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A recently asked question (linked here) deals with the remarkable identity $$ \sum_{n\in\mathbb Z} \mathrm{sinc}(n+x)= \pi,\quad x\in\mathbb R, $$ where $\mathrm{sinc}(x)=\sin(x)/x$.

It is easy to construct functions $f$ other than $\mathrm{sinc}(x)$ such that $\sum_{n\in\mathbb Z} f(n+x)$ is constant for all real $x$: define $f$ outside of $[0,1)$ to ensure convergence and then let $f(x)=C-\sum_{n\in\mathbb Z\setminus\{0\}}f(n+x)$ for $x\in[0,1)$. I wonder, however, whether there are analytic functions other than $\mathrm{sinc}(x)$ with this property? The set of such functions is a vector space over the complex numbers; is it finite-dimensional? If so, what is its dimension?

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    $\begingroup$ You can presumably just take derivatives of that identity to produce lots of other functions (I didn't carefully check how to justify this though). $\endgroup$
    – Christian Remling
    Commented Mar 6, 2019 at 19:56
  • $\begingroup$ Or take any translate of it. $\endgroup$
    – InfiniteLooper
    Commented Mar 7, 2019 at 8:53
  • $\begingroup$ What's precisely assumed on $f$? that $f$ is analytic, and that there exists a constant $c=c(f)$ such that for every $x$, the sum $\sum_{k=-n}^nf(x+n)$ converges to $c$? it's not obvious for me that a non-absolutely convergent sum over $\mathbf{Z}$ should be evaluated only on symmetric intervals, so I'm not sure what's meant. $\endgroup$
    – YCor
    Commented Mar 11, 2019 at 8:16

2 Answers 2

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If the Fourier transform $F(k)$ of $f(x)$ vanishes outside of the interval $(-1,1)$ then, by virtue of Poisson summation, $$\sum_{n=-\infty}^\infty f(x+n)=\sum_{n=-\infty}^\infty F(n)e^{2\pi inx}=F(0)$$ independent of $x$.

An example is $F(k)=k^2-a^2$ for $|k|<a$ and $F(k)=0$ for $|k|>a$, with $0<a<1$. Then $$f(x)=\frac{2}{\pi x^3}(a x \cos a x-\sin a x)\;\;\text{and}\;\;\sum_n f(x+n)=-a^2.$$

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    $\begingroup$ @LoïcTeyssier: Yes, the original function (Dirichlet kernel) is an example. But the Poisson summation formula comes with assumptions of course, and it would be good to state explicitly what else $f$ needs to satisfy for this to work. $\endgroup$
    – Christian Remling
    Commented Mar 6, 2019 at 20:05
  • $\begingroup$ @ChristianRemling: yes, I realized that ;) $\endgroup$
    – Loïc Teyssier
    Commented Mar 6, 2019 at 20:08
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    $\begingroup$ I wonder why the previous question on the same subject was removed, while this one is accepted so enthusiastically:-) $\endgroup$
    – Alexandre Eremenko
    Commented Mar 7, 2019 at 1:00
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    $\begingroup$ If $f$ is $L^1$ then the necessary and sufficient condition is that $F(k) = 0$ for every $k \ne 0$ $\endgroup$
    – reuns
    Commented Mar 7, 2019 at 4:46
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    $\begingroup$ To complete the answer to the question, it is easy to see that the family $(f_a)_{a>0}$, where $f_a(x)=\frac{2(ac\cos(ax)-\sin(ax)}{\pi x^3}$, spans a vector space of dimension $2^{\aleph_0}$ over $\mathbf{R}$, and hence the dimension of the considered vector space is $2^{\aleph_0}$. $\endgroup$
    – YCor
    Commented Mar 11, 2019 at 15:26
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A simplest case is a rectangular function defined as $${\text{rect}}\left( x \right) = \left\{ \begin{aligned} 1, & \quad - 1/2 < x < 1/2\hfill \\ 1/2, & \quad \left| x \right| = 1/2 \hfill \\ 0, & \quad {\text{otherwise}}{\text{.}} \hfill \\ \end{aligned} \right.$$ We can see that if $f(x) = \text{rect}(x/2)$, then: $$\sum_{n=-\infty}^\infty \text{rect}(x/2+n)=1$$ and the requirement: $$f(x)=C-\sum_{n\in\mathbb Z\setminus\{0\}}f(n+x), \qquad x\in[0,1)$$ is satisfied. The function $f(x) = \text{rect}(x/2)$ is analytic when $x\in[0,1)$.

The rectangular function is the Fourier transform of the sinc function.

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    $\begingroup$ The OP is asking for analytic functions satisfying the given identity $\endgroup$
    – Yemon Choi
    Commented Mar 11, 2019 at 7:28
  • $\begingroup$ Perhaps I am unaware of the definition you are using. I am using this definition en.wikipedia.org/wiki/Analytic_function and your example is not analytic in this sense at $x= \pm 1/2$ since it is not continuous on any neighbourhood of these points $\endgroup$
    – Yemon Choi
    Commented Mar 11, 2019 at 16:55
  • $\begingroup$ @YemonChoi: It was an error. I replaced $f(x) = \text{rect}(x)$ by $f(x) = \text{rect}(x/2)$. The function $f(x) = \text{rect}(x/2)$ is analytic within $x\in[0,1)$. $\endgroup$
    – u136536
    Commented Mar 11, 2019 at 18:25
  • $\begingroup$ Thank you, but my interpretation of the OP's original question is that he or she wants $f$ to be analytic everywhere. Note that sinc is analytic at the origin (removable singularity) $\endgroup$
    – Yemon Choi
    Commented Mar 11, 2019 at 18:29
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    $\begingroup$ @u136536 if you read entirely the question, you would see that your $f$ is a particular case of a more general observation by the OP. $\endgroup$
    – YCor
    Commented Mar 11, 2019 at 19:44

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