Strong convergence of a subsequence in $W^{1,p}$ Let $\{f_n\}$ be a sequence in $W^{1,p}(\mathbb{R}^2)$. 


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*If $\{f_n\}$ is bounded then there is a subsequence that converges weakly in $W^{1,p}(\mathbb{R}^2)$. Do we have strong convergence too in $W^{1,p}(\mathbb{R}^2)$?

*What are the weakest conditions required to have strong convergence?
 A: It's false even for $n=1$. Take $$f(x)=\begin{cases}0&\text{if }x\le -1\\x+1&\text{if }-1<x<0\\-x+1&\text{if }0\le x<1\\ 0&\text{if }x\ge 1\end{cases}.$$
The $f\in W^{1,p}(\mathbb{R})$. Now take $f_n(x):=f(x+n)$, so $f'_n(x+n)=f'(x+n)$. Then by the change of variables $y=x+n$ we have that $$\Vert f_n\Vert_{L^p}=\Vert f\Vert_{L^p},\quad \Vert f_n'\Vert_{L^p}=\Vert f'\Vert_{L^p}>0.$$
However, if you fix $x\in\mathbb{R}$ then for all $n\ge 1-x$ you get $f_n(x)=0\to 0$ as $n\to\infty$. So the sequence $\{f_n\}$ converges pointwise to zero in $\mathbb{R}$. If it converged strongly to some function $g$ in $W^{1,p}(\mathbb{R})$, then a subsequence would converge pointwise to $g$, which means that $g=0$. But $\Vert f_n'-0\Vert_{L^p}=\Vert f'\Vert_{L^p}>0$ does not go to zero. In dimension two you can do something similar. 
If you want strong convergence, for $p>1$ you should assume that $\Vert f_n\Vert_{W^{1,p}(\mathbb{R}^2)}\to \Vert g\Vert_{W^{1,p}(\mathbb{R}^2)}$, where $g$ is the weak limit and use the fact that the norm is strictly convex. See here Link
