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This question is motivated by my previous post in SE (math.stackexchange.com).

Prove or disprove that $\frac{\sin x}{x}$ is the only nonzero entire (i.e. analytic everywhere), or continuous, function, $f(x)$ on $\mathbb{R}$ such that $$\int_{-\infty}^\infty f(x) dx=\int_{-\infty}^\infty f(x)^2 dx=\sum_{-\infty}^\infty f(n) =\sum_{-\infty}^\infty f(n)^2 $$

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Why this question is interesting and non-trivial? In general, three equations define a set of codimension three. So, for a suitable space of functions the set of solutions should be infinite or empty. –  Petya Dec 9 '10 at 14:46
Since $\frac{\sin x}{x}$ is one such analytic function, so the set is not empty. Are you saying that there should be infinitely many analytic functions satisfying the identity? –  TCL Dec 9 '10 at 15:04
No. I say -- in general. There should be very specific reasons to have only one solution. I suggest to linearize these equations at your function and to understand the corresponding vector space - if all conditions are independent than one can apply the implicit function theorem and get an infinite number of solutions. –  Petya Dec 9 '10 at 15:12
@Shai. That is what I read from Borwein's article. Check one of the links in my previous post at SE mentioned above. –  TCL Dec 9 '10 at 16:16
It is an exercise - to prove that there exists a $C^\infty$-function $f$ which is supported in $[-3/4,3/4]$ and satisfies your conditions. –  Petya Dec 9 '10 at 21:13

7 Answers 7

up vote 7 down vote accepted

Seems like the Poisson formula is well-forgotten nowdays. Assume that $f$ is a real valued even Schwartz function and pass to its Fourier transform $g$ (with $2\pi$ in the exponent). Then the conditions are $$g(0)=\int_{\mathbb R} g^2=\sum_{k\in\mathbb Z} g(k)=\int_{[-1/2, 1/2]}G^2$$ where $G(x)=\sum_{k\in\mathbb Z} g(x+k)$ is the $1$-periodization of $g$. Not to bother about $1$-periodization too much, assume that $g$ is supported on $[-0.3,0.3]$ (that'll take care of analyticity of $f$ too). Then we just need a smooth real-valued even function on that interval whose value at $0$ equals the integral of its square. They are plenty.

This would be a nice question for an analysis qualifier exam. :) I'll try to propose it for the next one and see if it flies.

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Boy, am I glad you didn't help write the qualifying exams I had to take as a graduate student! –  Deane Yang Dec 10 '10 at 3:51
One more remark - that solution does not look flexible. Suppose you add more equations like $\int f^5dx=\int (\frac{sin x}{x})^5dx$ and $f(\pi)=0$ etc. What you will do in that case? Implicit function theorem will help in any such case. Just show that some determinants are non-zero. –  Petya Dec 10 '10 at 3:59
I want to say ninja'd, but you beat me to this by good 5 hours. :-) (Though I just spent the last hour trying to remember to whom to attribute Poisson's formula. As I only remembered the first letter of his name, I ended up going through en.wikipedia.org/wiki/List_of_mathematicians_(P) Man I'm horrible with names.) –  Willie Wong Dec 10 '10 at 10:57
@Petya: good point. But I don't think the situation is that easy. In the $C^\infty$ case, don't you need Nash-Moser inverse function theorem? So you need to show the determinants are non-zero in the entire neighborhood. And I may be just woefully ignorant, but is there a version of implicit function theorem that can be generally applied to $C^\omega$? A reference will be greatly appreciated. –  Willie Wong Dec 10 '10 at 12:19
@fedja. Can your answer be modified so it includes $\frac{\sin x}{x}$ as an example? –  TCL Dec 10 '10 at 18:23

Perhaps the most simple example of an analytic function $f$, other than $0$, satisfying all three equalities is $f(x) = \sin(\pi x)/ (\pi x)$. This is verified immediately by a change of variable $x'=\pi x$, giving $\int_{ - \infty }^\infty {f(x)\,{\rm d}x} = \int_{ - \infty }^\infty {f^2 (x)\,{\rm d}x} = 1 = \sum\nolimits_n {f(n)} = \sum\nolimits_n {f^2 (n)}$.

EDIT: TCL provided a (very interesting) generalization.

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Oh, what a shame! –  TCL Dec 12 '10 at 7:02
@Shai. The identity is true for constant other than $\pi$ too; see my answer above. –  TCL Dec 14 '10 at 0:32
Edited my answer accordingly. –  Shai Covo Dec 14 '10 at 1:45

Consider the function $f:\mathbb R\to\mathbb R$ which vanishes outside of $[-1,1]$ and such that for all $x \in [-1,1]$ has $$ f(x) = \frac{5 x^4}{4}+\frac{1}{2} \sqrt{\frac{37}{6}} x^3-\frac{9 x^2}{4}-\frac{1}{2} \sqrt{\frac{37}{6}} x+1. $$ Then $$\int_\mathbb Rf(x)\,\mathrm dx=\int_\mathbb Rf(x)^2\,\mathrm dx=\sum_{n\in\mathbb Z}f(n)=\sum_{n\in\mathbb Z}f(n)^2=1.$$

Moreover, the functions with the same support which on $[-1,1]$ coincide with the polynomials $$ \begin{array}{l} -\frac{1}{32} (x-1)^2 (x+1) \left(x \left(\left(11+\sqrt{3245}\right) x+\sqrt{3245}-29\right)-32\right); \\\\ \frac{(x-1)^3 (x+1) \left(x \left(7 \left(\sqrt{2657109}-923\right) x+5 \sqrt{2657109}-8431\right)-3392\right)} {3392}; \\\\ -\frac{1}{64} (x-1)^4 (x+1) \left(\left(3 \sqrt{3855}-175\right) x^2+2 \left(\sqrt{3855}-98\right) x-64\right); \\\\ \frac{(x-1)^5 (x+1) \left(3 \left(\sqrt{3040433}-2550\right) x^2+2 \left(\sqrt{3040433}-4571\right) x-3008\right)}{3008}; \\\\ -\frac{(x-1)^6 (x+1) \left(x \left(25 \left(3 \sqrt{622687}-6365\right) x+51 \sqrt{622687}-202885\right)-67328\right )}{67328}; \end{array} $$ have the same properties and are $C^1$, $C^2$, $C^3$, $\dots$, respectively, and it seems one can go on indefinitely. There are no other polynomials satisfying the conditions and of degree less than or equal to theirs. The unique $C^k$ polynomial of degree $4+k$ satisfying the conditions is the product of $(x-1)^{k+1}(x+1)$ and a degree $2$ factor: one should be able to describe this last factor explicitly somehow...

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This cannot be an analytic function (due to the compact support). –  Andrey Rekalo Dec 9 '10 at 14:13
The question does say "or continuous". –  Willie Wong Dec 9 '10 at 14:20
Andrey, of course they are not analytic! –  Mariano Suárez-Alvarez Dec 9 '10 at 14:24
So the harder question is for analytic functions. –  TCL Dec 9 '10 at 14:24
Your functions are not sufficiently smooth as you claim. The problem is at $x=-1$. The multiplicity of zero of the polynomial part at that point is at most $3$. –  Petya Dec 9 '10 at 20:34

Let me try to expand Petya comment in an explicit case, hoping that this can be useful.

Take $K >>0$ and consider the set of functions

$f_k:=e^{-kx^2}, \quad k=1,2, \ldots, K$.

Assume that we want to find an analytic function $f$ satisfying the requirement of the question and which is a linear combination of the $f_k$, namely

$f=\sum_{k=1}^K a_kf_k, \quad a_k \in \mathbb{R}$.

Then we must solve the following system of equations

$\sum_{k=1}^k A_ka_k =\sum_{i,j=1}^K A_{ij} a_ia_j$

$\sum_{k=1}^k B_ka_k =\sum_{i,j=1}^K A_{ij} a_ia_j$

$\sum_{k=1}^k B_ka_k =\sum_{i,j=1}^K B_{ij} a_ia_j$,


$A_k:=\int_{\mathbb{R}} f_k, \quad A_{ij}:=\int_{\mathbb{R}} f_i \cdot f_j$,

$B_k:=\sum_{n \in \mathbb{Z}} f_k(n), \quad B_{ij}:=\sum_{n \in \mathbb{Z}} f_i(n) \cdot f_j (n)$.

This system of equations geometrically describes the intersection of three non-empty quadrics hypersurfaces in $\mathbb{R}^K$, so one expects infinite solutions for $K$ big enough.

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The second equation can be made linear by replacing the right side with the left side of the first equation. In any case, $K=4$ seems to be large enough. –  S. Carnahan Dec 9 '10 at 16:34
It is not clear that three real quadrics have non-empty intersection. The situation when one solution is known should be studied in the linear approximation settings firstly (this idea goes back to Newton, or earlier, I think). I cannot see why differentials of equations are linearly dependent, though I did not prove it formally. If differentials are independent then any $k$-dimensional subspace of functions which contains $sin x/x$ will also contain an infinite number of solutions provided $k>3$. –  Petya Dec 9 '10 at 17:50

Here's a simple example of a smooth (rather than analytic) function $f$ satisfying the conditions. Define $f(x) = \exp [ - ax^2 /(1 - x^2 ) + bx]$ if $x \in (-1,1)$, and $f(x)=0$ otherwise. Here $a$ and $b$ are certain positive constants, to be evaluated below. That the function $f$ is smooth (i.e. infinitely differentiable) should be clear by comparison with the function $\exp(-1/x^2)$ (where the point $0$ corresponds to the points $\pm 1$). Now, since $f(0)=1$ and $f(n)=0$ for any integer $n \neq 0$, it remains to have $\int_{ - 1}^1 {f(x)\,{\rm d}x} = \int_{ - 1}^1 {f^2 (x)\,{\rm d}x} = 1$. Comparing the logs of $f$ and $f^2$, it is not surprising that there indeed exist $a$ and $b$ satisfying the two integral conditions. With some effort, one is likely to be able to prove this rigorously. However, for our purposes it is enough to be convinced by numerical results. Well, it is easy to get close to a solution. For example, letting $a=3.25247$ and $b=2.08761361$ gives $\int_{ - 1}^1 {f(x)\,{\rm d}x} \approx 0.999999999149$, $\int_{ - 1}^1 {f^2 (x)\,{\rm d}x} \approx 1.000000136$.

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'Convinced' isn't the same as proven. –  David Roberts Dec 10 '10 at 1:36
As I said - it is a level of an exercise. One can complete it (non be numerically convinced) as follows: 1. Fix a smooth function f such that $\int f=1$, $f(0)=1$, $supp(f)\subset[−3/4,3/4]$ and $0\le f(x)\le 1$ for every $x$. Note that $\int f^2$<1. 2. Fix now a non-negative non-zero smooth function $g$ supported in $[3/4,1]$. Let $g_1$ be an odd function, coinciding with $g$ for positive $x$. Than for any $c$ the function $f_c=f+cg_1$ satisfies almost all conditions. Note that $\int f^2_c=\int f^2+2c^2\int g^2$ and that $\int g^2 >0$. Hence for some $c$ $\int f^2_c=1$. That is it. –  Petya Dec 10 '10 at 2:04
@David Roberts: For the OP's purposes, it is enough to be convinced. –  Shai Covo Dec 10 '10 at 2:05
Also, it might be easy to prove my assertion rigorously, but I don't find it important in our context. –  Shai Covo Dec 10 '10 at 2:21
I do think - it is not easy to prove your assertion rigorously. At least, I am sure, it is harder then the initial statement. –  Petya Dec 10 '10 at 2:43

In fact, the identity is true for the function $f(x)=\frac{\sin ax}{ax}$ for each $0\lt a\le \pi$.

For the integral part, just apply substitution.

For the series part, use the fact that $\sum_{n=1}^\infty \frac{\sin nx}{n}=\frac{\pi-x}{2}$ for $0\lt x\lt 2\pi$, and that $\sum_{n=1}^\infty \frac{\sin^2 nx}{n^2}=\frac{\pi^2}{8}-\frac{1}{2}\left(x-\frac{\pi}{2}\right)^2$ for $0\lt x\le \pi$.

Thus $$\int_{-\infty}^\infty f(x) dx=\int_{-\infty}^\infty (f(x))^2 dx=\sum_{n=-\infty}^\infty f(n)=\sum_{n=-\infty}^\infty f(n)^2=\frac{\pi}{a}.$$

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Very interesting result. So, the choice $a = \pi$ is only essential for simplicity... (Some little error led me to conclude that $\pi$ and $1$ are the only suitable constants; I'll edit my answer accordingly.) –  Shai Covo Dec 14 '10 at 1:21

There is the characterization from http://www.math.caltech.edu/SimonPapers/324.pdf or the work of Lubinsky. Of course the conditions are slightly different then yours.

I am not sure of the meaning of requiring something at the integer points, but it seems to me that you miss an essential requirement on the growth.

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