Suppose that you have a function $f\in\mathcal{C}^2(\mathbb{R},\mathbb{R})$. Can one show that if the function $g$ defined by:
$$g(x):=\int_\mathbb{R}f(s)e^{-(s-x)^2/2}ds$$
has to zeros $x_1$ and $x_2$, the function $f$ must have at least two zeros ?
It is obvious that $f$ must have one zero, the difficult part is for the second one. I think that we can proceed by assuming that $f$ has one zeros and $g$ two zeros, then concluding by contradiction. (But without success for me until now...).
Thank you very much !
Suppose that $f$ has only one zero. Then, WLOG, $f<0$ on $(-\infty,0)$ and $f>0$ on $(0,+\infty)$.
Let $A(s)=\int_0^\infty f(t)e^{-(s-t)^2/2}\,dt$, $B(s)=\int_{-\infty}^0 (-f(t))e^{-(s-t)^2/2}\,dt$. Then $A,B>0$ and $g=A-B$.
We have $\frac d{ds}A(s)=\int_0^\infty (t-s)f(t)e^{-(s-t)^2/2}\,dt> -sA(s)$ and, similarly, $\frac{d}{ds}B(s)< -sB(s)$. Hence $\frac {A'}A>-s>\frac {B'}B$, so $\frac AB$ is strictly increasing and, thereby, can turn exactly $1$ at only one point, so $g$ can have only one zero then.