Distance between root of $f$ and its Gaussian convolution

Question

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, 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.

Answer 1

$\newcommand{\R}{\mathbb R}\newcommand{\la}{\lambda}\newcommand{\si}{\sigma}$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 $f'(0)\ne0$.

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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*}