# Exercise - Fourier Transform

I have got another question concerning the Fourier transform. I hope somebody can help me. Let $f \in L^1(\mathbb{R}^n)$.

• Prove that $\hat{f}$ is continous. (ok, I was able to show it)
• If $supp(f)$ is compact then $\hat{f}$ is infinitely often differentiable. (ok, I was able to show it)
• Argue that there are compactly supported $L^1$-functions $f$ with $\hat{f} \notin S(\mathbb{R^n})$ and $\hat{f} \notin L^1(\mathbb{R^n})$, respectively. (for the $L^1$-question you should use the explicit form of Fourier inversion).

The FIRST and SECOND point are easily proven, the first by dominated convergence for integrals and the second by differentiation of parameter dependent integrals.

Now, for the third point I do not how to proceed. In my notes, the fourier transform is defined for $f \in L^1(\mathbb{R^n})$ or $f\in S(\mathbb{R^n})$. However the recalled fourier inversion, is only valid between the Schwartz-space $S(\mathbb{R^n})$ and it states that we have $f(x)=\frac{1}{(2\pi)^{\frac{n}{2}}}\int_\mathbb{R^n} \hat{f(y)}e^{i<y,x>}dy$, where $\hat{f(y)}=\frac{1}{(2\pi)^{\frac{n}{2}}}\int_\mathbb{R^n} \hat{f(x)}e^{-i<x,y>}dy$. Moreover I know that the fourier transform of a compactly supported function, can not have compact support.

I have no clue how to prove this and any help is very appreciated! Thanks in advance!

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You can explicitly find an $f \in L^1$ with compact support, such that $\hat{f}$ is neither Schwarz nor in $L^1$. You should be able to write down such a function. – Christopher A. Wong Jan 24 '13 at 23:07

Hint: If $\hat{f}\in L^1$, then by Fourier Inversion, $f$ is continuous.
• The Fourier transform of a function $g\in\mathcal S(\Bbb R)$ is in $\mathcal S(\Bbb R)$. If the first assertion of the third point wasn't true, then each $L^1$ function with compact support would be represented by a Schwarz function.
• We can use Fourier inversion formula when the Fourier transform is integrable. With this, arguing by contradiction, we would have that each $L^1$ function vanishes at infinity, which is not the case.