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!

  • 1
    $\begingroup$ "$f$ is negative before $0$ and negative after" -- so $f$ is negative almost everywhere, so $E(x,\sigma)<0$ for all $x$. $\endgroup$ Oct 8, 2023 at 18:36
  • $\begingroup$ Thank you @user10354138 for noticing the error, I have edited the post. $f$ is negative before 0 and positive after. $\endgroup$
    – NancyBoy
    Oct 8, 2023 at 18:49
  • $\begingroup$ Is the constant $K$ allowed to depend on $f$? $\endgroup$
    – Adam
    Oct 15, 2023 at 14:25
  • $\begingroup$ Yes Adam it can. $\endgroup$
    – NancyBoy
    Oct 15, 2023 at 15:47


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