Understanding the definition of the order of an entire function in Ahlfors's Complex Analysis

Question

Let $f: \mathbb C \to \mathbb C$ be an entire function. The order of $f$ is defined by $$\lambda=\limsup_{r \to \infty} \frac{\log \log M(r)}{\log r}, $$ where $$M(r)=\max_{|z|=r} |f(z)| .$$

Ahlfors in his Complex Analysis claims that

"According to this definition $\lambda$ is the smallest number such that $$M(r)\leq e^{r^{\lambda+\varepsilon}} $$ for any given $\varepsilon > 0$ as soon as $r$ is sufficiently large."

Why is this true?

My attempt:

We know that $$\lambda=\lim_{\rho \to \infty} \sup_{r \geq \rho} \frac{\log \log M(r)}{\log r}. $$

From the definition of the limit we have that for any $\varepsilon>0$, there exists some $\rho_0>0$, such that $$\left\lvert \sup_{r \geq \rho} \frac{\log \log M(r)}{\log r}-\lambda \right\rvert \leq \varepsilon ,$$ for every $\rho \geq \rho_0$. In other words $$\frac{\log \log M(r)}{\log r} \leq \lambda+\varepsilon $$ for every $r \geq \rho_0$. From here it is easy to see that $$M(r)\leq e^{r^{\lambda+\varepsilon}}, $$ for all $r \geq \rho_0$. I cannot see why $\lambda$ is the smallest number with this property.

Thanks in advance.

Answer 1

Note that

$$F(\rho) = \sup_{r \geqslant \rho} \frac{\log\log M(r)}{\log r}$$

is a non-increasing function of $\rho$, hence $\lim\limits_{\rho\to\infty} F(\rho) = \inf\limits_{\rho > R} F(\rho).$

By the definition of the limes superior, for every $\mu < \lambda$, with $\varepsilon = \frac{\lambda-\mu}{3}$ there are arbitrarily large radii $r$ with

$$\frac{\log \log M(r)}{\log r} > \lambda - \varepsilon = \mu + 2\varepsilon,$$

and that inequality is equivalent to

$$M(r) > e^{r^{\mu+2\varepsilon}},$$

so for every $\mu < \lambda$, there is an $\varepsilon > 0$, such that there is no $\rho_0$ with

$$M(r) \leqslant e^{r^{\mu+\varepsilon}}$$

for all $r \geqslant \rho_0$, indeed

$$M(r)e^{-r^{\mu+\varepsilon}}$$ is unbounded for all small enough $\varepsilon > 0$ then.

Answer 2

I once stumbled upon this definition also. I would like to write up a little bit more details than Daniel's nice answer here.

For convenience, define (${\small \text{we will discuss this definition later}}$) for $r>0$ that $$ G(r):=\frac{\log\log M(r)}{\log r}. $$ We prove the following proposition:

With the definition $ \lambda:=\limsup_{r\to\infty}G(r), $ the following hold.

(1) For every $\epsilon>0$, there exists $R>0$ such that for every $r\geqslant R$, $$ M(r)\leqslant \exp(r^{\lambda+\epsilon})\tag{*} $$ (2) If $\mu>0$ is such that for every $\epsilon>0$, there exists $R>0$ such that for every $r\geqslant R$, $$ M(r)\leqslant \exp(r^{\mu+\epsilon}), $$ then $\lambda\leqslant\mu$. (Note that this is what "smallest" means in Ahlfors.)

Proof. By definition of the $\limsup$ we have $$\lambda=\inf_{R>0}\sup_{r\geqslant R}G(r),$$ which implies (by the definition of $\inf$) that for every $\epsilon>0$, there exists $R>0$ such that

• (i) $\sup_{r\geqslant R}G(r)<\lambda+\epsilon$;
• (ii) $\sup_{r\geqslant R}G(r)\geqslant\lambda$.

We will now prove that (i) implies (1) and (ii) implies (2).

Note that (i) in particular implies that $$ G(r)<\lambda+\epsilon,\quad r\geqslant R. $$ Taking the exponential function twice on both sides of $$ \log(r)G(r)<\log(r)(\lambda+\epsilon) $$ we get (1).

On the other hand, the assumption of (2) implies that for every $\epsilon>0$, $ \sup_{r>R}G(r)\leqslant \mu+\epsilon, $ and thus $$ \sup_{r>R}G(r)\leqslant \mu, $$ which together with (ii) yields $$ \lambda\leqslant\mu. $$

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Note: $G(r)$ is well-defined only for $M(r)>1$. But this is true for large enough $r$ since one could assume that the entire function is not constant.