Aug 152013
 

Harnack’s inequality 是关于调和函数的一个不等式, 被 A. Harnack 在 1887 年引进. 随后又被其他人重新发现, 比如 J. Serrin 和 J. Moser. 有不少重要的数学家对 Harnack’s inequality 做出各种推广. 通过Nash-Moser迭代, 人们发现在较为一般的散度型椭圆方程和抛物方程正解都具有这种性质. 从此, Harnack 不等式在偏微分方程解性质研究中发挥了巨大作用. 上世纪八十年代, P.Li 和丘成桐给出了Harnack不等式的另一种认识途径, 即所谓微分Harnack 估计. Li-Yau 对 Harnack不等式的新认识对 Ricci 流发展有重要影响. 经典椭圆型偏微分方程和抛物方程中的Harnack不等式在几何流中, 有很多应用. Perelman 证明 Poincaré conjecture, 就使用了 R. Hamilton 的一个 Harnack’s inequality 的推广形式. Harnack’s inequality 在偏微分方程有很多重要应用.

Harnack’s inequality  Let  \(D=\{z:|z|<1\}\). suppose \(f(z)\) is analytic on \(D\), \(\mathrm{Re}f(z)\geqslant0,\forall z\in D, f(0)>0\), then

  • \(|\mathrm{Im}f(z)|\leqslant f(0)\dfrac{2|z|}{1-{|z|^2}};\)
  • \(f(0)\dfrac{1-|z|}{1+|z|}\leqslant \mathrm{Re}f(z)\leqslant |f(z)| \leqslant f(0)\dfrac{1+|z|}{1-|z|},\)

and that equality holds if and only if  \(f(z)=w_0\dfrac{1+e^{i\alpha}z}{1-e^{i\alpha}z}\)(\(w_0,\alpha\in\Bbb R\), and \(w_0>0\)).

Proof   Without loss of generality, we assume \(f(0)=1\). Let

\[ h(z) =\frac{1+z}{1-z} \]

be the standard linear fractional map of \(D\) onto the right half plane. \(\forall r, 0\leqslant r<1\),  \(h(z)\) maps \(\{z: |z|\leqslant r\}\)  to the disc

\[E_r=\{w:|w-\frac{1+r^2}{1-r^2}|\leqslant\frac{2r}{1-r^2}\}.\]

According to Schwartz  lemma , it follows that

\[\left |\frac{f(z)-1}{f(z)+1}\right| \leqslant |z|,\]

Thus we are led to the conclusion that

\[ |h^{-1}(f(z))|\leqslant |z|.\]

this implies \(f(z)\in E_{|z|}\).        \(\Box\)

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