Prove that the support function of a closed non-empty set is convex

Question

Let $C$ be a closed non-empty set, but not necessarily convex. The support function of $C$ is given by $$S(z) = \sup_{c \in C} \langle z,c\rangle. $$ Prove that this is a convex function.

Attempt

Let $a \in [0,1]$. Then

$$S_C(ax + (1 - a)y) = \sup_{c \in C} \langle ax + (1- a)y,c\rangle \leq \sup_{c \in C } \langle a,c\rangle + \sup_{c \in C} \langle (1-a)y,c\rangle$$

Can I actually break up the sum as $\sup_{c \in C } \langle a,c\rangle + \sup_{c \in C} \langle (1-a)y,c\rangle$? If it is not bounded?

Answer 1

The inequality $$ \sup_{x\in A} (f(x)+g(x))\le \sup_{x\in A} f(x)+\sup_{x\in A} g(x)$$ holds in general.

Answer 2

Here is another way:

The supremum of an arbitrary collection of convex functions is convex (this can be seen by noting that the epigraph of the supremum is the intersection of the constituent epigraphs).

Since each function $z \mapsto \langle c, z \rangle$ is convex, it follows that the supremum over all $c \in C$ is also convex.