Prove that if f $c$-concave function and $c(x,y) = |x-y|^2/2$, then $|x|^2/2 - f(x)$ is convex

Question

Let $c$ be a symmetric function. A function $f:\mathbb R^d \to \mathbb R$ is $c$-concave if there exists a function $g$ such that $$f = \inf_y \left\{ c(x,y) - g(y) \right\}. $$

Now, let $c(x,y) = \frac{|x-y|^2}{2}$. How does one then prove that if $f$ is $c$-concave, then $h(x) = \frac{|x|^2}{2}-f(x)$ is convex? I saw this assertion in a book, but I've not been able to prove it.

Answer 1

Plugging in the formula for $f$, you get: \begin{align} h(x) &= \sup_y \left\{ 0.5x^2 - 0.5(x-y)^2 + g(y) \right\} \\ &= \sup_y \left\{ xy - 0.5y^2 + g(y) \right\}. \end{align} The supremum of affine functions is convex.