Fix \(\rho\ge0\), and let \(\mathcal O_\rho(\mathbb C)\) be the set of entire functions of order at most \(\rho\).
Show that \(\mathcal O_\rho(\mathbb C)\) is a ring.
Show that \(\mathcal O_\rho(\mathbb C)\) is integrally closed in \(\mathcal M(\mathbb C)\).
Solution
For an entire function \(h\), write \[ M_h(r)=\max_{|z|=r}|h(z)|. \]
1. If \(f,g\in\mathcal O_\rho(\mathbb C)\), then\(f+g\in\mathcal O_\rho(\mathbb C)\).
The maximum-modulus inequality gives \[ M_{f+g}(r)\le M_f(r)+M_g(r) \le2\max\{M_f(r),M_g(r)\}. \] Hence \[ \log\log M_{f+g}(r) \le \log\!\left(\log2+\max\{\log M_f(r),\log M_g(r)\}\right) \] for large \(r\). Dividing by \(\log r\) and taking limsup shows \[ \rho(f+g)\le\max\{\rho(f),\rho(g)\}\le\rho. \] The zero function causes no exception.
2. If \(f,g\in\mathcal O_\rho(\mathbb C)\), then\(fg\in\mathcal O_\rho(\mathbb C)\).
One has \[ M_{fg}(r)\le M_f(r)M_g(r). \] Thus \[ \log M_{fg}(r) \le\log M_f(r)+\log M_g(r) \le2\max\{\log M_f(r),\log M_g(r)\}. \] Taking another logarithm, dividing by \(\log r\), and passing to limsup gives \[ \rho(fg)\le\max\{\rho(f),\rho(g)\}\le\rho. \]
3. Therefore \(\mathcal O_\rho(\mathbb C)\) is a ring.
It contains the constant functions, is closed under additive inverses, and <1>1–<1>2 give closure under addition and multiplication.
4. Let \(f\in\mathcal M(\mathbb C)\) satisfy a monic relation\[ f^d+a_{d-1}f^{d-1}+\cdots+a_0=0, \qquad a_j\in\mathcal O_\rho(\mathbb C). \]Then \(f\) is entire.
Since every coefficient \(a_j\) is entire, this is in particular a monic integral relation over \(\mathcal O(\mathbb C)\). By E-HM213A17-73, \(\mathcal O(\mathbb C)\) is integrally closed in \(\mathcal M(\mathbb C)\). Hence \(f\in\mathcal O(\mathbb C)\).
5. For every \(z\in\mathbb C\),\[ |f(z)|\le1+\max_{0\le j<d}|a_j(z)|. \]
Set \[ A(z)=\max_{0\le j<d}|a_j(z)|. \] Suppose \(|f(z)|>1+A(z)\). Then \(|f(z)|>1\) and, using the monic relation, \[ |f(z)|^d \le A(z)\sum_{j=0}^{d-1}|f(z)|^j =A(z)\frac{|f(z)|^d-1}{|f(z)|-1}. \] Because \(|f(z)|-1>A(z)\), the right side is strictly smaller than \(|f(z)|^d\), a contradiction. This is the usual Cauchy bound for roots of a monic polynomial.
6. The entire function \(f\) has order at most \(\rho\).
Taking maxima on \(|z|=r\) in 5 gives \[ M_f(r) \le1+\max_{0\le j<d}M_{a_j}(r). \] There are only finitely many coefficients, each of order at most \(\rho\). Therefore the finite maximum on the right has order at most \(\rho\), and adding the constant \(1\) does not increase order. Hence \[ \rho(f)\le\rho. \] Thus \(f\in\mathcal O_\rho(\mathbb C)\).
7. Consequently \(\mathcal O_\rho(\mathbb C)\) is integrally closed in\(\mathcal M(\mathbb C)\).
Every meromorphic element integral over \(\mathcal O_\rho(\mathbb C)\) belongs to that ring by 4 and 6.