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Suppose $f$ is holomorphic on $D_{1}(0)$ the open unit disc. Let $\Gamma_{1} = \{z : |z| = 1, x>0, y>0\}$ where $z = x+iy$ and define $\Gamma_{2}, \Gamma_{3}, \Gamma_{4}$ similarly. On $\Gamma_{i}$, $|f(z)| \le M_{i}$. How can we show $|f(0)|\le (M_{1}M_{2}M_{3}M_{4})^{\frac{1}{4}}$?

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Perhaps I am missing some convention here, but why can I not define $f$ arbitrarily on the unit circle? There is no continuity assumption. So, choose $f(z) = \frac{1}{1-z}$ on the open unit disk, and $f(e^{i\theta}) = 0$, for $\theta \in [0,2\pi)$. – copper.hat May 13 '12 at 23:56

Hint: Consider $g(x) = f(x) f(ix) f(-x) f(-ix)$.

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Is there some implied connection between the behavior of $f$ on the unit circle with the behavior inside the unit disk? (I'm referring to the question, not your answer.) – copper.hat May 14 '12 at 1:00
However, I like your approach. – copper.hat May 14 '12 at 1:01
Yes, we do have to be careful. My solution would be correct if $f$ extends to a continuous function on the closed unit disk. If you only assume it's holomorphic in the open unit disk, there is no connection with the behaviour on the circle. – Robert Israel May 14 '12 at 1:15
Even if the statements about $f(z)$ for $z \in \Gamma_j$ refer to limits (but not as $z \to 1$, which is not in any $\Gamma_j$), it is not enough: consider $f(z) = e^{1/(1-z)}$. – Robert Israel May 14 '12 at 1:43
Is the hint suggesting that for all $x$ in $[0,1]$ we have $f(x)\leq M_1$, $f(ix)\leq M_2$ for all $x$ in $[0,1]$, etc.? – John Adamski May 14 '12 at 13:02

Jensen's formula states that either f(0)=0 or $$\log|f(0)|=\frac{1}{2\pi}\int_0^{2\pi}\log|f(re^{i\theta})|\,d\theta-\sum_{k=1}^n\log(\frac{r}{|a_k|})$$ for any function $f$ analytic on a region containing $\overline{B(0;r)}$ with zeros $a_1, a_2, \ldots, a_n$ in $B(0;r)$, repeated according to multiplicity.

Therefore, for any $0<r<1$ we have $$\log|f(0)|\leq\frac{1}{2\pi}\int_0^{2\pi}\log|f(re^{i\theta})|\,d\theta\leq\frac{1}{4}(\log(m_1)+\log(m_2)+\log(m_3)+\log(m_4),$$ where $m_1=\sup_{0<\theta<\pi/2}\{|f(re^{i\theta}|\}$, $m_2=\sup_{\pi/2<\theta<\pi}\{|f(re^{i\theta}|\}$, etc.

Letting $r\rightarrow1$, we have $$\log|f(0)|\leq\frac{1}{4}(\log(M_1)+\log(M_2)+\log(M_3)+\log(M_4))$$ $$|f(0)|\leq(M_1M_2M_3M_4)^\frac{1}{4}$$

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