Why $u_n \to u$ in $H^1$ implies that $u_n \to u$ in $L^2$?

Question

Why does a sequence $u_n \to u$ in $H^1$ imply that $u_n \to u$ in $L^2$? Is it because $H^1$ is continuously embedded inside $L^2$?

Answer 1

Essentially, yes: if $f$ is a continuous function, then in particular, it preserves convergent sequences (and their limits). In this case, $f$ is the natural embedding.

However, in this case, the result is a lot stronger: the norm of the Sobolev space $H^1$ of a function is strictly greater than the norm of the Lebesgue space $L^2$ (whenever the first one exists, possibly up to a constant, depending on your exact definition). In terms of convergence, it means that not only are convergent sequences still convergent, they converge (at least) as quickly.

Answer 2

I'd say that is because of the definition of the norms.

The norm $\|\cdot\|_{H^1}$ in $H^1$ is defined by $\|u\|_{H^1}^2=\|u\|_{L^2}^2+\|u'\|_{L^2}^2$ or $\|u\|_{H^1}=\|u\|_{L^2}+\|u'\|_{L^2}$.

In the fist case we have $$0\leq\|u_n-u\|_{L^2}^2\leq \|u_n-u\|_{L^2}^2+\|u_n'-u'\|_{L^2}^2=\|u_n-u\|_{H^1}^2$$ and, in the second, $$0\leq\|u_n-u\|_{L^2}\leq \|u_n-u\|_{L^2}+\|u_n'-u'\|_{L^2}=\|u_n-u\|_{H^1}.$$

So, the desired result can be seen as a consequence of the squeeze theorem.