Is it true that $L^2(\Omega)$, identified with its own dual, is dense in $H^{-1}(\Omega)$? $H^{-1}(\Omega)$ is the dual of $H^1_0(\Omega)$ and $H^1_0(\Omega)$ is the $H^1$-closure of smooth functions with compact support contained in $\Omega$. The $H^1$ inner product of a couple of functions is the sum of their $L^2$ inner product and the $L^2$ inner product of their derivatives of first order.
Ps 1: I am reading Richard Falk's 1974 paper (http://www.ams.org/journals/mcom/1974-28-128/S0025-5718-1974-0391502-8/S0025-5718-1974-0391502-8.pdf) and there, in section 3, by using his Theorem 1, he seems to use the "fact" that $L^2(\Omega)$ is dense in $H^{-1}(\Omega)$. I am having trouble verifying that. Can I get some help, please? Thanks,
Best Answer
Got it. I was trying to get this in an elementary way. Let's denote by $((u,v))$ the inner product in $H^1$, by $(u,v)$ the inner product in $L^2$, and by $<f,v>$ the action of $f$ on $v$ when $f$ is in $H^{-1}$ and $v$ is in $H_0^1$.
Let $f\in H^{-1}$. Since $((\cdot,\cdot))$ is an inner product on $H_0^1$ as well, by the Riesz representation theorem there exists $u\in H_0^1$ such that $\langle f,v\rangle = ((u,v))$ for all $v$ in $H_0^1$. $H_0^1$ is the closure of $C_0^\infty$ in the $H^1$-norm. Hence there is a sequence $\{u_n\}$ in $C_0^\infty$ that converges to $u$ in the $H^1$ norm. Let's study the term $((u_n, v))$: $$((u_n, v)) = \int u_n v + \int Du_n\cdot Dv$$ Since $u_n$ is smooth and has compact support, then $$((u_n, v)) = \int u_n v - \int (\Delta u_n)v$$ $$((u_n, v)) = (u_n - \Delta u_n, v)$$ Define $f_{u_n}:H_0^1 \to \mathbb{R}$ to be $\langle f_{u_n}, v\rangle:= ((u_n,v)) = (u_n - \Delta u_n, v)$. Then from its definition it follows that $f_{u_n}$ is in $H^{-1}$, and from the equality shown above that $f_{u_n}$ is also a linear continuous operator on $L^2$. Finally, $f_{u_n}$ converges to f in $H^{-1}$ by using the fact that $u_n$ converges to $u$ in the $H^1$ norm. Indeed,
$$|\langle f-f_{u_n}, v\rangle| = |((u,v)) - ((u_n, v))| = |((u-u_n,v))|$$ $$|\langle f-f_{u_n}, v\rangle| \leq ||u-u_n||_{H^1}||v||_{H^1}$$ Thus, by taking the sup over $||v||\leq 1$ we get $$||f-f_{u_n}||_{H^{-1}} \leq ||u-u_n||_{H^1}$$ which goes to zero as $n$ goes to infinity. Hence $L^2$ is dense in $H^{-1}$ (I would rather say $(L^2)'$ is dense in $H^{-1}$)