On the peripheral point spectrum and the asymptotic behavior of irreducible semigroups of Harris operators

Abstract

Given a positive, irreducible and bounded $C_0$-semigroup on a Banach lattice with order continuous norm, we prove that the peripheral point spectrum of its generator is trivial whenever one of its operators dominates a non-trivial compact or kernel operator. For a discrete semigroup, i.e. for powers of a single operator $T$, we show that the point spectrum of some power $T^k$ intersects the unit circle at most in $1$. As a consequence, we obtain a sufficient condition for strong convergence of the $C_0$-semigroup and for a subsequence of the powers of $T$, respectively.

Appendices

Appendix: Greiner’s zero-two law

We present the proof of Greiner’s zero-two law from [13] in a reformulation for Banach lattices with order continuous norm and without any continuity condition on the semigroup.

Throughout, let $\fancyscript{T}=(T(t))_{t\in R}$ be a positive and bounded semigroup on $E$, a Banach lattice with order continuous norm, and fix $\tau >0$.

Theorem 5.1

(Greiner’s zero-two law) Assume that $\mathrm {Fix} (\fancyscript{T})$ contains a quasi-interior point $e$ of $E_+$ and that there exists a strictly positive element in $\mathrm {Fix} (\fancyscript{T}^{\prime })$. Then

$$\begin{aligned} E_2 \mathrel {\mathop :}=\{ y\in E : (T(t)\wedge T(t+\tau ))\vert y\vert = 0\quad \text{ for} \text{ all} t\in R \} \end{aligned}$$

and $E_0\mathrel {\mathop :}=E_2^\bot$ are $\fancyscript{T}$-invariant bands. Moreover, if $P$ denotes the band projection onto $E_2$, then

$$\begin{aligned} \vert T(t) - T(t+\tau ) \vert Pe = 2Pe\quad for all t\in R \end{aligned}$$

and

$$\begin{aligned} \lim _{t\rightarrow \infty } \vert T(t) - T(t+\tau ) \vert (I-P)e =0. \end{aligned}$$

To simplify notation, for $t \in R$ we define the positive operators

$$\begin{aligned} S(t)\mathrel {\mathop :}=T(t)\wedge T(t+\tau ) \quad \text{ and} \quad D(t) \mathrel {\mathop :}=\vert T(t) - T(t+\tau ) \vert \end{aligned}$$

on $E$. It follows immediately that

$$\begin{aligned} S(t)x+\frac{1}{2}D(t)x = x \end{aligned}$$

(5.1)

for all $x\in \mathrm {Fix} (\fancyscript{T})$ and $t\in R$. Further properties of $S(t)$ and $D(t)$ are provided by the following lemma.

Lemma 5.2

Let $x^{\prime } \in \mathrm {Fix} (\fancyscript{T}^{\prime })$ be strictly positive and $\tau \in R$. Then the following assertions hold.

• (a) $D(t)T(s) \ge D(t+s)$ and $T(s)D(t)\ge D(t+s)$ for all $t,s\in R$. Moreover, if $x\in \mathrm {Fix} (\fancyscript{T})$, then $\lim _{t\rightarrow \infty } D(t)x \in \mathrm {Fix} (\fancyscript{T})$.

• (b) $S(t)T(s) \le S(t+s)$ and $T(s)S(t)\le S(t+s)$ for all $t,s\in R$. Moreover, if $x\in \mathrm {Fix} (\fancyscript{T})$, then $\lim _{t\rightarrow \infty } S(t)x \in \mathrm {Fix} (\fancyscript{T})$.

• (c) If $\lim _{t\rightarrow \infty } S(t)x > 0$ for all $0<x\in \mathrm {Fix} (\fancyscript{T})$, then $\lim _{t\rightarrow \infty } S(t)^m x > 0$ for all $m\in \mathbb{N }$ and $x\in \mathrm {Fix} (\fancyscript{T})$, $x>0$.

 

 

Proof

(a) For all $t, s\in R$ one observes that

$$\begin{aligned} D(t)T(s) = \vert T(t)-T(t+\tau )\vert \cdot \vert T(s)\vert \ge \vert (T(t)-T(t+\tau ))T(s)\vert = D(t+s) \end{aligned}$$

and similarly that $T(s)D(t)\ge D(t+s)$. Let $x\in \mathrm {Fix} (\fancyscript{T})$, $x\ge 0$. Then

$$\begin{aligned} D(t)x = D(t)T(s)x \ge D(t+s)x \end{aligned}$$

for all $t,s\in R$. Hence, by the order continuity of the norm, $y\mathrel {\mathop :}=\lim D(t)x$ exists in $E$ and

$$\begin{aligned} T(s)y = \lim _{t\rightarrow \infty } T(s)D(t)x \ge \lim _{t\rightarrow \infty } D(t+s)x = y \ge 0. \end{aligned}$$

Finally, we conclude from $\langle x^{\prime },T(s)y - y\rangle = 0$ that $y\in \mathrm {Fix} (\fancyscript{T})$ because $x^{\prime }$ is strictly positive.

(b) For all $t,s\in R$ one observes that

$$\begin{aligned} S(t)T(s)&= \frac{1}{2} (T(t+s)+T(t+\tau +s)-D(t)T(s) )\\&\le \frac{1}{2} (T(t+s)+T(t+\tau +s)-D(t+s)) = S(t+s) \end{aligned}$$

and similarly that $T(s)S(t) \le S(t+s)$. Hence, $S(t)x$ is increasing for all positive $x\in \mathrm {Fix} (\fancyscript{T})$. Since $0 \le S(t) \le \frac{1}{2}(T(t)+T(t+\tau ))$, we conclude as in the proof of part (a) that $\lim S(t)x$ exists in $\mathrm {Fix} (\fancyscript{T})$ for all $x\in \mathrm {Fix} (\fancyscript{T})$.

(c) Let $x\in \mathrm {Fix} (\fancyscript{T})$, $x>0$, and define recursively $x_k \mathrel {\mathop :}=\lim _{t\rightarrow \infty } S(t)x_{k-1}$ for all $k\in \mathbb{N }$ where $x_0 \mathrel {\mathop :}=x$. Then $x_k\in \mathrm {Fix} (\fancyscript{T})$ by part (b) and $x_k>0$ by assumption. It follows by induction that

$$\begin{aligned} S(t)^m x-x_m = \sum _{j=1}^m S(t)^{m-j}( S(t)x_{j-1} -x_j ) \end{aligned}$$

for all $m\in \mathbb N$ and $t\in R$. As $\Vert S(t) \Vert \le \sup _{t\in R}\Vert T(t)\Vert$, we conclude that

$$\begin{aligned} \lim _{t\rightarrow \infty } S(t)^m x = x_m>0 \end{aligned}$$

for all $m\in \mathbb{N }$. $\square$

The key for the proof of the zero-two law is the following combinatorial lemma.

Lemma 5.3

For every $m\in \mathbb{N }$

$$\begin{aligned} 2^{-m}\left( \sum _{j=1}^{m} \biggr \vert \genfrac(){0.0pt}{}{m}{j}-\genfrac(){0.0pt}{}{m}{j-1} \biggr \vert +2\right) \le \frac{2}{\sqrt{m}}. \end{aligned}$$

 

Proof

For $k\in \mathbb{N }$ and $m=2k-1$ we have

$$\begin{aligned} \sum _{j=1}^m \biggr \vert \genfrac(){0.0pt}{}{m}{j}-\genfrac(){0.0pt}{}{m}{j-1} \biggr \vert +2&= 2\sum _{j=1}^{k-1}\biggr (\genfrac(){0.0pt}{}{m}{j}-\genfrac(){0.0pt}{}{m}{j-1}\biggr )+2\\&= 2\genfrac(){0.0pt}{}{m}{k-1} = \genfrac(){0.0pt}{}{2k}{k} \\&= \sum _{j=1}^k \biggr (\genfrac(){0.0pt}{}{m+1}{j}-\genfrac(){0.0pt}{}{m+1}{j-1}\biggr )+1\\&= \frac{1}{2}\sum _{j=1}^{m+1}\biggr \vert \genfrac(){0.0pt}{}{m+1}{j}-\genfrac(){0.0pt}{}{m+1}{j-1} \biggr \vert +1. \end{aligned}$$

It follows from Stirling’s formula that the central binomial coefficient can be estimated by

$$\begin{aligned} \genfrac(){0.0pt}{}{2k}{k} \le \frac{2^{2k}}{\sqrt{2k}} \quad (k\in \mathbb{N }). \end{aligned}$$

Thus, we obtain that

$$\begin{aligned} 2^{-m}\left( \sum _{j=1}^m \biggr \vert \genfrac(){0.0pt}{}{m}{j}-\genfrac(){0.0pt}{}{m}{j-1}\biggr \vert +2\right)&= 2^{-(m+1)} \left(\sum _{j=1}^{m+1}\biggr \vert \genfrac(){0.0pt}{}{m+1}{j}-\genfrac(){0.0pt}{}{m+1}{j-1}\biggr \vert +2\right)\\&= 2^{-m} \genfrac(){0.0pt}{}{2k}{k} \le \frac{2}{\sqrt{m+1}} \le \frac{2}{\sqrt{m}}, \end{aligned}$$

which completes the proof. $\square$

 

Proof of Theorem

By Lemma 5.2 (b), we have

$$\begin{aligned} S(t)\vert T(s)y \vert \le S(t)T(s)\vert y\vert \le S(t+s)\vert y \vert = 0 \end{aligned}$$

for all $y\in E_2$ and $t, s\in R$, which shows that $E_2$ is $\fancyscript{T}$-invariant. Let $x^{\prime }\in \mathrm {Fix} (\fancyscript{T}^{\prime })$ be strictly positive. Since

$$\begin{aligned} 0 \le T(t)Pe = PT(t)Pe \le PT(t)e = Pe \end{aligned}$$

and $\langle x^{\prime },Pe - T(t)Pe\rangle =0$ for all $t\in R$, it follows that $Pe \in \mathrm {Fix} (\fancyscript{T})$. Define $e_0 \mathrel {\mathop :}=(I-P)e \in E_0$ and $e_2 \mathrel {\mathop :}=Pe\in E_2$. As $e_0 = e-e_2 \in \mathrm {Fix} (\fancyscript{T})$, we conclude that $E_0$, which equals the closure of the principle ideal generated by $e_0$, is $\fancyscript{T}$-invariant.

It follows immediately from (5.1) that $D(t)e_2 =2 e_2$ for all $t\in R$. Hence, it remains to show that $\lim D(t) e_0 = 0$. For simplicity, we omit the index $0$ and write $E=E_0$, $e=e_0$, $T(t)=T(t)|_{E_0}$ and so on. Now, $\lim S(t)y > 0$ for all $y\in \mathrm {Fix} (\fancyscript{T})$, $y>0$, by Lemma 5.2 (b) and the definition of $E_0$.

Assume that $h\mathrel {\mathop :}=\lim D(t)e >0$. As $h\le 2e$, there exists $m\in \mathbb{N }$ such that

$$\begin{aligned} k \mathrel {\mathop :}=\biggr ( h -\frac{2}{\sqrt{m}} e\biggr )^+>0. \end{aligned}$$

Moreover, $k\in \mathrm {Fix} (\fancyscript{T})$ since the fixed space is a sublattice. Indeed, if $y\in \mathrm {Fix} (\fancyscript{T})$, it follows from $T(t)\vert y\vert \ge \vert T(t)y\vert = \vert y\vert$ and $\langle x^{\prime },T(t)\vert y\vert - \vert y\vert \rangle = 0$ for all $t\in R$ that $\vert y\vert \in \mathrm {Fix} (\fancyscript{T})$ because $x^{\prime }$ is strictly positive. Now, Lemma 5.2 (c) yields that $S(t_0)^m k > 0$ for some $t_0\in R$. Let $t_1 \mathrel {\mathop :}=m(t_0 + \tau )$ and define the operator

$$\begin{aligned} U \mathrel {\mathop :}=T(t_1) - 2^{-m} S(t_0)^m \biggr (I+T(\tau )\biggr )^m. \end{aligned}$$

(5.2)

It follows from $S(t_0)(I+T(\tau )) \le T(t_0+\tau )+T(t_0)T(\tau ) = 2T(t_0+\tau )$ that $U$ is positive. Moreover,

$$\begin{aligned} T(n t_1) = U^n + R_n 2^{-m} \biggr ( I + T(\tau )\biggr )^m \end{aligned}$$

(5.3)

for all $n\in \mathbb{N }$ where $R_1 \mathrel {\mathop :}=S(t_0)^m$ and $R_{n+1} \mathrel {\mathop :}=U^nR_1 + R_nT(t_1)$. We infer from $e=U^ne+R_ne$ that $0\le R_n e \le e$ and $0\le U^n e \le e$ for all $n\in \mathbb{N }$. Now, by Lemma 5.2 (a) and Lemma 5.3, we obtain that

$$\begin{aligned} h \le D(nt_1)e&= \vert T(nt_1)(I-T(\tau ))\vert e \\&\le 2U^n e + R_n2^{-m} \left|\sum _{j=0}^m \genfrac(){0.0pt}{}{m}{j} T^j(\tau )(I-T(\tau ))\right|e \\&= 2U^n e + R_n 2^{-m} \left|\sum _{j=0}^m \genfrac(){0.0pt}{}{m}{j}T^j(\tau ) - \sum _{j=1}^{m+1}\genfrac(){0.0pt}{}{m}{j-1}T^j(\tau )\right|e \\&= 2U^n e + R_n2^{-m} \left( 2e + \sum _{j=1}^{m} \biggr \vert \genfrac(){0.0pt}{}{m}{j}-\genfrac(){0.0pt}{}{m}{j-1}\biggr \vert e\right)\\&\le 2 U^ne+R_n\frac{2}{\sqrt{m}}e \le 2 U^ne+\frac{2}{\sqrt{m}}e \end{aligned}$$

for every $n\in \mathbb{N }$. Let $y\mathrel {\mathop :}=\lim _{n\rightarrow \infty } U^n e \ge 0$. Then $h\le 2 y + \frac{2}{\sqrt{m}} e$ and hence

$$\begin{aligned} 0 < k = \biggr ( h -\frac{2}{\sqrt{m}} e\biggr )^+ \le 2y. \end{aligned}$$

Since $y$ is a fixed point of $U$, Eq. (5.3) yields that $T(n t_1)y \ge y \ge 0$ and we conclude from $\langle x^{\prime },T(nt_1)y - y\rangle =0$ that $T(n t_1)y=y$ for every $n\in \mathbb{N }$. By Eq. (5.2) we have

$$\begin{aligned} 0 = S(t_0)^m \biggr ( I+T(\tau ) \biggr )^m y \ge S(t_0)^my \ge 0. \end{aligned}$$

Therefore, $0 < S(t_0)^m k \le 2S(t_0)^m y = 0$ which contradicts the preceded observation that $S(t_0)^mk>0$. Hence, $h=\lim D(t)e = 0$. $\square$

 

Appendix: Axmann’s theorem

We give a proof Axmann’s theorem from [7], Satz 3.5] stating that not all powers of an irreducible Harris operator can be disjoint. A proof of this for $E=L^p$ can be found in [4], Sec. 6].

We start with a version for $L$-spaces and reduce the general case to it in what follows. Let us recall that a Banach lattice $E$ is said to be a $L$ -space if

$$\begin{aligned} ||x + y ||= ||x||+ ||y||\end{aligned}$$

holds for all $x,y \in E_+$.

Proposition 6.1

Let $T$ be a positive and irreducible operator on a $L$-space $E$ such that $T\not \in (E^{\prime }\otimes E)^\bot$. Then there is $n\in \mathbb N$, $n\ge 2$, such that $T\wedge T^n >0$.

 

Proof

Aiming for a contradiction, we assume that $T\wedge T^n =0$ for all $n \ge 2$. Since $E$ is a $L$-space, we may identify $E^{\prime }$ with $C(K)$ for some compact space $K$ by Kakutani’s theorem [16], Thm. 2.1.3]. For $n,m \in \mathbb N$, $n\ge 2$, define

$$\begin{aligned} A_n \mathrel {\mathop :}=\{ T^{\prime }h + T^{\prime n}(\mathbb 1 -h) : h\in C(K),\, 0\le h\le \mathbb 1 \} \subseteq C(K) \end{aligned}$$

and

$$\begin{aligned} O_{n,m} \mathrel {\mathop :}=\{t \in K : h(t) < 1/m \text{ for} \text{ some} h\in A_n\}. \end{aligned}$$

It follows from our assumption and Synnatschke’s theorem [16], Prop. 1.4.17] that

$$\begin{aligned} \inf A_n = (T^{\prime } \wedge T^{\prime n})\mathbb 1 = (T\wedge T^n)^{\prime } \mathbb 1 = 0 \end{aligned}$$

for all $n\ge 2$. Now, we show that each of the open sets $O_{n,m}$ is dense in $K$. Assume the opposite. Then there exists a non-empty open set $U\subseteq K\backslash O_{n,m}$ for some $n,m \in \mathbb{N }$. By Urysohn’s theorem, we can construct a continuous function $g:K\rightarrow [0,\frac{1}{m}]$ vanishing on $O_{n,m}$ such that $g(t_0)=\frac{1}{m}$ for some $t_0 \in U$. Hence, $g>0$ is a lower bound of $A_n$, which is impossible. Therefore, every $O_{n,m}$ and, by Baire’s theorem, also $G\mathrel {\mathop :}=\cap _{n,m} O_{n,m}$ is dense in $K$. Note that for all $t\in G$ and $n\ge 2$ one has that

$$\begin{aligned} \langle T^{\prime \prime }\delta _t \wedge T^{\prime \prime n} \delta _t,\mathbb 1 \rangle = \inf \{ h(t) : h\in A_n \} = 0 \end{aligned}$$

and consequently $T^{\prime \prime } \delta _t \wedge T^{\prime \prime n} \delta _t = 0$.

By assumption, there are $x\in E_+$ and $y^{\prime }\in E^{\prime }_+$ such that $T$ is not disjoint from $R = y^{\prime } \otimes x$. Then $R^{\prime }= x\otimes y^{\prime }$ corresponds to a rank-one operator $\mu \otimes g$ on $C(K)$ for some $\mu \in C(K)^{\prime }_+$ and $g\in C(K)_+$. Since $R^{\prime }\wedge T^{\prime } \ge (R\wedge T)^{\prime } >0$ there exists some $e\in C(K)_+$, such that $\varrho \mathrel {\mathop :}=(R^{\prime }\wedge T^{\prime })e>0$. For all $0\le h\le e$ and $t\in K$ it follows from

$$\begin{aligned} \varrho = (R^{\prime }\wedge T^{\prime })h + (R^{\prime }\wedge T^{\prime })(e-h) \le R^{\prime }h+T^{\prime }(e-h) \end{aligned}$$

that

$$\begin{aligned} \varrho (t) \le \langle R^{\prime }h,\delta _t\rangle +\langle e-h,T^{\prime \prime }\delta _t\rangle = g(t)\langle \mu ,h\rangle +\langle e-h,T^{\prime \prime }\delta _t\rangle . \end{aligned}$$

Taking the infimum over all $0\le h\le e$ yields that $\varrho (t) \le (g(t)\mu \wedge T^{\prime \prime }\delta _t)e$ for all $t\in K$. Now, fix $t\in \{ s\in K : \varrho (s)>0\} \cap G$ which exists since $G$ is dense in $K$. Then $\nu \mathrel {\mathop :}=g(t)\mu \wedge T^{\prime \prime }\delta _t >0$ because $\varrho (t)>0$. As $\nu$ is dominated by $g(t)\mu$ and $\mu$ corresponds to $x \in E$, $\nu$ itself corresponds to a vector $v\in E$, $v>0$, since $E$ is an ideal in $E^{\prime \prime }$. This vector $v$ satisfies

$$\begin{aligned} v \wedge T^n v \le T^{\prime \prime }\delta _t \wedge T^{\prime \prime n+1} \delta _t = 0 \end{aligned}$$

for all $n\in \mathbb{N }$ because $t\in G$.

Finally, we consider $w\mathrel {\mathop :}=Tv$. If $w=0$, then the closed ideal $\overline{E_v}$ is $T$-invariant and non-trivial since $T\ne 0$. If $w>0$, then the closure of the $T$-invariant ideal

$$\begin{aligned} J\mathrel {\mathop :}=\{ z\in E : |z|\le c(w+Tw+\dots +T^k w) \text{ for} \text{ some} c>0 \text{ and} m\in \mathbb{N }\} \end{aligned}$$

is non-trivial because $w\in \overline{J}$ and $v \in J^\bot$. In both cases, $T$ cannot be irreducible. Thus, we conclude that $T\wedge T^n > 0$ for some $n\ge 2$. $\square$

 

Theorem 6.2

Let $T$ be a positive and irreducible operator on $E$, a Banach lattice with order continuous norm, such that $T\not \in (E^{\prime }\otimes E)^\bot$. Then there is $n\in \mathbb{N }$, $n\ge 2$, such that $T\wedge T^n >0$.

 

Proof

Fix $\lambda > ||T ||$ and $y^{\prime }\in E^{\prime }_+$, $y^{\prime }\ne 0$. Then $z^{\prime } \mathrel {\mathop :}=(\lambda - T^{\prime })^{-1}y^{\prime }$ satisfies $\lambda z^{\prime } - T^{\prime }z^{\prime } = y^{\prime }$ and hence $T^{\prime }z^{\prime } \le \lambda z^{\prime }$. This implies that the closed ideal $\{ x\in E : \langle |x|,z^{\prime }\rangle =0\}$ is $T$-invariant and thus equal to $\{0\}$. Therefore, $||x||_{z^{\prime }} \mathrel {\mathop :}=\langle z^{\prime },|x|\rangle$ defines an order continuous lattice norm on $E$. Let $(F,||\cdot ||_F)$ be the completion of $(E,||\cdot ||_{z^{\prime }})$, i.e. the closure of $E$ in $(E,||\cdot ||_{z^{\prime }})^{\prime \prime }$. Then $F$ is an $L$-space and, since $||Tx ||_F \le \lambda ||T||_F$ for all $x\in E$, $T$ uniquely extends to a positive operator $\tilde{T}$ on $F$.

Now, it follows as in the proof of Proposition 3.2 that $\tilde{T}$ is irreducible and $\tilde{T} \not \in (F^{\prime }\otimes F)^{\bot }$.

Since the norm on $E$ is order continuous, so is $z^{\prime }$ and hence $E$ is an ideal in $F$ by [17], Lem. IV 9.3]. Thus, for $x\in E_+$ and $n\in \mathbb{N }$ we observe that

$$\begin{aligned} (\tilde{T}\wedge \tilde{T}^n)x&= \inf _F \{ \tilde{T}(x-y) +\tilde{T}^n y : y\in F,\, 0\le y\le x\}\\&= \inf _E \{ T(x-y)+T^n y : y\in E,\, 0\le y\le x\} \\&= (T\wedge T^n)x. \end{aligned}$$

As $E_+$ is dense in $F_+$, this shows that $\tilde{T} \wedge \tilde{T}^n = 0$ if and only if $T \wedge T^n = 0$. Thus, the assertion follows from Proposition 6.1. $\square$