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Let $f\in$$L^1(\mathbb{R})$ and $g(y)=\int_{\mathbb{R}}f(x)e^{-iyx}dx$. Show that if $f$ has compact support, then $g$ can not have compact support unless $f=0$.

First, I assume that $g$ has compact support. Since $f$ has compact support, there is an interval $I$ and a nonnegative integer $n$ such that I has length $2n\pi$ and $g(y)=\int_{\mathbb{R}}f(x)e^{-iyx}dx=\int_{I}f(x)e^{-iyx}dx$. Also, since $g$ has compact support, there is a positive integer $N$ such that $g(k)=0$ for all integer $k$ with $N\leq|k|$. This shows that the Fourrier coefficients $c_k$[$f$] of $f$ on $I$ is equal to zero if $N\leq|k|$. If I can show that $c_k$[$f$]$=0$ for $|k|<N$, then this question can be done. But I don't know how to show that $c_k$[$f$]$=0$ for $|k|<N$.

orangeskid
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Zank
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    I'm voting to close this question as a possible duplicate of http://math.stackexchange.com/questions/154454. – Watson Apr 01 '17 at 18:46

1 Answers1

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If $f$ has compact support, its Fourier transform $\hat{f}$ can be extended to a holomorphic function on a certain domain of $\mathbb{C}$ (you just have to set $g(z) = \int f(t)e^{-itz}dt$ and apply the holomorphy under integral sign theorem).

Therefore, if $\hat{f}$ has also compact support, it is zero on a set with an accumulation point, therefore it is zero everywhere by the isolated zeros theorem : $\hat{f}=0$. But then, as the fourier transform is injective, this implies $f=0$.

Tlön Uqbar Orbis Tertius
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    +1, great answer. I would say that $\hat f$ can be extended to all of $\mathbb{C}$. Moreover, since the function $\hat f$ is zero on an open subset of $\mathbb{R}$ it will be zero on $\mathbb{C}$ ( this argument will work for the multidimensional case too). – orangeskid Jun 12 '15 at 09:15