Let $f$ be an analytic function that bounded by $1$ in the unit disc and $f({1\over2})=0$. We need to estimate $|f({3\over4})|$. Since $f({1\over2})=0$, $$ g(z)=\left\{\begin{matrix} \frac{f(z)}{\bigl(\begin{smallmatrix}\frac{z-{1\over2}}{1-{1\over2}z}\\ \end{smallmatrix}\bigr)},\text{ if } \ne{1\over2}\\ {3\over4}f'\bigl(\begin{smallmatrix}{1\over2}\\ \end{smallmatrix}\bigr),\text{ if }z={1\over2} \end{matrix}\right.$$ is likewise analytic in $|z|<1$. Letting $z\to1$ we find that $|g|\leq1$` So that $$ |f(z)|\leq \begin{vmatrix} {z-{1\over2}\over 1-{1\over2}z}\\ \end{vmatrix}$$ throughout the disc. In particular $|f({3\over4})|\leq{2\over5}$. Note that the maximum value is achived by $$B_{1\over2}(z)={z-{1\over2}\over1-{1\over2}z}$$ when $B_a(z)={z-a\over1-\bar{a}z}$ with $|a|<1$.

[Edited from Complex Analysis by Bak & Newman, page 82].

The question is why is $g$ analytic? What is the motivation for this complicated $g$?


$\frac {f(z)} {z-\frac 1 2}=\frac {f(z)-f(\frac 1 2) } {z-\frac 1 2}\to f'(\frac 1 2)$ and $1-\frac 1 2 z \to \frac 3 4$ so $g$ is analytic. [ At points other than $\frac 1 2$ is its obviously analytic].

  • $\begingroup$ Why does $|z|\to1$ implies $|g|\leq1$? Indeed, when they say "$|z|\to1$" I guess they mean $|z|\nearrow1$, so that $|z|<1$, thus $|z-{1\over2}|<{1\over2}$ and $1-{1\over 2}z>{1\over2}$ so $\begin{vmatrix} \frac{z-{1\over2}}{1-{1\over2}z}\\ \end{vmatrix}<1$ and thus $|g|>1$ and not $|g|<1$ as they claim. $\endgroup$ – J. Doe Apr 25 at 10:40
  • $\begingroup$ @J.Doe Actually $|\frac {z-\frac 12} {1-\frac 1 2 z}| =1 \to 1$ as $|z| \to 1$. $\endgroup$ – Kavi Rama Murthy Apr 25 at 11:35
  • $\begingroup$ So how from that face we can deduce $|g|< 1$? $\endgroup$ – J. Doe Apr 25 at 13:45
  • 1
    $\begingroup$ Maximum modulus principle $\endgroup$ – Kavi Rama Murthy Apr 25 at 13:47

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