I want to show the equivalence between the two statements: ($\Omega$ is a bounded open subset of $\mathbb{R}^n$)
- $v\in L^2(\Omega)$ is in $H^1(\Omega)$,
- There is a constant $C$ s.t. $$ |\int_\Omega v\nabla \varphi dx| \leq C||\varphi||_{L^2(\Omega)},\ \forall \varphi\in C_c^\infty(\Omega).$$
$1\to2$ is clearly, but when I want to prove $2\to1$, I could find $\{\varphi_k\}_{k=1}^\infty\in C_c^\infty(\Omega)$ s.t. $\varphi_k\to v$ in $L^2$ norm, then from limits and Green's formula, $$ \lim_{k\to\infty}|\int_\Omega \nabla\varphi_k\varphi dx|\leq C||\varphi||_{L^2(\Omega)},\ \forall \varphi\in C_c^\infty(\Omega). $$ Then I have no idea for the inequality( if we let $\varphi$ take $\varphi_k$, then we would have $\lim_{k\to\infty} ||\nabla\varphi_k||_{L^2(\Omega)}\leq C$, but this couldn't prove the convergence in $L^2(\Omega)$. I want to ask the complete proof or some ideas aobut it.
Define $f(\varphi):=\int_\Omega v\nabla\varphi\,dx$. Then the condition implies that $f$ extends to a bounded linear functional on $L^2$. By Hilbert's representation theorem, we have $f(\varphi) = \langle \varphi,\psi\rangle = \int_\Omega\varphi\overline\psi\,dx$ with some $\psi\in L^2$. Can you go on?