Convex function can be written as supremum of some affine functions

Question

Let $\phi: \mathbb{R} \to \mathbb{R}$ be a convex function. Prove that $\phi$ can be written as the supremum of some affine functions $\alpha$, in the sense that $\phi(x) = \sup_\alpha \alpha(x)$ for every $x$, where each $\alpha$ is defined by$$\alpha: x \mapsto a_\alpha x + b_\alpha$$for some $a_\alpha$ and $b_\alpha$.

My progress is as follows. I can show that if $\phi$ is convex and $x \in \mathbb{R}$, there exists a real number $c$ such that$$\phi(y) \ge \phi(x) + c(y - x)$$for all $y \in \mathbb{R}$.

But I am at a loss on how to continue, how to finish. Could aybody help?

Answer 1

If $\phi$ is convex, for each point $(\alpha, \phi(\alpha))$, there exists an affine function $f_\alpha(x) = a_\alpha x + b_\alpha$ such that

• the line $L_\alpha$ corresponding to $f_\alpha$ passes through $(\alpha, \phi(\alpha))$;

• the graph $\phi$ lies above $L_\alpha$.

Let $A = \{f_\alpha: \alpha \in \mathbb{R}\}$ be the set of all such functions. We have

• $\sup_{f_\alpha \in A} f_\alpha(x) \geq f_x(x) = \phi(x)$ because $f_x$ passes through $(x, \phi(x))$;

• $\sup_{f_\alpha \in A} f_\alpha(x) \leq \phi(x)$ because all $f_\alpha$ lies below $\phi$.

We conclude that $\sup_{f_\alpha \in A} f_\alpha(x)= \phi(x)$.