9

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!

Sam
  • 15,908
  • 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
  • @Siminore: I think this cannot be the case. Otherwise $-u$ would have a maximum in the interior (at $x_0$) and - by the maximum modulus principle - would need to be constant on $B_R(x_0)$. – Sam Sep 01 '12 at 11:06
  • Yes, I think you are right. I am wondering if the whole thing is only a typo. Equation (2.31) of Gilbarg and Trudinger is essentially the case in which $u(x_0)=0$. I cannot understand why in (2.31) the absolute value is necessary but in this exercise it is not. – Siminore Sep 01 '12 at 12:01
  • @Siminore: Thanks for your response. I wondered whether it might be a typo too, but I doubt it can really be one, since the exercise continues on the next page: "If $u\ge 0$ in $\Omega$ infer that $$|Du(x_0)|\le \frac n{R} u(x_0)$$" This would be an easy consequence of the previous part, since substitution $u\mapsto -u$ gives $|Du(x_0)| \le \frac{n}{R} \sup_{\Omega}[u(x_0) - u] \le \frac{n}{R} u(x_0)$, in case $u\ge 0$. Also, as you say, the result with absolute values was essentially already obtained in the text. Normally the exercises are a bit trickier than only applying ...(contd) – Sam Sep 01 '12 at 22:52
  • (cont) ... a result with $u-u(x_0)$ in place of $u$. But still, I don't really see, how one might go from a vector to a scalar without ending up with some absolute values. – Sam Sep 01 '12 at 22:55
  • Well, one could of course take the inner product of both sides in the equation for $Du(x_0)$ with a unit vector pointing in the same direction as $Du(x_0)$, to obtain $$|Du(x_0)| = \frac nR {\int ! ! ! ! ! ! - }_{\partial B_R(x_0)} [u-u(x_0)]\langle Du(x_0), \nu \rangle$$ Maybe such a representation could help? – Sam Sep 01 '12 at 23:26
  • From the argument in the book, I can estimate $$\nabla u(y) = \frac{1}{\omega_n R^{n}}\int_B \nabla u , dx = \frac{1}{\omega_n R^{n}} \int_{\partial B} \left( u-u(x_0) \right) \nu , ds$$ and this is smaller, in modulus, than $$\frac{n}{R}\sup_{\partial B}|u|-\frac{C}{R}u(x_0) \leq \frac{C}{R} \left( \sup_{\partial B}|u|-u(x_0) \right).$$ But I cannot get rid of the absolute value of $u$. – Siminore Sep 02 '12 at 07:59

1 Answers1

5

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.