# On the order of growth of entire functions

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 $$$$O_G(f) := \inf \left\{r > 0: \exists A, B > 0 \,(\text{depending on } r) \text{ such that } |f(z)| \leq Ae^{B|z|^{r}} \text{ for all } z \in \mathbb{C} \right\}. \hspace{1cm} (1)$$$$

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?

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.
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