I have following definitions at disposal:
- exponential type $\tau$ of an entire function $f$ is defined as: \begin{align*}
\tau = \limsup\limits_{r \to \infty}^{} \frac{\log M(r)}{r}
\end{align*}where $M(r)=\max \{\:\lvert f(z) \rvert : \lvert z \rvert\ = r\}$ is the maximum modulus function. - (finite) order $\rho$ of an entire function $f$ is defined as the infimum of all numbers $a>0$ which satisfy \begin{align*}
M(r) \leq \exp(r^a)
\end{align*} for $r>R$ , or equivalently as: \begin{align*}
\rho = \limsup\limits_{r \to \infty} \frac{\log(\log M(r))}{\log r}
\end{align*}
It is often mentioned in literature that entire functions of order less than $1$ have the exponential type $0$:
\begin{align*}
\rho < 1 \implies \tau = 0
\end{align*}
Question: How does one formally prove this?
My attempt:
\begin{align*}
\frac{\log(\log M(r))}{\log r} < 1 \iff \log_{\,r} \log M(r) < 1 \iff \log M(r) <r \iff \frac{\log M(r) }{r} < 1
\end{align*}
which is the expression from the definition of $\tau$. From here I'm not sure what I should do or if I'm heading in the right direction.
I would appreciate any help!
Best Answer
Assume that $$ \rho = \limsup_{r \to \infty} \frac{\log(\log M(r))}{\log r} < 1 $$ and choose a number $a \in (\rho, 1)$. Then, for sufficiently large $r$, $$ \begin{align} &\frac{\log(\log M(r))}{\log r} \le a \\ \implies \quad &\log M(r) \le r^a \\ \implies \quad &\frac{\log M(r)}{r} \le r^{a-1} \, . \end{align} $$ It follows that $$ 0 \le \tau = \limsup_{r \to \infty} \frac{\log M(r)}{r} \le \lim_{r \to \infty} r^{a-1} = 0 \, . $$