11
$\begingroup$

Let $\mathbb{B}$ be the open unit ball in $\mathbb{C}^n, n\geq 1$ and let $H^\infty (\mathbb{B})$ be the space of all bounded holomorphic functions on $\mathbb{B}$. It is well known that $H^\infty (\mathbb{B})$ is a Banach algebra. I was searching for the description of its dual space, but I could not find anything. My question is if its dual space can be described in terms of some measures on the boundary. Can anyone give me some reference on it?

Thank you.


edit added (by YC): this question has also been posted on Math StackExchange: https://math.stackexchange.com/questions/2766383/dual-of-the-space-of-all-bounded-holomorphic-functions

$\endgroup$
6
  • 6
    $\begingroup$ The first thing to note is that for "big" spaces like $H^\infty$ (or $L^\infty$) elements of the dual will not always be measures on the space, but things like "finitely additive" measures. For instance, $L^\infty({\bf T})$ is $C(X)$ for some $X$ which is much more complicated that ${\bf T}$, and so the dual of $L^\infty({\bf T})$ will be the space of Radon measures on $X$. Of course, some measures on ${\bf T}$ will give you elements of $L^\infty({\bf T})$, namely the measures which are absolutely continuous with respect to Lebesgue measure on ${\bf T}$ $\endgroup$
    – Yemon Choi
    Commented May 4, 2018 at 13:12
  • $\begingroup$ Thank you for the quick response. It will be really helpful if you can give some reference on it. $\endgroup$ Commented May 4, 2018 at 13:16
  • $\begingroup$ On the other hand, if you ask about the dual of $A({\mathbb B})$, then some interesting things happen: for $n=1$ this is related to the F. and M. Riesz theorem, and there might be analogues in higher-dimensions but I don't recall the details right now. Maybe Rudin's book books.google.co.uk/books?id=wkjhBwAAQBAJ has something? $\endgroup$
    – Yemon Choi
    Commented May 4, 2018 at 13:17
  • 2
    $\begingroup$ Well, it depends what you mean by a "description". Abstract functional analsysis tells us that a bounded linear functional on $H^\infty({\mathbb B})$ can be extended to a bounded linear function on $L^\infty({\mathbb B})$, which can then be represented as a Radon measure on some complicated space $X$. Beyond that, it is difficult to say more unless you impose more restrictions on the original functional $\endgroup$
    – Yemon Choi
    Commented May 4, 2018 at 13:19
  • $\begingroup$ Could you clarify whether you are happy with a description in terms of finitely additive measures on $\partial {\mathbb B})$, or if you actually wanted the functionals to be represented by countably additive measures? The first case is possible, I apologize if I did not make that clear; it's just that I tend not to think of it as very useful. $\endgroup$
    – Yemon Choi
    Commented May 5, 2018 at 22:08

1 Answer 1

1
$\begingroup$

Contrary to some statements in the comments, the duals of $L^\infty$ are well-studied objects and can be represented as suitable spaces of measures on the underlying measure space---in your case the one-dimensional sphere (references Hildebrandt, in vol. 36 of the Transactions or Fichtenholz and Kantorovitch in vol. 5 of Studia Math). The nearest to a representation for the dual of $H^\infty$ you will find is as the quotient of this space of measures by the annihilator of $H^\infty$ therein---which follows from general abstract nonsense on duals of subspaces of Banach spaces.

$\endgroup$
4
  • 8
    $\begingroup$ I didn't say they weren't well studied, but I maintain that a finitely additive measure on the Borel sigma-algebra of the circle is a rather exotic beast compared to "usual" measures; if I recall correctly, one might lose the DCT, Fubini, and various other important tools of the trade. I understood the OP, when they spoke of measures, to be thinking of countably additive measures. (The Hildebrandt paper seems, on my quick reading, to give the same amount of information as found in e.g. Dunford--Schwarz.) $\endgroup$
    – Yemon Choi
    Commented May 5, 2018 at 22:05
  • $\begingroup$ @YemonChoi, DCT works for any measure space; it's Fubini that requires $\sigma$-finite measures. $\endgroup$
    – LSpice
    Commented Dec 27, 2019 at 3:19
  • 3
    $\begingroup$ @LSpice While what you say is true, I think you have confused $\sigma$-finite for $\sigma$-additive in Yemon's comment. I think Yemon's comment about the DCT was alluding to the fact that if a measure is not $\sigma$-additive, the DCT can be seen to fail by taking indicator functions of an increasing sequence of sets witnessing the failure of $\sigma$-additivity. $\endgroup$ Commented Dec 27, 2019 at 10:23
  • $\begingroup$ @RobertFurber, yes, you are right. $\endgroup$
    – LSpice
    Commented Dec 27, 2019 at 15:47

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .