# The Wiener algebra and Wiener’s lemma

Jordan Bell
January 17, 2015

## 1 Introduction

Let $\mathbb{T}=\mathbb{R}/2\pi\mathbb{Z}$. For $f\in L^{1}(\mathbb{T})$ we define

 $\left\|f\right\|_{L^{1}(\mathbb{T})}=\frac{1}{2\pi}\int_{\mathbb{T}}|f(t)|dt.$

For $f,g\in L^{1}(\mathbb{T})$, we define

 $(f*g)(t)=\frac{1}{2\pi}\int_{\mathbb{T}}f(\tau)g(t-\tau)d\tau,\qquad t\in% \mathbb{T}.$

$f*g\in L^{1}(\mathbb{T})$, and satisfies Young’s inequality

 $\left\|f*g\right\|_{L^{1}(\mathbb{T})}\leq\left\|f\right\|_{L^{1}(\mathbb{T})}% \left\|g\right\|_{L^{1}(\mathbb{T})}.$

With convolution as the operation, $L^{1}(\mathbb{T})$ is a commutative Banach algebra.

For $f\in L^{1}(\mathbb{T})$, we define $\hat{f}:\mathbb{Z}\to\mathbb{C}$ by

 $\hat{f}(k)=\frac{1}{2\pi}\int_{\mathbb{T}}f(t)e^{-ikt}dt,\qquad k\in\mathbb{Z}.$

We define $c_{0}(\mathbb{Z})$ to be the collection of those $F:\mathbb{Z}\to\mathbb{C}$ such that $|F(k)|\to 0$ as $|k|\to\infty$. For $f\in L^{1}(\mathbb{T})$, the Riemann-Lebesgue lemma tells us that $\hat{f}\in c_{0}(\mathbb{Z})$.

We define $\ell^{1}(\mathbb{Z})$ to be the set of functions $F:\mathbb{Z}\to\mathbb{C}$ such that

 $\left\|F\right\|_{\ell^{1}(\mathbb{Z})}=\sum_{k\in\mathbb{Z}}|F(k)|.$

For $F,G\in\ell^{1}(\mathbb{Z})$, we define

 $(F*G)(k)=\sum_{j\in\mathbb{Z}}F(j)G(k-j).$

$F*G\in\ell^{1}(\mathbb{Z})$, and satisfies Young’s inequality

 $\left\|F*G\right\|_{\ell^{1}(\mathbb{Z})}\leq\left\|F\right\|_{\ell^{1}(% \mathbb{Z})}\left\|G\right\|_{\ell^{1}(\mathbb{Z})}.$

$\ell^{1}(\mathbb{Z})$ is a commutative Banach algebra, with unity

 $F(k)=\begin{cases}1&k=0,\\ 0&k\neq 0.\end{cases}$

For $f\in L^{1}(\mathbb{T})$ and $n\geq 0$ we define $S_{n}(f)\in C(\mathbb{T})$ by

 $S_{n}(f)(t)=\sum_{|k|\leq n}\hat{f}(k)e^{ikt},\qquad t\in\mathbb{T}.$

For $0<\alpha<1$, we define $\mathrm{Lip}_{\alpha}(\mathbb{T})$ to be the collection of those functions $f:\mathbb{T}\to\mathbb{C}$ such that

 $\sup_{t\in\mathbb{T},h\neq 0}\frac{|f(t+h)-f(t)|}{|h|^{\alpha}}<\infty.$

For $f\in\mathrm{Lip}_{\alpha}(\mathbb{T})$, we define

 $\left\|f\right\|_{\mathrm{Lip}_{\alpha}(\mathbb{T})}=\left\|f\right\|_{C(% \mathbb{T})}+\sup_{t\in\mathbb{T},h\neq 0}\frac{|f(t+h)-f(t)|}{|h|^{\alpha}}.$

## 2 Total variation

For $f:\mathbb{T}\to\mathbb{C}$, we define

 $\mathrm{var}(f)=\sup\left\{\sum_{i=1}^{n}|f(t_{i})-f(t_{i-1})|:n\geq 1,0=t_{0}% <\cdots

If $\mathrm{var}(f)<\infty$ then we say that $f$ is of bounded variation, and we define $BV(\mathbb{T})$ to be the set of functions $\mathbb{T}\to\mathbb{C}$ of bounded variation. We define

 $\left\|f\right\|_{BV(\mathbb{T})}=\sup_{t\in\mathbb{T}}|f(t)|+\mathrm{var}(f).$

This is a norm on $BV(\mathbb{T})$, with which $BV(\mathbb{T})$ is a Banach algebra.11 1 N. L. Carothers, Real Analysis, p. 206, Theorem 13.4.

###### Theorem 1.

If $f\in BV(\mathbb{T})$, then

 $|\hat{f}(n)|\leq\frac{\mathrm{var}(f)}{2\pi|n|},\qquad n\in\mathbb{Z},n\neq 0.$
###### Proof.

Integrating by parts,

 $\hat{f}(n)=\frac{1}{2\pi}\int_{\mathbb{T}}f(t)e^{-int}dt=-\frac{1}{2\pi}\int_{% \mathbb{T}}\frac{e^{-int}}{-in}df(t)=\frac{1}{2\pi in}\int_{\mathbb{T}}e^{-int% }df(t),$

hence

 $|\hat{f}(n)|\leq\frac{1}{2\pi|n|}\mathrm{var}(f).$

## 3 Absolutely convergent Fourier series

Suppose that $f\in L^{1}(\mathbb{T})$ and that $\hat{f}\in\ell^{1}(\mathbb{Z})$. For $n\geq m$,

 $\left\|S_{n}(f)-S_{m}(f)\right\|_{C(\mathbb{T})}=\sup_{t\in\mathbb{T}}\left|% \sum_{m<|k|\leq n}\hat{f}(k)e^{ikt}\right|\leq\sum_{m<|k|\leq n}|\hat{f}(k)|,$

and because $\hat{f}\in\ell^{1}(\mathbb{Z})$ it follows that $S_{n}(f)$ converges to some $g\in C(\mathbb{T})$. We check that $f(t)=g(t)$ for almost all $t\in\mathbb{T}$.

We define $A(\mathbb{T})$ to be the collection of those $f\in C(\mathbb{T})$ such that $\hat{f}\in\ell^{1}(\mathbb{Z})$, and we define

 $\left\|f\right\|_{A(\mathbb{T})}=\left\|\hat{f}\right\|_{\ell^{1}(\mathbb{Z})}.$

$A(\mathbb{T})$ is a commutative Banach algebra, with unity $t\mapsto 1$, and the Fourier transform is an isomorphism of Banach algebras $\mathscr{F}:A(\mathbb{T})\to\ell^{1}(\mathbb{Z})$. We call $A(\mathbb{T})$ the Wiener algebra. The inclusion map $A(\mathbb{T})\subset C(\mathbb{T})$ has norm $1$.

###### Theorem 2.

If $f:\mathbb{T}\to\mathbb{C}$ is absolutely continuous, then

 $\hat{f}(k)=o(k^{-1}),\qquad|k|\to\infty.$
###### Proof.

Because $f$ is absolutely continuous, the fundamental theorem of calculus tells us that $f^{\prime}\in L^{1}(\mathbb{T})$. Doing integration by parts, for $k\in\mathbb{Z}$ we have

 $\displaystyle\mathscr{F}(f^{\prime})(k)$ $\displaystyle=\frac{1}{2\pi}\int_{\mathbb{T}}f^{\prime}(t)e^{-ikt}dt$ $\displaystyle=\frac{1}{2\pi}f(t)e^{-ikt}\Big{|}_{0}^{2\pi}-\frac{1}{2\pi}\int_% {\mathbb{T}}f(t)(-ike^{-ikt})dt$ $\displaystyle=ik\mathscr{F}(f)(k).$

The Riemann-Lebesgue lemma tells us that $\mathscr{F}(f^{\prime})(k)=o(1)$, so

 $\mathscr{F}(f)(k)=o\left(\frac{1}{k}\right),\qquad|k|\to\infty.$

###### Theorem 3.

If $f:\mathbb{T}\to\mathbb{C}$ is absolutely continuous and $f^{\prime}\in L^{2}(\mathbb{T})$, then

 $\left\|f\right\|_{A(\mathbb{T})}\leq\left\|f\right\|_{L^{1}(\mathbb{T})}+\left% (2\sum_{k=1}^{\infty}k^{-2}\right)^{1/2}\left\|f^{\prime}\right\|_{L^{2}(% \mathbb{T})}.$
###### Proof.

First,

 $|\hat{f}(0)|=\left|\frac{1}{2\pi}\int_{\mathbb{T}}f(t)dt\right|\leq\left\|f% \right\|_{L^{1}(\mathbb{T})}.$

Next, because $f$ is absolutely continuous, by the fundamental theorem of calculus we have $f^{\prime}\in L^{1}(\mathbb{T})$, and for $k\in\mathbb{Z}$,

 $\mathscr{F}(f^{\prime})(k)=ik\mathscr{F}(f)(k).$

Using the Cauchy-Schwarz inequality, and since $\mathscr{F}(f^{\prime})(0)=0$,

 $\displaystyle\left\|f\right\|_{A(\mathbb{T})}$ $\displaystyle=|\hat{f}(0)|+\sum_{k\neq 0}|\hat{f}(k)|$ $\displaystyle=|\hat{f}(0)|+\sum_{k\neq 0}|k|^{-1}|\mathscr{F}(f^{\prime})(k)|$ $\displaystyle\leq\left\|f\right\|_{L^{1}(\mathbb{T})}+\left(\sum_{k\neq 0}|k|^% {-2}\right)^{1/2}\left(\sum_{k\neq 0}|\mathscr{F}(f^{\prime})(k)|^{2}\right)^{% 1/2}$ $\displaystyle=\left\|f\right\|_{L^{1}(\mathbb{T})}+\left(2\sum_{k=1}^{\infty}k% ^{-2}\right)^{1/2}\left\|\mathscr{F}(f^{\prime})\right\|_{\ell^{2}(\mathbb{Z})}.$

By Parseval’s theorem we have $\left\|\mathscr{F}(f^{\prime})\right\|_{\ell^{2}(\mathbb{Z})}=\left\|f^{\prime% }\right\|_{L^{2}(\mathbb{T})}$, completing the proof. ∎

We now prove that if $\alpha>\frac{1}{2}$, then $\mathrm{Lip}_{\alpha}(\mathbb{T})\subset A(\mathbb{T})$, and the inclusion map is a bounded linear operator.22 2 Yitzhak Katznelson, An Introduction to Harmonic Analysis, third ed., p. 34, Theorem 6.3.

###### Theorem 4.

If $\alpha>\frac{1}{2}$, then $\mathrm{Lip}_{\alpha}(\mathbb{T})\subset A(\mathbb{T})$, and for any $f\in\mathrm{Lip}_{\alpha}(\mathbb{T})$ we have

 $\left\|f\right\|_{A(\mathbb{T})}\leq c_{\alpha}\left\|f\right\|_{\mathrm{Lip}_% {\alpha}(\mathbb{T})},$

with

 $c_{\alpha}=1+2^{1/2}\left(\frac{2\pi}{3}\right)^{\alpha}\frac{1}{1-2^{\frac{1}% {2}-\alpha}}.$
###### Proof.

For $f:\mathbb{T}\to\mathbb{C}$ and $h\in\mathbb{R}$, we define

 $f_{h}(t)=f(t-h),\qquad t\in\mathbb{T},$

which satisfies, for $n\in\mathbb{Z}$,

 $\displaystyle\mathscr{F}(f_{h})(n)$ $\displaystyle=\frac{1}{2\pi}\int_{\mathbb{T}}f(t-h)e^{-int}dt$ $\displaystyle=\frac{1}{2\pi}\int_{\mathbb{T}}f(t)e^{-in(t+h)}dt$ $\displaystyle=e^{-inh}\mathscr{F}(f)(n).$

Thus

 $\mathscr{F}(f_{h}-f)(n)=(e^{-inh}-1)\hat{f}(n),\qquad n\in\mathbb{Z}.$ (1)

For $m\geq 0$ and for $n\in\mathbb{Z}$ such that $2^{m}\leq|n|<2^{m+1}$, let

 $h_{m}=\frac{2\pi}{3}\cdot 2^{-m}.$

Then

 $\frac{2\pi}{3}=2^{m}\cdot\frac{2\pi}{3}\cdot 2^{-m}\leq|nh_{m}|<2^{m+1}\cdot% \frac{2\pi}{3}\cdot 2^{-m}=\frac{4\pi}{3}.$

If $n>0$ this implies that

 $\frac{\pi}{3}\leq\frac{nh_{m}}{2}<\frac{2\pi}{3}$

and so

 $|e^{-inh_{m}}-1|=2\sin\frac{nh_{m}}{2}\geq 2\sin\frac{\pi}{3}=\sqrt{3},$

and if $n<0$ this implies that

 $-\frac{2\pi}{3}<\frac{nh_{m}}{2}\leq-\frac{\pi}{3}$

and so

 $|e^{-inh_{m}}-1|\geq\sqrt{3}.$

This gives us

 $\displaystyle\sum_{2^{m}\leq|n|<2^{m+1}}|\hat{f}(n)|^{2}$ $\displaystyle\leq\sum_{2^{m}\leq|n|<2^{m+1}}3|\hat{f}(n)|^{2}$ $\displaystyle\leq\sum_{2^{m}\leq|n|<2^{m+1}}|e^{-inh_{m}}-1|^{2}|\hat{f}(n)|^{2}$ $\displaystyle\leq\sum_{n\in\mathbb{Z}}|e^{-inh_{m}}-1|^{2}|\hat{f}(n)|^{2}.$

Using (1) and Parseval’s theorem we have

 $\sum_{n\in\mathbb{Z}}|e^{-inh_{m}}-1|^{2}|\hat{f}(n)|^{2}=\left\|\mathscr{F}(f% _{h_{m}}-f)\right\|_{\ell^{2}(\mathbb{Z})}^{2}=\left\|f_{h_{m}}-f\right\|_{L^{% 2}(\mathbb{T})}^{2},$

and thus

 $\sum_{2^{m}\leq|n|<2^{m+1}}|\hat{f}(n)|^{2}\leq\left\|f_{h_{m}}-f\right\|_{L^{% 2}(\mathbb{T})}^{2}.$

Furthermore, for $g\in L^{\infty}(\mathbb{T})$ we have $\left\|g\right\|_{L^{2}(\mathbb{T})}\leq\left\|g\right\|_{L^{\infty}(\mathbb{T% })}$, so

 $\displaystyle\sum_{2^{m}\leq|n|<2^{m+1}}|\hat{f}(n)|^{2}$ $\displaystyle\leq\left\|f_{h_{m}}-f\right\|_{L^{\infty}(\mathbb{T})}^{2}$ $\displaystyle\leq\left\|f\right\|_{\mathrm{Lip}_{\alpha}(\mathbb{T})}^{2}\cdot h% _{m}^{2\alpha}$ $\displaystyle=\left(\frac{2\pi}{3\cdot 2^{m}}\right)^{2\alpha}\left\|f\right\|% _{\mathrm{Lip}_{\alpha}(\mathbb{T})}^{2}.$

By the Cauchy-Schwarz inequality, because there are $\leq 2^{m+1}$ nonzero terms in $\sum_{2^{m}\leq|n|<2^{m+1}}|\hat{f}(n)|$,

 $\displaystyle\sum_{2^{m}\leq|n|<2^{m+1}}|\hat{f}(n)|$ $\displaystyle\leq(2^{m+1})^{1/2}\left(\sum_{2^{m}\leq|n|<2^{m+1}}|\hat{f}(n)|^% {2}\right)^{1/2}$ $\displaystyle\leq 2^{\frac{m+1}{2}}\left(\frac{2\pi}{3\cdot 2^{m}}\right)^{% \alpha}\left\|f\right\|_{\mathrm{Lip}_{\alpha}(\mathbb{T})}$ $\displaystyle=2^{m\left(\frac{1}{2}-\alpha\right)}\cdot 2^{1/2}\left(\frac{2% \pi}{3}\right)^{\alpha}\cdot\left\|f\right\|_{\mathrm{Lip}_{\alpha}(\mathbb{T}% )}.$

Then, since $\alpha>\frac{1}{2}$,

 $\displaystyle\sum_{n\in\mathbb{Z}}|\hat{f}(n)|$ $\displaystyle=|\hat{f}(0)|+\sum_{m=0}^{\infty}\sum_{2^{m}\leq|n|<2^{m+1}}|\hat% {f}(n)|$ $\displaystyle\leq|\hat{f}(0)|+\sum_{m=0}^{\infty}2^{m\left(\frac{1}{2}-\alpha% \right)}\cdot 2^{1/2}\left(\frac{2\pi}{3}\right)^{\alpha}\cdot\left\|f\right\|% _{\mathrm{Lip}_{\alpha}(\mathbb{T})}$ $\displaystyle=|\hat{f}(0)|+2^{1/2}\left(\frac{2\pi}{3}\right)^{\alpha}\left\|f% \right\|_{\mathrm{Lip}_{\alpha}(\mathbb{T})}\sum_{m=0}^{\infty}2^{m\left(\frac% {1}{2}-\alpha\right)}$ $\displaystyle=|\hat{f}(0)|+2^{1/2}\left(\frac{2\pi}{3}\right)^{\alpha}\left\|f% \right\|_{\mathrm{Lip}_{\alpha}(\mathbb{T})}\frac{1}{1-2^{\frac{1}{2}-\alpha}}$

As

 $|\hat{f}(0)|\leq\left\|f\right\|_{L^{1}(\mathbb{T})}\leq\left\|f\right\|_{L^{% \infty}(\mathbb{T})}\leq\left\|f\right\|_{\mathrm{Lip}_{\alpha}(\mathbb{T})},$

we have for all $f\in\mathrm{Lip}_{\alpha}(\mathbb{T})$ that

 $\sum_{n\in\mathbb{Z}}|\hat{f}(n)|\leq c_{\alpha}\left\|f\right\|_{\mathrm{Lip}% _{\alpha}(\mathbb{T})},$

completing the proof. ∎

We now prove that if $\alpha>0$, then $BV(\mathbb{T})\cap\mathrm{Lip}_{\alpha}(\mathbb{T})\subset A(\mathbb{T})$.33 3 Yitzhak Katznelson, An Introduction to Harmonic Analysis, third ed., p. 35, Theorem 6.4.

###### Theorem 5.

If $\alpha>0$ and $f\in BV(\mathbb{T})\cap\mathrm{Lip}_{\alpha}(\mathbb{T})$, then

 $\left\|f_{h}-f\right\|_{L^{2}(\mathbb{T})}^{2}\leq\frac{1}{2\pi}h^{1+\alpha}% \left\|f\right\|_{\mathrm{Lip}_{\alpha}(\mathbb{T})}\mathrm{var}(f),\qquad h>0.$

and $f\in A(\mathbb{T})$.

###### Proof.

For $N\geq 1$ and $h=\frac{2\pi}{N}$,

 $\displaystyle\left\|f_{h}-f\right\|_{L^{2}(\mathbb{T})}^{2}$ $\displaystyle=\frac{1}{2\pi}\int_{0}^{2\pi}|f_{h}(t)-f(t)|^{2}dt$ $\displaystyle=\frac{1}{2\pi}\sum_{j=1}^{N}\int_{(j-1)h}^{jh}|f_{h}(t)-f(t)|^{2% }dt$ $\displaystyle=\frac{1}{2\pi}\sum_{j=1}^{N}\int_{0}^{h}|f_{jh}(t)-f_{(j-1)h}(t)% |^{2}dt$ $\displaystyle=\frac{1}{2\pi}\int_{0}^{h}\sum_{j=1}^{N}|f_{jh}(t)-f_{(j-1)h}(t)% |^{2}dt$ $\displaystyle\leq\frac{1}{2\pi}\left\|f_{h}-f\right\|_{L^{\infty}(\mathbb{T})}% \int_{0}^{h}\sum_{j=1}^{N}|f_{jh}(t)-f_{(j-1)h}(t)|dt$ $\displaystyle\leq\frac{1}{2\pi}\left\|f_{h}-f\right\|_{L^{\infty}(\mathbb{T})}% \int_{0}^{h}\mathrm{var}(f)dt.$

As $f\in\mathrm{Lip}_{\alpha}(\mathbb{T})$, $\left\|f_{h}-f\right\|_{L^{\infty}(\mathbb{T})}\leq h^{\alpha}\left\|f\right\|% _{\mathrm{Lip}_{\alpha}(\mathbb{T})}$, hence

 $\left\|f_{h}-f\right\|_{L^{2}(\mathbb{T})}^{2}\leq\frac{1}{2\pi}h^{1+\alpha}% \left\|f\right\|_{\mathrm{Lip}_{\alpha}(\mathbb{T})}\mathrm{var}(f).$

## 4 Wiener’s lemma

For $k\geq 1$, using the product rule $(fg)^{\prime}=f^{\prime}g+fg^{\prime}$ we check that $C^{k}(\mathbb{T})$ is a Banach algebra with the norm

 $\left\|f\right\|_{C^{k}(\mathbb{T})}=\sum_{j=0}^{k}\left\|f^{(j)}\right\|_{C(% \mathbb{T})}.$

If $f\in C^{k}(\mathbb{T})$ and $f(t)\neq 0$ for all $t\in\mathbb{T}$, then the quotient rule tells us that

 $\left(f^{-1}\right)^{\prime}(t)=-\frac{f^{\prime}(t)}{f(t)^{2}},$

using which we get $\frac{1}{f}\in C^{k}(\mathbb{T})$. That is, if $f\in C^{k}(\mathbb{T})$ does not vanish then $f^{-1}=\frac{1}{f}\in C^{k}(\mathbb{T})$.

If $B$ is a commutative unital Banach algebra, a multiplicative linear functional on $B$ is a nonzero algebra homomorphism $B\to\mathbb{C}$, and the collection $\Delta_{B}$ of multiplicative linear functionals on $B$ is called the maximal ideal space of $B$. The Gelfand transform of $f\in B$ is $\Gamma(f):\Delta_{B}\to\mathbb{C}$ defined by

 $\Gamma(f)(h)=h(f),\qquad h\in\Delta_{B}.$

It is a fact that $f\in B$ is invertible if and only if $h(f)\neq 0$ for all $h\in\Delta_{B}$, i.e., $f\in B$ is invertible if and only if $\Gamma(f)$ does not vanish.

We now prove that if $f\in A(\mathbb{T})$ and does not vanish, then $f$ is invertible in $A(\mathbb{T})$. We call this statement Wiener’s lemma.44 4 Yitzhak Katznelson, An Introduction to Harmonic Analysis, third ed., p. 239, Theorem 2.9.

###### Theorem 6 (Wiener’s lemma).

If $f\in A(\mathbb{T})$ and $f(t)\neq 0$ for all $t\in\mathbb{T}$, then $1/f\in A(\mathbb{T})$.

###### Proof.

Let $w:A(\mathbb{T})\to\mathbb{C}$ be a multiplicative linear functional. The fact that $w$ is a multiplicative linear functional implies that $\left\|w\right\|=1$. Define $u(t)=e^{it}$, $t\in\mathbb{T}$, for which $\left\|u\right\|_{A(\mathbb{T})}=1$. We define $\lambda=w(u)$, which satisfies

 $|\lambda|\leq\left\|w\right\|\left\|u\right\|_{A(\mathbb{T})}=1$

and because $\left\|u^{-1}\right\|_{A(\mathbb{T})}=1$ we have $\lambda^{-1}=w(u^{-1})$ and

 $|\lambda^{-1}|\leq\left\|w\right\|\left\|u^{-1}\right\|_{A(\mathbb{T})}=1,$

hence $|\lambda|=1$. Then there is some $t_{w}\in\mathbb{T}$ such that $\lambda=e^{it_{w}}$. For $n\in\mathbb{Z}$,

 $w(u^{n})=\lambda^{n}=e^{int_{w}}.$

If $P(t)=\sum_{|n|\leq N}a_{n}e^{int}$ is a trigonometric polynomial, then

 $w(P)=w\left(\sum_{|n|\leq N}a_{n}u^{n}\right)=\sum_{|n|\leq N}a_{n}w(u)^{n}=% \sum_{|n|\leq N}a_{n}e^{int_{w}}=P(t_{w}).$ (2)

For $g\in A(\mathbb{T})$, if $\epsilon>0$, then there is some $N$ such that $\left\|g-S_{N}(g)\right\|_{A(\mathbb{T})}<\epsilon$. Using (2) and the fact that $\left\|g\right\|_{C(\mathbb{T})}\leq\left\|g\right\|_{A(\mathbb{T})}$,

 $\displaystyle|w(g)-g(t_{w})|$ $\displaystyle\leq|w(g)-w(S_{N}(g))|+|w(S_{N}(g))-S_{N}(g)(t_{w})|$ $\displaystyle+|S_{N}(g)(t_{w})-g(t_{w})|$ $\displaystyle=|w(g-S_{N}(g))|+|S_{N}(g)(t_{w})-f(t_{w})|$ $\displaystyle\leq\left\|w\right\|\left\|g-S_{N}(g)\right\|_{A(\mathbb{T})}+% \left\|S_{N}(g)-g\right\|_{C(\mathbb{T})}$ $\displaystyle\leq\left\|w\right\|\left\|g-S_{N}(g)\right\|_{A(\mathbb{T})}+% \left\|g-S_{N}(g)\right\|_{A(\mathbb{T})}$ $\displaystyle<2\epsilon.$

Because this is true for all $\epsilon>0$, it follows that $w(g)=g(t_{w})$.

Let $\Delta$ be the maximal ideal space of $A(\mathbb{T})$. Then for $w\in\Delta$ there is some $t_{w}\in\mathbb{T}$ such that $w(f)=f(t_{w})$, hence, because $f(t)\neq 0$ for all $t\in\mathbb{T}$,

 $\Gamma(f)(w)=w(f)=f(t_{w})\neq 0.$

That is, $\Gamma(f)$ does not vanish, and therefore $f$ is invertible in $A(\mathbb{T})$. It is then immediate that $f^{-1}(t)=\frac{1}{f(t)}$ for all $t\in\mathbb{T}$, completing the proof. ∎

The above proof of Wiener’s lemma uses the theory of the commutative Banach algebras. The following is a proof of the theorem that does not use the Gelfand transform.55 5 Karlheinz Gröchenig, Wiener’s Lemma: Theme and Variations. An Introduction to Spectral Invariance and Its Applications, p. 180, §5.2.4, in Brigitte Forster and Peter Massopust, eds., Four Short Courses on Harmonic Analysis, pp. 175–234.

###### Proof.

Because $f\in A(\mathbb{T})$, $f^{*}$ defined by $f^{*}(t)=\overline{f(t)}$, $t\in\mathbb{T}$, belongs to $A(\mathbb{T})$. Let

 $g=\frac{|f|^{2}}{\left\|f\right\|_{C(\mathbb{T})}^{2}}=\frac{ff^{*}}{\left\|f% \right\|_{C(\mathbb{T})}^{2}}\in A(\mathbb{T}),$

which satisfies $0 for all $t\in\mathbb{T}$. As $\frac{1}{f}=\frac{f^{*}}{|f|^{2}}=\frac{f^{*}}{\left\|f\right\|_{C(\mathbb{T})% }^{2}g}$, to show that $1/f\in A(\mathbb{T})$ it suffices to show that $\frac{1}{g}\in A(\mathbb{T})$.

Because $g$ is continuous and $g(t)\neq 0$ for all $t\in\mathbb{T}$,

 $\delta=\inf_{t\in\mathbb{T}}g(t)>0;$

if $\delta=1$ then $g=1$, and indeed $\frac{1}{g}\in A(\mathbb{T})$. Otherwise, $\left\|g-1\right\|_{C(\mathbb{T})}=1-\delta<1$. This implies that $g$ is invertible in the Banach algebra $C(\mathbb{T})$ and that $g^{-1}=\sum_{j=0}^{\infty}(1-g)^{j}$ in $C(\mathbb{T})$. Let $h=1-g\in A(\mathbb{T})$.

For $\epsilon>0$, there is some $N$ such that $\left\|h-S_{N}(h)\right\|_{A(\mathbb{T})}<\epsilon$. Now, if $P$ is a trigonometric polynomial of degree $M$ then using the Cauchy-Schwarz inequality and Parseval’s theorem,

 $\displaystyle\left\|P\right\|_{A(\mathbb{T})}$ $\displaystyle=\left\|\hat{P}\right\|_{\ell^{1}(\mathbb{Z})}$ $\displaystyle\leq(2M+1)^{1/2}\left\|\hat{P}\right\|_{\ell^{2}(\mathbb{Z})}$ $\displaystyle=(2M+1)^{1/2}\left\|P\right\|_{L^{2}(\mathbb{T})}$ $\displaystyle\leq(2M+1)^{1/2}\left\|P\right\|_{L^{\infty}(\mathbb{T})}.$

Furthermore, for $j\geq 1$, $P^{j}$ is a trigonometric polynomial of degree $jM$. The binomial theorem tells us, with $P=S_{N}(h)$ and $r=h-P$,

 $h^{k}=(P+r)^{k}=\sum_{j=0}^{k}\binom{k}{j}P^{j}r^{k-j},$

and using this and $\left\|P^{j}\right\|_{A(\mathbb{T})}\leq(2jN+1)^{1/2}\left\|P^{j}\right\|_{L^{% \infty}(\mathbb{T})}$,

 $\displaystyle\left\|h^{k}\right\|_{A(\mathbb{T})}$ $\displaystyle\leq\sum_{j=0}^{k}\binom{k}{j}\left\|P^{j}\right\|_{A(\mathbb{T})% }\left\|r^{k-j}\right\|_{A(\mathbb{T})}$ $\displaystyle\leq\sum_{j=0}^{k}\binom{k}{j}\left\|P^{j}\right\|_{A(\mathbb{T})% }\left\|h-S_{N}(h)\right\|_{A(\mathbb{T})}^{k-j}$ $\displaystyle\leq\sum_{j=0}^{k}\binom{k}{j}(2jN+1)^{1/2}\left\|P^{j}\right\|_{% L^{\infty}(\mathbb{T})}\epsilon^{k-j}$ $\displaystyle\leq(2kN+1)^{1/2}\sum_{j=0}^{k}\binom{k}{j}\left\|P\right\|_{L^{% \infty}(\mathbb{T})}^{j}\epsilon^{k-j}$ $\displaystyle=(2kN+1)^{1/2}(\left\|P\right\|_{L^{\infty}(\mathbb{T})}+\epsilon% )^{k}.$

Because

 $\displaystyle\left\|P\right\|_{L^{\infty}(\mathbb{T})}$ $\displaystyle\leq\left\|h-S_{N}(h)\right\|_{L^{\infty}(\mathbb{T})}+\left\|h% \right\|_{L^{\infty}(\mathbb{T})}$ $\displaystyle\leq\left\|h-S_{N}(h)\right\|_{A(\mathbb{T})}+\left\|h\right\|_{L% ^{\infty}(\mathbb{T})}$ $\displaystyle<\epsilon+\left\|h\right\|_{L^{\infty}(\mathbb{T})},$

we have

 $\left\|h^{k}\right\|_{A(\mathbb{T})}\leq(2kN+1)^{1/2}(\left\|h\right\|_{L^{% \infty}(\mathbb{T})}+2\epsilon)^{k}=(2kN+1)^{1/2}(1-\delta+2\epsilon)^{k}.$

Take some $\epsilon<\frac{\delta}{2}$, so that $1-\delta+2\epsilon<1$. Then with $N=N(\epsilon)$,

 $\sum_{k=0}^{\infty}\left\|h^{k}\right\|_{A(\mathbb{T})}\leq\sum_{k=0}^{\infty}% (2kN+1)^{1/2}(1-\delta+2\epsilon)^{k}=\sqrt{2N}\Phi\left(1-\delta+2\epsilon,-% \frac{1}{2},\frac{1}{2N}\right)<\infty,$

where $\Phi$ is the Lerch transcendent. This implies that the the series $\sum_{k=0}^{\infty}h^{k}$ converges in $A(\mathbb{T})$. We check that $\sum_{k=0}^{\infty}h^{k}$ is the inverse of $1-h$, namely, $g=1-h$ is invertible in $A(\mathbb{T})$, proving the claim. ∎

## 5 Spectral theory

Suppose that $A$ is a commutative Banach algebra with unity $1$. We define $U(A)$ to be the collection of those $f\in A$ such that $f$ is invertible in $A$. It is a fact that $U(A)$ is an open subset of $A$. We define

 $\sigma_{A}(f)=\{\lambda\in\mathbb{C}:f-\lambda\not\in U(A)\},$

called the spectrum of $f$. It is a fact that $\sigma_{A}(f)$ is a nonempty compact subset of $\mathbb{C}$.

If $A\subset B$ are Banach algebras with unity $1$, we say that $A$ is inverse-closed in $B$ if $f\in A$ and $f^{-1}\in B$ together imply that $f^{-1}\in A$.66 6 Karlheinz Gröchenig, Wiener’s Lemma: Theme and Variations. An Introduction to Spectral Invariance and Its Applications, p. 183, §5.2.5, in Brigitte Forster and Peter Massopust, eds., Four Short Courses on Harmonic Analysis, pp. 175–234.

###### Lemma 7.

Suppose that $A\subset B$ are Banach algebras with unity $1$. The following are equivalent:

1. 1.

$A$ is inverse-closed in $B$.

2. 2.

$\sigma_{A}(f)=\sigma_{B}(f)$ for all $f\in A$.

###### Proof.

Assume that $A$ is inverse-closed in $B$ and let $f\in A$. If $\lambda\not\in\sigma_{A}(f)$ then $f-\lambda\in U(A)\subset U(B)$, hence $\lambda\not\in\sigma_{B}(f)$. Therefore $\sigma_{B}(f)\subset\sigma_{A}(f)$. If $\lambda\not\in\sigma_{B}(f)$ then $f-\lambda\in U(B)$. That is, $(f-\lambda)^{-1}\in B$. Because $A$ is inverse-closed in $B$ and $f-\lambda\in A$, we get $(f-\lambda)^{-1}\in A$. Thus $\lambda\not\in\sigma_{A}(f)$, and therefore $\sigma_{A}(f)\subset\sigma_{B}(f)$. We thus have obtained $\sigma_{A}(f)=\sigma_{B}(f)$.

Assume that for all $f\in A$, $\sigma_{A}(f)=\sigma_{B}(f)$. Suppose that $f\in A$ and $f^{-1}\in B$. That is, $f\in U(B)$, so $0\not\in\sigma_{B}(f)$. Then $0\not\in\sigma_{A}(f)$, meaning that $f\in U(A)$. ∎

$A(\mathbb{T})\subset C(\mathbb{T})$ are Banach algebras with unity $1$. Wiener’s lemma states that $A(\mathbb{T})$ is inverse-closed in $C(\mathbb{T})$. It is apparent that for $f\in C(\mathbb{T})$, $\sigma_{C(\mathbb{T})}(f)=f(\mathbb{T})\subset\mathbb{C}$. Therefore, Lemma 7 tells us for $f\in A(\mathbb{T})$ that $\sigma_{A(\mathbb{T})}(f)=f(\mathbb{T})$.

The Wiener-Lévy theorem states that if $f\in A(\mathbb{T})$, $\Omega\subset\mathbb{C}$ is an open set containing $f(\mathbb{T})$, and $F:\Omega\to\mathbb{C}$ is holomorphic, then $F\circ f\in A(\mathbb{T})$.77 7 Karlheinz Gröchenig, Wiener’s Lemma: Theme and Variations. An Introduction to Spectral Invariance and Its Applications, p. 187, Theorem 5.16, in Brigitte Forster and Peter Massopust, eds., Four Short Courses on Harmonic Analysis, pp. 175–234; Walter Rudin, Fourier Analysis on Groups, Chapter 6; N. K. Nikolski (ed.), Functional Analysis I, p. 235; V. P. Havin and N. K. Nikolski (eds.), Commutative Harmonic Analysis II, p. 240, §7.7. In particular, if $f\in A(\mathbb{T})$ does not vanish, then $\Omega=\mathbb{C}\setminus\{0\}$ is an open set containing $f(\mathbb{T})$ and $F(z)=\frac{1}{z}$ is a holomorphic function on $\Omega$, and hence $F\circ f(t)=\frac{1}{f(t)}$ belongs to $A(\mathbb{T})$, which is the statement of Wiener’s lemma.