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 PaleyWiener theorem that is even on the real line and whose restriction to $[0,\infty)$ is monotone?
The answer is yes :
Let $h$ be an even real valued Schwartz function whose Fourier
transform has compact support. Then choose $f(y) = \int_{\infty}^y
x h(x)^{2} dx$ .

$\begingroup$ Great! Could you possibly add some extra details? $\endgroup$ – 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 nonnegative even PaleyWiener function $f$ with $f'$ nonnegative 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 $$\Bigg[{\cos(xy)\over y^{2n+1}}\big(x^{2n}g(x)\big)^{(2n)}\Bigg]_0^\inftyy^{2n1}\int_0^\infty \big(x^{2n}g(x)\big)^{(2n+1)}\cos(xy)\; dx$$ The last integral is the Fourier transform of a smooth compactlysupported function, so decays in $y$. Thus we have, as $y\rightarrow\infty$, $$f^{(2n)}(y)=y^{2n1}(2n)!g(0)\big(1+h(y)\big)$$ 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 $f^{(n1)}$ 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, $$f(x)=\int_{\infty}^x {\sin^m(\pi t/m)\over (\pi t/m)^{m1}}\ dt$$ is a positive PW function (at least with the more relaxed definition of $L^2$ on the line), with $f'$ nonnegative on $\mathbb R_{>0}$. Since the integrand is odd, $f$ is even as $$f(x)=\int_{\infty}^{x} {\sin^m(\pi t/m)\over (\pi t/m)^{m1}}\ dt\int_{x}^x {\sin^m(\pi t/m)\over (\pi t/m)^{m1}}\ dt=f(x)$$

$\begingroup$ @BR: Interesting... doesn't this solve the problem then? $\endgroup$ – Alan Haynes Dec 10 '11 at 16:26

$\begingroup$ 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!). $\endgroup$ – B R Dec 10 '11 at 17:30