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CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-opus-5[1m])audited 2026-08-24
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A nonconstant Blaschke factor has constant boundary modulus

Statement refuted

A function holomorphic on the unit disc and continuous on its closure must be constant whenever its modulus is constant on the unit circle.

Facts & Assumptions

Given: A parameter aC with a<1 and the Blaschke factor Ba(z):=za1az. Complex conjugation and modulus obey zw=zw and z2=zz (Conjugation is an involutive real-field automorphism, zz=z2, and modulus is definite, multiplicative, and subadditive), and C is a field (C=R[x]/(x2+1) is a field, every element is uniquely a+bi, and every nonzero element has inverse (abi)/(a2+b2)).

[L1]

If a holomorphic function has constant modulus on the boundary of a bounded domain, then it is constant or has a zero in the domain (Constant boundary modulus forces an interior zero or constancy).

[L2]

A quotient of holomorphic functions is holomorphic wherever its denominator is nonzero (Linearity, product, reciprocal, and quotient rules for complex derivatives).

Counterexample

technique · direct
1.1

If a=0, the denominator is 1. If a0, choose R with 1<R<1/a; for z<R one has az<1, so 1az0. Thus [L2] makes Ba holomorphic on a neighbourhood of the closed unit disc, and hence continuous there.

L2givenalgebra
2.1

When z=1, direct expansion gives za2=1zaaz+a2=1az2. The denominator is nonzero by step 1.1, so Ba(z)=1 on the entire unit circle.

step 1.1givenalgebra
3.1

Since Ba(a)=0 and its boundary modulus is 1, the function is nonconstant. It therefore realizes the zero alternative in [L1] and refutes the proposed implication, including the case a=0, where B0(z)=z.

step 1.1step 2.1L1algebra

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