It is well known that a convex function defined on $\mathbb{R}$ is continuous (it is even left and right differentiable. Now you can define a convex function for any normed vector space $E$ : $f : E\mapsto \mathbb{R}$ is convex iff $$f\big(\lambda x + (1-\lambda)y\big) \le \lambda f(x)+(1-\lambda)f(y)$$
I know that such a function is not necessarily continuous if $E$ has infinite dimension: $f$ can be a discontinuous linear form. For instance, if $E = \ell^2(\mathbb{N})$ the space of square summable sequences (endowed with the supremum norm $||\cdot||_{\infty}$ instead of its natural norm), and $f(u) = \sum \limits_{i \ge 1} \frac{u_i}{i}$, then $f$ is linear, thus convex, yet it is well-known that $f$ is not continuous.
Now my question is: what about finite dimensions? Does there exist a convex function $f : \mathbb{R}^2 \to \mathbb{R}$ which is not continuous?
I know that there are discontinuous functions from $\mathbb{R}^2$ to $\mathbb{R}$ that have derivatives in every direction (that's a good start since this is a necessary condition !) but I don't know any that is convex.
Best Answer
Corollary 10.1.1 of Convex Analysis by Rockafellar says all convex functions from $\mathbb R^{n}$ to $\mathbb R$ are continuous. The proof is very long and it is not worth reproducing the complete proof here. In the infinite dimensional case there are are discontinuous linear functionals.