CounterexampleConstruction: AI-generatedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-21

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$z^{n}$ tends locally uniformly to zero on the unit disc but not uniformly on the closed disc

Statement refuted

Refuted claim: Local uniform convergence on the open unit disc forces uniform convergence on the closed unit disc.

For $f_{n} (z) = z^{n}$ , the sequence $(f_{n})$ converges locally uniformly to $0$ on $D = {z : ∣ z ∣ < 1}$ , but it does not converge to $0$ uniformly, or even pointwise, on $\overline{D}$ .

Facts & Assumptions

Given: The functions $f_{n} (z) = z^{n}$ and the local-uniform convention of Locally uniform convergence on an open subset of the complex plane is compact convergence.

[L1]

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).

[L2]

If a real $q$ satisfies $∣ q ∣ < 1$ , then $q^{n} \to 0$ (For $∣ r ∣ < 1$ the sequence $r^{k}$ is null, and for $∣ r ∣ > 1$ the sequence $∣ r ∣^{k}$ diverges to $+ \infty$ ).

[L3]

Uniform convergence to zero requires that for every $ε > 0$ , all sufficiently late functions have modulus below $ε$ at every point of the domain (Uniform convergence and the uniformly Cauchy condition for complex-valued functions, with the componentwise dictionary).

[L4]

Complex modulus satisfies $∣ ∣ z ∣ - ∣ w ∣ ∣ \leq ∣ z - w ∣$ , so $z \mapsto ∣ z ∣$ is continuous (Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive).

Counterexample

technique · direct

1.1

Let $K \subseteq D$ be compact. If $K = \emptyset$ , uniform convergence on $K$ is vacuous. Otherwise [L4] and [L1] give $q = \max_{z \in K} ∣ z ∣$ , and $q < 1$ because the maximum is attained at a point of $K \subseteq D$ ; then $∣ f_{n} (z) ∣ = ∣ z ∣^{n} \leq q^{n} \to 0$ by [L2], uniformly for $z \in K$ .

L1L2L4

1.2

At the boundary point $z = 1$ , one has $f_{n} (1) = 1$ for every natural $n$ , including $n = 0$ , so the sequence does not converge pointwise to zero there and fails the uniform condition [L3] on $\overline{D}$ .

L3algebra

2.1

Step 1.1 proves local uniform convergence on the open disc, while step 1.2 proves failure on its closure, so the claimed implication is false.

step 1.1step 1.2∎

Depends on

Used by

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Sources

Lars Ahlfors, Complex Analysis, 3rd ed., Ch. 5 §1.1 (standard reference, not scraped)
E. Stein and R. Shakarchi, Complex Analysis, Ch. 2 §5.2 (standard reference, not scraped)
Matthias Weber, Complex Analysis, §2.4 (standard reference, not scraped)

znzn tends locally uniformly to zero on the unit disc but not uniformly on the closed disc