This from the book Elliptic Partial Differential equations by Han and Lin.

Lemma 1.32 Suppose $u$ is a non-negative harmonic function in $B_1$. Then there holds $$\sup\limits_{B_{1/2}}|D\log{u}|\leq C$$ where $C=C(n)$ is a positive constant.

Proof: We may assume $u>0$ in $B_1$. Set $v=\log{u}$. Then direct calculation shows $$\Delta v=-|Dv|^2.$$

We need the interior gradient estimate on $v$. Set $w=|Dv|^2$. Then we get $$\Delta w+2\sum\limits_{i=1}^{n}D_ivD_iw=2\sum\limits_{i,j=1}^n(D_{ij}v)^2.$$

We need a cutoff function. First note

$$\sum\limits_{i,j}^n (D_{ij}v)^2\geq\sum\limits_{i=1}^n(D_{ii}v)^2\geq \frac{1}{n}(\Delta v)^2=\frac{|Dv|^4}{n}=\frac{w^2}{n}$$

Taking a nonnegative function $\phi\in C_0^1(B_1)$. We obtain by the Holder inequality \begin{align} &\Delta(\phi w)+\sum\limits_{i=1}^nD_ivD_i(\phi w)\\ &=2\phi\sum\limits_{i,j=1}^n(D_{ij}v)^2+4\sum\limits_{i,j=1}^nD_i\phi D_jvD_{ij}v+2w\sum\limits_{i=1}^nD_i\phi D_iv+(\Delta\phi)w\\ & \geq \phi\sum\limits_{i,j=1}^n(D_{ij}v)^2-2|D\phi||Dv|^3-\left(|\Delta \phi|+C\frac{|D\phi|^2}{\phi}\right)|Dv|^2 \end{align} if $\phi$ is chosen such that $|D\phi|^2/\phi$ is bounded in $B_1$. Choose $\phi=\eta^4$ for some $\eta \in C_0^1(B_1)$. Hence for such fixed $\eta$ we obtain

\begin{align} &\Delta(\eta^4w)+ 2\sum\limits_{i=1}^n D_ivD_i(\eta^4w)\\ &\geq \frac{1}{n} \eta^4|Dv|^4-C\eta^3|D\eta||Dv|^3-4\eta^2(\eta\Delta\eta+C|D\eta|^2)|Dv|^2\\ &\geq \frac{1}{n}\eta^4|Dv|^4-C\eta^3|Dv|^3-C\eta^2|Dv|^2 \end{align} where $C$ is a positive constant depending only on $n$ and $\eta$. Hence we get by the Holder inequality $$\Delta(\eta^4w)+2\sum\limits_{i=1}^nD_ivD_i(\eta^4w)\geq\frac{1}{n}\eta^4w^2-C$$ where $C$ is a positive constant depending only on $n$ and $\eta$. Suppose $\eta^4w$ attains its maximum at $x_0\in B_1$. Then $D(\eta^4w)=0$ and $\Delta\eta^4w\leq0$ at $x_0$. Hence there holds $$\eta^4w^2(x_0)\leq C(n,\eta).$$ If $w(x_0)\geq 1$, then $\eta^4w(x_0)\leq C(n)$. Otherwise $\eta^4w(x_0)\leq w(x_0)\leq \eta^4(x_0)$. In both cases we conclude $$\eta^4w\leq C(n,\eta)$$ in $B_1$.

My questions are

  1. How is Holder inequality used here (the two instances mentioned in the proof)?

  2. How do we infer the result from the last inequality?


The Holder inequality is

$$ 2ab \le a^2 + b^2.$$

A more useful one is

$$ 2ab = 2(\sqrt\epsilon a)(b/\sqrt\epsilon) \le \epsilon a^2 + b^2 /\epsilon. $$

for all $\epsilon >0$. Then one has

\begin{align*} 4 D_i\phi D_jvD_{ij}v &= 2\left( 2 D_i\phi D_jv/\sqrt{\phi}\right) (\sqrt\phi D_{ij}v) \\ &\le \epsilon \phi |D_{ij } v|^2 + \frac{4}{\epsilon}\frac{|D_i\phi|^2}{\phi} |D_jv|^2 \\ \Rightarrow 4 \sum_{i,j=1}^n D_i\phi D_jvD_{ij}v &\ge - \epsilon \phi \sum_{i,j=1}^n |D_{ij } v|^2 - \frac{4}{\epsilon}\frac{|D\phi|^2}{\phi} |Dv|^2 \end{align*}

Now choose $\epsilon = 1$. Also use

$$2|D v|^2 \sum_{i=1}^n D_i\phi D_i v= 2|D v|^2 D\phi \dot Dv \ge -2 |Dv|^2 |D\phi| |Dv|.$$

The second one could be done this way: use

\begin{align} \eta^2 |Dv|^2 &\le \frac 12(\epsilon_1 \eta^4|Dv|^4 +1/\epsilon_1),\\ \eta^3 |Dv|^3 &= (\eta^2 |Dv|^2)(\eta |Dv|) \\ &\le \frac 12 (\epsilon_2 \eta^4 |Dv|^4 + \eta^2|Dv|^2/\epsilon_2) \\ &\le \frac 12 (\epsilon_2 \eta^4 |Dv|^4 + \frac 1{2\epsilon_2} (\epsilon_3\eta^4|Dv|^4+ 1/\epsilon_3)) \end{align}

Then we choose $\epsilon_1, \epsilon_2, \epsilon_3$ small so that $$-C\eta^3|Dv|^3-C\eta^2|Dv|^2 \ge -\frac {1}{2n} \eta^4 |Dv|^4-C$$ and this gives $$\frac {1}{n} \eta^4 |Dv|^4-C\eta^3|Dv|^3-C\eta^2|Dv|^2 \ge \frac {1}{2n} \eta^4 \omega^2-C.$$

This is slightly different from your inequality, but the argument is the same.

The last inequality follows from the fact that $x_0$ is the maximum of $\eta^4 \omega^2$. So

$$ \eta^4 \omega^2 (x) \le \eta^4 \omega(x_0)$$

for all $x$.


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