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 [Reference Mouton6]) and the exponential spectrum $\varepsilon (a):=\{\lambda \in \mathbb C: a\lambda \not \in \mathrm {Exp}(A)\}$ (see [Reference Harte4]) of a in A. Both these spectra are nonempty compact subsets of the complex plane and, for every $a \in A$ ,
Applications of the boundary and exponential spectra can be found in [Reference Mouton7,Reference Harte4], 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 [Reference Mouton and Harte8, Definition 3.2 and Theorem 3.5], the restricted topology (or the Btopology) 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 $aU$ 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 Btopology 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
In addition, for an element $a \in A$ , the restricted boundary and the restricted connected hull are defined in [Reference Mouton and Harte8] as
and
respectively.
We now recall the following results from [Reference Mouton and Harte8].
Theorem 1.1 [Reference Mouton and Harte8, 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 [Reference Mouton and Harte8, 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:
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 [Reference Mouton and Harte8, 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 [Reference Aupetit1, 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 [Reference Mouton and Harte8, 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 squaresummable 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 [Reference Douglas2, 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 [Reference Halmos3, 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 $aG = \mathrm { Comp}^B({\lambda }_0, B\backslash \sigma (a))$ by [Reference Mouton and Harte8, 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 [Reference Harte and Wickstead5]) if $\omega (f(a)) = f(\omega (a))$ :

(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

(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 [Reference Harte and Wickstead5]), 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 [Reference Mouton and Harte8, Proposition 5.11] that f is continuous in the Btopology. Therefore, we have a continuous function f in the Btopology 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
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
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 [Reference Mouton and Harte8, 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 [Reference Taylor and Lay9, Problem 9, page 399] and it was shown in [Reference Mouton and Harte8, 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.