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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 $u$.

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    $\begingroup$ Yes, you got it. $\endgroup$ – Giuseppe Negro Mar 14 '14 at 17:02

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