Wiener's lemma for absolutely convergent Fourier series

Let $\chi$ be a character of $A(\mathbb{T})$ and put $z=\chi (e_1)$. Since $e_n{e}_{-n}=1$ and $\mathrm{\parallel }{e}_{\pm n}{\mathrm{\parallel }}_A=1$, boundedness from [L4] applied for every $n\ge 1$ gives $\mathrm{\mid }z{\mathrm{\mid }}^n\le \mathrm{\parallel }\chi \mathrm{\parallel }$ and $\mathrm{\mid }z{\mathrm{\mid }}^{-n}\le \mathrm{\parallel }\chi \mathrm{\parallel }$, hence $\mathrm{\mid }z\mathrm{\mid }=1$. By [L2], every $f\in A(\mathbb{T})$ is the $A$-norm limit of its finite Fourier sums, so continuity from [L4] gives $\chi (f)={\sum }_{k\in \mathbb{Z}}\widehat{f}(k)z^k\mathrm{.}$ Thus the characters of $A(\mathbb{T})$ are exactly evaluations at points of $\mathbb{T}$.

If $f$ were a nonunit, the ideal $(f)$ would be proper, so [L3] would give a character $\chi$ with $\chi (f)=0$. By step 1.1 this character is evaluation at a point of $\mathbb{T}$, contradicting the hypothesis. Conversely, a character vanishing on an element rules out a multiplicative inverse. Thus $f$ is a unit exactly when it has no zero on $\mathbb{T}$.

The hypothesis makes $f$ a unit, so its algebra inverse is the pointwise reciprocal $1\mathrm{/}f$.