# Lower Semicontinuity Equivalence

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 $$x_0 \in X$$).

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.

• Ok, but if the sequentially lower semicontinuous definition remains (not with nets or filters) the sufficiency fails in the $T_2$ topological spaces right? Apr 27 '20 at 20:25
• 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. Apr 27 '20 at 23:36
• I mean, only the Hausdorff separability is not enough for use sequences instead nest or filters. Apr 28 '20 at 14:57
• That is what I said earlier. For general Hausdorff spaces, if one used nets in place of sequences, the statement holds true. Apr 28 '20 at 17:03

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 $$x_0\in f^{-1}(-\infty,c]$$.