Weak convergence and strong convergence

Question

Suppose $f_i$ is uniformly bounded in $W^{1,p}$ for some $+\infty>p>1$, then by passing to a sub-sequence, we can suppose $f_i$ is weakly convergent to $f$ in $W^{1,p}$. Assume furthermore that $2p > n$, then we have $W^{1,p}$ is compactly embedding into $L^{2p}$. My question is, can we conclude $f_i$ convergent to $f$ in $L^{2p}$?

Answer 1

We know that $f_i \to f$ weakly in $W^{1,p}(\Omega)$ and $f_i \to g$ strongly in $L^{2p}$, by the compact embedding. Note that $g\in L^p$ (I assume $\Omega$ is bounded).

If we can show that $g\in W^{1,p}(\Omega)$, and $f_i \to g$ weakly in $W^{1,p}(\Omega)$, then $g=f$ since weak limit is unique. This will then answer your question.

Claim 1: $g\in W^{1,p}(\Omega)$.

Using that $f_i \to g$ strongly in $L^{2p}$. For any compactly supported $C^1$ test function $\phi$, as $f_i \in W^{1,p}(\Omega)$,

$$\int_\Omega \frac{\partial f_i}{\partial x_j} \phi = -\int_\Omega f_i \frac{\partial \phi}{\partial x_j} \to -\int_\Omega g \frac{\partial \phi}{\partial x_j}. $$

On the other hand, $$\bigg|\bigg|\frac{\partial f_i}{\partial x_j} \bigg|\bigg|_{L^p}\leq K \Rightarrow \{\frac{\partial f_i}{\partial x_j}\} \to g_j $$

weakly in $L^p$. Thus

$$\int_\Omega \frac{\partial f_i}{\partial x_j} \phi \to \int_\Omega g_j \phi\ .$$

Then

$$\int_\Omega g_j \phi = -\int_\Omega g \frac{\partial \phi}{\partial x_j}$$

for all test function $\phi$. As $g_j \in L^p$, $g\in W^{1,p}(\Omega)$ and $g_j$ is the weak derivative of $g$. Thus claim 1 is shown.

Claim 2: $f_i \to g$ weakly in $W^{1,p}(\Omega)$

From the proof of claim 1 we see that

$$\int_\Omega \frac{\partial f_i}{\partial x_j} \phi \to \int_\Omega g_j \phi$$

for all test function. Thus $\nabla f_i \to \nabla g$ weakly in $L^p$. Also $f_i \to g$ weakly in $L^p$ (as $2p>p$ and strong implies weak), we see that $f_i \to g$ weakly in $W^{1,p}(\Omega)$.

Answer 2

There is a more general result that we can state:

Let $X,Y$ be Banach spaces and $T:X\to Y$ a compact operator. Then, for each sequence $x_n\in X$ such that $x_n\to x$ weakly, we have that $Tx_n\to Tx$ strongly.

Proof: Because $x_n$ converge weakly, we can assume that $x_n$ is bounded. By definition of compactness, we have that $\overline{T(\{x,x_1,...,x_n,...\})}$ is compact.

We conclude that there exist a subsequence (not relabeld) such that $Tx_n\to v$ strongly. Now, note that $Tx_n\to Tx$ weakly in $Y$, hence $v=Tx$.

To finish, you have to prove that $Tx_n\to Tx$ strongly (remember that we proved it only for a subsequence). This follows from the following topological result:

Assume that $X$ is a metric space. Suppose that $x_n\in X$ is a sequence such that every subsequence has a convergent subsequence that converge to $x$ then, $x_n\to x$.

Now, to answer your qeustion, note that when we say that $W^{1,p}$ is compactly embedded in $L^{2p}$ then, we are saying that $I: W^{1,p}\to L^{2p}$ defined by $Ix=x$ is compact.