Lower Semicontinuity Equivalence

Question

Let be $(X , ||\cdot||)$ a normed space and $f:X \longrightarrow \mathbb{R}$ a function. $f$ is lower semicontinuous if $\{x \in X:f(x) \leq c \}$ is closed $\forall c \in \mathbb{R}$, and is said to be sequentially lower semicontinuous at $x_0$ if for each sequence $\{x_n\}_{n \in \mathbb{R}}$ that converges to $x_0$ is verified that:
\begin{equation*}
f(x_0) \leq \liminf_{n \to \infty}f(x_n)
\end{equation*}
How to prove that in a normed space (the statement is valid for a first countable space) both conditions are equivalent, (f is a lower semicontinuous function if and only if is sequantially lower semicontinuous at each $x_0 \in X$).

Answer 1

Necessity: Suppose $f$ is lower semicontinuous and let $\{x_n:n\in \mathbb{N}\}$ be a sequence that converges to $x$. For any $\alpha < f(x)$ the set $V=\{f>\alpha\}$ is an open neighborhood of $x$. Hence there is $n_0\in \mathbb{N}$ such that $n\geq n_0$ implies that $f(x_n)>\alpha$; this implies that $\alpha\leq\liminf_nf(x_n)$. The conclusion follows by letting $\alpha\rightarrow f(x)$.

Sufficiency: It is enough to show that $F_\alpha:=\{f\leq \alpha\}$ is closed for any $\alpha\in\mathbb{R}$. Let $\{x_n:n\in \mathbb{N}\}$ be a sequence in $F_\alpha$ that converges to a point $x\in X$. Then $f(x_n)\leq \alpha$ for all $n\in \mathbb{N}$, and so \begin{aligned} f(x)\leq \liminf_nf(x_n)= \sup_{n\in \mathbb{N}}\inf_{m\in \mathbb{N}: m\geq n}f(x_m)\leq \alpha. \end{aligned} Therefore $x\in F_\alpha$.

Note: For general Hausdorff topological spaces, the statement holds after substituting sequences by nets.