# The Fourier transform of a compactly supported, absolutely integrable function is real-analytic

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?

• I did not check the details, but I'd write $e^{ix\xi}$ as a power series and then try to interchange integration and summation. – Thomas Feb 27 '16 at 13:59
• it reduces to showing $F(s) = \int_a^b f(x) e^{-s x} dx$ is an entire function – reuns Feb 28 '16 at 7:41
• @user1952009 Not suitable for higher dimensions. – Xiang Yu Feb 28 '16 at 7:50
• of course it is @Xiang – reuns Feb 28 '16 at 7:51

## 1 Answer

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, $\infty$.

• 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 7:57
• @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 8:03
• Yes, the flaw can be easily remedied. – Xiang Yu Feb 28 '16 at 8:05
• 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 6:55