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I would like some help with the following problem (Gilbarg/Trudinger, Ex. 2.13):

Let $u$ be harmonic in $\Omega \subset \mathbb R^n$. Use the argument leading to (2.31) to prove the interior gradient bound, $$|Du(x_0)|\le \frac{n}{d_0}[\sup_{\Omega} u - u(x_0)], \quad d_0 = \mathrm{dist}(x_0, \partial \Omega)$$

The argument mentioned in the exercise is the following: Since $u$ is harmonic, it follows that also $Du$ is harmonic. Hence if $B_R = B_R(x_0)\subset \subset \Omega$ we obtain

$$Du(x_0) = \frac{1}{\omega_n R^n} \int_{B_R} Du = \frac{1}{\omega_n R^n} \int_{\partial B_R} u\nu = \frac{n}{R} \frac{1}{n\omega_n R^{n-1}}\int_{\partial B_R} u\nu$$

where $\nu$ is the outward pointing unit normal vector (and the integration on the RHS is supposed to happen componentwise). Subtituting $u \mapsto u-u(x_0)$, this shows $$Du(x_0) = \frac{n}{R} \frac{1}{n\omega_n R^{n-1}}\int_{\partial B_R} [u-u(x_0)]\nu$$ But I don't see how to go on from here...

Of course, if $\sup_{B_R(x_0)} |u-u(x_0)| = \sup_{B_R(x_0)} [u-u(x_0)]$, then there is no problem, since then one can just take norms on both sides. I don't think this needs to be the case in general, however. (integrating a suitable function against the Poisson kernel.)

Your help would be appreciated, thanks!

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But $\sup_\Omega u - u(x_0) \geq 0$, I think. –  Siminore Aug 29 '12 at 14:37
    
@Siminore: Yes, this is true. Could you maybe elaborate on what implications can be drawn from this? I don't see it. –  Sam Aug 29 '12 at 14:51
    
Probably I cannot understand. If I take absolute values, I find $$|Du(x_0)| \leq \frac{C}{R} \left( \sup_\Omega u - u(x_0) \right).$$ Now let $R \to d_0$ and you should find the estimate. Am I wrong? –  Siminore Aug 29 '12 at 14:57
    
I don't see how you can take absolute values and end up with the $\sup_\Omega u - u(x_0)$ part. I can see how you would obtain $$|Du(x_0)| \le \frac{n}{R} \sup_{\partial B_R} |u-u(x_0)| |\nu| \le \frac{n}{R} \sup_{B_R} |u-u(x_0)|$$ but not what you have written, unfortunately. –  Sam Aug 29 '12 at 15:08
    
Probably $u(x) \geq u(x_0)$ for every $x \in \partial B_R$, by the maximum principle. Then you can remove the first absolute value. –  Siminore Aug 29 '12 at 16:22

1 Answer 1

up vote 3 down vote accepted

By rotational symmetry we may assume that $Du(x_0)$ points in the direction of the first basis vector $e_1$. We must prove that $${\int\!\!\!\!\!\!-}_{\partial B_R} [u-u(x_0)]\nu_1\le \sup u-u(x_0)$$ where $\nu_1$ is the first component of the unit normal vector. By the mean value property $u-u(x_0)$ has zero mean on $\partial B_R$. Thus, we can add any number to $\nu_1$ without changing the integral. Let's add $1$: $${\int\!\!\!\!\!\!-}_{\partial B_R} [u-u(x_0)]\nu_1={\int\!\!\!\!\!\!-}_{\partial B_R} [u-u(x_0)](\nu_1+1)$$ Now that the factor $\nu_1+1$ is nonnegative, we use a one-sided bound on $u-u(x_0)$: $${\int\!\!\!\!\!\!-}_{\partial B_R} [u-u(x_0)](\nu_1+1)\le (\sup u-u(x_0)){\int\!\!\!\!\!\!-}_{\partial B_R}(\nu_1+1)$$ Finally, $${\int\!\!\!\!\!\!-}_{\partial B_R}(\nu_1+1)=1$$ because $\nu_1$ has zero mean.

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Oh, yeah! Very nice. =) Thank you so much! –  Sam Sep 16 '12 at 3:42

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