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.

  • $\begingroup$ 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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.