# Extension of single zero-crossing property

Let $$f\in\mathscr{C}^2(\mathbb{R},\mathbb{R})$$ a strictly increasing function, striclty convex on $$(-\infty,0)$$, strictly concave on $$(0,\infty)$$ and let $$\sigma_1>\sigma_2>0$$ be two real numbers. Suppose that the function $$g$$ defined as:

$$g_{\sigma_1,\sigma_2}(x) \triangleq\int_\mathbb{R} (f(\sigma_1 s + x)-f(\sigma_2 s + x))e^{-s^2/2}ds$$

has only one zero in $$\mathbb{R}$$. Formally: $$~\exists! x^*\in\mathbb{R},~g(x^*) = 0$$.

1. I am numericaly quite sure that then, for all $$\sigma_1', \sigma_2' >0$$, the function $$g_{\sigma_1',\sigma_2'}$$ has only one zero in $$\mathbb{R}$$ (the zero location might obviously be different).

2. I am also very interested in weaker version of (1), if $$\alpha>0$$, the function $$g_{\alpha\sigma_1,\alpha\sigma_2}$$ has a unique zero.

I don't know how to prove this kind of "invariance" property...

## Possible hints

This question arises from signal processing. The function $$g$$ can be seen has a $$\text{DoG}$$ function (Difference of Gaussian).

Any hints, solution or even counter-examples will be highly appreciated. Thank you very much!