Factorization in the Wiener algebra on the unit disc.

2010-10-06T18:55:56Z

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 $W^+$?

Answer by Sergei for Factorization in the Wiener algebra on the unit disc.

2010-10-21T03:28:27Z

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.

Factorization in the Wiener algebra on the unit disc. - MathOverflow

Factorization in the Wiener algebra on the unit disc.

Answer by Sergei for Factorization in the Wiener algebra on the unit disc.