Use the Cauchy inequalities or the maximum modulus principle to solve the following problems:
Prove that if \(f\) is an entire function that satisfies \[ \sup _{|z|=R}|f(z)| \leq A R^{k}+B \] for all \(R>0\), some integer \(k\geq 0\), and some constants \(A, B > 0\), then \(f\) is a polynomial of degree \(\leq k\).
Show that if \(f\) is holomorphic in the unit disc, is bounded, and converges uniformly to zero in the sector \(\theta < \arg(z) < \phi\) as \(\abs{z} \to 1\), then \(f \equiv 0\).
Let \(w_1, \cdots w_n\) be points on \(S^1 \subset \CC\). Prove that there exists a point \(z\in S^1\) such that the product of the distances from \(z\) to the points \(w_j\) is at least 1. Conclude that there exists a point \(w\in S^1\) such that the product of the above distances is exactly 1.
Show that if the real part of an entire function is bounded, then \(f\) is constant.
Solution: Part 1
Fix an integer \(m>k\). By Cauchy’s estimate on the circle \(|z|=R\), \[ |f^{(m)}(0)| \le {m!\over R^m}\max_{|z|=R}|f(z)| \le m!\qty(AR^{k-m}+BR^{-m}). \] Letting \(R\to\infty\) gives \(f^{(m)}(0)=0\). This holds for every \(m>k\), so \[ f(z)=\sum_{m=0}^k {f^{(m)}(0)\over m!}z^m. \]
Solution: Part 2
Let \(S=\{z\in\DD:\theta<\arg z<\phi\}\). Choose \(N\) so large that the rotated sectors \[ e^{2\pi i j/N}S, \qquad 0\le j<N, \] cover every angular direction. Since \(f\) is bounded, put \[ M=\sup_{z\in\DD}|f(z)|<\infty \] and define \[ g(z)=\prod_{j=0}^{N-1} f\qty(e^{2\pi i j/N}z). \]
Fix \(\varepsilon>0\). By uniform convergence to \(0\) in \(S\) as \(|z|\to1\), there is \(\rho<1\) such that \[ |f(z)|<\varepsilon \qquad(z\in S,\ \rho<|z|<1). \] For every \(r\in(\rho,1)\) and every \(|z|=r\), at least one rotated point \(e^{2\pi i j/N}z\) lies in \(S\). Hence \[ |g(z)|\le \varepsilon M^{N-1} \qquad(|z|=r). \] The maximum modulus principle gives the same bound on \(|z|\le r\). Since \(\varepsilon\) is arbitrary, \(g\equiv0\) on \(\DD\).
The ring of holomorphic functions on the connected disk is an integral domain, so one factor \(f(e^{2\pi i j/N}z)\) is identically zero. Rotation is a bijection of \(\DD\), hence \(f\equiv0\).
Solution: Part 3
Consider \[ f(z) \da \prod_{1\leq k \leq n} (w_k - z) .\] Then \(f\) is holomorphic and nonconstant on \(\DD\), so attains a maximum \(M\) on \(S^1\). Moreover, \(\abs{f(z)} = \prod \abs{w_k-z}\) is exactly the product of distances from \(z\) to the \(w_k\). Moreover, since \(\abs{f(0)} = \prod\abs{w_k} = 1\), we must have \(M>1\).
Now note that \(f(w_k) = 0\) and \(f\) is continuous in \(\DD\). So \(\abs{f(z)} \in [0, M] \subseteq \RR\) where \(M>1\), so by the intermediate value theorem, \(\abs{f(z)} = 1\) for some \(z\).
Solution: Part 4
Suppose \(|\Re f(z)|\le M\) on \(\CC\) and define \(g(z)=e^{f(z)}\). Then \(g\) is entire and \[ |g(z)|=e^{\Re f(z)}\le e^M. \] By Liouville’s theorem, \(g\) is constant. Since \[ g'(z)=f'(z)e^{f(z)} \] and \(e^{f(z)}\neq0\), we get \(f'\equiv0\). Thus \(f\) is constant.