TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-16

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A uniform limit of continuous complex-valued functions is continuous

Statement

Let $(X, d)$ be a metric space. If continuous functions $f_{n} : X \to C$ converge uniformly to $f : X \to C$ in the sense of Uniform convergence and the uniformly Cauchy condition for complex-valued functions, with the componentwise dictionary, then $f$ is continuous.

Facts & Assumptions

Given: Continuous $f_{n} : X \to C$ with $f_{n} \to f$ uniformly.

[L1]

A uniform limit of continuous real-valued functions is continuous (The uniform limit of continuous real-valued functions on a metric space is continuous).

[L2]

For $m \geq 1$ , a map into $R^{m}$ is continuous if and only if each component is continuous (A vector-valued function has a limit, or is continuous, if and only if each of its components does; with the algebra of continuous vector-valued functions).

Proof

technique · direct

1.1L2

Write $f_{n} = u_{n} + i v_{n}$ and $f = u + i v$ . The componentwise dictionary in Uniform convergence and the uniformly Cauchy condition for complex-valued functions, with the componentwise dictionary shows that $u_{n} \to u$ and $v_{n} \to v$ uniformly; [L2] shows each $u_{n}, v_{n}$ is continuous.

2.1step 1.1L1

By [L1], both $u$ and $v$ are continuous, including when $X$ is empty.

3.1step 2.1L2∎

The componentwise continuity criterion [L2] now makes $f = u + i v$ continuous.

Depends on

Used by

Dependency tree · two levels

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Sources

E. Stein and R. Shakarchi, Complex Analysis, Ch. 1 (standard reference, not scraped)