Lemma 1

— Let $\overline{u}$ be the lower envelope of the continuous affine linear functions on E that are $\geq u$ on K. Then S is the set of points $(a, b) \in E \times \mathbf{R}$ such that $a \in K$ and $u(a) \leq b \leq \overline{u}(a)$.

By the Hahn-Banach theorem, for $(a, b)$ to belong to S, it is necessary and sufficient that $h(a, b) \geq 0$ for every continuous affine linear function $h$ on $E \times \mathbf{R}$ such that $h(x, u(x)) \geq 0$ for $x \in K$. Setting $h(x, t) = f(x) - \lambda t$, where $f$ is a continuous affine linear function on E, we see that the relation $(a, b) \in S$ is equivalent to the following property: the relation

implies

(5)

First set $\lambda = 0$; the fact that (4) implies (5) for every continuous affine linear function $f$ on $E$ is equivalent to the relation $a \in K$ by the Hahn-Banach theorem. Next, set $\lambda = -1$ and replace $f$ by $-f$; then the relation $f(x) \leq u(x)$ in $K$ must imply $f(a) \leq b$. But since $u$ is convex and continuous on $K$, $u(a)$ is equal to the supremum of the $f(a)$ for the continuous affine linear functions $f$ on $E$ such that $f(x) \leq u(x)$ on $K$ (TVS, II, §5, No. 4, Prop. 5); we thus obtain the relation $b \geq u(a)$. Finally, set $\lambda = 1$; to say that (4) implies (5) then means, by definition, that $b \leq \overline{u}(a)$. This proves the lemma, since the relation (4) (resp. (5)) is equivalent to the one obtained by multiplying both sides by a scalar $> 0$.