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A holomorphic function equals its average on every circle inside a larger concentric holomorphy disc

Statement

Let f be holomorphic on D(a,R) and let 0<r<R. Then

f(a)=12π∫02πf(a+rexp⁡(iθ)) dθ.

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 f holomorphic on D(a,R) and a radius 0<r<R.

[L1]

Under these hypotheses, Cauchy's circle formula gives f(z)=(2πi)−1∫γf(ζ)/(ζ−z) dζ for ∣z−a∣<r, where γ(θ)=a+rexp⁡(iθ) is positively oriented (Cauchy's integral formula on a circle compactly contained in a disc of holomorphy).

[L2]

For a piecewise-C1 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).

[L3]

Every holomorphic function is continuous (Complex differentiability at a point implies continuity there).

Proof

technique · direct
1.1L1

Apply [L1] at the centre z=a to obtain f(a)=(2πi)−1∫γf(ζ)/(ζ−a) dζ.

2.1step 1.1L2L3algebra

By [L3], f is continuous; on the circle the denominator ζ−a is nonzero because its modulus is r>0, so elementary complex division makes ζ↦f(ζ)/(ζ−a) continuous on the trace. With ζ=γ(θ)=a+rexp⁡(iθ), [L2] gives dζ=irexp⁡(iθ) dθ while ζ−a=rexp⁡(iθ), so step 1.1 becomes f(a)=(2πi)−1∫02πf(a+rexp⁡(iθ)) i dθ.

3.1step 2.1algebra∎

Cancelling the nonzero factor i in step 2.1 yields the stated circular average; the calculation requires r>0 and also covers every constant or zero function.

Depends on

Used by

Dependency tree · two levels

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Sources