# Bound for Gaussian convolution roots

I am struggling on the following problem.

### The problem

Let $$f$$ be a real function, smooth, Lipschitz continuous and bounded. Let assume that $$f$$ is negative before 0 and positive after. For $$\sigma>0$$, we define the gaussian convolution $$E$$ by :

$$E(x,\sigma)\triangleq \frac{1}{\sigma\sqrt{2\pi}}\int_\mathbb{R}f(s)e^{-\frac{(s-x)^2}{2\sigma^2}}ds.$$

We know that under these conditions, the function $$E(\cdot,\sigma)$$ has only one zero-crossing denoted $$x_\sigma$$.

The problem is: Can we found a constant $$K$$ such that $$|x_\sigma|\leq K\sigma^2$$ ?

### What we know

• In this post, it is shown that this approximation exists for $$\sigma$$ tending to zero.

• It is perhaps useful to note that $$E$$ is the solution of the heat equation problem $$\mathcal{P}$$:

$$\mathcal{P} : \begin{cases}\partial_\sigma E(x,\sigma) = \sigma\partial_{xx}E(x,\sigma) \\ E(x,0) = f(x)\end{cases}.$$

• One can notice that the function is equal to: $$E(x,\sigma) = \mathbb{E}(f(x+\sigma Z)), ~Z\sim\mathcal{N}(0,1).$$

Thank you very much!

• "$f$ is negative before $0$ and negative after" -- so $f$ is negative almost everywhere, so $E(x,\sigma)<0$ for all $x$. Oct 8, 2023 at 18:36
• Thank you @user10354138 for noticing the error, I have edited the post. $f$ is negative before 0 and positive after. Oct 8, 2023 at 18:49
• Is the constant $K$ allowed to depend on $f$?