I am trying to verify that the Laplacian operator is closed on $D(-\Delta)$. For this I do the following
Let $-\Delta:D(-\Delta)\subset L^2\to L^2$ where $D(\Delta):=\left\{u\in L^2:-\Delta u\in L^2\right\}$
Let $\left\{u_n\right\}\subset D(-\Delta)$ with $u_n\to u$ in $D(-\Delta)$ and $-\Delta u_n\to v$ in $L^2$. I want proves that $u\in D(-\Delta)$ and $-\Delta u_n\to -\Delta u$ in $L^2$.
\begin{align} |-\Delta u|_{2}&\leq |-\Delta u_n-\Delta u|_2+|-\Delta u_n|_2\\ &=|-\Delta (u_n-u)|_{2}+|-\Delta u_n|_{2}\\ &\leq |-\Delta| |u_n-u|_{2}+|-\Delta u_n|_{2}\\ &\to c \end{align} because $-\Delta u_n\to v\text{ in } L^2\Rightarrow |-\Delta u_n|\leq c$, $n\to \infty$. Therefore $u\in D(-\Delta)$.
By the other hand, $|-\Delta u_n-\Delta u|_{2}\leq |-\Delta| |u_n-u|_{2}\to 0$ therefore $-\Delta u_n\to -\Delta u$ in $L^2.$
The problem is that in the above, I used that $|-\Delta (u_n-u)|_{2}\leq |-\Delta| |u_n-u|_{2}$ but I don't know if that is true.
Actualization 1. Let $H^2:=\left\{u\in L^2: (Id-\Delta)u\in L^2\right\}$ with the norm $|u|_{H^2}=|(Id-\Delta)u|_{2}$ and $D(-\Delta)=\left\{u\in L^2: -\Delta u \in L^2\right\}$.
Affirmation 1. $D(-\Delta)=H^2$
Affirmation 2. $-\Delta:D(-\Delta)\subset L^2\to L^2$ is closed in $H^2$
Proof: Let $\left\{u_n\right\}\subset D(-\Delta)$ with $u_n\xrightarrow[ ]{H^2}u$ and $-\Delta u_n\xrightarrow[]{L^2} v$ then \begin{align} |u|_{H^2}^{2}&\leq |u_n-u|_{H^2}^2+|u_n|_{H^2}^2\\ &\leq |u_n-u|_{H^2}^2+c\to c \end{align} In the above use the following: $u_n\xrightarrow[ ]{H^2}u$ then exists $c>0$ such that for all $n$, $|u_n|_{H^2}\leq c$? is this true?
Moreover, \begin{align} |-\Delta u_n-(-\Delta u)|_{2}&=|-\Delta(u_n-u)|_{2}\\ &\leq |u_n-u|_{H^2}\to 0 \end{align}
Therefore, $-\Delta$ is closed in $H^{2}$