# about a theorem of weakly lower semicontinuous functions

I am studying the proof of the following theorem

Theorem: Let $$E$$ a Hilbert space and suppose that $$\varphi :E \rightarrow R$$ is a weakly lower semicontinuous functional. Suppose that $$\varphi$$ is coercive (that is $$\varphi(u) \rightarrow + \infty$$ if $$|| u|| \rightarrow +\infty$$). Then $$\varphi$$ bounded below and exists $$u_0\in E$$ such that $$\varphi(u_0) = \inf_E \varphi (u)$$ .

Proof:

By the coercivity exist $$R>0$$ such that $$\varphi (u)\geq \varphi(0) , \forall u \in E$$ with $$|| u|| \geq R$$. Because $$E$$ is Hilbert the ball $$\overline{B_{R}(0)} = \{ x \in E ; || x|| \leq R\}$$ is compact in the weak topology of $$E$$ (by the Kakutani theorem). Consider the restriction $$\varphi : \overline{B_{R}(0)} \rightarrow R$$. this restriction is lower semicontinuous in the weak topology. and the author continues the proof..

I not seeing how to prove the affirmation "this restriction is lower semicontinuous in the weak topology." someone can give me a help to prove this affirmation?

I don't know if this justifies the affirmation :

Let $$u_n$$ in $$\overline{B_{R}(0)}$$ a sequence converging to $$u$$ in the topology of $$\overline{B_{R}(0)}$$ ($$\overline{B_{R}(0)}$$ is equipped with the topology given by the weak topology of $$E$$). Then $$u_n$$ converges weakly to $$u$$ in $$E$$. By the hypothesis on $$\varphi$$ we have $$\varphi(u ) \leq \liminf \ \varphi(u_n)$$. Then $$\varphi : \overline{B_{R}(0)} \rightarrow R$$ is lower semicontinuous in the weak topology.

Some important definitions:

Definition 1 : Let $$E$$ a Hilbert space a functional $$\varphi : E \rightarrow R$$ is weakly lower semicontinuous if $$\varphi(u) \leq \liminf \ \varphi(u_n)$$ for all sequence $$u_n$$ converging weakly to $$u$$.

Definition 2 : Let $$X$$ a topologial space satisfying the first axiom of countability. The functional $$\varphi$$ is lower semicontinuous if and only if $$\varphi(u) \leq \liminf \ \varphi(u_n)$$ for all sequence $$u_n$$ converging to $$u$$.

• Yes, you got it. – Giuseppe Negro Mar 14 '14 at 17:02