Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adapted
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The fundamental inequality: the modulus of the integral is at most the absolute line integral for rectifiable contours

Statement

For a continuous f on the trace of a rectifiable contour γ, ∣∫γf(z) dz∣≤∫γ∣f(z)∣ ∣dz∣.

Facts & Assumptions

Given: A rectifiable contour γ and a continuous integrand f.

[L2]

Complex modulus satisfies ∣z+w∣≤∣z∣+∣w∣ and ∣zw∣=∣z∣∣w∣ (Conjugation is an involutive real-field automorphism, zz‾=∣z∣2, and modulus is definite, multiplicative, and subadditive).

[L3]

Apply Every endpoint chord is no longer than the arc: ∥γ(b)−γ(a)∥2≤L(γ) to Γ=Φ∘γ. Chord and subpath lengths equal the complex lengths by Complex path length equals coordinate length and piecewise speed integral. With sγ=sΓ as in The absolute line integral over a rectifiable path using its arc-length function, the chord bound is therefore ∣γ(v)−γ(u)∣≤sγ(v)−sγ(u).

Proof

technique · direct
1.1L2

For a tagged partition P, the complex polygonal sum SP=∑f(γ(ξj))(γ(tj+1)−γ(tj)) satisfies ∣SP∣≤∑∣f(γ(ξj))∣ ∣γ(tj+1)−γ(tj)∣ by [L2].

2.1step 1.1L1L3

By [L3], each chord in step 1.1 is at most sγ(tj+1)−sγ(tj). The resulting tagged arc-length sums converge to the existing absolute line integral from [L1].

3.1step 1.1step 2.1L1∎

Letting the mesh tend to 0, [L1] identifies the limit of SP with the complex line integral and step 2.1 identifies the majorant limit with the absolute integral, proving the inequality with sharp constant 1. The same argument includes a constant contour: every chord and arc-length increment is 0, so both sides vanish.

Depends on

Dependency tree · 2 levels

Sources