How to show that $f$ and $\mathcal{F}(f)$ can't both be compactly supported?

Question

Let $f : \mathbb{R}\to \mathbb{R}$ be continuous, I want to show that it can't happen that both $f$ and its Fourier transform $\mathcal{F}(f)$ are compactly supported unless $f = 0$.

That means that I want to show that if both $f$ and $\mathcal{F}(f)$ are compactly supported, $f = 0$.

I've seem some questions like this here, but I believe this is not duplicate. That because on those questions I've seem people using the Fourier series, while here I'm talking about the Fourier transform.

In the Fourier series we require $f$ to be periodic. Here $f$ needs not to be periodic.

In that case, I can't restrict $f$ to $[-\pi,\pi]$ and talk about its Fourier series, because it need not be the case that $f(-\pi)=f(\pi)$, since $f$ is arbitrary.

Now, if $f$ is compactly supported, there's an interval $[a,b]$ such that $f(x) = 0$ for all $x\notin [a,b]$. If $\mathcal{F}(f)$ also is compactly supported there's $[c,d]$ such that $\mathcal{F}(f)(\xi)=0$ if $\xi\notin [c,d]$.

I've then thought of two approaches:

1. Try to use this together with the Fourier inversion formula, to show that $f(x) = 0$.

2. Try to use this together with Plancherel's theorem to show that $\|f\|_2 =0$ and hence $f = 0$.

Now, in the first case I get

$$f(x)=\int_{c}^{d}\mathcal{F}(f)(\xi)e^{2\pi i x\xi}d\xi = \int_c^d \int_a^b f(y)e^{-2\pi i y\xi}e^{2\pi ix\xi} dyd\xi$$

Or yet

$$f(x)=\int_a^b f(y)\int_c^d e^{2\pi i\xi (x-y)}d\xi dy,$$

but this doesn't seem to lead anywhere.

Plancherel's theorem also doesn't seem of great aid here. We have

$$\int_a^b |f(x)|^2 dx = \int_c^d |\mathcal{F}(f)(\xi)|^2 d\xi,$$

which doesn't help much indeed.

How can I show that if both $f$ and its Fourier transform are compactly supporetd then $f = 0$ in the context of the Fourier transform and not Fourier series?

Answer 1

If $f$ is compactly supported then \begin{align} \hat{f}(\xi) & =\frac{1}{\sqrt{2\pi}}\int_{-M}^{M}f(x)e^{-ix\xi}dx \\ & = \frac{1}{\sqrt{2\pi}}\int_{-M}^{M}f(x)\sum_{n=0}^{\infty}\frac{(-ix\xi)^n}{n!}dx \\ & = \sum_{n=0}^{\infty}\left[\frac{(-i)^n}{\sqrt{2\pi}n!}\int_{-M}^{M}f(x)x^{n}dx\right]\xi^n \end{align} This power series converges everywhere. A power series cannot have an infinite set of zeros with a finite accumulation point, unless it is identically $0$. So $\hat{f}$ cannot be compactly supported if $f$ is compactly supported, unless $f$ is identically $0$.