Let $L^2=L^2(\mathbb{R}^n)$
$-\Delta:D(-\Delta)\subset L^2\to\subset L^2$ by $-\Delta u=\mathcal{F}^{-1}(|\xi|^2\widehat{u}(\xi))$ and $D(-\Delta)=H^{2,2}=\left\{u\in L^2: \mathcal{F}^{-1}((1+|\xi|)^2\widehat{u}(\xi))\in L^2\right\}$ ($L^2$-Sobolev space of order 2 [Wong Introduction pseudo differential operator])
I have read that the Laplacian is bounded in the Sobolev space (for example, I read it here Laplace operator defined on a Sobolev space) but I cannot see this in a simple way. I don't know if I'm misunderstanding something about the norms. My question is:
Question 1. Is $-\Delta$ a bounded operator? i.e. $\left\|-\Delta u\right\|_{L^2}\leq C\left\|u\right\|_{L^2}$?
No, it is not bounded with respect to $L^2$ norms. For example, fix a non-trivial, say Schwartz or a smooth compactly supported function $f$, and consider for each $r>0$, the rescaling $f_r(x)=f(rx)$. Then, by a change of variables, you see that \begin{align} \|f_r\|_{L^2(\Bbb{R}^n)}=\frac{1}{r^{n/2}}\|f\|_{L^2(\Bbb{R}^n)}. \end{align} Also, by the chain rule, $\Delta(f_r)(x)=r^2\cdot(\Delta f)(rx)=r^2\cdot (\Delta f)_r(x)$, so \begin{align} \|\Delta(f_r)\|_{L^2(\Bbb{R}^n)}=r^2\cdot\frac{1}{r^{n/2}}\|\Delta f\|_{L^2(\Bbb{R}^n)}. \end{align} So, if $-\Delta$ was bounded with respect to $L^2$ norms, then the inequality would have to hold for each $f_r$, and hence for each $r>0$, \begin{align} r^2\|\Delta f\|_{L^2(\Bbb{R}^n)}&\leq C\|f\|_{L^2(\Bbb{R}^n)}. \end{align} By taking $r\to\infty$, you see that this is a contradiction.
On the other hand, it is bounded $H^2(\Omega)\to L^2(\Omega)$ when you use the Sobolev norm rather than merely the $L^2$ norm on $H^2(\Omega)$.