The goal is to prove that if epigraph of a function $f:X \rightarrow \mathbb{R}$ is closed then it is lower semicontinuous. The epigraph of $f$, $\operatorname{epi} f$ is given as
$$
\operatorname{epi} f = \{ (x, r): f(x) \leq r \}
$$
while a lower semicontinuous function is defined as a function for which:
$$
f(\bar{x}) \leq \liminf_{u \rightarrow \bar{x}} f(u)
$$
for all $\bar{x} \in X$, where $X$ is a normed space. I have been able to prove the converse but need some pointers on how to proceed in this direction.
Best Answer
Take a sequence $\left \{ x_{n} \right \}$ such that $x_{n }\to x_0\in X$ and suppose $\liminf f(x_n)<f(x_0).$ Then, there is a $\gamma$ and a subsequence $\left \{ x_{n_k} \right \}$ such that $f(x_{n_k})\le \gamma<f(x_0).$ Therefore, $(x_{n_k},\gamma )\in \text{epi} f.$ But now we have a contradiction, because since $x_{n_k}\to x_0$ and the epigraph is closed, we should have $f(x_0)\le \gamma.$