I am self-reading Levin's "Lectures on Entire Functions". There are two questions, I hope I they are not too much.
1) On page 3 (bottom), the author says "If the same inequality holds for some sequence of values $r_n\to \infty$", does it mean that $h(r_n)<\phi(r_n)$ for all $n$?
2) On page 4 (above), the author says that "it follows from the definition of the order" that $\exp(r^{\rho-\epsilon})\stackrel{n}<M_f(f)\stackrel{as}<\exp(r^{\rho+\epsilon})$ for every $\epsilon>0$. Why is this true?
Note: Here is $M_f(r):=\max_{|z|=r}|f(z)|$.
-- James
$h \stackrel{n} < \phi$ means by definition that there is $r_n \to \infty, h(r_n) < \phi(r_n)$ so yes you understood 1 right Note that by the two definitions involved, the negation of $h \stackrel{as} < \phi$ is $\phi \stackrel{n} \le h$
(note that $h \stackrel{as} < \phi$ just means $h(r) < \phi (r)$ for $r \ge r_0$, where $h,\phi$ are real functions defined on the positive axis (or an infinite segment $(a,\infty)$)
For 2, by the definition of growth order, $M_f(r) \stackrel{as}<\exp(r^{\rho+\epsilon})$ for any $\epsilon >0$ as $\rho+\epsilon >\rho$
On the other hand, since $\rho$ is the infimum of such, we cannot have $M_f(r) \stackrel{as}<\exp(r^{\rho-\epsilon})$ for any $\epsilon >0$.
Hence by point 1 above, we get $\exp(r^{\rho-\epsilon}) < \exp(r^{\rho-\frac{\epsilon}{2}}) \stackrel{n} \le M_f(r)$, when we apply the above observation with $\frac{\epsilon}{2}$