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.
Part (b) requires the limit \(\abs z\to1\). With \(\abs z\to0\) the conclusion fails: \(f(z)=z\) is bounded on \(\DD\) and tends to \(0\) uniformly as \(\abs z\to0\).
Solution
1. (a) \(f^{(n)}(0) = 0\) for all \(n > k\).
By the Cauchy estimates on \(\abs{z} = R\), \(\abs{f^{(n)}(0)} \leq \frac{n!}{R^n}\qty(A R^k + B) = n!\qty(A R^{k-n} + B R^{-n})\) for every \(R > 0\). For \(n > k\) the right-hand side tends to \(0\) as \(R \to \infty\).
2. (a) \(f\) is a polynomial of degree \(\leq k\).
The Taylor series of the entire function \(f\) about \(0\) converges on \(\CC\): \(f(z) = \sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!} z^n\). By step 1, only the terms with \(n \leq k\) are nonzero.
3. (b) Let \(S=\{z\in\DD:\theta<\arg z<\phi\}\) and choose an integer \(N\) with \(2\pi/N<\phi-\theta\). Then \(g(z)\coloneqq\prod_{j=0}^{N-1} f\qty(e^{2\pi i j/N}z)\) is identically zero on \(\DD\).
Put \(M=\sup_{\DD}\abs f<\infty\). For every \(z\neq0\), some rotation \(e^{2\pi ij/N}z\) has argument in \((\theta,\phi)\), because consecutive rotations differ in argument by \(2\pi/N<\phi-\theta\). Fix \(\varepsilon>0\). By hypothesis there is \(\rho<1\) with \(\abs{f(z)}<\varepsilon\) for \(z\in S\) and \(\rho<\abs z<1\). Hence \(\abs{g(z)}\le\varepsilon M^{N-1}\) for \(\rho<\abs z<1\). For each \(r\in(\rho,1)\), the maximum modulus principle on \(\abs z\le r\) extends this bound to \(\abs z\le r\), so it holds on all of \(\DD\). Since \(\varepsilon\) is arbitrary, \(g\equiv0\).
4. (b) \(f \equiv 0\).
The holomorphic functions on the connected disc \(\DD\) form an integral domain, so by step 3 one factor \(f(e^{2\pi ij/N}z)\) is identically zero on \(\DD\). Rotation is a bijection of \(\DD\), hence \(f\equiv0\).
5. (c) There exists \(z \in S^1\) with \(\prod_{j=1}^n \abs{z - w_j} \geq 1\).
The polynomial \(P(z) = \prod_{j=1}^n (z - w_j)\) is holomorphic, and \(\abs{P(0)} = \prod_j \abs{w_j} = 1\). By the maximum modulus principle on \(\overline\DD\), \(\max_{\abs z = 1}\abs{P(z)} \geq 1\).
6. (c) There exists \(w \in S^1\) with \(\prod_j \abs{w - w_j} = 1\).
5. By the intermediate value theorem along an arc of \(S^1\) from \(w_1\) to \(z\), \(\varphi\) takes the value \(1\).
7. (d) If \(\Re f\) is bounded, then \(f\) is constant.
Suppose \(\Re f \leq M\). Then \(g(z) = e^{f(z)}\) is entire and \(\abs{g(z)} = e^{\Re f(z)} \leq e^M\), so \(g\) is constant by Liouville’s theorem. Hence \(0=g' = f'e^{f}\), and \(e^f\neq0\), so \(f'\equiv0\) and \(f\) is constant. (If instead \(\Re f\) is bounded below, apply this to \(-f\).)
8. Q.E.D.
Step 2 proves (a), step 4 proves (b), steps 5 and 6 prove (c), and step 7 proves (d).