Given $X\subseteq \Bbb R^m, f:X\to\Bbb R$ and $x\in X$, we say $f$ is lower semicontinous (l.s.c for short) at x if
$\forall \varepsilon>0\ \exists\ \delta >0\ \forall \in B(\delta,x), \ f(x)\le f(y)+\varepsilon$. I wish to show:
If $X$ is closed, then $f$ is l.s.c if and only if the set $f^{-1}((-\infty,r]):=\{a\in X:f(a)\le r\}$ is closed for each $r\in \Bbb R$.
I wonder how I could use the requirement that the set $X$ is closed.

  • $\begingroup$ No they are the same problem problem... $\endgroup$
    – Tomás
    Jan 18 '13 at 0:18
  • 1
    $\begingroup$ Please modify the original question rather than starting a new question asking the same thing. $\endgroup$
    – robjohn
    Jan 18 '13 at 0:27

Let's begin with the "if" part: I'm gonna show that if $y_n\rightarrow y$ then $f(y)\leq\liminf f(y_n)$. Let $\epsilon_n=\frac{1}{n}$; by hypothesis, for each $n$ we can find some $\delta_n$ such that $f(y)\leq f(y_n)+\epsilon_n$. Now you take the inferior limit.

To finish, take a sequence $(y_n)\subset f^{-1}((-\infty,r])$ such that $y_n\rightarrow y$. By using what we have just proved, we have that $f(y)\leq\liminf f(y_n)$. Can you conclude?

On the other and, suppose that for all $r$, the set $f^{-1}((-\infty,r])$ is closed and suppose ad absurdum that there exist $\epsilon>0$ such that for all $\delta_n=\frac{1}{n}$, we can find $y_n\in B(\delta_n,x)$ with $$f(y)>f(y_n)+\epsilon$$

The last inequality implies that $y_n\in f^{-1}((-\infty,f(y)-\epsilon])$. Can you use the fact that $f^{-1}((-\infty,r])$ is closed and $y_n\rightarrow y$ to conclude?

Note: The hypothesis of $X$ being closed is unnecessary.

  • $\begingroup$ No this is not true @qua. You can take the interval $(0,1)$ on the line. It is an open set in $\mathbb{R}$ and you can find a lot of closed sets in it, for example, the set $\{\frac{1}{2}\}$ is closed. $\endgroup$
    – Tomás
    Jan 18 '13 at 11:22
  • $\begingroup$ I think there is a problem with this solution: it is not guaranteed that the "n" of $y_n$ and the "n" of $\epsilon_n$ are the same. $\endgroup$
    – perlman
    Sep 18 '17 at 20:24

First Direction: Suppose that $f$ is l.s.c. To show that $A_r = \{a\in X:f(a)\le r\}$ is closed, let $\{a_n\}$ be a sequence in $A_r$, converging to $a_\infty$, which we want to show is in $A_r$. Since $f$ is l.s.c., $$ f(a_\infty) \leq \liminf f(a_n) \leq a, $$ since $a_n \in A_r$. Therefore, $a_\infty \in A_r$.

Second Direction: The argument is basically identical.

It's much more convenient to use the equivalent sequential definition of lower semi-continuity

  • $\begingroup$ But the problem here is we do not have the equivalent sequential definition of the lower semi continuity. What I only have is just the definition in the first paragraph of my question. I am afraid that I cannot use what you said. $\endgroup$ Jan 17 '13 at 23:55
  • $\begingroup$ Their equivalence is trivial, it's the same as the two equivalent definitions of continuity. $\endgroup$
    – quasi
    Jan 18 '13 at 1:40

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