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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 $x_0 \in X$).

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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.

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  • $\begingroup$ Ok, but if the sequentially lower semicontinuous definition remains (not with nets or filters) the sufficiency fails in the $T_2$ topological spaces right? $\endgroup$ Apr 27 '20 at 20:25
  • $\begingroup$ Not quite sure follow your question. If the space (besides Hausdorff separability) is also first countable, then it is enough to consider sequences; otherwise nets are a good device to probe continuity-like properties. In you case, sequences are enough because your space id a normed space. $\endgroup$ Apr 27 '20 at 23:36
  • $\begingroup$ I mean, only the Hausdorff separability is not enough for use sequences instead nest or filters. $\endgroup$ Apr 28 '20 at 14:57
  • $\begingroup$ That is what I said earlier. For general Hausdorff spaces, if one used nets in place of sequences, the statement holds true. $\endgroup$ Apr 28 '20 at 17:03
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I'll call $f$ "set-lower semicontinuous" if $f^{-1}(-\infty,c]$ is closed for all $c\in\mathbb{R}$, and "net-lower semicontinuous" if $f(x_0)\leq\liminf f(x_n)$ whenever $x_n\to x_0$.

First assume that $f$ is set-lower semicontinuous, and suppose $x_n\to x_0$. Given $\epsilon>0$, let $c_\epsilon=f(x_0)-\epsilon$. Then $f^{-1}(-\infty,c_\epsilon]$ is closed, thus $f^{-1}(c_\epsilon,\infty)$ is open, and hence contains a tail $\left\{x_n:n\geq N\right\}$ of the sequence $\left\{x_n\right\}$. Thus $x_n\in f^{-1}(c_\epsilon,\infty)$ for $n\geq N$, i.e. $$c_\epsilon\leq\inf_{n\geq N}f(x_n)\leq\liminf f(x_n).$$

Taking $\epsilon\to 0$ we have $c_\epsilon\to f(x_0)$, so we conclude that $f$ is net-lower semicontinuous.


In the other direction, suppose $f$ is net lower-semicontinuous. Let $c\in\mathbb{R}$ and $x_n\to x_0$ with $x_n\in f^{-1}(-\infty,c]$. We need to prove that $x_0\in f^{-1}(-\infty,c]$. The net-lower semicontinuous condition yields $$f(x_0)\leq\liminf_n f(x_n)=\sup_N\inf_{n\geq N}f(x_n)\leq\sup_N c=c,$$ so $x_0\in f^{-1}(-\infty,c]$.

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