# Analyticity of the convolution of two functions

Does there exist a function $$f\in C^\infty(\Bbb R)$$ not identically zero such that:

1. $$f$$ is supported on $$[-1,1]$$,
2. $$f$$ is analytic on $$(-1,1)$$,
3. 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 $$A$$.

• Do you know whether $e^{-1/(1-t^2)}\chi_{[-1,1]}$ satisfies 3.? – md2perpe Aug 24 '18 at 19:38
• No, I don't. I have tried examples that are not $C^\infty$, like $(1-x^2)\chi_{[-1,1]}$, but none of them satisfy 3. – Julián Aguirre Aug 24 '18 at 21:48
• what does analytic at $0$ mean? – zhw. Aug 25 '18 at 1:06
• It can be expanded as a power series on a neighborhood of $0$. – Julián Aguirre Aug 25 '18 at 10:47
• Doesn't $f = 0$ work? – mathworker21 Feb 11 at 21:53

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.
• Your $f$ is real. And how do you deal with the case $f_o \ast f_e$ – reuns Feb 12 at 1:53
• @reuns: I assumed the question was about real-valued $f$. But you are correct that the general case does not follow as easily from the odd/even case as I thought. – Lukas Geyer Feb 12 at 3:46
• Actually, I just noticed that this proof really does not use the analyticity of $f$, only of $F$ near $0$. (Also, analyticity would automatically imply that $a=-1$.) However, if $f$ is not analytic, there is an easy example of such a function, namely one that has a support contained in say $[1/2,1]$, for which $F$ would have support contained in $[1,2]$, so it would be trivially analytic near $0$. At least this shows that the general proof would need some substantial new argument. – Lukas Geyer Feb 13 at 1:14