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? 
 A: 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, $\infty$.
