Let $f$ be a bounded measurable function with support on the unit disk $\mathbb D \subset \mathbb R^2$ and let $g$ be an analytic function on $\mathbb R^2$. Is it true that the convolution $h = f \star g$ is again analytic?
Since $g$ is of compact support one can see that $\partial_x(f \star g) = f \star \partial_xg$ and the same holds for $\partial_y$, showing that $h$ is $\mathcal C^\infty$. But how can one deduce analyticity form this fact?
Best Answer
Probably the simplest way to see it is to use the fact that an analytic function on $\mathbb{R}^n$ extends to a holomorphic function in a neighbourhood of $\mathbb{R}^n \subset \mathbb{C}^n$. That extension is given locally by the power series expansion of $g$, which converges in the Reinhardt domain determined by its domain of convergence in $\mathbb{R}^n$.
Then, since $f$ has compact support, you can extend the convolution to a (possibly smaller) neighbourhood of $\mathbb{R}^n$: For a given $x_0 \in \mathbb{R}^n$ consider the compact set $K = \overline{B_1(x_0)} - \operatorname{supp} f$. ($A-B = \{a - b : a \in A, b\in B\}$ here, not the set difference.) For every $x \in K$, there is an $\varepsilon_x > 0$ such that the power series of $g$ with centre $x$ converges in $B_{\varepsilon_x}(x) + iB_{\varepsilon_x}(0)$. Since $K$ is compact, there are finitely many points $x_1, \dotsc, x_k \in K$ such that $K \subset \bigcup\limits_{i=1}^k B_{\varepsilon_{x_i}}(x_i)$. Let $\varepsilon = \min \{\varepsilon_{x_i} : 1 \leqslant i \leqslant k\}$. Then $\varepsilon > 0$ and $g$ is holomorphic at least in $K + iB_\varepsilon(0)$. For $x \in B_1(x_0)$ and $y \in B_\varepsilon(0)$, the convolution
$$(f\ast g)(x+iy) = \int_{\operatorname{supp} f} f(t)g(x+iy-t)\,dt$$
is thus well-defined.
Differentiating under the integral, we see that $f\ast g$ is holomorphic in $B_1(x) + iB_\varepsilon(0)$. Since $x_0 \in \mathbb{R}^n$ was arbitrary, $f\ast g$ is holomorphic in a neighbourhood of $\mathbb{R}^n$, whence its restriction to $\mathbb{R}^n$ is real-analytic.