$L^2$-convergence implies convergence in $H_0^k$

Question

Suppose that $Mx_n$ converges to $M x$ in $L^2([0,1], \mathbb{R}^m)$ for $n \to \infty$ and $M x_n \in H_0^k([0,1], \mathbb{R}^m)$, for $M \in \mathbb{R}^{m\times m}$ and a $k \in \mathbb{N}$. The space $H_0^k([0,1], \mathbb{R}^m)$ denotes the Sobolev space $W_0^{k,2}([0,1], \mathbb{R}^m)$ endowed with the standard Sobolev norm. How can I show that $M x \in H_0^k([0,1], \mathbb{R}^m)$? I think, by the Gagliardo-Nirenberg inequality, it suffices to show that $$\lim_{n\to \infty} \left|\left|\frac{\partial^k}{\partial \omega^k} M (x_n-x)\right|\right|_{L^2([0,1], \mathbb{R}^m)}^2=0,$$ but I don't see this. I would be very grateful for help or hints.

Answer 1

Call $f_n:=M x_n$, $f:=M x$. You have that $f_n\to f$ in $L^2$, and you know that $f_n\in H^k_0$ for all $n$.

The above is not enough to prove that $f\in H^k_0$. You need more assumptions. The reason is that $H^k_0$ is dense in $L^2$.

There is a simple assumption one could make here that ensures $f\in H^k_0$: it is enough to assume that the sequence $(f_n)_n$ is bounded in $H^k_0$, i.e., $$ \|f_n\|_{H^k_0}\leq C<\infty, $$ for some constant $C$ that does not depend on $n$. In fact, with this assumption, any subsequence $f_{n_k}$ has a weakly convergent sub-subsequence $f_{n_{k_j}}\to g$ in $H^k_0$, due to the fact that $H^k_0$ is reflexive. The function $g$ might in principle depend on the chosen sub-subsequence. However, $f_{n_{k_j}}\to g$ weakly in $L^2$ (because the $L^2$ norm is weaker than the $H^k_0$ norm) and $f_{n_{k_j}}\to f$ weakly in $L^2$ by assumption, so by the uniqueness of weak limits, $f=g$ a.e.. It follows that $f\in H^k_0$ and $f_n\to f$ weakly in $H^k_0$. In particular, by weak convergence, you have $$ \|f\|_{H^k_0}\leq \liminf_{n\to\infty} \|f_n\|_{H^k_0}. $$