3
$\begingroup$

Let $(M,\delta)$ be Metric space, $f$ be a real function on $M$. Suppose $\{x \in M: f(x) > \alpha\}$ is open set in $M$ forall $\alpha \in \mathbb{R}$. Prove $f$ is a lower semicontinuous function.

Definition: Let $(M,\delta)$ be Metric space. $f$ is a real function on $M$.We say, $f$ is a lower semicontinuous function if:

$\forall$ $\{x_m\}\longrightarrow x \in M$, we have $f(x)\leq \liminf_{m\rightarrow \infty} f(x_m)$

Please help me!

$\endgroup$
2
$\begingroup$

Hint: Let $\epsilon>0$. Show that $$\liminf_{m \to \infty} f(x_m) \ge f(x)-\epsilon$$ using the fact that $x \in \{y \in M: f(y) > f(x)-\epsilon\}$, which is an open set by assumption.

$\endgroup$
1
  • $\begingroup$ My exercise sheet asks one to prove this in the case of a first countable topological space. We do we use first countability? $\endgroup$ Apr 14 '20 at 14:00
1
$\begingroup$

Example of semicontinuous function :

$$ f(x) = \begin{cases} x^2,\ x\neq 0 \\ -1,\ x=0 \end{cases} $$

$\endgroup$
2
  • 1
    $\begingroup$ How do you think this example? $\endgroup$ Nov 28 '13 at 12:05
  • $\begingroup$ Consider the condition $f(x) < {\rm lim\ inf}\ f(x_m)$ when $x_m\rightarrow x$. for instance $x_m=\frac{1}{m},\ x=0$. $\endgroup$
    – HK Lee
    Nov 28 '13 at 12:12

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.