# Proof of $\Phi^\alpha =\{x\in\mathbb R\mid f(x)\leq \alpha \}$ bounded imply $f$ lower semi continuous.

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 $$\liminf_{n\to \infty }f(x_n). Set $$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 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, $$f(x_{k_n})\leq \alpha for all $$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 ?

• How do you define “lower semi continuous”? Besides, the question from the title is not the same as the question from the body. Jul 15, 2019 at 10:43
• @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 $x_n\to x_0$, $\liminf_{n\to \infty }f(x_n)\geq f(x_0)$.
– Surb
Jul 15, 2019 at 10:45
• It is possible. But it could also be$$(\forall a\in\mathbb R)(\forall\varepsilon>0)(\exists\delta>0):\lvert x-a\rvert<\delta\implies f(x)>f(a)-\varepsilon.$$ Jul 15, 2019 at 10:49
• @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).
– Surb
Jul 15, 2019 at 10:51
• The title sounds pretty bad. OP has to edit the title. Jul 15, 2019 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...