- Here are some examples of analytic functions on the unit disc that cannot be extended analytically past the unit circle. The following definition is needed. Let \(f\) be a function defined in the unit disc D, with boundary circle C. A point w on C is said to be regular for \(f\) if there is an open neighborhood U of w and an analytic function \(g\) on \(U _ { i }\) , so that \(f = g\) on \(\mathbb { D } \cap U\) . A function f defined on D cannot be continued analytically past the unit circle if no point of C is regular for \(f .\)
- Let
\[ f (z) = \sum_ {n = 0} ^ {\infty} z ^ {2 ^ {n}} \quad \text { for } | z | < 1. \]
Notice that the radius of convergence of the above series is 1. Show that f cannot be continued analytically past the unit disc. [Hint: Suppose \(\theta = { 2 \pi p } / { 2 ^ { k } }\) , where \(p\) and k are positive integers. Let \(z = r e ^ { i \theta }\) ; then \(| f ( r e ^ { i \theta } ) | \longrightarrow \infty \mathrm { \ a s \ } r \longrightarrow 1 . ]\)
- \(^\ast\) Fix \(0 < \alpha < \infty\) . Show that the analytic function f defined by
\[ f (z) = \sum_ {n = 0} ^ {\infty} 2 ^ {- n \alpha} z ^ {2 ^ {n}} \quad \text { for } | z | < 1 \]
extends continuously to the unit circle, but cannot be analytically continued past the unit circle. [Hint: There is a nowhere diferentiable function lurking in the background. See Chapter 4 in Book I.]
Solution
Throughout, a dyadic point is \(e^{i\theta}\) with \(\theta = 2\pi p / 2^k\), \(p,k\) positive integers; dyadic points are dense in \(C\). The set of regular points of a function on \(\mathbb D\) is open in \(C\), so if it contains no dyadic point, it is empty.
1. (1a) \(f(z)=\sum_{n\ge0} z^{2^n}\) cannot be continued analytically past \(C\).
Let \(\theta = 2\pi p / 2^k\). For \(n \ge k\), \(2^n\theta\) is a multiple of \(2\pi\), so \((re^{i\theta})^{2^n} = r^{2^n}\) and \[f(re^{i\theta}) = \sum_{n=0}^{k-1} (re^{i\theta})^{2^n} + \sum_{n=k}^{\infty} r^{2^n}.\] Each term of the second sum tends to \(1\) as \(r\to1^-\), and the terms are nonnegative, so the second sum tends to \(\infty\); hence \(\abs{f(re^{i\theta})}\to\infty\). A function regular at a boundary point is bounded near it, so no dyadic point is regular, and no point of \(C\) is regular.
2. (1b) \(f(z)=\sum_{n\ge0} 2^{-n\alpha} z^{2^n}\) extends continuously to \(\overline{\mathbb D}\) and cannot be continued analytically past \(C\).
Since \(\sum 2^{-n\alpha} < \infty\), the Weierstrass \(M\)-test shows that the series converges uniformly on \(\overline{\mathbb D}\), so \(f\) extends continuously. Suppose a dyadic point \(e^{i\theta}\), \(\theta = 2\pi p/2^k\), were regular. Then \(f\) agrees near \(e^{i\theta}\) with a function analytic on a neighborhood of \(e^{i\theta}\), so for every \(m\) the derivative \(\frac{d^m}{dr^m}f(re^{i\theta})=e^{im\theta}f^{(m)}(re^{i\theta})\) stays bounded as \(r\to1^-\). But \(f(re^{i\theta})=P(r)+\sum_{n\ge k}2^{-n\alpha}r^{2^n}\) with \(P\) a polynomial, and for an integer \(m\ge\alpha\), \[\frac{d^m}{dr^m}\sum_{n\ge k}2^{-n\alpha}r^{2^n}\ge\sum_{n\ge n_0}2^{-n\alpha}\bigl(2^{n-1}\bigr)^m r^{2^n-m},\] where \(n_0\ge k\) is chosen with \(2^{n_0}\ge2m\), so that \(2^n(2^n-1)\cdots(2^n-m+1)\ge(2^{n-1})^m\). As \(r\to1^-\) the right side tends to \(2^{-m}\sum_{n\ge n_0}2^{n(m-\alpha)}=\infty\). Hence no dyadic point is regular, and no point of \(C\) is regular.