# Review Yau

1. Introduction

Assume that ${\boldmath{M}}$ is a noncompact Riemann manifold and ${f:B(R)\rightarrow\mathbb{R}}$ harmonic and positive if the Ricci curvature on ${B(R)}$ has a lower bound ${\mathrm{Ric}(\boldmath{M})\ge-(n-1)K}$ for some ${ K\ge 0}$, then

$\displaystyle \sup_{B(R/2)}|\nabla\log f|\le (n-1)K+\frac{C}{R} \ \ \ \ \ (1)$

We are going to prove the case ${\boldmath{M}=\mathbb{R}^n}$, i.e, ${\mathrm{Ric}(\boldmath{M})=0}$. Set ${h=\log f}$.

2. Proof

Consider and ${\phi}$ defined on ${\mathbb{R}}$ which satisfies that ${\phi}$ is supported on ${B(R)}$.
Set ${G=\phi^2|\nabla h|^2}$, then

1. G is non-negative on ${B(R)}$ and ${G=0}$ on ${\partial B(R)}$. Thus G attains its maximum in ${B(R)}$, say at ${x_0}$. Then

$\displaystyle |\nabla G(x_0)|=0\,\,\,\,\,,\,\,\,\Delta G(x_0)\le 0 \ \ \ \ \ (2)$

2. Consider function ${h}$,we have

$\displaystyle \Delta h=-|\nabla h|^2 \ \ \ \ \ (3)$

$\displaystyle \frac{1}{2}\Delta|\nabla h|^2=\sum_{i,j}|h_{ij}|^2-\langle\nabla h,\nabla|\nabla h|^2\rangle \ \ \ \ \ (4)$

3. Choose orthogonal frame ${\{e_i\}}$, s.t.
• ${h_{\alpha}=0}$, whenever ${\alpha\neq 0}$.
• ${|h_1|=|\nabla h|}$, which means ${e_1=\displaystyle\frac{\nabla h}{|\nabla h|}}$.
• Because of the special frame, we get that
1. ${\displaystyle\langle\nabla h,\nabla|\nabla h|^2\rangle=2h_ih_jh_{ij}=2h_1^2h_{11}=2|\nabla h|^2h_{11}}$
2. ${|\nabla|\nabla h|^2|^2=4\sum_{i,j}|h_{ij}h_j|^2=4\sum_{i}|h_{1i}|^2|\nabla h|^2}$
• Consider the equations above, then

$\displaystyle \begin{array}{lcl} \displaystyle\sum_{i,j}|h_{ij}|^2&\ge& \displaystyle|h_{11}|^2+\sum_{\alpha\ge 2}|h_{\alpha\alpha}|^2+2\sum_{\alpha\ge 2}|h_{1\alpha}|^2\\ &\ge&\displaystyle |h_{11}|^2+2\sum_{\alpha\ge 2}|h_{1\alpha}|^2+\displaystyle\frac{1}{n-1}|\sum_{\alpha\ge 2}h_{\alpha\alpha}|^2\\ &=&\displaystyle|h_{11}|^2+2\sum_{\alpha\ge 2}|h_{1\alpha}|^2+\displaystyle\frac{1}{n-1}|\Delta h-h_{11}|^2\\ &=&\displaystyle|h_{11}|^2+2\sum_{\alpha\ge 2}|h_{1\alpha}|^2+\displaystyle\frac{1}{n-1}||\nabla h|^2+h_{11}|^2\\ &\ge& \displaystyle\frac{n}{n-1}|h_{11}|^2+2\sum_{\alpha\ge 2}|h_{1\alpha}|^2+\frac{1}{n-1}|\nabla h|^4+\frac{2}{n-1}h_{11}|\nabla h|^2 \end{array} \ \ \ \ \ (5)$

Here we used Cauchy-Schwarz Inequality and (3).

• Because ${n\ge 2}$, we obtain:

$\displaystyle \begin{array}{lcl} \sum_{i,j}|h_{ij}|^2&\ge&\displaystyle\frac{n}{n-1}|h_{11}|^2+\frac{n}{n-1}\sum_{\alpha\ge 2}|h_{1\alpha}|^2+\frac{1}{n-1}|\nabla h|^4+\frac{2}{n-1}h_{11}|\nabla h|^2\\ &=&\displaystyle\frac{n}{4(n-1)}\frac{1}{|\nabla h|^2}|\nabla|\nabla h|^2|^2+\frac{1}{n-1}|\nabla h|^4+\frac{2}{n-1}h_{11}|\nabla h|^2 \end{array} \ \ \ (6)$

• Thus by (4),

$\displaystyle \begin{array}{lcl} \displaystyle\frac{1}{2}\Delta|\nabla h|^2&\ge& \displaystyle\frac{n}{4(n-1)}\frac{1}{|\nabla h|^2}|\nabla|\nabla h|^2|^2+\frac{1}{n-1}|\nabla h|^4+\frac{2}{n-1}h_{11}|\nabla h|^2-2|\nabla h|^2h_{11}\\ &=&\displaystyle\frac{n}{4(n-1)}\frac{1}{|\nabla h|^2}|\nabla|\nabla h|^2|^2+\frac{1}{n-1}|\nabla h|^4-\frac{2n-4}{n-1}h_{11}|\nabla h|^2 \end{array} \ \ \ \ \ (7)$

• Substitute ${|\nabla h|^2=\displaystyle\frac{G}{\phi^2}}$ in the last inequality:

$\displaystyle \displaystyle\frac{1}{2}\Delta(\frac{G}{\phi^2})\ge \frac{n}{4(n-1)}\frac{\phi^2}{G}|\nabla\frac{G}{\phi^2}|^2+\frac{1}{n-1}|\frac{G}{\phi^2}|^2-\frac{n-2}{n-1}\langle\nabla\frac{G}{\phi^2},\nabla h\rangle \ \ \ \ \ (8)$

• Before we continue:

$\displaystyle \begin{array}{lcl} \displaystyle\Delta(\phi^2\cdot\frac{G}{\phi^2})&=&\displaystyle\nabla\cdot[\nabla(\phi^2\cdot\frac{G}{\phi^2})]\\ &=&\displaystyle\nabla\cdot[(\nabla\phi^2)\frac{G}{\phi^2}+\phi^2\nabla(\frac{G}{\phi^2})]\\ &=&\displaystyle(\Delta \phi^2)\frac{G}{\phi^2}+2\nabla\phi^2\cdot\nabla(\frac{G}{\phi^2})+\Delta(\frac{G}{\phi^2})\phi^2 \end{array} \ \ \ \ \ (9)$

• Therefore we multiply (8) with ${\phi^2}$. Then by (9),

$\displaystyle \begin{array}{lcl} \displaystyle\frac{1}{2}\Delta(\frac{G}{\phi^2})\phi^2&=&\displaystyle\frac{1}{2}\Delta G-\frac{1}{2}\Delta\phi^2\cdot\frac{G}{\phi^2}-\nabla\phi^2\cdot\nabla(\frac{G}{\phi^2})\\ &\ge&\displaystyle \frac{n}{4(n-1)}\frac{\phi^4}{G}|\nabla(\frac{G}{\phi^2})|^2+\frac{1}{n-1}\frac{G^2}{\phi^2}-\frac{n-2}{n-1}\phi^2\langle\nabla(\frac{G}{\phi^2}),\nabla h\rangle \end{array} \ \ \ \ \ (10)$

• Let’s put this at the point ${x_0}$. We have:
• ${\displaystyle|\nabla(\frac{G}{\phi^2})|^2=|(\nabla G)\frac{1}{\phi^2}+G\nabla(\frac{1}{\phi^2})|^2=G^2|\nabla(\frac{1}{\phi^2})|^2=4G^2\frac{|\nabla\phi|^2}{\phi^6}}$
• ${\displaystyle\phi^2\langle\nabla\phi^2,-\frac{2\nabla\phi}{\phi^3}\cdot G\rangle=\phi^2\langle2\phi\nabla\phi,-\frac{2\nabla\phi}{\phi^3}\cdot G\rangle=-4G\langle\nabla\phi,\nabla\phi\rangle=-4G|\nabla\phi|^2}$
• Then at ${x_0}$,

$\displaystyle \begin{array}{lcl} 0\ge\displaystyle\frac{1}{2}\Delta G\ge \frac{n}{4(n-1)}\frac{\phi^4}{G}\cdot G^2\frac{4|\nabla\phi|^2}{\phi^2}&+&\displaystyle\frac{1}{n-1}\frac{G^2}{\phi^2}-\frac{n-2}{n-1}\phi^2\langle\nabla(\frac{G}{\phi^2}),\nabla h\rangle\\&+&\displaystyle\frac{1}{2}\Delta\phi^2\cdot\frac{G}{\phi^2}+\nabla\phi^2\cdot\nabla(\frac{G}{\phi^2}) \end{array} \ \ \ \ \ (11)$

• Mutiply (11)with ${\phi^2}$:

$\displaystyle \begin{array}{lcl} 0\ge \displaystyle\frac{n}{n-1}G|\nabla\phi|^2+\frac{1}{n-1}G^2&+&\displaystyle\frac{n-2}{n-1}\phi^4\langle G\frac{2\nabla\phi}{\phi^3},\nabla h\rangle\\ &+&\displaystyle\frac{1}{2}\Delta\phi^2\cdot G+\phi^2\nabla\phi^2\cdot\nabla(\frac{G}{\phi^2}) \end{array} \ \ \ \ \ (12)$

which is:

$\displaystyle \begin{array}{lcl} 0&\ge& \displaystyle\frac{n}{n-1}G|\nabla\phi|^2+\frac{1}{n-1}G^2+\frac{1}{2}\Delta \phi^2\cdot G-4G|\nabla \phi|^2+\frac{2n-4}{n-1}G\phi\langle\nabla\phi,\nabla h\rangle\\ &\ge&\displaystyle\frac{n}{n-1}G|\nabla\phi|^2+\frac{1}{n-1}G^2+\frac{1}{2}\Delta \phi^2\cdot G-4G|\nabla \phi|^2-\frac{2n-4}{n-1}G\phi|\nabla \phi||\nabla h|\\ &=&\displaystyle\frac{n}{n-1}G|\nabla\phi|^2+\frac{1}{n-1}G^2+\frac{1}{2}\Delta \phi^2\cdot G-4G|\nabla \phi|^2-\frac{2n-4}{n-1}G^{3/2}|\nabla\phi| \end{array} (13)$

which means:

$\displaystyle 0\ge \displaystyle\frac{n}{n-1}|\nabla \phi|^2+\frac{1}{n-1}G+\frac{1}{2}\Delta\phi^2-\frac{2n-4}{n-1}G^{1/2}|\nabla\phi| \ \ \ \ \ (14)$

And we can simplify it to:

$\displaystyle (2n-3)|\nabla\phi|^2-(n-1)\phi\Delta\phi+(2n-4)|\nabla\phi|G^{1/2}\ge G \ \ \ \ \ (15)$

for any cutoff function ${\phi}$ at ${x_0}$.

• Consider ${\phi=R^2-\rho^2}$, where ${\rho=\sqrt{\sum_jx_j^2}}$, then
• ${|\nabla \rho|=1}$
• ${\Delta \rho^2=2n}$

Take it into (15),

$\displaystyle 4(2n-3)\rho^2+2n(n-1)(R^2-\rho^2)+2(2n-4)\rho G^{1/2}\ge G \ \ \ \ \ (16)$

Thus

$\displaystyle 4(2n-3)R^2+2n(n-1)R^2+2(2n-4)R G^{1/2}\ge G \ \ \ \ \ (17)$

Then

$\displaystyle G^{1/2}\le \{2(n-2)+\sqrt{6n^2-10n+4}\}R \ \ \ \ \ (18)$

Because ${x_0}$ is the maximum point, then

$\displaystyle \sup_{B(R)}G^{1/2}\le C(n)\cdot R$

restrict this to ${B_{R/2}}$, then we have:

$\displaystyle \sup_{B(R/2)}(R^2-\rho^2)|\nabla h|\le C(n)\cdot R$

therefore

$\displaystyle \frac{3}{4}R^2\sup_{B(R/2)}|\nabla h|\le C(n)\cdot R$

i.e,

$\displaystyle \sup_{B(R/2)}|\nabla h|\le \frac{4}{3}C(n)\cdot R$

2.1. Remark

• This estimate is sharp, when we come to the example of linear functions.
• We can replace the cutoff function ${\phi}$ by other cutoff functions.

3. Application

1. If ${f}$ is positive in the whole space, then ${f}$ is a constant.
2. What if when the domain is $\mathbb{R}^n-\{0\}$?
3. When $\displaystyle\frac{f(x)}{x}\rightarrow 0$, then $f$ is a constant.
4. Prove Harnack Inequality.