I'm reading below result from a lecture note.
Corollary 0.18. Let $X$ be a normed space, $f: X \rightarrow(-\infty,+\infty]$ a convex function, $A \subset X$ an affine set, and $a: A \rightarrow \mathbb{R}$ a continuous affine function such that $a \leq f$ on $A$. Assume that $$ \operatorname{int}(\operatorname{dom}(f)) \cap A \neq \emptyset, $$ and either $f$ is continuous at some point of $\operatorname{dom}(f)$, or $f$ is l.s.c. and $X$ a Banach space. Then there exists a continuous affine extension $\hat{a}: X \rightarrow \mathbb{R}$ of a, such that $\hat{a} \leq f$.
Below I show that the continuity of $a$ is implied by that of $f$ and the hypothesis $a \le f$ on $A$. Could you have a check on my attempt?
Assume that $f$ is continuous at $x_0 \in \operatorname{dom}(f)$. Then $f$ is bounded on a neighborhood $U$ of $x_0$. Then $a$ is bounded from above on $U \cap A$ which is a neighborhood of $a$ in the subspace topology of $A$. This combining with the fact that $a$ is affine (a translation of a linear map) shows that $a$ is continuous at $x_0$ and thus on $A$.
