Proving that $|f(z)| \leq 1$ in the domain $\Omega := \{z = x + iy \in \mathbb{C} : |y| < x\}$ 
Let $\Omega = \{z = x + iy \in \mathbb{C} : |y| < x\}$. Let $f$ be a holomorphic function in $\Omega$ which is continuous in $\bar{\Omega}$. Assume that $|f(z)| \leq 1$ in the boundary of $\Omega$. Assume also that $|f(z)| \leq e^\sqrt{|z|}$ for $z \in \bar{\Omega}$. Show that $|f(z)| \leq 1$ for all $z \in \Omega$.


The first thing I thought of when I saw this question was the maximum modulus theorem. However, I can't simply assume that $|f|$ attains maximum somewhere as it is unbounded, so I thought of considering the following sub-domain:
$$
\Omega_r := \{z = x + iy \in \mathbb{C} : x > 0, |y| < x, |z| < r\}
$$
If I can show that for an arbitrary $r$, $|f(z)| \leq 1$ in $\Omega_r$, then we're done as we have $\Omega = \bigcup_{r > 0} \Omega_r$.
Now its closure is compact, so $|f|$ must attain a maximum somewhere in $\overline{\Omega_r}$. If it attains in the interior, then maximum modulus theorem applies. If it attains it at the top or bottom boundary, then immediately $|f(z)| \leq 1$ everywhere (as it is part of the boundary of $\Omega$). The issue I'm facing now is what if $|f|$ attains its maximum at the arc.
Any insight on this problem is appreciated.
 A: Defining $z^a=e^{a\log z}, -\frac{\pi}{4}  \le \arg z \le \frac{\pi}{4}, z \ne 0$, so on $\bar \Omega - ${$0$}, we notice that $|e^{-\epsilon z^a}|=e^{-\Re {(
\epsilon z^a)}}$ so in particular if $0<a<2$, $\Re {(\epsilon z^a)}=\epsilon|z|^a\cos{\arg az} \ge \epsilon|z|^a\cos{\frac{a\pi}{4}}$, hence $|e^{-\epsilon z^a}| \le e^{-\epsilon |z^a|\cos{\frac{a\pi}{4}}}$.
After these preliminaries, we can choose $\frac{1}{2}<a<2$ (for example $a=1$ will do) and considering the function $g(z)=e^{-\epsilon z}f(z)$ we notice that for large $|z|=R>0$ in our domain and fixed $\epsilon>0$ we get $|g(z)| \le e^{-\epsilon R\cos{\frac{\pi}{4}}}|f(z)| \le e^{-(\epsilon R\cos{\frac{\pi}{4}}-\sqrt R)} \le 1$, so by maximum modulus on our domain cut by the circle of radius $R$ first, and then letting $R \to \infty$ we get $|g(z)| \le 1$ since obviously the mutliplier $e^{-\epsilon z}$ is $\le 1$ in absolute value on the line boundaries too. So we get $|f(z)| \le |e^{\epsilon z}|$ on our domain for arbitrary $\epsilon >0$. But now fixing $z$ and letting $\epsilon \to 0$ we get the required inequality $|f(z)| \le 1$ so we are done!
Note that since here $a=1$ works we can dispense with the preliminaries above, but if the estimate of $|f|$ would increase to $|f(z)| \le e^{|z|^{\beta}}, 1 \le \beta <2$, we would need them. More generally if we have the similar problem with an angle $\alpha < 2\pi$ (here $\alpha = \frac{\pi}{2}$) between the bounding lines, we need a power estimate of the type $|f(z)| \le e^{|z|^{\frac{\pi}{\alpha}-\delta}}$ for some fixed $\delta >0$ for the method to apply.
