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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).
Every chord of a rectifiable path is at most the length of the corresponding subpath (Every endpoint chord is no longer than the arc: ).
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
- Continuous integrands have complex and absolute line integrals along every rectifiable path
- Conjugation is an involutive real-field automorphism, $z\overline z=|z|^2$, and modulus is definite, multiplicative, and subadditive
- Every endpoint chord is no longer than the arc: $\lVert\gamma(b)-\gamma(a)\rVert_2\le L(\gamma)$
Used by
Dependency tree · next 3 levels
Direct dependencies and their dependencies through the next three levels: 52 results over 14 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)