Give an example of an entire function of finite order \(\rho\) such that \[ \frac{\log\log M(r)}{\log r} \] does not converge to \(\rho\) as \(r\to\infty\), where \[ M(r)=\max_{|z|=r}|f(z)|. \]
Solution
Choose a strictly increasing sequence of positive integers \(N_k\) so rapidly that \[ N_{k+1}>e^{N_k^2} \qquad(k\ge1), \] and define \[ \boxed{ f(z)=\sum_{k=1}^{\infty}\frac{z^{N_k}}{N_k!}.} \] Then \(f\) has order \(1\), but \[ \frac{\log\log M(r)}{\log r} \] has limsup \(1\) and liminf \(0\).
1. The series defining \(f\) converges to an entire function.
The series is a subseries of the exponential power series. For \(|z|\le R\), \[ \sum_{k=1}^{\infty} \left|\frac{z^{N_k}}{N_k!}\right| \le \sum_{n=0}^{\infty}\frac{R^n}{n!} =e^R. \] Thus the series converges uniformly on every compact disk, so \(f\) is entire.
2. The order of \(f\) is at most \(1\).
All coefficients of \(f\) are nonnegative. Hence for \(|z|=r\), \[ |f(z)| \le\sum_k\frac{r^{N_k}}{N_k!} \le e^r. \] Therefore \[ M(r)\le e^r, \] and so \[ \limsup_{r\to\infty} \frac{\log\log M(r)}{\log r} \le1. \]
3. Along \(r=2N_k\), the logarithmic growth quotient tends to \(1\).
At \(r=2N_k\), the \(k\)-th term gives \[ M(2N_k) \ge\frac{(2N_k)^{N_k}}{N_k!}. \] Since \(N_k!\le N_k^{N_k}\), \[ M(2N_k)\ge2^{N_k}. \] Consequently \[ \log\log M(2N_k) \ge\log(N_k\log2), \] and therefore \[ \liminf_{k\to\infty} \frac{\log\log M(2N_k)}{\log(2N_k)}\ge1. \] Together with 2, this shows that the order of \(f\) is exactly \[ \rho(f)=1. \]
4. Set\[ r_k=\sqrt{N_{k+1}}. \]The contribution to \(M(r_k)\) from terms with index \(j\le k\) satisfies\[ \sum_{j\le k}\frac{r_k^{N_j}}{N_j!} \le k r_k^{N_k}. \]
For large \(k\) we have \(r_k>1\), and \(N_j\le N_k\) for \(j\le k\). Since \(N_j!\ge1\), \[ \frac{r_k^{N_j}}{N_j!} \le r_k^{N_j} \le r_k^{N_k}. \] Summing the first \(k\) terms gives the bound.
5. The tail with \(j\ge k+1\) is bounded independently of \(k\) for allsufficiently large \(k\).
We use the elementary factorial estimate \[ n!\ge (n/e)^n. \] For \(j\ge k+1\) we have \(N_j\ge N_{k+1}=r_k^2\), and hence \[ \frac{r_k^{N_j}}{N_j!} \le\left(\frac{e r_k}{N_j}\right)^{N_j} \le\left(\frac{e}{r_k}\right)^{N_j}. \] For large \(k\), \(e/r_k<1/2\), so \[ \sum_{j\ge k+1}\frac{r_k^{N_j}}{N_j!} \le\sum_{j\ge k+1}2^{-N_j}<1. \]
6. Along \(r=r_k\), the logarithmic growth quotient tends to \(0\).
By 4 and 5, \[ M(r_k)\le k r_k^{N_k}+1. \] Thus, for large \(k\), \[ \log M(r_k) \le N_k\log r_k+O(\log k), \] and therefore \[ \log\log M(r_k) \le \log N_k+\log\log r_k+O(1). \]
But \[ \log r_k=\frac12\log N_{k+1}>\frac12N_k^2. \] Hence \[ \frac{\log N_k}{\log r_k}\longrightarrow0, \qquad \frac{\log\log r_k}{\log r_k}\longrightarrow0. \] It follows that \[ \frac{\log\log M(r_k)}{\log r_k}\longrightarrow0. \]
7. The required quotient does not converge to the order.
By 3, \[ \limsup_{r\to\infty} \frac{\log\log M(r)}{\log r}=1=\rho(f), \] while 6 gives a sequence \(r_k\to\infty\) along which the quotient tends to \(0\). Therefore the quotient has no limit, as required.