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I want prove that the Wiener algebra, $W(\mathbb{T})$, does not coincide with the disc algebra $A(\mathbb{D})$. I know that there are some ways to do this. For example, one smart way is use Rudin-Shapiro sequence, see this post.

Note that $W(\mathbb{T})$ is an isometry to $\ell^1$, with pre-dual $c$ (or $c_0$, even something else). However, I never hear story about the pre-dual of disc algebra.

Question: Does $A(\mathbb{D})$ has pre-dual? What it is?

Landau
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    The first naive question would be: Has the unit ball of the disk algebra enough extreme points? Have you thought about this? – MaoWao Jan 14 '22 at 09:30
  • Yes. But it seems not directly to draw conclusion by applying Krein-Miliman theorem. – Landau Jan 14 '22 at 13:08
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    For information about extreme points, see this post: https://math.stackexchange.com/questions/3934564/extreme-points-of-disc-algebra. This is non-trivial, and discussed in Hoffman's book. – Landau Jan 15 '22 at 05:00

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As you noticed, the Wiener algebra is isometric to $\ell_1$. In particular, it has the Schur property (weakly convergent sequences converge in the norm). However, this is very much not the case for the disc algebra as it, for example contains a sequence equivalent to the standard basis of $c_0$. It is a nice exercise to find explicitly a sequence in the disk algebra that converges weakly but not in the norm.

However, since $A(\mathbb D)$ is separable and contains a copy of $c_0$, by Sobczyk's theorem, this copy is complemented. No dual space contains a complemented copy of $c_0$ (Theorem 2.4.15). Hence $A(\mathbb D)$ is not isomorphic to a dual space.

For more details, see this paper.

Tomasz Kania
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  • Interesting answer. But it seems to me that the main question is whether the disk algebra is a dual space. This is not covered by your answer, right? – MaoWao Jan 14 '22 at 13:10
  • @MaoWao, it is not. Please find my updated answer. – Tomasz Kania Jan 14 '22 at 13:26
  • Even better! Thanks for the insights. – MaoWao Jan 14 '22 at 13:27
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    @Tomasz Kania, nice answer. I reconsiderd your exercise: If $(a_n)$ is a real sequence in $\mathbb{D}$ with limit $1$ and $f_n(z):=\left(\frac{z-a_n}{1-a_n z}\right)^2$, then $(f_n)$ is weakly convergent to $1$, but not in norm. Correct? – Gerd Jan 15 '22 at 10:00
  • @Gerd Indeed you can use Hahn-Banach theorem, Riesz representation theorem, and DCT to get $f_n$ converge to $1$ weakly. But it seems $f_n\to 1$ in norm... – Landau Jan 17 '22 at 04:12
  • @Landau Since $f_n(a_n)=0$ it is impossible that $f_n \to 1$ in norm, On the other hand $(f_n)$ is norm bounded and pointwise convergent to $1$. Thus it should be weakly convergent to $1$. – Gerd Jan 17 '22 at 08:10
  • @Gerd You are right and your example indeed work. – Landau Jan 17 '22 at 12:02