Ngô Quốc Anh

May 22, 2010

The third and fouth fundamental results in the calculus of variation

Filed under: Giải tích 8 (MA5206), PDEs — Ngô Quốc Anh @ 17:56

Followed by this entry where the following questions have been discussed

1. Boundedness implies weakly convergence: 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 $X$.
2. Weakly convergence becomes strongly convergence via compact operator: A compact operator $C : E \to X$ between Banach spaces maps every weakly convergent sequence in $E$ into one that converges strongly in $X$.

Now I shall discuss more results which appear frequently in the calculus of variation.

Let us recall over a minifold $M$, the Sobolev norm $H^1(M)$ (or $W^{1,2}(M)$) is defined by $\displaystyle \|u\|_{H^1}^2=\|u\|_{L^2}^2+\|\nabla u\|_{L^2}^2$.

It is immediately to see that if $u_n \to u$ strongly in $H^1$, i.e. $\|u_n-u\|_{H^1}\to 0$ then $u_n \to u$ strongly in $L^2$ and $\nabla u_n \to \nabla u$ strongly in $L^2$.

It turns out to discuss what happen to weakly convergence. Actually, we shall prove the following important result, called the third fundamental result.

Weakly convergence in Sobolev spaces implies weakly convergence in $L^p$ spaces. We assume $u_n \rightharpoonup u$ in $H^1$. We shall prove both $u_n$ and $\nabla u_n$ converge weakly to $u$ and $\nabla u$ in $L^2$, respectively.

By the principle of uniform boundedness, any weakly convergence sequence is bounded in the norm. Consequently, $\{u_n\}_n$ and $\{\nabla u_n\}_n$ are bounded in $L^2$. By the weak compactness of balls in $L^2$, there is a subsequence $n_k$ such that $\displaystyle u_{n_k} \rightharpoonup v, \quad \nabla u_{n_k} \rightharpoonup w$

in $L^2$ (i.e., both $v, w$ are in $L^2$). Since the weak convergence in $L^2$ implies the convergence in $\mathcal D'$ the dual space of $\mathcal D$-the space of test functions. It follows that $w = \nabla v$ and, hence, $v \in H^1$. It follows that $u\equiv v$ and thus $\displaystyle u_{n_k} \rightharpoonup u, \quad \nabla u_{n_k} \rightharpoonup \nabla u$

in $L^2$ as desired.

Now we consider the reverse case. We shall prove the following

Weakly convergence in $L^p$ spaces plus the boundedness implies weakly convergence in Sobolev spaces. We assume $u_n \rightharpoonup u \in L^2$ in $L^2$ and $\|u_n\|_{H^1}$ is bounded. We shall prove that $u \in H^1$ and $u_n \rightharpoonup u$ in $H^1$.

Since $\{u_n\}$ is bounded in $H^1$, by the first fundamental result, $u_n \rightharpoonup v$ in $H^1$ for some $v \in H^1$. By the third fundamental result above, $u_n \rightharpoonup v$ in $L^2$. It follows from the uniqueness of weak limit that $u \equiv v$ which implies $u \in H^1$.

In order to prove $u_n \rightharpoonup u$ in $H^1$, we shall use the following result whose proof is based on the simple contradiction argument.

Let $X$ be a topological space. A sequence $\{x_n\} \subset X$  converges to $x \in X$ (in the topological of $X$) if and only if any subsequence of $\{x_n\}$ contains a sub-subsequence that converges to $x$.

Let us pick a particular subsequence of $\{u_n\}$ and rename it back to $\{u_n\}$ for simplicity. It suffices to prove that $\{u_n\}$ contains a subsequence that converges to $u$ weakly in $H^1$. Followed the proof of the third fundamental result, there is a subsequence of $\{u_n\}$ and a function $v \in H^1$ such that $\displaystyle u_{n_k} \rightharpoonup v, \quad \nabla u_{n_k} \rightharpoonup v$

both in $L^2$. It follows from the definition of weakly convergence in $H^1$ that in fact we get $\displaystyle u_{n_k} \rightharpoonup v$

in $H^1$. The reason is the following: $\displaystyle (u_{n_k},\varphi)_{L^2}+(\nabla u_{n_k},\nabla\varphi)_{L^2} \to (v,\varphi)_{L^2} + (\nabla v, \varphi)_{L^2}$

for any $\varphi \in H^1$. Having this and the fact that weak limit is unique we deduce that $u \equiv v$. The latter now implies $u_{n_k} \rightharpoonup u$

in $H^1$. The proof follows.