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Is there any even Schwartz function whose restriction to $[0,\infty)$ is monotone and whose Fourier transform is compactly supported? In other words, is there any entire analytic function satisfying the condition of the Paley-Wiener theorem that is even on the real line and whose restriction to $[0,\infty)$ is monotone?

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2 Answers 2

up vote 5 down vote accepted

The answer is yes :
Let $h$ be an even real valued Schwartz function whose Fourier transform has compact support. Then choose $f(y) = \int_{-\infty}^y x h(x)^{2} dx$ .

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Great! Could you possibly add some extra details? –  B R Dec 11 '11 at 15:55

I think your question is true or nearly true. I interpet it as "Can we find a non-negative even Paley-Wiener function $f$ with $f'$ non-negative on $\mathbb R_{>0}$?" As in AH's answer, $$f(y)=\int_0^\infty g(x)\cos(xy)\ dx$$ where $g$ is a smooth function of compact support. I'll make a general statement that is not quite what you want, then give an example that is nearly (at least) what you want.

It has been proven (see here) that if $g(0)\ne 0$, then $f^{(n)}$ has only finitely many zeros. (Conversely, if $g^{(n)}(0)=0$ for all $n$, $f$ changes sign infinitely often, see here.)

This isn't too hard to prove. Take $n\ge 1$, then $$f^{(2n)}(y)=(-1)^{n}\int_0^\infty x^{2n}g(x)\cos(xy)\; dx$$ Integrating by parts $2n$ times makes this $$\Bigg[{\cos(xy)\over y^{2n+1}}\big(x^{2n}g(x)\big)^{(2n)}\Bigg]_0^\infty-y^{-2n-1}\int_0^\infty \big(x^{2n}g(x)\big)^{(2n+1)}\cos(xy)\; dx$$ The last integral is the Fourier transform of a smooth compactly-supported function, so decays in $|y|$. Thus we have, as $|y|\rightarrow\infty$, $$|f^{(2n)}(y)|=y^{-2n-1}(2n)!|g(0)|\big(1+h(y)\big)$$ where $h(y)\in o(1)$. Since $|h(y)|<1$ for $y$ sufficiently large, $f^{(2n)}$ can only have finitely many zeros. Noting that if $f^{(n)}$ has infinitely many zeros, Rolle's Theorem would imply that $f^{(n-1)}$ has infinitely many zeros, we see that $f^{(n)}$ only has finitely many zeros for all $n\ge 0$.

So you can easily find $f$ with every derivative having finitely many zeros. But, it turns out, $f^{(n)}$ always has at least $n$ zeros. This isn't as bad as it sounds, as $f$ being even implies that $f'$ has a zero. And just because a function has zero doesn't mean it is negative. Unfortunately, I can't find anything that gives a definitive answer here.


From here (p. 82), for $m\ge 4$ an even integer, $$f(x)=-\int_{-\infty}^x {\sin^m(\pi t/m)\over (\pi t/m)^{m-1}}\ dt$$ is a positive PW function (at least with the more relaxed definition of $L^2$ on the line), with $f'$ non-negative on $\mathbb R_{>0}$. Since the integrand is odd, $f$ is even as $$f(x)=-\int_{-\infty}^{-x} {\sin^m(\pi t/m)\over (\pi t/m)^{m-1}}\ dt-\int_{-x}^x {\sin^m(\pi t/m)\over (\pi t/m)^{m-1}}\ dt=f(-x)$$

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@BR: Interesting... doesn't this solve the problem then? –  Alan Haynes Dec 10 '11 at 16:26
    
AH, maybe! I'm not sure that the example is the Fourier transform of a smooth function, so it might not be Schwartz on the line. The paper says "The Fourier transform is simply related to, and indeed has the same support as, that of $[(\pi t/m)^{-1}\sin(\pi t/m)]^m$". The polynomial decay of the latter function guarantees that the Fourier transform will not be smooth at the endpoints. Perhaps the example turns out better, but I don't know (and I need to move on, too!). –  B R Dec 10 '11 at 17:30

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