Lemma 2

If $u$ is strictly convex on $K$, then $u(x) < \overline{u}(x)$ for every non-extremal point $x$ of $K$.

For, there then exist two distinct points $y, z$ of $K$ such that $x = (y + z)/2$, whence $u(x) < (u(y) + u(z))/2$ since $u$ is strictly convex. If $f$ is an affine linear function on $E$, then $f(x) = (f(y) + f(z))/2$; applying this relation to the continuous affine linear functions $f$ that are $\geq u$ on $K$, we obtain

These lemmas established, we therefore (Lemma 1) have $(a, \overline{u}(a)) \in S$ for all $a \in K$. Since $G$ is compact, being the image of $K$ under the continuous mapping $x \mapsto (x, u(x))$, there exists, by Prop. 1 of No. 1, a positive measure $\nu$ on $G$, of total mass 1 , having $(a, \overline{u}(a))$ as barycenter. Since the restriction of the projection $pr_1$ to $G$ is a homeomorphism of $G$ onto $K$, the measure $\nu$ may be transported by means of this homeomorphism, which yields a measure $\mu$ on $K$ (positive and of total mass 1 ) such that

(6)

The first of the relations (6) means that $a$ is the barycenter of $\mu$. The function $\overline{u}$ is upper semi-continuous and bounded on $K$, hence is $\mu$-integrable (§4, No. 4, Cor. 1 of Prop. 5); moreover, since the function $-\overline{u}$ is by definition convex, we have, by the Cor. of Prop. 2 of No. 1,

(7)

whence, on comparing with the second of the formulas (6),

But since $u(\mathbf{x}) \leq \overline{u}(\mathbf{x})$ on K, the relation (8) implies that $u(\mathbf{x}) = \overline{u}(\mathbf{x})$ almost everywhere for $\mu$. Taking into account Lemma 2, one sees that Theorem 1 will be proved once the following lemma has been established: