# Explicit Paley-Wiener function

By a Paley-Wiener function I mean a function $f(z)$ that is the Fourier image of a test function. Equivalently, by Paley-Wiener theorem, $f(z)$ is an entire function that is of rapid decay on the real line and has a finite exponential type, $$|f(z)|\le\frac{C_ne^{B|\mathrm{Im}z|}}{1+|\mathrm{Re}z|^n},\quad\forall n\in\mathbb{N}.$$

Question: What is the simplest explicit formula for a non-zero Paley-Wiener function?

The best I can think of right now is $$f(z)=\int_0^1dte^{-\imath zt-\frac1{t(1-t)}}$$ which is not really explicit. Thanks.

• The awkwardness that seems necessary to express a test function seems inescapable, indeed, ... – paul garrett Mar 25 '18 at 0:47

In the standard terminology, a Paley-Wiener function is a Fourier transform of a function from $L^2$ with compact support.
The simplest formula for a Paley-Wiener function is $\phi(x)=(\sin x)/x$. This is the Fourier transform of the characteristic function of an interval.
But you seem to have some unusual definition of a Paley Wiener function, though you do not state precisely what your "test functions" are. If you mean infinitely differentiable functions with compact support, take an infinite product of the form $\prod \phi(a_kx)$ where $a_k$ is an appropriately chosen sequence of real numbers tending to $0$. I don't think there is a simpler form, without an infinite product or an integral.
• Although I've not taken the trouble to "check around", I am mildly surprised at the usage of "Paley-Wiener function" for Fourier transform of compactly-supported $L^2$ function. Namely, for me, it means "FT of test function". (And, yes, at another extreme, there is Schwartz' result about compactly-supported distributions.) So, I ask a sincere question about how widespread this usage is... – paul garrett Mar 25 '18 at 0:46