Alphabeta Math
CorollaryStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-16
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ML estimate: a contour integral is bounded by a supremum bound times path length

Statement

If ∣f(z)∣≤M on the trace of a rectifiable contour γ, with M≥0, then ∣∫γf(z) dz∣≤ML(γ).

Facts & Assumptions

Given: A continuous f with ∣f∣≤M on a rectifiable contour γ.

[L1]

The fundamental inequality bounds the complex integral by the absolute line integral (The fundamental inequality: the modulus of the integral is at most the absolute line integral for rectifiable contours).

[L2]

The arc-length function satisfies sγ(b)−sγ(a)=L(γ) (The arc-length function sγ(t)=L(γ∣[a,t]) of a rectifiable path).

[L3]

The Stieltjes integral bound gives ∣∫f dα∣≤MVar⁡(α) under ∣f∣≤M (The total-variation bound for a Riemann–Stieltjes integral).

[L4]

For piecewise-C1 paths, published scalar and vector line integrals obey the bound ML(γ) (Line-integral estimates by arc length and the supremum of the field).

Proof

technique · direct
1.1L2L3

Apply [L3] to ∣f∘γ∣ and the nondecreasing sγ; by [L2], ∫γ∣f∣ ∣dz∣≤ML(γ).

2.1step 1.1L1

Combine step 1.1 with [L1].

3.1step 2.1L4∎

This agrees with the published piecewise-C1 estimate [L4] on its exact domain and extends it to rectifiable contours. The cases M=0 and L(γ)=0 give zero directly.

Depends on

Dependency tree · 2 levels

Sources