On the order of growth of entire functions

Question

Definition. Let $f: \mathbb{C} \to \mathbb{C}$ be an entire function. The order of growth of $f$, denoted by $O_G(f)$, is defined as \begin{equation} O_G(f) := \inf \left\{r > 0: \exists A, B > 0 \,(\text{depending on } r) \text{ such that } |f(z)| \leq Ae^{B|z|^{r}} \text{ for all } z \in \mathbb{C} \right\}. \hspace{1cm} (1) \end{equation}

My question: If $f$ is entire and $O_G(f) = \rho$, do there exist constants $A,B > 0$ such that \begin{align*} \hspace{2cm} |f(z)| \leq Ae^{|z|^{\rho}} \quad \text{ for all } z \in \mathbb{C} \;? \hspace{2cm} (2) \end{align*}

(In other words: does the set on the RHS of (1) contain its infimum?)

My thinking so far: For each $n \in \mathbb{N}$ there exist constants $A_n > 0$ and $B_n > 0$ such that \begin{align*} |f(z)| \leq A_n e^{B_n |z|^{\rho + \frac{1}{n}}} \quad \text{for all } z \in \mathbb{C}. \end{align*}

Now if the sequences $(A_n)_{n=1}^{\infty}$ and $(B_n)_{n=1}^{\infty}$ are bounded, then we can let $A := \sup_{n} A_n$ and $B := \sup_n B_n$, and we will have \begin{align*} \hspace{2cm} |f(z)| \leq A e^{B |z|^{\rho + \frac{1}{n}}} \quad \text{for all } z \in \mathbb{C}, \hspace{2cm} (3) \end{align*}

and then taking $n \to \infty$ gives (2). But must there exist bounded sequences $(A_n)_{n=1}^{\infty}$ and $(B_n)_{n=1}^{\infty}$ satisfying (2)? If not, what would be a counterexample?

Answer 1

No, the infimum may not be achieved. Take $f$ to be any non-constant polynomial.

Then, if $r>0$ we can choose $A>0$ such that $|f(z)| \leq Ae^{|z|^{r}}$ (this can be argued using that every real exponential with positive exponent grows faster than every polynomial, for this and other arguments see this).

Thus, we have $O_G(f)=0$. Nonetheless, there are no constants $A,B>0$ such that $|f(z)|\leq Ae^{B|z|^0}= Ae^B$ for all $z\in\mathbb{C}$, as that would imply that $f$ is constant by Liouville's Theorem.

Answer 2

The answer is no unless the type of the function of the given order is finite (eg $\sin, \exp$ are order $1$ and finite type); an easy counterexample is given by Weierstrass $\Delta(z)=1/\Gamma(z)$ which is easily seen to be of order $1$ from its Weierstrass factorization $$\Delta(z)=ze^{\gamma z}\Pi_{k \ge 1}(1+z/k)e^{-z/k}$$ but Jensen theorem shows that $$\int_0^{2\pi}\log |\Delta(Re^{it})|dt >> R\log R$$ since the roots in the disc of radius $R$ are $0,-1,..-[R]$ etc

In general both the order and type of an entire function can be determined form the Taylor coefficients by some well known formulas, so one can construct examples for any combination order, type easily