TITLE: Does the rate of decay of an entire function dictate the global growth rate?
QUESTION [5 upvotes]: If $f$ is entire and we assume a decay rate on $\mathbb{R}^+$, does this dictate a growth rate on $\mathbb{C}$?
For example, is there a function that decays as $e^{-x^2}$ on $\mathbb{R}^+$ and is of exponential type?
What I am actually interested in is whether there exists an entire function such that $|f(z)|\le Me^{-a z}$ for all $z>0$ and $ |f(z)| \le Me^{b |z|} $ for all $z$, with $b<a$.

REPLY [2 votes]: The answer to your question is no. More generally, for functions of exponential type $b$ (which means $|f(z)|\leq Me^{b|z|}$), define the indicator:
$$h(\theta)=\limsup_{r\to\infty}\frac{\log|f(re^{i\theta})|}{r}.$$
Then there is a description of all possible indicators: they are trigonometrically convex:
$$h''+h\geq 0,$$
where derivatives should be understood in the generalized sense where they do not exist. Equivalently and more intuitively: possible indicators are exactly the support functions
of compact convex sets. From which follows that $\min h\geq -\max h$.
There are many deep generalizations of these things to functions which are not
necessarily of exponential type, and for decrease not necessarily on a straight line. For example, there is a theorem of Hayman and Kjellberg that for every
curve $\gamma$ tending to infinity,
$$\limsup_{z\to\infty,\;z\in\gamma}\frac{\log|f(z)|}{M(|z|)}\geq -1.$$
This holds for every non-constant entire function (not necessarily of exponential type).