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][1], 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.


  [1]: https://math.stackexchange.com/questions/4745372/zero-crossing-for-convolution/4745403?noredirect=1#comment10066982_4745403