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.