Weak Convergence in $W^{1,2}(D)$ implies strong convergence in $L^{2}(D)$

Question

Let $D$ is some bounded domain in $\mathbb{R^n}$. At the bottom of page 20/top of page 21 in this book, they observe that a sequence of functions $\{u_{\epsilon}\} \in W^{1,2}(D)$ is uniformly bounded so that $ u_{\epsilon} \rightarrow u$ weakly in $W^{1,2}(D)$ and strongly in $L^2(D)$. I understand the weak convergence in $W^{1,2}(D)$ follows from a theorem functional analysis (In reflexive Banach spaces, every bounded sequence has a weakly convergent subsequence). What fact does the strong convergence in $L^2(D)$ rely on?

Answer 1

That follows from Rellich–Kondrachov_theorem, which says that

$$ W^{1, p} (\Omega) \to L^{q} (\Omega)$$

is a compact embedding when $1 \le q < p^* = \frac{np}{n-p}$.

Since when $p=2$, $p^* = 2n/(n-2)>2$. The boundedness of $\Omega$ implies that

$$ L^q (\Omega ) \to L^2(\Omega)$$

is a bounded embedding when $2\le q< p^*$ (see here). Thus the composition is also compact. Since compact operator sends weakly convergent sequence to strong convergent sequence (see here), $u_\epsilon \to u$ strongly in $L^2$.

Remark: When $n=2$ it is still true. The boundedness of $\Omega$ again implies

$$ W^{1, 2} (\Omega) \to W^{1, p}(\Omega)$$

for all $1\le p<2$. Then the Rellich-Kondrachov theorem implies that

$$ W^{1, p} (\Omega) \to L^{p^*} (\Omega)$$

is compact. Since we want $2p/(2-p) = p^*=2$, we choose $p = 1$. Indeed by choosing $p\to 2$, one can show similarly that $W^{1, 2}(\Omega)$ compactly embed into $L^q$ for all $q\ge 1$, when $n=2$.

Lastly, when $n=1$, $W^{1, 2}$ compactly embed into $C^{0, \alpha}$ for $\alpha <1/2$, and in particular it embeds compactly into $L^2$.