Let
\[ F(z)=\sum_{n=0}^{\infty}a_nz^n \]
be an entire function of finite order. The growth order of \(F\) is linked with the growth of the coefficients \(a_n\) as \(n\to\infty\).
Suppose that
\[ |F(z)|\leq Ae^{a|z|^\rho}. \]
Show that
\[ \limsup_{n\to\infty}|a_n|^{1/n}n^{1/\rho}<\infty. \tag{8} \]
Conversely, if (8) holds, show that
\[ |F(z)|\leq A_\varepsilon e^{a_\varepsilon|z|^{\rho+\varepsilon}} \]
for every \(\varepsilon>0\).
Hint. For part 1, use Cauchy’s inequality
\[ |a_n|\leq \frac{A}{r^n}e^{ar^\rho} \]
and the fact that \(u^{-n}e^{u^\rho}\), for \(u>0\), has minimum
\[ e^{n/\rho}\left(\frac{\rho}{n}\right)^{n/\rho} \]
at \(u=n^{1/\rho}/\rho^{1/\rho}\). Choose \(r\) in terms of \(n\) to attain this minimum.
For part 2, note that for \(|z|=r\),
\[ |F(z)| \leq \sum_n\frac{c^nr^n}{n^{n/\rho}} \leq \sum_n\frac{c^nr^n}{(n!)^{1/\rho}} \]
for some constant \(c\), since \(n^n\geq n!\). This reduces the problem to Problem 3.
Solution
For part 1, Cauchy’s inequality gives, for every \(r>0\), \[ |a_n|\le \frac{A}{r^n}e^{ar^\rho}. \] Choose \[ r=\left(\frac{n}{a\rho}\right)^{1/\rho}. \] Then \(ar^\rho=n/\rho\), and therefore \[ |a_n|^{1/n} \le A^{1/n}e^{1/\rho}\left(\frac{a\rho}{n}\right)^{1/\rho}. \] Hence \[ \limsup_{n\to\infty}|a_n|^{1/n}n^{1/\rho} \le e^{1/\rho}(a\rho)^{1/\rho}<\infty. \]
Conversely, assume (8). Then there is \(c>0\) such that, for all sufficiently large \(n\), \[ |a_n|\le \frac{c^n}{n^{n/\rho}}. \] After increasing a constant to absorb finitely many initial terms, \[ |F(z)|\le C\sum_{n=0}^\infty \frac{(c|z|)^n}{n^{n/\rho}}. \] Since \(n!\le n^n\), \[ \frac1{n^{n/\rho}}\le \frac1{(n!)^{1/\rho}}. \] Thus \[ |F(z)|\le C\sum_{n=0}^\infty\frac{(c|z|)^n}{(n!)^{1/\rho}}. \] By Problem 3, the entire function on the right has order \(\rho\). Hence for every \(\varepsilon>0\) there are constants \(A_\varepsilon,a_\varepsilon>0\) such that \[ |F(z)|\le A_\varepsilon e^{a_\varepsilon|z|^{\rho+\varepsilon}}. \] This proves the converse.