The Hadamard three-line theorem states:
Let $f: \mathbb R + i(0,1) \to \mathbb C$ and $M_y := \sup_{x \in > \mathbb R} |f(x + iy)|$. If
- f is analytic on $\mathbb R + i(0,1)$ and continuous on $\mathbb R + i[0,1]$,
- $\sup_{y \in (0,1)} M_y < B$ for some $B > 0$ fixed, then $M_y \leq M_0^{1-y} M_1^y$.
Q: Is it possible to relax the second condition to $M_y < \frac{B}{(y(1-y))^n}$ for some $n \in \mathbb N$.
My thoughts:
The supremum is still bounded on every subset, $\sup_{y \in (\epsilon,1-\epsilon)} M_y < \infty$ for any $\epsilon > 0$ fixed.
Is it possible to estimate $M_y$ close to the boundary by $M_0$ and $M_1$?
This is related to the question whether $y \mapsto M_y$ is continuous. This is certainly not true for any analytic function, e.g. $f(z) = e^{iz}$ satisfies $M_0 = 1$ by $M_y = \infty$ for any $y>0$ fixed.
However, we additionally have the property that $M_y < \infty$ for any $y>0$. Then, it seems obvious that $\lim_{y \searrow 0} M_y = M_0$ (at least no counter-example comes to my mind). But I am unsure how this can be proven.