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 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 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 $\mathbb{T}$.

The standard maximal-ideal/Gelfand--Mazur criterion for a unital commutative complex Banach algebra says that an element is invertible exactly when no character vanishes on it: under the Axiom of Choice a nonunit lies in a maximal ideal, whose quotient character vanishes there; conversely a vanishing character rules out a multiplicative inverse. Applying this criterion and step 1.1, $f$ 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 $1\mathrm{/}f$.