On the generalized Fredholm spectrum in a complex semisimple banach algebra

Abstract

We shall show a spectral mapping property of the generalized Fredholm spectrum in the more general context of Banach algebras. This is an extension of (Schmoeger in Demonstr Math XXXI(4):723–733, 1998, Theorem 3.3). More precisely, for $a \in {\mathcal{A}}$ which is a complex semisimple Banach algebra with identity, we shall prove that if g is a analytic function on a neighbourhood of the spectrum of a and if it has no zero in the generalized Fredholm spectrum of a then g(a) is a generalized Fredholm element in ${\mathcal{A}}$.

1 Introduction and main results

The present paper is inspired by the paper [14], so that the extension of [14, Theorem 3.3] to complex semisimple Banach algebras will be precisely given in Main Theorem. In order to prove this we first need some ideas and results on this context; thus in Sect. 2, we will expose them in the form of lemmas. Finally, the last section will be devoted to the proofs of the main results.

Next, To prepare this task we first exhibit some basic notations and definitions (also their properties that will be needed later) which are taken from [3, 4, 10, 13]. ${\mathcal{A}}$ always denotes an infinite-dimensional complex semisimple Banach algebra with identity 1. We use the following notations and appellations:

• $I({\mathcal{A}})\,{:=}\,\{e \in {\mathcal{A}}\backslash \{0,1\}, e^{2} = e \}$ which is the set of all nontrivial idempotents of ${\mathcal{A}}$,

• $I_{m}({\mathcal{A}})\,{:=}\, \{ e \in I({\mathcal{A}}), e {\mathcal{A}} e = {\mathbb {C}}e \}$ which is the set of all minimal idempotents of ${\mathcal{A}}$,

• $soc({\mathcal{A}}) \,{:=}\, \{ s \in {\mathcal{A}}: (\exists n \in {\mathbb {N}}) (\exists a_{1},\ldots ,a_{n} \in {\mathcal{A}})(\exists e_{1},\ldots ,e_{n} \in I_{m}({\mathcal{A}})): s = \sum \limits _ {k=1}^{k=n} a_{k}e_{k}\}$ which is the socle of ${\mathcal{A}}$,

• $\Phi ({\mathcal{A}}) \,{:=}\, \{ a \in {\mathcal{A}}: (\exists b \in {\mathcal{A}}): ab-1, ba -1 \in soc({\mathcal{A}}) \}$ which is the set of all Fredholm elements of ${\mathcal{A}}$,

• $\Phi _{g}({\mathcal{A}}) \,{:=}\, \{ a \in {\mathcal{A}}: (\exists b \in {\mathcal{A}}); aba=a, 1-ab-ba \in \Phi ({\mathcal{A}}) \}$ which is the set of all generalized Fredholm elements of ${\mathcal{A}}$.

Recall that in the special case where ${\mathcal{A}}\,{:=}\, {\mathcal{B}}({\mathcal{X}})$ which is the algebra of all bounded operators on a Banach space ${\mathcal{X}}$, we have $soc({\mathcal{A}}) = {\mathcal{F}}({\mathcal{X}})$ (respectively, $\Phi ({\mathcal{A}}) = \Phi ({\mathcal{X}})$, $\Phi _{g}({\mathcal{A}}) = \Phi _{g}({\mathcal{X}})$) which is the set of all finite rank (respectively, Fredholm, generalized Fredholm) operators on ${\mathcal{X}}$. Note that the role of the ideal $soc({\mathcal{A}})$, as the class of finite rank operators in the Banach space context, is vital in many theories, for instance abstract Fredholm theory, belonging to the general case of semisimple Banach algebra. Several works have been studied the socle; see [2, 4, 5, 8, 9, 11, 12, 15] and the references therein.

Now if $a \in {\mathcal{A}}$ we set $\rho _{{\mathcal{A}}}(a) \,{:=}\, \{\lambda \in {\mathbb {C}}: \lambda - a \in {\mathcal{A}}^{-1} \}$, where ${\mathcal{A}}^{-1}$ is the set of all invertible elements in ${\mathcal{A}}$, $\rho _{\Phi _{g}({\mathcal{A}})}(a) \,{:=}\, \{ \lambda \in {\mathbb {C}}: \lambda - a \in \Phi _{g}({\mathcal{A}}) \}$, $\sigma _{{\mathcal{A}}}(a) \,{:=}\, {\mathbb {C}} \backslash \rho _{{\mathcal{A}}}(a)$ which is the spectrum of a and $\sigma _{\Phi _{g}({\mathcal{A}})}(a) \,{:=}\, {\mathbb {C}} \backslash \rho _{\Phi _{g}({\mathcal{A}})}(a)$ which is the generalized Fredholm spectrum of a. Finally, Hol(a) denotes the algebra of all functions that are analytic in a neighborhood of the spectrum $\sigma _{{\mathcal{A}}}(a)$, where $a \in {\mathcal{A}}$ [6, Page 201].

For some studies of $\Phi _{g}({\mathcal{A}})$ and $\sigma _{\Phi _{g}({\mathcal{A}})}(.)$ we can see (e.g. [10] and the references therein).

The aim of this paper is to show the following result which is similar to the well known spectral mapping property that $g(a) \in {\mathcal{A}}^{-1}$ when $a \in {\mathcal{A}}$ and $g \in Hol(a)$ having no zero in $\sigma _{{\mathcal{A}}}(a)$.

Main Theorem 1.1

Let $a \in {\mathcal{A}}$. Then

• (1) If $g \in Hol(A)$ has no zero in $\sigma _{\Phi _{g}({\mathcal{A}})} (a)$ then $g(a) \in \Phi _{g}({\mathcal{A}})$;

• (2) $\sigma _{\Phi _{g}({\mathcal{A}})}(f(a)) \subseteq f(\sigma _{\Phi _{g}({\mathcal{A}})}(a))$ for all $f \in Hol(A)$.

Remark 1.2

• (1) A proof of Theorem 1.1 is given in [14] when ${\mathcal{A}}$ is the algebra ${\mathcal{B}}({\mathcal{X}})$ of all bounded operators on a Banach space ${\mathcal{X}}$.

• (2) The proof of Theorem 1.1 will be given in Sect. 3 of this paper. The method adopted in this proof is inspired by Schmoeger [14].

• (3) It is known that the inclusion in Theorem1.1 can generally be strict in the Banach algebra setting (e.g. [14, Example page 731]).

As in [14, Corollary 3.4], an sufficient and necessary condition for the equality $\sigma _{\Phi _{g}({\mathcal{A}})}(f(a)) = f(\sigma _{\Phi _{g}({\mathcal{A}})}(a))$ for all $f \in Hol(a)$ is given in the following result.

Corollary 1.3

Let $a \in {\mathcal{A}}$. The following conditions are equivalent:

• (i) $\sigma _{\Phi _{g}({\mathcal{A}})}(f(a)) = f(\sigma _{\Phi _{g}({\mathcal{A}})}(a))$ for all $f \in Hol(a)$;

• (ii) $\sigma _{\Phi _{g}({\mathcal{A}})}(a) = \emptyset$.

2 Lemmas

As in the proof of [14, Proposition 1.5], the following result will have an important role in the proof of Lemma 2.3.

Lemma 2.1

[10, Theorem 5.4]

For $x \in {\mathcal{A}}$ the following assertions are equivalent:

• (i) $x \in \Phi _{g}({\mathcal{A}})$;

• (ii) there is $y \in {\mathcal{A}}$ such that $xyx = x$ and $xy-yx \in soc({\mathcal{A}})$.

The following result is crucial in the our study concerning the general context of Banach algebras.

Lemma 2.2

Let $e \in {\mathcal{A}}$ be a nontrivial idempotent. Then we have

• (i) $e{\mathcal{A}}e$ is a semisimple Banach algebra with identity e;

• (ii) For all $y \in e{\mathcal{A}}e$ we have $y = eye$;

• (iii) $I_{m}(e{\mathcal{A}}e) \subseteq I_{m}({\mathcal{A}})$;

• (iv) $soc(e{\mathcal{A}}e) \subseteq soc({\mathcal{A}})$;

• (v) $e (soc({\mathcal{A}}))e \subseteq soc(e{\mathcal{A}}e)$.

Proof

• (i) See [15, Lemma 3.1].

• (ii) It is immediate.

• (iii) It is trivial by Rickart [13, Corollary 2.1.6].

• (iv) is a trivial consequence of (iii).

• (v) From [1, Theorem 5.29] and [1, Corollary 5.33] we obtain $(e{\mathcal{A}}e) \bigcap soc({\mathcal{A}}) \subseteq soc(e{\mathcal{A}}e)$ since $e{\mathcal{A}}e$ is a closed (semisimple) subalgebra of the semisimple Banach algebra ${\mathcal{A}}$. Moreover, we have $e(soc({\mathcal{A}}))e \subseteq (e{\mathcal{A}}e) \bigcap soc({\mathcal{A}})$. Therefore $e(soc({\mathcal{A}}))e \subseteq soc(e{\mathcal{A}}e)$.

$\square$

The following result, in this general setting, may be viewed as an analogue of [14, Proposition 1.5] and [14, Corollary 1.5]. His proof of this general case is based considerably on 2.2. Recall that $S^{c}$ denotes the commutant of $S \subseteq {\mathcal{A}}$, that is the set of all $x \in {\mathcal{A}}$ such that x commutes with every element in S.

Lemma 2.3

Let $e_{1}, e_{2} \in {\mathcal{A}}$ two nontrivial idempotents such that $1 = e_{1}+ e_{2}$. For all $a \in \{e_{1},e_{2}\}^{c}$ we have the following:

• (i) $a \in \Phi _{g}({\mathcal{A}})$ $\Leftrightarrow$ $e_{k}a \in \Phi _{g}({\mathcal{A}}_{k})$ for $k=1,2$, where ${\mathcal{A}}_{k}\,{:=}\,e_{k}{\mathcal{A}}e_{k}$;

• (ii) $\rho _{\Phi _{g}({\mathcal{A}})} (a) = \rho _{\Phi _{g}({\mathcal{A}}_{1})} (ae_{1}) \bigcap \rho _{\Phi _{g}({\mathcal{A}}_{2})} (ae_{2})$;

• (iii) $\sigma _{\Phi _{g}({\mathcal{A}})} (a) = \sigma _{\Phi _{g}({\mathcal{A}}_{1})} (ae_{1}) \bigcup \sigma _{\Phi _{g}({\mathcal{A}}_{2})} (ae_{2})$.

Proof

• (i) Assume that $a \in \Phi _{g}({\mathcal{A}})$. By Lemma 2.1, there exists $b \in {\mathcal{A}}$ such that

  $$\begin{aligned} aba=a , ab-ba \in soc({\mathcal{A}}). \end{aligned}$$

  (1)

  Then, $eae = (eae)(ebe)(eae)$ and $(eae)(ebe)-(ebe)(eae) \in soc(e{\mathcal{A}}e)$ by part (v) in Lemma 2.2, where $e \in \{ e_{1}, e_{2} \}$. Therefore $eae \in \Phi _{g}(e{\mathcal{A}}e)$ from Lemma 2.1. Conversely, again by Lemma 2.1, for $k=1,2$ take $u_{k} \in e_{k}{\mathcal{A}}e_{k}$ such that $ae_{k} u_{k}ae_{k}=ae_{k}$ and $ae_{k}u_{k}-u_{k}ae_{k} \in soc(e_{k}{\mathcal{A}}e_{k})$. Put $u\,{:=}\, e_{1}u_{1}e_{1} + e_{2}u_{2}e_{2}$. It is easy to verify that $aua =a$ and $au-ua \in soc(e_{1}{\mathcal{A}}e_{1}) + soc(e_{2}{\mathcal{A}}e_{2})$. Thus, by part (iv) in Lemmas 2.2 and 2.1, $a \in \Phi _{g}({\mathcal{A}})$.

• (ii) It is trivial if we apply the part (i) to $\lambda - a$ which belongs to $\{ e_{1},e_{2}\}^{c}$, where $\lambda \in {\mathbb {C}}$.

• (iii) It is now immediate.

$\square$

The following result is just an immediate adaptation, without any new technique, to our context of [14, Proposition 3.2].

Lemma 2.4

Let $a \in {\mathcal{A}}$ and $g \in Hol(a)$. Suppose that g has at most a finite number of zeros in $\sigma _{{\mathcal{A}}}(a)$ and that each such zero belongs to $\rho _{\Phi _{g}({\mathcal{A}})}(a)$. Then $g(a) \in \Phi _{g}({\mathcal{A}})$.

Proof

If g has no zero in $\sigma _{{\mathcal{A}}}(a)$ then g(a) is invertible in ${\mathcal{A}}$ by the Spectral Mapping Theorem, hence $g(a) \in \Phi _{g}({\mathcal{A}})$.

Otherwise, we would have

$$\begin{aligned} f(\lambda ) = (\lambda - \lambda _{1})^{n_{1}}(\lambda - \lambda _{2})^{n_{2}}\ldots (\lambda - \lambda _{k})^{n_{k}}h(\lambda ) \end{aligned}$$

with h has no zero in $\sigma _{{\mathcal{A}}}(a)$, $h \in Hol(a)$ and $\lambda _{1},\ldots ,\lambda _{k} \in \sigma _{{\mathcal{A}}}(a) \bigcap \rho _{\Phi _{g}({\mathcal{A}})}(a)$ and $n_{1},\ldots ,n_{k} \in {\mathbb {N}}\backslash \{0\}$. Thus, from [10, Theorem 5.2] we derive $g(a) \in \Phi _{g}({\mathcal{A}})$. $\square$

3 Proof of main results

After having prepared the appropriate ingredients, we are now at the point of giving the proof of Main Theorem 1.1.

Proof

(1) Assume that $g \in Hol(a)$ sucht that

$$\begin{aligned} Z_{\sigma _{{\mathcal{A}}}(a)}(g) \bigcap \sigma _{\Phi _{g}({\mathcal{A}})} (a) = \emptyset \end{aligned}$$

(2)

where $Z_{G}(g)\,{:=}\, \{z : z \in G, g(z)=0\}$ for all $G \subseteq D$ and D is an open neighborhood of $\sigma _{{\mathcal{A}}}(a)$ on which g is analytic. We will show that $g(a) \in \Phi _{g}({\mathcal{A}})$. The proof is divided in tree steps.

• (Case 1) If g has at most a finite number of zero in $\sigma _{{\mathcal{A}}}(a)$ then $g(a) \in \Phi _{g}({\mathcal{A}})$.

• (Case 2) Suppose now that g has infinitely many zeros on $\sigma _{{\mathcal{A}}}(a)$.

Step 1 There are two closed disjoint subsets $\sigma _{1}$ and $\sigma _{2}$ and two disjoint open subsets $D_{1}$, $D_{2}$ such that $\sigma _{{\mathcal{A}}}(a)= \sigma _{1} \bigcup \sigma _{2}$, $\sigma _{k} \subseteq D_{k}$ for $k= 1, 2$ , $D_{1} \bigcup D_{2} \subseteq D$, g vanishes on $D_{1}$ and g has only finitely many zeros on $\sigma _{2}$. Indeed, let $\sigma$ be the set of all $\lambda \in \sigma _{{\mathcal{A}}}(a)$ such that g is locally zero at $\lambda$, that is

$$\begin{aligned} \sigma \,{:=}\, \{ \lambda \in \sigma _{{\mathcal{A}}}(a): (\exists \mu _{\lambda }> 0): g_{|D(\lambda , \mu _{\lambda })} = 0 \}. \end{aligned}$$

Note that $\sigma$ is a nontrivial subset of $\sigma _{{\mathcal{A}}}(a)$ since g has infinitely many zeros on $\sigma _{{\mathcal{A}}}(a)$ and has no zero in $\sigma _{\Phi _{g}({\mathcal{A}})}(a)$. For every $\lambda \in \sigma$, let $\mu _{\lambda } > 0$ such that g vanishes on $D(\lambda ,\mu _{\lambda })$. Then $\sigma = (\bigcup \nolimits _{\lambda \in \sigma }D(\lambda ,\mu _{\lambda })) \bigcap \sigma _{{\mathcal{A}}}(a)$. Thus $\sigma$ is open in $\sigma _{{\mathcal{A}}}(a)$. Let now $\lambda _{0} \in \overline{\sigma }$. Take $\delta > 0$ such that $D(\lambda _{0}, \delta ) \subseteq D$ and $\lambda _{\delta } \in \sigma \bigcap D(\lambda _{0}, \delta )$. Then g vanishes on $D(\lambda _{0}, \delta )$. Therefore $\lambda _{0} \in \sigma$. Thus $\sigma$ is a spectral set of a (that is, a subset of $\sigma _{{\mathcal{A}}}(a)$ which is open and closed in $\sigma _{{\mathcal{A}}}(a)$ [7, Page 643]). Put $\sigma _{1} \,{:=}\, \sigma$ and $\sigma _{2} \,{:=}\, \sigma _{{\mathcal{A}}}(a) {\setminus } \sigma$. Then there are two disjoint open subsets $V_{1}$, $V_{2}$ such that $\sigma _{k} \subseteq V_{k}$ for $k=1,2$. Let $D_{1} \,{:=}\, (\bigcup \nolimits _{\lambda \in \sigma }D(\lambda , \mu _{\lambda })) \bigcap V_{1}$ and $D_{2} = V_{2} \bigcap D {\backslash } \sigma$. This trivially implies that g vanishes on $D_{1}$. Let now $\{\lambda _{n}\}_{n\ge 1} \subseteq \sigma _{2}$ on which g vanishes. There exists an subsequence $\{\lambda _{n_{k}}\}_{k\ge 1}$ which converges to $\lambda \in \sigma _{2}$. Then $g(\lambda )=0$, so there exists $\mu$ such that g vanishes on $D(\lambda , \mu )$. Therefore $\lambda \in \sigma _{1} \bigcap \sigma _{2} = \emptyset$. Thus g has at most a finite number of zeros in $\sigma _{2}$.

Step 2 There are nontrivial idempotents $e_{1}$ and $e_{2}$ as in Lemma 2.3 such that $g(a) e_{k} = e_{k} g(a), (k=1,2)$.

Indeed, by Conway [6, Proposition 4.11] there is a nontrivial idempotent e, which is called the spectral idempotent associated to a and $\sigma$, in ${\mathcal{A}}$ such that

• (i) If $ba=ab$ then $be=eb$; so $ae=ea$ and $g(a)e=eg(a)$;

• (ii) Let $e_{1}\,{:=}\,e$ and $e_{2}\,{:=}\,1-e$, then $\sigma _{{\mathcal{A}}}(ae_{1}) = \sigma _{1} \bigcup \{0\}$ and $\sigma _{{\mathcal{A}}}(ae_{2}) = \sigma _{2} \bigcup \{0\}$.

Step 3 We have g(a) is generalized Fredholm element in ${\mathcal{A}}$.

In fact, since g vanishes on $\sigma$ then $g(a)e_{1} = 0$ which is a generalized Fredholm element in $e_{1}{\mathcal{A}}e_{1}$. Also, $g(a)e_{2}$ is a generalized Fredholm element in $e_{2}{\mathcal{A}}e_{2}$ since, on the one hand, $g(ae_{2})=g(a)e_{2}$ which arises from the fact that $\sigma _{{\mathcal{A}}_{2}}(ae_{2}) \subseteq \sigma _{{\mathcal{A}}}(a)$; and, on the other hand, $g(ae_{2})$ is a generalized Fredholm element in $e_{2}{\mathcal{A}}e_{2}$ by using Lemmas 2.3(iii), 2.4 and Relation 2. Thus, the part (i) in Lemma 2.3. forces g(a) to be in $\Phi _{g}({\mathcal{A}})$.

(2) Let D be an open neighborhood of $\sigma _{{\mathcal{A}}}(a)$ on which f is analytic. Take $\mu \in {\mathbb {C}} \backslash f(\sigma _{\Phi _{g}({\mathcal{A}})} (a))$ and define $g \in Hol(a)$ by $g \,{:=}\, \mu - f$. It follows that $g \in Hol(a)$ which has no zero in $\sigma _{\Phi _{g}({\mathcal{A}})}$. Then, by part (1), $g(a) \in \Phi _{g}({\mathcal{A}})$. Thus $\mu - f(a) \in \Phi _{g}({\mathcal{A}})$, hence $\mu \in \rho _{\Phi _{g}({\mathcal{A}})}(f(a))$. $\square$

Finally, we give a proof of Corollary 1.3. It is similar to that of [14, Corollary 3.4]. Here, we use the convention that $f(\emptyset )=\emptyset$.

Proof

$(ii) \Rightarrow (i)$ It is inferred from Theorem 1.1.

$(i) \Rightarrow (ii)$ Take $f=1$ on a neighbourhood D of $\sigma _{{\mathcal{A}}}(a)$. Then, $f(a)=1$.

Hence $\sigma _{\Phi _{g}}(1)=\emptyset$. We derive $f(\sigma _{\Phi _{g}({\mathcal{A}})}(a)) = \emptyset$, thus $\sigma _{\Phi _{g}({\mathcal{A}})}(a)=\emptyset$. $\square$