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I try to understand the last part of Interior Regularity theorem. The theorem says:

if $u\in W^{1,2}(\Omega)$ , where $\Omega$ is in $R^n$ and bounded, and $u$ is the weak solution of $\Delta u = f $ then, for every $\Omega'\subset \subset \Omega $, $u \in W^{2,2}(\Omega')$ and $\Vert u \Vert_{W^{2,2}(\Omega')} \leq C (||u||_{L^2(\Omega)} + ||f||_{L^2(\Omega)})$.

Now, I am in the last lines of proof and the question is:

Why $\int_{\Omega '}|D_i \nabla u|^2 \leq C(||u||_{L^2(\Omega)} + ||f||_{L^2(\Omega)}) $ implies that $\Vert u \Vert_{W^{2,2}(\Omega')} \leq C (||u||_{L^2(\Omega)} + ||f||_{L^2(\Omega)})$ ?

Maybe I don't understand how the Sobolev norm works? There are some properties that escapes me? I mean, I have the estimation for $||D^2 u||_{L^2(\Omega')}$ but what about $||u||_{L^2(\Omega')}$ and $||Du||_{L^2(\Omega')}$?

Thank u.

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We know that $u\in W^{2,2}(\Omega')$ because $\|D^2 u\|_{L^2(\Omega')}<\infty$ and $u\in W^{1,2}(\Omega')$.

Now, take a cut-off function $\eta\in C_c^{\infty}(\Omega)$ such that $0\leq \eta \leq 1$, $\eta=1$ in $\Omega'$, $\|\nabla \eta\|_{L^{\infty}(\Omega)}\leq C_1(\Omega',\Omega)$, $\|D^2 \eta\|_{L^{\infty}(\Omega)}\leq C_2(\Omega',\Omega)$.

Then $\eta u \in W_0^{2,2}(\Omega)$. By Poincare's theorem you will have an estimate of the form $$\|\eta u\|_{W^{2,2}(\Omega)} \leq C(\Omega)\|D^2(\eta u)\|_{L^2(\Omega)}$$ On the other hand, by the properties of $\eta$, you also have $$ \|D^2(\eta u)\|_{L^2(\Omega)}\leq C(\Omega',\Omega)\|D^2 u\|_{L^2(\Omega')} $$ and $$\|u\|_{W^{2,2}(\Omega')}\leq \|\eta u\|_{W^{2,2}(\Omega)} $$ so putting together the above inequalities you obtain the desired result.

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  • $\begingroup$ Thanks for your answer. I don't understand how you can use the Poincarè Inequality in the space $W_0^{2,2}$...I knew that is valid in the space $W_0^{1,p}$ and furthermor the estimation is valid in $L^p$ spaces.. $\endgroup$ – Giovanni Febbraro May 30 at 10:35
  • $\begingroup$ Now I had this idea: if I simply add $||u||_L^2$ and $||\nabla u||_L^2$ in the inequality $\int_{\Omega '}|D_i \nabla u|^2 \leq C(||u||_{L^2(\Omega)} + ||f||_{L^2(\Omega)})$ ? Then I use a Lemma that says me $||\nabla u||_{L^2(\Omega')}^2 \leq C (||u||_{L^2(\Omega)}^2 + ||f||_{L^2(\Omega)}^2)$ $\endgroup$ – Giovanni Febbraro May 30 at 10:49
  • $\begingroup$ @GiovanniFebbraro 1. You just apply the Poincarè inequality you know to $\nabla v$ since $v\in W_0^{2,2}(\Omega)$ implies $\nabla v \in W_0^{1,2}(\Omega)$. 2. Once you bound $\|v\|_{L^p}$ by $\|\nabla v\|_{L^p}$ you automatically also have a bound for $\| v\|_{W^{1,p}}$ by $\|\nabla v\|_{L^p}$. Above I took $v=\eta u$. $\endgroup$ – Lorenzo Quarisa May 30 at 10:51
  • $\begingroup$ Now I had this idea: if I simply add $||u||_L^2$ and $||\nabla u||_L^2$ in the inequality $\int_{\Omega '}|D_i \nabla u|^2 \leq C(||u||_{L^2(\Omega)} + ||f||_{L^2(\Omega)})$ ? Then I use a Lemma that says me $||\nabla u||_{L^2(\Omega')}^2 \leq C (||u||_{L^2(\Omega)}^2 + ||f||_{L^2(\Omega)}^2)$. Can it works? $\endgroup$ – Giovanni Febbraro May 30 at 10:53

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