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:

  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 $A$.

  • 3
    $\begingroup$ Do you know whether $e^{-1/(1-t^2)}\chi_{[-1,1]}$ satisfies 3.? $\endgroup$ – md2perpe Aug 24 '18 at 19:38
  • $\begingroup$ 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. $\endgroup$ – Julián Aguirre Aug 24 '18 at 21:48
  • 2
    $\begingroup$ what does analytic at $0$ mean? $\endgroup$ – zhw. Aug 25 '18 at 1:06
  • 1
    $\begingroup$ It can be expanded as a power series on a neighborhood of $0$. $\endgroup$ – Julián Aguirre Aug 25 '18 at 10:47
  • $\begingroup$ Doesn't $f = 0$ work? $\endgroup$ – 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.

  • $\begingroup$ Your $f$ is real. And how do you deal with the case $f_o \ast f_e$ $\endgroup$ – reuns Feb 12 at 1:53
  • $\begingroup$ @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. $\endgroup$ – Lukas Geyer Feb 12 at 3:46
  • $\begingroup$ Very nice argument. I have been trying to extend it to the general case without succes so far. $\endgroup$ – Julián Aguirre Feb 12 at 14:51
  • $\begingroup$ 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. $\endgroup$ – Lukas Geyer Feb 13 at 1:14

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