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As we know, the fourier transform is a map $\mathcal{F}:L^1\rightarrow C_0$ (all with domain $\mathbb{R}$). Can one characterize the space of $f\in L^1$ such that $\mathcal{F}$ has compact support, i.e. is in $C_c\subset C_0$? In particular, is this space dense in $L^1$? I believe this should be true, but I find it hard to come up with nontrivial examples of such $f$. I guess the formula $\mathcal{F}(f\ast g)=\mathcal{F}(f)\mathcal{F}(g)$ should be useful?

Alexander
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  • One can characterize the space by a Paley-Weiner type condition: $\hat f$ has compact support if and only if $f$ is the restriction to $\Bbb R$ of an entire function $g\in H(\Bbb C)$ such that $|g(x+iy)|\le c\exp(A|y|)$. I don't see how this characterization helps showing the space is dense. Your "guess" sounds like a hint from somewhere, making this sound like a homework problem... Have you covered approximate identities? The Schwarz space? – David C. Ullrich Dec 12 '15 at 14:48
  • I don't have a copy of that book. Have you covered approximate identities? (Those $(\phi_n)$ things where $\phi_n*f\to f$.) Have you covered the Schwarz space? – David C. Ullrich Dec 12 '15 at 15:34
  • @DavidC.Ullrich yeah, all this stuff is known to me. How can we use these things here though? I guess we need to consider schwartz functions and cut these off using the formula mentioned in my question? – Alexander Dec 12 '15 at 16:52
  • Answered in answer. – David C. Ullrich Dec 12 '15 at 17:55

1 Answers1

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One can characterize the space of integrable functions such that the Fourier transform has compact support via a Paley-Wiener sort of argument; in fact $\hat f$ has compact support if and only if $f$ is (almost everywhere) the restriction to $\Bbb R$ of an entire function $g$ with $$|g(x+iy)|\le ce^{A|y|}.$$

I don't see how this characterization helps in showing that the space is dense. But that's not hard. Since the Fourier transform maps the Schwarz space to itself there exists a Schwarz function $\phi$ with $\int\phi=1$ such that $\hat\phi$ has compact support. (You can make $\hat\phi$ be any infinitely differentiable function with compact support.)

Define $$\phi_\delta(t)=\frac1\delta\phi\left(\frac t\delta\right).$$Then $\hat\phi_\delta$ has compact support, so if $f\in L^1$ then $\widehat{\phi_\delta*f}$ has compact support, and $||\phi_\delta*f-f||_1\to0$ as $\delta\to0$.

Edit: Come to think of it, it's even easier than this. For some reason I wanted a $\phi$ that gave pointwise convergence. All we need is convergence in norm, and that's much easier: We can let $\phi$ be any $L^1$ function such that $\hat \phi$ has compact support.

(So for example $\phi=\hat\psi$ where $\psi$ is just $C^2_c$, or where $\psi$ is the convolution of any two $L^2$ functions with compact support...)

David C. Ullrich
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  • nice! I somehow tried to come with an explicit example of such an $\phi$, but it seems not so easy to construct. – Alexander Dec 12 '15 at 21:41
  • Well, there are various ways of constructing an infinitely differentiable function $\psi$ with compact support, some more "explicit" than others; then $\phi=\hat\psi$ works. Meanwhile see edit... – David C. Ullrich Dec 13 '15 at 02:48
  • Is it possible to construct a sequence whose convergence to f is faster than the rate their supports gets larger? – Carl Lincoln Mar 01 '21 at 20:42
  • @CarlLincoln Exactly what do you mean by "whose convergence to f is faster than the rate their supports gets larger"? – David C. Ullrich Mar 02 '21 at 11:46
  • The size of the support of $\hat{\phi}_\delta$ relative to size of support of $\hat{\phi}$ is about $1/\delta$, so as $\delta$ goes to zero the size of supports increase with the rate $1/\delta$. My question is how about the convergence rate of L1 norm as $\delta$ does to zero in terms of $\delta$? – Carl Lincoln Mar 02 '21 at 21:29
  • @CarlLincoln Regarding "the size of the support of $\hat\phi$ relative to the size of the support of $\phi$": in fact $\phi$ and $\hat\phi$ cannot both have compact support unless $\phi=0$. – David C. Ullrich Mar 03 '21 at 09:45