Convergence in $H^1(\Omega)$ and $L^2(\Omega)$

Question

Let $\Omega$ be a bounded domain (maybe that doesn't matter), if $f_n\rightarrow f$ in $H^1(\Omega)$, does it follow $f_n\rightarrow f$ in $L^2(\Omega)$ since $H^1$ is dense in $L^2$? Is it true that $|\cdot |_{H^1}\leq C|\cdot |_{L^2}$ for some constant. Thanks very much for the help. Also, what's the definition of $H^1_0(\Omega)$? My professor used this notation but without specify the definition.

Answer 1

Take a look at the norm of $H^1(\Omega)$:

$$\|u\|_{H^1} = \|u\|_{L^2} + \| Du\|_{L^2} \ge \|u\|_{L^2}.$$

Thus if $\|f_n -f\|_{H^1} \to 0$, then $\|f_n - f\|_{L^2} \to 0$. In terms of operator it just says that the inclusion $H^1 (\Omega) \to L^2(\Omega)$ is a bounded linear operator. It is true that $H^1(\Omega)$ is dense in $L^2(\Omega)$, but we are not using it here.

$H^1_0(\Omega)$ is a subspace of $H^1(\Omega)$ which is generated by $C^\infty_c(\Omega)$.

The reverse inequality $\|\cdot \|_{H^1} \le C \|\cdot \|_{L^2}$ is not true in general. For example, $f_n (x) = \sin nx$ is a sequence in $H^1(0,2\pi)$ with bounded $L^2$ norm but unbounded $H^1$ norm. So much a $C$ cannot be found.