The version of Phragmen Lindelof theorem we work with saying that for $f$ holomorphic on the complex sector: $S=(z|-\pi/4 < arg(z) < pi/4)$ and continuous
on $\bar S$.
If $|f(z)|<1$ on $\partial S $ and there existing such constants $c,C >0 $ , $|f(z)|<Ce^{c|z|}$
So $\forall z \in S $ $|f(z)|<1$
I need to find counterexample to the case when the last condition:"there existing such constants $c,C >0$ , $|f(z)|<Ce^{c|z|}$" not happening, and so the argument that $\forall z \in S $ $|f(z)|<1$ would be false.
Best Answer
One simple counterexample is $$\frac{1}{2}e^{z^2}.$$ We can observe that if $z=r(1\pm i),$ then $\lvert f(z)\rvert <1$ but $f(r)\to\infty$ as $r\to\infty, r\in\mathbb{R}$.