# Ununderstood estimating of $|f({3\over 4})|$

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].
• 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. – J. Doe Apr 25 at 10:40
• @J.Doe Actually $|\frac {z-\frac 12} {1-\frac 1 2 z}| =1 \to 1$ as $|z| \to 1$. – Kavi Rama Murthy Apr 25 at 11:35
• So how from that face we can deduce $|g|< 1$? – J. Doe Apr 25 at 13:45