- 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 \(R > 0\) , and for 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 \(\theta < \arg z < \varphi { \mathrm { ~ a s ~ } } | z | \to 1\) , then \(f = 0\)
Let \(w _ { 1 } , \ldots , w _ { n }\) be points on the unit circle in the complex plane. Prove that there exists a point z on the unit circle such that the product of the distances from z to the points \(w_j\), \(1 \leq j \leq n _ { \mathrm { : } }\) , is at least 1. Conclude that there exists a point w on the unit circle such that the product of the distances from w to the points \(w_j\) , \(1 \leq j \leq n\) , is exactly equal to 1.
Show that if the real part of an entire function f is bounded, then f is constant.
Solution
Write \(f(z)=\sum_{n\ge0}a_nz^n\). Cauchy’s estimate on \(|z|=R\) gives \[ |a_n|\le \frac{AR^k+B}{R^n}. \] For \(n>k\), letting \(R\to\infty\) yields \(a_n=0\). Hence \(f\) is a polynomial of degree at most \(k\).
By hypothesis there is an open boundary arc \[ I=\{e^{it}:\theta<t<\varphi\} \] on which \(f(z)\to0\) uniformly as \(|z|\to1\) inside the sector. Thus defining \(f=0\) on \(I\) gives a continuous extension to that arc. Fix a smaller open subarc \(I'\Subset I\). After rotating coordinates, a neighborhood of each point of \(I'\) is conformally equivalent to a half-disc whose diameter lies on the real axis, and the boundary values of \(f\) there are identically \(0\), hence real. By the Schwarz reflection principle, \(f\) extends holomorphically across \(I'\), with the extension vanishing on \(I'\). The identity theorem then forces the extension, and hence \(f\) on \(\mathbb D\), to vanish identically.
Let \[ P(z)=\prod_{j=1}^n(z-w_j). \] Since \(P\) is monic, \(P(0)=(-1)^n\prod_jw_j\), so \(|P(0)|=1\). By the maximum modulus principle applied on the closed unit disc, \[ 1=|P(0)|\le \max_{|z|=1}|P(z)|. \] Hence some \(z\) on the unit circle satisfies \[ \prod_{j=1}^n|z-w_j|=|P(z)|\ge1. \] On the other hand, \(P(w_1)=0\). The continuous function \(|P|\) on the connected unit circle therefore takes every value between \(0\) and its maximum, in particular the value \(1\). Thus there is \(w\) on the unit circle with \[ \prod_{j=1}^n|w-w_j|=1. \]
Suppose \(|\Re f|\le M\). Then \[ g(z)=e^{f(z)} \] is entire and satisfies \(|g(z)|=e^{\Re f(z)}\le e^M\). By Liouville, \(g\) is constant. Differentiating gives \[ 0=g'(z)=f'(z)e^{f(z)}, \] so \(f'\equiv0\) and \(f\) is constant.