I would like to know why on $W^{2,2}\cap W^{1,2}_0$ the norms

$$ ||u|| _{W^{2,2}}=\sum_{|\alpha|\leq 2}||D^\alpha u||_{L^2}$$ and $$||\Delta u||_{L^2}$$ are equivalent.

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    $\begingroup$ Are you working on a bounded domain? $\endgroup$ Nov 28 '12 at 20:43

Im gonna suppose that $\Omega$ is bounded because i dont know if $\|\Delta u||_{L^2}$ is a norm if $\Omega$ is not bounded. If $\Omega$ is bounded then $||\Delta u||_{L^2}$ is a norm. Let $H=H_0^1(\Omega)\cap H^2(\Omega)$, $\|u\|_1=\sum_{|\alpha|\leq 2}||D^\alpha u||_{L^2}$ and $\|u\|_2=||\Delta u||_{L^2}$.

Note that (trivially) $\|u\|_2\leq C\|u\|_1$, hence, if you know that $H$ is a Banach space for both norms then, you can conclude that the norms are equivalents by using a standard application of the Open Mapping Theorem, therefore, im gonna prove that $H$ is a Banach space for the norm $\|u\|_2$.

Let $(u_{n})\subset H$ be a Cauchy sequence, i.e. $$\|u_{n}-u_{m}\|_{2}\rightarrow 0\ \mbox{if}\ n,m\rightarrow\infty.$$

Let $w_{nm}\equiv u_{n}-u_{m}$ and $-\Delta w_{nm}\equiv f_{nm}\in L^{2}(\Omega)$. Therefore, $$\left\{ \begin{array}{ll} -\Delta w_{nm}=f_{nm}, & \hbox{em $\Omega$,} \\ w_{nm}\in H_{0}^{1}(\Omega). & \hbox{} \end{array} \right. $$

By using a priori estimates, we can find a constant $C>0$ such that \begin{equation}\|w_{nm}\|_{H^{2}(\Omega)}\leq C\|f_{nm}\|_{L^{2}(\Omega)}\ \forall\ n,m\in\mathbb{N}.\end{equation}

On the other hand \begin{eqnarray*} % \nonumber to remove numbering (before each equation) \|w_{nm}\|_{2}^{2} &=& \int_{\Omega}|\Delta w_{nm}|^{2}\\ &=& \|\Delta w_{nm}\|_{L^{2}(\Omega)}^{2}, \\ &=& \|f_{nm}\|_{L^{2}(\Omega)}^{2}, \end{eqnarray*}

From the last inequality we can conclud that $\|f_{nm}\|_{L^{2}(\Omega)}\rightarrow 0$ if $n,m\rightarrow \infty$, because $w_{nm}$ is a Cauchy sequence.

By using the a priori estimate we get that $$\|w_{nm}\|_{H^{2}}\rightarrow 0,\ se\ n,m\rightarrow\infty.$$

As $H^{2}(\Omega)$ is complete, there exists $u\in H^{2}(\Omega)$ such that $$u_{n}\rightarrow u$$

To prove that $u\in H_{0}^{1}(\Omega)$ we use trace theory. From trace theory, there exists $T:H^{2}(\Omega)\rightarrow L^{2}(\partial\Omega)$ bounded linear operator. As $(u_{n})\subset H_{0}^{1}(\Omega)$, we have that $$T(u_{n})=0,\ \forall\ n\in\mathbb{N}.$$

By using the continuity of $T$, we have $$\|T(u_{n})-T(u)\|_{L^{2}(\partial\Omega)}\leq C\|u_{n}-u\|_{H^{2}(\Omega)}\ \rightarrow\ 0,\ se\ n\rightarrow\infty,$$ hence, $$\|T(u)\|_{L^{2}(\partial\Omega)}\leq C\|u_{n}-u\|_{H^{2}(\Omega)}\ \rightarrow\ 0,\ se\ n\rightarrow\infty,$$ therefore, $$\|T(u)\|_{L^{2}(\partial\Omega)}=0,$$ which implies, $$T(u)=0.$$

From the last equality we conclude that $u\in H$.

The proof that $H^2$ is Banach with the norm $\|\ \|_1$ is standard. Once $H$ is a closed subset of $H^2$ you can conclude that $H$ is Banach with the norm $\|\ \|_1$.

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    $\begingroup$ This answer only holds if $\Omega$ is 'sufficiently smooth' (convex or $C^1$ should be enough). Otherwise the a-priori estimate fails (e.g. in the L-shape domain). $\endgroup$
    – gerw
    Mar 25 '19 at 7:16

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