On powers of operators with spectrum in cantor sets and spectral synthesis

TL;DR

This paper investigates the spectral properties of operators with spectra in certain Cantor sets, establishing conditions under which the growth of inverse powers is polynomial, and explores the implications for spectral synthesis in Beurling algebras.

Contribution

It proves that operators with spectrum in specific Cantor sets and polynomial growth conditions have polynomial inverse growth, and demonstrates the sharpness of these results, linking to spectral synthesis in weighted function spaces.

Findings

Operators with spectrum in $E_{1/q}$ and polynomial growth have polynomial inverse growth.

The result fails if $1/\xi$ is not a Pisot number.

The constant $b(1/q)$ is optimal for the growth condition.

Abstract

For $\xi \in \big( 0, \frac{1}{2} \big)$, let $E_{\xi}$ be the perfect symmetric set associated with $\xi$, that is $$E_{\xi} = \Big\{ \exp \Big( 2i \pi (1-\xi) \sum_{n = 1}^{+\infty} \epsilon_{n} \xi^{n-1} \Big) : \, \epsilon_{n} = 0 \textrm{ or } 1 \quad (n \geq 1) \Big\}$$ and $$b(\xi) = \frac{\log{\frac{1}{\xi}} - \log{2}}{2\log{\frac{1}{\xi}} - \log{2}}.$$ Let $q\geq 3$ be an integer and $s$ be a nonnegative real number. We show that any invertible operator $T$ on a Banach space with spectrum contained in $E_{1/q}$ that satisfies \begin{eqnarray*} & & \big\| T^{n} \big\| = O \big( n^{s} \big), \,n \rightarrow +\infty \\ & \textrm{and} & \big\| T^{-n} \big\| = O \big( e^{n^{\beta}} \big), \, n \rightarrow +\infty \textrm{ for some } \beta < b(1/q),\end{eqnarray*} also satisfies the stronger property $\big\| T^{-n} \big\| = O \big( n^{s} \big), \, n \rightarrow +\infty.$ We also show that this result is false for $E_\xi$ when $1/\xi$ is not a Pisot number and that the constant $b(1/q)$ is sharp. As a consequence we prove that, if $\omega$ is a submulticative weight such that $\omega(n)=(1+n)^s, \, (n \geq 0)$ and $C^{-1} (1+|n|)^s \leq \omega(-n) \leq C e^{n^{\beta}},\, (n\geq 0)$, for some constants $C>0$ and $\beta < b( 1/q),$ then $E_{1/q}$ satisfies spectral synthesis in the Beurling algebra of all continuous functions $f$ on the unit circle $\mathbb{T}$ such that $\sum_{n = -\infty}^{+\infty} | \widehat{f}(n) | \omega (n) < +\infty$.

Equations

$$E_{\xi}=\Big{\{}\exp\Big{(}2i\pi(1-\xi)\displaystyle\sum_{n=1}^{+\infty}\epsilon_{n}\xi^{n-1}\Big{)}:\,\epsilon_{n}=0\textrm{ or }1\quad(n\geq 1)\Big{\}}$$

$$b(\xi )=\frac{\log \frac{1}{\xi }-\log 2}{2\log \frac{1}{\xi }-\log 2}.$$

$$\displaystyle\big{\|}T^{n}\big{\|}=O\big{(}n^{s}\big{)},\,n\rightarrow+\infty$$

$$E_{\xi}=\Big{\{}\exp\Big{(}2i\pi(1-\xi)\displaystyle\sum_{n=1}^{+\infty}\epsilon_{n}\xi^{n-1}\Big{)}:\,\epsilon_{n}=0\textrm{ or }1\quad(n\geq 1)\Big{\}},$$

$$b(\xi )=\frac{\log \frac{1}{\xi }-\log 2}{2\log \frac{1}{\xi }-\log 2}.$$

$$\big{\|}T^{n}\big{\|}=O\big{(}n^{s}\big{)},\,n\rightarrow+\infty,$$

$$\big{\|}T^{-n}\big{\|}=O\big{(}e^{n^{\beta}}\big{)},\,n\rightarrow+\infty\textrm{ for some }\beta<b(1/q),$$

$$\big{\|}T^{-n}\big{\|}=O\big{(}n^{s}\big{)},\,n\rightarrow+\infty.$$

$$\big{\|}T^{-n}\big{\|}=O\big{(}n^{r}\big{)},\,n\rightarrow+\infty\ {\text{ for every }}\ r>s+1/2.$$

$$A_{\omega}(\mathbb{T})=\Big{\{}f\in\mathcal{C}(\mathbb{T}):\,\big{\|}f\big{\|}_{\omega}=\sum_{n=-\infty}^{+\infty}|\widehat{f}(n)|\omega(n)<+\infty\Big{\}},$$

$$I_{\omega}(E)=\Big{\{}f\in A_{s}(\mathbb{T}):\,f=0\ {\text{ on }}\ E\Big{\}},$$

$$\Big{\{}f\in A_{\omega}(\mathbb{T}):\,f=0\ {\text{on a neighborhood of}}\ E\Big{\}}.$$

$$h(I)=\{z\in \mathbb{T}:f(z)=0,\forall f\in I\}.$$

$$\omega (n)=(1+n{)}^s(n\ge 0),$$

$$C^{-1}(1+|n|{)}^s\le \omega (-n)\le C{e}^{n^{\beta }},(n\ge 0),$$

$$\sum _{n\ge 0}|L_n{|}^{\gamma }=\sum _{k\ge 1}{2}^{k-1}{\left(2\pi {\xi }^{k-1}(1-2\xi )\right)}^{\gamma }=\frac{(2\pi (1-2\xi ){)}^{\gamma }}{2{\xi }^{\gamma }}\sum _{k\ge 1}(2{\xi }^{\gamma }{)}^k.$$

$${\int }_0^{2\pi }\frac{1}{d(e^{it},E_{\xi }{)}^{\delta }}dt<+\infty \text{ if and only if }\delta <1-\frac{\log 2}{\log 1/\xi }.$$

$$\omega (n)=(1+n{)}^s,n\ge 0,$$

$$\omega (-n)=O(e^{n^{\beta }}),n\to +\infty ,\text{for some}\beta <b(1/q).$$

$${\omega }_{\gamma }(n)=(1+n{)}^s,n\ge 0\mathrm{and}{\omega }_{\gamma }(-n)=e^{n^{\gamma }},n>0.$$

$$|f(z)|\le e^{-d(z,E_{1/q}{)}^{-\delta }},z\in \mathbb{T}\setminus E_{1/q}.$$

$$\Vert f_n(z^{q^m})-f(z^{q^m}){\Vert }_{\omega }$$

$$\le \underset{k\in \mathbb{Z}}{\sup }\frac{\omega (q^{m}k)}{{\omega }_{\gamma }(k)}\Vert f_n-f{\Vert }_{{\omega }_{\gamma }}$$

$$\le C\Vert f_n-f{\Vert }_{{\omega }_{\gamma }},$$

$$\big{\|}T^{n}\big{\|}=O\big{(}n^{s}\big{)},\,n\rightarrow+\infty,$$

$$\big{\|}T^{-n}\big{\|}=O\big{(}e^{n^{\beta}}\big{)},\,n\rightarrow+\infty\textrm{ for some }\beta<b(1/q),$$

$$\big{\|}T^{-n}\big{\|}=O\big{(}n^{s}\big{)},\,n\rightarrow+\infty.$$

$$f(T)=\sum _{n=-\infty }^{+\infty }\hat{f}(n)T^n.$$

$$T^{-n}=-\sum _{k=1}^{+\infty }\hat{f}(k)T^{k{q}^{m+1}-n},$$

$$\Vert T^{-n}\Vert$$

$$\le \sum _{k=1}^{+\infty }|\hat{f}(k)|(1+k{q}^{m+1}-n{)}^s$$

$$\le q^s{n}^s\sum _{k=1}^{+\infty }|\hat{f}(k)|(1+k{)}^s.$$

Full text

[TABLE]

and

[TABLE]

Let $q\geq 3$ be an integer and $s$ be a nonnegative real number. We show that any invertible operator $T$ on a Banach space with spectrum contained in $E_{1/q}$ that satisfies

[TABLE]

also satisfies the stronger property \displaystyle\big{\|}T^{-n}\big{\|}=O\big{(}n^{s}\big{)},\,n\rightarrow+\infty. We also show that this result is false for $E_{\xi}$ when $1/\xi$ is not a Pisot number and that the constant $b(1/q)$ is sharp. As a consequence we prove that, if $\omega$ is a submulticative weight such that $\displaystyle\omega(n)=(1+n)^{s},\,(n\geq 0)$ and $\displaystyle C^{-1}(1+|n|)^{s}\leq\omega(-n)\leq Ce^{n^{\beta}},\,(n\geq 0)$, for some constants $\displaystyle C>0$ and $\displaystyle\beta<b(1/q),$ then $E_{1/q}$ satisfies spectral synthesis in the Beurling algebra of all continuous functions $f$ on the unit circle $\mathbb{T}$ such that $\sum_{n=-\infty}^{+\infty}|\widehat{f}(n)|\omega(n)<+\infty$.

1. Introduction.

Let $T$ be an invertible operator defined on a Banach space with spectrum ($\mathrm{Sp}(T)$) contained in the unit circle $\mathbb{T}$. Several results show that growth conditions on $\displaystyle(\|T^{n}\|)_{n\geq 0}$ and geometric or arithmetic conditions on $\displaystyle\mathrm{Sp}(T)$ induce a certain behavior on the negative powers $\displaystyle(\|T^{-n}\|)_{n\geq 0}$. For example, it is shown by the author in [15], that if $\displaystyle\sup_{n\geq 0}\|T^{n}\|<+\infty$ and if $\displaystyle\mathrm{Sp}(T)$ is countable then $\displaystyle\sup_{n\geq 0}\|T^{-n}\|<+\infty$ or $\displaystyle\limsup_{n\to+\infty}\frac{\log\|T^{-n}\|}{\sqrt{n}}>0$. See also [6, Theorem 5.6] for a similar result on representations of groups by operators.

For \xi\in\big{(}0,\frac{1}{2}\big{)}, Let $E_{\xi}$ be the perfect symmetric set associated with $\xi$, that is

[TABLE]

and let

[TABLE]

Notice that $E_{1/3}$ is the classical triadic Cantor set. Let $q\geq 3$ be an integer. J. Esterle showed in [8, p. 79] that if $\displaystyle\sup_{n\geq 0}\|T^{n}\|<+\infty$ and $\displaystyle\mathrm{Sp}(T)\subset E_{1/q}$ then $\displaystyle\sup_{n\geq 0}\|T^{-n}\|<+\infty$ or $\displaystyle\limsup_{n\to+\infty}\frac{\log\|T^{-n}\|}{{n}^{\beta}}=+\infty$, for every $\beta<b(1/q)$. On the other hand it is proved in [9, Theorem 2.2 ] that the constant $b(1/q)$ is sharp. For further results of this type see [1], [2], [4], [7], [8], [14], [15]. In this paper we are interested by operators with spectrum contained in $E_{\xi}$ and such that $\displaystyle(\|T^{n}\|)_{n\geq 0}$ has a polynomial growth. Let $s$ be a nonnegative real number. We show (Theorem 2.2) that if

[TABLE]

[TABLE]

and if $\displaystyle\mathrm{Sp}(T)\subset E_{1/q}$, then $T$ satifies the stronger condition

[TABLE]

This extends the result in [8, p. 79] for the case $s=0$ and improves significantly [2, Corollary 2.5], where it is proved that, under the same assumptions,

[TABLE]

On the other hand we show (Propositions 2.3 and 2.4) that the above result is false for $E_{\xi}$ when $1/\xi$ is not a Pisot number and that the constant $b(1/q)$ is sharp.

In section 3, we apply Theorem 2.2 to show that the sets $E_{1/q}$ satisfy spectral synthesis in certain Beurling weighted algebras. First, we recall some notations and definitions. Let \omega=\big{(}\omega(n)\big{)}_{n\in\mathbb{Z}} be a weight, that is a real sequence such that $\omega(n)\geq 1$ and $\omega(n+m)\leq\omega(n)\omega(m)$ for $m,n\in\mathbb{Z}$. We define the Banach algebra

[TABLE]

where $\widehat{f}(n)$ is the $\textrm{n}^{\textrm{th}}$ Fourier coefficient of $f$, and $\mathcal{C}(\mathbb{T})$ the space of all continuous functions on $\mathbb{T}$. In this paper, we will consider only regular algebras, that is algebras $A_{\omega}(\mathbb{T})$ for which the weight satisfies $\displaystyle\sum_{n\in\mathbb{Z}}\frac{\log{\omega(n)}}{1+n^{2}}<+\infty$ (see [13, ex. 7, p. 118]).

Let $E$ be a closed subset of $\mathbb{T}$. We define the closed ideal

[TABLE]

and the ideal $J_{\omega}(E)$ to be the closure in $A_{\omega}(\mathbb{T})$ for the norm $\|\,\|_{\omega}$ of the set

[TABLE]

We say that a function $f$ satisfies spectral synthesis for $E$ in $A_{\omega}(\mathbb{T})$ if $f\in J_{\omega}(E)$, and that $E$ satisfies spectral synthesis in $A_{\omega}(\mathbb{T})$ if $J_{\omega}(E)=I_{\omega}(E).$

Let $I$ be a closed ideal of $A_{\omega}(\mathbb{T})$ such that $h(I)=E$, where

[TABLE]

Then $J_{\omega}(E)\subset I\subset I_{\omega}(E)$ (see [13, p. 224]). Thus the set $E$ satisfies spectral synthesis in $A_{\omega}(\mathbb{T})$ if and only if every closed ideal $I$ of $A_{\omega}(\mathbb{T})$ such that $h(I)=E$ is equal to $I_{\omega}(E)$.

In the case where the weight $\omega$ is defined by $\omega(n)=(1+|n|)^{s}$ for all $n\in\mathbb{Z}$, with $s$ a nonnegative real number, we denote by \big{(}A_{s}(\mathbb{T}),\|\,\|_{s}\big{)} the algebra \big{(}A_{\omega}(\mathbb{T}),\|\,\|_{\omega}\big{)}, and by $I_{s}(E)$ and $J_{s}(E)$ the closed ideals $I_{\omega}(E)$ and $J_{\omega}(E)$. We remark that $A_{0}(\mathbb{T})=A(\mathbb{T})$ is in fact the classical Wiener algebra.

In [10], C. S. Herz proved that $E_{1/q}$ satisfies spectral synthesis in the Wiener algebra $A(\mathbb{T})$ (see also [11, p. 58–59]). In [8], J. Esterle showed that $E_{1/q}$ satisfies spectral synthesis in $A_{\omega}(\mathbb{T})$ for all weights $\omega$ such that $\omega(n)=1\,(n\geq 0)$ and \omega(-n)=O\big{(}e^{n^{\beta}}\big{)}, $n\to+\infty$, for some $\displaystyle\beta<b(1/q).$ Here we extend this result (Theorem 3.5) to a larger class of weights, that is, weights satisfying

[TABLE]

and

[TABLE]

for some constants $s\geq 0$, $\displaystyle C>0$ and $\displaystyle\beta<b(1/q).$

2. Behavior of powers of operators with spectrum in Cantor sets.

Let \xi\in\big{(}0,\frac{1}{2}\big{)}. We write $\mathbb{T}\setminus E_{\xi}=\bigcup_{n\geq 0}L_{n}$, where the $L_{n}$ are the contiguous arcs to $E$. We know from the construction of $E_{\xi}$ (see [12, Chapitre I]) that for every $k\geq 1$, there are exactly $2^{k-1}$ arcs among the $L_{n}$, which are of length $2\pi\xi^{k-1}(1-2\xi)$. Hence, for a real number $\gamma$,

[TABLE]

So, the sum $\sum_{n\geq 0}|L_{n}|^{\gamma}$ is finite if and only if $\gamma>{{\log 2}\over{\log{1/\xi}}}$. On the other hand, by a simple computation, we can see that for a real number $\delta$, the integral $\displaystyle\int_{0}^{2\pi}{{1}\over{d(e^{it},E_{\xi})^{\delta}}}dt$ converges if and only if the series $\sum_{n\geq 0}|L_{n}|^{1-\delta}$ converges. Therefore

[TABLE]

We denote by $A^{\infty}(\mathbb{D})$ the set of all functions holomorphic in the unit disk $\mathbb{D}$ and infinitely differentiable in the closed disk $\overline{\mathbb{D}}$.

Lemma 2.1**.**

Let $s$ be a nonnegative real, $q\geq 3$ be an integer and $\omega$ be a weight such that

[TABLE]

and

[TABLE]

Then there exists an outer fonction $f\in A^{\infty}(\mathbb{D})$ such that $f(0)=1$ and for every nonnegative integer $m$, $f(z^{q^{m}})\in J_{\omega}(E_{1/q})$.

Proof.

Let $\gamma$ be a real number such that $\beta<\gamma<b(1/q)$. There exists a real number $\delta$ such that $\frac{\gamma}{1-\gamma}<\delta<1-{{\log 2}\over{\log{q}}}$. We set

[TABLE]

Since $\delta<1-{{\log 2}\over{\log{q}}}$, it follows from (1) and [8, Lemma 7.2] that there exists a nonzero outer function $f\in A^{\infty}(\mathbb{D})$ which vanishes exactly on $E_{1/q}$ and which satisfies

[TABLE]

We may assume that $f(0)=1$. Now since $\frac{\gamma}{1-\gamma}<\delta$, by [8, Lemma 7.1], we have $f\in J_{\omega_{\gamma}}(E_{1/q})$.

Let $m$ be a nonnegative integer. Notice that for every $z\in E_{1/q}$, $z^{q^{m}}\in E_{1/q}$. Take a sequence $(f_{n})_{n}\subset A_{\omega_{\gamma}}(\mathbb{T})$, of functions vanishing on a neighborhood of $E_{1/q}$ and such that $\|f_{n}-f\|_{\omega_{\gamma}}\to 0$, $n\to\infty$. For every $n$, $f_{n}(z^{q^{m}})$ vanishes on a neighborhood of $E_{1/q}$ and we have

[TABLE]

where C=\sup_{k\in\mathbb{Z}}\big{(}\omega(k)/\omega_{\beta}(k)\big{)}\max\left(q^{ms},\sup_{k\geq 0}\big{(}e^{q^{\beta m}k^{\beta}-k^{\gamma}}\big{)}\right)<+\infty, where $\omega_{\beta}$ is defined similarly to $\omega_{\gamma}$. Thus $\|f_{n}(z^{q^{m}})-f(z^{q^{m}})\|_{\omega}\to 0$, as $n\to\infty$. ∎

The following theorem is the main result of this section.

Theorem 2.2**.**

Let $s$ be a nonnegative real number, $q\geq 3$ an integer and $T$ an invertible operator on a Banach space with spectrum contained in $E_{1/q}$ . If

[TABLE]

and

[TABLE]

Then $T$ satisfies the stronger property

[TABLE]

Proof.

We take the weight $\omega(n)=(1+n)^{s}$ if $n\geq 0$ and $\omega(n)=e^{|n|^{\beta}}$ if $n<0$. For $f\in A_{\omega}(\mathbb{T})$, we set

[TABLE]

Let $I$ be the set of functions $f\in A_{\omega}(\mathbb{T})$ such that $f(T)=0$. Then $I$ is a closed ideal of $A_{\omega}(\mathbb{T})$ and $h(I)=\mathrm{Sp}(T)$ (see [15, Théorème 2.4]). Therefore $J_{\omega}(E_{1/q})\subset I$. Now, let $f$ the outer function given by Lemma 2.1. Then, for every nonnegative integer $m$, $f(T^{{q^{m}}})=0$.

Let $n$ be an integer greater or equal to 1 and let $m$ be the unique integer such that $q^{m}\leq n<q^{m+1}$. It follows from the last equalities that

[TABLE]

and

[TABLE]

∎

The remainder of this section is devoted to discussing the sharpness of the hypothesis in Theorem 2.2. In the following proposition we show that the constant $b(\xi)$ is the best possible.

Proposition 2.3**.**

For every $\xi\in(0,1/2)$ there exists a contraction $T$ on a Hilbert space with spectrum contained in $E_{\xi}$, sucth that

[TABLE]

and

[TABLE]

Proof.

Let $L$ be the classical Lebesgue function associated to $E_{\xi}$ and let $dL$ be the unique positive measure on $[0,2\pi]$ such that $\int_{a}^{b}dL=L(b)-L(a)$ for $0\leq a\leq b\leq 2\pi$ (see [12, Chapter 1] and [9]). We set

[TABLE]

the inner function associated to the measure $dL$. Denote by $H^{2}$ and $H^{\infty}$ the usual Hardy spaces on the unit disk $\mathbb{D}$. Set ${\mathcal{H}}=H^{2}\circleddash VH^{2}$. Let $P_{{\mathcal{H}}}$ be the orthogonal projection from $H^{2}$ onto ${\mathcal{H}}$ and $T:{\mathcal{H}}\to{\mathcal{H}}$, $g\to P_{{\mathcal{H}}}(\alpha g)$, where $\alpha$ is the identity function $z\to z$; T is a contraction and its spectrum is contained in $E_{\xi}$. It is shown in [9, Theorem 2.2] that \big{\|}T^{-n}\big{\|}=O\big{(}n^{b(\xi)}\big{)},\,n\rightarrow+\infty and $\limsup_{n\to\infty}\|T^{-n}\|=\infty$.

Assume that for somme $\beta<b(\xi)$, $\sup_{n\geq 0}\|T^{-n}\|/e^{n^{\beta}}<+\infty$. We set

[TABLE]

As in the proof of Lemma 15, Using (1), [8, Lemma 7.2 and Lemma 7.1] we show that there exists a nonzero outer fonction $f\in A^{\infty}(D)$ such that $f\in J_{\omega_{\beta}}(E_{1/q})$. By the same argument as in the proof of Theorem 2.2, we get that $f(T)=0$. On the other hand the operator $f(T)$ is defined by $g\to P_{{\mathcal{H}}}(fg)$. Finaly, for $g=\frac{V-V(0)}{z}\in\mathcal{H}$, the equality $f(T)g=0$ implies that $f\in VH^{2}$.

∎

Now we will prove that Theorem 2.2 cannot be extended to all the sets $E_{\xi}$. More precisely we show that it is false for all $E_{\xi}$ such that $1/\xi$ is not a Pisot number. For this we introduce the definition below. We assume that $s$ is a nonnegative integer. We set $A_{s}^{+}(\mathbb{T})=\{f\in A_{s}(\mathbb{T}):\ \hat{f}(n)=0,\ n<0\}$. A closed subset $E$ of $\mathbb{T}$ is said to be an interpolating set of order $s$ for $A_{s}(\mathbb{T})$, if for every function $f\in A_{s}(\mathbb{T})$, there exists $g\in A_{s}^{+}(\mathbb{T})$ such that

[TABLE]

It is shown in [16], that $E_{\xi}$ is an interpolating set of order $s$ for $A_{s}(\mathbb{T})$ if and only if ${{1}/{\xi}}$ is a Pisot number.

Proposition 2.4**.**

Assume that $s$ is a nonnegative integer and that $1/\xi$ is not a Pisot number. Then there exists a bounded operator $T$ on a Banach space such that

[TABLE]

[TABLE]

and

[TABLE]

Proof.

We denote by $\pi$ the canonical surjection from $A_{p}^{+}(\mathbb{T})$ onto $A_{p}^{+}(\mathbb{T})/I_{s}^{+}(E_{\xi})$, where $I_{s}^{+}(E_{\xi})=\{f\in A_{s}(\mathbb{T}),\ f=0\ \mathrm{on}\ E_{\xi}\}$. Note that since $E_{\xi}$ is a perfect set $I_{s}^{+}(E_{\xi})$ is also the set of all functions $f$ in $A_{p}^{+}(\mathbb{T})$ such that for all $0\leq k\leq s$, $f^{(k)}=0$ on $E_{\xi}$. Let $\alpha$ be the identity function $z\to z$. It is easily seen that $E_{\xi}$ is an interpolating set of order $s$ for $A_{s}^{+}(\mathbb{T})$ if and only if we have

[TABLE]

Let $T$ be the operator defined on $A_{p}^{+}(\mathbb{T})/I_{s}^{+}(E_{\xi})$ by $\pi(f)\to\pi(\alpha f)$. We have $\mathrm{Sp}(T)=E_{\xi}$ and \big{\|}T^{n}\big{\|}=\big{\|}\pi(\alpha)^{n}\big{\|}=O\big{(}n^{s}\big{)},\,n\rightarrow+\infty.

It is shown in [16, Theorem 6] that $E_{\xi}$ is not an interpolating set of order $s$ for $A_{s}^{+}(\mathbb{T})$. Thus

[TABLE]

Since $E_{\xi}$ satisfies condition (1) it follows from the proof of [16, Theorem 8] that for every $\delta<1-\frac{\log 2}{\log{1/\xi}}$ and every $r>s+\frac{3}{2}(1-\delta)$,

[TABLE]

Thus for every $r>s+{\frac{3}{2}}{\frac{\log 2}{\log{1/\xi}}}$, \big{\|}T^{-n}\big{\|}=O\big{(}n^{r}\big{)},\,n\rightarrow+\infty.

∎

3. Spectral synthesis for Cantor sets in weighted Beurling algebras.

Let $s$ be a nonnegative real number. We denote by $[s]$ the integral part of $s$, that is the nonnegative integer such that $[s]\leq s<[s]+1$. For $f$ in $A_{s}(\mathbb{T})$, we denote by $f^{(j)}$, $0\leq j\leq[s]$, the $j^{th}$–derivative of $f$ with respect to the variable $t$, that is f^{(j)}\big{(}e^{it}\big{)}=\frac{\partial^{j}}{\partial t^{j}}\big{(}f\big{(}e^{it}\big{)}\big{)}.

To prove the main theorem of this section (Theorem 3.6), we show first that $E_{{1}/{q}}$ satisfies spectral synthesis in $A_{s}(\mathbb{T})$. To do this, we approximate any function $f$ in $A_{s}(\mathbb{T})$ such that $f^{(j)}(1)=0$ $(0\leq j\leq[s])$, by a sequence of functions $(f_{N,[s]})_{N\geq 1}$ in $A_{s}(\mathbb{T})$ which are piecewise polynomial and such that each function $f_{N,[s]}^{([s])}$ interpolates $f^{([s])}$ at the points $\displaystyle e^{\frac{2ik\pi}{N}}$, for all $k\in\{0,\ldots,N-1\}$ (for $s=0$, see the Herz criterion in [10] and [11, p. 58–59]). The functions $f_{N,[s]}$ are defined by

[TABLE]

where $\Delta_{[s],\varepsilon}$ (for $0<\varepsilon\leq\pi$) is the $2\pi$-periodical function given on $[-\pi,\pi]$ by the formula

[TABLE]

and

[TABLE]

Lemma 3.1**.**

Assume that $0\leq s<1$. There exists a positive constant $K$ such that for every function $f\in A_{s}(\mathbb{T})$ and every integer $N\geq 1$,

[TABLE]

Proof.

First, we will show this inequality for $f(e^{it})=e^{int}\,(n\in\mathbb{Z})$. We have $[s]=0$ and

[TABLE]

The Fourier coefficients of $\Delta_{0,\varepsilon}$ are :

[TABLE]

Let $f(e^{it})=e^{int}$. For $N\geq 1$, we set

[TABLE]

where $\delta_{a}$ denotes the Dirac measure at $a$. We have

[TABLE]

Therefore

[TABLE]

We will consider two cases :

Case 1: $N\geq 2|n|$.

Let $m=n+kN$ for some $k\in\mathbb{Z}$. Note that since $N\geq 2|n|$, $m\neq 0$. Thus we have

[TABLE]

Therefore

[TABLE]

Case 2: $N\leq 2|n|$.

For $|m|<N$,

[TABLE]

We remark that there are at most two integers $m$ of the form $m=n+kN$ with $|m|<N$. Thus

[TABLE]

Assume now that $|m|\geq N$ with $m=n+kN$ and $k\in\mathbb{Z}$. We note that in this case \displaystyle\big{|}\frac{n}{N}+k\big{|}=\frac{|m|}{N}\geq 1. We have

[TABLE]

Then we get

[TABLE]

Combining (12) and (13) we obtain

[TABLE]

Finally, we deduce from (11) and (14) that there exists a positive constant $K$ such that for all integers $N\geq 1$ and $n\in\mathbb{Z}$,

[TABLE]

where $f(e^{it})=e^{int}$. We deduce easily that this inequality is still true for all trigonometric polynomials, then, by density, for all functions $f$ in $A_{s}(\mathbb{T})$.

∎

Lemma 3.2**.**

Assume that $0\leq s<1$. Then for every function $f\in A_{s}(\mathbb{T})$, we have

[TABLE]

Proof.

First, we prove the proposition for $f(e^{it})=e^{int}$ ($n\in\mathbb{Z}$). If $n=0$, we deduce from (9) that for all $N\geq 1$, $\widehat{f_{N,0}}(0)=1$ and $\widehat{f_{N,0}}(m)=1$ for $m\neq 0$, so that $f_{N,0}=f$.

Now, we suppose that $n\neq 0$. Let $N\geq 2|n|$ be fixed and $m=n+kN$ ($k\in\mathbb{Z}$). If $k\neq 0$, then \displaystyle|m|\geq\big{(}|k|-\frac{1}{2}\big{)}N. We have

[TABLE]

On the othar hand, if $k=0$, we have $m=n$ and

[TABLE]

Thus

[TABLE]

It follows from (15) and (16) that

[TABLE]

which implies

[TABLE]

Now we consider the general case: let $f$ be in $A_{s}(\mathbb{T})$ and $\varepsilon>0$. There exists $M_{0}>0$ such that \displaystyle\big{\|}f-g\big{\|}_{s}\leq\varepsilon, where $\displaystyle g=\sum_{m=-M_{0}}^{M_{0}}\widehat{f}(m)e^{imt}$. Using the above, we can find $N_{0}>0$ such that for all $N\geq N_{0}$, \displaystyle\big{\|}g-g_{N,0}\big{\|}_{s}\leq\varepsilon. For $N\geq N_{0}$, we deduce from lemma 3.1 that

[TABLE]

which proves that

[TABLE]

∎

Lemma 3.3**.**

Let $s$ be a nonnegative real number and $p=[s]$. If $p\geq 1$, then for every function $f\in A_{s}(\mathbb{T})$, we have

[TABLE]

Proof.

For $f\in A_{s}(\mathbb{T})$, we have

[TABLE]

and

[TABLE]

Thus

[TABLE]

Integration by parts gives

[TABLE]

and

[TABLE]

Now the lemma follows from (17) and (18). ∎

Lemma 3.4**.**

Let $s$ be a nonnegative real number and $f$ a function in $A_{s}(\mathbb{T})$ such that $f^{(j)}(1)=0$ for $0\leq j\leq[s]$. Then

[TABLE]

Proof.

We set $p=[s]$. The case $p=0$ is exactly Lemma 3.2. So assume that $p\geq 1$. Let $f\in A_{s}(\mathbb{T})$ such that $f^{(j)}(1)=0$ for $0\leq j\leq p$. We note that for every $N\geq 1$, $(f_{N,s})^{(p)}=(f^{(p)})_{N,0}$ and $(f_{N,s})^{(j)}(1)=0$ for $0\leq j\leq p$. By Lemma 3.3,

[TABLE]

It follows from Lemma 3.2 that $\lim_{N\rightarrow+\infty}\|f^{(p)}-(f^{(p)})_{N,0}\|_{s-p}=0$, which finishes the proof.

∎

Let $s$ be a nonnegative real number, $E$ be a closed set of $\mathbb{T}$ whose boundary ($\mathrm{bdry}(E)$), is finite or countable and $f$ be a function in $A_{s}(\mathbb{T})$ such that $f=f^{\prime}=\ldots=f^{(p)}=0$ on $E$, where $p=[s]$. We set $I=\{g\in A_{s}(\mathbb{T});\ gf\in J_{s}(E)\};$ $I$ is a closed ideal of $A_{s}(\mathbb{T})$, $J_{s}(E)\subset I$ and $h(I)\subset h(J_{s}(E))=E$. Let $\zeta\in\overset{\circ}{E}$, where $\overset{\circ}{E}$ is the interior of $E$. Since $A_{s}(\mathbb{T})$ is a regular algebra there exists a function $g\in A_{s}(\mathbb{T})$ whose support is contained in $\overset{\circ}{E}$ and such that $f(\zeta)=1$. We have $fg=0$ which implies that $g\in I$ and then $\zeta\notin h(I)$. It follows that $h(I)\subset\mathrm{bdry}(E)$. Assume that $h(I)$ is nonempty and take $\zeta$ an isolated point in $h(I)$. There exists a function $h\in A_{s}(\mathbb{T})$ such that $h(\zeta)=1$ and $h=0$ on a neighborhood of $h(I)\setminus\{\zeta\}$. For $n\geq 1$, we set

[TABLE]

According to [1, Proposition 2.4], we have $\displaystyle\lim_{n\rightarrow+\infty}\|u_{n}hf-hf\|_{s}=0$, since $(hf)^{(j)}(\zeta)=0$ for $0\leq j\leq[s]$. Moreover, for each $n\geq 1$, $u_{n}\in J_{s}(\{\zeta\}$ (see [5, Proposition 6] or [1, Théoème 3.2]). So there exists a function $v_{n}\in A_{s}(\mathbb{T})$, vanishing in a neighborhood of $\zeta$ and such that $\|u_{n}-v_{n}\|_{s}\leq 1/n$. Now $v_{n}h$ vanishes on a neighborhood of $h(I)$ and then $v_{n}h\in I$. Therefore $v_{n}hf\in J_{s}(E)$ and since $\displaystyle\lim_{n\rightarrow+\infty}\|v_{n}hf-hf\|_{s}=0$, we have $hf\in J_{s}(E)$ and then $h\in I$. This contradicts that $h(\zeta)\neq 0$, which implies that $h(I)=\emptyset$ and $I=A_{s}(\mathbb{T})$. We deduce that $f\in J_{s}(E)$. Notice that for $s=0$, it is well known that in the Wiener algebra $A(\mathbb{T})$, every closed subset of $\mathbb{T}$ whose boundary is a finite or countable set, satisfies the spectral synthesis (see [13, Theorem, p. 245]).

Suppose now that $E$ is a finite union of closed arcs (not reduced to a single point). Then $\mathrm{bdry}(E)$ is a finite set and if $f$ is a function in $A_{s}(\mathbb{T})$ which vanishes on $E$, then all its derivatives of order less or equal to $[s]$, also vanish on $E$. We deduce that $E$ satisfies spectral synthesis in $A_{s}(\mathbb{T})$. This allows us to prove the following lemma, which was obtained in [3] by intricate computations.

Lemma 3.5**.**

Let $s$ be a nonnegative real number and $q\geq 3$ an integer. Then $E_{1/q}$ satisfies the spectral synthesis in $A_{s}(\mathbb{T})$.

Proof.

For all $n\geq 1$, we set

[TABLE]

where $L_{\epsilon_{1},\ldots,\epsilon_{n}}$ is the closed arc \Big{\{}e^{it}:\,t\in\big{[}\displaystyle 2\pi(q-1)\sum_{k=1}^{n}\epsilon_{k}q^{-k},2\pi(q-1)\big{(}\sum_{k=1}^{n}\epsilon_{k}q^{-k}+q^{-n}\big{)}\big{]}\Big{\}}. We have $\displaystyle E_{1/q}=\bigcap_{n\geq 1}F_{n}$ (see [12, Chapter I]).

Let $f\in I_{s}(E_{1/q})$, we have to show that $f\in J_{s}(E_{1/q})$. Notice that since $E_{1/q}$ is a perfect set, $f$ vanishes, as well as all its derivatives of order less or equal to $[s]$, on $E_{1/q}$. We observe that for each $n\geq 1$ and each $(\epsilon_{1},\ldots,\epsilon_{n})\in\{0,1\}^{n}$, the end points of the arc $L_{\epsilon_{1},\ldots,\epsilon_{n}}$ are of the form $\frac{2\pi k}{q^{n}}$ and $\frac{2\pi(k+1)}{q^{n}}$ for some integer $k\in\{0,1,\ldots,q^{n}-1\}$, and since $(f_{q^{n},\epsilon})^{([s])}=f^{([s])}=0$ at these end points, $f_{q^{n},\epsilon}$ vanishes on $L_{\epsilon_{1},\ldots,\epsilon_{n}}$. Thus $f_{q^{n},\epsilon}$ vanishes on $F_{n}$. Since $F_{n}$ is a finite union of arcs, it satisfies spectral synthesis in $A_{s}(\mathbb{T})$. Therefore, each function $f_{q^{n}}$ is in $J_{s}(E_{{1}/{q}})$. As $J_{s}(E_{1/q})$ is closed in $A_{s}(\mathbb{T})$, we deduce from Lemma 3.4 that $f\in J_{s}(E_{1/q})$. ∎

Now we are able to prove the main result of this section.

Theorem 3.6**.**

Let $s$ be a nonnegative real number, $q\geq 3$ an integer and $\omega$ a weight such that

[TABLE]

[TABLE]

and

[TABLE]

Then $E_{{1}/{q}}$ satisfies spectral synthesis in $A_{\omega}(\mathbb{T})$.

Proof.

We denote by $\pi_{\omega}$ the canonical surjection from $A_{\omega}(\mathbb{T})$ to $A_{\omega}(\mathbb{T})/J_{\omega}(E_{1/q})$, and by $T$ the operator of multiplication by $\pi_{\omega}(\alpha)$ on $A_{\omega}(\mathbb{T})/J_{\omega}(E_{1/q})$. It is easy to see that $T$ satisfies the hypothesis of Theorem 2.2. It follows that \big{\|}\pi_{\omega}(\alpha)^{-n}\big{\|}=\big{\|}T^{-n}\big{\|}=O\big{(}n^{s}\big{)},\,n\rightarrow+\infty. Thus we can define a continuous morphism $\Theta$ from $A_{{s}}(\mathbb{T})$ into $A_{\omega}(\mathbb{T})/J_{\omega}(E_{1/q})$ by

[TABLE]

We note that $\textrm{Ker}\,\Theta$ is a closed ideal of $A_{{s}}(\mathbb{T})$ and that $J_{{\omega}}(E_{1/q})=\textrm{Ker}\,\Theta\cap A_{\omega}(\mathbb{T})$, since for $f\in A_{\omega}(\mathbb{T})$, $\Theta(f)=\pi_{\omega}(f)$. It follows that $h(\textrm{Ker}\,\Theta)\subset E_{1/q})$ and that $J_{{s}}(E_{1/q})\subset\textrm{Ker}\,\Theta$. By Lemma 3.5, $E_{1/q}$ satisfies spectral synthesis in $A_{{s}}(\mathbb{T})$, that is $I_{{s}}(E_{\frac{1}{q}})=J_{{s}}(E_{1/q})$. Therefore $I_{{\omega}}(E_{1/q})=I_{{s}}(E_{1/q})\cap A_{\omega}(\mathbb{T})=J_{{s}}(E_{1/q})\cap A_{\omega}(\mathbb{T})\subset\textrm{ker}\,\Theta\cap A_{\omega}(\mathbb{T})=J_{\omega}(E_{1/q})$, which completes the proof. ∎