1

I am following the notes of my professor but I am unsure if I am following it correctly. Here's the problem:

Suppose that for $1\leq i \leq n, |u_{x_i}(\vec{x_0})| = C\left| \displaystyle\int_{B(\vec{x_0})} \nabla u\cdot e_i\,d\vec{x} \right|$ where $e_i$'s are the standard basis vectors in $\mathbb{R}^n$. How can I show that $|\nabla u(\vec{x_0})|\leq C\left|\displaystyle \int_{B(\vec{x_0})}\nabla \cdot u\,d\vec{x}\right|$ where $C$ is the same constant and that $B(\vec{x_0})$ is just a ball centered at $\vec{x_0}$?

I am confused because from what I know, $|\nabla u(\vec{x_0})| = \sqrt{u_{x_1}^2+\dots+u_{x_n}^2} \leq |u_{x_1}|+\dots+|u_{x_n}|.$

and from the given, $|u_{x_1}|+\dots+|u_{x_n}| = C\left(\left| \displaystyle\int_{B(\vec{x_0})} \nabla u\cdot e_1\,d\vec{x} \right| + \dots +\left| \displaystyle\int_{B(\vec{x_0})} \nabla u\cdot e_n\,d\vec{x} \right|\right)$ which is the reverse direction of what I intend to show...

any help?

Tomas Jorovic
  • 3,983
  • 3
  • 27
  • 38
  • Are you sure about your first equality? What is $x_0$?. Does this hold for every point? In the second integral you find a divergence instead of a gradient. – Quickbeam2k1 Sep 22 '13 at 19:53
  • $x_0$ is just a point in the interior or $\vec{x_0} \in U$. I am considering a ball centered at $x_0$ with radius $r$ inside the region $U$ here on which $u(\vec{x})$ is harmonic. Yes the first inequality is accurate, it can be found in Evans PDE book p.29 equation 20. My professor is doing it differently which is why I asked. – Tomas Jorovic Sep 22 '13 at 20:07
  • and yes, it has to be a divergence inside the integral because if it were a gradient, $\nabla u$ would be a vector and it would not make much sense to perform the integration. I am sure it's the divergence operator in there because the divergence theorem is used subsequently (the remaining parts I fully understand. just this step I am unsure of!) – Tomas Jorovic Sep 22 '13 at 20:08
  • In that theorem you find nowhere the divergence. It is also important that $C$ depends on the measure of $B$. Why do you want an estimate in terms of divergence? The theorem (7) gives you an estimate of the $L^1$ norm of $u$.There is no derivative on the right hand side – Quickbeam2k1 Sep 23 '13 at 08:42
  • Consider the function $f(x,y)=x-y$. This function is harmonic, the gradient is $(1,-1)$ but the divergence is zero. Hence you will never be able to find such an estimate. – Quickbeam2k1 Sep 23 '13 at 17:43

0 Answers0