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 $x_\sigma^*$ 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 $x^*$ 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.

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.

Let $f$ be a real infinitely differentiable function, negative before 0, positive after. 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 $x_\sigma^*$ 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 $x^*$ 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.

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 $x_\sigma^*$ 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 $x^*$ 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.