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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.

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    $\begingroup$ The even functions $f(z)=g(z^2)$ look weak $*$ closed, and then just take a suitable $M$ that contains these. $\endgroup$ Nov 12 '15 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 '15 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 '15 at 5:39
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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 $H^{\infty}(\mathbb{D}).$

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

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