Interior regularity of elliptic equation (proof)

Question

In the proof of interior regularity of second order elliptic equation in Evans, he use directly $v=u \eta^2$ where $\eta$ is a cutoff function.
But by the definition of weak solution $v$ should be in $H_0^1(\Omega)$.
So I just wonder why $u\eta^2$ is actually in $H_0^1(\Omega)$ i.e. why it can be approached by a sequence of test functions.

Answer 1

Since $u \in H^1(\Omega)$, there exists a sequence $\{u_k\}\subset C^\infty (\Omega)\cap H^1(\Omega)$ such that $u_k \to u$ in $H^1(\Omega)$. Let $v_k := u_k \eta^2$ and $v=u\eta^2$. Since $\eta \in C^\infty_0(\Omega)$ we have that $\{v_k\}\subset C^\infty_0(\Omega)$. Then since $\eta$ is bounded, \begin{align*} \| v_k-v\|_{L^2(\Omega)}^2 &= \int_\Omega \eta^4 \vert u_k - u \vert^2 \,dx \leqslant C \| u_k-u\|_{L^2(\Omega)}^2. \end{align*} Moreover, $D(\eta^2 u) = 2 \eta u D\eta + \eta^2 Du$, we have that \begin{align*} \| Dv_k-Dv\|_{L^2(\Omega)}^2 &= \int_\Omega \vert D(\eta^2u_k)-D(\eta^2u)\vert\,dx \\ &=\int_\Omega \vert 2 \eta u_k D\eta + \eta^2 Du_k-2 \eta u D\eta - \eta ^2Du\vert^2\,dx \end{align*} Using the Cauchy-Schwarz inequality and that $\eta ,D\eta$ are bounded, we obtain \begin{align*} \| Dv_k-Dv\|_{L^2(\Omega)}^2&\leqslant 4 \int_\Omega \eta^2 \vert D\eta\vert^2 \vert u_k-u \vert ^2\,dx + \int_\Omega \eta^4 \vert D u_k - D u\vert^2 \, dx \\ &\leqslant C \| u_k - u \|_{H^1(\Omega)}. \end{align*} Combining these inequalities we have that $$ \| v_k - v \|_{H^1(\Omega)} \leqslant C \| u_k - u \|_{H^1(\Omega)} \to 0 $$ as $k\to \infty$. Thus, $v_k \to v$ in $H^1(\Omega)$ so $v$ is in closure of $C^\infty_0(\Omega)$ in $H^1(\Omega)$, that is, $H^1_0(\Omega)$.