Domination by ergodic elements in ordered Banach algebras

Abstract

We recall the definition and properties of an algebra cone in an ordered Banach algebra (OBA) and continue to develop spectral theory for the positive elements. An element $a$ of a Banach algebra is called ergodic if the sequence of sums $\sum _{k=0}^{n-1} \frac{a^k}{n}$ converges. If $a$ and $b$ are positive elements in an OBA such that $0\le a\le b$ and if $b$ is ergodic, an interesting problem is that of finding conditions under which $a$ is also ergodic. We will show that in a semisimple OBA that has certain natural properties, the condition we need is that the spectral radius of $b$ is a Riesz point (relative to some inessential ideal). We will also show that the results obtained for OBAs can be extended to the more general setting of commutatively ordered Banach algebras (COBAs) when adjustments corresponding to the COBA structure are made.

1 Introduction

In [3, 4, 6–12, 14, 16], spectral theory for positive elements in ordered Banach algebras (OBAs) was developed. The results in these papers were extended to the more general setting of commutatively ordered Banach algebras (COBAs) in [13]. Some of the results in [3, 7, 14, 16] were related to the following problem: if $a$ and $b$ are positive elements in an OBA such that $0\le a\le b$, under what conditions are certain properties of $b$ inherited by $a$? This is a classical problem for positive operators on Banach lattices; a survey of some of the results for the problem is given in [18, Chapter 18]. An element $a$ of a Banach algebra is said to be ergodic if the sequence of sums $\sum _{k=0}^{n-1} \frac{a^k}{n}$ converges. In this paper we will obtain results giving conditions under which a dominated positive element in an OBA or COBA is ergodic, given that the dominating element is ergodic. A corresponding problem for uniformly ergodic operators on Banach lattices was studied in [15].

In Sect. 2 we provide some preliminary notation and definitions and in Sect. 3 we review some of the main concepts about OBAs and COBAs. In Sect. 4 the main aim is to establish the ergodic theorem (Theorem 4.10), which will serve as a crucial tool in proving the main results in Sect. 5. This result is a Banach algebra version of part of Dunford’s result in [5, Theorem 3.16]. To prove Theorem 4.10 several Banach algebra results, which are of interest in themselves, will be obtained. A number of these results are Banach algebra versions of results in [5] (see Theorems 2.19, 2.20, 2.21, 3.6). While these are typically operator theoretic results, we will obtain ours by purely algebraic means. In Sect. 5 we show that under conditions similar to those of [15, Theorem 4.5], a dominated positive element in an OBA or COBA is ergodic if the dominating element is ergodic. The main results are Theorems 5.1, 5.2 and especially Theorem 5.5.

2 Preliminaries

Throughout $A$ will be a complex Banach algebra with unit 1. The spectrum and spectral radius of an element $a$ in $A$ will be denoted by $\sigma (a)$ and $r(a)$, respectively. The set of isolated points of $\sigma (a)$ will be denoted by $\mathbf iso \ \sigma (a)$. For any $\alpha \in \mathbf iso \ \sigma (a)$, we will denote by $p(a,\alpha )$ the spectral idempotent corresponding to $a$ and $\alpha$. A point $\alpha \in \sigma (a)$ is said to be a pole of order $k$ of the resolvent function $\lambda \mapsto R(\lambda , a)=(\lambda 1-a)^{-1}$, $\lambda \in \mathbb C \setminus \sigma (a)$, if $\alpha \in \mathbf iso \ \sigma (a)$ and $k$ is the smallest natural number such that $(\alpha 1-a)^kp(a, \alpha )=0$.

In this paper all the ideals will be assumed to be two-sided. Let $F$ be an ideal in $A$. A point $\alpha \in \mathbf iso \ \sigma (a)$ is said to be a Riesz point relative to $F$ if the corresponding spectral projection $p(a, \alpha )$ belongs to $F$. An ideal $I$ in $A$ is called inessential whenever the spectrum in $A$ of every element in $I$ is either finite or a sequence converging to zero. We will need the following characterization of Riesz points in a semisimple Banach algebra:

Lemma 2.1

([10], Lemma 2.1) Let $A$ be a semisimple Banach algebra, $I$ an inessential ideal of $A$, and $a\in A$. Then a point $\alpha$ in $\sigma (a)$ is a Riesz point of $\sigma (a)$ relative to $I$ if and only if $\alpha$ is a pole of the resolvent of $a$ and $p(a,\alpha )\in I$.

A point $\alpha \in \sigma (a)$ is called an eigenvalue of $a$ if there exists a $0\ne u\in A$ such that $au=\alpha u$ or $ua=\alpha u$. Then $u$ is said to be an eigenvector corresponding to $\alpha$.

For part of the material in Sect. 3 we will need to recall the following. Let $E$ be a complex Banach lattice and denote by $\mathcal L (E)$ the space of bounded linear operators on $E$. An operator $T:E\rightarrow E$ is regular if it can be written as a linear combination over $\mathbb C$ of positive operators. The space of all regular operators on $E$ is denoted by $\mathcal L ^r(E)$ and it is a subspace of $\mathcal L (E)$. When $\mathcal L ^r(E)$ is provided with the $r$-norm

$$\begin{aligned} ||T||_r=\inf \{||S||:S\in \mathcal L (E),|Tx|\le S|x|\quad \text{ for } \text{ all } x\in E\}, \end{aligned}$$

it becomes a Banach algebra which contains the unit of $\mathcal L (E)$ ([17, IV §1] and [1]). We denote by $\mathcal K ^r(E)$ the closure in $\mathcal L ^r(E)$ of the ideal of finite rank operators on $E$. This is a closed, inessential ideal.

3 Ordered Banach algebras

In [16, Sect. 3] an algebra cone $C$ of a Banach algebra $A$ was defined and it was shown that $C$ induces an ordering on $A$ which is compatible with the algebraic structure of $A$. Such a Banach algebra is called an ordered Banach algebra (OBA). In [13, Sect. 3], we defined an algebra $c$-cone, which is more general than an algebra cone, and showed that a Banach algebra can be ordered by an algebra $c$-cone, and is then called a commutatively ordered Banach algebra (COBA). We now recall those definitions and the additional properties that algebra cones and algebra $c$-cones may have.

A nonempty subset $C$ of a Banach algebra $A$ is called a cone if $C$ satisfies the following:

• (i) $C+C\subseteq C$,

• (ii) $\lambda C \subseteq C$ for all $\lambda \ge 0$.

If $C$ also satisfies the property $C\cap -C =\{0\}$, then it is called a proper cone. We say that $C$ is closed if it is a closed subset of $A$.

Every cone $C$ in a Banach algebra $A$ induces an ordering $\le$ defined by $a\le b$ if and only if $b-a\in C$, for $a,b\in A$. This ordering is reflexive and transitive. In addition, $C$ is proper if and only if the ordering is antisymmetric. In view of the fact that $C$ induces an ordering on $A$, we find that $C=\{a\in A:a\ge 0\}$. Therefore the elements of $C$ are called positive.

A cone $C$ in a Banach algebra $A$ is called an algebra cone if it satisfies the following:

• (i) $ab\in C$ for all $a,b\in C$,

• (ii) $1\in C$, where $1$ is the unit of $A$.

Following [13], we call a cone $C$ that satisfies the weaker conditions $ab\in C$ for all $a,b\in C$ such that $ab=ba$ and $1\in C$ an algebra $c$-cone. Obviously, every algebra cone is an algebra $c$-cone.

A Banach algebra ordered by an algebra cone (algebra $c$-cone) is called an ordered Banach algebra (OBA) (commutatively ordered Banach algebra (COBA)). Clearly, every OBA is a COBA. We will denote by $(A, C)$ a Banach algebra $A$ ordered by an algebra cone (algebra $c$-cone) $C$.

The following result, which can easily be established by induction, plays an important role in Sect. 5:

Proposition 3.1

Let $(A, C)$ be an OBA and let $a,b\in A$. If $0\le a\le b$, then $0\le a^n\le b^n$ for any $n\in \mathbb N$.

In order to obtain the COBA analogue of Proposition 3.1, the additional assumption $ab=ba$ is required. As the following counter example shows, this condition cannot be dropped.

Example 3.2

([13], Example 3.13) Let $A=M_2(\mathbb C )$ and $C=\{a\in A:a=a^*$ and $\sigma (a)\subseteq [0,\infty )\}$, where $a^*$ denotes the complex conjugate transpose of $a$. Then $C$ is a closed algebra $c$-cone of $A$. If $a=\left( \begin{array}{rr} 1 &{} -1 \\ -1 &{} 1 \end{array}\right)$, $b=\left( \begin{array}{rr} 2 &{} -1 \\ -1 &{} 1 \end{array}\right) \in A$, then $0\le a\le b$ but $a^2\le b^2$ does not hold.

The following result, which can be deduced from the proof of [14, Theorem 3.2], will also be a useful tool.

Proposition 3.3

Let $A$ be an OBA with a closed algebra cone $C$ and let $0\ne a\in C$ such that $r(a)>0$. If $r(a)$ is a simple pole of the resolvent of $a$, then $p=p(a, r(a))\in C$, $ap=pa=r(a)p$ and $apa=r(a)^2p$.

The result in Proposition 3.3 also holds in a COBA and has the same proof.

An algebra cone or algebra $c$-cone $C$ is said to be inverse-closed if it has the property that if $a\in C$ and $a$ is invertible, then $a^{-1}\in C$. We will need the following result, which is a COBA analogue of [4, Proposition 4.2] and has the same proof:

Proposition 3.4

Let $(A, C)$ be a COBA with $C$ closed and inverse-closed. If $a\in C$, then $0\le a\le r(a)1$.

Suppose that $(A, C)$ is an OBA or COBA. We say that $C$ is normal if there exists a scalar $\alpha >0$ such that $0\le a\le b$ relative to $C$ implies that $||a||\le \alpha ||b||$. If $C$ has the weaker property that there exists a scalar $\alpha >0$ such that $0\le a\le b$ and $ab=ba$ imply that $||a||\le \alpha ||b||$, then $C$ is said to be $c$-normal. Clearly, every normal algebra cone or $c$-cone is $c$-normal. Also, it can easily be shown that a normal or $c$-normal algebra cone or $c$-cone is necessarily proper.

If $0\le a\le b$ relative to $C$ implies that $r(a)\le r(b)$ then we say that the spectral radius is monotone w.r.t. $C$. If we have the weaker property that $0\le a\le b$ relative to $C$ and $ab=ba$ imply that $r(a)\le r(b)$, then we say that the spectral radius is $c$-monotone w.r.t. $C$. Obviously, monotonicity implies $c$-monotonicity.

It is well known that if $C$ is a normal algebra cone in a Banach algebra $A$, then the spectral radius in $(A, C)$ is monotone (see [16, Theorem 4.1]). Similarly, if $C$ is a $c$-normal algebra $c$-cone in a Banach algebra $A$, then the spectral radius is $c$-monotone in $(A, C)$ (see [13, Theorem 4.2]).

If $(A, C)$ is an OBA, $F$ a closed ideal in $A$ and $\pi :A\rightarrow A/F$ the canonical homomorphism, then $(A/F, \pi C)$ is an OBA. However, if $(A, C)$ is a COBA, $\pi C$ is in general only a cone and not an algebra $c$-cone. Therefore although $A/F$ can be ordered by $\pi C$ in the usual way, the structure $(A/F, \pi C)$ is not a COBA. The cone $\pi C$ however does have the property that it contains the unit of $A/F$ as well as all powers of elements of $\pi C$. Such a cone is called an algebra $c^{\prime }$-cone. A Banach algebra ordered by an algebra $c^{\prime }$-cone is called a $C^{\prime }$OBA. Obviously, every COBA is a $C^{\prime }$OBA.

We will say that the spectral radius in an OBA or $C^{\prime }$OBA $(A/F, \pi C)$ is monotone if $0\le a\le b$ in $A$ relative to $C$ implies that $r(a+F, A/F)\le r(b+F, A/F)$. If we have the weaker property that $0\le a\le b$ relative to $C$ and $ab=ba$ imply that $r(a+F, A/F)\le r(b+F, A/F)$, we will say that the spectral radius in $(A/F, \pi C)$ is $c$-monotone.

The following are examples of an OBA, COBA and $C^{\prime }$OBA.

Example 3.5

([10], Example 3.2) Let $E$ be a Dedekind complete Banach lattice, $C=\{x\in E:x\ge 0\}$ and $K=\{T\in \mathcal L (E):TC\subseteq C\}$. Then $(\mathcal L ^r(E), K)$ is an OBA with a closed, normal algebra cone and $(\mathcal L ^r(E)/ \mathcal K ^r(E), \pi K)$ is an OBA such that the spectral radius in $(\mathcal L ^r(E)/ \mathcal K ^r(E), \pi K)$ is monotone.

Example 3.6

([13], Examples 3.2 and 3.17) Let $A$ be a $C^*$-algebra and $C=\{a\in A:a=a^*$ and $\sigma (a)\subseteq [0,\infty )\}$. Then $C$ is a closed, normal algebra $c$-cone of $A$. Therefore $(A, C)$ is a COBA. Suppose that $F$ is a closed ideal in $A$ and for each $a+F$ in $A/F$, define $(a+F)^*=a^*+F$. Then $\pi C$ is a normal algebra $c^{\prime }$-cone in $A/F$. Therefore $(A/F, \pi C)$ is a $C^{\prime }$OBA.

If $A$ in Example 3.6 is commutative, then $(A, C)$ is an OBA.

For more examples of OBAs and COBAs see [3, 7–11, 13, 16].

We will need the following result, which follows immediately from [10, Theorem 4.4].

Corollary 3.7

Let $(A, C)$ be an OBA with $C$ closed and the spectral radius in $(A, C)$ monotone. Let $I$ be a closed inessential ideal of $A$ such that the spectral radius in $(A/I, \pi C)$ is monotone. Suppose that $a,b\in A$ with $0\le a\le b$ and $r(a)=r(b)$. If $r(b)$ is a Riesz point of $\sigma (b)$, then $r(a)$ is a Riesz point of $\sigma (a)$.

If $(A, C)$ is a COBA, we can establish the analogue of Corollary 3.7 with the weaker property of $c$-monotonicity in $(A, C)$ and in the $C^{\prime }$OBA $(A/I, \pi C)$, provided that $ab=ba$ is also assumed. If we replace $c$-monotonicity with monotonicity in the $C^{\prime }$OBA $(A/I, \pi C)$, then the condition $ab=ba$ can be dropped.

4 The ergodic theorem

Let $(f_n)$ be the sequence of functions defined by $f_n(\lambda )=\sum _{k=0}^{n-1}\frac{\lambda ^k}{n}$ ($\lambda \in \mathbb C$). If $a$ is an element of a Banach algebra, then the terms of the sequence $(f_n(a))$ are called ergodic sums of $a$, and $a$ is said to be ergodic if its sequence of ergodic sums converges.

The aim of this section is to establish the ergodic theorem (Theorem 4.10), which will be useful in the proofs of results in the next section. This result is a generalization of part of a result of Dunford ([5, Theorem 3.16]) to Banach algebras. The results from Theorem 4.1 through Proposition 4.9, some of which are Banach algebra versions of results in [5], lead to Theorem 4.10. The proofs of Dunford’s results rely partly on operator theory, while our proofs are completely algebraic.

We start with the following theorem, which gives conditions under which a Banach algebra element obtained through the holomorphic functional calculus will be the zero element.

Theorem 4.1

Let $A$ be a Banach algebra and let $a\in A$. Suppose that $f$ is a complex valued function analytic on a neighbourhood $\Omega$ of $\sigma (a)$ such that

• (i) for every pole $\lambda$ of the resolvent of $a$ of order $k$, $f^{(j)}(\lambda )=0$ ($j=0,1,\ldots ,k-1$) and

• (ii) $\sigma (a)$ contains at least one pole $\lambda _1$ of the resolvent of $a$, and there exists a neighbourhood $U$ of $\sigma (a){\setminus } \{\lambda _1, \lambda _2,\ldots ,\lambda _n\}$ for some $n\ge 1$, where $\lambda _1,\lambda _2,\ldots ,\lambda _n$ are poles of the resolvent of $a$, such that $f(\lambda )=0$ for all $\lambda \in U$.

Then $f(a)=0$.

Proof

If $\Gamma$ is a smooth contour surrounding $\sigma (a)$, then $f(a)=\frac{1}{2\pi i}\int _{\Gamma }f(\lambda )(\lambda 1-a)^{-1}d \lambda$. We may take $\Gamma$ to be the union of $\Gamma _1$ and $C_1,C_2,\ldots ,C_n$, where $\Gamma _1$ is a smooth contour contained in $U$ and surrounding $\sigma (a){\setminus } \{\lambda _1,\lambda _2,\ldots ,\lambda _n\}$, and $C_i$ is a small circle centered at the pole $\lambda _i$ of the resolvent of $a$ and separating $\lambda _i$ from the rest of $\sigma (a)$. Therefore

$$\begin{aligned} f(a)=\frac{1}{2\pi i}\int \limits _{\Gamma _1}f(\lambda )(\lambda 1-a)^{-1}d\lambda + \sum _{i=1}^{n}\frac{1}{2\pi i}\int \limits _{C_i}f(\lambda )(\lambda 1-a)^{-1}d\lambda . \end{aligned}$$

From assumption (ii) we have that $\frac{1}{2\pi i}\int _{\Gamma _1} f(\lambda )(\lambda 1-a)^{-1}d\lambda =0$. Now, since $f^{(j)}(\lambda _i)=0$ for $i=1,2,\ldots ,n$ and for $j=0,1,\ldots ,k_i-1$ (with $k_i$ the order of the pole $\lambda _i$) by assumption (i), there exist analytic functions $g_i$ ($i=1,2,\ldots ,n$) on $\Omega$ such that $f(\lambda )=(\lambda - \lambda _i)^{k_i}g_i(\lambda )$ on a neighbourhood of $\lambda _i$ including $C_i$. Since $\lambda _i$ is a pole of order $k_i$ of the resolvent of $a$, this resolvent has a Laurent series expansion $(\lambda 1-a)^{-1}= \frac{a_{-k_i}}{(\lambda -\lambda _i)^{k_i}}+ \frac{a_{-k_i+1}}{(\lambda -\lambda _i)^{k_i-1}}+\cdots +a_0+a_1(\lambda -\lambda _i)+\cdots$, so that $f(\lambda )(\lambda 1-a)^{-1}=(\lambda -\lambda _i)^{k_i}g_i(\lambda )(\lambda 1-a)^{-1}=g_i(\lambda )[a_{-k_i}+a_{-k_i+1}(\lambda -\lambda _i)+ \cdots +a_0(\lambda -\lambda _i)^{k_i}+a_1(\lambda -\lambda _i)^{k_i+1}+\cdots ]$, on a deleted neighbourhood of $\lambda _i$ which includes $C_i$. Since $g_i$ is analytic on a neighbourhood of $\sigma (a)$, it has no singularities on or inside $C_i$, and so $f(\lambda )(\lambda 1-a)^{-1}$ has no singularities on or inside $C_i$. It follows from Cauchy’s theorem that $\frac{1}{2\pi i}\int _{C_i}f(\lambda )(\lambda 1-a)^{-1}d\lambda =0$ for $i=1,2,\ldots ,n$. Hence $f(a)=0$. $\square$

The following result is an immediate consequence of Theorem 4.1.

Corollary 4.2

Let $A$ be a Banach algebra and let $a\in A$. Suppose that $f,g$ are complex valued functions analytic on a neighbourhood of $\sigma (a)$ such that

• (i) for every pole $\lambda$ of the resolvent of $a$ of order $k$, $f^{(j)}(\lambda )=g^{(j)}(\lambda )$ ($j=0,1,\ldots ,k-1$) and

• (ii) $\sigma (a)$ contains at least one pole $\lambda _1$ of the resolvent of $a$, and there exists a neighbourhood $U$ of $\sigma (a) {\setminus } \{\lambda _1,\lambda _2,\ldots ,\lambda _n\}$ for some $n\ge 1$, where $\lambda _1,\lambda _2,\ldots ,\lambda _n$ are poles of the resolvent of $a$, such that $f(\lambda )=g(\lambda )$ for all $\lambda \in U$.

Then $f(a)=g(a)$.

The following theorem is a very important application of Corollary 4.2, and forms the basis for the rest of our work.

Theorem 4.3

Let $A$ be a Banach algebra and let $a\in A$. Suppose that $f$ is a complex valued function analytic on a neighbourhood of $\sigma (a)$. If $\alpha$ is a pole of order $k$ of the resolvent of $a$, then $f(a)=f(a)(1-p)+\sum _{n=0}^{k-1}\frac{(a-\alpha 1)^n}{n!}f^{(n)}(\alpha )p$, where $p= p(a,\alpha )$.

Proof

Since $\alpha$ is an isolated point in $\sigma (a)$, we can take two open sets $U_0$ and $U_1$ such that $\sigma (a){\setminus }\{\alpha \} \subseteq U_0$, $\{\alpha \}\subseteq U_1$, $U_0\cap U_1=\emptyset$ and $f$ is analytic on $U=U_0\cup U_1$. Let $\chi :U\rightarrow \mathbb C$ be the function defined by $\chi (\lambda ) =0$ if $\lambda \in U_0$ and $\chi (\lambda )=1$ if $\lambda \in U_1$. Then $\chi$ is analytic on $U$ and $p=\chi (a)$. Now let $g:U\rightarrow \mathbb C$ be the function defined by $g(\lambda )=f(\lambda )(1-\chi (\lambda )) +\sum _{n=0}^{k-1}\frac{(\lambda -\alpha )^n}{n!}f^{(n)}(\alpha )\chi (\lambda )$. We show that $f$ and $g$ satisfy conditions (i) and (ii) of Corollary 4.2. If $\lambda \in U_0$ then $\chi (\lambda )=0$ and so $g(\lambda )=f(\lambda )$. Hence $f$ and $g$ satisfy condition (ii) of Corollary 4.2. Since $g(\lambda )= f(\lambda )$ for all $\lambda \in U_0$, we have that $g^{(j)}(\lambda _i)=f^{(j)}(\lambda _i)$ $(j=0,1,\ldots ,k_i-1)$ for every pole $\lambda _i\in U_0$ of order $k_i$ of the resolvent of $a$. We show that $g^{(j)}(\alpha )=f^{(j)}(\alpha )$ for $j=0,1,\ldots ,k-1$. We restrict the functions $f$ and $g$ to the set $U_1$. For $j=0$, it is clear that $f(\alpha )=g(\alpha )$. For $j=1$, we have that

$$\begin{aligned} g^{\prime }(\lambda )&= f^{\prime }(\alpha )+ \frac{2(\lambda -\alpha )}{2!}f^{\prime \prime }(\alpha )+ \frac{3(\lambda - \alpha )^2}{3!}f^{\prime \prime \prime }(\alpha )+\cdots \\&\quad + \frac{(k-1)(\lambda -\alpha )^{k-2}}{(k-1)!}f^{(k-1)}(\alpha ). \end{aligned}$$

Therefore $g^{\prime }(\alpha )=f^{\prime }(\alpha )$. Next,

$$\begin{aligned} g^{\prime \prime }(\lambda )&= f^{\prime \prime }(\alpha )+(\lambda - \alpha )f^{\prime \prime \prime }(\alpha )+\frac{3\cdot 4(\lambda - \alpha )^2}{4!}f^{(4)}(\alpha )+\cdots \\&\quad +\frac{(k-2)(k-1)(\lambda - \alpha )^{k-3}}{(k-1)!}f^{(k-1)}(\alpha ). \end{aligned}$$

Hence $g^{\prime \prime }(\alpha )=f^{\prime \prime }(\alpha )$. Continuing in this way, we obtain

$$\begin{aligned} g^{(k-1)}(\lambda )\!=\!\frac{1\cdot 2\cdots (k\!-\!1)}{(k\!-\!1)!} f^{(k-1)}(\alpha )\!=\!f^{(k-1)}(\alpha ), \end{aligned}$$

and so $g^{(k-1)}( \alpha )\!=\!f^{(k-1)}(\alpha )$. Therefore condition (i) of Corollary 4.2 is satisfied and so the result follows. $\square$

The following three corollaries are consequences of Theorem 4.3.

Corollary 4.4

Let $A$ be a Banach algebra and let $a\in A$. Let $(f_n)$ be a sequence of complex valued functions analytic on a neighbourhood of $\sigma (a)$. Suppose that $\alpha \ne 0$ is a pole of order $k$ of the resolvent of $a$ such that $f_n(\alpha )\rightarrow 1$ and $f_n^{(j)}(\alpha )\rightarrow 0$ ($j=1,2,\ldots ,k-1$) as $n\rightarrow \infty$. If $(\alpha 1-a)f_n(a)\rightarrow 0$ as $n\rightarrow \infty$ then $f_n(a)\rightarrow p$ as $n\rightarrow \infty$, where $p=p(a, \alpha )$.

Proof

Since $\alpha$ is an isolated point in $\sigma (a)$, we can take two open sets $U_0$ and $U_1$ such that $\sigma (a){\setminus } \{\alpha \}\subseteq U_0$, $\{\alpha \}\subseteq U_1$, $U_0\cap U_1= \emptyset$ and $f_n$ is analytic on $U=U_0\cup U_1$ for all $n\in \mathbb N$. By [2, Theorem 3.3.4], we have that $\sigma ((1-p)a)= (\sigma (a)\cap U_0)\cup \{0\}$. Since $\alpha \ne 0$, it follows that $\alpha \notin \sigma ((1-p)a)$, so that $b=\alpha 1-(1-p)a$ is invertible. Then

$$\begin{aligned} f_n(a)(1-p)&= f_n(a)(1-p)(\alpha 1-(1-p)a)b^{-1}= f_n(a)(1-p)(\alpha 1-a)b^{-1}\\&= f_n(a)(\alpha 1-a)(1-p)b^{-1}, \end{aligned}$$

since $(1-p)p=0$. From the assumption $(\alpha 1-a)f_n(a)\rightarrow 0$ as $n\rightarrow \infty$, it follows that $f_n(a)(1-p)\rightarrow 0$ as $n\rightarrow \infty$. Also, from Theorem 4.3, we get that $f_n(a)=f_n(a)(1-p) +\sum _{j=0}^{k-1}\frac{(a-\alpha 1)^j}{j!}f_n^{(j)}(\alpha )p$. Together with the assumptions $f_n(\alpha )\rightarrow 1$ and $f_n^{(j)}(\alpha )\rightarrow 0$ ($j=1,2,\ldots ,k-1$) as $n\rightarrow \infty$, it follows that $f_n(a)\rightarrow p$ as $n\rightarrow \infty$. $\square$

In the case where $\alpha$ in Corollary 4.4 is a simple pole, we get the following result.

Corollary 4.5

Let $A$ be a Banach algebra and $a\in A$. Let $(f_n)$ be a sequence of complex valued functions analytic on a neighbourhood of $\sigma (a)$. Suppose that $\alpha \ne 0$ is a simple pole of the resolvent of $a$ such that $f_n(\alpha )\rightarrow 1$ as $n\rightarrow \infty$. Then $f_n(a)\rightarrow p$ as $n\rightarrow \infty$ if and only if $(\alpha 1-a)f_n(a)\rightarrow 0$ as $n\rightarrow \infty$, where $p=p(a, \alpha )$.

Proof

Suppose that $f_n(a)\rightarrow p$ as $n\rightarrow \infty$. Then $(\alpha 1-a)f_n(a)\rightarrow (\alpha 1-a)p$ as $n\rightarrow \infty$. Since $\alpha$ is a simple pole of the resolvent of $a$, we have that $(\alpha 1-a)p=0$. Hence $(\alpha 1-a)f_n(a)\rightarrow 0$ as $n\rightarrow \infty$. The converse follows from Corollary 4.4. $\quad \square$

In Corollary 4.4 we made the assumption that $\alpha$ is a pole of arbitrary order $k$. The following result shows that the other conditions then force $\alpha$ to be a simple pole.

Corollary 4.6

Let $A$ be a Banach algebra, $a\in A$ and suppose that $\alpha$ is a pole of order at most $k\ge 1$ of the resolvent of $a$. Suppose that $(f_n)$ is a sequence of complex valued functions analytic on a neighbourhood of $\sigma (a)$. If $(a-\alpha 1)f_n (a)\rightarrow 0$ and $f_n(\alpha ) \rightarrow 1$ as $n\rightarrow \infty$, then $\alpha$ is a simple pole of the resolvent of $a$.

Proof

Let $p=p(a, \alpha )$. By Theorem 4.3 we may write

$$\begin{aligned} f_n(a)=f_n(a)(1-p)+\sum _{j=0}^{k-1}\frac{(a-\alpha 1)^j}{j!}f_n^{(j)}(\alpha )p. \end{aligned}$$

(*)

For $k=1$, the result is trivial. For $k=2$ we have from (*) that $f_n(a)=f_n(a)(1-p)+f_n(\alpha )p+f_n^{\prime }(\alpha )(a-\alpha 1)p$. Therefore

$$\begin{aligned} (a-\alpha 1)f_n(a)=(a-\alpha 1)f_n(a)(1-p)+(a-\alpha 1)f_n(\alpha )p+(a-\alpha 1)^2f_n^{\prime }(\alpha )p. \end{aligned}$$

(**)

Since $k=2$, we have that $\alpha$ is a pole of order at most $2$ of the resolvent of $a$. Therefore $(a-\alpha 1)^2f_n^{\prime } (\alpha )p=0$. Using the assumptions $(a-\alpha 1)f_n(a)\rightarrow 0$ and $f_n(\alpha )\rightarrow 1$ as $n\rightarrow \infty$, it then follows from (**) that $(a-\alpha 1)p=0$. Hence $\alpha$ is a simple pole of the resolvent of $a$. For any $k>2$, the general procedure is as follows: In the first step multiply both sides of (*) by $(a-\alpha 1)^{k-1}$. Since $\alpha$ is a pole of order at most $k$, this makes all but the first two terms of the expression on the right hand side of (*) zero. On the resulting equation, we take limits as $n\rightarrow \infty$ and use the assumptions $(a-\alpha 1)f_n(a)\rightarrow 0$ and $f_n(\alpha )\rightarrow 1$ as $n\rightarrow \infty$ to obtain that $(a-\alpha 1)^{k-1}p=0$. In the second step multiply both sides of (*) by $(a-\alpha 1)^{k-2}$. Using arguments similar to the first step and the fact that $(a-\alpha 1)^{k-1}p=0$, we get that $(a-\alpha 1)^{k-2}p=0$. After $k-1$ steps it follows that $(a-\alpha 1)p=0$. $\square$

In the following proposition we establish that if $\sigma (a)$ contains only one element, then a stronger form of the reverse implication in Corollary 4.5 holds. This result is not used further; it is included for the sake of interest.

Proposition 4.7

Let $A$ be a Banach algebra and $a\in A$. Suppose that $\sigma (a)=\{\alpha \}$ and $\alpha$ is a simple pole of the resolvent of $a$. If $(f_n)$ is a sequence of complex valued functions analytic on a neighbourhood of $\sigma (a)$ and if $f_n(\alpha )\rightarrow 1$, then $f_n(a)\rightarrow p(a, \alpha )$.

Proof

Let $\Gamma$ be a small circle centred at $\alpha$. Then $f_n(a)=\frac{1}{2\pi i}\int _{\Gamma }f_n(\lambda )(\lambda 1-a)^{-1}d \lambda$. Since $\alpha$ is a simple pole of the resolvent of $a$, we obtain the Laurent series expansion $(\lambda 1-a)^{-1}=\frac{a_{-1}}{\lambda -\alpha }+a_0+ a_1(\lambda -\alpha )+a_2(\lambda -\alpha )^2+\cdots$ ($a_j\in A, j=-1,0,1,2,\ldots$) on a deleted neighbourhood $N_0$ of $\alpha$ which contains $\Gamma$. Let $S(\lambda )$ be the sum of the power series $a_0+a_1 (\lambda -\alpha )+a_2(\lambda -\alpha )^2+\cdots$. Then $S$ is analytic on $N_0$ and then $(\lambda 1-a)^{-1}=S(\lambda )+ \frac{a_{-1}}{\lambda -\alpha }$. Clearly, $f_nS$ is analytic on $N_0$. Since $\Gamma$ is contained in $N_0$, it follows that

$$\begin{aligned} f_n(a) = \frac{1}{2\pi i}\int \limits _{\Gamma }f_n(\lambda ) S(\lambda )d\lambda +\frac{a_{-1}}{2\pi i}\int \limits _{\Gamma }\frac{f_n(\lambda )}{ \lambda -\alpha }d\lambda = f_n(\alpha )a_{-1}. \end{aligned}$$

Since $a_{-1}=p(a,\alpha )$ and $f_n(\alpha )\rightarrow 1$, it follows that $f_n(a)\rightarrow p(a, \alpha )$. $\square$

In the rest of this section, we will consider only sequences of analytic functions of the form $f_n(\lambda )=\sum _{k=0}^{n-1} \frac{\lambda ^k}{n}$ ($\lambda \in \mathbb C$), as they are of particular interest for the problem at hand. Note that $f_n(1)=1$ for all $n\in \mathbb N$. The next proposition shows that if we take this sequence of functions and $\alpha =1$ in Corollary 4.5, then we obtain a stronger form of the forward implication in Corollary 4.5. To prove this result we will use the following lemma.

Lemma 4.8

Let $A$ be a Banach algebra and $a\in A$. Then $(1-a)\sum _{k=0}^{n-1}\frac{a^k}{n}\rightarrow 0$ if and only if $\frac{a^{n}}{n} \rightarrow 0$.

Proof

We have that $(1-a)\sum _{k=0}^{n-1}\frac{a^k}{n}=\sum _{k=0}^{n-1}\frac{a^k}{n}-\sum _{k=0}^{n-1}\frac{a^{k+1}}{n}=\frac{1}{n}- \frac{a^n}{n}$. Now, $\frac{1}{n}-\frac{a^n}{n}\rightarrow 0$ if and only if $\frac{a^n}{n}\rightarrow 0$. $\square$

Proposition 4.9

Let $A$ be a Banach algebra and $a\in A$. If $(f_n(a))$ converges, where $f_n(a)=\sum _{k=0}^{n-1}\frac{a^k}{n}$, then $(1-a)f_n(a)\rightarrow 0$.

Proof

Suppose that $(f_n(a))$ converges, say $f_n(a)\rightarrow b$. We have that $(n+1)f_{n+1}(a)-nf_n(a)=a^n$, and so $\frac{a^n}{n}=\frac{1}{n}((n+1)f_{n+1}(a)-nf_n(a))=\frac{n+1}{n}f_{n+1}(a)-f_n(a)\rightarrow b-b=0$. It follows from Lemma 4.8 that $(1-a)f_n(a)\rightarrow 0$. $\square$

The following theorem, which we will call the ergodic theorem, is our key result corresponding to part of [5, Theorem 3.16].

Theorem 4.10

Let $A$ be a Banach algebra and $a\in A$. Suppose that $1\in \mathbf{iso} \ \sigma (a)$. Let $(f_n)$ be the sequence of functions $f_n(\lambda )= \sum _{k=0}^{n-1} \frac{\lambda ^k}{n}$ ($\lambda \in \mathbb C$). Then the following statements are equivalent:

• (i) $(f_n(a))$ converges, i.e. $a$ is ergodic, with $f_n(a)\rightarrow p(a, 1)$.

• (ii) $(1-a)f_n(a)\rightarrow 0$ as $n\rightarrow \infty$ and $1$ is a simple pole of the resolvent of $a$.

Proof

(i) $\Rightarrow$ (ii): Suppose that $f_n(a)\rightarrow p(a, 1)$. Then by Proposition 4.9, we have that $(1-a)f_n(a)\rightarrow 0$. Since $(1-a)f_n(a)\rightarrow (1-a)p(a, 1)$, by uniqueness of limits, $(1-a)p(a, 1)=0$. Therefore $1$ is a simple pole of the resolvent of $a$.

(ii) $\Rightarrow$ (i): If $(1-a)f_n(a)\rightarrow 0$ as $n\rightarrow \infty$ and $1$ is a simple pole of the resolvent of $a$, then since $f_n(1)=1$, it follows from Corollary 4.5 that $(f_n(a))$ is convergent, with $f_n(a)\rightarrow p(a, 1)$. $\square$

5 Domination by ergodic elements

In this section we will obtain results for the problem of domination by ergodic elements. As stated earlier, a corresponding problem for uniformly ergodic operators on Banach lattices was studied in [15] and a result was obtained (see [15, Theorem 4.5]). The proof of this result depends on a theorem by Dunford (see [5, Theorem 3.16]). The results we will obtain here are OBA and COBA analogues (although not generalizations in the strict sense) of [15, Theorem 4.5] and these are Theorems 5.1, 5.2 and 5.5. For a Dedekind complete Banach lattice Theorem 5.5 generalizes [15, Theorem 4.5] in the special case of the dominated operator being positive. A key result that will be used to establish these theorems is Theorem 4.10.

If $a$ is an element of a Banach algebra $A$, then, in this section, $f_n(a)$ will always indicate the element $\sum _{k=0}^{n-1}\frac{a^k}{n}\in A$.

We start with the following result.

Theorem 5.1

Let $(A, C)$ be an OBA with $C$ closed and normal, and let $a,b\in A$ such that $0\le 1\le a\le b$. Suppose that $r(b)=1\in \mathbf{iso} \ \sigma (a)$ and that $1$ is a simple pole of the resolvent of $b$. If $b$ is ergodic with $f_n(b)\rightarrow p(b, 1)$ and if $p(a, 1)=p(b, 1)$, then $a$ is ergodic, with $f_n(a)\rightarrow p(a, 1)$.

Proof

Since $1$ is a simple pole of the resolvent of $b$, by Proposition 3.3, it is an eigenvalue of $b$ with positive corresponding eigenvector $p(b, 1)$. Since $p(a, 1)=p(b, 1)$, the assumption $0\le 1\le a\le b$ implies that $0\le (a-1)p(a, 1)\le (b-1) p(b, 1)=0$. From the fact that $C$ is normal, and hence proper, it follows that $(1-a)p(a, 1)=0$, so that $1$ is a simple pole of the resolvent of $a$. Now since $b$ is ergodic, Proposition 4.9 and Lemma 4.8 imply that $\frac{b^n}{n}\rightarrow 0$ as $n\rightarrow \infty$. It follows from Proposition 3.1 and the normality of $C$ that $\frac{a^n}{n}\rightarrow 0$ as $n\rightarrow \infty$. Lemma 4.8 then implies that $(1-a)f_n(a)\rightarrow 0$ as $n\rightarrow \infty$. From Theorem 4.10, it follows that $a$ is ergodic, with $f_n(a)\rightarrow p(a, 1)$. $\square$

The remaining results about domination by ergodic elements will be proved under conditions similar to those of [15, Theorem 4.5]. We start with Theorem 5.2, which is the basic result from which the others will be obtained.

Theorem 5.2

Let $A$ be an OBA with a closed, normal algebra cone $C$ and let $a,b\in A$ such that $0\le a\le b$. Suppose that $1\in \mathbf{iso} \ \sigma (a)$ is a pole of the resolvent of $a$. If $b$ is ergodic, then $a$ is ergodic.

Proof

From Proposition 3.1 we have that $0\le \frac{a^n}{n}\le \frac{b^n}{n}$ for all $n\in \mathbb N$. Since $b$ is ergodic, $\frac{b^n}{n}\rightarrow 0$ as $n\rightarrow \infty$ by Proposition 4.9 and Lemma 4.8, and since $C$ is normal, $\frac{a^n}{n}\rightarrow 0$ as $n\rightarrow \infty$. Lemma 4.8 then implies that $(1-a)f_n(a)\rightarrow 0$ as $n\rightarrow \infty$. Since $1$ is a pole of the resolvent of $a$, it follows from Corollary 4.6 that $1$ is a simple pole of the resolvent of $a$. From Theorem 4.10 it follows that $a$ is ergodic, with $f_n(a)\rightarrow p(a, 1)$ as $n\rightarrow \infty$. $\square$

To prove our main theorem, the following two lemmas will be required in addition to the theory we have developed so far.

Lemma 5.3

Let $A$ be a Banach algebra and $a\in A$. If $a$ is ergodic, then $r(a)\le 1$.

Proof

Since $a$ is ergodic, $(f_n(a))$ converges. It follows from Proposition 4.9 that $(1-a)f_n(a)\rightarrow 0$ as $n\rightarrow \infty$. Lemma 4.8 then implies that $\frac{||a^n||}{n}\rightarrow 0$ as $n\rightarrow \infty$. Therefore there exists a constant $c>0$ such that $\frac{||a^n||}{n}\le c$ for all $n\in \mathbb N$, so that $||a^n||\le cn$ for all $n\in \mathbb N$. Hence $r(a)=\lim _{n\rightarrow \infty }||a^n||^{\frac{1}{n}}\le \lim _{n\rightarrow \infty } c^{\frac{1}{n}}n^{\frac{1}{n}}=1$. $\square$

Lemma 5.4

Let $A$ be a Banach algebra and $a\in A$. If $\frac{a^n}{n}\rightarrow 0$ as $n\rightarrow \infty$ and $1\notin \sigma (a)$, then $\sum _{k=0}^{n-1}\frac{a^k}{n}$ converges to $0$ as $n\rightarrow \infty$.

Proof

If $\frac{a^n}{n}\rightarrow 0$ as $n\rightarrow \infty$, then by Lemma 4.8 $(1-a)\sum _{k=0}^{n-1}\frac{a^k}{n}\rightarrow 0$ as $n\rightarrow \infty$. If also $1\notin \sigma (a)$, then $1-a$ is invertible, which yields the result. $\square$

Theorem 5.5

Let $(A, C)$ be a semisimple OBA with $C$ closed and normal, and let $a, b\in A$ such that $0 \le a\le b$. Let $I$ be a closed inessential ideal of $A$ such that the spectral radius in $(A/I, \pi C)$ is monotone. If $b$ is ergodic and if $r(b)$ is a Riesz point of $\sigma (b)$, then $a$ is ergodic.

Proof

Since $C$ is normal, the spectral radius in $(A, C)$ is monotone. This, together with Lemma 5.3 and the ergodicity of $b$, implies that $r(a)\le r(b)\le 1$. Then we have four cases: $r(a)<r(b)<1$, $r(a)<r(b)=1$, $r(a)=r(b)<1$ and $r(a)=r(b)=1$. Now from Proposition 3.1, we have that $0\le \frac{a^n}{n}\le \frac{b^n}{n}$ for all $n\in \mathbb N$. Since $b$ is ergodic, $\frac{b^n}{n}\rightarrow 0$ as $n\rightarrow \infty$ by Proposition 4.9 and Lemma 4.8, and then, by the normality of $C$, $\frac{a^n}{n}\rightarrow 0$ as $n\rightarrow \infty$. In the first three cases, we get that $1\notin \sigma (a)$. Therefore $\sum _{k=0}^{n-1}\frac{a^k}{n}\rightarrow 0$ as $n\rightarrow \infty$ by Lemma 5.4, so that $a$ is ergodic. To deal with the last case suppose that $r(a)=r(b)=1$. Since $r(b)$ is a Riesz point of $\sigma (b)$, by Corollary 3.7, we have that $r(a)$ is a Riesz point of $\sigma (a)$. Lemma 2.1 then implies that $r(a)$ is a pole of the resolvent of $a$. From Theorem 5.2, it follows that $a$ is ergodic. $\square$

To obtain COBA versions of Theorems 5.1, 5.2 and 5.5, the additional condition $ab=ba$ is assumed, since this assumption is needed in the COBA version of Proposition 3.1, and normality can be replaced with $c$-normality. For the quotient algebra in Theorem 5.5, we then work with the $C^{\prime }$OBA $(A/F, \pi C)$. With $ab=ba$, monotonicity in $(A/F, \pi C)$ can be replaced with $c$-monotonicity. Whether the assumption $ab=ba$ can be dropped under any circumstances in the COBA case remains an open problem. Example 3.2 shows that it is essential for our proof.

We end this section with the following observation:

Proposition 5.6

Let $A$ be a COBA with a closed and inverse-closed algebra $c$-cone $C$. If $a\in C$ and $a$ is ergodic, then $a\le 1$.

Proposition 5.6 follows from Proposition 3.4 and Lemma 5.3.