Observe that \{x,t \mid f(x) \leq t\} \subseteq X \oplus \mathbb {R} is a closed convex set. Hence there is a hyperplane through (0,f(0)) so that the entire set is in one half-space. This means a nontrivial affine equation A(x,t) \geq b which is satisfied whenever t \geq f(x), and where A(0,f(0)) = b.

Clearly A(x,t) = \alpha _0(x) - at for some \alpha _0 \in X^*, a \in \mathbb {R}. If a = 0 we have \alpha _0(x) \leq b for all x, which impossible. So by normalizing let's set a = 1. This means \alpha _0 (x) + f(x) \geq b = f(0).

Recall that the minmax problem f|_0 is given by (X,X^*,L(x,\alpha ) \mapsto f(x) - \alpha (x)). Strong duality means \inf _x\sup _\alpha L(x,\alpha ) = \sup _\alpha \inf _x L(x,\alpha ). We always have the inequality \geq , so it suffices to identify an \alpha ^* so that \inf _x \sup _\alpha L(x,\alpha ) \leq \inf _x L(x,\alpha ^*)

Clearly, for our L, we have \inf _x\sup _\alpha L(x,\alpha ) = f(0), since the supremum is \infty unless x = 0. On the other hand, taking \alpha ^* = -\alpha _0, we have f(0) \leq f(x) - \alpha ^*(x) for all x by construction, finishing the proof.