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This question is motivated by my previous post about sinc function.

Prove or disprove that $\frac{\sin x}{x}$ is the only nonzero entire (i.e. analytic everywhere) function $f(x)$ on $\mathbb{R}$ such that $\int_0^\infty f(x) dx=\int_0^\infty f(x)^2 dx$ or $\int_{-\infty}^\infty f(x) dx=\int_{-\infty}^\infty f(x)^2 dx.$

If $f$ is required only to be continuous, then other examples are possible, e.g. the even extension of the following function: $ f(x)=\left\{\begin{array}{ll} -2(5+\sqrt{65})x^2+(7+\sqrt{65})x-1 & 0\le x\le \frac{1}{2}\\ 2(5+\sqrt{65})x^2-(13+3\sqrt{65})x+4+\sqrt{65} & \frac{1}{2}\le x\le 1\\ \frac{1}{x^2} & x\ge 1 \end{array}\right. $

As commented below, it turns out that there are easy answers to the above question. AD also showed a function below that also satisfies $ \sum_{n=1}^\infty f(n)=\sum_{n=1}^\infty f(n)^2=0. $ In view of these answers, my question is now revised to:

Prove or disprove that $\frac{\sin x}{x}$ is the only nonzero entire 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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    SEe the word "nonzero"?2010-12-08

3 Answers 3

1

The revised question has been answered in this post at MO. In particular, it was shown that $\frac{\sin ax}{ax}$ satisfies this equality for each $0.

3

Similar to the suggestion of Zaricuse, that is take an entire function $f$ such that $\int_{-\infty}^\infty f(x) dx \ne0$ then solve for $\int_{-\infty}^\infty af(x)dx = \int_{-\infty}^\infty (af(x))^2dx$ Then $g(z)=af(z)$ solves half of the problem. To reach $\sum g(n)=\sum g(n)^2$ we may for example start with $f(z)=\sin (\pi z) \cdot h(z)$ where $h$ is an other integrable entire function.

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    @Ross Millikan: Maybe, but it is a bit cheap.2010-12-09
2

In the spirit of Zaricuse's solution without the sum requirement, take any sufficiently well behaved functions f and g. Then you should be able to find a linear combination af+bg that satisfies both equations. If you let

$if=\int_{-\infty}^\infty f(x) dx$ $if2=\int_{-\infty}^\infty f(x)^2 dx$ $sf=\sum_{n=1}^\infty f(n)$ $sf2=\sum_{n=1}^\infty f(n)^2$

and similarly for fg and g, we have

$a*if + b*ig=a^2*if2+2ab*ifg+b^2ig2$ and
$a*sf + b*sg=a^2*sf2+2ab*sfg+b^2sg2$

which can be solved for a and b in most cases.

Added In response to the new request that the values of the two integrals and two sums all match, I just need enough knobs to turn. Define $g(k,x)=\exp(-kx^2)$ and take $f(x)=g(1,x)+ag(2,x)+bg(3,x)+cg(4,x)$ The nice thing about this $f$ is that $f^2$ is written in terms of $g(k,x)$, though k goes up to 8. The integral of $g(k,x)$ is just $\sqrt{\frac{\pi}{k}}$ and the sum is calculated by Wolfram Alpha as $\vartheta_3(0,\exp(-k))$. We can make a table:

$\begin{array}{ccc}k&\int g(k,x)&\sum g(k,x)\\1&1.772453851&1.77264\\2&1.253314137&1.27134\\3&1.023326708&1.09959\\4&0.886226925&1.03663\\5&0.79266546&1.01348\\6&0.723601255&1.00496\\7&0.669924586&1.00182\\8&0.626657069&1.00067\end{array}$

So the integral of f is $\sqrt{\pi}(1+a/\sqrt{2}+b/\sqrt{3}+c/\sqrt{4})$ The integral of f^2 is $\sqrt{\pi}(1/\sqrt{2}+2a/\sqrt{3}+(a^2+2b)/\sqrt{4}+(2c+2ab)/\sqrt{5}+(b^2+2ac)/\sqrt{6}+2bc/\sqrt{7}+c^2/\sqrt{8})$ with similar expressions for the sum in terms of theta3. We want to find a,b,c so that the integrals and sums all match. Unless my matrix of coefficients has a very unlikely dependence it will be available. Excel claims $a=-3.782590725, b=4.503400057, c=-1.83137936$ is very close to a solution, and there should be more.

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    *Mathematica* 's `NSolve[]` returns four solutions to this problem. I did not have the patience to wait for the `Solve[]` version, but from the looks of things the exact results are too complicated to be easily manipulable.2010-12-10