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Let $f:\mathbb{R}^d\to\mathbb{C}$ be a compactly supported, absolutely integrable function. Show that the function $\widehat{f}$ is real-analytic.

Since $f$ is compactly supported and absolutely integrable, then we have the estimate: $$\int_{\mathbb{R}^d} |x_j f(x)|\,dx\leq C\int_{\mathbb{R}^d} |f(x)|\,dx=C\|f\|_{L^1(\mathbb{R}^d)},$$ where $x_j$ is the $j$-th coordinate function, thus $x_jf$ lies in $L^1(\mathbb{R}^d)$. Notice that $$\frac{\partial }{\partial \xi_j}\widehat{f}(\xi)=-2\pi i\widehat{x_jf}(\xi),$$ it follows that $f$ is differentiable. Using induction, we can show that $f$ is $n$-th differentiable for all $n$, thus smooth, but how to show that $f$ is real-analytic?

Xiang Yu
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1 Answers1

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Since $f$ has compact support so has $x\mapsto f(x)\exp(ix\xi)$ (with $\xi$ fixed), so, up to a constant, the Fourier transform of $f$ is equal to $$ \begin{eqnarray} \int_{\mathbb{R}^n } f(x)\exp(ix\xi)\, dx & = & \int_{\mathbb{R}^n } f(x)\sum_{k=0}^\infty \frac{(ix\xi)^k}{k!} \, dx \\ & = & \sum_{k=0}^\infty \int_{\mathbb{R}^n } f(x) \frac{(ix\xi)^k}{k!} \, dx \\ & = & \sum_{k=0}^\infty \left(\int_{\mathbb{R}^n } f(x) \frac{(ix)^k}{k!} \, dx \,\right)\xi^k = \sum_{k=0}^\infty a_k \xi^k \end{eqnarray} $$

The second equality is justified by $f$ being continuous with compact support, see e.g. the answers to this question .

If you have doubts regarding the convergence of the last sum note that $$ \begin{eqnarray} \left|\int_{\mathbb{R}^n } f(x) \frac{(ix)^k}{k!} \, dx\right| &\le& \int_{\mathbb{R}^n }\left| f(x) \frac{(ix)^k}{k!}\right| \,dx \\ &\le& C \frac{R^{nk}}{k!} \end{eqnarray} $$ where $R$ is the radius of some ball which contains the support of $f$, so the radius of convergence of the series in the last line of the preceding calculation is, in fact, $\infty$.

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Thomas
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  • I think the Fourier transform on higher dimensions is a little subtle. The equality in the first display should be $$\sum_{k=0}^\infty \int_{\mathbb{R}^n} f(x)\frac{(i\xi\cdot x)^k}{k!}\ dx=\sum_{k=0}^\infty \int_{\mathbb{R}^n} f(x) \frac{(i(\xi_1 x_1+\cdots+\xi_nx_n))^k}{k!}\ dx=\sum \text{polynomial of}\ \xi_1,\dots,\xi_n$$. – Xiang Yu Feb 28 '16 at 07:57
  • @XiangYu Yes you are right, my bad. But the main point are the questions regarding convergence, which rely on estimates, not necessarily equlities. Will think of writing an improved answer if I find the time. – Thomas Feb 28 '16 at 08:03
  • Yes, the flaw can be easily remedied. – Xiang Yu Feb 28 '16 at 08:05
  • This function is not real-valued. – Ice sea Aug 16 '19 at 22:59
  • @Icesea which function is not real-valued? – Thomas Aug 17 '19 at 06:55
  • @Thomas The Fourier transform is not real-valued. Usually, real-analytic is an notion for real-valued functions https://www.encyclopediaofmath.org/index.php/Real_analytic_function. However, $\int_{\mathbb{R}^n}f(x)\exp{i x\xi}dx$ is not usually real-valued. – Ice sea Aug 17 '19 at 11:11
  • @Icesea that's right. You may consider this an abuse of notation, but the point here (and also in case of real analytic functions) is that the Fourier transform (which has a real domain, like $f$) admits, in the case considered, locally a representation through it's Taylor series (which is not an obious fact). – Thomas Aug 19 '19 at 13:04
  • @Icesea (and of course you can define vector valued (real) analytic maps by requiring that each component is real analytic. The word 'real' refers to the domain, not the target space. And so this can be applied, too, to complex valued functions, of course). – Thomas Aug 19 '19 at 13:21