How does this inequality follow? 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 $u&gt;0$ 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}
&amp;\Delta(\phi w)+\sum\limits_{i=1}^nD_ivD_i(\phi w)\\
&amp;=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\\
&amp; \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}
&amp;\Delta(\eta^4w)+ 2\sum\limits_{i=1}^n D_ivD_i(\eta^4w)\\
&amp;\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\\
&amp;\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


How is Holder inequality used here (the two instances mentioned in the proof)?
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 $\epsilon &gt;0$. Then one has 

\begin{align*}
4 D_i\phi D_jvD_{ij}v &amp;= 2\left( 2 D_i\phi D_jv/\sqrt{\phi}\right) (\sqrt\phi D_{ij}v) \\
&amp;\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 &amp;\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 &amp;\le \frac 12(\epsilon_1 \eta^4|Dv|^4 +1/\epsilon_1),\\
\eta^3 |Dv|^3 &amp;= (\eta^2 |Dv|^2)(\eta |Dv|) \\
&amp;\le \frac 12 (\epsilon_2 \eta^4 |Dv|^4 + \eta^2|Dv|^2/\epsilon_2) \\
&amp;\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 $x$.
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 $\epsilon &gt;0$. Then one has 

\begin{align*}
4 D_i\phi D_jvD_{ij}v &amp;= 2\left( 2 D_i\phi D_jv/\sqrt{\phi}\right) (\sqrt\phi D_{ij}v) \\
&amp;\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 &amp;\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 &amp;\le \frac 12(\epsilon_1 \eta^4|Dv|^4 +1/\epsilon_1),\\
\eta^3 |Dv|^3 &amp;= (\eta^2 |Dv|^2)(\eta |Dv|) \\
&amp;\le \frac 12 (\epsilon_2 \eta^4 |Dv|^4 + \eta^2|Dv|^2/\epsilon_2) \\
&amp;\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 $x$.