Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passaudited 2026-08-16
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A uniformly convergent sequence of continuous integrands on a fixed contour permits passage of the limit through the complex line integral

Statement

Let γ be a fixed rectifiable contour. If continuous functions fn on its trace converge uniformly to a continuous f, then ∫γfn(z) dz⟶∫γf(z) dz.

Facts & Assumptions

Given: A rectifiable contour γ and uniformly convergent continuous functions fn→f on its trace.

[L1]

Continuous integrands have complex line integrals along every rectifiable path (Continuous integrands have complex and absolute line integrals along every rectifiable path).

[L2]

If ∣g∣≤M on the trace, then ∣∫γg dz∣≤ML(γ) (ML estimate: a contour integral is bounded by a supremum bound times path length).

Proof

technique · cases
1.1assume-case zeroL2

If L(γ)=0, [L2] gives ∫γ(fn−f) dz=0 for every n.

1.2assume-case positivechoose

If L(γ)>0, given ε>0 choose N such that ∣fn−f∣<ε/L(γ) on the trace for n≥N.

2.1step 1.2L1L2

By [L1] all integrals exist, and [L2] applied to fn−f gives ∣∫γfn dz−∫γf dz∣<ε for n≥N.

3.1step 1.1step 2.1cases-exhaustive∎

The two length cases are exhaustive and prove convergence without ever dividing by zero.

Depends on

Used by

Dependency tree · two levels

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Sources