Let $f(z)$ be a holomorphic function in the angle $A=\{0<\arg z<\frac{\pi}2\}$, continuous in $\bar A$, satisfying $|f(z)|\le M$ on $\partial A$ and satysfying the following growth condition:
$$
|f(x+i y)|\le Ce^{y^2}\quad\hbox{in $A$}. \label{1}\tag{$\ast$}
$$
**Question**. Does it follow that $f$ is bounded in $A$?


If in \eqref{1} we consider $y^\rho$ with $\rho<2$, then we have $|f(z)|\le C e^{|z|^\rho}$ and by the [Phragmén–Lindelöf principle](https://en.wikipedia.org/wiki/Phragm%C3%A9n%E2%80%93Lindel%C3%B6f_principle#Phragm%C3%A9n%E2%80%93Lindel%C3%B6f_principle_for_a_sector_in_the_complex_plane) $f$ would be bounded in $A$. For $\rho=2$ it's not the case as the example $f(z)=e^{-i z^2}$, where $|f(x+i y)|=e^{2xy}$ shows. But for this function the condition \eqref{1} doesn't hold.



Phragmén–Lindelöf principle for the critical exponent
For a small $\alpha > 0$, write $$\beta = \frac{\sin^2 \alpha}{\sin(2 \alpha)} = \frac{\tan \alpha}{2} ,$$ and define $$g(z) = f(z) \exp(i \beta z^2).$$ Then $$|g(z)| \leqslant |f(z)| \leqslant C \exp(|z|^2)$$ for $z \in A$, $$|g(i r)| = |f(i r)| \leqslant M$$ for $r > 0$, and $$\begin{aligned}|g(r e^{i \alpha})| & = |f(r e^{i \alpha}))| \exp(-\beta r^2 \sin(2 \alpha)) \\ & \leqslant C \exp(r^2 \sin^2 \alpha - \beta r^2 \sin(2 \alpha)) = C \end{aligned}$$ for $r > 0$. By the Phragmén–Lindelöf principle, $g$ is bounded by $C + M$ in $A\_\alpha = \{z \in \mathbb C : \alpha < z < \tfrac\pi2\}$. Passing to a limit as $\alpha \to 0^+$, we find that $f$ is bounded by $C + M$ in $A$, and by another application of Phragmén–Lindelöf principle, $f$ is in fact bounded by $M$.


