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

Suppose $f$ is supported by the compact set $K$. Then $$ \hat{f}(\xi) =\frac{1}{\sqrt{2\pi}}\int_Kf(x)e^{-ix\xi}dx $$ is an entire function of the complex variable $\xi$ (by Morera's theorem). So (unless $\hat{f}(\xi)$ is identically zero) it follows (by the identity theorem) that the set of zeros $\xi \in \mathbb C$ of $\hat{f}$ has no accumulation point in the finite plane. So that set of zeros certainly does not contain the complement of a compact set.

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Morera argument:
Let $\Gamma$ be a closed contour in $\mathbb C$. For each $x \in K$, the function $\frac{1}{\sqrt{2\pi}}f(x) e^{-ix\xi}$ is an entire function of $\xi$, so $$ \int_\Gamma \frac{1}{\sqrt{2\pi}}f(x) e^{-ix\xi} d\xi = 0 $$ (by the Cauchy integral theorem).
By Fubini, \begin{align} \int_\Gamma \hat{f}(\xi)\,d\xi &= \int_\Gamma\frac{1}{\sqrt{2\pi}}\int_Kf(x)e^{-ix\xi}dx\,d\xi \\ &= \int_K\int_\Gamma\frac{1}{\sqrt{2\pi}}f(x)e^{-ix\xi}\,d\xi\,dx \\ &= \int_K 0\,dx = 0. \end{align} This is true for every closed contour $\Gamma$, so by Morera's theorem we conclude $\hat{f}$ is entire.

Answer 2

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 $0$.

Answer 3

Suppose $f$ is supported in $[-B,B]$ and $\hat{f}$ is supported in $[-A,A]$ which we may take to satisfy $A>B$. Create a periodic version $\tilde{f}$ of $f$ in $[-A,A]$ such that it is zero in $[-A,-B]\cup [B,A]$.

Explictly we may write:

$$\tilde{f}(x)=\begin{cases}f(x) \quad,\: x\in [-B,B]\\ 0 \quad, \:x\in [-A,-B]\cup [B,A]\\ \text{Defined by being periodic of period $2A$ elsewhere.} \end{cases}$$

In this case, we have:

$$\tilde{f}(x)\sim\sum_{n\in \mathbb{Z}}\left(\frac{1}{2A}\int_{-A}^{A } f(x) e^{-2\pi i \frac{n}{2A}x}dx\right) e^{2\pi i \frac{n}{2A}x}= \sum_{n\in \mathbb{Z}}\frac{1}{2A}\hat{f}\left(\frac{n}{2A}\right) e^{2\pi i n\frac{x}{2A}}$$

Indeed, integrating over $[-A,A]$ is the same as integrating over $\mathbb{R}$ due to our support of $f$. If we remember $\hat{f}$ is also compactly supported, this sum turns out to be actually finite. A final observation is that $\tilde{f}-\sum_{n=-N}^N\frac{1}{2A}\hat{f}\left(\frac{n}{2A}\right) e^{2\pi i n\frac{x}{2A}}$ is continuous in the circle and has zero Fourier coefficients, so we must have equality, namely:

$$\tilde{f}(x)=\sum_{n=-N}^N\frac{1}{2A}\hat{f}\left(\frac{n}{2A}\right) e^{2\pi i n\frac{x}{2A}}$$

Furthermore, because this is identically zero in $[B,A]$, we may take $x_o$ an interior point there. This means every derivative is zero in this point aswell and we need:

$$\sum_{n=-N}^N\frac{1}{2A}\hat{f}\left(\frac{n}{2A}\right) \left(\frac{2\pi i n}{2A}\right)^je^{2\pi i n\frac{x_o}{2A}}=0 \quad \quad \forall j\in \mathbb{N}$$

Equivalently we need: $$ \begin{bmatrix} 1 & 1 & 1 & ... & 1\\ 0 & \frac{2\pi i n_1}{2A} & \frac{2\pi i n_2}{2A} & \dots & \frac{2\pi i n_{2N-1}}{2A} \\ 0 & \left(\frac{2\pi i n_1}{2A}\right)^2 & \left(\frac{2\pi i n_2}{2A}\right)^2 & \dots & \left(\frac{2\pi i n_{2N-1}}{2A}\right)^{2} \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & \left(\frac{2\pi i n_1}{2A}\right)^{2N} & \left(\frac{2\pi i n_2}{2A}\right)^{2N} & \dots & \left(\frac{2\pi i n_{2N-1}}{2A}\right)^{2N} \end{bmatrix}\begin{bmatrix}\hat{f}(0)\\ \hat{f}(n_1/2A)e^{2\pi i n_1 x_o/(2A)}\\ \vdots \end{bmatrix}=0\Leftrightarrow \hat{f}\left(\frac{n}{2A}\right)=0 \quad \forall n \in [-N,N]\cap\mathbb{Z}$$

This follows from the fact linear combinations of these columns are linear combinations of the lines of a Vandermonde matrix. Therefore, this matrix is invertible and we need every coefficient to be zero in ther Fourier expansion of $\tilde{f}$ from which we conclude that $\tilde{f}\equiv 0\Rightarrow f\equiv 0$.