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Let $X$ a topological space satisfying the first countability axiom. I want to prove the following result:

$\varphi : X \rightarrow R $ is lower semicontinuous (this mean that $\varphi^{-1} (a, + \infty)$ is open for all $a \in R$) if and only if $ \varphi (\hat{u}) \leq \liminf \varphi (u_n) $ (for all sequence $u_n$ converging to $x$).

For the first part my proof is :

Let $u_n \subset X$ with $u_n \rightarrow \hat{u}$. Write $a = \varphi (\hat{u})$. Let $\epsilon >0$. We have $\hat{u} \in \varphi^{-1}(a - \epsilon , + \infty)$ . This set is open in X, then exists $n_0$ a natural number such that

$$u_n \in \varphi^{-1}(a - \epsilon , + \infty), \forall n \geq n_0$$ then

$$ \varphi (\hat{u}) \leq \inf_{n > n_0} \varphi(u_n) + \epsilon \leq \inf_{n > k} \varphi(u_n) + \epsilon \ \forall k \geq n_0$$

then $ \varphi (\hat{u}) \leq \liminf_{n > n_0} \varphi(u_n) + \epsilon $ Then follow the first part.

The proof of the first part is correct? Someone can give me a help with the other part?

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  • $\begingroup$ You have written "converging to $x$", but you called the point $\hat{u}$. And at the end, you should say something along the lines of "That holds for all $\epsilon>0$, hence $\varphi(\hat{u})\leqslant\liminf\varphi(u_n)$" [and you forgot that $\varphi$ in $\varphi(u_n)$ several times]. Apart from these cosmetics and small glitches, it is correct. You need the first countability only for the other direction. Hint for that: indirect proof, suppose $\varphi$ is not lower semicontinuous. Show that then there is an $\hat{u}$ and a sequence $u_n\to\hat{u}$ with $\varphi(\hat{u})>\liminf\varphi(u_n)$ $\endgroup$ – Daniel Fischer Mar 13 '14 at 22:48
  • $\begingroup$ @DanielFischer Thanks for the comment. Following your idea, my best is this : (indirect proof ). supoose that exists $a \in R$ such that $\varphi^{-1}(a, + \infty)$ is not open in X. then for all open $V$, exist $u_V \in V - \varphi^{-1} (a, + \infty)$. I am not seeing how to use the axiom to construct the sequence ... . can you give one more hint ? $\endgroup$ – math student Mar 13 '14 at 23:48
  • $\begingroup$ Continue with choosing a good $\hat{u}$. What sort of point should that probably be then? $\endgroup$ – Daniel Fischer Mar 13 '14 at 23:52
  • $\begingroup$ I did ! Your suggestions helped me a lot ! Thanks ! $\endgroup$ – math student Mar 14 '14 at 0:39

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