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An infinite Blaschke product whose zeros accumulate at the boundary

Example

Let an:=1−1n2 for n≥2. Then ∑n≥2(1−∣an∣)=∑n≥21n2<∞, so (an) is a Blaschke sequence and the Blaschke product B=∏n≥2ban converges normally on D to a holomorphic function with ∣B∣≤1, whose zeros are exactly the points 1−1/n2, each simple, and whose boundary function is unimodular m-almost everywhere. The value at the origin is B(0)=∏n≥2(1−1n2)=lim⁡N→∞N+12N=12. The zeros accumulate at 1∈T, so B has no continuous extension to D‾ and is not a finite product. The associated H2 norm is ∥B∥H2=1, while the boundary function has modulus 1 almost everywhere.

Facts & Assumptions

Given: The sequence an=1−1/n2 (n≥2), its Blaschke product B=∏n≥2ban, and the partial products BN=∏n=2Nban.

[F1]

A Blaschke sequence ∑n(1−∣an∣)<+∞ has a normally convergent Blaschke product B, holomorphic with ∣B∣≤1, whose zeros are exactly the an with multiplicity and whose boundary function satisfies ∣B∗∣=1 m-almost everywhere; each ba has ba(0)=∣a∣ (Blaschke factors and Blaschke products, Boundary values and zeros of a Blaschke product).

[F2]

For rational p>0 the p-series ∑k≥11/kp converges if and only if p>1; in particular ∑k≥11/k2 converges at p=2, and deleting the first term preserves convergence (For rational p>0, ∑1/kp converges iff p>1).

[F3]

If ∣gr∣≤1 for all r and gr→g∗ m-almost everywhere as r↑1, then ∫Tgr dm→∫Tg∗ dm; the radial means ∫T∣g(rζ)∣2 dm are nondecreasing in r and their supremum is ∥g∥H22 (Dominated convergence, Radial p-means of a holomorphic function are nondecreasing, Analytic Hardy spaces on the unit disc, The one-dimensional torus and its normalized Haar integral).

Verification

1.1givenF1F2algebra

The sequence is Blaschke. Since ∣an∣=1−1/n2, one has 1−∣an∣=1/n2 and ∑n≥21/n2<∞ by [F2]; by [F1] the product B converges normally, is holomorphic with ∣B∣≤1, has exactly the simple zeros 1−1/n2, and has ∣B∗∣=1 m-almost everywhere.

2.1step 1.1F1algebra

Value at the origin. By [F1], B(0)=∏n≥2∣an∣=∏n≥2(1−1n2), and the finite products telescope: using 1−1n2=(n−1)(n+1)n2, ∏n=2N(1−1n2)=(N−1)!⋅(N+1)!/2(N!)2=N+12N⟶12.

2.2step 1.1F1algebra

No continuous extension. The zeros an=1−1/n2 converge to 1∈T. If B had a continuous extension to D‾, then along an→1 one would get ∣B(1)∣=lim⁡n∣B(an)∣=0, while on the other hand the boundary values of the extension agree m-almost everywhere with B∗, so the continuous function ∣B∣ restricted to T equals 1 on a set of full measure, hence equals 1 everywhere by continuity; at ζ=1 this gives ∣B(1)∣=1, a contradiction. Thus B has no continuous extension to D‾; in particular B is not a finite Blaschke product, since a finite product would extend continuously.

3.1step 1.1F3algebra∎

The H2 norm. Since ∣B∣≤1 and B(rζ)→B∗(ζ) for m-almost every ζ (the nontangential limits of [F1] in particular give radial limits almost everywhere), [F3] gives ∫T∣B(rζ)∣2 dm(ζ)→∫T∣B∗∣2 dm=1 as r↑1. The radial means are nondecreasing in r by [F3], so their supremum is the limit, that is, ∥B∥H22=1 and hence ∥B∥H2=1.

Depends on

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