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.

About YiMin

This is just a nerd PhD student of Math@UT Austin.

One thought on “Review Yau

  1. YiMin says:

    latex2wp is not as good as I have expected.

Comments are closed.