if f is in $H^{1}$ the Hardy space with integrable norm, and if f is not identically zero, then the zeroes of f (certainly countable in number) satisfy the Blaschke condition.Can any "Blaschke condition" be defined if hardy space is considered on the upper-half plane instead of unit disk. Curious to know if this carries over by the conformal equivalence that maps upper-half plane to unit disk and vice-versa.
$\begingroup$
$\endgroup$
2
-
$\begingroup$ The short answer is "yes." The longer answer is either 1) do the calculation yourself, or 2) look in Garnett's {\it Bounded Analytic Functions} book. $\endgroup$– Mike JuryCommented Feb 22, 2013 at 1:15
-
1$\begingroup$ Garnett's book is a great text and a good suggestion, but I can't resist putting a word in for thevrecent introductory text of Mashreghi books.google.ca/books/about/… $\endgroup$– Yemon ChoiCommented Feb 22, 2013 at 6:00
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
4
Blaschke condition in the upper half-plane is $$\sum\left|\Im\frac{1}{z_k}\right|<\infty.$$ For the proof follow the advice given above, that is prove it yourself, or look in a book. In addition to the books mentioned above, I recommend Koosis Hardy spaces, or Levin, Distribution of values of entire functions, or de Branges, Hilbert spaces of entire functions.
answered Feb 22, 2013 at 6:49
-
$\begingroup$ @ Alexandre Eremenko. Are you sure this is right? I get the condition that the sum $$\sum 1 - \left | \frac {z_k-i}{z_k+i}\right |$$ be finite and this only agrees with your one under a priori assumptions on the $z_k$ $\endgroup$– jbcCommented Mar 7, 2013 at 13:27
-
$\begingroup$ Yes, I am sure. See Levin, Chap V, section 3. I slightly edited the formula. $\endgroup$ Commented Mar 7, 2013 at 22:37
-
$\begingroup$ @ Alexandre Eremenko ): I could not found the deduction for the upper half plane in the suggested site. Can you please show the deduction or suggest any other link for this ? $\endgroup$– EmptyCommented Jun 3, 2015 at 12:38
-