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 $b<a$.



Does the rate of decay of an entire function dictate the global growth rate?
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 $|z|=r$.)


