Are convex function from a convex, bounded and closed set in $\mathbb{R}^n$ continuous? If I have a convex function $f:A\to \mathbb{R}$, where $A$ is a convex, bounded and closed set in $\mathbb{R}^n$, for example $A:=\{x\in\mathbb{R}^n:\|x\|\le 1\}$ the unit ball. Does this imply that $f$ is continuous? I've searched the web and didn't found a theorem for this setting (or which is applicable in this case). If the statement is true, a reference would be appreciated.
 A: No. A convex function is continuous in the interior of its domain, but it need not be continuous on the boundary.
For example, with $A = \{ x \in \mathbb{R}^n : \lVert x\rVert \leqslant 1\}$, where $\lVert \cdot\rVert$ is the Euclidean norm (or any strictly convex norm), the function
$$f(x) = \begin{cases}0 &, \lVert x\rVert < 1\\ g(x) &, \lVert x\rVert = 1 \end{cases}$$
is convex for every $g \colon \partial A \to [0,\infty)$.
A: The continuity  of convex functions defined on  topological vectors spaces   is rather well understood. For functions defined on  a finite dimensional Banach space, i.e., $\mathbb{R}^n$,  the classical monograph Convex Analysis by  R.T. Rockafellar  is a good place to check.  Let me first point out a simple trick used in convex analysis. $\newcommand{\bR}{\mathbb{R}}$ 
If $C\subset \bR^n$ is a convex set and $f: C\to \bR$ is  a convex function, then   we can define an extension
$$ \hat{f}:\bR^n\to (-\infty,\infty],\;\; \hat{f}(x)=\begin{cases} f(x), \;\; x\in C\\ 
\infty, &x\in\bR^n\setminus C.\end{cases} $$ 
The above extension is obviously convex. Hence  we may as well work from the very beginning  with convex functions $f:\bR^n\to (-\infty,\infty]$. The set  where $f<\infty$ is called the domain of $f$  and it is denoted by $D(f)$. The domain  is a convex  subset of $\bR^n$. It carries an induced  topology, and the interior of $C$ with respect to the induced topology is called the relative interior.   
Theorem  10.1  in the above mentioned book of Rockafellar shows that  the restriction of a convex function to the relative interior of its domain is a continuous function.  
For example, any convex function defined on the closed unit ball in $\bR^n$, must be continuous in the interior of the ball. Daniel Fisher's example shows that's the best one could hope for.
