# True or False: $f$ analytic in the right plane, continuous on the boundary, $|f| \le 1$ on the boundary, then $|f| \le 1$

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.

-

The problem is that $\Omega$ is unbounded, otherwise the maximum modulus principle would indeed be useful.

A counterexample is $f(z) = \exp(-z^4)$. If $z = iy$, then $f(z) = \exp(-y^4)$, so $|f(z)| \le 1$.

(Note that for this example, it's even the case that $|f(iy)| \to 0$ as $y \to \pm\infty$.)

-
$$f(z)=e^z$$
Then $|f| = 1$ on $i \mathbb R$.