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In the Theorem $3.9$ (page $41$) of the book: Gilbarg Trudinger, we have the estimate: $$\displaystyle\sup_\Omega\left[\operatorname{dist}(x,\partial\Omega)\cdot|\nabla u|\right]\leq C\left(\sup_\Omega|u|+\sup_\Omega\left[\operatorname{dist}^2(x,\partial\Omega)\cdot|f(x)|\right]\right),$$ where $u$ is a Poisson equation solution, $\Delta u=f$ in $\Omega$. Why in compact subsets, this estimate implies in the uniform boundedness of the gradient $|\nabla u|$? How can I separate the gradient from product in the supremum?

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If $x$ is restricted to a compact subset $K\subset \Omega$, then $\operatorname{dist}(x,\partial\Omega)\ge \operatorname{dist}(K,\partial\Omega)>0$. (Recall that the distance between sets is attained when one is closed and the other is compact.) This implies $$ \sup_K \, \operatorname{dist}(x,\partial\Omega) |\nabla u(x)| \ge \sup_K \, \operatorname{dist}(K,\partial\Omega) |\nabla u(x)| = \operatorname{dist}(K,\partial\Omega) \sup_K |\nabla u| $$ where the first inequality holds pointwise, and the second step is pulling out a positive constant. Hence, $$ \sup_K |\nabla u|\leq C(\operatorname{dist}(K,\partial\Omega))^{-1} \left(\sup_\Omega|u|+\sup_\Omega\left[\operatorname{dist}^2(x,\partial\Omega)\cdot|f(x)|\right]\right) \tag{1} $$ Sometimes you may be worried that the suprema on the right might be infinite. Then pick a domain $\Omega'$ such that $K\subset \Omega'$ and $\overline{\Omega'}$ is a compact subset of $\Omega$. Apply (1) to $\Omega'$ instead of $\Omega$.

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  • $\begingroup$ But in the left hand side, I think you are using $$\operatorname{dist}(K,\partial\Omega)\cdot\sup_K\cdot|\nabla u|\leq\sup_\Omega\operatorname{dist}(x,\partial\Omega)\cdot|\nabla u|.$$ This is true? Because on the left side we have a product and in the right side we have a supremum of product. $\endgroup$ – José Carlos Feb 5 '13 at 20:35
  • $\begingroup$ One more time, thank you very much. I just have one last doubt, is the $C^2$-convergence that you explained some days ago in math.stackexchange.com/questions/284506/… If you could be help me there... ome more time, thank you very much!! $\endgroup$ – José Carlos Feb 5 '13 at 20:58

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