Alphabeta Math
CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-08-21
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Uniform convergence on the closed unit disc does not give a holomorphic extension to a larger disc

Statement refuted

Refuted claim: If a complex power series converges uniformly on a closed disc, its sum extends holomorphically to some larger centred disc.

The series

F(z):=∑n≥1znn2

converges uniformly on ∣z∣≤1, but its sum on ∣z∣<1 has no holomorphic extension to any disc centred at 0 with radius greater than 1.

Facts & Assumptions

Given: The complex power series defining F, with its partial sums and convergence interpreted as in Complex series, absolute convergence, complex power series, and radius of convergence.

[L1]

If ∣fn(x)∣≤Mn and the real series ∑Mn converges, then the complex function series ∑fn converges absolutely pointwise and uniformly (Weierstrass M-test for complex-valued function series).

[L2]

The series ∑n≥11/n2 converges, while the harmonic series ∑n≥11/n diverges (For rational p>0, ∑1/kp converges iff p>1).

[L3]

Inside the radius of convergence, a complex power series may be differentiated term by term: (∑n≥0cnzn)′=∑n≥1ncnzn−1 (Inside its disc of convergence a complex power series is holomorphic and may be differentiated term by term).

[L4]

A continuous real-valued function on a nonempty compact metric space is bounded and attains a maximum (A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value).

[L5]

A complex differentiable function is continuous (Complex differentiability at a point implies continuity there).

[L8]

Complex modulus is multiplicative and satisfies the triangle inequality, hence ∣∣u∣−∣v∣∣≤∣u−v∣ (Conjugation is an involutive real-field automorphism, zz‾=∣z∣2, and modulus is definite, multiplicative, and subadditive).

[L9]

Every holomorphic function has complex derivatives of every order locally (All higher complex derivatives exist and satisfy Cauchy's integral formula on an interior circle).

Counterexample

technique · contradiction
1.1L1L2

On ∣z∣≤1, one has ∣zn/n2∣≤1/n2, so [L1] and the convergent series in [L2] give absolute pointwise and uniform convergence of the displayed series.

1.2L3

For real 0<r<1, [L3] gives F′(r)=∑n≥1rn−1/n.

1.3L4L5L7L8L9assume-contra

Suppose, for contradiction, that a function G holomorphic on D(0,S) for some S>1 agrees with F on ∣z∣<1. By [L9], G′ is holomorphic near [0,1], hence continuous by [L5]; [L8] makes ∣G′∣ continuous, [L7] makes [0,1] compact, and [L4] bounds ∣G′∣ there.

2.1step 1.2L2L5L6choose

Given B>0, divergence in [L2] supplies N≥2 with ∑n=1N1/n>2B; by [L6] and [L5], the finitely many monomials rn−1 are continuous at r=1, so choose 0<r<1 with rn−1>1/2 for every 1≤n≤N, and step 1.2 then gives F′(r)>12∑n=1N1/n>B.

3.1step 2.1step 1.3discharge-contradiction∎

Step 2.1 makes ∣F′(r)∣=∣G′(r)∣ exceed every proposed bound for points 0<r<1, contradicting step 1.3; no such extension exists, and the refuted claim is false.

Depends on

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Sources