THE RESTRICTED CONNECTED HULL: FILLING THE HOLE

Abstract

By defining and applying the restricted topology, we have investigated certain connections between the boundary spectrum, the exponential spectrum, the topological boundary of the spectrum and the connected hull of the spectrum (see Mouton and Harte [‘Linking the boundary and exponential spectra via the restricted topology’, J. Math. Anal. Appl. 454 (2017), 730–745]). We now solve a remaining problem regarding the restricted connected hull.

1 Introduction and preliminaries

For a complex Banach algebra A with unit $1$ , let $A^{-1}$ denote the set of all invertible elements. We will indicate elements of the form $\lambda 1$ in A by $\lambda $ . We denote the spectrum $\{\lambda \in \mathbb C: a-\lambda \not \in A^{-1}\}$ of an element a in A by $\sigma (a)$ (or by $\sigma (a, A)$ , if necessary, to avoid confusion). The symbols $\partial \sigma (a)$ and $\eta \sigma (a)$ will denote the boundary and the connected hull, respectively, of $\sigma (a)$ . (As usual, the connected hull of a set K in $\mathbb C$ is the union of K with its ‘holes’, where a hole of K is a bounded component of $\mathbb C\backslash K$ .)

With $\partial _A (A\backslash A^{-1})$ denoting the topological (norm) boundary of $A\backslash A^{-1}$ in A and $\mathrm {Exp}(A)$ the set of all (finite) products of exponentials of elements in A, we will consider the boundary spectrum $S_{\partial }(a):= \{\lambda \in \mathbb C: a-\lambda \in \partial _A (A\backslash A^{-1})\}$ (see [6]) and the exponential spectrum $\varepsilon (a):=\{\lambda \in \mathbb C: a-\lambda \not \in \mathrm {Exp}(A)\}$ (see [4]) of a in A. Both these spectra are nonempty compact subsets of the complex plane and, for every $a \in A$ ,

(1.1) $$ \begin{align} \partial \sigma(a) \subseteq S_{\partial}(a) \subseteq \sigma(a) \subseteq \varepsilon(a) \subseteq \eta \sigma(a). \end{align} $$

Applications of the boundary and exponential spectra can be found in [7,4], respectively.

Throughout this note, A will be a complex Banach algebra with unit 1 and B a closed subalgebra of A such that $1 \in B$ .

In [8, Definition 3.2 and Theorem 3.5], the restricted topology (or the B-topology) is defined via the restricted closure $\mathrm {cl}^B(K)$ of an arbitrary subset K of A, where $\mathrm {cl}^B(K)$ is the set of all elements $a \in A$ with the property that $a-U$ and K have nonempty intersection, for every neighbourhood U of $0$ in B in the relative (norm) topology. The set $K=A\backslash A^{-1}$ being of particular interest in this context, the restricted boundary of $A\backslash A^{-1}$ and the component of $A^{-1}$ containing $1$ in the B-topology are denoted by $\partial ^B(A\backslash A^{-1})$ and $\mathrm {Comp}^B(1, A^{-1})$ , respectively. The restricted connected hull ${\eta }^B(A\backslash A^{-1})$ of $A\backslash A^{-1}$ in A is given by

(1.2) $$ \begin{align} {\eta}^B(A\backslash A^{-1}) := A\backslash \mathrm{Comp}^B(1, A^{-1}). \end{align} $$

In addition, for an element $a \in A$ , the restricted boundary and the restricted connected hull are defined in [8] as

(1.3) $$ \begin{align} \partial_{B}(a):= \partial_B(a,A) = \{\lambda \in \mathbb C:a-\lambda \in \partial^{B}(A\backslash A^{-1})\} \end{align} $$

and

(1.4) $$ \begin{align} \eta_{B}(a):= \eta_B(a, A) = \{\lambda \in \mathbb C:a-\lambda \in \eta^{B}(A\backslash A^{-1})\}, \end{align} $$

respectively.

We now recall the following results from [8].

Theorem 1.1 [8, Corollary 6.4].

For any $a \in A$ , $\partial _{\mathbb C}(a) \subseteq \partial _{B}(a) \subseteq \partial _{A}(a)$ , with $\partial _{\mathbb C}(a) = \partial \sigma (a)$ and $\partial _{A}(a)= S_{\partial }(a)$ .

Theorem 1.2 [8, Corollary 6.9].

For any $a \in A$ , $\eta _A(a) \subseteq \eta _B(a) \subseteq \eta _{\mathbb C}(a)$ , with $\eta _A(a)=\varepsilon (a)$ . In addition, if $a \notin \mathbb C$ , then $\eta _{\mathbb C}(a) = \mathbb C$ , while if $a \in \mathbb C$ , then $\eta _{\mathbb C}(a) = \eta \sigma (a)$ .

We note that there is a type of duality between the boundary and exponential spectra:

$$ \begin{align*}S_{\partial}(a)=\partial_A(a) \quad\mbox{and}\quad \varepsilon(a)=\eta_A(a).\end{align*} $$

The last equation, together with (1.1), implies that $\eta _A(a) \subseteq \eta \sigma (a)$ , and since $\partial _{\mathbb C}(a) = \partial \sigma (a)$ , we have $\partial \sigma (a) \subseteq \partial _B(a)$ . However, it is clear from the last part of Theorem 1.2 that we do not, in general, have $\eta _{\mathbb C}(a)=\eta \sigma (a)$ or that the inclusion $\eta _B(a) \subseteq \eta \sigma (a)$ holds (as in the case $B=A$ ). In Section 2, we ‘fill the hole’ in this theory by establishing exactly how far the inclusion $\eta _A(a) \subseteq \eta \sigma (a)$ can be generalised (see Theorems 2.2 and 2.4).

Applying the results in Section 2, we then establish certain mapping properties of $\eta _B$ in Section 3.

2 The restricted connected hull relative to the connected hull of the spectrum

We first observe the following result.

Lemma 2.1. $\mathrm {Exp}(B) \subseteq \mathrm {Comp}^B(1, A^{-1})$ .

This follows trivially, since $\mathrm {Exp}(B) = \mathrm {Comp}^B(1,B^{-1})$ , but can alternatively be obtained by considering a map similar to that in the proof of Lemma 3.2 and applying [8, Proposition 5.11].

The next result follows easily.

Theorem 2.2. If $a \in B$ , then $\eta _B(a) \subseteq \eta \sigma (a)$ .

Proof. Let $\lambda \notin \eta \sigma (a):=\eta \sigma (a,A)$ . Then $a-\lambda \in B$ and, since $\eta \sigma (b,A) = \eta \sigma (b,B)$ for all $b \in B$ , we have $0 \notin \eta \sigma (a-\lambda , B)$ . It follows from [1, Theorem 3.3.6] that $a-\lambda \in \mathrm {Exp}(B)$ , and hence $a-\lambda \in \mathrm { Comp}^B(1,A^{-1})$ , by Lemma 2.1. Therefore, (1.2) implies that $a-\lambda \notin \eta ^B(A\backslash A^{-1})$ , so that $\lambda \notin \eta _B(a)$ , by (1.4).

Although it is possible to have equality in Theorem 2.2 (see [8, Example 6.11]), the inclusion is, in general, proper. This is shown in the following example, where $\mathbb T$ indicates the unit circle and $\mathbb D$ the open unit disk in $\mathbb C$ .

Example 2.3. Let $l^2(\mathbb Z)$ be the Hilbert space of all bilateral square-summable sequences, A the Banach algebra ${\mathcal L}(l^2(\mathbb Z))$ of all bounded linear operators on $l^2(\mathbb Z)$ and $a \in A$ the bilateral shift. Then $\eta _A(a) \subsetneq \eta \sigma (a)$ .

Proof. By [2, Corollary 5.30], $A^{-1}=\mathrm {Exp}(A)$ , so that $\sigma (b) = \varepsilon (b)$ for all $b \in A$ . Since $\sigma (a)=\mathbb T$ (see [3, Problem 84]), it follows that $\varepsilon (a)=\mathbb T$ and $\eta \sigma (a)=\overline {\mathbb D}$ . However, $\varepsilon (a) = \eta _A(a)$ , by Theorem 1.2, and hence the result follows.

Finally, we have the following result.

Theorem 2.4. If $a \notin B$ , then $\eta _B(a) = \mathbb C$ .

Proof. By Theorem 1.2, $\varepsilon (a) = \eta _A(a) \subseteq \eta _B(a)$ for any $a \in A$ , and hence it suffices to prove that $\mathbb C\backslash \varepsilon (a) \subseteq \eta _B(a)$ whenever $a \notin B$ . So suppose that $a \notin B$ and let ${\lambda }_0 \in \mathbb C\backslash \varepsilon (a)$ . Then ${\lambda }_0 \notin \sigma (a)$ . If $G=\mathrm {Comp}^B(a-{\lambda }_0, A^{-1})$ , then $a-G = \mathrm { Comp}^B({\lambda }_0, B\backslash \sigma (a))$ by [8, Proposition 5.10], and since $a \notin B$ , it follows that $1 \notin G$ . Therefore, $G\neq \mathrm {Comp}^B(1,A^{-1})$ , so that $a-\lambda _0 \notin \mathrm {Comp}^B(1,A^{-1})$ . By (1.2), we then have $a-\lambda _0 \in \eta ^B(A\backslash A^{-1})$ , so that $\lambda _0 \in \eta _B(a)$ , by (1.4).

3 Mapping properties of the restricted connected hull

Let $K(\mathbb C)$ denote the set of all nonempty, compact subsets of $\mathbb C$ . A mapping $\omega :A \rightarrow K(\mathbb C)$ is said to be a Mobius spectrum on A (see [5]) if $\omega (f(a)) = f(\omega (a))$ :

1. (a) for all $a \in A$ and for all functions f of the form $f(\lambda )=\alpha \lambda + \beta $ ( $\alpha , \beta \in \mathbb C$ ); and

2. (b) for all $a \in A$ and $f(\lambda )={1}/{\lambda }$ , such that f is well defined on $\omega (a) \cup \sigma (a)$ .

We immediately have the following result.

Proposition 3.1. $\eta _B$ is a Mobius spectrum on B.

Proof. Since $\varepsilon $ is a Mobius spectrum on any Banach algebra (see [5]), we have $\varepsilon (f(a),B)=f(\varepsilon (a,B))$ for all $a \in B$ and for all functions f of the form $f(\lambda )=\alpha \lambda + \beta $ ( $\alpha , \beta \in \mathbb C$ ), and that $\varepsilon (a^{-1},B) = (\varepsilon (a,B))^{-1}$ for all $a \in B$ such that $0 \notin \varepsilon (a,B) \cup \sigma (a,B)$ . Then the result follows since $\eta _B(x,B)=\{\lambda \in \mathbb C: x-\lambda \in \eta ^{B}(B\backslash B^{-1})\}=\varepsilon (x,B)$ for any $x \in B$ .

Returning to $\eta _{B}(a):= \eta _B(a, A)$ , we now develop some more mapping properties, starting with the following lemma.

Lemma 3.2. If $a \in \mathrm {Exp}(B)$ and $b \in B \cap \mathrm {Comp}^B(1,A^{-1})$ , then $ab \in \mathrm {Comp}^B(1,A^{-1})$ .

Proof. Let $a=e^{b_1}e^{b_2}\cdots e^{b_n}$ with $b_1, b_2, \ldots, b_n \in B$ , and define $f:[0,1] \rightarrow A$ by $f(t) = e^{tb_1}e^{tb_2}\cdots e^{tb_n}b$ . Then $f(0)=b$ , $f(1)=ab$ and f is continuous in the norm topology of A. Since $f([0,1]) \subseteq B$ , it follows from [8, Proposition 5.11] that f is continuous in the B-topology. Therefore, we have a continuous function f in the B-topology from $[0,1]$ to $A^{-1}$ joining b and $ab$ , and so $ab \in \mathrm {Comp}^B(1,A^{-1})$ .

Corollary 3.3. If $a \in B \cap \mathrm {Comp}^B(1,A^{-1})$ and $0 \neq \lambda \in \mathbb C$ , then $\lambda a \in \mathrm {Comp}^B(1,A^{-1})$ .

It follows from Corollary 3.3 that, for $a \in B$ , (1.4) may equivalently be written as

$$ \begin{align*}\eta_{B}(a)= \{\lambda \in \mathbb C:\lambda -a \in \eta^{B}(A\backslash A^{-1})\}.\end{align*} $$

Theorem 3.4. Let $a \in A$ and $f(\lambda )=\alpha \lambda + \beta $ with $\alpha , \beta \in \mathbb C$ . Then $\eta _B(f(a)) = f(\eta _B(a))$ .

Proof. Let $a \in A$ . If $\alpha = 0$ (that is, f is constant), then since

$$ \begin{align*}\eta_B(f(a))=\{\lambda \in \mathbb C: \beta-\lambda \notin \mathrm{ Comp}^B(1,A^{-1})\}\end{align*} $$

and $f(\eta _B(a))=\{\beta \}$ , it is clear that $f(\eta _B(a)) \subseteq \eta _B(f(a))$ . Now let $\lambda \in \eta _B(f(a))$ , that is, $\beta - \lambda \notin \mathrm {Comp}^B(1,A^{-1})$ . By Corollary 3.3, $\mathrm {Comp}^B(1,A^{-1})$ contains all nonzero complex scalar multiples of 1, and so $\lambda = \beta \in f(\eta _B(a))$ . Hence, $\eta _B(f(a)) \subseteq f(\eta _B(a))$ .

If $\alpha \neq 0$ and $a \notin B$ , then $f(a)=\alpha a + \beta \notin B$ , and so $\eta _B(a) = \mathbb C = \eta _B(f(a))$ , by Theorem 2.4. Since $f(\eta _B(a))=f(\mathbb C)=\mathbb C$ , we have $\eta _B(f(a))=f(\eta _B(a))$ . For the case $\alpha \neq 0$ and $a \in B$ , let $\lambda \in \eta _B(f(a))$ , so that $\alpha a + \beta - \lambda \notin \mathrm { Comp}^B(1,A^{-1})$ . If $\mu ={\alpha ^{-1}} (\lambda - \beta )$ , then $\lambda = f(\mu )$ and $a-\mu = {\alpha ^{-1}}(\alpha a + \beta - \lambda ) \notin \mathrm { Comp}^B(1,A^{-1})$ by Corollary 3.3, since $\alpha a + \beta - \lambda \in B$ . Therefore, $\mu \in \eta _B(a)$ , so that $\lambda \in f(\eta _B(a))$ . The other inclusion is obtained similarly.

Turning to the inverse function, we first observe the following result.

Proposition 3.5. Let $a \in B$ with $0 \in \sigma (a,B)\backslash \sigma (a,A)$ . If $f(\lambda )={1}/{\lambda }$ , then $f(\eta _B(a)) \subseteq \eta _B(f(a))$ .

Proof. This is obvious, since $\eta _B(a^{-1})=\mathbb C$ , by Theorem 2.4.

We can now prove the following mapping property of $\eta _B$ .

Theorem 3.6. Let $a \in B$ such that $0 \notin \eta \sigma (a)$ . If $f(\lambda ) = {1}/{\lambda }$ , then $\eta _B(f(a))=f(\eta _B(a))$ .

Proof. Let $\lambda \in \eta _B(f(a)) = \eta _B(a^{-1})$ , so that $a^{-1} - \lambda \notin \mathrm {Comp}^B(1,A^{-1})$ . If $0 \notin \eta \sigma (a):=\eta \sigma (a,A)$ , then since $\varepsilon (a,B) \subseteq \eta \sigma (a,B) = \eta \sigma (a,A)$ , it follows that $0 \notin \varepsilon (a,B)$ . Therefore, $a \in \mathrm {Exp}(B)$ , so that $a^{-1} \in \mathrm {Exp}(B)$ . By Lemma 2.1, $a^{-1} \in \mathrm { Comp}^B(1,A^{-1})$ , so that $\lambda \neq 0$ . If $a-{\lambda ^{-1}} \in \mathrm { Comp}^B(1,A^{-1})$ , then $({\lambda ^{-1}}-a)\lambda \in \mathrm {Comp}^B(1,A^{-1})$ , by Corollary 3.3. Since $a^{-1} \in \mathrm { Exp}(B)$ , it follows from Lemma 3.2 that $a^{-1}-\lambda = a^{-1}({\lambda ^{-1}}-a)\lambda \in \mathrm {Comp}^B(1,A^{-1})$ ; which is a contradiction. Hence, $a-{\lambda ^{-1}} \notin \mathrm {Comp}^B(1,A^{-1})$ , so that ${\lambda ^{-1}} \in \eta _B(a)$ , and hence $\lambda \in f(\eta _B(a))$ . The other inclusion is obtained similarly.

In the following example, let $\mathbb T$ indicate the unit circle and $\mathbb D$ the open unit disk in $\mathbb C$ , as before.

Example 3.7. Let $A={\mathcal C}(\mathbb T)$ and $B={\mathcal A}(\overline {\mathbb D})$ , as in [8, Example 6.11]. Let $a \in B$ denote the identity function $f(\lambda )=\lambda $ on $\overline {\mathbb D}$ . Then, $\sigma (a,A)=\mathbb T$ , $\sigma (a,B)=\overline {\mathbb D} = \eta _B(a)$ and $\eta _B(a^{-1})=\mathbb C$ .

Proof. The first two statements are given by [9, Problem 9, page 399] and it was shown in [8, Example 6.11] that $\eta _B(a)=\overline {\mathbb D}$ . Since $0 \in \sigma (a,B)\backslash \sigma (a,A)$ , we have $a^{-1} \in A\backslash B$ , and hence $\eta _B(a^{-1}) = \mathbb C$ , by Theorem 2.4.

Example 3.7 shows that we do not, in general, have equality in Proposition 3.5, and also that the condition $0 \notin \eta \sigma (a)$ in Theorem 3.6 cannot be omitted.