Absolute Convergence and the Wiener Algebra

Since $\mathrm{\mid }{a}_k{e}_k(x)\mathrm{\mid }=\mathrm{\mid }{a}_k\mathrm{\mid }$ and ${\sum }_k\mathrm{\mid }{a}_k\mathrm{\mid }<\mathrm{\infty }$, the Weierstrass M-test gives absolute uniform convergence to a continuous $g$.

Uniform convergence permits integration term by term against $e_{-m}$; character orthogonality gives $\widehat{g}(m)=a_m$.

For $a_k=\widehat{f}(k)$, the Abel means are ${\sum }_k{r}^{\mathrm{\mid }k\mathrm{\mid }}{a}_k{e}_k$. They converge uniformly to $g$ by dominated tail control, while [L1] says they converge to $f$ in $L^1$; hence $f=g$ almost everywhere.

For finite Fourier sums, multiplying and collecting equal frequencies gives $\widehat{fg}(n)={\sum }_k\widehat{f}(k)\widehat{g}(n-k)$.

Truncate both coefficient series. By [L1] the truncations converge uniformly, and their convolution coefficients converge in ${\mathrm{\ell }}^1$ because ${\mathrm{\ell }}^1\ast {\mathrm{\ell }}^1\subseteq {\mathrm{\ell }}^1$; thus the formula in step 1.1 holds for $f,g$.

Tonelli's theorem for the nonnegative double series gives $\mathrm{\parallel }fg{\mathrm{\parallel }}_A\le {\sum }_{n,k}\mathrm{\mid }\widehat{f}(k)\mathrm{\mid }\mathrm{\mid }\widehat{g}(n-k)\mathrm{\mid }=\mathrm{\parallel }f{\mathrm{\parallel }}_A\mathrm{\parallel }g{\mathrm{\parallel }}_A\mathrm{.}$

The coefficient map is an isometric bijection from $A(\mathbb{T})$ to ${\mathrm{\ell }}^1(\mathbb{Z})$ by [L1]; completeness, commutativity, and the unit therefore follow from those of ${\mathrm{\ell }}^1$ and ${\delta }_0$.

The series ${\sum }_k(1+\mathrm{\mid }k\mathrm{\mid }{)}^{-2s}$ converges, by comparison with the integral of $x^{-2s}$ on $[1,\mathrm{\infty })$.

Cauchy--Schwarz yields ${\sum }_k\mathrm{\mid }{a}_k\mathrm{\mid }\le {({\sum }_k(1+\mathrm{\mid }k\mathrm{\mid }{)}^{2s}\mathrm{\mid }{a}_k{\mathrm{\mid }}^2)}^{1\mathrm{/}2}{({\sum }_k(1+\mathrm{\mid }k\mathrm{\mid }{)}^{-2s})}^{1\mathrm{/}2}<\mathrm{\infty }\mathrm{.}$

Put $h=(4N{)}^{-1}$. For $N\le \mathrm{\mid }k\mathrm{\mid }<2N$, $\mathrm{\mid }{e}^{-2\pi ikh}-1\mathrm{\mid }\ge \sqrt{2}$, since $2\pi \mathrm{\mid }k\mathrm{\mid }h\in [\pi \mathrm{/}2,\pi )$.

For $u_h(x)=f(x-h)-f(x)$, translation in the coefficient integral gives ${\widehat{u}}_h(k)=(e^{-2\pi ikh}-1)\widehat{f}(k)$.

Finite character orthogonality applied to the block and steps 1.1--1.2 gives $2{\sum }_{N\le \mathrm{\mid }k\mathrm{\mid }<2N}\mathrm{\mid }\widehat{f}(k){\mathrm{\mid }}^2\le {\int }_0^1\mathrm{\mid }{u}_h(x){\mathrm{\mid }}^{2}dx\mathrm{.}$

The Hölder bound gives $\mathrm{\mid }{u}_h(x)\mathrm{\mid }\le C(4N{)}^{-\alpha }$, so step 2.1 proves the assertion.

On $2^j\le \mathrm{\mid }k\mathrm{\mid }<2^{j+1}$, $(1+\mathrm{\mid }k\mathrm{\mid }{)}^{2s}\le 3^{2s}{2}^{2js}$; [L1] therefore bounds that weighted block by $O(2^{-2j(\alpha -s)})$.

Since $\alpha -s>0$, these bounds form a convergent geometric series. The $k=0$ term is finite because $f$ is bounded and integrable.

Choose $s$ with $1\mathrm{/}2<s<\alpha$.

By [L1] the weighted ${\mathrm{\ell }}^2$ hypothesis at this $s$ holds, and [L2] makes $(\widehat{f}(k))$ summable. This is precisely $f\in A(\mathbb{T})$.

Take the smooth periodic test function $\phi =e_{-k}$ in the defining identity. Since ${\phi }^{\mathrm{\prime }}=-2\pi ik{e}_{-k}$, it gives $-2\pi ik\widehat{f}(k)=-\widehat{g}(k)$.

Rearranging proves the formula; for $k=0$ it says $\widehat{g}(0)=0$, which is also the same test-function identity with $\phi =1$.

For a finite $E\subset \mathbb{Z}$, orthogonality of the characters gives ${\sum }_{k\in E}\mathrm{\mid }\widehat{g}(k){\mathrm{\mid }}^2\le \mathrm{\parallel }g{\mathrm{\parallel }}_2^2$.

By [L1], ${\sum }_{k\in E}(1+\mathrm{\mid }k\mathrm{\mid }{)}^2\mathrm{\mid }\widehat{f}(k){\mathrm{\mid }}^2\le \mathrm{\mid }\widehat{f}(0){\mathrm{\mid }}^2+{(1+\frac{1}{2\pi })}^2\mathrm{\parallel }g{\mathrm{\parallel }}_2^2\mathrm{.}$ Taking increasing finite $E$ gives the weighted ${\mathrm{\ell }}^2$ hypothesis.

Apply [L2] with $s=1$.

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 $1\mathrm{/}f$.

Let $\mathrm{\Omega }$ be an open set on which $\mathrm{\Phi }$ is holomorphic and which contains $f(\mathbb{T})$. The compact-neighbourhood lemma A compact subset of an open Euclidean set has a compact Jordan neighborhood inside that open set gives a finite union $J$ of closed grid rectangles with $f(\mathbb{T})\subseteq \mathrm{int}J\subseteq J\subseteq \mathrm{\Omega }$. Orient the frontier edges of the constituent grid cells positively and cancel each internal edge against its reverse. The resulting polygonal chain $\mathrm{\Gamma }$ is a cycle in $\mathrm{\Omega }\setminus f(\mathbb{T})$, with $n(\mathrm{\Gamma },w)=1$ for $w\in f(\mathbb{T})$ and $n(\mathrm{\Gamma },w)=0$ for $w\mathrm{\notin }\mathrm{\Omega }$: summing the cell indices first proves this away from the grid lines, and local constancy of the cycle index extends it to every point off the frontier. Thus $\mathrm{\Gamma }$ is null-homologous in $\mathrm{\Omega }$. For $z\in {\mathrm{\Gamma }}^{\ast }$, $z-f$ has no zero on $\mathbb{T}$, so $(z-f{)}^{-1}\in A(\mathbb{T})$ by [L1].

The map $z\mapsto \mathrm{\Phi }(z)(z-f{)}^{-1}$ is continuous into $A(\mathbb{T})$ (the inverse identity follows from $(z-f{)}^{-1}-(w-f{)}^{-1}=(w-z)(z-f{)}^{-1}(w-f{)}^{-1}$). Hence its normalized chain integral $F:=\frac{1}{2\pi i}{\int }_{\mathrm{\Gamma }}\mathrm{\Phi }(z)(z-f{)}^{-1}dz$ is an $A(\mathbb{T})$ element, as the finite sum of norm-limits of edgewise Riemann sums by [L2].

Evaluation at $x$ commutes with those norm-limits. Since step 1.1 gives $n(\mathrm{\Gamma },f(x))=1$ and makes $\mathrm{\Gamma }$ null-homologous in $\mathrm{\Omega }$, [L3] applied to $\mathrm{\Phi }$ gives $F(x)=\mathrm{\Phi }(f(x))$. Thus $F=\mathrm{\Phi }\circ f$ and belongs to $A(\mathbb{T})$.