I am trying to prove directly that the Laplacian operator is closed on its maximal set. I have the following with some doubts.
Considering the operator $-\Delta$ defined on $D(-\Delta)=\left\{u\in L^2: -\Delta u\in L^2\right\}$, let $(u_n)\subset D(-\Delta)$ such that $u_n\to u$ in $L^2$, $-\Delta u_n\to v$ for some $v\in L^2$. The operator will be closed if $u\in D(-\Delta)$ and $-\Delta u=v$.
By triangle inequality, $\left\|u\right\|_{2}\leq \left\|u-u_n\right\|_{2}+\left\|u_n\right\|_{2}$, then $\left\|u\right\|_{2}\leq \lim_n \left\|u_n\right\|_{2}$. Assuming that the limit exists, we would have $u\in L^2$.
Question 1. Is there a limit $\lim_n \left\|u_n\right\|_{2}$?
Similarly, I run into the same difficulty proving that $-\Delta u\in L^2$ as $\left\|-\Delta u\right\|_{2}\leq \left\|-\Delta(u-u_n)\right\|_{2}+\left\|-\Delta u_n\right\|_{2}$.
Also, $\left\|-\Delta u-v\right\|_{2}\leq \left\|-\Delta(u-u_n)\right\|_{2}+\left\|-\Delta u_n-v\right\|_{2}$.
Question 2. Is it valid $\left\|-\Delta(u-u_n)\right\|_{2}\leq \left\|-\Delta\right\|\left\|u-u_n \right\|_{2}$?
Thank you.
Actualization. Apparently, $-\Delta u\in L^2$ must be considered in the distributional sense. Attached image.
