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Page 141, Question 3:

Let $G=\{z:|\operatorname{Im} z| < \pi/2\}$ and suppose $f:G\rightarrow C$ and $\limsup|f(z)| \leq M$ on $w$ in the boundary of $G$. Also, suppose $A < \infty$ and $a < 1$ can be found such that $|f(z)| < \exp[A \exp(a|\operatorname{Re} z|)]$ for all $z$ in $G$. Show that $|f(z)|\leqslant M$ for all $z$ in $G$.

Thanks in advance.

  • 0
    I think my problem is finding \phi(z) that satispies the second condition in Phragmen-Lindelof theorem: limsup |f(z)|*|\phi(z)|^\epsilon <= M, for every \epsilon > 0.2012-12-07

1 Answers 1

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Consider $f\circ\log (z)$ and apply Corol. 4.2 of Chapter 6 in Conway's book.

Corollary 4.2 Let $a\geq \frac{1}{2}$ and put $G=\{z: |arg z|<\frac{\pi}{2a}\}$. Suppose that $f$ is analytic on $G$ and there is a constant $M$ s.t. $\limsup_{z\to w}|f(z)|\leq M$ for all $w$ in $\partial G$. If there are positive constants $P$ and $b s.t. $|f(z)|\leq P \exp(|z|^b)$ for all z with $|z|$ sufficiently large, then $|f(z)|\leq M$ for all $z$ in $G$.

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    Just as the OP provided the problem from the book, you should add the corollary to your answer2014-03-21