Let $f$ be an analytic function in $\Omega = \{z : \mathcal R z > 0\}$.
Assume $f$ is continuous on $\bar \Omega$ and $|f(z)| \le 1$, when $z \in i\Bbb R$.
Is $|f(z)| \le 1$ for all $z \in \Omega$?
I think the answer is yes, but I don't see how to prove it.
As $f$ is only continuous on $\bar \Omega$, the maximum modulus principle doesn't apply.