$\mathbf{Problem:}$ Let stripe $S=\{z\in \mathbb{C}|0<\Re(z)<1\}$, $f$ is analytic in $S$ and continuous to the boundary of $S$. Suppose $|f(z)|\leq e^{C|z|^{2-\delta}}$ for some $C>0$ and $\delta>0$. Define $\displaystyle M(x)=\sup\limits_{y}|f(x+iy)|$. Then $M(t)\leq M(0)^{1-t}M(1)^{t}$ for $0\leq t\leq 1$, which means $logM(t)$ is a convex function.

$\mathbf{Proof:}$ First let us prove if $M(0)\leq 1$ and $M(1)\leq 1$, then $|f|\leq 1$.

Fix $\epsilon>0$, define $F(z)=e^{\epsilon z^2}f(z)$ on $S$, then $|F(z)|\to 0$ as $z\to \infty$. By the maximum principle, $\sup_S|F(z)|$ must occur on the edge of $S$.

$|F(z)|\leq e^{\epsilon z^2}$ for $z\in S$

Let $\epsilon \to 0$, we get $|f(z)|\leq 1$ for $z\in S$.

For the general case, define $F(z)=f(z)M(0)^{z-1}M(1)^{-z}$, then $|F(z)|\leq 1$ on $\partial S$. And $F$ has growth rate no more than some $e^{C|z|^{2-\delta}}$, so apply the above result, we know that $\sup_S |F(z)|\leq 1$, which means $\sup_S |f(z)|\leq M(0)^{1-z}M(1)^{z}$.

Let $z=t\in [0,1]$, we get the conclusion.

$\text{Q.E.D}\hfill \square$

$\mathbf{Remark:}$ $f$ can not grow too fast, otherwise this theorem fails.

Consider $\displaystyle f(z)=e^{e^{\pi(z-\frac 12)i}}$ on $S$. Easy to verify that $|f(z)|\leq 1$ on $\partial S$. But on $\Re(z)=\frac 12$, $f$ grows extremely fast.