Let $f$ be a :


1. $f\in\mathcal{C}^\infty(\mathbb{R},\mathbb{R})$,
2. for all $x> 0,~f(x)>0$,
3. for all $x< 0,~f(x)<0$,


I am struggling to find a bound for the distance between the root of $f$ and the root of its Gaussian convolution. Formaly,



> 
> Let $f$ satisfies the previous properties and define $g$ as:
> $$g\_\sigma(x):=\int\_\mathbb{R}f(s)e^{-\frac{(x-s)^2}{2\sigma^2}}ds.$$
> From [this post](https://math.stackexchange.com/questions/4745372/zero-crossing-for-convolution/4745403?noredirect=1#comment10066982_4745403), we know that $g\_\sigma$ has a unique 0, denoted ${\lambda\_\sigma}$. How can we bound the convolution's zero $\lambda\_\sigma$ depending on $\sigma$ and the derivatives of $f$ ?
> 
> 
> 


Numericaly, I think that $\lambda\_\sigma$ is linear in $\sigma$. But I have no clue to for the derivatives...


Any hint, ideas or solution will be highly appreciated ! Thank you very much !


Since $\lambda\_\sigma$ is a function of $f$ and $\sigma$, I have thought the implicit function theorem but I don't know how to apply it in this case.



Distance between root of $f$ and its Gaussian convolution
$\newcommand{\R}{\mathbb R}\newcommand{\la}{\lambda}\newcommand{\si}{\sigma}$After some simple rewriting, we see that $\la\_\si$ is the root $x=x\_\si$ of the equation $G(\si,x)=0$, where
\begin{equation\*}
 G(\si,x):=Ef(x+\si Z)[=\si\sqrt{2\pi}\,g\_\si(x)\text{ if }\si>0] 
\end{equation\*}
and $Z$ is a standard normal random variable.
To ensure that $G(\si,x)$ or, equivalently, $g\_\si(x)$ is finite for all real $x$ and all small enough $\si>0$, assume that
\begin{equation\*}
 |f(x)|\le Ce^{Cx^2/2}
\end{equation\*}
for some real $C>0$ and all real $x$. We will also require that $f$ does not vanish at $\pm\infty$, so that
\begin{equation\*}
 f\_a:=\inf\{|f(x)|\colon|x|\ge a\}>0 
\end{equation\*}
for each real $a>0$ and
\begin{equation\*}
 |f'(x)|+|f''(x)|\le Ce^{Cx^2/2}
\end{equation\*}
for the same real $C>0$ and all real $x$.


In view of your conditions on $f$, for each real $a>0$ and all real $x\ge a$ we have
\begin{equation\*}
\begin{aligned}
 G(\si,x)&\ge Ef(x+\si Z)1(Z>0)+Ef(x+\si Z)1(Z<-x/\si) \\ 
 &\ge f\_a/2-CEe^{Cx^2+C\si^2Z^2}1(Z<-x/\si)\to f\_a/2>0 
\end{aligned}
\tag{10}\label{10}
\end{equation\*}
uniformly in real $x\ge a$ as $\si\downarrow0$. Similarly, for each real $a>0$ and all small enough $\si>0$ we have $G(\si,x)<0$ for all real $x\le-a$. Thus,
\begin{equation\*}
 x\_\si\to0\quad\text{as } \si\downarrow0. 
\end{equation\*}


Let $G\_1(\si,x)$ and $G\_2(\si,x)$ denote, respectively, the first and second partial derivatives of $G(\si,x)$ w.r.t. $\si$, so that
\begin{equation\*}
 G\_1(0,x)=EZf'(x)=0 
\end{equation\*}
and
\begin{equation\*}
 \text{$G\_2(\si,x)=EZ^2f''(x+\si Z)\to f''(x)$ as $x\to0$ and $\si\downarrow0$.}
\end{equation\*}
So, by Taylor's expansion, for some $c\_\si$ between $0$ and $x\_\si$,
\begin{equation\*}
\begin{aligned}
 0=G(\si,x\_\si)&=G(0,x\_\si)+G\_1(0,x\_\si)\si+G\_2(c\_\si,x\_\si)\si^2/2 \\ 
 &=f(x\_\si)+0\,\si+(f''(0)+o(1))\si^2/2 \\ 
 &=(f'(0)+o(1))x\_\si+0\,\si+(f''(0)+o(1))\si^2/2 
\end{aligned}
\end{equation\*}
as $\si\downarrow0$.


Thus,
\begin{equation\*}
 x\_\si=-\frac{f''(0)+o(1)}{2f'(0)}\,\si^2
\end{equation\*}
as $\si\downarrow0$, provided that $f'(0)\ne0$.




---


**Details on \eqref{10}:** For each real $a>0$, $\si\in(0,\min(a,(4C)^{-1/2}))$, and real $x\ge a$,
\begin{equation\*}
\begin{aligned}
 G(\si,x)&=Ef(x+\si Z)1(x+\si Z>0)+Ef(x+\si Z)1(x+\si Z<0) \\ 
 &\ge Ef(x+\si Z)1(Z>0)+Ef(x+\si Z)1(Z<-x/\si) \\ 
 &\ge Ef(x)1(Z>0)-CEe^{C(x+\si Z)^2/2}1(Z<-x/\si) \\ 
 &\ge f(x)P(Z>0)-CEe^{Cx^2+C\si^2Z^2}1(Z<-x/\si) \\ 
 &\ge f\_a/2-Ce^{Cx^2}\int\_{-\infty}^{-x/\si}e^{(C\si^2-1/2)z^2}\,dz \\ 
 &\ge f\_a/2-Ce^{Cx^2}\int\_{-\infty}^{-x/\si}e^{-z^2/4}\,dz \\ 
 &\ge f\_a/2-C\sqrt2\,e^{Cx^2}e^{-(x/\si)^2/4} \\ 
 &\ge f\_a/2-C\sqrt2\,e^{(C-1/(4\si^2))a^2} \underset{\si\downarrow0}\longrightarrow f\_a/2>0. 
\end{aligned} 
\end{equation\*}


