# Zero-crossing for convolution

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:

$$g(x):=\int_\mathbb{R}f(s)e^{\frac{-(s-x)^2}{2}}ds$$

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 !

• Not clear what the domain of $f$ is. Is in in 2d or 1d? Jul 31, 2023 at 13:41
• Sorry for the confusion but I thought it was clear by the $\mathcal{C}^2(\mathbb{R},\mathbb{R})$ definition. It is in one dimension. Jul 31, 2023 at 13:43
• For $g$ to have a zero the integral on the RHS must be equal to zero somewhere, which means the integrand is either identically zero or has equal positive and negative parts. Since $\mathrm e^{(...)}$ is always positive, $f$ therefore must have negative parts. Jul 31, 2023 at 13:52
• Thank you @K.defaoite, this proves the first point that claims that $f$ has a least one zero. But how do you prove that $f$ has at least 2 zeros ? (I might have misunderstood your argument) Jul 31, 2023 at 13:55
• Indeed this is more challenging than I originally gave it credit for. I think the way to go is the other way, i.e, to let $z$ be the only zero of $f$, then show that $f$ only having one zero means that $g$ can also only have one zero. Jul 31, 2023 at 14:28

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.
• Thank you very much @fedja, but how do you have that $A/B$ is increasing ? Jul 31, 2023 at 15:02
• @Gaetano Erm... We've just shown that $\frac{d}{ds}(\frac AB)=\frac{A'}A-\frac{B'}B>0$, haven't we? Jul 31, 2023 at 15:07
• @Gaetano You are welcome! For some reason I forgot to type $\log$ in my response (so the LHS should actually be $\frac d{ds}[\log\frac AB]$), but it changes nothing :-) Jul 31, 2023 at 15:55