# 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

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