The Wiener algebra and Wiener’s lemma

1 Introduction

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

$${\parallel f\parallel }_{L^1(𝕋)}=\frac{1}{2\pi }{\int }_{𝕋}|f(t)|𝑑t.$$

For $f,g\in L^1(𝕋)$, we define

$$(f*g)(t)=\frac{1}{2\pi }{\int }_{𝕋}f(\tau )g(t-\tau )𝑑\tau ,t\in 𝕋.$$

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

$${\parallel f*g\parallel }_{L^1(𝕋)}\le {\parallel f\parallel }_{L^1(𝕋)}{\parallel g\parallel }_{L^1(𝕋)}.$$

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

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

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

We define $c_0(\mathbb{Z})$ to be the collection of those $F:\mathbb{Z}\to ℂ$ such that $|F(k)|\to 0$ as $|k|\to \mathrm{\infty }$. For $f\in L^1(𝕋)$, the Riemann-Lebesgue lemma tells us that $\widehat{f}\in c_0(\mathbb{Z})$.

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

$${\parallel F\parallel }_{{\mathrm{\ell }}^1(\mathbb{Z})}=\sum _{k\in \mathbb{Z}}|F(k)|.$$

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

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

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

$${\parallel F*G\parallel }_{{\mathrm{\ell }}^1(\mathbb{Z})}\le {\parallel F\parallel }_{{\mathrm{\ell }}^1(\mathbb{Z})}{\parallel G\parallel }_{{\mathrm{\ell }}^1(\mathbb{Z})}.$$

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

For $f\in L^1(𝕋)$ and $n\ge 0$ we define $S_n(f)\in C(𝕋)$ by

$$S_n(f)(t)=\sum _{|k|\le n}\widehat{f}(k)e^{ikt},t\in 𝕋.$$

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

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

$${\parallel f\parallel }_{{\mathrm{Lip}}_{\alpha }(𝕋)}={\parallel f\parallel }_{C(𝕋)}+\underset{t\in 𝕋,h\ne 0}{sup}\frac{|f(t+h)-f(t)|}{{|h|}^{\alpha }}.$$

2 Total variation

For $f:𝕋\to ℂ$, we define

If then we say that $f$ is of bounded variation, and we define $BV(𝕋)$ to be the set of functions $𝕋\to ℂ$ of bounded variation. We define

$${\parallel f\parallel }_{BV(𝕋)}=\underset{t\in 𝕋}{sup}|f(t)|+\mathrm{var}(f).$$

This is a norm on $BV(𝕋)$, with which $BV(𝕋)$ is a Banach algebra.

Theorem 1.

If $f\in BV(𝕋)$, then

$$|\widehat{f}(n)|\le \frac{\mathrm{var}(f)}{2\pi |n|},n\in \mathbb{Z},n\ne 0.$$

Proof.

Integrating by parts,

$$\widehat{f}(n)=\frac{1}{2\pi }{\int }_{𝕋}f(t)e^{-int}𝑑t=-\frac{1}{2\pi }{\int }_{𝕋}\frac{e^{-int}}{-in}𝑑f(t)=\frac{1}{2\pi in}{\int }_{𝕋}{e}^{-int}𝑑f(t),$$

hence

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

∎

3 Absolutely convergent Fourier series

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

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

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

$${\parallel f\parallel }_{A(𝕋)}={\parallel \widehat{f}\parallel }_{{\mathrm{\ell }}^1(\mathbb{Z})}.$$

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

Theorem 2.

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

$$\widehat{f}(k)=o(k^{-1}),|k|\to \mathrm{\infty }.$$

Proof.

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

$ℱ(f^{\prime })(k)$ $={\displaystyle \frac{1}{2\pi }}{\displaystyle {\int }_{𝕋}}{f}^{\prime }(t)e^{-ikt}𝑑t$

$={{\displaystyle \frac{1}{2\pi }}f(t)e^{-ikt}|}_0^{2\pi }-{\displaystyle \frac{1}{2\pi }}{\displaystyle {\int }_{𝕋}}f(t)(-ik{e}^{-ikt})𝑑t$

$=ikℱ(f)(k).$

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

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

∎

Theorem 3.

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

$${\parallel f\parallel }_{A(𝕋)}\le {\parallel f\parallel }_{L^1(𝕋)}+{\left(2\sum _{k=1}^{\mathrm{\infty }}{k}^{-2}\right)}^{1/2}{\parallel f^{\prime }\parallel }_{L^2(𝕋)}.$$

Proof.

First,

$$|\widehat{f}(0)|=\left|\frac{1}{2\pi }{\int }_{𝕋}f(t)𝑑t\right|\le {\parallel f\parallel }_{L^1(𝕋)}.$$

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

$$ℱ(f^{\prime })(k)=ikℱ(f)(k).$$

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

${\parallel f\parallel }_{A(𝕋)}$ $=|\widehat{f}(0)|+{\displaystyle \sum _{k\ne 0}}|\widehat{f}(k)|$

$=|\widehat{f}(0)|+{\displaystyle \sum _{k\ne 0}}{|k|}^{-1}|ℱ(f^{\prime })(k)|$

$\le {\parallel f\parallel }_{L^1(𝕋)}+{\left({\displaystyle \sum _{k\ne 0}}{|k|}^{-2}\right)}^{1/2}{\left({\displaystyle \sum _{k\ne 0}}{|ℱ(f^{\prime })(k)|}^2\right)}^{1/2}$

$={\parallel f\parallel }_{L^1(𝕋)}+{\left(2{\displaystyle \sum _{k=1}^{\mathrm{\infty }}}{k}^{-2}\right)}^{1/2}{\parallel ℱ(f^{\prime })\parallel }_{{\mathrm{\ell }}^2(\mathbb{Z})}.$

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

We now prove that if $\alpha >\frac{1}{2}$, then ${\mathrm{Lip}}_{\alpha }(𝕋)\subset A(𝕋)$, and the inclusion map is a bounded linear operator.

Theorem 4.

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

$${\parallel f\parallel }_{A(𝕋)}\le c_{\alpha }{\parallel f\parallel }_{{\mathrm{Lip}}_{\alpha }(𝕋)},$$

with

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

Proof.

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

$$f_h(t)=f(t-h),t\in 𝕋,$$

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

$ℱ(f_h)(n)$ $={\displaystyle \frac{1}{2\pi }}{\displaystyle {\int }_{𝕋}}f(t-h)e^{-int}𝑑t$

$={\displaystyle \frac{1}{2\pi }}{\displaystyle {\int }_{𝕋}}f(t)e^{-in(t+h)}𝑑t$

$=e^{-inh}ℱ(f)(n).$

Thus

$$ℱ(f_h-f)(n)=(e^{-inh}-1)\widehat{f}(n),n\in \mathbb{Z}.$$ (1)

For $m\ge 0$ and for $n\in \mathbb{Z}$ such that , let

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

Then

If $n>0$ this implies that

and so

$$|e^{-in{h}_m}-1|=2\sin \frac{n{h}_m}{2}\ge 2\sin \frac{\pi }{3}=\sqrt{3},$$

and if this implies that

and so

$$|e^{-in{h}_m}-1|\ge \sqrt{3}.$$

This gives us

$\le {\displaystyle \sum _{n\in \mathbb{Z}}}{|e^{-in{h}_m}-1|}^2{|\widehat{f}(n)|}^2.$

Using (1) and Parseval’s theorem we have

$$\sum _{n\in \mathbb{Z}}{|e^{-in{h}_m}-1|}^2{|\widehat{f}(n)|}^2={\parallel ℱ(f_{h_m}-f)\parallel }_{{\mathrm{\ell }}^2(\mathbb{Z})}^2={\parallel f_{h_m}-f\parallel }_{L^2(𝕋)}^2,$$

and thus

Furthermore, for $g\in L^{\mathrm{\infty }}(𝕋)$ we have ${\parallel g\parallel }_{L^2(𝕋)}\le {\parallel g\parallel }_{L^{\mathrm{\infty }}(𝕋)}$, so

$\le {\parallel f_{h_m}-f\parallel }_{L^{\mathrm{\infty }}(𝕋)}^2$

$\le {\parallel f\parallel }_{{\mathrm{Lip}}_{\alpha }(𝕋)}^2\cdot h_m^{2\alpha }$

$={\left({\displaystyle \frac{2\pi }{3\cdot 2^m}}\right)}^{2\alpha }{\parallel f\parallel }_{{\mathrm{Lip}}_{\alpha }(𝕋)}^2.$

By the Cauchy-Schwarz inequality, because there are $\le 2^{m+1}$ nonzero terms in ,

$\le 2^{\frac{m+1}{2}}{\left({\displaystyle \frac{2\pi }{3\cdot 2^m}}\right)}^{\alpha }{\parallel f\parallel }_{{\mathrm{Lip}}_{\alpha }(𝕋)}$

$=2^{m\left(\frac{1}{2}-\alpha \right)}\cdot 2^{1/2}{\left({\displaystyle \frac{2\pi }{3}}\right)}^{\alpha }\cdot {\parallel f\parallel }_{{\mathrm{Lip}}_{\alpha }(𝕋)}.$

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

${\displaystyle \sum _{n\in \mathbb{Z}}}|\widehat{f}(n)|$

$\le |\widehat{f}(0)|+{\displaystyle \sum _{m=0}^{\mathrm{\infty }}}{2}^{m\left(\frac{1}{2}-\alpha \right)}\cdot 2^{1/2}{\left({\displaystyle \frac{2\pi }{3}}\right)}^{\alpha }\cdot {\parallel f\parallel }_{{\mathrm{Lip}}_{\alpha }(𝕋)}$

$=|\widehat{f}(0)|+2^{1/2}{\left({\displaystyle \frac{2\pi }{3}}\right)}^{\alpha }{\parallel f\parallel }_{{\mathrm{Lip}}_{\alpha }(𝕋)}{\displaystyle \sum _{m=0}^{\mathrm{\infty }}}{2}^{m\left(\frac{1}{2}-\alpha \right)}$

$=|\widehat{f}(0)|+2^{1/2}{\left({\displaystyle \frac{2\pi }{3}}\right)}^{\alpha }{\parallel f\parallel }_{{\mathrm{Lip}}_{\alpha }(𝕋)}{\displaystyle \frac{1}{1-2^{\frac{1}{2}-\alpha }}}$

As

$$|\widehat{f}(0)|\le {\parallel f\parallel }_{L^1(𝕋)}\le {\parallel f\parallel }_{L^{\mathrm{\infty }}(𝕋)}\le {\parallel f\parallel }_{{\mathrm{Lip}}_{\alpha }(𝕋)},$$

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

$$\sum _{n\in \mathbb{Z}}|\widehat{f}(n)|\le c_{\alpha }{\parallel f\parallel }_{{\mathrm{Lip}}_{\alpha }(𝕋)},$$

completing the proof. ∎

We now prove that if $\alpha >0$, then $BV(𝕋)\cap {\mathrm{Lip}}_{\alpha }(𝕋)\subset A(𝕋)$.

Theorem 5.

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

$${\parallel f_h-f\parallel }_{L^2(𝕋)}^2\le \frac{1}{2\pi }{h}^{1+\alpha }{\parallel f\parallel }_{{\mathrm{Lip}}_{\alpha }(𝕋)}\mathrm{var}(f),h>0.$$

and $f\in A(𝕋)$.

Proof.

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

${\parallel f_h-f\parallel }_{L^2(𝕋)}^2$ $={\displaystyle \frac{1}{2\pi }}{\displaystyle {\int }_0^{2\pi }}{|f_h(t)-f(t)|}^2𝑑t$

$={\displaystyle \frac{1}{2\pi }}{\displaystyle \sum _{j=1}^N}{\displaystyle {\int }_{(j-1)h}^{jh}}{|f_h(t)-f(t)|}^2𝑑t$

$={\displaystyle \frac{1}{2\pi }}{\displaystyle \sum _{j=1}^N}{\displaystyle {\int }_0^h}{|f_{jh}(t)-f_{(j-1)h}(t)|}^2𝑑t$

$={\displaystyle \frac{1}{2\pi }}{\displaystyle {\int }_0^h}{\displaystyle \sum _{j=1}^N}{|f_{jh}(t)-f_{(j-1)h}(t)|}^{2}dt$

$\le {\displaystyle \frac{1}{2\pi }}{\parallel f_h-f\parallel }_{L^{\mathrm{\infty }}(𝕋)}{\displaystyle {\int }_0^h}{\displaystyle \sum _{j=1}^N}|f_{jh}(t)-f_{(j-1)h}(t)|dt$

$\le {\displaystyle \frac{1}{2\pi }}{\parallel f_h-f\parallel }_{L^{\mathrm{\infty }}(𝕋)}{\displaystyle {\int }_0^h}\mathrm{var}(f)𝑑t.$

As $f\in {\mathrm{Lip}}_{\alpha }(𝕋)$, ${\parallel f_h-f\parallel }_{L^{\mathrm{\infty }}(𝕋)}\le h^{\alpha }{\parallel f\parallel }_{{\mathrm{Lip}}_{\alpha }(𝕋)}$, hence

$${\parallel f_h-f\parallel }_{L^2(𝕋)}^2\le \frac{1}{2\pi }{h}^{1+\alpha }{\parallel f\parallel }_{{\mathrm{Lip}}_{\alpha }(𝕋)}\mathrm{var}(f).$$

∎

4 Wiener’s lemma

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

$${\parallel f\parallel }_{C^k(𝕋)}=\sum _{j=0}^k{\parallel f^{(j)}\parallel }_{C(𝕋)}.$$

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

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

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

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

$$\mathrm{\Gamma }(f)(h)=h(f),h\in {\mathrm{\Delta }}_B.$$

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

We now prove that if $f\in A(𝕋)$ and does not vanish, then $f$ is invertible in $A(𝕋)$. We call this statement Wiener’s lemma.

Theorem 6 (Wiener’s lemma).

If $f\in A(𝕋)$ and $f(t)\ne 0$ for all $t\in 𝕋$, then $1/f\in A(𝕋)$.

Proof.

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

$$|\lambda |\le \parallel w\parallel {\parallel u\parallel }_{A(𝕋)}=1$$

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

$$|{\lambda }^{-1}|\le \parallel w\parallel {\parallel u^{-1}\parallel }_{A(𝕋)}=1,$$

hence $|\lambda |=1$. Then there is some $t_w\in 𝕋$ such that $\lambda =e^{i{t}_w}$. For $n\in \mathbb{Z}$,

$$w(u^n)={\lambda }^n=e^{in{t}_w}.$$

If $P(t)={\sum }_{|n|\le N}{a}_n{e}^{int}$ is a trigonometric polynomial, then

$$w(P)=w\left(\sum _{|n|\le N}{a}_n{u}^n\right)=\sum _{|n|\le N}{a}_{n}w{(u)}^n=\sum _{|n|\le N}{a}_n{e}^{in{t}_w}=P(t_w).$$ (2)

For $g\in A(𝕋)$, if $ϵ>0$, then there is some $N$ such that . Using (2) and the fact that ${\parallel g\parallel }_{C(𝕋)}\le {\parallel g\parallel }_{A(𝕋)}$,

$|w(g)-g(t_w)|$ $\le |w(g)-w(S_N(g))|+|w(S_N(g))-S_N(g)(t_w)|$

$+|S_N(g)(t_w)-g(t_w)|$

$=|w(g-S_N(g))|+|S_N(g)(t_w)-f(t_w)|$

$\le \parallel w\parallel {\parallel g-S_N(g)\parallel }_{A(𝕋)}+{\parallel S_N(g)-g\parallel }_{C(𝕋)}$

$\le \parallel w\parallel {\parallel g-S_N(g)\parallel }_{A(𝕋)}+{\parallel g-S_N(g)\parallel }_{A(𝕋)}$

Because this is true for all $ϵ>0$, it follows that $w(g)=g(t_w)$.

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

$$\mathrm{\Gamma }(f)(w)=w(f)=f(t_w)\ne 0.$$

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

Proof.

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

$$g=\frac{{|f|}^2}{{\parallel f\parallel }_{C(𝕋)}^2}=\frac{f{f}^{*}}{{\parallel f\parallel }_{C(𝕋)}^2}\in A(𝕋),$$

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

Because $g$ is continuous and $g(t)\ne 0$ for all $t\in 𝕋$,

$$\delta =\underset{t\in 𝕋}{inf}g(t)>0;$$

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

For $ϵ>0$, there is some $N$ such that . Now, if $P$ is a trigonometric polynomial of degree $M$ then using the Cauchy-Schwarz inequality and Parseval’s theorem,

${\parallel P\parallel }_{A(𝕋)}$ $={\parallel \widehat{P}\parallel }_{{\mathrm{\ell }}^1(\mathbb{Z})}$

$\le {(2M+1)}^{1/2}{\parallel \widehat{P}\parallel }_{{\mathrm{\ell }}^2(\mathbb{Z})}$

$={(2M+1)}^{1/2}{\parallel P\parallel }_{L^2(𝕋)}$

$\le {(2M+1)}^{1/2}{\parallel P\parallel }_{L^{\mathrm{\infty }}(𝕋)}.$

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

$$h^k={(P+r)}^k=\sum _{j=0}^k\left(\genfrac{}{}{0pt}{}{k}{j}\right)P^j{r}^{k-j},$$

and using this and ${\parallel P^j\parallel }_{A(𝕋)}\le {(2jN+1)}^{1/2}{\parallel P^j\parallel }_{L^{\mathrm{\infty }}(𝕋)}$,

${\parallel h^k\parallel }_{A(𝕋)}$ $\le {\displaystyle \sum _{j=0}^k}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{j}}\right){\parallel P^j\parallel }_{A(𝕋)}{\parallel r^{k-j}\parallel }_{A(𝕋)}$

$\le {\displaystyle \sum _{j=0}^k}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{j}}\right){\parallel P^j\parallel }_{A(𝕋)}{\parallel h-S_N(h)\parallel }_{A(𝕋)}^{k-j}$

$\le {\displaystyle \sum _{j=0}^k}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{j}}\right){(2jN+1)}^{1/2}{\parallel P^j\parallel }_{L^{\mathrm{\infty }}(𝕋)}{ϵ}^{k-j}$

$\le {(2kN+1)}^{1/2}{\displaystyle \sum _{j=0}^k}\left({\displaystyle \genfrac{}{}{0pt}{}{k}{j}}\right){\parallel P\parallel }_{L^{\mathrm{\infty }}(𝕋)}^j{ϵ}^{k-j}$

$={(2kN+1)}^{1/2}{({\parallel P\parallel }_{L^{\mathrm{\infty }}(𝕋)}+ϵ)}^k.$

Because

${\parallel P\parallel }_{L^{\mathrm{\infty }}(𝕋)}$ $\le {\parallel h-S_N(h)\parallel }_{L^{\mathrm{\infty }}(𝕋)}+{\parallel h\parallel }_{L^{\mathrm{\infty }}(𝕋)}$

$\le {\parallel h-S_N(h)\parallel }_{A(𝕋)}+{\parallel h\parallel }_{L^{\mathrm{\infty }}(𝕋)}$

we have

$${\parallel h^k\parallel }_{A(𝕋)}\le {(2kN+1)}^{1/2}{({\parallel h\parallel }_{L^{\mathrm{\infty }}(𝕋)}+2ϵ)}^k={(2kN+1)}^{1/2}{(1-\delta +2ϵ)}^k.$$

Take some , so that . Then with $N=N(ϵ)$,

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

$${\sigma }_A(f)=\{\lambda \in ℂ:f-\lambda \notin U(A)\},$$

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

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 $f^{-1}\in A$.

Lemma 7.

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

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

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 $\lambda \notin {\sigma }_A(f)$ then $f-\lambda \in U(A)\subset U(B)$, hence $\lambda \notin {\sigma }_B(f)$. Therefore ${\sigma }_B(f)\subset {\sigma }_A(f)$. If $\lambda \notin {\sigma }_B(f)$ 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 $\lambda \notin {\sigma }_A(f)$, 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 $0\notin {\sigma }_B(f)$. Then $0\notin {\sigma }_A(f)$, meaning that $f\in U(A)$. ∎

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

The Wiener-Lévy theorem states that if $f\in A(𝕋)$, $\mathrm{\Omega }\subset ℂ$ is an open set containing $f(𝕋)$, and $F:\mathrm{\Omega }\to ℂ$ is holomorphic, then $F\circ f\in A(𝕋)$. In particular, if $f\in A(𝕋)$ does not vanish, then $\mathrm{\Omega }=ℂ\setminus \{0\}$ is an open set containing $f(𝕋)$ and $F(z)=\frac{1}{z}$ is a holomorphic function on $\mathrm{\Omega }$, and hence $F\circ f(t)=\frac{1}{f(t)}$ belongs to $A(𝕋)$, which is the statement of Wiener’s lemma.