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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.

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up vote 7 down vote accepted

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 $g \colon \partial A \to [0,\infty)$.

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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.

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