Let $E$ be the strip $\{z\in\mathbb{C}:0<\Re z <1\}$. Let $f$ be analytic on $E$ and continuous on $\bar{E}$. Show that if $f$ is bounded on $E$ and $|f|\leq1$ on the boundary of $E$, then $|f|\leq1$ on $E$.
The hint that comes with the problem says to consider the analytic function $f_{\epsilon}(z)=(1+\epsilon z)^{-1}f(z)$ on open set $\{z\in\mathbb{C}:0<\Re z<1, -M<\Im z<M\}$ for $M$ large.
From some searching, it looks like I need to use Möbius transformation but I'm not sure how to integrate that in this case. Any input is appreciated.
This is the Phragmen-Lindelof principle. Note that $$ \lim_{|y|\to\infty} |f_\epsilon(x+iy)| = 0 $$ uniformly on $x\in [0,1]$. If we apply maximum modulus principle to the region $$ \{x+iy\;|\;0\leq x\leq 1,\;|y|\leq M\} $$ for large $M>0$, we can see that the maximum modulus of $|f_\epsilon|$ cannot occur on $y=\pm M$. Thus it must occur on the sides $x=0$ or $x=1$. This gives $$ |f_\epsilon(x+iy)|\leq 1,\quad\forall x\in[0,1],\;|y|\leq M, $$ for all sufficiently large $M>0$ and thus $$ |f_\epsilon(x+iy)|\leq 1,\quad\forall x\in [0,1], y\in \mathbb{R}. $$ Finally take $\epsilon\to 0$ to get the desired bound $$ |f(x+iy)|\leq 1,\quad\forall x\in [0,1], y\in \mathbb{R}. $$