Pointwise convergence and Sobolev bounded If $\{f_n\}$ is a sequence of function on $\Omega$, and $\lim_{n\to\infty}f_n=f$ on $\Omega$. $\|f_n\|_{W^{k,p}(\Omega)}$ is bounded (or uniformly bounded), then whether $f$ is in $W^{k,p}(\Omega)$. If not, are there some counterexample?  
 A: Counter example in $W^{1,1}([-1,1])$
$$
    f_n(x)=\left\{
                \begin{array}{ll}
                  0 &  x\in [-1,0]\\
                  nx & x\in \left(0,\frac{1}{n}\right]\\
                  1 &  x\in \left(\frac{1}{n},1\right]\\
                \end{array}
              \right.
$$
$f_n \rightarrow \chi_{(0,1]}$ pointwise.
A: Not sure about the general case, but it's true at least when $\Omega$ is bounded, has Lipschitz boundary and $p>1$. First let $k =  1$, then a bounded sequence $\{f_n \}$ in $W^{1, p}(\Omega)$ (by passing to subsequence) converges to $g \in L^p(\Omega)$ strongly, by Rellich–Kondrachov theorem. By picking again a subsequence, we can choose $f_n \to g$ almost everwhere. Thus $f = g \in L^p(\Omega)$. 
For any $i$, by weak compactness of $L^p$ space, there are $f_i \in L^p(\Omega)$ so that 
$$\frac{\partial f_n}{\partial x^i} \to f_i$$
weakly in $L^p(\Omega)$. Now for any test function $\phi$, we have 
$$\int_\Omega f \partial_i \phi = \lim_{n\to \infty} \int_\Omega f_n \phi = - \lim_{n\to \infty} \int_\Omega \frac{\partial f_n}{\partial x^i} \phi = -\int_\Omega f_i \phi \ .$$
Thus $f \in W^{1, p}(\Omega)$.
The general case ($k\geq 1$) follows from this one by considering $\partial^\alpha f_n \in W^{1, p}(\Omega)$, where $|\alpha| = k-1$.  
