Laplace operator defined on a Sobolev space

Question

Consider the Laplace operator $$A:W^{2,2}(\mathbb{R})\to L^2(\mathbb{R})\;\;\\A u = -u^{\prime \prime}$$ I want to know why this operator is closed (I'm using the closed graph theorem):
Let $(u_n)\subseteq W^{2,2}(\mathbb{R})$ be a sequence such that $u_n\to x$ in $L^2(\mathbb{R})$ and $Au_n=-u_n^{\prime \prime}\to y$ in $L^2(\mathbb{R})$. Now, why is $x\in W^{2,2}(\mathbb{R})$? (It must be proven that $Ax=y$, but this is ok for me..)

I'm stuck on: Is $(u_n)$ a cauchy sequence $W^{2,2}(\mathbb{R})$ with respect to $\|u\|_{2,2}^2=(\|u\|_{L^2}^2+\sum\limits_{|\alpha|\le 2}\|D^{\alpha}u\|_{L^2}^2)$? I know that $(u_n)$ and $(u_n)$ are cauchy sequences in $L^2(\mathbb{R})$ but I know nothing about $(u_n')$. Regards

Answer 1

In some sense the norm you are using is the graph norm of the Laplacian, so the operator is bounded, since

$$||-\Delta u||_{L^2} \le ||u||_{2,2} $$

and since continuity implies closedness, you are done.

Answer 2

The operator is continuous from $W^{2,2}(\mathbb R)$ to $L^2(\mathbb R)$, hence closed.

It is closed as operator from $D(A)\subset L^2(\mathbb R)$ to $L^2(\mathbb R)$: The key to proceed with your proof is the inequality $$ \|u'\|_{L^2(\mathbb R)}^2\le \|u\|_{L^2(\mathbb R)} \|u''\|_{L^2(\mathbb R)} . $$ This inequality can be proven for smooth functions with compact support, and then by density arguments it holds for $W^{2,2}(\mathbb R)$-functions.

Now take sequence $u_n\to u$, $Au\to v$ in $L^2$. The inequality above implies that $u_n'$ is a Cauchy-sequence in $L^2$, so $u_n'\to w$. It remains to show $u'=w$, $w'=v$, which can be proven by passing to the limit in the definition of the weak derivative.