An example of infimal convolution for nonconvex functions.

Question

Let $E$ be a normed vectorial space. Given two functions $\phi,\psi:E\to(−\infty, +\infty]$, the inf-convolution of $\phi$ and $\psi$ is defined as follows: for every $x\in E$, let $$ (\phi\nabla\psi)(x)=\inf_{x\in E}\{\phi(x-y)+\psi(y)\}. $$ In the particular case where $\psi(x)=\alpha\|x\|^2$, $\alpha>0$, we have that \begin{eqnarray} (\alpha\|\cdot\|^2\nabla\psi)(x)=\inf_{x\in E}\{\alpha\|x-y\|^2+\psi(y)\}. \end{eqnarray} It is a well-known result that if $\psi$ is convex, then $\alpha\|\cdot\|^2\nabla\psi$ is convex. I am looking for an example where that infimal convolution $\alpha\|\cdot\|^2\nabla\psi$ still convex but the function $f$ it is not convex. I have made some research, but I found results only when both functions are convex and I wasn't able to provide an example. So, there is an example when $\alpha\|\cdot\|^2\nabla\psi$ still convex even when $f$ is not?

Answer 1

The simplest example is $\psi(x) = -\alpha \, \|x\|^2$.