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A holomorphic function equals its average on every circle inside a larger concentric holomorphy disc
Statement
Let be holomorphic on and let . Then
Thus a holomorphic function equals its average on every positive-radius circle lying with a larger concentric disc inside its holomorphy domain.
Facts & Assumptions
Given: A function holomorphic on and a radius .
Under these hypotheses, Cauchy's circle formula gives for , where is positively oriented (Cauchy's integral formula on a circle compactly contained in a disc of holomorphy).
For a piecewise- contour and an integrand continuous on its trace, the complex contour integral equals the parameter integral of the pulled-back integrand multiplied by the contour derivative (For piecewise-C1 contours the Riemann–Stieltjes integral agrees with the parametric complex integral and the published real line integrals).
Every holomorphic function is continuous (Complex differentiability at a point implies continuity there).
Proof
Apply [L1] at the centre to obtain .
By [L3], is continuous; on the circle the denominator is nonzero because its modulus is , so elementary complex division makes continuous on the trace. With , [L2] gives while , so step 1.1 becomes .
Cancelling the nonzero factor in step 2.1 yields the stated circular average; the calculation requires and also covers every constant or zero function.
Depends on
Used by
Nothing in the library uses this result yet.
Dependency tree · two levels
18 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.
Sources
- Matthias Weber, Complex Analysis, Corollary 2.2.1 (standard reference, not scraped)