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In the paper Monotonicity for Elliptic Equations in Unbounded Lipschitz Domains - Berestycki - Caffarelli - Niremberg, they use the estimate "obtained by interior elliptic regularity": if $B=B_\rho(y)\subset\Omega$ and $|x-y|\leq\rho/2$, then $$|u(x)-u(y)|\leq \widehat C\max_Bu\cdot\left(\frac{|x-y|}{\rho}\right)^\gamma+\widehat C\rho^{2-\gamma}|x-y|^\gamma,$$ for $\gamma\in(0,1]$. Someone can help me to obtain this estimate? Thank you.

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I don't know why they write $\gamma$ there. Observe that $\gamma\to a^\gamma$ is a decreasing function of $\gamma$ for any fixed base $0<a<1$. Here we have $$ \frac{|x-y|}{\rho}\le \frac12 $$ and since the right hand side is $$ \widehat C\max_Bu\cdot\left(\frac{|x-y|}{\rho}\right)^\gamma+\widehat C\rho^{2}\left(\frac{|x-y|}{\rho}\right)^\gamma $$ it is a decreasing function of $\gamma$. Thus, it suffices to prove the estimate for $\gamma=1$. Theorem 3.9 in Gilbarg-Trudinger (used with $\Omega=B$) gives the gradient bound $$ \sup_{B_{\rho/2}(y)} |\nabla u| \le \frac{C}{\rho}\left( \sup_B |u| + \rho^2 \sup_B |\Delta u|\right) $$ Recalling that $\Delta u$ is uniformly bounded, and using the mean value theorem, we obtain $$ |u(x)-y(y)| \le \frac{C}{\rho}|x-y|\left( \sup_B |u| + \rho^2 \sup_B |\Delta u|\right) $$ for $x\in B_{\rho/2}(y)$.

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The $\gamma$ is used to prove that $u$ is Hölder Continuous. In the paper, we use this estimate to prove this. Can I use this estimate argument for each $\gamma\in(0,1]$? 5PM, you answered an another question to me here: math.stackexchange.com/questions/284506/… but I have one more doubt, about the uniform convergence of the Laplacian (the $C^2$-convergence). Do you know what implies this convergence? Thank you very much!!! –  José Carlos Feb 4 '13 at 18:35
    
I didn't understand why the right hand side is a decreasing function of $\gamma$. Could you explain it? –  José Carlos Feb 4 '13 at 19:29
    
@JoséCarlos I updated my answer with more detail. I'll get back to the convergence question.. –  user53153 Feb 4 '13 at 19:52
    
Thank you very much @5PM. –  José Carlos Feb 4 '13 at 19:53
    
I am waiting your help in the other question @5PM! Thank you! –  José Carlos Feb 4 '13 at 20:40

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