Convex Functions: Lower Semicontinuity and Epigraph

Question

In the lecture notes by Christian Clason, on page 24, there is an interesting Theorem he proves (he calls it "Lemma 3.1"), namely:

Let $F: X \to \overline{\mathbb R}$. Then $\text{epi} F$ is

[...]

(iii) (weakly) closed if and only if $F$ is (weakly) lower semicontinuous.

There is also a proof given, I don't understand a detail in the direction $\text{epi}(F)$ is closed $\Rightarrow F$ is lower semicontinuous. From the lecture notes:

Let conversely $\text{epi}(F)$ be closed and assume that $F$ is not lower semicontinuous. Then there exists a sequence $\{x_n\}_{n\in \mathbb N}\subset X$ with $x_n\to x\in X$ and $F(x) > \lim\inf_{n\to\infty} F(x_n) =: M \in \left[-\infty, \infty \right)$.

We now distinguish two cases:

a) $x\in \text{dom}(F)$: In this case, we can select a subsequence, again denoted by $\{x_n\}_{n\in\mathbb N}$, such that there exists an $\epsilon > 0$ with $F(x_n)\leq F(x)-\epsilon$.

I have huge problems understanding why we can choose such a subsequence as described if $x\in \text{dom}(F) := \{x\in X \ : \ F(x) < \infty\}$.

Answer 1

We can always choose such a subsequence, this isn't where they use the fact that $x\in\mathrm{dom}F$. The fact that $x\in\mathrm{dom}F$ is used when they write "and thus $(x_n, F(x) - \epsilon)\in\mathrm{epi}F$ ". If $x\not\in\mathrm{dom}F$, then $F(x) - \epsilon = \infty$ and $(x_n, F(x) - \epsilon)\not\in\mathrm{epi}F$.