• $d\in\mathbb N$
  • $\Lambda\subseteq\mathbb R^d$ be open
  • $\mathcal V:=\left\{\phi\in C_c^\infty(\Lambda)^d:\nabla\cdot u=0\right\}$
  • $\tilde V:\left\{u\in H_0^1(\Lambda,\mathbb R^d):\nabla\cdot u=0\right\}$

It's easy to see that $$V:=\overline{\mathcal V}^{\left\|\;\cdot\;\right\|_{H_0^1(\Lambda,\:\mathbb R^d)}}\subseteq\tilde V\;.\tag 1$$ Under the assumption that $\Lambda$ is bounded $\partial\Lambda$ is Lipschitz, I was able to show that for each $\eta\in\tilde V'$ with $\left.\eta\right|_V=0$, we've got $\eta=0$.

Why can we conclude that $V=\tilde V$?

Maybe it's a consequence by the Hahn-Banach theorem. If we equip $\tilde V$ with the norm induced by $H_0^1(\Lambda,\mathbb R^d)$, it should be a normed space and $V$ should be a subspace of $\tilde V$. By the statement about the $\eta\in\tilde V'$ above, the Hahn-Banach theorem would yield that $V$ is dense in $\tilde V$.

Is that the correct argumentation? And if so, why does the density of $V$ in $\tilde V$ yield $V=\tilde V$?


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