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CorollaryStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck 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 M0, then γf(z)dzML(γ).

Facts & Assumptions

Given: A continuous f with fM 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 fdαMVar(α) under fM (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
1.1

Apply [L3] to fγ and the nondecreasing sγ; by [L2], γfdzML(γ).

L2L3
2.1

Combine step 1.1 with [L1].

step 1.1L1
3.1

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.

step 2.1L4

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 87 results over 21 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