Analyticity of the convolution of two functions This question came about when trying to answer this one.
Does there exist a function $f\in C^\infty(\Bbb R)$ not identically zero such that:


*

*$f$ is supported on $[-1,1]$,

*$f$ is analytic on $(-1,1)$,

*The convolution $f*f$ is analytic at $0$?


Typical example of an $f$ satisfying 1. and 2. is $e^{-1/(1-t^2)}\chi_{[-1,1]}$, where $\chi_A$ is the characteristic function of the set $A$.
 A: Here is an argument that such a function does not exist under the additional assumption that $f$ is odd or even. Let $F = f \ast f$. Then for any $n \ge 0$ we have that
$$
F^{(2n)}(0) = (f^{(n)} \ast f^{(n)})(0) = \pm \int_{-1}^1 (f^{(n)}(t))^2 \, dt = \pm \| f^{(n)} \|_2^2.
$$
Assuming that $F$ is analytic in a neighborhood of $0$ implies that there exists some $C$ such that
$$
\| f^{(n)} \|_2^2 = |F^{(2n)}(0)| \le C^n (2n)!
$$
for all $n \ge 0$. Now let $a = \inf \{ x : f(x) \ne 0 \}$ be the left endpoint of the support. Applying Taylor's theorem with the integral form of the remainder around $a$ (where the Taylor series is zero) we get that for any $x \ge a$:
$$
\begin{split}
|f(x)|^2 & = \frac{1}{(n!)^2}\left( \int_{a}^x (x-t)^n f^{(n+1)}(t) \, dt  \right)^2 \le \frac{1}{(n!)^2} \int_{a}^x (x-t)^{2n} \, dt  \cdot \int_{a}^x (f^{(n+1)}(t))^2 \, dt \\
 & \le \frac{1}{(n!)^2} \frac{(x-a)^{2n+1}}{2n+1} \| f^{(n+1)} \|_2^2 \le \frac{(2n+2)!}{(n!)^2} \cdot \frac{C^{n+1} (x-a)^{2n+1}}{(2n+1)} \\
& = (2n+2) \binom{2n}{n} C^{n+1} (x-a)^{2n+1} \le (2n+2) \cdot 4^n \cdot C^{n+1} (x-a)^{2n+1},
\end{split}
$$
using Cauchy-Schwarz as well as the estimate $\frac{(2n)!}{(n!)^2} = \binom{2n}{n} \le 4^n$ for the central binomial coefficient. This shows that for all $|x-a|$ sufficiently small we get that $f(x) =0$, i.e., $f$ vanishes in a neighborhood of $a$, contradicting the definition of $a$.
Additional remarks about the general case: Note that this argument does not use the assumption (2) of analyticity of $f$. However, if one drops both the requirements that $f$ is odd or even and that it is analytic in $(-1,1)$, then there are simple examples of such functions $f$. E.g., any smooth function supported on $[1/2,1]$ will have the property that $f \ast f$ is supported on $[1,2]$, so it is trivially analytic near $0$. This indicates that the argument above does not easily apply to the general case of a function which is not odd or even.
