About Wiener algebra

Question

Let $W$ be the space of all analytic functions on the set $D=\{z\in\mathbb{C}: |z|<1\}$ of the form $$f(z)=\sum_{n=0}^{\infty}c_nz^n$$ with $$\|f\|_w=\sum_{n=0}^{\infty}|c_n|<\infty.$$ Now $(W,\|.\|_w)$ is a Banach algebra. Can anyone tell why $W$ is not complete under the norm $$\|f\|_\infty=\sup_{z\in D}\{|f(z)|\}?$$

Answer 1

Let $A$ be the algebra of functions continuous on $\overline D$ and holomorphic in $D$.

Lemma. The polynomials are dense in $A$.

Proof: Say $f\in A$. For $0<r<1$ let $f_r(z)=f(rz)$. Since $f$ is uniformly continuous there exists $r$ such that $||f-f_r||_\infty<\epsilon$. But $f_r$ is holomorphic in the disk of radius $1/r>1$, so the power series for $f_r$ converges to $f_r$ uniformly on $\overline D$. qed.

So if $W$ were complete in the sup norm then $A=W$. The Rudin-Shapiro polynomials allow you to construct an $f\in A\setminus W$.

Or: If $W=A$ then the Closed Graph Theorem shows that $||f||_w\le c||f||_\infty$ for $f\in A$; the Rudin-Shapiro polynomials show that this is not so.