Suppose that $f$ is an entire fucntion, $|f(z)|\leq e^{|z|}$ for all $z$, and $|f(x)|\leq 1$ for all $x\in \mathbb{R}$. Is it true that $|f(z)|\leq e^{|y|}$ for all $z=x+iy$?
What I've got so far was that $x\leq 0$ using the second and the third condition but I do not know if this is even relevant to the question. Can you please help me out of this one?
Yes, I believe that this is true.
Let $g(z) = f(z)e^{iz}.$ We have that $g$ is continuous on $\Omega = \{x+iy | x\geq 0,y \geq 0\}$ and that $g$ is holomorphic in the interior of $\Omega.$ Further, $f(z) \leq e^{2|z|}$ in $\Omega,$ and $$g(z) \leq 1$$ on the boundary of $\Omega.$ An application of the Phragmen-Lindelöf principle then shows that $g(z) \leq 1$ in $\Omega.$ We thus know that $|f(z)| \leq e^{|y|}$ in the first quadrant. A similar argument applies to each quadrant, so we are done.