Direct Method: How to show that F is weakly lower semicontinuous?

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Let $1<q<\infty$ be given and minimize

$F(v):= \int\limits_a^b \lvert v'(x)\rvert^q\, dx$

in the class

$K(\alpha,\beta)=\left\{v\in H^{1,q}((a,b),\mathbb{R}^M : v(a)=\alpha, v(b)=\beta\right\}$

Show that the minimum exists. Use the direct method of variation.


Hello!

I have already shown that the space is reflexive. And I have shown that F is coercive.

But how can I show that F is weakly low semicontinious?

Let $(x_n)$ with $x_n\to x$ weakly. Then I have to show:

$F(x)\leq\liminf F(x_n)$.

How can I show that?

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Show that $F$ is convex. Convex closed sets are also weakly closed, hence the sublevels $\{F \le t \}$ are weakly closed, which means that $F$ is weakly lower semicontinuous.