Proposition 21

Let $f$ be a convex function, defined over the non-empty open convex set X in the topological vector space E. Then $f$ is continuous if, and only if, it is bounded above when restricted to some non-empty open subset U of X.

The condition is obviously necessary, we prove that it is sufficient. Let $x_0 \in X$ be a point such that $f$ is bounded above in a neighbourhood V of $x_0$; we show firstly that $f$ is continuous at $x_0$. By translation, we can restrict ourselves to the case when $x_0 = 0$ and $f(x_0) = 0$; moreover we can suppose that the neighbourhood V is balanced (I, p. 7, prop. 4). Suppose that $f(x) \leq a$ in V; for every $\varepsilon, 0 < \varepsilon < 1$, we observe that if $x \in \varepsilon V$, then $x/\varepsilon \in V$ and $-x/\varepsilon \in V$. Applying inequality (1) of II, p. 16 to the points $x/\varepsilon$ and 0 and to the number $\lambda = \varepsilon$, we see that $f(x) \leq \varepsilon f(x/\varepsilon) \leq \varepsilon a$; applying it to points $x$ and $-x/\varepsilon$ and the number $\lambda = 1/(1 + \varepsilon)$, gives $f(x) \geq -\varepsilon f(-x/\varepsilon) \geq -\varepsilon a$. Thus $f(x)$ is arbitrarily small in $\varepsilon V$, if $\varepsilon$ is sufficiently small, and $f$ is continuous at $x = 0$.

Now let $y$ be some point of X; since X is open, there is a number $\rho > 1$ such that $z = \rho y$ belongs to X. Let $g$ be the homothety $x \mapsto \lambda x + (1 - \lambda)z$ of centre $z$ and ratio $\lambda = 1 - \frac{1}{\rho}$, which transforms 0 into $y$; for every point $g(x) \in g(V)$, we have from (1)

Thus $f$ is bounded above in a neighbourhood of $y$ and hence, by the first part, is continuous at $y$. The proposition is proved.