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An infinite Blaschke product whose zeros accumulate at the boundary
Example
Let for . Then , so is a Blaschke sequence and the Blaschke product converges normally on to a holomorphic function with , whose zeros are exactly the points , each simple, and whose boundary function is unimodular -almost everywhere. The value at the origin is The zeros accumulate at , so has no continuous extension to and is not a finite product. The associated norm is , while the boundary function has modulus almost everywhere.
Facts & Assumptions
Given: The sequence (), its Blaschke product , and the partial products .
A Blaschke sequence has a normally convergent Blaschke product , holomorphic with , whose zeros are exactly the with multiplicity and whose boundary function satisfies -almost everywhere; each has (Blaschke factors and Blaschke products, Boundary values and zeros of a Blaschke product).
For rational the -series converges if and only if ; in particular converges at , and deleting the first term preserves convergence (For rational , converges iff ).
If for all and -almost everywhere as , then ; the radial means are nondecreasing in and their supremum is (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
The sequence is Blaschke. Since , one has and by [F2]; by [F1] the product converges normally, is holomorphic with , has exactly the simple zeros , and has -almost everywhere.
Value at the origin. By [F1], , and the finite products telescope: using ,
No continuous extension. The zeros converge to . If had a continuous extension to , then along one would get , while on the other hand the boundary values of the extension agree -almost everywhere with , so the continuous function restricted to equals on a set of full measure, hence equals everywhere by continuity; at this gives , a contradiction. Thus has no continuous extension to ; in particular is not a finite Blaschke product, since a finite product would extend continuously.
The norm. Since and for -almost every (the nontangential limits of [F1] in particular give radial limits almost everywhere), [F3] gives as . The radial means are nondecreasing in by [F3], so their supremum is the limit, that is, and hence .
Depends on
- Blaschke factors and Blaschke products
- Boundary values and zeros of a Blaschke product
- The zero set of a Hardy function satisfies the Blaschke condition
- For rational $p > 0$, $\sum 1/k^p$ converges iff $p > 1$
- Analytic Hardy spaces on the unit disc
- Radial p-means of a holomorphic function are nondecreasing
- Dominated convergence
- The one-dimensional torus and its normalized Haar integral
Used by
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Sources
- R. K. Srivastava, Lecture Notes on Hardy Spaces (MA650, IIT Guwahati), §5.8 (standard reference, not scraped)
- J. B. Garnett, Bounded Analytic Functions, revised first edition, Chapter II §2 (standard reference, not scraped)