# Fourier transform of a compactly supported function

In which space does the Fourier transform of a smooth compactly supported function $\phi$ lie? I would not say it lies in $\mathcal{S}$, heuristically as one can approximate the step function which is $1$ in $[0,1]$ and $0$ outside by smooth functions, and the Fourier transform of that function decays very slowly.

What if I add the requirement that the integral average of $\phi$ is $0$? I would expect cancellation in the phase space, the higher the frequency, the higher the cancellation.

Any hint would be appreciated!

I think that $\hat{\phi}$ should lie in $\mathcal{S}$ actually. To prove this, one must show that $\xi^n (D^m\hat{\phi})(\xi)$ is a bounded function for all $m,n$, where here $D^m$ denotes the $m$-th derivative. To bound this function, first we use the fact that $$D^m\hat{\phi}(\xi) = \mathcal{F}\left\{(-2\pi ix)^m \phi(x)\right\},$$ where $\mathcal{F}$ denotes the Fourier transform $$\mathcal{F}(\psi)(\xi) = \hat{\psi}(\xi) := \int_\mathbb{R}\psi(x)e^{-2\pi ix\xi}\,dx.$$ We may therefore write $$\xi^nD^m\hat{\phi}(\xi) = \xi^n\mathcal{F}\left\{(-2\pi ix)^m\phi(x)\right\} = \frac{(2\pi i\xi)^n}{(2\pi i)^n}\mathcal{F}\left\{(-2\pi ix)^m\phi(x)\right\}.$$ If we next use the identity $$(2\pi i\xi)^n\mathcal{F}(\psi) = \mathcal{F}(D^n\psi),$$ we see that $$\xi^nD^m\hat{\phi}(\xi) = \frac{1}{(2\pi i)^n}\mathcal{F}\left\{D^n[(-2\pi ix)^m\phi(x)]\right\}.$$ This proves that $$|\xi^nD^m\hat{\phi}(\xi)|\leq \frac{1}{(2\pi)^n}\|D^n[(-2\pi ix)^m\phi(x)]\|_{L^1}<\infty$$ for all $\xi$.

It indeed lies in the Schwartz space, which is the space $$\mathcal{S} := \left\{f: \mathbb{R} \to \mathbb{R} \,:\, \forall \alpha,\beta \in \mathbb{N}\, \sup_{x\in\mathbb{R}} \left|x^\alpha f^{(\beta)}(x)\right| < \infty \right\}$$ where $f^{(n)}$ denotes the $n$-th derivative of $f$. This definition looks a bit intimidating at first, but it's easy to understand. A function $f$ lies in $\mathcal{S}$ if it decays rapidly, i.e. faster than any polynomial grows (thus $x^nf(x)$ is bounded for every n), and if the same is true for every derivative of $f$.

This space is closed under fourier transform, i.e. if $f \in S$, then so is the fourier transform $\mathcal{F}(f)$. The same is true from the reverse fourier transform. The space of compactly supported $C^\infty$ functions is quite obviously a subspace of $\mathcal{S}$, hence it's image under $\mathcal{F}$ is thus also a subspace of $\mathcal{S}$.

The problem with your reasoing is that you're thinking about pointwise limits. Under those, $\mathcal{S}$ isn't complete, so knowing something about the limit doesn't tell you much about $\mathcal{S}$. There are topologies on $\mathcal{S}$ which make it a complete space, and under those you indeed cannot approximate the step function by functions in $\mathcal{S}$. The "trick" of those topologies is basically to force convergence of all the derivatives, which e.g. prevents you from approximating any discontinous functions.

• Yes, I was just being stupid. Thanks! – Second May 5 '13 at 13:05
• @Second I added a paragraph explaining why pointwise limits don't say much here – fgp May 5 '13 at 13:11
• But if the function is smooth compactly supported, then more is true. See chapter 29, Treves "Topological vector spaces, distributiona and Kernels" about Paley-Wiener theorem. – TCL May 5 '13 at 13:12
• @fgp the transform $\hat{f}$ of fourier on a function $f\in C^{\infty}_c$ with support compact, is $\hat{f}\in C^{\infty}$ ? thanks – user89940 May 7 at 7:33

In addition to being decaying rapidly (as a function of real variable), it lies in the space of entire analytic functions (as a function of complex variable).