# Estimation in Sobolev norm

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.

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.
• 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.. – Giovanni Febbraro May 30 at 10:35
• 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 $||\nabla u||_{L^2(\Omega')}^2 \leq C (||u||_{L^2(\Omega)}^2 + ||f||_{L^2(\Omega)}^2)$ – Giovanni Febbraro May 30 at 10:49
• @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 $v=\eta u$. – Lorenzo Quarisa May 30 at 10:51
• 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? – Giovanni Febbraro May 30 at 10:53