If $v$ is a bounded linear functional on $C^{\infty}_0$, then $v \in H^1$.

Question

In a proof I am reading, we are given a function $v \in L^2(\omega)$, and we must show that $v \in H^1(\omega).$

They argue that "since $|\int_{\omega} v \partial_i \psi dx |\leq C||\psi||_{L^2(\omega)}$", we get that $v \in H^1(\omega)$ where $\psi \in C^{\infty}_0(\omega).$

I tried to make sense of this through the definition of weak derivatives, but could not progress.

Answer 1

Since $|\int_\omega v \partial_i \psi \, dx| \leq C \|\psi\|_{L^2(\omega)}$ for all $\psi \in C_0^\infty(\omega)$ and as $C_0^\infty(\omega)$ is dense in $L^2(\omega)$, by Riesz representation theorem there exists $w_i \in L^2(\omega)$ s.t. $\int_\omega v \partial_i \psi \, dx = \int_\omega w_i \psi \, dx$, i.e., $-w_i$ is the $i$-th weak partial derivative of $v$. So $v$ is a function in $L^2$ admitting all first-order weak partial derivatives which are all in $L^2$, i.e., $v \in H^1(\omega)$.