Theorem If $u\in C(\overline{B_R(x_0)})$ and is harmonic in $B_R(x_0)$, then $$|D^mu(x_0)|\leq\frac{n^m\exp(m-1)m!}{R^m}\max_\limits{\overline{B_R~(x_0)}}|u|$$
We can prove the theorem by induction, but I am stuck at the first step of the proof. When $m=1$, we have
$$\triangledown u(x_0)=\frac{n}{\omega_nR^n}\int_{B_R~(x_0)}\triangledown udx\cdots(1)$$
$$=\frac{n}{\omega_nR^n}\int_{\partial B_R~(x_0)}u\cdot vdS\cdots(2)$$
then
$$|\triangledown u(x_0)|\leq\frac{n}{\omega_nR^n}\int_{\partial B_R~(x_0)}|u|dS$$
$$=\frac{n}{R}\max_\limits{\overline{B_R~(x_0)}}|u|$$
I am confused with the equalities $(1)$ and $(2)$.
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In $(1)$, I think the gradient of a function is a vector field, so why can we use the $\triangledown u$ in our integral?
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And how can we transform $(1)$ to $(2)$?
Any advice is helpful. Thank you.
Best Answer
(1) is just an identity between two vectors. It is clear what is on the LHS, and on the RHS it is an integral of a vector field, whose components are just the integral of the corresponding component, so (1) holds because it holds componentwise by the mean value theorem.
(2) is the vector field version of the divergence theorem. As before you can do it componentwise, but I find it neater to think in the following way. Two vectors are identical iff their dot products with any fixed vector are identical. So let $X$ be a fixed vector, and we dot product both sides of (2) with $X$. We are then left to show
$$ \int_B \nabla u\cdot Xdx=\int_{\partial B} uX\cdot vdS. $$
This is simply the divergence theorem applied to the vector field $uX$, so this is true for any vector $X$. Hence both sides agree as vectors.