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.
Real Analysis – Continuity of Convex Functions from Convex, Bounded, and Closed Sets
convex-analysisreal-analysis
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Best Answer
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 $g \colon \partial A \to [0,\infty)$.