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 $b<a$.
This is a typical walk-to-the-library problem. I used Boas, but probably other standard books would have worked too.
Boas proves the following results: (1) if $f$ is of order $1$, then $\limsup m(r)M(r)^{1+\epsilon}=\infty$ (Theorem 3.3.1); (2) if $f$ is of order $0<\rho<1$, then we obtain $m(r_n)\gtrsim M(r_n)^{-a}$ along a suitable sequence $r_n\to\infty$ from Theorem 3.2.11; (3) finally, if $f$ is of order $0$, then $\log m(r)\simeq \log M(r)$ on a suitable sequence again (Theorem 3.6.2).
So there is no such function.
(I've used the usual notations $m(r), M(r)$ for the min and max, respectively, of $|f|$ on $|z|=r$.)
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).
I don't have the answer to your question but on top of my head : consider an entire function $f$ that goes to zero on $\mathbb R^+$, the function $z\mapsto f(1/z)$ is holomorphic on $\mathbb C^*$. Since $f$ is going to zero via real values, we know this function is not meromorphic with a pole . Furthermore, it cannot be bounded otherwise it would be holomorphic on the riemann sphere thus constant. Finally, $f$ has an essential singularity at infinity. However this does not answer your question.
