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I am struggling with the following problem. Let $f$ be a real smooth function:

  • strictly convex on $(-\infty,0)$,
  • strictly concave on $(0,\infty)$,
  • strictly increasing.

For $\sigma>0$, how can one prove that the function $\Delta$ defined by:

$$\Delta(x) := f(x) - \frac{1}{\sqrt{2\pi}\sigma}\int_\mathbb{R}f(s)e^{-\frac{(s-x)^2}{2\sigma^2}}ds\quad\forall x\in\mathbb{R}$$

has a single zero on $\mathbb{R}$ ?

I am pretty sure that the statement is true (numerically). Any hints, solutions or counter-examples will be highly appreciated! Thank you very much.

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1 Answer 1

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This problem appears to be related to scale-space theory, which uses the diffusion-operator interpretation of Gaussian smoothing. Here is an interesting example of a false proof:

Assume there is $x<y$ with $\Delta(x)=\Delta(y)=0$, then $f(x)-f(y)<0$, but also by commutativity of convolution $$ f(x)-f(y) ~=~ \frac{1}{\sqrt{2\,\pi}\,\sigma}\,\int \limits_\mathbb{R} \left(f(s-x)-f(s-y)\right)\,e^{-\frac{s^2}{2\,\sigma^2}}\,ds~>~0.$$ The last inequality is true because both factors in the integral are positive. Contradiction!

Unfortunately this 'proof ' only uses the third requirement and thus fails for the counterexample $f(x) ~=~ 3\,x+\sin x,$ e.g. with the points $x=0$ and $y=k\,\pi,~k\in \mathbb{N}$. Probably one can fix this with the conditions 1) and 2) because the counterexample is based on the periodicity of sin and the fact that linear functions smooth to themselves.

This also implies that condition 3) is superfluous because any smooth function fulfilling 1) and 2) can be modified by adding $\alpha x$ with $\alpha> |f'(0)|$ to fulfill 3) without changing $\Delta$.

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  • $\begingroup$ Thank you @Karl Fabian. As you said, this function is more or less a smoothed Laplacian. It should detect only one zero if the original function $f$ has only one zero. But I have to admit that I can't manage to proove this property, do you have a sketch of a how to prove this ? Thank you ! $\endgroup$
    – NancyBoy
    Aug 20, 2023 at 9:27
  • $\begingroup$ Thank you for noticing me that the third condition was not necessary. $\endgroup$
    – NancyBoy
    Aug 20, 2023 at 11:48

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