Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

A Blaschke product is a function of the form $$B(z):=z^k\prod_{n=1}^{\infty}\frac{a_n-z}{1-\overline{a_n}z}\frac{|a_n|}{a_n}$$ where the $a_n$ are the non-zero zeros of $B$, and satisfie $\sum_{n=1}^{\infty}(1-|a_n|) < \infty$.

Blashke products are holomorphic and bounded by 1 on the unit disk. A well known theorem asserts that $B$ has radial limits almost everywhere on the unit circle, i.e. that the limit $$\lim_{r \rightarrow 1} B(re^{i \theta})$$ exist for almost every $\theta$. I'm looking for an example of Blashke product such that the radial limit does not exist at a certain point, say $1$ for example. In particular, a Blaschke product with zeros in $(0,1)$ such that $$\limsup_{r \rightarrow 1}|B(r)| =1$$ would work.

Does anyone have a construction or reference?

Thank you, Malik

share|improve this question
    
Have you checked with the book "$H^p$-spaces" by P. Duren? –  AD. Sep 23 '10 at 20:35
    
Yes, I have, but I couldn't find anything about this. –  Malik Younsi Sep 24 '10 at 0:19

2 Answers 2

up vote 3 down vote accepted

There is an exercise in Rudin's Real and complex analysis whose solution would answer your question, #14 in Chapter 15:

Prove that there is a sequence $\{\alpha_n\}$ with $0\lt\alpha_n\lt1$, which tends to $1$ so rapidly that the Blaschke product with zeros at the points $\alpha_n$ satisfies the condition $$\limsup_{r\to1}|B(r)|=1.$$ Hence this $B$ has no radial limit at $z=1$.

(The previous exercise says that the limit is $0$ if $\alpha_n=1-n^{-2}$.)

Instead of trying to solve it (with the guess of something like $\alpha_n=1-4^{-n}$), I found the article "On functions with Weierstrass boundary points everywhere" by Campbell and Hwang, which says on page 510 (page 4 of the pdf):

Let $B(z)$ be the Blaschke product with zeros at $z=1-\exp(-n)$, for $n=1,2,\ldots$. Then $B(z)$ has no radial limit at $z=1$....

The authors cite page 12 of "Sequential and continuous limits of meromorphic functions" by Bagemihl and Seidel for this fact, but I do not currently have access to that article. Hopefully you can track it down to get your question answered, or perhaps someone will take up the challenge of solving Rudin's problem.

share|improve this answer
    
Another relevant reference is "Radial limits and star invariant subspaces of bounded mean oscillation" by William S. Cohn, which contains necessary and sufficient conditions for radial limits to exist at a particular point for a Blaschke product and all of its subproducts. However, I do not see any conditions there to guarantee that a particular Blaschke product does not have a radial limit at a particular point. –  Jonas Meyer Jan 13 '11 at 22:25
    
Thank you very much, I will take a look at those very interesting references. –  Malik Younsi Jan 21 '11 at 1:15
    
@Malik: You're welcome. If you find a more complete answer, e.g., in the Bagemihl and Seidel reference, please share. –  Jonas Meyer Jan 21 '11 at 5:52

Let $c_n$ be a sequence dense in $S^1$ and define $a_n = \left ( 1 + \frac{1}{n^2} \right ) c_n$. Then $1 - |z_n| = \frac{1}{n^2}$ so this are the zeros of a Blaschke product.

So, let $c_n = r e^{i \phi_n}$ and fill this in in the Blaschke product. Then we if $B_n(z,r)$ is the term inside the Blaschke product, then we can try to evaluate if $\sum 1 - B_n(z,r)$ converges. We can evaluate the convergence of this with the integral test and then we see the limit does not exist as $r \to 1^{-}$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.