1 Introduction

In this paper we give sufficient conditions for the property \(r(a)\in \sigma (a)\) where a is an element of a Banach algebra \(\mathcal{A}\) with spectrum \(\sigma (a)\) and spectral radius r(a). That the spectral radius belongs to the spectrum is an aspect of what is usually subsumed under the term “Perron–Frobenius theory”, classically for positive matrices on \(\mathbb {R}^n\) or positive linear operators on Banach lattices. More general, we consider here elements of a Banach algebra that is ordered by a cone. We fix the relevant notation.

Let \(\mathcal{A}\) be a complex Banach algebra with unit \(\textbf{1}\). A set \(\emptyset \not = K \subseteq \mathcal{A}\) is called a cone if K is closed, \(K+K \subseteq K\), \(\lambda K \subseteq K\) \((\lambda \ge 0)\) and \(K \cap (-K) = \{0\}\). By setting \(a \le b\) \(:\Leftrightarrow \) \(b-a \in K\) we obtain a partial order on \(\mathcal{A}\).

In the following we always assume that K is a normal cone, that is

$$\begin{aligned} \exists \gamma \ge 1 ~ \forall a,b \in \mathcal{A}: ~ 0\le a \le b ~ \Rightarrow ~ \Vert a\Vert \le \gamma \Vert b\Vert , \end{aligned}$$

and we fix a \(\gamma \) with this property.

Following the notation in Raubenheimer and Rode [10] we call K an algebra cone if K satisfies in addition \(\textbf{1} \in K\) and

$$\begin{aligned} a,b \in K ~ \Rightarrow ~ ab \in K. \end{aligned}$$

In contrast to the usual definition of ordered Banach algebras we do not assume in general that K is an algebra cone, that is in our setting \(\mathcal{A}\) is a Banach algebra and \(\mathcal{A}\) is an ordered Banach space.

The problem of developing a Perron–Frobenius theory in Banach algebras ordered by a cone that is not fully invariant under multiplication was also addressed by Mouton and Muzundu in [7, 9]: In a Commutatively Ordered Banach Algebra (COBA) one only assumes that \(\textbf{1} \in K\) and that

$$\begin{aligned} a,b \in K, ~ ab=ba ~ \Rightarrow ~ ab \in K. \end{aligned}$$

This includes the important special case of a \(C^*\)-algebra ordered by the cone of positive semidefinite self-adjoint elements.

Let \(\sigma (a)\), \(\rho (a)\) and r(a) denote the spectrum, resolvent set and spectral radius of \(a \in \mathcal{A}\), respectively. For \(r(a)>0\), the property \(r(a)\in \sigma (a)\) is invariant under scaling and we can always resort to \(r(a)=1\) by considering a/r(a). Following the notation in [4] also in our setting, we call \(a \in \mathcal{A}\) asymptotically positive if \(r(a)> 0\) and

$$\begin{aligned} \lim _{k \rightarrow \infty } d_k(a) = 0 ~ \text {where} ~ d_k(a):= \text{ dist }\left( \frac{a^k}{r(a)^k},K\right) ~ (k \in \mathbb {N}_0). \end{aligned}$$

Let us call a condition C(a) a Perron–Frobenius condition if C(a) implies \(r(a)\in \sigma (a)\) for all \(a \in \mathcal{A}\) with \(r(a)>0\). Note that \(C(a)=\)[\(a \in K\) for a normal cone K] is not a Perron–Frobenius condition, as can be seen by the trivial example \(\mathcal{A}=\mathbb {C}\), \(K=\{\lambda i: \lambda \ge 0\}\) and \(a=i\).

There are many known Perron–Frobenius conditions in matrix algebras, operator algebras and Banach algebras. Without claiming completeness we refer to

  1. 1.

    \(C(a)=[a\ge 0]\), i.e. a is positive, for general Banach algebras ordered by an algebra cone, see Raubenheimer and Rode [10, Theorem 5.2],

  2. 2.

    \(C(a)=[\exists k_0 \forall k \ge k_0: a^k \ge 0]\), i.e. a is eventually positive, for matrix algebras ordered by the cone of matrices with nonnegative entries, see Chaysri and Noutsos [2, Theorem 2.5],

  3. 3.

    \(C(a)=[\lim _{k \rightarrow \infty } d_k(a)= 0]\), i.e. a is asymptotically positive, for operator algebras on ordered Banach lattices, see Glück [4, Theorem 4.1],

  4. 4.

    \(C(a)=[\lim _{k \rightarrow \infty } \sqrt{k}d_k(a)= 0]\), i.e. a is asymptotically positive with a rate of convergence, for general Banach algebras ordered by an algebra cone, see [5],

and the references given in the cited literature. Note that the setting in [5] comprises the one in [4].

For a survey on spectral theory in ordered Banach algebras we refer to the paper of Mouton and Raubenheimer [8].

Our main result generalizes the above mentioned results from [4] and [5] in two ways. First, we relax the condition that K is an algebra cone, which already gives a new result for the matrix case. Second, we relax the condition of asymptotic positivity by showing that \(C(a)=[\limsup _{k \rightarrow \infty } d_k(a)\) is “sufficiently small”] is a Perron–Frobenius condition in our setting.

2 Results

Theorem 1

Let \(a \in \mathcal {A}\) with \(r(a)>0\) such that

$$\begin{aligned} \limsup _{k \rightarrow \infty }d_k(a) < \frac{1}{\pi \gamma +1}. \end{aligned}$$

Then \(r(a)\in \sigma (a)\).

Remark 1

An inspection of the proofFootnote 1 shows, that the weaker but less manageable condition on Abel means of the sequence \((d_k(a))_{k \in \mathbb {N}_0}\),

$$\begin{aligned} AM(a):=\limsup _{t \rightarrow 1 +} \,(t-1)\sum _{k=0}^\infty \frac{d_k(a)}{t^{k+1}}<\frac{1}{\pi \gamma +1}, \end{aligned}$$

is still sufficient for \(r(a) \in \sigma (a)\). This condition is weaker by Lemma 3.

Aiming at optimality of the constant in Remark 1 we can define \(\tau (\mathcal{A})\) as the maximum of all \(\tau >0\) such that

$$\begin{aligned} \forall a \in \mathcal{A}: ~ r(a)>0, ~ AM(a) < \tau ~~ \Longrightarrow ~~ r(a) \in \sigma (a). \end{aligned}$$

Remark 1 shows \(\tau (\mathcal{A}) \ge (\pi \gamma +1)^{-1}\). For \(a= -\textbf{1}\) we have \(1=r(a) \notin \sigma (a)=\{-1\}\) and, for any cone K, we have \(\text{ dist }(\pm \textbf{1},K)\le 1\), hence \(AM(-\textbf{1}) \le 1\). This gives the trivial bound \(\tau (\mathcal{A}) \le 1\). If either \(\textbf{1} \in K\) or \(-\textbf{1} \in K\), then

$$\begin{aligned} d_k(-\textbf{1}) = \text{ dist }((-1)^k \textbf{1},K) \end{aligned}$$

is alternatingly 0 and \(\le 1\). Thus \(AM(-\textbf{1}) \le 1/2\) and we have \(\tau (\mathcal{A}) \le 1/2\).

As an introductory example consider the matrix-algebra \(\mathcal{A}=\mathbb {C}^{m \times m}\) endowed with the row-sum norm \(\Vert \cdot \Vert _\infty \) and ordered by the algebra cone \(K:= [0,\infty )^{m \times m}\) of all matrices with nonnegative entries. Here \(\gamma =1\). Thus, for \(A \in \mathbb {C}^{m \times m}\), Theorem 1 reads

$$\begin{aligned} \limsup _{k \rightarrow \infty }d_k(A) < \frac{1}{\pi +1} ~ \Rightarrow ~ r(A) \in \sigma (A), \end{aligned}$$

and we have

$$\begin{aligned} \forall m\in \mathbb {N}: ~ 0,2415\approx \frac{1}{\pi +1} \le \tau (\mathbb {C}^{m \times m}) \le \frac{1}{2}. \end{aligned}$$

For the matrix

$$\begin{aligned} A=\frac{1}{2}\left( \begin{array}{cc} -1 &{} 1\\ 1 &{} -1 \end{array}\right) \in \mathbb {C}^{2 \times 2} \end{aligned}$$

we have \(\sigma (A)=\{-1,0\}\) and \(d_k(A)=1/2\) \((k \in \mathbb {N})\). Hence

$$\begin{aligned} \limsup _{k \rightarrow \infty } d_k(A) = AM(A)= \frac{1}{2}. \end{aligned}$$

In case \(m=1\) we have for \(a \in \mathbb {C}\) that \(r(a)=1\) if and only if \(a=e^{it}\) for some \(t \in [0,2\pi )\). For \(t \in (0,2\pi )\) one can check that

$$\begin{aligned} \limsup _{k \rightarrow \infty } d_k(e^{it})=1, \quad AM(e^{it}) \ge \frac{1}{2}. \end{aligned}$$

In particular \(\tau (\mathbb {C})=1/2\).

The matrix

$$\begin{aligned} A=\frac{1}{5+2\sqrt{5}}\left( \begin{array}{cc} 9 &{} -2\\ -2 &{} 1 \end{array}\right) \in \mathbb {C}^{2 \times 2} \end{aligned}$$

is an example where, in this setting, A is not asymptotically positive but Theorem 1 applies. We have

$$\begin{aligned} \lim _{k \rightarrow \infty } A^k = \left( \begin{array}{cc} \frac{1}{2}+ \frac{1}{\sqrt{5}} &{} -\frac{1}{2\sqrt{5}}\\ -\frac{1}{2\sqrt{5}} &{} \frac{1}{2}- \frac{1}{\sqrt{5}} \end{array}\right) , \end{aligned}$$

thus

$$\begin{aligned} \lim _{k \rightarrow \infty } d_k(A) = \frac{1}{2\sqrt{5}} < \frac{1}{\pi +1}, \end{aligned}$$

and \(1 \in \sigma (A)\).

3 Proof

We will use the following lemmas. For Lemma 1 see, e.g., [1, Theorem 3.3.5.]. A result related to Lemma 2, though with the worse constant \(4\gamma \), can be found in the preprint [6, Lemma 4.6] by Huang, Jaffe, Liu, and Wu.

Lemma 1

Let \(a \in \mathcal {A}\). Then

$$\begin{aligned} r((\lambda \textbf{1} -a)^{-1}) = \frac{1}{\text{ dist }(\lambda ,\sigma (a))} \quad (\lambda \in \rho (a)). \end{aligned}$$

Lemma 2

Let \((a_k)_{k \in \mathbb {N}_0}\) and \((\lambda _k)_{k \in \mathbb {N}_0}\) be sequences in K and \(\mathbb {C}\), respectively, such that the series

$$\begin{aligned} \sum _{k=0}^\infty \lambda _k a_k \end{aligned}$$

is absolutely convergent. Then we have

$$\begin{aligned} \left\| \sum _{k=0}^\infty \lambda _k a_k\right\| \le \pi \gamma \left\| \sum _{k=0}^\infty |\lambda _k| a_k\right\| . \end{aligned}$$

Proof

W.l.o.g. we may first assume \(\lambda _k \not =0\) \((k \in \mathbb {N}_0)\) and then \(|\lambda _k| =1\) \((k \in \mathbb {N}_0)\) since the modulus of \(\lambda _k\) can be absorbed into the vector \(a_k\). Set \(S^1:= \{z \in \mathbb {C}:|z|=1\}\) and let \(h: S^1 \rightarrow \mathbb {R}\) be defined as

$$\begin{aligned} h(z) = \left\{ \begin{array}{cc} 1, &{} \Re (z) \ge 0 \\ 0, &{} \Re (z) < 0 \end{array} \right. . \end{aligned}$$

First note that

$$\begin{aligned} \forall z \in S^1: ~ z = \frac{1}{2} \int _{-\pi }^\pi h(ze^{-it}) e^{it} dt. \end{aligned}$$

Thus, for \(l \ge 0\),

$$\begin{aligned} \left\| \sum _{k=0}^l \lambda _k a_k\right\|= & {} \frac{1}{2} \left\| \sum _{k=0}^l \left( \int _{-\pi }^\pi h\left( \lambda _k e^{-it}\right) e^{it} dt\right) a_k\right\| \\= & {} \frac{1}{2} \left\| \int _{-\pi }^\pi \left( \sum _{k=0}^l h\left( \lambda _k e^{-it}\right) e^{it} a_k \right) dt \right\| \\\le & {} \frac{1}{2} \int _{-\pi }^\pi \left\| \sum _{k=0}^l h \left( \lambda _k e^{-it}\right) e^{it} a_k \right\| dt \\= & {} \frac{1}{2} \int _{-\pi }^\pi \left\| \sum _{k=0}^l h \left( \lambda _k e^{-it}\right) a_k \right\| dt \\\le & {} \frac{\gamma }{2} \int _{-\pi }^\pi \left\| \sum _{k=0}^l a_k \right\| dt = \pi \gamma \left\| \sum _{k=0}^l a_k\right\| . \end{aligned}$$

Now \(l \rightarrow \infty \) proves the desired inequality. \(\square \)

Lemma 3

Let \((t_n)_{n \in \mathbb {N}}\) be a sequence in \((1,\infty )\) with limit 1, and let \((\beta _k)_{k \in \mathbb {N}_0}\) be a bounded sequence in \([0, \infty )\) and \(s:=\limsup _{k \rightarrow \infty } \beta _k\). Then

$$\begin{aligned} \limsup _{n \rightarrow \infty }(t_n-1)\sum _{k=0}^\infty \frac{\beta _k}{t_n^{k+1}} \le s. \end{aligned}$$

Proof

Let \(\varepsilon > 0\). Then we find \(k_0 \in \mathbb {N}_0\) such that \(\beta _k < s + \varepsilon \) \((k > k_0)\). Thus, for \(n \in \mathbb {N}\),

$$\begin{aligned} (t_n-1)\sum _{k=0}^\infty \frac{\beta _k}{t_n^{k+1}}\le & {} (t_n-1)\sum _{k=0}^{k_0}\frac{\beta _k}{t_n^{k+1}} + (s+\varepsilon ) (t_n-1)\sum _{k=k_0+1}^{\infty }\frac{1}{t_n^{k+1}} \\\le & {} (t_n-1)\sum _{k=0}^{k_0}\frac{\beta _k}{t_n^{k+1}} +(s+\varepsilon ) \frac{(t_n-1)}{t_n}\sum _{k=0}^{\infty }\frac{1}{t_n^{k}} \\= & {} (t_n-1)\sum _{k=0}^{k_0}\frac{\beta _k}{t_n^{k+1}} + s+\varepsilon . \end{aligned}$$

We obtain

$$\begin{aligned} \limsup _{n \rightarrow \infty }(t_n-1)\sum _{k=0}^\infty \frac{\beta _k}{t_n^{k+1}} \le s + \varepsilon , \end{aligned}$$

and \(\varepsilon \rightarrow 0+\) proves the assertion. \(\square \)

Proof of Theorem 1

We may assume \(r(a)=1\). Let \(\lambda _0 \in \sigma (a)\) with \(|\lambda _0|=1\). By assumption there is a sequence \((b_k)_{k \in \mathbb {N}_0}\) in \(\mathcal {A}\) such that \(a^k+b_k \ge 0\) \((k \in \mathbb {N}_0)\) and

$$\begin{aligned} s:=\limsup _{k \rightarrow \infty } \Vert b_k\Vert < \frac{1}{\pi \gamma +1}. \end{aligned}$$

For \(|\lambda | > 1\) we have

$$\begin{aligned} (\lambda \textbf{1} -a)^{-1} = \sum _{k=0}^\infty \frac{a^k}{\lambda ^{k+1}}. \end{aligned}$$

We choose a sequence \((t_n)\) in \((1,\infty )\) with limit 1. Now Lemma 2 yields

$$\begin{aligned} \Vert ((t_n \lambda _0) \textbf{1} -a)^{-1}\Vert= & {} \left\| \sum _{k=0}^\infty \frac{a^k}{(t_n \lambda _0)^{k+1}} \right\| \le \left\| \sum _{k=0}^\infty \frac{a^k+b_k}{(t_n \lambda _0)^{k+1}} \right\| + \left\| \sum _{k=0}^\infty \frac{b_k}{(t_n \lambda _0)^{k+1}} \right\| \\\le & {} \pi \gamma \left\| \sum _{k=0}^\infty \frac{a^k+b_k}{t_n^{k+1}} \right\| + \left\| \sum _{k=0}^\infty \frac{b_k}{(t_n \lambda _0)^{k+1}} \right\| \\\le & {} \pi \gamma \left\| \sum _{k=0}^\infty \frac{a^k}{t_n^{k+1}} \right\| + \pi \gamma \left\| \sum _{k=0}^\infty \frac{b_k}{t_n^{k+1}} \right\| + \left\| \sum _{k=0}^\infty \frac{b_k}{(t_n \lambda _0)^{k+1}} \right\| \\= & {} \pi \gamma \Vert (t_n \textbf{1} -a)^{-1}\Vert + \pi \gamma \left\| \sum _{k=0}^\infty \frac{b_k}{t_n^{k+1}} \right\| + \left\| \sum _{k=0}^\infty \frac{b_k}{(t_n \lambda _0)^{k+1}} \right\| \\\le & {} \pi \gamma \Vert (t_n \textbf{1} -a)^{-1}\Vert + (\pi \gamma +1)\sum _{k=0}^\infty \frac{\Vert b_k\Vert }{t_n^{k+1}}. \end{aligned}$$

By Lemma 3 with \(\beta _k:= \Vert b_k\Vert \) \((k \in \mathbb {N}_0)\) we have

$$\begin{aligned} \limsup _{n \rightarrow \infty } (t_n-1)\sum _{k=0}^\infty \frac{\Vert b_k\Vert }{t_n^{k+1}} \le s < \frac{1}{\pi \gamma +1}. \end{aligned}$$

Moreover, by Lemma 1,

$$\begin{aligned} (t_n-1) \Vert ((t_n \lambda _0) \textbf{1} -a)^{-1}\Vert\ge & {} (t_n-1) r(((t_n \lambda _0) \textbf{1} -a)^{-1}) \\= & {} \frac{t_n-1}{\text{ dist }(t_n\lambda _0,\sigma (a))}=1. \end{aligned}$$

If we now assume that \(1 \notin \sigma (a)\) we obtain

$$\begin{aligned} 1\le & {} \limsup _{n \rightarrow \infty } (t_n-1)\Vert ((t_n \lambda _0) \textbf{1} -a)^{-1}\Vert \\\le & {} \limsup _{n \rightarrow \infty }(t_n-1) \left[ \pi \gamma \Vert (t_n \textbf{1} -a)^{-1}\Vert + (\pi \gamma +1)\sum _{k=0}^\infty \frac{\Vert b_k\Vert }{t_n^{k+1}}\right] \\= & {} (\pi \gamma +1)\limsup _{n \rightarrow \infty }(t_n-1) \sum _{k=0}^\infty \frac{\Vert b_k\Vert }{t_n^{k+1}} < 1, \end{aligned}$$

a contradiction. \(\square \)

4 Applications

1. Let \(\mathcal{A}\) be a Banach algebra and let \(\phi :\mathcal{A} \rightarrow \mathbb {C}\) be a continuous linear functional on \(\mathcal{A}\) (with \(\Vert \phi \Vert \ge 1\) to avoid the trivial case). Then

$$\begin{aligned} K:=\{c \in \mathcal{A}: \Vert c\Vert \le \Re (\phi (c))\} \end{aligned}$$

is a normal cone with \(\gamma =\Vert \phi \Vert \). By application of Theorem 1 we get the following corollaries.

Corollary 1

Let \(a \in \mathcal{A}\) with \(r(a)=1\). If there exists a sequence \((c_k)\) in \(\mathcal{A}\) and \(k_0 \in \mathbb {N}\) with

$$\begin{aligned} \forall k\ge k_0: ~ \Vert c_k\Vert \le \Re (\phi (c_k)), \quad \limsup _{k \rightarrow \infty } \Vert a^k-c_k\Vert < \frac{1}{\pi \Vert \phi \Vert +1}, \end{aligned}$$

then \(1 \in \sigma (a)\).

A special case of the situation in Corollary 1 is that \((a^k)\) is eventually in K. Then we can eventually choose \(c_k=a^k\), and obtain:

Corollary 2

Let \(a \in \mathcal{A}\) with \(r(a)=1\). If

$$\begin{aligned} \exists k_0 \in \mathbb {N}~ \forall k\ge k_0: ~ \Vert a^k\Vert \le \Re (\phi (a^k)), \end{aligned}$$

then \(1 \in \sigma (a)\).

2. Let \(\mathcal{A}\) be a \(C^*\)-algebra and let

$$\begin{aligned} K:=\{c \in \mathcal{A}: c=c^*, \sigma (c) \subseteq [0,\infty )\}. \end{aligned}$$

Then K is a normal cone with \(\gamma =1\) [3, 1.6.9] and we have the following.

Corollary 3

Let \(a \in \mathcal{A}\) with \(r(a)=1\). If

$$\begin{aligned} \limsup _{k \rightarrow \infty } \left\| a^{2k}-\frac{1}{4}(a^k+(a^*)^k)^2\right\| < \frac{1}{\pi +1} \end{aligned}$$

then \(\sigma (a) \cap \{-1,1\} \not = \emptyset \).

Proof

Since \((b+b^*)^2 \ge 0\) for each \(b \in \mathcal{A}\) we have

$$\begin{aligned} d_k(a^2) \le \left\| a^{2k}-\frac{1}{4}(a^k+(a^*)^k)^2\right\| \quad (k \in \mathbb {N}_0). \end{aligned}$$

Application of Theorem 1 to \(a^2\) yields \(1 \in \sigma (a^2)\), hence \(\sigma (a) \cap \{-1,1\} \not = \emptyset \). \(\square \)

Remark 2

Of course, other quite natural assumptions are possible, such as

$$\begin{aligned} \limsup _{k \rightarrow \infty } \left\| a^{2k}-(aa^*)^k\right\| < \frac{1}{\pi +1} \end{aligned}$$

or

$$\begin{aligned} \limsup _{k \rightarrow \infty } \left\| a^{2k}-\frac{1}{2}(a^k(a^*)^k+(a^*)^ka^k )\right\| < \frac{1}{\pi +1}. \end{aligned}$$

However, numerical experiments with random matrices indicate that the assumption in Corollary 3 is fulfilled more easily.

3. Let \(a \in \mathcal{A}\) with \(r(a)=1\). Assume that \(\{a^k:k \in \mathbb {N}\}\) is relatively compact, and that the sequence \((a^k)\) has a finite number of accumulation points \(b_1,\dots , b_m\) which are linearly independent. To apply Theorem 1 set

$$\begin{aligned} K:= \left\{ \sum _{j=1}^m \alpha _j b_j: \alpha _j \ge 0 ~ (j=1,\dots ,m)\right\} . \end{aligned}$$

Then K is a cone since \(b_1,\dots , b_m\) are linearly independent, and K is normal since it is contained in a finite dimensional subspace of \(\mathcal{A}\). Now a is asymptotically positive with respect to this cone, hence \(1 \in \sigma (a)\). Without the assumption of linear independence this is trivially wrong as can be seen by considering \(a=-\textbf{1}\).