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

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?

• Do you know a Paley-Wiener-theorem that states that a $\mathcal F(f)$ is the restriction of an entire function if $f$ is compactly supported? – Tim B. Dec 11 '16 at 15:05
• No, although I've heard of this result, I'm trying to prove the claim without using it. – Gold Dec 11 '16 at 15:27

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 $0$.
• @user1620696 : If you expand the power series about a zero, then there's a first coefficient that is non-zero, allowing you to factor out $(\xi-\xi_0)^{n}$ and end up with a power series that is not zero at $\xi=\xi_0$. Then, by continuity, the power series cannot vanish in a neighborhood of $\xi_0$, which proves that zeros of a power series are isolated in the interior of the region of convergence. – Disintegrating By Parts Dec 11 '16 at 22:25
• @user1620696 : Yes, you are correct that the power series converges uniformly, which allows you to interchange integration and summation over the finite interval $[-M,M]$. – Disintegrating By Parts Dec 11 '16 at 22:27