On a theorem of Reiter and spectral synthesis

Abstract

Motivated by and extending a theorem of Reiter on sets of synthesis in $\mathbb {R}^N$, we establish a general result for Fourier algebras of locally compact groups, even in the wider context of regular, semisimple and Tauberian commutative Banach algebras, which contains Reiter’s theorem as a special case and explains why it holds. In addition, we give a number of examples and several further results on weak spectral sets.

1 Introduction

By a classical result due to L. Schwartz the unit sphere $S^{N-1}$ in $\mathbb {R}^N$ fails to be a set of synthesis for the Fourier algebra $A(\mathbb {R}^N)$ when $N\ge 3$. Subsequently, Reiter [20] proved an intriguing theorem saying that if F is any closed subset of $\mathbb {R}^N$, $N\ge 3$, then the set $F\cup S^{N-1}$ cannot be a set of synthesis unless F contains $S^{N-1}$. However, the sole fact that $S^{N-1}$ disobeys synthesis does not explain why Reiter’s theorem holds.

In this paper we prove a general result for the Fourier algebra A(G) of an arbitrary locally compact group G, satisfying Ditkin’s condition at infinity, which contains Reiter’s theorem as a special case and at the same time explains why it holds (Sect. 3). Actually, our result is valid for general semisimple, regular, Tauberian commutative Banach algebras, which satisfy Ditkin’s condition at infinity (Sect. 4). Moreover, we deal with weak spectral sets rather than just sets of synthesis. In Sect. 5 we present a number of examples illustrating the results of Sect. 4. These examples concern, in particular, conjugacy classes and double cosets of compact subgroups in G. Finally, in Sect. 6 we add a number of further results on weak spectral sets.

2 Preliminaries

Let A be a regular and semisimple commutative Banach algebra with structure space $\Delta (A)$ and Gelfand transform $a \rightarrow \widehat{a}$. For any subset M of A, the hull h(M) of M is defined by $h(M) = \{\varphi \in \Delta (A): \varphi (M) = \{0\}\}$. Associated to each closed subset E of $\Delta (A)$ are two distinguished ideals with hull equal to E, namely

$$\begin{aligned} k(E) = \{a \in A: \widehat{a}(\varphi ) = 0\quad \text{ for } \text{ all }\,\,\varphi \in E\} \end{aligned}$$

and

$$\begin{aligned} j(E) = \{a \in A: \widehat{a}\,\,\text{ has } \text{ compact } \text{ support } \text{ disjoint } \text{ from }\,E\}. \end{aligned}$$

Then k(E) is the largest ideal with hull equal to E and j(E) is the smallest such ideal, and consequently $J(E) = \overline{j(E)}$ is the smallest closed ideal with hull E. The set E is called a set of synthesis or spectral set if $k(E) = J(E)$. If $a\in \overline{a j(E)}$ for every $a\in k(E)$, then E is called a Ditkin set. Finally, A is said to satisfy Ditkin’s condition at infinity if $\emptyset$ is a Ditkin set. As general references to spectral synthesis we mention [1, 10, 21, 22].

For any Banach space X, the duality between X and its dual Banach space $X^*$ is written as $\langle f,x\rangle$ or $\langle x, f\rangle$, $x\in X$, $f\in X^*$. For a subset M of X, $M^\perp$ will denote the annihilator of M in $X^*$.

Throughout Sects. 4 and 6 of the paper A will denote a commutative, regular, semisimple and Tauberian Banach algebra, which satisfies Ditkin’s condition at infinity. We recall that A is said to be Tauberian if the set of all $a\in A$ such that $\widehat{a}$ has compact support is dense in A.

For $a\in A$ and $f\in A^*$, the functional $a\cdot f$ is defined by $\langle a\cdot f, b \rangle = \langle f, ab\rangle$, $b\in A$. It is clear that $\Vert a\cdot f\Vert \le \Vert a\Vert \cdot \Vert f\Vert$ and, for $\gamma \in \Delta (A)$ and $a\in A$, $a\cdot \gamma = \widehat{a}(\gamma )\gamma = \langle \gamma , a\rangle \gamma$. As $\emptyset$ is a Ditkin set, $a\in \overline{aA}$ for each $a\in A$. This in turn implies that $\langle a,f\rangle = 0$ whenever $a\cdot f = 0$.

For $f \in A^*$, the spectrum $\sigma (f)$ of f is defined to be

$$\begin{aligned} \sigma (f) = \{\gamma \in \Delta (A): \text{ for } \text{ each }\,a\in A, a\cdot f = 0\,\,\text{ implies }\,\, \widehat{a}(\gamma ) = 0\}. \end{aligned}$$

The following properties of the spectrum $\sigma (f)$ are used throughout the paper:

• (1) $\sigma (f) = \emptyset$ if and only if $f=0$.

• (2) For any $f\in A^*$ and $a\in A$, $\sigma (a\cdot f) \subseteq \sigma (f) \cap {\mathrm{supp}} \ \, \widehat{a}$.

• (3) For any closed subset E of $\Delta (A)$, $\sigma (f)\subseteq E$ if and only if $f\in J(E)^\perp$.

• (4) If $(f_\alpha )_\alpha$ is a net in $A^*$ converging to f in the $w^*$-topology and $\sigma (f_\alpha ) \subseteq E$ for some closed subset E of $\Delta (A)$, then $\sigma (f)\subseteq E$.

For closed ideals I and J of A, IJ denotes the closed ideal

$$\begin{aligned} \overline{\left\{ \sum _{i=1}^n a_ib_i: a_i \in I, \,b_i\in J,\,1\le i\le n, \,n\in \mathbb {N}\right\} }. \end{aligned}$$

Accordingly, for $k\in \mathbb {N}$, $k\ge 2$, $I^k$ is inductively defined by $I^k = I^{k-1}I$. Using identities such as $4 ab = (a+b)^2 - (a-b)^2$, a straightforward inductive argument shows that $I^k$ is also the closed linear span of all the elements of the form $a^k$, $a\in I$.

3 Reiter’s theorem for the Fourier algebra

As mentioned in the introduction, according to a classical result due to L. Schwartz, the unit sphere $S^{N-1}$ in $\mathbb {R}^N = \Delta (L^1(\mathbb {R}^N))$ fails to be a set of synthesis for $N\ge 3$ (see [22, 7.3.1 and 7.3.2]). In an attempt to construct examples of functions in $k(S^{N-1}){\setminus } J(S^{N-1})$, Reiter proved the intriguing result that if F is any closed subset of $\mathbb {R}^N$ which does not contain $S^{N-1}$, then $S^{N-1} \cup F$ fails to be a set of synthesis [20, Theorem 2]. In this section, we present a generalization of Reiter’s theorem to Fourier algebras of arbitrary locally compact groups and explain why Reiter’s theorem holds.

Let G be a locally compact group and A(G) the Fourier algebra of G as introduced and extensively studied by Eymard [4]. This is a commutative, semisimple, regular and Tauberian Banach algebra of functions on G, which has since become one of the main objects of study in abstract harmonic analysis. The Gelfand spectrum of A(G) can be identified with G, via the evaluation functionals. If G is abelian, then A(G) is isometrically isomorphic to the $L^1$-algebra of the dual group $\widehat{G}$ of G. Recall that $A(G)^*$ is isomorphic to the VN(G), the von Neumann algebra generated by the left regular representation of G on the Hilbert space $L^2(G)$. We will throughout assume that $u\in \overline{u A(G)}$ for every $u\in A(G)$. No locally compact group seems to be known for which this condition is not satisfied.

To every closed subset E of $G = \Delta (A(G))$ we associate the closed subset

$$\begin{aligned} \sigma _1(E) = \overline{\textstyle \bigcup \{\sigma (u\cdot f): u\in k(E), f\in J(E)^\perp \}} \end{aligned}$$

of E and the closed ideal

$$\begin{aligned} I_1(E) = \{u\in A(G): u k(E)\subseteq J(E)\} \end{aligned}$$

of A(G). The set $\sigma _1(E)$, which has been introduced in [24] and used there to study synthesis problems, is the hull of $I_1(E)$ and always contained in the boundary $\partial (E)$ of E. Moreover, by its very definition, $\sigma _1(E) = \emptyset$ if and only if the set E is a set of synthesis.

Let $\Phi$ be a group of homeomorphisms of G that acts continuously on A(G) in the following sense.

• (1) For each $u\in A(G)$ and $\phi \in \Phi$, the function $\phi (u)$ defined by $\phi (u)(x) = u(\phi ^{-1}(x))$, $x\in G$, belongs to A(G).

• (2) If a sequence $(u_n)_n$ in A(G) converges to u for some $u\in A(G)$, then $\phi (u_n)\rightarrow \phi (u)$ for every $\phi \in \Phi$.

Recall that $\Phi$ acts transitively on a subset E of G if for some (and hence every) $x\in E$, $E = \{\phi (x): \phi \in \Phi \}$. In this case, $\phi (E) = E$ for each $\phi \in \Phi$.

The following theorem is a dichotomy result for such a set E. Concerning synthesibility, it behaves either very good or as bad as possible.

Theorem 3.1

Let G be a locally compact group and suppose that $\Phi$ is a group of homeomorphisms of G which acts continuously on A(G). Let E be a closed subset of G such that $\Phi$ acts transitively on E. Then either E is a set of synthesis or $\sigma _1(E) = E$. Moreover, E is a set of synthesis if and only if E contains a set of synthesis F whose relative interior is nonempty (i.e. $\overline{E{\setminus } F} \ne E$).

Let $G = \mathbb {R}^N$, $N\ge 3$, $\Phi = SO(N)$, acting on $\mathbb {R}^N$ by rotation, and $E = S^{N-1}$. Then $\Phi$ acts transitively on E. Since $S^{N-1}$ is not of synthesis for $A(\mathbb {R}^N)$, it follows from the preceding theorem that $\sigma _1(S^{N-1}) = S^{N-1}$. Note that in [13, Example 6.6] this was obtained as a consequence of Varopoulos’s work [25]. Since, as shown by Herz [8], the unit circle is a set of synthesis for $A(\mathbb {R}^2)$, we see that both alternatives in Theorem 3.1 occur.

We now present a class of sets E for which $\sigma _1(E) = E$. Contrary to the sphere, these are thin sets. For any $f\in VN(G)$, set

$$\begin{aligned} X_f = \{ u\cdot f: u\in A(G)\}. \end{aligned}$$

Then $X_f$ is a (not necessarily closed) A(G)-invariant subspace of VN(G).

Proposition 3.2

Let F be a closed subset of G and let $f\in VN(G)$ be such that $J(F)^\perp \cap X_f \ne \{0\}$, but $k(F)^\perp \cap X_f = \{0\}$. Then F contains a nonempty set E such that $\sigma _1(E) = E$.

Proof

Let $g\in J(F)^\perp \cap X_f$, $g\ne 0$, and set $E = \sigma (g)$. As

$$\begin{aligned} J(\sigma _1(E)) k(E) \subseteq I_1(E)k(E) \subseteq J(E)\subseteq J(\sigma _1(E)), \end{aligned}$$

we have $J(\sigma _1(E))k(E) = J(E)$. This equality implies that, for any $u\in J(\sigma _1(E))$, $u\cdot g \in k(E)^\perp \cap X_f$. However, since $E\subseteq F$, $k(E)^\perp \cap X_f = \{0\}$ and therefore $u\cdot g = 0$. As $u\in \overline{uA(G)}$, we get that $\langle u,g\rangle = 0$. This proves that $g\in J(\sigma _1(E))^\perp$, so that $E = \sigma (g)\subseteq \sigma _1(E)$ and hence $\sigma _1(E) = E$. $\square$

In the context of metrizable locally compact abelian groups, for the existence of sets F satisfying the hypothesis of the preceding proposition, we refer the reader to [23].

Theorem 3.3

Let E be a closed subset of G and suppose that $\sigma _1(E) = E$. Then, for any closed subset F of G, $E\cup F$ is a set of synthesis (if and) only if $E\subseteq F$ and F is a set of synthesis. Moreover, in this case, $\overline{F{\setminus } E} = F$.

Reiter’s theorem is now an immediate consequence of Theorem 3.3 since $\sigma _1(S^{N-1}) = S^{N-1}$, and we note that this equality is the reason for Reiter’s theorem to hold. The next corollary follows from Theorem 3.1.

Corollary 3.4

Let G, $\phi$ and E be as in Theorem 3.1. If E fails to be a set of synthesis, then E cannot be written as a union of countably many sets of synthesis.

As do the previous results, the following corollary, which is an immediate consequence of Theorems 3.1 and 3.3, also holds in a more general setting (see Corollary 4.4)

Corollary 3.5

Let E and F be closed subsets of $\mathbb {R}^N$, $N\ge 3$, such that $E\subseteq S^{N-1}\subseteq F$.

• (i) If E is a set of synthesis for $A(\mathbb {R}^N)$, then $\overline{S^{N-1}{\setminus } E} = S^{N-1}$.

• (ii) If F is a set of synthesis for $A(\mathbb {R}^N)$, then $\overline{F{\setminus } S^{N-1}} = F$.

We refrain from presenting proofs of Theorems 3.1 and 3.3 and Corollaries 3.4 and 3.5 here because in the next section we shall prove abstract versions of these results in the general setting of semisimple, regular and Tauberian commutative Banach algebras. The purpose of stating the above results separately is to emphasize the motivation for what will be accomplished in Sect. 4. We continue with an example taken from [14].

Example 3.6

Let $T: \mathbb {R}^N \rightarrow \mathbb {R}^N$ be a linear map and let $\Phi = \{e^{ s T}: s\in \mathbb {R}\}$ be the associated one-parameter group. Suppose that $\Phi$ is closed in $GL(N, \mathbb {R})$. Then, as proved in [14], for any compact connected subset H of $\Phi$ and any $x\in \mathbb {R}^N$, the set H(x) is a set of synthesis for $A(\mathbb {R}^N)$ (more generally, for the so-called Figà-Talamanca-Herz algebras $A_p(\mathbb {R}^N)$, $1< p <\infty$). Now, let $H_n = \{e^{s T}: - n \le s \le n\}$, $n\in \mathbb {N}$. Then, since $\Phi$ is closed in $GL(N, \mathbb {R})$, each orbit $\Phi (x)$ is closed in $\mathbb {R}^N$ (for instance, see [7, p.150, Exercise 5]). It now follows from Corollary 3.4 that $\Phi (x) = \bigcup _{n=1}^\infty H_n(x)$ is a set of synthesis.

4 An abstract version of Reiter’s theorem and some consequences

From now on we allow the more general setting of a regular, semisimple and Tauberian commutative Banach algebra A with structure space $\Delta (A)$ and Gelfand transform $a \rightarrow \widehat{a}$. We assume in addition that $\emptyset$ is a Ditkin set. We also consider weak spectral sets rather than just sets of synthesis. These sets were introduced and first studied by Warner [26] in connection with the union problem for sets of synthesis. A closed subset E of $\Delta (A)$ is called a weak spectral set or set of weak synthesis if there exists $n \in \mathbb {N}$ such that $a^n \in J(E)$ for every $a \in k(E)$. The smallest such number n are denoted $\xi (E)$ and called the characteristic of E. Thus $\xi (E) = 1$ if and only if E is a set of synthesis. Using this terminology, the main result of [25] says that $S^{N-1}$ is a weak spectral set for $A(\mathbb {R}^N)$ with $\xi (S^{N-1}) = \lfloor \frac{N+1}{2}\rfloor$.

We say that weak spectral synthesis holds for A if $\xi (E) < \infty$ for every closed subset E of $\Delta (A)$. Warner proved that the union of two weak spectral sets E and F is again a weak spectral set and that $\xi (E\cup F) \le \xi (E)+ \xi (F)$. Subsequently, weak spectral sets and the weak synthesis problem gained considerable attention and have been studied by several authors [9, 11–13, 16, 17, 19, 26, 27]. There are several important Banach algebras for which weak synthesis holds, whereas spectral synthesis fails. In contrast, as independently shown in [9, 17], for a locally compact group G, weak spectral synthesis holds for the Fourier algebra A(G) if and only if G is discrete.

We define, for each $n\in \mathbb {N}$, a closed ideal $I_n(E)$ of A by

$$\begin{aligned} I_n(E) = \{a\in A: a k(E)^n\subseteq J(E)\}. \end{aligned}$$

Then $I_n(E)$ is the largest closed ideal I of A such that $I k(E)^n\subseteq J(E)$. Moreover, for each n we define a subset $\sigma _n(E)$ of $\Delta (A)$ by

$$\begin{aligned} \sigma _n(E) = \overline{\bigcup \{\sigma (a\cdot f): a\in k(E)^n, f\in J(E)^\perp \}}\,^{w^*}, \end{aligned}$$

where the $w^*$-closure is taken in $\Delta (A)$. The set $\sigma _1(E)$ was introduced in [24] to study sets of synthesis, and subsequently the decreasing sequence of sets $\sigma _n(E)$ and the increasing sequence of ideals $I_n(E)$, $n\in \mathbb {N}$, have been defined and employed in [13]. For some examples of Banach algebras they have been determined explicitly [13, Section 6]. Moreover, for every $n\in \mathbb {N}$, $\sigma _n(E)\subseteq \overline{\Delta _n(E)}$, where $\Delta _n(E)$ denotes the n-difference spectrum, which was introduced in [16] as a tool to study weak spectral synthesis. The following facts will be used several times in the sequel.

• (i) For any closed subset E of $\Delta (A)$,

  $$\begin{aligned} J(\sigma _n(E)) \subseteq I_n(E) \subseteq k(\sigma _n(E)) \end{aligned}$$

  and hence $h(I_n(E)) = \sigma _n(E)$ [13, Lemma 2.2].

• (ii) The set E is a weak spectral set if and only if $\sigma _n(E) = \emptyset$ for some $n\in \mathbb {N}$. Moreover, in this case $\xi (E)$ is the smallest such number n. In particular, E is a set of synthesis if and only if $\sigma _1(E) = \emptyset$ [13, Proposition 2.3]

Let now $\Phi$ be a group of homeomorphisms of $\Delta (A)$ such that

• (1) for each $a\in A$ and $\phi \in \Phi$, there exists an element $\phi (a)$ of A such that $\widehat{\phi (a)}(\gamma ) = \widehat{a}(\phi ^{- 1}(\gamma ))$ for all $\gamma \in \Delta (A)$;

• (2) if $\Vert a_n -a\Vert \rightarrow 0$, then $\Vert \phi (a_n) - \phi (a)\Vert \rightarrow 0$ for every $\phi \in \Phi$.

We then say that $\Phi$ acts continuously on A. Note that, since A is semisimple, $\phi (a)$ is uniquely determined and the map $a\rightarrow \phi (a)$ is an automorphism of A.

The following theorem is the abstract version of Theorem 3.1.

Theorem 4.1

Let $\Phi$ be as above and let E be a closed subset of $\Delta (A)$ such that $\Phi$ acts transitively on E. Then either E is a weak spectral set or $\sigma _n(E) = E$ for every $n\in \mathbb {N}$. Furthermore, the set E is a weak spectral set with $\xi (E) \le m$ for some $m\in \mathbb {N}$ if and only if E contains a weak spectral set F with $\xi (F)\le m$ whose relative interior is nonempty.

Proof

For $a\in k(E)$, $\phi \in \Phi$ and $\gamma \in E$, we have $\widehat{\phi (a)}(\gamma ) = \widehat{a}(\phi ^{-1}(\gamma )) = 0$ since E if $\Phi$-invariant. Thus the ideal k(E) is $\Phi$-invariant and hence so is $k(E)^m$ for every $m\in \mathbb {N}$ because each $\phi \in \Phi$ is a continuous automorphism of A. Moreover, if C is a compact subset of $\Delta (A)$ such that $C\cap E = \emptyset$, then $\phi (C)\cap E = \phi (E\cap C) = \emptyset$. It follows that j(E), and consequently J(E), is $\Phi$-invariant. Now, for every $\phi \in \Phi$,

$$\begin{aligned} k(E)^m \phi (I_m(E)) = \phi (k(E)^m I_m(E)) \subseteq \phi (J(E)) = J(E). \end{aligned}$$

Since $I_m(E)$ is the largest closed ideal I of A such that $k(E)^m I \subseteq J(E)$, we conclude that $\phi (I_m(E)) = I_m(E)$ for every $\phi \in \Phi$.

Now assume that $\sigma _m(E) \ne \emptyset$ and choose $\gamma \in \sigma _m(E)$. Then, since $\Phi$ acts transitively on E, $E = \{\phi (\gamma ): \phi \in \Phi \}$. Finally, since $h(I_m(E)) = \sigma _m(E)$ and the ideal $I_m(E)$ is $\Phi$-invariant, it follows that $\widehat{a}$ vanishes on E for every $a\in I_m(E)$ and therefore $\sigma _m(E) = E$.

For the last assertion, assume that E fails to be a weak spectral set with $\xi (E)\le m$. If $F\subseteq E$ is any weak spectral set with $\xi (F)\le m$, then $E = \sigma _m(E)\subseteq \overline{E{\setminus } F}$ by [13, Lemma 3.1], whence F has empty interior in E. $\square$

The next theorem is an abstract version of Theorem 3.3 and therefore of Reiter’s theorem.

Theorem 4.2

Let E be a closed subset of $\Delta (A)$ and suppose that, for some $m\in \mathbb {N}$, $\sigma _m(E) = E$. Then, for any closed subset F of $\Delta (A)$, $E\cup F$ is a weak spectral set with $\xi (E\cup F)\le m$ (if and) only if $E\subseteq F$ and F is a weak spectral set with $\xi (F) \le m$. Moreover, in this case $\overline{F{\setminus } E} = F$.

Proof

The first inclusion below being obvious, by hypothesis we have

$$\begin{aligned} k(E)^m k(F{\setminus } E)^m \subseteq k(E\cup F)^m \subseteq J(E\cup F)\subseteq J(E). \end{aligned}$$

Hence, since $I_m(E)$ is the largest ideal I of A with $I k(E)^m \subseteq J(E)$,

$$\begin{aligned} k(F{\setminus } E)^m \subseteq I_m(E) \subseteq k(\sigma _m(E)) = k(E). \end{aligned}$$

Thus $E\subseteq \overline{F{\setminus } E}$, and since $E\cup F = E \cup (F{\setminus } E) \subseteq \overline{F{\setminus } E}$, we conclude that $F = \overline{F{\setminus } E}$. It follows that

$$\begin{aligned} k(F)^m = k(\overline{F{\setminus } E})^m = k(E\cup F)^m \subseteq J(E\cup F)\subseteq J(F). \end{aligned}$$

This proves that $\xi (F)\le m$ and also that $\overline{F{\setminus } E} = F$. $\square$

Corollary 4.3

Let E and $\Phi$ be as in Theorem 4.1, and let F and S be closed subsets of $\Delta (A)$ such that $E \subseteq S \subseteq E\cup F$. Suppose that S is a weak spectral set and that $\xi (E) = \infty$ or $\xi (E) > \xi (S)$. Then $E\subseteq \overline{F{\setminus } E}$ and hence $F = \overline{F{\setminus } E}$.

Proof

Let $m = \xi (S)$. Since $\xi (E) > m$, $\sigma _m(E) \ne \emptyset$ and hence $\sigma _m(E) = E$ by Theorem 4.1 . The statement now follows from Theorem 4.2. $\square$

Corollary 4.4

Let $\Phi$ be as above and let E be a closed subset of $\Delta (A)$ such that $\Phi$ acts transitively on E. Then the following two conditions are equivalent.

• (i) E is a set of weak synthesis.

• (ii) E can be written as a countable union of weak spectral sets.

Moreover, if $E = \bigcup _{n=1}^\infty E_n$, where each $E_n$ is a weak spectral set, and if $m\in \mathbb {N}$ is such that $E_m$ has nonempty relative interior $E_m^\circ$ in E, then $\xi (E)\le \xi (E_m)$.

Proof

(i) $\Rightarrow$ (ii) is trivial. For (ii) $\Rightarrow$ (i), assume that $E = \bigcup _{n=1}^\infty E_n$, where each $E_n$ is weak spectral set, and nevertheless E is not a set of weak synthesis. Then $\sigma _m(E) = E$ for all $m\in \mathbb {N}$ by Theorem 4.1. Since each $E_n$ is of weak synthesis, by [13, Lemma 3.1] we have $\sigma _{\xi (E_n)}(E) \subseteq \overline{E{\setminus } E_n}$ for every n. Since $E = \bigcup _{n=1}^\infty E_n$ and E is a Baire space, at least one of the sets $E_n$ has nonempty interior in E. However, this contradicts $E = \sigma _{\xi (E_n)}(E) \subseteq \overline{E{\setminus } E_n}$.

For the additional statement, note that we have $\xi (\phi (E_m)) = \xi (E_m)$ for each $\phi \in \Phi$, and since $\Phi$ acts transitively on E, $E = \bigcup _{\phi \in \Phi } \phi (E_m^\circ )$ whenever $E_m^\circ \ne \emptyset$. Thus every point of E has a closed relative neighbourhood in E with characteristic equal to $\xi (E_m)$. The statement now follows from [9, Proposition 1.6 and the remark following it]. $\square$

Corollary 4.5

Let E, S and F be closed subsets of $\Delta (A)$ such that $E\subseteq S \subseteq F$, and suppose that $\sigma _m(S) = S$ for some $m\in \mathbb {N}$.

• (i) If E is a weak spectral set with $\xi (E)\le m$, then $\overline{S{\setminus } E} = S$.

• (ii) If F is a weak spectral set with $\xi (F)\le m$, then $\overline{F{\setminus } S} = F$.

Proof

• (i) Since $\sigma _m(S) = S$, S is not a weak spectral set with $\xi (S)\le m$ and therefore, by Theorem 4.1, the relative interior of the subset E of S must be empty.

• (ii) follows from Theorem 4.2 taking $E = S$ since $F\supseteq S$.

5 Examples

In this section we exhibit a number of examples in which Theorem 4.1 can be used to conclude that $\sigma _n(E) = E$. These examples concern closures of conjugacy classes and double cosets KaK, $a\in G$, where K is a compact subgroup of G, in $G = \Delta (A(G))$ and also the Fourier algebra of coset spaces G / K.

Let K be a compact subgroup of G. Then $K\times K$ acts on G by $(x,y)\cdot z = x^{-1}zy$, $x,y \in K$, $z\in G$, and it acts transitively on each double coset K z K. It is clear that $K\times K$ acts continuously on A(G) by $(x,y)(u)(z) = u(x^{-1} z y)$.

Example 5.1

Consider $SO(N), N \ge 3$, and identify $SO(N-1)$ with the subgroup of all elements of SO(N) which fix the vector $(1,0,\ldots , 0)^t \in \mathbb {R}^N$. Note that there is a bijection

$$\begin{aligned}{}[-1,1] \rightarrow SO(N)//SO(N-1),\quad t \rightarrow SO(N-1) A(t) SO(N-1), \end{aligned}$$

where

$$\begin{aligned} A(t) = \left( \begin{array}{l@{\quad }l}\left( \begin{array}{l@{\quad }l} t &{} -\sqrt{1-t^2}\\ \sqrt{1-t^2} &{} t \end{array}\right) &{} 0_{2,d-2} \\ 0_{d-2,2} &{} 1_{d-2,d-2}\end{array}\right) \end{aligned}$$

and $\mathrm{O}_{n,m}$ and $1_{n,m}$ denotes the zero and identity matrix of size $n\times m$, respectively. Let $E_t = SO(N-1) A(t)SO(N-1)$. Then, by [3, Corollary 3.4], $\xi (E_t) \ge \lfloor \frac{N+1}{2}\rfloor$ for every $t\in \,]-1,1[\,$. Since $SO(N-1)\times SO(N-1)$ acts transitively on $E_t$, we conclude that $\sigma _n(E_t) = E_t$ for all $n < \lfloor \frac{N+1}{2}\rfloor$ and $t\in \,]-1,1[\,$. However, we do not know whether these sets $E_t$ are weak spectral sets. In contrast, the sets $E_1$ and $E_{-1}$ are sets of synthesis, because $E_1 = SO(N-1)$ and $E_{-1} = A(-1)SO(N-1)$ since $A(-1)$ commutes with all matrices in $SO(N-1)$.

In the next example, for any $a\in G$, let $C(a) = \overline{\{x a x^{-1}: x\in G\}}$, the closure of the conjugacy class of a.

Example 5.2

• (1) Let G be a 2-step nilpotent locally compact group and let Z denote its centre. For $x,y\in G$, let $[y,x] = y x y^{-1}x^{-1}$ denote the commutator of y and x. Then, for $a\in G$, $C(a) = a\{[a^{-1},x]: x\in G\}$ and the map $x\rightarrow [a^{-1},x]$ is a homomorphism from G into Z. Thus $\overline{C(a)}$ is a coset of some closed subgroup of G and hence a set of synthesis.

• (2) In [15] Meaney has investigated the problem of whether conjugacy classes in compact connected Lie groups are sets of synthesis. We remind the reader that an element of a compact connected Lie group G is called regular if it is not contained in two distinct maximal tori and that the set of regular elements of G has full Haar measure. One of the main results of [15] says that if G is semisimple and a is a regular element of G, then C(a) fails to be a set of synthesis [15, Theorem 3.3]. From this it can be deduced that if G is any nonabelian compact connected Lie group, then there exist elements of G whose conjucacy classes are not of synthesis [15, Corollary 3.6].

• (3) Let G be a semisimple compact connected Lie group and fix a maximal torus T of G. Then, for any regular element a of T, $\xi (C(a)) > \frac{1}{2} \dim C(a)$ [15, Corollary 3.5]. Since G acts transitively on C(a), we get $\sigma _n(C(a)) = C(a)$ for all $n \le \frac{1}{2} \dim C(a)$ (Theorem 4.1). We note that the regular elements a of T are characterized by the equation $\dim C(a) = \dim G - \dim T$ [2, Theorem 2.11].

Now consider the special case $G =SO(N)$, $N\ge 3$, and let $m = \lfloor \frac{N}{2}\rfloor$. Then a maximal torus T(N) of G is given by $T(N) = SO(2) \times \cdots \times SO(2)$ (m-fold direct product), where T(N) acts on $\mathbb {R}^{2m} = \mathbb {R}^2\times \cdots \times \mathbb {R}^2$ and $\mathbb {R}^{2m + 1} = \mathbb {R}^2\times \cdots \times \mathbb {R}^2\times \mathbb {R}$ in the obvious manner [2, Theorem 3.4]. Since $\dim SO(N) = \frac{1}{2} N(N-1)$, we obtain

Therefore, for any regular element $a\in T(N)$, $\sigma _n(C(a)) = C(a)$ at least for all $n\le \lfloor \frac{N}{2}\rfloor \left( \lfloor \frac{N}{2}\rfloor - 1\right)$.

Remark 5.3

Let G be a locally compact group and K a compact subgroup of G. The Fourier algebra A(G / K) of the space of left cosets of K in G was introduced by Forrest [6]. It can be identified with the subalgebra of A(G) consisting of all functions in A(G) which are constant on left cosets of K, and then point evaluations provide a homeomorphism between G / K and $\Delta (A(G/K))$. For further results on A(G / K) see [18]. Note that in general A(G / K) is much smaller than A(G). In fact, $u\in A(G)$ belongs to A(G / K) if and only if u can be represented as $u = f*\check{g}$, where $f, g \in L^2(G)$ and $L_kg = g$ for all $k\in K$ [12, Lemma 5.1].

Let H be a closed subgroup of G. Then H acts continuously on G / K and on A(G / K) by $h(u)(xK) = u(h^{-1}xK)$. Let $x\in G$ and $E = H(xK) = \{hxK: h\in H\}$. Then H acts transitively on E and hence, for each $m\in \mathbb {N}$, either $\sigma _m(E) = E$ or $\sigma _m(E)= \emptyset$ by Theorem 4.1.

Now suppose that G is a semidirect product $G = N\rtimes K$, and identify N with the normal subgroup $N\times \{e_K\}$ and K with the subgroup $\{e_N\}\times K$ of G. Then A(G / K) can equally well be viewed as a subalgebra of A(N), and we have $\Delta (A(G/K)) = N$. Indeed, the restriction map $u\rightarrow u|_N$ is an isometric isomorphism of A(G / K) onto its range B, say. For $\Vert u|_N\Vert = \Vert u\Vert$, it suffices to show that $\Vert u\Vert \le \Vert u|_N\Vert$. There exists $v\in A(G)$ such that $v|_N = u$ and $\Vert v\Vert = \Vert u|_N\Vert$. Then define $w\in A(G/K)$ by $w(x) = \int _K v(xk)dk$, where dk is normalized Haar measure of K, and notice that $\Vert w\Vert \le \Vert v\Vert$ and $w|_N = v|_N = u|_N$, whence $w = u$.

In our final example, we take for G the motion group of $\mathbb {R}^N$ and for both, K and H, the subgroup SO(N).

Example 5.4

Let $G_N = \mathbb {R}^N \rtimes SO(N)$, $N\ge 2$, and let B denote the algebra $A(G_N/SO(N))$, viewed as a subalgebra of $A(\mathbb {R}^N)$ (Remark 5.3). We want to determine the sets $\sigma _n(S^{N-1})$ for $S^{N-1}\subseteq \Delta (B)$. This will be done by exploiting the main result of [25].

Let R denote the subalgebra of all radial functions in $A(\mathbb {R}^N)$, and for any $u\in R$ define $\widetilde{u}$ on $G_N$ by $\widetilde{u}(x,T) = u(x)$. If u is positive definite, then so is $\widetilde{u}$. In fact, for any $x_1, \ldots , x_n \in \mathbb {R}^N$, $T_1, \ldots , T_n\in SO(N)$ and $\lambda _1, \ldots , \lambda _n\in \mathbb {C}$, we have

$$\begin{aligned} \sum _{i,j = 1}^n \lambda _i\overline{\lambda _j}\, \widetilde{u}((x_j, T_j)^{-1} (x_i, T_i))= & {} \sum _{i,j = 1}^n \lambda _i\overline{\lambda _j} \,\widetilde{u}(T_j^{-1}(x_i - x_j), T_j^{-1}T_i)\\= & {} \sum _{i,j = 1}^n \lambda _i\overline{\lambda _j}\, u(T_J^{-1}(x_i - x_j)) \\= & {} \sum _{i,j = 1}^n \lambda _i\overline{\lambda _j}\, u(x_i - x_j) \ge 0. \end{aligned}$$

Since every function in R is a finite linear combination of positive definite functions in R, it follows that the map $u\rightarrow \widetilde{u}$ is an embedding of R into $A(G_N/SO(N))$. In this manner, we view R as a subalgebra of B. Then, for each $n\in \mathbb {N}$,

$$\begin{aligned} (k(S^{N-1}) \cap R)^n \subseteq (k(S^{N-1}) \cap B)^n\subseteq k(S^{N-1}). \end{aligned}$$

Now, by [25, Theorem 3 and Lemma 2(iv)], each inclusion in the following decreasing chain of ideals

$$\begin{aligned} k(S^{N-1}) \cap R \supseteq (k(S^{N-1})\cap R)^2 \supseteq \cdots \supseteq (k(S^{N-1}) \cap R)^{\lfloor \frac{N+1}{2}\rfloor } = J(S^{N-1})\cap R \end{aligned}$$

is proper. It follows from this that the chain of ideals $(k(S^{N-1})\cap B)^n$ has the same length. Indeed, if $(k(S^{N-1})\cap B)^m = (k(S^{N-1})\cap B)^{m+1}$ for some $1\le m < \lfloor \frac{N+1}{2}\rfloor$, then

$$\begin{aligned} (k(S^{N-1})\cap B)^m = (k(S^{N-1})\cap B)^{\lfloor \frac{N+1}{2}\rfloor }\subseteq k(S^{N-1})^{\lfloor \frac{N+1}{2}\rfloor } = J(S^{N-1}), \end{aligned}$$

because the characteristic of $S^{N-1} \subseteq \Delta (A(\mathbb {R}^N)$ equals $\lfloor \frac{N+1}{2}\rfloor$. Hence $(k(S^{N-1}) \cap R)^m \subseteq J(S^{N-1})$, which is a contradiction. Since SO(N) acts transitively on $S^{N-1}\subseteq \Delta (B)$, we conclude that

$$\begin{aligned} \sigma _n(S^{N-1}) = S^{N-1}\quad \text{ for } 1 \le n < \left\lfloor \frac{N+1}{2}\right\rfloor \quad \text{ and }\quad \sigma _{\lfloor \frac{N+1}{2}\rfloor }(S^{N-1}) = \emptyset . \end{aligned}$$

6 Further results on weak spectral sets in the spectrum of a commutative Banach algebra

In this section, as an application of our tools, $\sigma _n(E)$ and $I_n(E)$, we present a series of results on weak spectral sets which do not follow from those in the previous sections. The algebra A is assumed to have the properties specified at the beginning of Sect. 4.

Lemma 6.1

Let E and D be closed subsets of $\Delta (A)$ such that $D\subseteq E$ and D is a Ditkin set. Then, for every $n\in \mathbb {N}$,

$$\begin{aligned} \sigma _n(E) \subseteq \sigma _n(\overline{E{\setminus } D})\quad \mathrm{and}\quad \overline{\sigma _n(E){\setminus } D} = \sigma _n(E). \end{aligned}$$

In particular, if there is a family of Ditkin sets $D_\lambda \subseteq E$, $\lambda \in \Lambda$, such that $\bigcap _{\lambda \in \Lambda } \sigma _n(\overline{E {\setminus } D_\lambda }) = \emptyset$, then E is of weak synthesis with $\xi (E)\le n$.

Proof

Since D is a Ditkin set, $k(E) = k(\overline{E{\setminus } D})k(D)$ (see Corollary 6.4 below). So $k(E)^n = k(\overline{E{\setminus } D})^nk(D)^n = k(\overline{E{\setminus } D})^n k(D)$. For the first asserted inclusion it suffices to show that $\sigma ((ab)\cdot f) \subseteq \sigma _n(\overline{E{\setminus } D})$ for every $a \in k(\overline{E{\setminus } D})^n$, $b\in k(D)$ and $f\in J(E)^\perp$. Fix such a, b and f, and observe first that $\sigma (b\cdot f) \subseteq \overline{E{\setminus } D}$. In fact, since D is a set of synthesis, there exists a sequence $(b_j)_j$ in j(D) converging to b. Then

$$\begin{aligned} \sigma (b_j\cdot f) \subseteq \sigma (f) \cap \mathrm{supp}\widehat{b_j} \subseteq E \cap (\Delta (A){\setminus } D) = E{\setminus } D \end{aligned}$$

and therefore $\sigma (b\cdot f) \subseteq \overline{E{\setminus } D}$. As $a\in k(\overline{E{\setminus } D})^n$, it follows that $\sigma ((ab)\cdot f) \subseteq \sigma _n(\overline{E{\setminus } D})$.

For $\sigma _n(E)\subseteq \overline{\sigma _n(E){\setminus } D}$, it is enough to verify that if $a\in k(E)^n$ and $f\in J(E)^\perp$, then $\sigma (a\cdot f) \subseteq \overline{\sigma _n(E){\setminus } D}$. Since $a\in k(D)$ and D is a Ditkin set, $a = \lim _{i \rightarrow \infty } (a b_i)$, where $b_i\in j(D)$. Then, because $\sigma (a\cdot f) \subseteq \sigma _n(E)$, for each i,

$$\begin{aligned} \sigma ((a b_i)\cdot f) \subseteq \sigma (a\cdot f) \cap \mathrm{supp}\widehat{b_i} \subseteq \sigma _n(E)\cap (\Delta (A){{\setminus }} D) = \sigma _n(E) {\setminus } D. \end{aligned}$$

Consequently, $\sigma (a\cdot f) \subseteq \overline{ \sigma _n(E) {\setminus } D}$. This proves $\sigma _n(E)\subseteq \overline{ \sigma _n(E) {\setminus } D}$ and hence equality. $\square$

Next we give a sufficient condition for $k(E\cup F) = k(E)k(F)$ to hold.

Lemma 6.2

Let E and F be closed subsets of $\Delta (A)$ and suppose that there exists a Ditkin set D such that $E\cap F \subseteq D \subseteq E\cup F$. Then

$$\begin{aligned} k(E\cup F) = k(E)k(F). \end{aligned}$$

Proof

It suffices to show that $k(E\cup F) \subseteq k(E)k(F)$. To that end, note first that, since D is a set of synthesis,

$$\begin{aligned} k(D) = J(D) \subseteq J(E\cap F) = \overline{J(E) + J(F)} \subseteq \overline{k(E) + k(F)}. \end{aligned}$$

Thus using that D is a Ditkin set,

$$\begin{aligned} a \in \overline{ a \overline{( k(E) + k(F))}} = \overline{a (k(E) + k(F))} \end{aligned}$$

for every $a\in k(D)$. Therefore, given $a\in k(E\cup F)\subseteq k(D)$, there are sequences $(a_i)_i \subseteq k(E)$ and $(b_i)_i\subseteq k(F)$ such that $\Vert a - a(a_i + b_i)\Vert \rightarrow 0$. As $a a_i \in k(E)k(F)$ and $a b_i\in k(E)k(F)$, we get that $a\in k(E)k(F)$. The reverse inclusion being trivial, we get $k(E\cup F) = k(E)k(F)$, . $\square$

Remark 6.3

The conclusion of Lemma 6.2 also holds for closed subsets E and F of $\Delta (A)$ satisfying $k(E\cup F)^2 = k(E\cup F)$. To see this, let $a\in k(E\cup F)$ be given. Then, given $\epsilon > 0$, there exist $a_i, b_i \in k(E\cup F)$, $1 \le i \le n$, such that $\Vert a - \sum _{i=1}^n (a_i b_i)\Vert \le \epsilon$. Since $k(E\cup F) = k(E)\cap k(F)$, each $a_ib_i$ belongs to k(E)k(F). Since $\epsilon > 0$ is arbitrary, it follows that $a\in k(E)k(F)$. So $k(E\cup F) \subseteq k(E)k(F)$, as was to be shown.

We also note that if one of the ideals k(E), k(F) or $k(E\cup F)$ has an approximate identity, then $k(E\cup F)^2 = k(E\cup F)$.

As an immediate consequence of Lemma 6.2 we obtain

Corollary 6.4

For any closed subset E of $\Delta (A)$ and Ditkin set $D\subseteq \Delta (A)$, $k(E\cup D) = k(E)k(D)$.

Theorem 6.5

Let E and F be closed subsets of $\Delta (A)$. If F is a weak spectral set and $m = \xi (F)$, then $E\cup F$ is a weak spectral set with $\xi (E\cup F)\le m$ if and only if $\sigma _m(E) \subseteq F$ and $k(E\cup F)^m = k(E)^m k(F)^m$.

Proof

Suppose first that $E\cup F$ is a weak spectral set with $n = \xi (E\cup F) \le m$. Then

$$\begin{aligned} k(E\cup F)^m \subseteq k(E\cup F)^n = J(E\cup F) = J(E)J(F) \subseteq k(E)^m k(F)^m \end{aligned}$$

and hence $k(E\cup F)^m = k(E)^m k(F)^m$. Moreover,

$$\begin{aligned} J(F)k(E)^m \subseteq k(F)^m k(E)^m = k(E\cup F)^m = J(E\cup F) \subseteq J(E). \end{aligned}$$

By the very definition of the ideal $I_m(E)$, this means that $J(F)\subseteq I_m(E)$. This implies $\sigma _m(E) = h(I_m(E)) \subseteq h(J(F)) = F$.

Conversely, assume that $\sigma _m(E) \subseteq F$ and $k(E\cup F)^m = k(E)^m k(F)^m$. Then $J(F)k(E)^m \subseteq J(E)$ and hence

$$\begin{aligned} k(E\cup F)^m= & {} k(E)^m k(F)^m = k(E)^m J(F) = k(E)^m J(F)^2\\\subseteq & {} J(E)J(F) = J(E\cup F). \end{aligned}$$

Thus $E\cup F$ is of weak synthesis with $\xi (E\cup F) \le m$. $\square$

It is worth pointing out that the condition $k(E)^n = k(E)$ for all $n\in \mathbb {N}$ does not imply that E is of weak synthesis. In fact, in [5, Theorem 5.3] an example is given of a regular uniform algebra on a compact Hausdorff space X such that $\Delta (A) = X$ and there exists a point $x\in X$ such that the singleton $\{x\}$ fails to be a set of synthesis, but $k(\{x\})^n = k(\{x\})$ holds for all $n\in \mathbb {N}$.

Corollary 6.6

Let E and D be closed subsets of $\Delta (A)$ such that D is a Ditkin set. Then $E\cup D$ is a set of synthesis if and only if $\sigma _1(E) \subseteq D$.

Proof

Taking $F = D$ in Theorem 6.5, we see that $E\cup D$ being a set of synthesis forces $\sigma _1(E) \subseteq D$. Conversely, suppose that $\sigma _1(E)\subseteq D$. Since D is a Ditkin set, we have $k(E\cup D) = k(E)k(D)$ by Corollary 6.4. Another application of Theorem 6.5 shows that $E\cup D$ is a set of synthesis. $\square$

Another immediate consequence of Theorem 6.5 is the following

Corollary 6.7

Let E and F be closed subsets of $\Delta (A)$ such that F is of synthesis. If k(E) has an approximate identity and $\sigma _1(E)\subseteq F$, then $E\cup F$ is a set of synthesis.

This result shows that synthesibility of $E\cup F$ depends substantially on the position of both sets to each other.

It is well-known that if E is a closed subset of $\Delta (A)$ and if there exists a Ditkin set D such that $\partial (E) \subseteq D\subseteq E$, then E is a set of synthesis. In fact, this is a simple application of the local membership principle. We continue with a generalization in the context of weak spectral synthesis.

Theorem 6.8

Let E be a closed subset of $\Delta (A)$ and $m\in \mathbb {N}$. Suppose that there are countably many Ditkin sets $D_n$, $n\in \mathbb {N}$, such that $\sigma _m(E) \subseteq \bigcup _{n=1}^\infty D_n \subseteq E$. Then E is a set of weak synthesis with $\xi (E) \le m$.

Proof

By Lemma 6.1, each of the sets $V_n = \sigma _m(E) {\setminus } (\sigma _m(E)\cap D_n)$ is open and dense in $\sigma _m(E)$. Since $\sigma _m(E)$ is a Baire space, $\bigcap _{n=1}^\infty V_n$ is dense in $\sigma _m(E)$. However, since $\bigcup _{n=1}^\infty D_n \supseteq \sigma _m(E)$, $\bigcap _{n= 1}^\infty V_n = \emptyset$. Thus $\sigma _m(E) = \emptyset$, whence E is a weak spectral set with $\xi (E) \le m$.

If $D = \bigcup _{n=1}^\infty D_n$ is closed, then it is also a Ditkin set. Therefore, the main issue of the preceding theorem is that D is not assumed to be closed in $\Delta (A)$. The following corollary is a considerable extension of a result due to Warner [26].

Corollary 6.9

Let E and F be closed subsets of $\Delta (A)$ and suppose that there exists a sequence $(D_n)_n$ of Ditkin sets such that $\partial (E)\cap F \subseteq \bigcup _{n=1}^\infty D_n \subseteq E$.

• (i) If, for some $m\in \mathbb {N}$, $k(E)^m J(F)\subseteq J(E)$, then E is a weak spectral set with $\xi (E)\le m$.

• (ii) If $E\cup F$ is a weak spectral set, then so is E and $\xi (E)\le \xi (E\cup F)$.

Proof

• (i) Since $k(E)^m J(F)\subseteq J(E)$, we have $\sigma _m(E) \subseteq F$, and hence by hypothesis

  $$\begin{aligned} \sigma _m(E) \subseteq \partial (E)\cap F \subseteq \bigcup _{n=1}^\infty D_n \subseteq E. \end{aligned}$$

  It follows now from Theorem 6.8 that E is of weak synthesis with $\xi (E)\le m$.

• (ii) Let $m = \xi (E\cup F)$. Then

  $$\begin{aligned} k(E)^mJ(F) \subseteq k(E)^m k(F)^m \subseteq k(E\cup F)^m = J(E\cup F) \subseteq J(E), \end{aligned}$$

and hence the statement follows from (i) $\square$