Is the convex conjugate of a strictly convex function still strictly convex?

Question

Let $f:\mathbb{R}^n \to \mathbb{R}$ be a strictly convex function, i.e., for any $\lambda\in(0,1)$ and $x\neq y \in \mathbb{R}^n$, $f(\lambda x + (1-\lambda) y) < \lambda f(x) + (1-\lambda) f(y)$. Let $f^\ast$ be the convex conjugate of $f$, i.e., $f^\ast(x^\ast) = \max_{\mathbb{R}^n} \Big\{{x^\ast}^\top x - f(x)\Big\}$. Is $f^\ast$ still a strictly convex function on its domain?

Answer 1

The answer is NO.

Let $C$ be a compact convex set containing more than just one point.
Set $$f(x)=\sigma_C(x)+\tfrac{1}{2}\|x\|^2,$$ where $\sigma_C$ is the support function of $C$.
Note that $\sigma_C$ is convex and has full domain.
Hence $$f\; \text{is strictly convex}$$ because $\tfrac{1}{2}\|x\|^2$ is strictly convex.
On the other hand, $$f^* = \tfrac{1}{2}d_C^2,$$ where $d_C$ is the distance function of the set $C$. Note that $f^*(c)=0$ for all $c\in C$, so $f^*$ is not strictly convex because $C$ contains more than one point.