Proposition 5

Let $X$ be a closed convex set in a locally convex space $E$. Then every lower semi-continuous convex function $f$ defined in $X$ is the upper envelope of a family of functions that are the restrictions to $X$ of continuous affine linear functions in $E$.

For, the set $A \subset E \times \mathbf{R}$ of points $(x, t)$ such that $x \in X$ and $t \geqslant f(x)$ is convex (II, p. 17, prop. 19) and closed, since the function $(x, t) \mapsto f(x) - t$ is lower semi-continuous. Then let $x$ be any point of $X$ and let $a \in \mathbf{R}$ be such that $a < f(x)$. By cor. 1 of II, p. 38, there exists a closed hyperplane $H$ in $E \times \mathbf{R}$, that contains $(x, a)$ and does not meet $A$. Every linear continuous form on $E \times \mathbf{R}$ being of the form

where $\lambda \in \mathbf{R}$ and $u$ is a continuous linear form on $E$, it follows that $H$ has an equation of the form $u(z) + \lambda t = \alpha$, and as $H$ contains $(x, a)$ we have $\alpha = u(x) + \lambda a$. Now the point $(x, f(x)) \in A$ does not belong to $H$ and therefore $\lambda \neq 0$. Dividing by $-\lambda$, if necessary, we can write the equation of $H$ as $t - a = u(z - x)$. As $f(x) - a > 0$, we have, therefore, $f(z) > u(z - x) + a$ for all $z \in X$ and this proves the proposition.