Why in Corollary 2.1 on page 10 (see the picture) from Ekeland and Temam book Convex Analysis and Variational Problems there is equality in (2.11), i.e why $$\forall u\in V,\quad \overline F(u)=\liminf\limits_{v\to u}{F(v)}$$
I only see the obvious relation $\overline F(u)\leq \liminf\limits_{v\to u}{F(v)}$. This follows from the fact that $\overline F(v)\leq F(v),\,\forall v\in V$ and thus $$\liminf\limits_{v\to u}{\overline F(v)}\leq \liminf\limits_{v\to u}{F(v)}$$ But also $\overline F$ is lower semi-continuous, so we get $$\overline F(u)\leq \liminf\limits_{v\to u}{\overline F(v)}\leq \liminf\limits_{v\to u}{F(v)}$$
Another observation: If $F$ is convex and l.s.c, then $F$ coincides with its $\Gamma$-regularization, i.e $\Gamma(F)=F$. Then we have $\overline F=F$. This follows from the inequality $\Gamma(F)\leq \overline F\leq F$. Here, the $\Gamma$-regularization is the pointwise supremum of all affine minorants of $F$. The above observation means that for convex and l.s.c functionals, the lower semicontinuity property is no longer inequality, but equality.
I think the line "Let $G$ be a function such that ..." should continue with "... $\text{epi} G = \overline{\text{epi} F}$".
Take this $G$. Its epigraph is the closure of $\text{epi} F$, i.e. the smallest closed set to contain $\text{epi} F$. Thus there cannot be any l.s.c. function $\bar{F} \leq F$ with $G(v) < \bar{F}(v)$ in some point $v$. Therefore $G = \bar{F}$.
To check (2.11), let's define $H(u) := \liminf_{v \to u} F(v)$. Then $G \leq H$, because the liminf is taken over a larger set to define $G$. Moreover, $H \leq F$ and $H$ is "obviously" l.s.c. But $G$ is the largest one with these properties, hence $G = H$.
(Note that the closure of the epigraph means the topological closure and in liminf you use filters. If you take sequential closure instead, you can/should use sequences.)