Let $H^{\infty}(\mathbb{D})$ denotes the Banach space of bounded holomorphic functions in the unit disc. Consider the weak* topology on $L^{\infty}(\mathbb{T})$ that it inherits as the dual of $L^{1}(\mathbb{T}).$ Under this topology $H^{\infty}(\mathbb{D})$ is a weak* closed subspace of $L^{\infty}(\mathbb{T}).$ I have the following question:

Does there exist a subspace $M\subseteq H^{\infty}(\mathbb{D})$ which is not weak* closed and contains a nontrivial weak* closed unital subalgebra of $H^{\infty}(\mathbb{D})?$

By nontrivial I mean it contains a nonconstant holomorphic function.

  • 1
    $\begingroup$ The even functions $f(z)=g(z^2)$ look weak $*$ closed, and then just take a suitable $M$ that contains these. $\endgroup$ Nov 12, 2015 at 5:25
  • $\begingroup$ yes, it is weak* closed. But I'm not able to see how to choose $M?$ For example disc algebra $A(\mathbb{D})$ is not weak* closed but it also does not contain the weak* algebra generated by $z^2.$ $\endgroup$
    – vikram
    Nov 12, 2015 at 5:35
  • $\begingroup$ Well, just take an $M\supseteq A$, with $A$ even functions, that is not weak $*$ closed (for example, not norm closed would work). $\endgroup$ Nov 12, 2015 at 5:39

1 Answer 1


I discussed with some of my friends and found this example. Let $\mathcal{P}$ denotes the set of all polynomials, and let $\mathcal{U}:=\{\varphi(z^2):\varphi\in H^{\infty}(\mathbb{D})\}.$ Set $$M=\mathcal{P}+\mathcal{U}.$$ Note that $M$ contains $\mathcal{U}$ which is weak* closed. Then as the weak* closure of polynomials is dense in $H^{\infty}(\mathbb{D}),$ the weak* closure of $M$ is $H^{\infty}(\mathbb{D}).$ But $M$ is not $H^{\infty}(\mathbb{D}).$ To see that note $\sin{z}\in H^{\infty}(\mathbb{D})$ but $\sin{z}$ can not be written as a sum of a polynomial and an even function in $H^{\infty}(\mathbb{D}).$

  • $\begingroup$ This is essentially the same idea that @ChristianRemling suggested in his comment above $\endgroup$
    – Yemon Choi
    Nov 15, 2015 at 15:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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