# Fourier transform of compactly supported distribution is smooth

My advisor made the comment that if $u\in \mathcal{E}'$ is a compactly supported distribution, then $\hat{u}(\xi)\in C^{\infty}(\mathbb{R}^n)$ is actually a smooth function (not merely a distribution or generalized function).

I'm trying to prove this, but I'm realizing I'm not even sure how we should properly define the F.T. of a distribution in $\mathcal{E}'$. When $u\in \mathcal{S}'$, then the standard definition $$\left<\hat{u},\phi \right> := \left<u,\hat{\phi} \right>$$ makes perfect sense, since $\phi\in\mathcal{S}\Rightarrow \hat{\phi}\in\mathcal{S}$, and so pairing $\hat{\phi}$ with an element of $\mathcal{S}'$ is well defined.

The problem I'm having with applying this same definition to $u\in\mathcal{E}'$ is that while $\phi\in C^{\infty}$, we don't necessarily have $\hat{\phi}\in C^{\infty}$; and so the statement $\left<u,\hat{\phi} \right>$ might not make sense, as $u$ should only be acting on test functions from $C^{\infty}$.

I've been toying around with the idea of a smooth cutoff function $\rho \equiv 1$ one on $\text{supp}(u)$, and taking $$\left<\hat{u},\phi \right> := \left<u,\widehat{\rho\phi} \right>,$$ to ensure that $\widehat{\rho\phi}$ is at least well-defined. But I'm not sure if that is even close to the right idea, though. So some clarification on this matter would be much appreciated.

That aside, considering that $e^{-ix\cdot\xi}\in C^{\infty}(\mathbb{R}^n_x)$, I'm at least able to show that $\left<u,e^{-ix\cdot\xi} \right>$ is a function of $\xi$ (and not a generalized function). This of course heuristically matches up with the idea that $$\left<u,e^{-ix\cdot\xi} \right> = \int u(x)e^{-ix\cdot\xi}\;dx = \hat{u}(\xi),$$ if $u(x)$ had in fact been a function. Since $\exists\{u_k(x)\}\subset C^{\infty}_0(\mathbb{R}^{n})$ that converge weakly to $u\in\mathcal{E}'$, then we have

\begin{align*} \mathbb{C} \ni \left<u,\phi \right> &= \lim_{k\to\infty}\left<u_k,\phi \right>\\ &= \lim_{k\to\infty}\int u_k(x)e^{-ix\cdot\xi}\;dx\\ &= \lim_{k\to\infty}\hat{u_k}(\xi)\\ &= U(\xi), \end{align*} some function of $\xi$. But when it comes to showing that $U = \hat{u}$ as distributions, we would need to prove that $$\left<U(\xi),\psi(\xi) \right> = \int U(\xi)\psi(\xi) \;d\xi= \left<u,\hat{\psi} \right>,$$ for all $\psi$ in some appropriate space of test functions, which leads to the same issues with $\hat{\psi}$ I was having before.

I also haven't been able to figure out how I would prove that $U(\xi)$ is smooth. So any help with that would also be welcomed.

## 2 Answers

The Fourier transform is the same as the FT $\mathcal{S}'\to\mathcal{S}'$ since $\mathcal{E}'$ is canonically contained in $\mathcal{S}'$ (which boils down to $\mathcal{S}$ being dense in $\mathcal{E}$). It is therefore perfectly well-defined to talk about Fourier transforms of compactly supported distributions.

One way of making sense the statement $\mathcal{F}(u) = \xi\mapsto\langle{u,e^{-i\xi x}}\rangle$ would therefore be to find an approximation of $e^{-i\xi x}$ with Schwartz functions. Another is the way with cutoff functions you described. Note that the support of a distribution is defined exactly so that "$f_1 = f_2$ in an open set containing $supp(u)$ $\implies \langle u,f_1 \rangle = \langle u,f_2\rangle$" holds. This proves independence of the cutoff-function.

Actually proving it is another question. There is an ugly (but standard) way with approximations and an elegant way. The elegant way is this proof: $$\langle \mathcal{F}u,\phi\rangle = \langle u,\mathcal{F}\phi\rangle = \langle u(x),\int \phi(\xi)e^{-i\xi x}d\xi\rangle = \int \langle u(x),\phi(\xi)e^{-i\xi x}\rangle d\xi = \int \phi(\xi) \langle u(x),e^{-i\xi x}\rangle d\xi$$ To make this work one has to show that the first integral exists as a $\mathcal{S}$-valued Pettis-integral and that this integral equals the Fourier transform. It is probably easiest to do this if $\phi$ has compact support because there are general existence theorem for the Pettis integral of continuous functions with compact support.

Now the smoothness is a thing that can either be done by brute force or with elegance (i.e. some more functional analysis). My favorite way: Show that $\xi \mapsto (x\mapsto e^{-i\xi x})$ is sufficiently smooth as a map $\mathbb{R}^n \to \mathcal{E}(\mathbb{R}^n)$. Because $\langle u,-\rangle: \mathcal{E}(\mathbb{R}^n) \to \mathbb{C}$ is analytic (because it is linear and continuous) you then find that $\mathcal{F}(u)$ is at least as smooth as you have proved before. Now what kind of smoothness you can easily prove certainly depends on your previous knowledge.

Say you prove it continuous (should be possible directly with the definition) then you can use the formulas for derivatives of fourier transform to prove continuity of all derivatives. If you can prove $C^k$ (also possible with careful application of the definition and some Taylor-ing) then you get directly that $\mathcal{F}(u)$ is $C^k$.

But in fact more is true: $\mathcal{F}(u)$ is real analytic! It is in fact the restriction of the complex analytic map $\mathbb{C}^n \to \mathbb{C}, \zeta\mapsto \langle u,e^{-i \zeta x}\rangle$ to $\mathbb{R}^n$. So one should really show that $\zeta \mapsto (x\mapsto e^{-i \zeta x})$ is complex analytic as a map $\mathbb{C}^n\to\mathcal{E}(\mathbb{R}^n)$. This can be done by showing that the usual power series converges w.r.t. the $\mathcal{E}$-topology.

And the cherry on top is the Paley-Wiener theorem that gives you the converse: Every holomorphic function $\mathbb{C}^n \to \mathbb{C}$ with a certain asymptotic behaviour (which is also satisfied by $\mathcal{F}(u)$) is the Fourier transform of some distribution with compact support. This can even be generalized to a similar characterizations about distributions with support contained in some fixed convex subset $K\subseteq\mathbb{R}^n$.

• Thanks, everyone. It's nice to know that I wasn't just missing something trivial. This is a good theorem to know! – Patch Feb 5 '15 at 17:46

If $u \in \mathcal E'$ is compactly supported, then your smooth cutoff extends it to a tempered distribution.

The fact that the Fourier transform of a compactly supported distribution is analytic (not just $C^\infty$) is part of Schwartz's version of the Paley-Wiener theorem.