My professor often hints that the Fourier Transform of a compactly supported $L^1$ function is of class $C^{\infty}$. Is this always true? How can I see this?


Yes, it is. You can see this by differentiating under the integral, directly from the definition of the Fourier transform, i.e., $$ \partial_\xi^\alpha \widehat{f}(\xi) = \int f(x) \partial^\alpha_\xi e^{-2\pi i x \xi } dx = \int f(x) (-2\pi i x)^\alpha e^{-2\pi i x \xi} dx, $$ where $\alpha$ is any multiindex.

You can use a standard theorem (see e.g. Difference of differentiation under integral sign between Lebesgue and Riemann to justify the differentiation under the integral). Just use that if $f$ has support contained in $B_R (0)$, then $|f(x) (-2\pi i x)^\alpha e^{-2\pi i x\xi}| \leq (2\pi R)^{|\alpha|} \cdot |f(x)|$, which is integrable if $f$ is.

Note: Of course, you need $f$ to be integrable, but this is part of your assumptions already.

EDIT: As noted by @Julián Aguirre, the Fourier transform of a compactly supported function is actually analytic, i.e., it can be extended to an entire function on $\Bbb{C}^d$. To see this, one can use the power series expansion of $e^x$. For simplicity, I am here only giving the proof in the one-dimsensional case, but the claim is true also in $\Bbb{R}^d$. We have $$ \widehat{f}(\xi) = \int f(x) e^{-2\pi i x \xi} dx = \int f(x) \sum_{n=0}^\infty \frac{(-2\pi i x \xi)^n}{n!} dx = \sum_{n=0}^\infty \xi^n \cdot \underbrace{\int f(x) \frac{(-2\pi i x)^n}{n!}\, dx}_{=:a_n}, $$ where we still have to justify the interchange of integration and summation. But for $\xi \in \mathbb{C}$ with $|\xi| \leq T$ and for $\mathrm{supp} f \subset [-R,R]$, we have $$ \int \sum_{n=0}^\infty |f(x)| \left|\frac{(-2\pi i x \xi)^n}{n!}\right| dx \leq \sum_{n=0}^\infty \frac{(2\pi R T)^n}{n!} \| f \|_{L^1} = e^{2\pi RT} \cdot \|f\|_{L^1} < \infty. $$ We can then use dominated convergence to justify the interchange. Also, we get (absolute) convergence of the power series $\sum_n a_n \xi^n$. Since absolutely convergent power series are smooth, this provides another proof for the smoothness of $\widehat{f}$.

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  • $\begingroup$ But why can I differentiate under the integral? Also, how does this help me to conclude? $\endgroup$ – john melon May 18 '17 at 7:35
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    $\begingroup$ @johnmelon: I added some details. Does this help? $\endgroup$ – PhoemueX May 18 '17 at 7:37
  • $\begingroup$ yes perfect. thank you $\endgroup$ – john melon May 18 '17 at 7:41
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    $\begingroup$ I would like to add that $\widehat f$ is not only $C^\infty$, but an entire function, when $\xi$ is considered a complex variable. $\endgroup$ – Julián Aguirre May 18 '17 at 11:00
  • $\begingroup$ @JuliánAguirre: Thanks, I expanded my answer to account for your comment :) $\endgroup$ – PhoemueX May 22 '17 at 7:51

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