (2k-1)$$, $$n=1,2,\ldots$$. Fixed $$n\ge 1$$, then it is clear that $$b_j=a_j e^{\langle r_j,x_n \rangle i}$$ for every $$j=1,2,\ldots ,n$$, i.e.
\begin{aligned} -1=e^{i\left( 2j-1+\frac{1}{2(2j-1)}\right) 2\pi \prod _{k=1}^n(2k-1)},\quad \ j=1,2,\ldots ,n. \end{aligned}
Nevertheless, as it is deduced from Bohr’s paper [2], $$B_1(p)$$ and $$B_2(p)$$ are not equivalent according to old Definition 3.
Now, mutatis mutandi, the proofs of Lemma 1, Lemma 2 and Proposition 3 can be rewritten. Specifically, the modifications are as follows.
• In the proof of Lemma 1: “taking $$\mathbf {x}_0=\tau \mathbf {g}$$” must be changed by “taking $$\mathbf {x}_n=\tau \mathbf {g}$$ for any integer value $$n\ge 1$$”.

• In the proof of Lemma 2: “there exists $$\mathbf {x}_0\in \mathbb {R}^{\sharp \varLambda }$$ such that $$b_j=a_j e^{\langle \mathbf {r}_j,\mathbf {x}_0\rangle i}$$ for each $$j\ge 1$$” must be changed by “for any integer value $$n\ge 1$$, with $$n\le \sharp \varLambda$$, there exists $$\mathbf {x}_n\in \mathbb {R}^{\sharp \varLambda }$$ such that $$b_j=a_j e^{\langle \mathbf {r}_j,\mathbf {x}_n\rangle i}$$ for each $$j=1,\ldots ,n$$”.

• In the proof of Lemma 2: “define the sequence of trigonometric polynomials
\begin{aligned} Q_k(t):=\sum _{j\ge 1}p_{j,k}a_je^{\langle \mathbf {r}_j,\mathbf {x}_0\rangle i}e^{i\lambda _jt},\quad \ k=1,2,\ldots \hbox {''} \end{aligned}
must be changed by “define the sequence of trigonometric polynomials
\begin{aligned} Q_k(t):=\sum _{j\ge 1}p_{j,k}b_je^{i\lambda _jt},\quad \ k=1,2,\ldots , \end{aligned}
where, fixed k, we can take $$b_j=a_je^{\langle \mathbf {r}_j,\mathbf {x}_{n_k}\rangle i}$$ with $$n_k$$ the greatest integer value j such that $$p_{j,k}\ne 0$$.”
• In the proof of Lemma 2: “However, note that
\begin{aligned} M\{|Q_{k_1}(t)-Q_{k_2}(t)|^2\}= & {} \displaystyle \sum _{j\ge 1}(p_{j,k_1}-p_{j,k_2})^2\left| e^{\langle \mathbf {r}_j,\mathbf {x}_0\rangle i}\right| ^2|a_j|^2\\= & {} M\{|P_{k_1}(t)-P_{k_2}(t)|^2\}\},\hbox {''} \end{aligned}
must be changed by “However, note that
\begin{aligned} M\{|Q_{k_1}(t)-Q_{k_2}(t)|^2\}&=\displaystyle \sum _{j\ge 1}(p_{j,k_1}-p_{j,k_2})^2\left| e^{\langle \mathbf {r}_j,\mathbf {x}_{n_{k_2}}\rangle i} \right| ^2|a_j|^2\\&=M\{|P_{k_1}(t)-P_{k_2}(t)|^2\}, \end{aligned}
where we suppose that $$n_{k_2}\ge n_{k_1}$$.”
• In the proof of Lemma 2: “we have $$f_2(t)\in AP(\mathbb {R},\mathbb {C})$$. Moreover, $$\{Q_k(t)\}_{k\ge 1}$$ also converges formally to the series $$\sum _{j\ge 1}a_j e^{\langle \mathbf {r}_j,\mathbf {x}_0\rangle i}e^{i\lambda _jt},$$ which, by [1, p. 21], represents the Fourier series of $$f_2(t)$$. Finally, by taking into account Definition 5 (in terms of Proposition 1) we have $$f_1\sim f_2$$.” must be changed by “we have $$f_2(t)\in AP(\mathbb {R},\mathbb {C})$$ and, by [1, p. 21], $$\sum _{j\ge 1}b_je^{i\lambda _jt}$$ represents its Fourier series. Finally, by taking into account Definition 5 (in terms of Proposition 1$$^\prime$$) we have $$f_1\sim f_2$$.”

• In the proof of Proposition 3: “Since $$f_1\sim f_l$$ for each $$l=1,2,\ldots$$, we deduce from Proposition 1 that there exists $$\mathbf {x}_l=(x_{l,1},x_{l,2},\ldots )\in \mathbb {R}^{\sharp G_\varLambda }$$ such that
\begin{aligned} a_{l,j}=a_{1,j}e^{i\langle \mathbf {r}_j,\mathbf {x}_l\rangle },\ j=1,2\ldots ,\ \text{ with } \lambda _j\in \varLambda .\hbox {''} \end{aligned}
(15)
must be changed by “Since $$f_1\sim f_l$$ for each $$l=1,2,\ldots$$, we deduce from Proposition 1$$^\prime$$ that for any integer value $$n\ge 1$$, with $$n\le \sharp \varLambda$$, there exists $$\mathbf {x}_{l,n}=(x_{l,n,1},x_{l,n,2},\ldots )\in \mathbb {R}^{\sharp G_\varLambda }$$ such that
\begin{aligned} a_{l,j}=a_{1,j}e^{i\langle \mathbf {r}_j,\mathbf {x}_{l,n}\rangle },\ j=1,2\ldots ,n\ \text{ with } \lambda _j\in \varLambda .\hbox {''} \end{aligned}
(15)
• In the proof of Proposition 3: “and, since (16) is satisfied for any $$k=1,2,\ldots$$, we can also construct a vector $$\mathbf {x}_0\in \mathbb {R}^{\sharp G_{\varLambda }}$$ such that, by taking into account remarks 1 and 2, verifies
\begin{aligned} a_{j}=a_{1,j}e^{i\langle \mathbf {r}_j,\mathbf {x}_0\rangle },\ j=1,2\ldots , \text{ with } \lambda _j\in \varLambda .\hbox {''} \end{aligned}
must be changed by “and, since (16) is satisfied for any $$k=1,2,\ldots$$, we can construct, for any integer value $$n\ge 1$$ with $$n\le \sharp \varLambda$$, a vector $$\mathbf {x}_{0,n}\in \mathbb {R}^{\sharp G_{\varLambda }}$$ such that, by taking into account remarks 1 and 2, verifies
\begin{aligned} a_{j}=a_{1,j}e^{i\langle \mathbf {r}_j,\mathbf {x}_{0,n}\rangle },\ j=1,2\ldots ,n \text{ with } \lambda _j\in \varLambda .\hbox {''} \end{aligned}
The rest of the article remains unchanged.

Finally, the authors wish to thank Mattia Righetti for his quote to Bohr’s example given by the set of exponents $$\varLambda _0$$ considered above [2, 3].

## References

1. 1.
Besicovitch, A.S.: Almost Periodic Functions. Dover, New York (1954)Google Scholar
2. 2.
Bohr, H.: Zur Theorie der allgemeinen Dirichletschen Reihen. Math. Ann. (German) 79, 136–156 (1918)
3. 3.
Righetti, M.: On Bohr’s equivalence theorem. J. Math. Anal. Appl. (Corrigendum Ibid.) 449, 939–940 (2017)
4. 4.
Sepulcre, J.M., Vidal, T.: Almost periodic functions in terms of Bohr’s equivalence relation. Ramanujan J. 46(1), 245–267 (2018)

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## Authors and Affiliations

1. 1.Department of MathematicsUniversity of AlicanteAlicanteSpain