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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 on the trace of a rectifiable contour ,
Facts & Assumptions
Given: A rectifiable contour and a continuous integrand .
Both the complex and absolute line integrals exist (Continuous integrands have complex and absolute line integrals along every rectifiable path).
Complex modulus satisfies and (Conjugation is an involutive real-field automorphism, , and modulus is definite, multiplicative, and subadditive).
Apply Every endpoint chord is no longer than the arc: to . Chord and subpath lengths equal the complex lengths by Complex path length equals coordinate length and piecewise speed integral. With as in The absolute line integral over a rectifiable path using its arc-length function, the chord bound is therefore .
Proof
For a tagged partition , the complex polygonal sum satisfies by [L2].
By [L3], each chord in step 1.1 is at most . The resulting tagged arc-length sums converge to the existing absolute line integral from [L1].
Letting the mesh tend to , [L1] identifies the limit of with the complex line integral and step 2.1 identifies the majorant limit with the absolute integral, proving the inequality with sharp constant . The same argument includes a constant contour: every chord and arc-length increment is , so both sides vanish.
Depends on
Complex path length equals coordinate length and piecewise speed integral
The absolute line integral over a rectifiable path using its arc-length function
Continuous integrands have complex and absolute line integrals along every rectifiable path
Conjugation is an involutive real-field automorphism, , and modulus is definite, multiplicative, and subadditive
Every endpoint chord is no longer than the arc: