# Ngô Quốc Anh

## July 16, 2009

### A beautiful inequality regarding complex variables

The following inequality

$\displaystyle \left|{\frac{{{z_{1}}-{z_{2}}}}{{1-\overline{{z_{1}}}{z_{2}}}}}\right|\geqslant\frac{{\left|{{z_{1}}}\right|-\left|{{z_{2}}}\right|}}{{1-\left|{{z_{1}}}\right|\left|{{z_{2}}}\right|}},\quad\forall{z_{1}},{z_{2}}\in D\left({0,1}\right)$

holds true. To prove, we do as follows: By a direct computation, we get

$\displaystyle {\left|{\frac{{{z_{1}}-{z_{2}}}}{{1-\overline{{z_{1}}}{z_{2}}}}}\right|^{2}}= 1-\frac{{\left({1-{{\left|{{z_{1}}}\right|}^{2}}}\right)\left({1-{{\left|{{z_{2}}}\right|}^{2}}}\right)}}{{{{\left|{1-\overline{{z_{1}}}{z_{2}}}\right|}^{2}}}}$

and

$\displaystyle 1-\frac{{\left({1-{{\left|{{z_{1}}}\right|}^{2}}}\right)\left({1-{{\left|{{z_{2}}}\right|}^{2}}}\right)}}{{{{\left|{1-\overline{{z_{1}}}{z_{2}}}\right|}^{2}}}}\geqslant 1-\frac{{\left({1-{{\left|{{z_{1}}}\right|}^{2}}}\right)\left({1-{{\left|{{z_{2}}}\right|}^{2}}}\right)}}{{{{\left|{1-\left|{{z_{1}}}\right|\left|{{z_{2}}}\right|}\right|}^{2}}}}.$

Similarly,

$\displaystyle 1-\frac{{\left({1-{{\left|{{z_{1}}}\right|}^{2}}}\right)\left({1-{{\left|{{z_{2}}}\right|}^{2}}}\right)}}{{{{\left|{1-\left|{{z_{1}}}\right|\left|{{z_{2}}}\right|}\right|}^{2}}}}=\frac{{{{\left({\left|{{z_{1}}}\right|-\left|{{z_{2}}}\right|}\right)}^{2}}}}{{{{\left|{1-\left|{{z_{1}}}\right|\left|{{z_{2}}}\right|}\right|}^{2}}}}.$

Thus,

$\displaystyle\left|{\frac{{{z_{1}}-{z_{2}}}}{{1-\overline{{z_{1}}}{z_{2}}}}}\right|\geqslant\frac{{\left|{{z_{1}}}\right|-\left|{{z_{2}}}\right|}}{{1-\left|{{z_{1}}}\right|\left|{{z_{2}}}\right|}}.$

As an application we can prove the following Lindelof theorem. It says that if $f$ is assumed to be holomorphic and bounded by $1$ in $D(0, 1)$. Then

$\displaystyle\left|{f\left( z\right)}\right|\leqslant\frac{{\left|{f\left( 0\right)}\right|+\left| z\right|}}{{1+\left|{f\left( 0\right)}\right|\left| z\right|}},\quad\forall z\in D\left({0,1}\right).$

To prove, we firstly see by the Schwarz-Pick theorem that

$\displaystyle\left|{\frac{{f\left( z\right)-f\left( 0\right)}}{{1-\overline{f\left( 0\right)}f\left( z\right)}}}\right|\leqslant\left|{\frac{{z-0}}{{1-\overline 0 z}}}\right| =\left| z\right|.$

On the other hand, from above we have

$\displaystyle\frac{{\left|{f\left( z\right)}\right|-\left|{f\left( 0\right)}\right|}}{{1-\left|{f\left( 0\right)}\right|\left|{f\left( z\right)}\right|}}\leqslant\left|{\frac{{f\left( z\right)-f\left( 0\right)}}{{1-\overline{f\left( 0\right)}f\left( z\right)}}}\right|.$

Combining all above yields

$\displaystyle\frac{{\left|{f\left( z\right)}\right|-\left|{f\left( 0\right)}\right|}}{{1-\left|{f\left( 0\right)}\right|\left|{f\left( z\right)}\right|}}\leqslant\left| z\right|$

which implies that

$\displaystyle\left|{f\left( z\right)}\right|-\left|{f\left( 0\right)}\right|\leqslant\left| z\right|\left({1-\left|{f\left( 0\right)}\right|\left|{f\left( z\right)}\right|}\right).$

Thus,

$\displaystyle\left|{f\left( z\right)}\right|\left({1+\left|{f\left( 0\right)}\right|\left| z\right|}\right)\leqslant\left|{f\left( 0\right)}\right|+\left| z\right|$

which gives

$\displaystyle\left|{f\left( z\right)}\right|\leqslant\frac{{\left|{f\left( 0\right)}\right|+\left| z\right|}}{{1+\left|{f\left( 0\right)}\right|\left| z\right|}}$

for all $z \in D(0,1)$. A careful reader may continue as following

$\displaystyle \left|{\frac{{f\left( 0\right)-f\left( z\right)}}{{1-\overline{f\left( 0\right)}f\left( z\right)}}}\right|\leqslant\left|{\frac{{0-z}}{{1-\overline 0 z}}}\right| =\left| z\right|.$

Therefore,

$\displaystyle\frac{{\left|{f\left( 0\right)}\right|-\left|{f\left( z\right)}\right|}}{{1-\left|{f\left( 0\right)}\right|\left|{f\left( z\right)}\right|}}\leqslant\left| z\right|$

which implies

$\displaystyle \left|{f\left( 0\right)}\right|-\left|{f\left( z\right)}\right|\leqslant\left| z\right|\left({1-\left|{f\left( 0\right)}\right|\left|{f\left( z\right)}\right|}\right)$

and hence,

$\displaystyle\frac{{\left|{f\left( 0\right)}\right|-\left| z\right|}}{{1-\left|{f\left( 0\right)}\right|\left| z\right|}}\leqslant\left|{f\left( z\right)}\right|.$

Finally, we obtain

$\displaystyle \frac{{\left|{f\left( 0\right)}\right|-\left| z\right|}}{{1-\left|{f\left( 0\right)}\right|\left| z\right|}}\leqslant\left|{f\left( z\right)}\right|\leqslant\frac{{\left|{f\left( 0\right)}\right|+\left| z\right|}}{{1+\left|{f\left( 0\right)}\right|\left| z\right|}}$

for all $z \in D(0,1)$.