A Power Series with a Natural Boundary
The lacunary series z + z^2 + z^4 + ... blows up at every dyadic root of unity, so no arc of the unit circle can be crossed.
1 Statement
Let \(\mathbb {D} = \set {z : \abs {z} < 1}\) and \(\mathbb {T} = \set {z : \abs {z} = 1}\), and
Theorem 1.1 (Natural boundary). The series \(f\) has radius of convergence \(1\), and every point of \(\mathbb {T}\) is a singular point of \(f\). So \(f\) cannot be continued analytically across any arc of the unit circle.
2 Setting
Definition 2.1 (Regular and singular points). A point \(\zeta \in \mathbb {T}\) is regular for \(f\) if there are an open disc \(D\) centred at \(\zeta \) and a function \(g\) analytic on \(D\) with \(g = f\) on \(D \cap \mathbb {D}\). Otherwise \(\zeta \) is singular. If every point of \(\mathbb {T}\) is singular, \(\mathbb {T}\) is a natural boundary for \(f\).
This is the standard notion [1, ch. 16]. A power series with radius of convergence \(1\) always has at least one singular point on \(\mathbb {T}\); for instance \(1/(1 - z)\) has exactly one. The theorem says that \(f\) is as bad as possible.
Definition 2.2 (Dyadic roots of unity). \(\omega \in \mathbb {T}\) is a dyadic root of unity if \(\omega ^{2^m} = 1\) for some \(m \geq 0\), that is, \(\omega = e^{2\pi i j / 2^m}\) for some integers \(j\) and \(m \geq 0\).
3 Idea of the proof
Along the radius towards a dyadic root of unity \(\omega \), all but finitely many terms \((r\omega )^{2^n}\) equal \(r^{2^n}\), which is positive. So \(f\) blows up along that radius, and \(\omega \) cannot be regular. Such roots are dense in the circle, and the regular points form an open set, so there are no regular points at all.
4 Proof
Proof. For \(\abs {z} < 1\) the series is dominated by \(\sum _k \abs {z}^k < \infty \). At \(z = 1\) its terms do not tend to \(0\), so it diverges, and the radius is exactly \(1\).
Lemma 4.2 (Blow-up at dyadic roots). If \(\omega ^{2^m} = 1\), then \(\abs {f(r\omega )} \to \infty \) as \(r \to 1^-\).
Proof. For \(n \geq m\), \(\omega ^{2^n} = \bigl (\omega ^{2^m}\bigr )^{2^{n-m}} = 1\). Hence, for \(0 < r < 1\),
The first sum has modulus at most \(m\). For any \(N > m\), the second sum is at least \(\sum _{n=m}^{N} r^{2^n}\), which tends to \(N - m + 1\) as \(r \to 1^-\). So
for every \(N\), and the limit is infinite.
Proof. Given \(e^{i\theta }\) and \(m \geq 0\), choose an integer \(j\) with \(\abs {\theta - 2\pi j / 2^m} \leq \pi / 2^m\). Since \(\abs {e^{is} - e^{it}} \leq \abs {s - t}\), the root \(e^{2\pi i j / 2^m}\) lies within \(\pi / 2^m\) of \(e^{i\theta }\).
Lemma 4.4 (Regular points form an open set). If \(\zeta \) is regular, then so is every point of \(\mathbb {T}\) close enough to \(\zeta \).
Proof. Let \(D\) and \(g\) be as in Definition 2.1. For \(\zeta ' \in \mathbb {T} \cap D\), choose an open disc \(D'\) centred at \(\zeta '\) with \(D' \subseteq D\). Then \(g\) is analytic on \(D'\) and equals \(f\) on \(D' \cap \mathbb {D}\).
Proof of Theorem 1.1. A dyadic root \(\omega \) is singular. If it were regular, with \(g\) analytic on a disc \(D\) about \(\omega \), then \(g\) would be bounded near \(\omega \). But \(f(r\omega ) = g(r\omega )\) for \(r\) close to \(1\), which is unbounded by Lemma 4.2.
By Lemma 4.4, the set of regular points is an open subset of \(\mathbb {T}\), and it contains no dyadic root. By Lemma 4.3, every non-empty open subset of \(\mathbb {T}\) contains a dyadic root. Hence the set of regular points is empty.
5 Remarks
Remark 5.1. The functional equation \(f(z) = z + f(z^2)\) explains the result. Singularities of \(f(z^2)\) include every square root of a singularity of \(f\), so once \(1\) is singular, so are \(-1\), then \(\pm i\), and so on through all dyadic roots of unity.
Remark 5.2. The real reason is the size of the gaps between exponents. By Hadamard’s gap theorem, any power series \(\sum c_k z^{n_k}\) with radius of convergence \(1\) and \(n_{k+1} / n_k \geq \lambda > 1\) for all \(k\) has \(\mathbb {T}\) as a natural boundary. Our proof uses the special structure \(n_k = 2^k\) to avoid that theorem.