Factorization in the Wiener algebra on the unit disc. - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T22:47:27Z http://mathoverflow.net/feeds/question/41315 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/41315/factorization-in-the-wiener-algebra-on-the-unit-disc Factorization in the Wiener algebra on the unit disc. AD 2010-10-06T18:55:56Z 2010-10-21T03:28:27Z <p>Consider the Banach algebra $W^+=\ell^1(\mathbb{Z}^+)$, viewed upon as the analytic functions $f$ on the unit disc $\mathbb{D}$ such that $$\|f\|=\sum_{k\ge0}|a_k|&lt;\infty$$ 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 $W^+$? </p> http://mathoverflow.net/questions/41315/factorization-in-the-wiener-algebra-on-the-unit-disc/42996#42996 Answer by Sergei for Factorization in the Wiener algebra on the unit disc. Sergei 2010-10-21T03:28:27Z 2010-10-21T03:28:27Z <p>There is no factorization in Wiener algebra, it is easy to construct a counterexample. </p> <p>Namely, if $B$ is a Blaschke factor with zeroes $z_n$, $z_n\to 1$ (of course $\sum(1-|z_n|) &lt;\infty$) and $g= (z-1)^2$ then $f= Bg$ has $C^1$ boundary values, and so is in the Wiener algebra. </p> <p>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 $1$. </p> <p>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. </p>