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Consider the Banach algebra $W^+=\ell^1(\mathbb{Z}^+)$, viewed upon as the analytic functions $f$ on the unit disc $\mathbb{D}$ such that $$\|f\|=\sum_{k\ge0}|a_k|<\infty$$ where $$f(z)=\sum a_kz^k$$ is the Taylor expansion of $f$. Clearly, $W^+\subset H^\infty(\mathbb{D})$. Now, it is well known that $H^\infty$ admits inner-outer factorization. Is there some similar factorization theorem for $W^+$?

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Very interesting question, but could you possibly write $W_+$ instead of $A$, since $A$ usually means the disc algebra instead, and it's destroying my brain every time I see $A$ here? It looks very difficult; do you know what happens for the disc algebra? I don't know, but for many properties the Wiener and disc algebra are much "closer" and more similar to each other than they are to $H^\infty$, so it could be relevant. Do you know anything about Blaschke products in the Wiener algebra? I don't, but it could be another good thing to look at; I haven't managed to find very much yet. –  Zen Harper Oct 9 '10 at 14:22
@Zen Harper: I hope this is better. I have no solution, we would have to use a different technique in order to do something (in $W^+$ or in $A(\mathbb{D})$). –  Mr AD Oct 9 '10 at 21:41
The analytic Wiener algebra and the disc algebra might be closer in spirit than either is to $H^\infty$, but they are very different from each other. For instance, take the Blaschke product $b(z)$ with one simple zero at $a$; then generically the norm of $b^n$ inside $W_+$ is $O(\sqrt{n})$ if I recall correctly (Katznelson?). Have you tried looking at some of the papers Jean Esterle cowrote concerning zero sets in $W_+$? –  Yemon Choi Oct 9 '10 at 22:05
@Yemon Choi: Yes, the disc algebra might be more promising in the sense it is more similar to $H^\infty$. I do not know about Esterle's work in $W^+$ - do you have a ref. where I can start? –  Mr AD Oct 10 '10 at 21:34
@Yemon Choi: No, I can not MathSciNet any more. Having a factorization theorem I would hope to get more info on how the functions without zeros look like. My main interest is to get a better understand of the invisible spectrum (see aif.cedram.org/item?id=AIF_1999__49_6_1925_0). –  Mr AD Oct 12 '10 at 4:44

1 Answer 1

up vote 2 down vote accepted

There is no factorization in Wiener algebra, it is easy to construct a counterexample.

Namely, if $B$ is a Blaschke factor with zeroes $z_n$, $z_n\to 1$ (of course $\sum(1-|z_n|) <\infty$) and $g= (z-1)^2$ then $ f= Bg$ has $C^1$ boundary values, and so is in the Wiener algebra.

On the other hand, $B$ is an inner part of $f$ (in $H^\infty$), and it is not in $W$, because it is not even continuous at $1$.

On the other hand, if $f(z)\ne 0$ for all $z:|z|=1$, then $f$ has only finitely many zeroes in the unit disc, so the factorization is trivial: the inner part is a finite Blaschke product.

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Thank you Sergei, that was indeed very simple. –  Mr AD Oct 21 '10 at 7:55

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