3
$\begingroup$

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 !

$\endgroup$
5
  • $\begingroup$ Not clear what the domain of $f$ is. Is in in 2d or 1d? $\endgroup$
    – user619894
    Jul 31, 2023 at 13:41
  • $\begingroup$ 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. $\endgroup$
    – NancyBoy
    Jul 31, 2023 at 13:43
  • $\begingroup$ 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. $\endgroup$
    – K.defaoite
    Jul 31, 2023 at 13:52
  • $\begingroup$ 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) $\endgroup$
    – NancyBoy
    Jul 31, 2023 at 13:55
  • $\begingroup$ 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. $\endgroup$
    – K.defaoite
    Jul 31, 2023 at 14:28

1 Answer 1

2
$\begingroup$

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.

$\endgroup$
6
  • $\begingroup$ Thank you very much @fedja, but how do you have that $A/B$ is increasing ? $\endgroup$
    – NancyBoy
    Jul 31, 2023 at 15:02
  • $\begingroup$ @Gaetano Erm... We've just shown that $\frac{d}{ds}(\frac AB)=\frac{A'}A-\frac{B'}B>0$, haven't we? $\endgroup$
    – fedja
    Jul 31, 2023 at 15:07
  • $\begingroup$ Absolutely, sorry I was confusing. Thank you very much for the clever answer ! $\endgroup$
    – NancyBoy
    Jul 31, 2023 at 15:12
  • $\begingroup$ @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 :-) $\endgroup$
    – fedja
    Jul 31, 2023 at 15:55
  • $\begingroup$ Hi @fedja, I posted a related question here if you have some time to spend on it :) $\endgroup$
    – NancyBoy
    Aug 6, 2023 at 13:16

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .