Long-term analysis of positive operator semigroups via asymptotic domination

Abstract

We consider positive operator semigroups on ordered Banach spaces and study the relation of their long time behaviour to two different domination properties. First, we analyse under which conditions almost periodicity and mean ergodicity of a semigroup $\mathcal {T}$ are inherited by other semigroups which are asymptotically dominated by $\mathcal {T}$. Then, we consider semigroups whose orbits asymptotically dominate a positive vector and show that this assumption is often sufficient to conclude strong convergence of the semigroup as time tends to infinity. Our theorems are applicable to time-discrete as well as time-continuous semigroups. They generalise several results from the literature to considerably larger classes of ordered Banach spaces.

Appendix A: Examples of AN–spaces

Motivated by the discussion in Sects. 4.1 and 4.2 we present several non-trivial examples of AN–spaces in this “Appendix”. First we show by a rather general construction that there exist, on the one hand, infinite-dimensional examples of such spaces which are reflexive and that there exist, on the other hand, examples of such spaces where not every order interval is weakly compact (see Propositions A.1 and A.2). Furthermore we demonstrate how the cone of an AL-Banach lattice can be adapted to yield an AN–space which is no longer lattice ordered and not reflexive, but in which all order intervals are weakly compact (Example A.4).

Let X be a Banach space and let $u \in X$, $u' \in X'$ such that $\langle u', u\rangle = 1$. Set $P = u' \otimes u$, $Q = I-P$. Then the set

$$\begin{aligned} X_{+,u,u'} := \{y \in X: \langle u', y\rangle \ge \Vert Qy\Vert \}. \end{aligned}$$

is a closed generating cone in X that renders X an ordered Banach space. We call $X_{+,u,u'}$ a centered cone with parameters u and $u'$. Some properties of those cones were studied in detail in [22, Section 4.1]; in particular, it is shown in [22, Proposition 4.1.4] that the cone $X_{+,u,u'}$ is normal and has non-empty interior. Related constructions had already occurred earlier in the literature on several occasions, as for instance in [34, Section 1] and [18, 27, 29]; see also the last part of Section 2.6 in [2].

Let us now show that every Banach space which is endowed with a centred cone can be given an equivalent norm which is additive on the positive cone.

Proposition A.1

Let X be a Banach space ordered by the centred cone $X_{+,u,u'}$. Then

$$\begin{aligned} \Vert z\Vert _1 := \max \{|\langle u',z \rangle |,\Vert Qz\Vert \} \qquad \text {for all } z \in X \end{aligned}$$

defines a norm on X which is equivalent to the original norm $\Vert \cdot \Vert$ and which is additive on $X_{+,u,u'}$. In particular $(X, X_{+,u,u'}, \Vert \cdot \Vert _1)$ is an AN–space.

Proof

The fact that $\Vert \cdot \Vert _1$ is equivalent to the given norm follows from $P+Q = I$, and the fact that $\Vert \cdot \Vert _1$ is additive on the cone follows from $\Vert z\Vert _1 = \langle u',z \rangle$ for all $z \in X_{+,u,u'}$. $\square$

This proposition shows that every Banach space X can be endowed with a cone and an equivalent norm such that it becomes an AN-space (and thus, after a further renorming, a base norm space, cf. Proposition 4.4); a related observation is briefly discussed in [31, Example 4 on p. 27]. On the other hand, the following general result implies that a Banach space which is ordered by a centred cone is reflexive if and only if each order interval is weakly compact:

Proposition A.2

Let X be an ordered Banach space with normal cone and assume that $X_+$ has non-empty interior. Then the following assertions are equivalent:

• (i) The space X is reflexive.

• (ii) Every order interval in X is weakly compact.

Proof

“(i) $\Rightarrow$ (ii)” Every ordered interval in X is weakly closed and, by the normality of $X_+$, norm bounded.

“(ii) $\Rightarrow$ (i)” Let x be an interior point of $X_+$. Then $-x$ is an interior point of $-X_+$, and thus x is an interior point of $2x- X_+$. Thus, the order interval

$$\begin{aligned}{}[0,2x] = X_+ \cap (2x-X_+) \end{aligned}$$

has non-empty interior. Hence, [0, 2x] contains a closed ball B of strictly positive radius. Since [0, 2x] is weakly compact, so is B. Hence, X is reflexive. $\square$

Propositions A.1 and A.2 have the following interesting consequence:

Remark A.3

Let X be a non-reflexive Banach space; we endow X with a centred cone $X_{+,u',u}$ and with the norm $\Vert \cdot \Vert _1$ from Proposition A.1. Then X is an AN–space, but it follows from Proposition A.2 that not every order interval in X is weakly compact. In particular, X cannot be a projection band in its bidual according to Proposition 2.6.

This is in remarkable contrast to the Banach lattice case since every AL-Banach lattice is a projection band in its bidual.

In view of the results in Sect. 3, ordered Banach spaces with weakly compact order intervals are of particular interest. AL-Banach lattices and more generally preduals of von Neumann-algebras have this property. Naturally, the question arises whether one can find further examples of AN–spaces in which the order intervals are weakly compact. Trivial examples of such spaces are given by direct sums of AL-Banach lattices with reflexive spaces that are endowed with a centred cone (and an appropriate norm, see Proposition A.1). Let us close this “Appendix” with a less trivial example; our construction takes the positive cone in an AL-Banach lattice and makes it somewhat smaller.

Example A.4

Let X be an AL-Banach lattice with positive cone $X_+$ and let $\dim X \ge 3$. Take $\varphi \in X'$ such that $\varphi _+,\ \varphi _- \not = 0$ (note that $X'$ is also a Banach lattice, so $\varphi _+$ and $\varphi _-$ are well-defined). Set

$$\begin{aligned} K = \{x \in X_+: \langle \varphi , x\rangle \ge 0\} =\{ x \in X_+: \langle \varphi _+,x\rangle \ge \langle \varphi _-, x\rangle \}. \end{aligned}$$

Then K is a closed, convex and proper cone. We denote the order on X with respect to the original cone $X_+$ by $\le$, while we denote the order with respect to the new cone K by $\le _K$.

Let us now show that the new cone K is also generating. Consider

$$\begin{aligned} F = \{x \in X: \; |x| \wedge |y| = 0 \text { for all } y \in X \text { satisfying } \langle \varphi _+,|y|\rangle = 0\}, \end{aligned}$$

where $\wedge$ denotes the infimum with respect to the original order $\le$ on X. The set F is a non-zero band in X and the functional $\varphi _+$ is strictly positive on F while $\varphi _-$ vanishes on F. Hence $F_+ \subset K$. Choose $y \in F_+$ such that $\langle \varphi _+,y \rangle = 1$; then $\langle \varphi _{-},y\rangle = 0$ holds. For each $x \in X$ we have

$$\begin{aligned} x= x_+ - x_- = \underbrace{(x_+ +\langle \varphi _-, x_+\rangle y)}_{\in K}-\underbrace{\langle \varphi _-, x_+\rangle y}_{\in K} - \underbrace{(x_-+\langle \varphi _-, x_-\rangle y)}_{\in K}+\underbrace{\langle \varphi _-, x_-\rangle y}_{\in K}. \end{aligned}$$

Hence, K is indeed generating in X.

Next we note that the order intervals on X with respect to $\le _K$ are contained in the order intervals with respect to $\le$. Indeed, for all $x,y \in X$ we have

$$\begin{aligned}{}[x,y]_{\le _K} = (x+K) \cap (y-K) \subseteq (x+X_+) \cap (y - X_+) = [x,y]_{\le }. \end{aligned}$$

Hence, all order intervals with respect to $\le _K$ are weakly compact. Clearly, the norm is additive on K, as it is additive on the larger cone $X_+$.

Finally, we show that (X, K) is not a lattice if $\dim F \ge 2$. Let P be the band projection onto F, and set $Q := I-P$; we have $P,Q \not = 0$ since $\varphi _+,\varphi _- \not = 0$. Choose $v \in F_+ = PX_+$ with $\langle \varphi _+,v\rangle =1$ as well as $w \in Q X_+$ with $\langle \varphi _-, w\rangle = 1$ and define $u = -v+w$.

Let Y be the set of all upper bounds of 0 and u with respect to the order $\le _K$. A vector $y \in X$ is contained in Y if and only if the following three conditions are satisfied:

• (i) $y \ge 0$ and $y \ge u$,

• (ii) $\langle \varphi _+,y\rangle \ge \langle \varphi _-,y\rangle$,

• (iii) $\langle \varphi _+,y-u\rangle \ge \langle \varphi _-, y-u\rangle$.

Now we note that condition (i) is equivalent to $y \ge 0 \vee u$ (where $\vee$ denotes the supremum with respect to the order $\le$); since the vector $0 \vee u$ equals w, condition (i) is satisfied if and only if $y \ge w$. Moreover, condition (iii) is equivalent to $\langle \varphi _+,y\rangle \ge \langle \varphi _-,y\rangle - 2$; hence, condition (iii) is automatically satisfied in case that (ii) is satisfied.

Summing up, we conclude that $y \in Y$ if and only if $y \ge w$ and $\langle \varphi _+,y\rangle \ge \langle \varphi _-,y\rangle$, which is in turn satisfied if and only if $y \ge w$ and $\langle \varphi _+,y\rangle \ge \langle \varphi _-,y\rangle \ge 1$ (since $\langle \varphi _-,w\rangle = 1$).

Define $Y_1 = \{y\ge 0: \langle \varphi _+,y\rangle = 1, \ Qy = w\}$. Then $Y_1 \subseteq Y$. Moreover, every element $y_1 \in Y_1$ is minimal in Y with respect to the order $\le _K$. Indeed, let $y_1 \in Y_1$ and let $y_0 \in Y$ such that $y_0 \le _K y_1$. Then, in particular, $y_0 \le y_1$. Hence we have $Qy_0 \le Qy_1$ but, on the other hand, $Qy_1 = w = Qw \le Qy_0$, so $Qy_0 = Qy_1$. Moreover, we have $Py_0 \le Py_1$ and thus $\langle \varphi _+, Py_0\rangle \le \langle \varphi _+, Py_1\rangle$ but, on the other hand, $\langle \varphi _+, Py_1\rangle = \langle \varphi _+, y_1\rangle = 1 \le \langle \varphi _+, y_0\rangle = \langle \varphi _+, Py_0\rangle$, so actually $\langle \varphi _+,Py_0\rangle = \langle \varphi _+, Py_1\rangle$. Since $\varphi _+$ is strictly positive on $F = PX$, we conclude that $Py_0 = Py_1$. We have thus proved that $y_0 = y_1$.

Therefore, $Y_1$ consists of minimal upper bounds of $\{0,u\}$ with respect to $\le _K$. However, one readily checks that $Y_1$ has more than one element if $\dim F \ge 2$. Our assertion is proved.

Note that, if X is infinite dimensional and $\dim F \ge 2$ in the above example, then $(X,\le _K)$ is an AN–space with weakly compact order intervals which is neither reflexive nor a lattice.