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
$a \in A$
,
$$ \begin{align} \partial \sigma(a) \subseteq S_{\partial}(a) \subseteq \sigma(a) \subseteq \varepsilon(a) \subseteq \eta \sigma(a). \end{align} $$
Applications of the boundary and exponential spectra can be found in [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 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
$$ \begin{align} {\eta}^B(A\backslash A^{-1}) := A\backslash \mathrm{Comp}^B(1, A^{-1}). \end{align} $$
In addition, for an element
$a \in A$
, the restricted boundary and the restricted connected hull are defined in [Reference Mouton and Harte8] as
$$ \begin{align} \partial_{B}(a):= \partial_B(a,A) = \{\lambda \in \mathbb C:a-\lambda \in \partial^{B}(A\backslash A^{-1})\} \end{align} $$
and
$$ \begin{align} \eta_{B}(a):= \eta_B(a, A) = \{\lambda \in \mathbb C:a-\lambda \in \eta^{B}(A\backslash A^{-1})\}, \end{align} $$
respectively.
We now recall the following results from [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:
$$ \begin{align*}S_{\partial}(a)=\partial_A(a) \quad\mbox{and}\quad \varepsilon(a)=\eta_A(a).\end{align*} $$
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 [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 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 [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
$a-G = \mathrm { Comp}^B({\lambda }_0, B\backslash \sigma (a))$
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 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 [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.
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