Fourier transforms having compact support

Question

As we know, the fourier transform is a map $\mathcal{F}:L^1\rightarrow C_0$ (all with domain $\mathbb{R}$). Can one characterize the space of $f\in L^1$ such that $\mathcal{F}$ has compact support, i.e. is in $C_c\subset C_0$? In particular, is this space dense in $L^1$? I believe this should be true, but I find it hard to come up with nontrivial examples of such $f$. I guess the formula $\mathcal{F}(f\ast g)=\mathcal{F}(f)\mathcal{F}(g)$ should be useful?

Answer 1

One can characterize the space of integrable functions such that the Fourier transform has compact support via a Paley-Wiener sort of argument; in fact $\hat f$ has compact support if and only if $f$ is (almost everywhere) the restriction to $\Bbb R$ of an entire function $g$ with $$|g(x+iy)|\le ce^{A|y|}.$$

I don't see how this characterization helps in showing that the space is dense. But that's not hard. Since the Fourier transform maps the Schwarz space to itself there exists a Schwarz function $\phi$ with $\int\phi=1$ such that $\hat\phi$ has compact support. (You can make $\hat\phi$ be any infinitely differentiable function with compact support.)

Define $$\phi_\delta(t)=\frac1\delta\phi\left(\frac t\delta\right).$$Then $\hat\phi_\delta$ has compact support, so if $f\in L^1$ then $\widehat{\phi_\delta*f}$ has compact support, and $||\phi_\delta*f-f||_1\to0$ as $\delta\to0$.

Edit: Come to think of it, it's even easier than this. For some reason I wanted a $\phi$ that gave pointwise convergence. All we need is convergence in norm, and that's much easier: We can let $\phi$ be any $L^1$ function such that $\hat \phi$ has compact support.

(So for example $\phi=\hat\psi$ where $\psi$ is just $C^2_c$, or where $\psi$ is the convolution of any two $L^2$ functions with compact support...)