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Nevanlinna’s First Main Theorem with exact centre constant

Statement

Write Mrh=(2π)−1∫02πh(reit) dt for a circular mean whenever it exists.

Let f be a nonconstant meromorphic function on C and let a∈C^. For every r>0, m(r,a;f)+N(r,a;f)=T(r,f)+C(f,a). For a=∞, set C(f,∞)=0. For finite a, write the first nonzero Laurent term at the centre as f(z)−a=cazka+higher Laurent terms,ca≠0, and set C(f,a)=12log⁡(1+∣a∣2)−log⁡∣ca∣. In particular, the difference m(r,a;f)+N(r,a;f)−T(r,f) is Of,a(1) as r→∞.

Facts & Assumptions

Given: A nonconstant meromorphic f on C and a target a∈C^.

[F1]

The normalized chordal distance is δ(w,a)=∣w−a∣/(1+∣w∣21+∣a∣2) for finite w,a, and δ(w,∞)=1/1+∣w∣2 (Counting, chordal proximity and characteristic).

[F2]

m(r,a;f) is the circular mean of log⁡(1/δ(f,a)), and T(r,f)=m(r,∞;f)+N(r,∞;f) (Counting, chordal proximity and characteristic).

[F3]

For finite a, the centre-divisor Jensen identity is Mrlog⁡∣f−a∣=log⁡∣ca∣+N(r,a;f)−N(r,∞;f), with divisor-circle means interpreted by continuous radial limits (Meromorphic Jensen identity with a zero or pole at the centre).

[F4]

The divisor counts are finite on bounded discs and N(r,a;f) and m(r,a;f) are finite and continuous for every r>0, including divisor radii (Well-definedness and radius conventions for Nevanlinna quantities).

Proof

technique · use the pointwise chordal identity for finite targets, average it on a regular circle, and substitute the centre-divisor Jensen identity; handle infinity directly from the definition
1.1F1given

Fix finite a and a regular radius r whose circle contains no pole and no a-point. From [F1], at every point of that circle, log⁡(1/δ(f,a))=12log⁡(1+∣f∣2)+12log⁡(1+∣a∣2)−log⁡∣f−a∣.

2.1F2step 1.1

Averaging the identity in step 1.1 and using [F2] gives m(r,a;f)−m(r,∞;f)=12log⁡(1+∣a∣2)−Mrlog⁡∣f−a∣.

3.1F2F3step 2.1algebra

Substitute [F3] into step 2.1 to obtain m(r,a;f)−m(r,∞;f)=12log⁡(1+∣a∣2)−log⁡∣ca∣−N(r,a;f)+N(r,∞;f). Rearranging and using [F2] yields the claimed formula for finite a at each regular radius.

4.1F2F4step 3.1

Regular radii are dense because [F4] makes the divisor sets finite on bounded discs. The finite-target quantities in the formula are continuous by [F4], so the equality from step 3.1 on that dense set extends to every r>0. For a=∞, C(f,∞)=0 and the asserted identity is exactly the defining equality in [F2].

5.1step 4.1algebra∎

For each fixed f,a, the constant C(f,a) in step 4.1 is finite and independent of r; therefore the difference in the statement is bounded as r→∞.

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