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A divergent Blaschke sum: no nonzero Hardy function has these zeros

Statement refuted

"Every sequence $(a_{n})$ of distinct points of $D$ without accumulation point in $D$ is the zero sequence of some nonzero function in $H^{p} (D)$ , $0 < p \leq \infty$ , and its formal Blaschke product $\prod_{n} b_{a_{n}}$ converges normally on $D$ ."

Facts & Assumptions

Given: The sequence $a_{n} : = 1 - \frac{1}{n}$ for $n \geq 2$ , the normalized Blaschke factors $b_{a_{n}}$ , and the partial products $B_{N} = \prod_{n = 2}^{N} b_{a_{n}}$ .

[F1]

For a nonzero $f \in H^{p} (D)$ , $0 < p \leq \infty$ , the zero sequence satisfies the Blaschke condition $\sum_{n} (1 - ∣ a_{n} ∣) < + \infty$ (The zero set of a Hardy function satisfies the Blaschke condition, Analytic Hardy spaces on the unit disc).

[F2]

Normal convergence of a product on $D$ means that for every compact $K \subseteq D$ there is $N$ with $b_{a_{n}}$ zero-free on $K$ for $n \geq N$ and $\sum_{n \geq N} \sup_{K} ∣ 1 - b_{a_{n}} ∣ < + \infty$ ; the zeros of $b_{a}$ are exactly $a$ , and $∣ b_{a} (z) ∣ = ∣ a - z ∣ / ∣ 1 - \overline{a} z ∣$ for every $a \in D$ and $z \in D$ (Blaschke factors and Blaschke products, Boundary values and zeros of a Blaschke product).

[F3]

For $a, z \in D$ , $1 - ∣ b_{a} (z) ∣^{2} = \frac{(1 - ∣ a ∣^{2}) (1 - ∣ z ∣^{2})}{∣ 1 - \overline{a} z ∣^{2}}$ ; moreover $∣ 1 - \overline{a} z ∣ \leq 1 + ∣ z ∣$ and $1 - ∣ b_{a} ∣ \geq \frac{1}{2} (1 - ∣ b_{a} ∣^{2})$ because $1 + ∣ b_{a} ∣ \leq 2$ (Blaschke factors and Blaschke products, Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive).

[F4]

For rational $p > 0$ the $p$ -series $\sum_{k \geq 1} 1 / k^{p}$ converges if and only if $p > 1$ ; in particular the harmonic series $\sum_{k \geq 1} 1 / k$ diverges at $p = 1$ (For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ ).

[F5]

For every set $A \subseteq Ω$ whose intersection with each compact subset of a plane domain $Ω$ is finite, and with prescribed positive finite integer multiplicities, there is a nonzero holomorphic function on $Ω$ with exactly those zeros and multiplicities (Every locally finite effective divisor on a plane domain is a holomorphic zero divisor, Identity theorem for holomorphic functions).

Take the distinct points $a_{n} = 1 - \frac{1}{n} \in D$ , $n \geq 2$ , with assigned multiplicity one; they are discrete in $D$ with the only accumulation point $1$ on the boundary.

Counterexample

1.1givenF4algebra

The Blaschke sum diverges. Here $1 - ∣ a_{n} ∣ = 1 / n$ , so by [F4] $\sum_{n \geq 2} (1 - ∣ a_{n} ∣) = \sum_{n \geq 2} 1 / n = + \infty$ ; thus $(a_{n})$ is not a Blaschke sequence.

1.2givenF2F3algebra

The formal Blaschke product is not normally convergent. Fix a nonempty compact $K \subseteq D$ and let $ρ : = \sup_{z \in K} ∣ z ∣ < 1$ . For every $z \in K$ and $n \geq 2$ , [F3] gives $1 - ∣ b_{a_{n}} (z) ∣^{2} = \frac{(1 - ∣ a_{n} ∣^{2}) (1 - ∣ z ∣^{2})}{∣ 1 - \overline{a_{n}} z ∣^{2}} \geq \frac{(1 / n) (1 - ρ^{2})}{4},$ because $1 - ∣ a_{n} ∣^{2} = (1 - ∣ a_{n} ∣) (1 + ∣ a_{n} ∣) \geq 1 / n$ and $∣ 1 - \overline{a_{n}} z ∣^{2} \leq (1 + ρ)^{2} \leq 4$ ; hence $1 - ∣ b_{a_{n}} (z) ∣ \geq (1 - ρ^{2}) / (8 n)$ uniformly on $K$ . Since $∣ 1 - b_{a_{n}} (z) ∣ \geq 1 - ∣ b_{a_{n}} (z) ∣$ , the series $\sum_{n} \sup_{K} ∣ 1 - b_{a_{n}} ∣$ diverges, so the normal-convergence criterion of [F2] fails.

2.1step 1.1F1

No nonzero $H^{p}$ function has these zeros. Suppose $f \in H^{p} (D)$ , $0 < p \leq \infty$ , is nonzero with zero sequence $(a_{n})$ . By [F1] its zeros satisfy $\sum_{n} (1 - ∣ a_{n} ∣) < + \infty$ , contradicting step 1.1. Hence no such $f$ exists, refuting the first half of the quoted statement.

2.2step 1.2F2F4algebra

The partial products converge locally uniformly to $0$ . If $z$ equals one of the prescribed zeros, the products are eventually zero. Otherwise, for fixed $z \in D$ , using $\log t \leq t - 1$ for $t > 0$ and step 1.2 with $K = {z}$ , $\log ∣ B_{N} (z) ∣ = \sum_{n = 2}^{N} \log ∣ b_{a_{n}} (z) ∣ \leq - \sum_{n = 2}^{N} (1 - ∣ b_{a_{n}} (z) ∣) \leq - \frac{1 - ∣ z ∣^{2}}{8} \sum_{n = 2}^{N} \frac{1}{n} ⟶ - \infty,$ so $B_{N} (z) \to 0$ ; the bound $\frac{1 - ∣ z ∣^{2}}{8} \geq \frac{1 - ρ^{2}}{8}$ uniform on a compact $K$ makes the convergence locally uniform on $D$ . Hence the partial products converge to the zero function, and the formal product has no nonzero holomorphic limit; combined with step 1.2 this refutes the second half of the quoted statement.

3.1step 2.1F5∎

A holomorphic function with exactly these zeros nevertheless exists. The set $A = {a_{n} : n \geq 2}$ is locally finite in the plane domain $D$ : for a nonempty compact $K \subseteq D$ , set $ρ = \max_{K} ∣ z ∣ < 1$ ; the condition $a_{n} \in K$ implies $1 - 1 / n \leq ρ$ , hence $n \leq 1 / (1 - ρ)$ , so only finitely many terms meet $K$ . Thus by [F5] with multiplicity one there is a holomorphic $f$ on $D$ whose zeros are exactly the points $a_{n}$ , all simple, and which has no other zeros. By step 2.1 this $f$ lies in no $H^{p} (D)$ , $0 < p \leq \infty$ : it exhibits the failure of the Blaschke factorization and of the $H^{p}$ zero-set condition outside the Blaschke regime.

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Sources

R. K. Srivastava, Lecture Notes on Hardy Spaces (MA650, IIT Guwahati), §5.8 and Remark 5.18 (standard reference, not scraped)
J. B. Garnett, Bounded Analytic Functions, revised first edition, Chapter II §2 (standard reference, not scraped)