Problem statement:
Let $\Omega=\{z\in\mathbb{C}: 0<\text{Re} z<1\}.$ Assume $f$ is bounded and continuous on $\bar{\Omega}$ and holomorphic on $\Omega$. Show that $$\sup_\Omega |f|=\sup_{\partial\Omega} |f|.$$
I was given the following hint: For $\epsilon>0$, consider $f_\epsilon (z)=f(z)e^{\epsilon z^2}.$ Relax the hypothesis that $f$ is bounded.
I haven't been able to make much headway, so I was hoping for a further hint (not a solution). I'm not clear on exact what the hint is saying when it says to relax the hypothesis. I think it's saying that if I try a proof by contradiction, I'll get that $f$ is unbounded.
If $f$ is constant, then the result is clear, so suppose not. Evidently, $f_\epsilon$ is holomorphic on $\Omega$ and continuous on $\bar{\Omega}.$ It is also non-constant (as we assume $f$ is non-constant), so by the contrapositive of the maximum modulus principal, there is no $z_0\in\Omega$ for which $|f_\epsilon(z_0)|\geq |f_\epsilon(z)|,$ where $z\in\Omega$ in arbitrary. I'm not really sure what, if anything, to do with this information.
This is a Phragmén–Lindelöf type result.
Assume that $|f(z)| \le K$ for $z \in \Omega$ and $|f(z)| \le M$ for $z \in \partial \Omega$. The goal is to show that $|f(z)| \le M$ for all $z \in \Omega$.
Hints (as requested): For $\epsilon > 0$ define $f_\epsilon (z)=f(z)e^{\epsilon z^2}$. Then $$ \tag{*} |f_\epsilon (z)|=|f(z)|e^{\operatorname{Re}\epsilon z^2} = |f(z)| e^{\epsilon (x^2 - y^2)} \le |f(z)| e^{\epsilon (1 - y^2)} $$ for $z = x + iy$. Now investigate $f_\epsilon$ on a rectangle $$ R_s = \{ z \mid 0 \le \operatorname{Re} z \le 1, -s \le \operatorname{Im} z \le s \} $$ for fixed $s > 0$. Use $(*)$ to obtain estimates for $f_\epsilon$ on the boundary of $R$, and apply the maximum principle to conclude that $$ |f_\epsilon (z)| \le \max(M e^{\epsilon}, K e^{\epsilon(1-s^2)}) $$ in the rectangle $R_s$. For sufficiently large $s$ it follows that $|f_\epsilon (z)| \le M e^{\epsilon}$ in $R_s$ and thus in $\Omega$.
The conclusion then follows by letting $\epsilon \to 0$.
Finally, the hypothesis that $f$ is bounded can be relaxed by a condition that $f(x+iy)$ does now grow "too fast" for $|y| \to \infty$.