# Ngô Quốc Anh

## May 15, 2010

### Two fundamental results in the calculus of variation

Filed under: Giải tích 8 (MA5206), PDEs — Ngô Quốc Anh @ 20:28

I suddenly think that I should post this entry ‘cos sometimes I don’t remember these stuffs. These results appear frequently in solving PDEs especially when using the direct method. For example, the simplest case is the following eigenvalue problem

$\displaystyle -{\rm div}(|\nabla u|^{p-2}\nabla u)=\lambda |u|^{p-2}u$

over a bounded domain $\Omega \subset \mathbb R^n$ with Dirichlet boundary condition. We assume $p. Our aim is to show the existence of the first eigenvalue $\lambda_1>0$. Obviously, our problem is to solve the following optimization

$\displaystyle\mathop {\inf }\limits_{u \in W^{1,2}_0(\Omega)} \left\{ {\int_\Omega {|\nabla u{|^p}dx} :\int_\Omega {|u{|^p}dx} = 1} \right\}$.

The direct method says that we firstly select a minimizing sequence, say $\{u_n\}_n$, then we need to prove $\{u_n\}$ is convergent. There are two steps in the above argument which lead to this entry. Our first claim is the following.

Boundedness implies weakly convergence. The first result says that

If $E$ is a reflexive Banach space and $\{x_n\}_n \subset E$ is a bounded sequence. Then up to a subsequence $x_n$ converges weakly to some $x$ in $X$.

The proof of this claim can be found in a book due to Brezis (Theorem III.27). Interestingly, its converse also holds by the Eberlein-Šmulian theorem.

Theorem (Eberlein-Šmulian). Suppose $E$ is a Banach space such that every bounded sequence $\{x_n\}_n$ contains a weakly convergent subsequence. Then $E$ is reflexive.

There was an elementary proof of this theorem. We refer the reader to a paper due to Whitley [here]. Let us get back to our optimization problem. Once we have a minimizing sequence $\{u_n\}_n \subset W^{1,p}_0(\Omega)$ it is clear to see that $\{u_n\}_n$ is bounded in $W^{1,p}_0(\Omega)$ since

$\displaystyle\int_\Omega {|{u_n}{|^p}dx} = 1$

and

$\displaystyle\int_\Omega {|\nabla {u_n}{|^p}dx} \leqslant C, \quad \forall n$.

By using the first claim, $u_n$ converges weakly to some $u \in W^{1,p}_0(\Omega)$. It is worth noticing that by saying $u_n \rightharpoonup u$ in $W^{1,p}_0(\Omega)$ we mean $u_n \rightharpoonup u$ in $L^p(\Omega)$ and $Du_n \rightharpoonup Du$ in $L^p(\Omega, \mathbb R^n)$. Now we need further argument

Weakly convergence becomes strongly convergence via compact operator. This second result says that

A compact operator $C : E \to X$ between Banach spaces maps every weakly convergent sequence in $E$ into one that converges strongly in $X$.

The proof of this relies on the contradiction argument and the fact that once a sequence converges strongly to some limit, this limit is unique. However, the converse is no long true. For example, by the Schur theorem, a sequence $\{x_n\}$ in $\ell^1$ converges weakly, it also converges strongly. We take $C$ to be the identity $I$ in $\ell^1$. Since $\ell^1$ has infinity dimensional, by using the Riesz theorem, $I$ cannot be compact.

Using this claim we deduce that $u_n$ converges strongly to $u$ in $L^q$ for any $1\leqslant q. By using the Minkowski and Holder inequalities we can show that $u$ satisfies the constraint. It now follows from the weakly lower semi-continuous of norm that $u$ indeed satisfies the equation. The proof follows.