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 $L^{2p}$?
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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.
Tomás
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Good generalization, deep answer, but it seems that your notation of $x_n$ and its subsequence $u_n$ be confused, hope you can make it more clear. – van abel Nov 13 '13 at 00:15
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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 $W^{1,p}(\Omega)$.
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It seems your answer is not related to my question? could you explain it a litter more? – van abel Nov 12 '13 at 04:16
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@vanabel: I am sorry that the answer is not even clear to me for a first scan! I have edited the post. – Nov 12 '13 at 07:12
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Thanks a lot. I believe you were right. But there are still two point I want make it more clear to me: 1. The test function $\phi$, in my opinion, shall be $C_0^\infty$, rather than $C^1$, which in turn makes all your argument valid for $\Omega$ not bounded. 2. The weak convergence in $W^{1,p}$ ($f_i\rightharpoonup f$ in $W^{1,p}$) is equivalent to the weak convergence of each $D^\alpha f_i \rightharpoonup D^\alpha f$ in $L^p$ for $|\alpha|\leq1$. I feel this is quite natural but how to make it a rigorous argument? – van abel Nov 12 '13 at 12:54
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For the first question, I mean to use $C^1_0(\Omega)$ instead of $C^1(\Omega)$. It works as $C^1_0(\Omega)$ is dense in $L^q$ for all $1< q<\infty$. For the second one, that's actually a good question to me and I am also not sure about that. I need to think more on that. – Nov 12 '13 at 20:50
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@vanabel, to answer your second question, note that $T: W^{1,p}(\Omega)\to (L^p(\Omega))^{n+1}$ defined by $$Tu=\left(u,\frac{\partial u}{\partial x_1},\cdots,\frac{\partial u}{\partial x_n}\right)$$ is an isometry (choose suitable norms). How is weak convergence characterized in $(L^p(\Omega))^{n+1}$? – Tomás Nov 17 '13 at 19:13
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Why do you care if $f_i$ converges strongly rather than just a subsequence of it (which is trivial) when $f_i$ is a subsequence itself in the first place?
Bananach
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