I am reading Evans' PDE proof of theorem 6, section 7.4.3, page 444 where the following is said to be "clear"
$[H^2(\Omega) \cap H^1_0(\Omega)] \times H^1_0(\Omega)$ is dense in
$H^1_0(\Omega) \times L^2(\Omega) =: X$in the norm of $X$ which is defined to be
$$ || (u,v) ||_X = \left( B(u,u) + || v ||^2_{L^2(\Omega)} \right)^{1/2}$$
where the bilinear form $B(u,u) = (Lu,u)_{L^2}$, and $L$ is an elliptic operator
$$ Lu = – \nabla \cdot (A \nabla u) + cu$$
for some symmetric matrix $A$ and $\Omega \subset \mathbb{R}^n$
I know that $Y$ is dense in $X$ with respect to norm $N$ if the closure of $Y$ in the norm $N$ is the entirety of $X$. But I am not sure how to apply it to the proposition given above.
Best Answer
Let $(u,v)\in H^1_0(\Omega) \times L^2(\Omega)$.
Since $C^\infty_0(\Omega)$ is dense in $L^2(\Omega)$ with respect to $\| \cdot \|_{L^2(\Omega)}$, there exists $\{v_k\}\subset C^\infty_0(\Omega)$ such that $v_k \to v$ in $L^2(\Omega)$. Moreover, $C^\infty_0(\Omega)$ is dense in $H^1_0(\Omega)$ with respect to $\| \cdot \|_{H^1(\Omega)}$ (by definition), so there exists $\{u_k\}\subset C^\infty_0(\Omega)$ such that $u_k \to u$ in $H^1(\Omega)$.
Since $(u_k,v_k) \in C^\infty_0(\Omega) \times C^\infty_0(\Omega) \subset (H^2(\Omega) \cap H^1_0(\Omega) )\times H^1_0(\Omega) $ all that is left to show is that $(u_k,v_k) \to (u,v)$ in the norm $\|\cdot \|_X$.
Observe that if $w\in H^2 (\Omega) \cap H^1_0(\Omega)$ then integrating by parts gives \begin{align*} B(w,w) &= -\int_\Omega w \, \mathrm{div} (A\nabla w) \, dx + \int_\Omega cw^2 \, dx \\ &= \sum_{i,j=1}^n\int_\Omega a^{ij} (\partial_iw)(\partial_jw) \, dx + \int_\Omega cw^2 \, dx \\ &\leqslant C (\| w\|_{L^2(\Omega)}^2 + \| \nabla w \|_{L^2(\Omega)}^2) \\ &\leqslant C \|w\|_{H^1(\Omega)}^2. \end{align*} Here I've used that $A=(a^{ij})$ and $c$ are bounded (Evans always assumes this) and the Hölder inequality. Thus, \begin{align*} \| (u_k,v_k)-(u,v)\|_X &= \big ( B(u_k-u,u_k-u)+ \|v_k - v\|_{L^2(\Omega)}^2 \big )^{1/2} \\ &\leqslant C \big ( \|u_k-u\|_{H^1(\Omega)} + \|v_k - v\|_{L^2(\Omega)}\big ) \to 0 \end{align*} as $k\to \infty$. This shows that $(u_k,v_k) \to (u,v)$ in the norm $\|\cdot \|_X$ which completes the proof.