TheoremStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck 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 $f_{n}$ on its trace converge uniformly to a continuous $f$ , then $\int_{γ} f_{n} (z) d z ⟶ \int_{γ} f (z) d z .$

Facts & Assumptions

Given: A rectifiable contour $γ$ and uniformly convergent continuous functions $f_{n} \to f$ 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 $∣ g ∣ \leq M$ on the trace, then $∣ \int_{γ} g d z ∣ \leq M L (γ)$ (ML estimate: a contour integral is bounded by a supremum bound times path length).

Proof

technique · cases

1.1

If $L (γ) = 0$ , [L2] gives $\int_{γ} (f_{n} - f) d z = 0$ for every $n$ .

assume-case zeroL2

1.2

If $L (γ) > 0$ , given $ε > 0$ choose $N$ such that $∣ f_{n} - f ∣ < ε / L (γ)$ on the trace for $n \geq N$ .

assume-case positivechoose

2.1

By [L1] all integrals exist, and [L2] applied to $f_{n} - f$ gives $∣ \int_{γ} f_{n} d z - \int_{γ} f d z ∣ < ε$ for $n \geq N$ .

step 1.2L1L2

3.1

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

step 1.1step 2.1cases-exhaustive∎

Depends on

Used by

The unit-circle integral of exp(z)/z is 2 pi i by uniform termwise integration Example

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 41 results over 10 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

L. Ahlfors, Complex Analysis, 3rd ed., Ch. 4 (standard reference, not scraped)