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In class we have seen that if $u$ is harmonic, $\Omega \subset \mathbb R^N$ and $\Omega' \subset \subset \Omega$ is a subdomain, then $$ \|Du\| _{L^\infty(\Omega')} \leq \frac{N}{d(\Omega', \partial \Omega)}\|u\|_{L^\infty(\Omega)}, $$ which is Theorem 2.10 in Gilbarg and Trudinger for the case $|\alpha| = 1$. The proof I can see follows from the bound $$ \left| \frac{\partial u}{\partial x_i}(x_0) \right| \leq \frac Nr \|u\|_{L^\infty(B_r(x_0))} \quad \forall x_0 \in \Omega, \forall 0 < r < d(x_0, \partial \Omega). $$ (which in turn follows from $u_{x_i}$ being harmonic and the Mean Value Property).

Now, as I see it: $$ \|Du\|_{L^\infty(\Omega')} = \sup_{\Omega'} |Du| \leq \sqrt{N \sup_{i, \Omega'}\left| \frac{\partial u}{\partial x_i}(x_0) \right|^2} \leq \sqrt{N \frac{N^2}{d(\Omega', \partial \Omega)^2}\|u\|_{L^\infty(\Omega)}^2} \\ = \frac{N^{3/2}}{d(\Omega', \partial \Omega)}\|u\|_{L^\infty(\Omega)}. $$

Why is this exponent $3/2$ appearing? What am I missing? Maybe I should have considered the maximum norm in $\mathbb R^N$ instead of the cannonical one, but wouldn't this be a problem?

Thanks in advance and kind regards.

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No, there is no power missing. Using the Mean-Value Property and the Gauss-Green-Theorem (see Proof of the Gauss-Green Theorem) for each component of the gradient, we have

\begin{align} Du(x_0)=\frac{1}{\omega_Nr^N}\int_{B_{r}(x_0)} Du(x)dx=\frac{1}{\omega_Nr^N}\int_{\partial B_r(x_0)}u(x)\nu(x)dS(x), \end{align}

where $\omega_N$ represents the volume of the unit ball and $\nu$ the outward unit normal. If we denote the standard inner product in $\mathbb{R}^N$ by a dot "$\cdot $", then we have

\begin{align} \big|Du(x_0)\big|^2&=|Du(x_0)\cdot Du(x_0)|\\ &=\frac{1}{\omega_Nr^N}\bigg|\int_{\partial B_r(x_0)}u(x)\nu(x)\cdot Du(x_0)dS(x)\bigg|\\ &\leq \frac{1}{\omega_Nr^N}\int_{\partial B_r(x_0)}|u(x)|\big|\nu(x)\cdot Du(x_0)\big|dS(x)\\ &\leq \frac{1}{\omega_Nr^N}\int_{\partial B_r(x_0)}|u(x)|\big|\nu(x)\big|\big|Du(x_0)\big|dS(x)\\ &\leq \frac{N\omega_Nr^{N-1}}{\omega_Nr^N}||u||_{L^{\infty}(B_r(x_0))}\big|Du(x_0)\big|, \end{align} where we used the triangle-inequality in the third line, Cauchy-Schwartz in the fourth line and the fact that the outward unit normal has euclidean norm one in the last line. If you divide by the Euclidean norm of the gradient, then you get your desired estimate without the power.

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  • $\begingroup$ Thanks to you, now I understand Gilbarg & Trudinger's proof. Thanks a lot! $\endgroup$ Dec 16 '20 at 22:30

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