Let $f:\mathbb R\to \mathbb R$. We suppose that for all $\alpha \in\mathbb R$, $\Phi^\alpha =\{x\in\mathbb R\mid f(x)\leq \alpha \}$ closed. Prove that $f$ is Lower semi continuous. I think that my proof is very complicated, and I was wondering if there is easier or not ?

Proof Let $(x_n)$ a sequence that converge to $x_0$. Suppose by contradiction that $\liminf_{n\to \infty }f(x_n)<f(x_0)$. Set $y_n=\inf_{k\geq n}x_k$ and $\ell=\liminf_{n\to \infty }f(x_n)$. Let $\alpha \in\mathbb R$ s.t. $$\ell<\alpha <f(x_0).$$ By definition of $y_n$, for all $n\in\mathbb N^*$ there is $k_n\geq n$ s.t. $$f(x_{k_n})\leq y_n+\frac{1}{n}\leq \ell+\frac{1}{n},$$ where the last inequality comes from the fact that $(y_n)$ is increasing. So, let $N$ s.t. $\frac{1}{n}\leq \alpha -\ell$ for all $n\geq N$. In particular, $$f(x_{k_n})\leq \alpha <f(x_0),$$ for all $n\geq N$, and thus $(x_{k_n})_{n\geq N}$ is a sequence of $\Phi^\alpha $ that doesn't converges in $\Phi^\alpha $. Contradiction.

I think my proof is quite complicate. Is there easier ?

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    $\begingroup$ How do you define “lower semi continuous”? Besides, the question from the title is not the same as the question from the body. $\endgroup$ – José Carlos Santos Jul 15 '19 at 10:43
  • $\begingroup$ @JoséCarlosSantos: In $\mathbb R$ there are not millions of way to define lower semi continuity... Here it look to be $\liminf_{x\to x_0}f(x)\geq f(x_0)$ or equivalently, for all $x_n\to x_0$, $\liminf_{n\to \infty }f(x_n)\geq f(x_0)$. $\endgroup$ – Surb Jul 15 '19 at 10:45
  • $\begingroup$ It is possible. But it could also be$$(\forall a\in\mathbb R)(\forall\varepsilon>0)(\exists\delta>0):\lvert x-a\rvert<\delta\implies f(x)>f(a)-\varepsilon.$$ $\endgroup$ – José Carlos Santos Jul 15 '19 at 10:49
  • $\begingroup$ @JoséCarlosSantos: This is equivalent to the previous definitions. But indeed, the OP could have mention which definition he used (even if it looks quite obvious that he use the sequentially definition). $\endgroup$ – Surb Jul 15 '19 at 10:51
  • $\begingroup$ The title sounds pretty bad. OP has to edit the title. $\endgroup$ – Kavi Rama Murthy Jul 15 '19 at 12:02

A very little simplification could be to remark that $$y_n+\frac{1}{n}\underset{n\to \infty }{\longrightarrow } \ell<\alpha,$$ and thus, there is $N$ s.t. $$f(x_{k_n})\leq y_n+\frac{1}{n}<\alpha $$ for all $n\geq N$. So no need to use the fact that $(y_n)$ is increasing. But except this point, I'm not sure there is an easier proof. Well, I don't think that it's a very complicate proof neither...


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