Comparison of spectral radii and Collatz-Wielandt numbers for homogeneous maps, and other applications of the monotone companion norm on ordered normed vector spaces
aa r X i v : . [ m a t h . F A ] M a r Comparison of spectral radii andCollatz-Wielandt numbers for homogeneousmaps, and other applications of the monotonecompanion norm on ordered normed vectorspaces
Horst R. Thieme ∗ School of Mathematical and Statistical SciencesArizona State University, Tempe, AZ 85287-1804, USAMarch 7, 2014
Abstract
It is well known that an ordered normed vector space X withnormal cone X + has an order-preserving norm that is equivalent tothe original norm. Such an equivalent order-preserving norm is givenby ♯x♯ = max { d ( x, X + ) , d ( x, − X + ) } , x ∈ X. (0.1)This paper explores the properties of this norm and of the half-norm ψ ( x ) = d ( x, − X + ) independently of whether or not the cone is normal.We use ψ to derive comparison principles for the solutions of abstractintegral equations, derive conditions for point-dissipativity of nonlin-ear positive maps, compare Collatz-Wielandt numbers, bounds, andorder spectral radii for bounded homogeneous maps and give condi-tions for a local upper Collatz-Wielandt radius to have a lower positiveeigenvector. ∗ ( [email protected] ) eywords: Ordered normed vector space, homogeneous map, normalcone, cone spectral radius, Collatz-Wielandt numbers, Krein-Rutman theo-rem, nonlinear eigenvectors, comparison theorems, basic reproduction num-ber
For models in the biological, social, or economic sciences, there is a naturalinterest in solutions that are positive in an appropriate sense, i.e., they taketheir values in the cone of an ordered normed vector space.
A closed subset X + of a normed real vector space X is called a wedge if(i) X + is convex,(ii) αx ∈ X + whenever x ∈ X + and α ∈ R + .A wedge is called a cone if(iii) X + ∩ ( − X + ) = { } .Nonzero points in a cone or wedge are called positive .A wedge is called solid if it contains interior points.A wedge is called generating if X = X + − X + , (1.1)and total if X is the closure of X + − X + .We introduce the properties of cones that will be needed in this expos´e.Details and more properties will be discussed in Section 2.A cone X + is called normal , if there exists some δ > k x + z k ≥ δ whenever x ∈ X + , z ∈ X + , k x k = 1 = k z k . (1.2)Equivalent conditions for a cone to be normal are given in Theorem 2.1.2 + is called an inf -semilattice [1] (or minihedral [20]) if x ∧ y = inf { x, y } exist for all x, y ∈ X + . X + is called a sup -semilattice if x ∨ y = sup { x, y } exist for all x, y ∈ X + . X + is called a lattice if x ∧ y and x ∨ y exist for all x, y ∈ X + . X is called a lattice if x ∨ y exist for all x, y ∈ X .Since x ∧ y = − (( − x ) ∨ ( − y )), also x ∧ y exist for all x, y in a lattice X .In function spaces, typical cones are formed by the nonnegative functions.If X + is a cone in X , we introduce a partial order on X by x ≤ y if y − x ∈ X + for x, y ∈ X and call X an ordered normed vector space .Every ordered normed vector space carries the monotone companion half-norm ψ ( x ) = d ( x, − X + ), ψ ( x ) ≤ ψ ( y ) , x, y ∈ X, x ≤ y, and the monotone companion norm ♯x♯ = max { ψ ( x ) , ψ ( − x ) } . See [2], [21,(4.2)], [9, L.4.1]. The cone X + is normal if and only if the monotone compan-ion norm is equivalent to the original norm. By the open mapping theorem, X cannot be complete with respect to both norms unless X + is normal. Theproperties of the companion (half) norm are studied in Section 3. Throughout this paper, let X be an ordered normed vector space with cone X + . We use the notation˙ X = X \ { } and ˙ X + = X + \ { } . Definition 1.1.
Let X and Z be ordered vector space with cones X + and Z + and U ⊆ X .Let B : U → Z . B is called positive if B ( U ∩ X + ) ⊆ Y + . B is called order-preserving (or monotone or increasing) if Bx ≤ By whenever x, y ∈ U and x ≤ y .Positive linear maps from X to Y are order-preserving. In the follow-ing, we describe some applications of the monotone companion (half-) norm.Other applications can be found in [2].3 .3 Positivity of solutions to abstract integral inequal-ities We consider integral inequalities of the following kind on an interval [0 , b ],0 < b < ∞ , u ( t ) ≥ Z t K ( t, s ) u ( s ) ds, t ∈ [0 , b ] . (1.3)Here u : [0 , b ] → X is a continuous function, K ( t, s ), 0 ≤ s ≤ t ≤ b , arebounded linear positive operators such that, for each x ∈ X , K ( t, s ) x is acontinuous function of ( t, s ), 0 ≤ s ≤ t ≤ b . The monotone companion half-norm makes it possible to prove the following result without assuming thatthe cone X + is normal (Section 6). Theorem 1.2.
Let X be an ordered Banach space. Let u : [0 , b ] → X be acontinuous solution of the inequality (1.3). Then u ( t ) ∈ X + for all t ∈ [0 , b ] . The monotone companion half-norm can also be useful for normal cones. Weuse it to prove the following result which is important for discrete dynamicalsystems on cones (Section 7).
Theorem 1.3.
Let X + be a normal cone and F : X + → X + . Assumethat there exist real constants c, ˜ c > and a linear, bounded, positive map A : X → X with spectral radius r ( A ) < and these properties:For any x ∈ X + with k x k ≥ c there exists some y ∈ X with k y k ≤ ˜ c and F ( x ) ≤ A ( x ) + y. Then, for any bounded subset B of X + there exists a bounded subset ˜ B of X + such that F n ( B ) ⊆ ˜ B for all n ∈ N .Further, F is point-dissipative: There exists some ˆ c > such that lim sup n →∞ k F n ( x ) k ≤ ˆ c, x ∈ X + . If, in addition, F is continuous and power compact (or, more generally,asymptotically smooth [35, Def.2.25]), then there exists a compact subset K of X + such that F ( K ) = K and, for every bounded subset B of X + , d ( F n ( x ) , K ) → uniformly for x ∈ B . F is power compact if F m is a compact map for some m ∈ N .4 .5 Spectral radii for homogeneous, bounded, order-preserving maps For a linear bounded map B on a complex Banach space, the spectral radiusof B is defined as r ( B ) = sup {| λ | ; λ ∈ σ ( B ) } , (1.4)where σ ( B ) is the spectrum of B , σ ( B ) = C \ ρ ( B ) , (1.5)and ρ ( B ) the resolvent set of B , i.e., the set of those λ ∈ C for which λ − B has a bounded everywhere defined inverse. The following alternative formulaholds, r ( B ) = inf n ∈ N k B n k /n = lim n →∞ k B n k /n , (1.6)which is also meaningful in a real Banach space. If B is a compact linear mapon a complex Banach space and r ( B ) >
0, then there exists some λ ∈ σ ( B )and v ∈ X such that | λ | = r ( B ) and Bv = λv = 0. Such an λ is called aneigenvalue of B . This raises the question whether r ( B ) could be an eigenvalueitself. There is a positive answer, if B is a positive operator and satisfies somegeneralized compactness assumption.Positive linear maps are order-preserving. They have the remarkableproperty that their spectral radius is a spectral value [8] [33, App.2.2] if X is a Banach space and X + a normal generating cone.The celebrated Krein-Rutman theorem [23], which generalizes parts of thePerron-Frobenius theorem to infinite dimensions, establishes that a compactpositive linear map B with r ( B ) > X withtotal cone X + has an eigenvector v ∈ ˙ X + such that Bv = r ( B ) v and apositive bounded linear eigenfunctional v ∗ : X → R , v ∗ = 0, such that v ∗ ◦ B = r ( B ) v ∗ .This theorem has been generalized into various directions by Bonsall [7]and Birkhoff [5], Nussbaum [29, 30], and Eveson and Nussbaum [15] (seethese papers for additional references). In the following, X , Y and Z are ordered normed vector spaces with cones X + , Y + and Z + respectively. 5 efinition 1.4. B : X + → Y is called (positively) homogeneous (of degreeone) , if B ( αx ) = αB ( x ) for all α ∈ R + , x ∈ X + .Since we do not consider maps that are homogeneous in other ways, wewill simply call them homogeneous maps. If follows from the definition that B (0) = 0 . Homogeneous maps are not Frechet differentiable at 0 unless B ( x + y ) = B ( x ) + B ( y ) for all x, y ∈ X + . For the following holds. Proposition 1.5.
Let B : X + → Y be homogeneous. Then the directionalderivatives of B exist at 0 in all directions of the cone and ∂B (0 , x ) = lim t → B ( tx ) − B (0) t = B ( x ) , x ∈ X + . There are good reasons to consider homogeneous maps. Here is a math-ematical one.
Theorem 1.6.
Let F : X + → Y and u ∈ X . Assume that the directionalderivatives of F at u exist in all directions of the cone. Then the map B : X + → Y + , B = ∂F ( u, · ) , B ( x ) = ∂F ( u, x ) = lim t → F ( u + tx ) − F ( u ) t , x ∈ X + , is homogeneous.Proof. Let α ∈ R + . Obviously, if α = 0, B ( αx ) = 0 = αB ( x ). So we assume α ∈ (0 , ∞ ). Then F ( u + t [ αx ]) − F ( u ) t = α F ( u + [ tα ] x ) − F ( u ) tα . As t →
0, also αt → αx existsand ∂F ( u, αx ) = αF ( u, x ) . Another good reason are mathematical population models that take intoaccount that, for many species, reproduction involves a mating process be-tween two sexes. The maps involved therein are not only homogeneous butalso order-preserving. 6ctually, the eigenvector problem for homogeneous maps is quite differ-ent from the one for linear maps. Consider the following simple two sexpopulation model x n = Bx n − where x n = ( f n , m n ) and Bx = ( B x, B x ) B ( f, m ) = p f f + β f f mf + m ,B ( f, m ) = p m m + β m f mf + m . (1.7)This system models a population of females and males which reproduce oncea year with f n and m n representing the number of females and males at thebeginning of year n . The numbers p f , p m are the respective probabilities ofsurviving one year. The harmonic mean describes the mating process andthe parameters β f and β m scale with the resulting amount of offspring permated pair. B always has the eigenvectors (1 ,
0) and (0 ,
1) associated with p f and p m respectively. Looking for a different eigenvalue, λ , of B , we can assume that f + m = 1 and obtain λ − p f β f = m, λ − p m β m = f, m + f = 1 . We add and solve for λ , λ = β f β m + β m p f + β f p m β f + β m and then for m and f , m = β m + p m − p f β f + β m , f = β f + p f − p m β f + β m . The eigenvector ( m, f ) lies in the biological relevant positive quadrant if andonly if the eigenvalue λ is larger than the two other eigenvalues p f and p m .We notice that we generically have three linearly independent eigenvectors(rather than at most two as for a 2 × .5.2 Cone norms for homogeneous bounded maps For a homogeneous map B : X + → Y , we define k B k + = sup {k Bx k ; x ∈ X + , k x k ≤ } (1.8)and call B bounded if this supremum is a real number. Since B is homoge-neous, k Bx k ≤ k B k + k x k , x ∈ X + . (1.9)Let H ( X + , Y ) denote the set of bounded homogeneous maps B : X + → Y and H ( X + , Y + ) denote the set of bounded homogeneous maps B : X + → Y + and HM( X + , Y + ) the set of those maps in H ( X + , Y + ) that are also order-preserving. H ( X + , Y ) is a real vector space and k · k + is a norm on H ( X + , Y ) calledthe cone-norm. H ( X + , Y + ) and HM( X + , Y + ) are cones in H ( X + , Y ). We write H ( X + ) = H ( X + , X + ) and HM( X + ) =HM( X + , X + ).It follows for B ∈ H ( X + , Y + ) and C ∈ H ( Y + , Z + ) that CB ∈ H ( X + , Z + )and k CB k + ≤ k C k + k B k + . Let B ∈ H ( X + ) and define φ : Z + → R by φ ( n ) = ln k B n k + . Then φ ( m + n ) ≤ φ ( m ) + φ ( n ) for all m, n ∈ Z + , and a well-known result implies thefollowing formula for the cone spectral radius r + ( B ) := inf n ∈ N k B n k /n + = lim n →∞ k B n k /n + , (1.10)which is analogous to (1.6). Mallet-Paret and Nussbaum [26, 27] suggest analternative definition of a spectral radius for homogeneous (not necessarilybounded) maps B : X + → X + . First, define asymptotic least upper boundsfor the geometric growth factors of B -orbits, γ ( x, B ) := γ B ( x ) := lim sup n →∞ k B n ( x ) k /n , x ∈ X + , (1.11)and then r o ( B ) = sup x ∈ X + γ B ( x ) . (1.12)8ere γ B ( x ) := ∞ if the sequence ( k B n ( x ) k /n ) is unbounded and r o ( B ) = ∞ if γ B ( x ) = ∞ for some x ∈ X + or the set { γ B ( x ); x ∈ X + } is unbounded.The number r + ( B ) has been called partial spectral radius by Bonsall [8], X + spectral radius by Schaefer [32, 33], and cone spectral radius by Nussbaum[30]. Mallet-Paret and Nussbaum [26, 27] call r + ( B ) the Bonsall cone spectralradius and r o ( B ) the cone spectral radius. For x ∈ X + , the number γ B ( x )has been called local spectral radius of B at x by F¨orster and Nagy [16].We will follow Nussbaum’s older terminology which shares the spirit withSchaefer’s [32] term X + spectral radius and stick with cone spectral radius for r + ( B ) and call r o ( B ) the orbital spectral radius of B . Later, we will alsointroduce a Collatz-Wielandt spectral radius. We refer to all of them as orderspectral radii .One readily checks that r + ( αB ) = α r + ( B ) , α ∈ R + , r + ( B m ) = ( r + ( B )) m , m ∈ N . (1.13)The same properties hold for r o ( B ) though proving the second property takessome more effort [26, Prop.2.1]. Actually, as we show in Section 3 for bounded B , γ ( x, B m ) = ( γ ( x, B )) m , m ∈ N , x ∈ X + , (1.14)which readily implies r o ( B ) = ( r o ( B )) m , m ∈ N . (1.15)The cone spectral radius and the orbital spectral radius are meaningful if B is just positively homogeneous and bounded, but as in [26, 27] we will bemainly interested in the case that B is also order-preserving and continuous.Though the two concepts coincide for many practical purposes, they areboth useful. Theorem 1.7.
Let X be an ordered normed vector with cone X + and B : X + → X + be continuous, homogeneous and order-preserving.Then r + ( B ) ≥ r o ( B ) ≥ γ B ( x ) , x ∈ X + .Further r o ( B ) = r + ( B ) if one of the following hold:(i) X + is complete and normal.(ii) B is power compact.(iii) X + is normal and a power of B is uniformly order-bounded. iv) X + is complete and B is additive ( B ( x + y ) = B ( x ) + B ( y ) for all x ∈ X + ). The inequality is a straightforward consequence of the respective defini-tions. For the concepts and the proof of (iii) see Section 10 and Theorem12.9. The other three conditions for equality have been verified in [26, Sec.2],(the overall assumption of [26] that X is a Banach space is not used in theproofs.) Statement (i) also follows from Theorem 9.4.For a bounded positive linear operator on an ordered Banach space, thespectral radius and cone spectral radius coincide provided that the cone isgenerating [27, Thm.2.14]. This is not true if the cone is only total [8, Sec.2,8]. For x ∈ ˙ X + , the lower Collatz-Wielandt number of x is defined as [16][ B ] x = sup { λ ≥ Bx ≥ λx } . (1.16)By (8.5), [ B ] x is a lower eigenvalue, B ( x ) ≥ [ B ] x x, x ∈ ˙ X + . (1.17)The lower local Collatz-Wielandt radius of x is defined as η x ( B ) =: sup n ∈ N [ B n ] /nx . (1.18)This implies η x ( B n ) ≤ ( η x ( B )) n , n ∈ N . (1.19)The lower Collatz-Wielandt bound is defined as cw ( B ) = sup x ∈ ˙ X + [ B ] x , (1.20)and the Collatz-Wielandt radius of B is defined as r cw ( B ) = sup x ∈ ˙ X + η x ( B ) . (1.21)From the definitions, cw ( B ) ≤ r cw ( B ) and, by (1.19), r cw ( B n ) ≤ ( r cw ( B )) n for all n ∈ N . 10 homogeneous, bounded, order-preserving map B : X + → X + is alsobounded with respect to the monotone companion norm and ♯B♯ + ≤ k B k + .See Section 5. So we can define the companion cone spectral radius, r ♯ + ( B ),and the companion orbital spectral radius, r ♯o ( B ), in full analogy to (1.10)and (1.12). If X + is normal (and the original and the companion norm areequivalent), the companion radii coincide with the original ones.The monotonicity of the companion norm makes it possible to connect theCollatz-Wielandt radius and the other spectral radii by inequalities (Section9). Theorem 1.8.
Let B : X + → X + be homogeneous, bounded, and order-preserving. Then cw ( B ) ≤ r cw ( B ) ≤ r ♯o ( B ) ≤ (cid:26) r o ( B ) r ♯ + ( B ) (cid:27) ≤ r + ( B ) . If X + is complete and B is continuous (all with respect to the original norm),then r ♯ + ( B ) ≤ r o ( B ) . If B is also compact, the following progressively more general results havebeen proved over the years which finally lead to an extension of the Krein-Rutman theorem from linear to homogeneous maps. Theorem 1.9.
Let B be compact, continuous, homogeneous and order-preservingand let r be any of the numbers cw ( B ) , r cw ( B ) , r + ( B ) . Then, if r > , thereexists some v ∈ ˙ X + with B ( v ) = rv . More specifically, r = cw ( B ) : Krein-Rutman 1948 [23] , r = r cw ( B ) : Krasnosel’skii 1964 [20, Thm.2.5] , r = r + ( B ) : Nussbaum 1981 [29], Lemmens Nussbaum [25] .Actually, if r = r + ( B ) , then cw ( B ) = r cw ( B ) = r + ( B ) . A proof for r = cw ( B ) can also been found in [9]. [20, Thm.2.5] is onlyformulated for the case that B is defined and linear on X , but the proof alsoworks under the assumptions made above. The case r = r cw ( B ) can alsobeen found in [29, Cor.2.1]. The case r = r + ( B ) is basically proved in [29,Thm.2.1], but some finishing touches are contained in the introduction of[25]. If B ( v ) = rv with v ∈ ˙ X + and r = r + ( B ), then r ≤ [ B ] v ≤ cw ( B ) andequality of all three numbers follows.11t is well-known that the compactness of B can be substantially relaxedthough not completely dropped if B is linear [29]. This is also possible (toa lesser degree) if B is just homogeneous using homogeneous measures ofnoncompactness [26, 27]. We only mention two special cases of [26, Thm.3.1]and [27, Thm.4.9], respectively. Theorem 1.10.
Let X + be complete. Let B : X + → X + and B = K + H where K : X + → X + is homogeneous, continuous, order preserving and com-pact and H : X + → X + is homogeneous, continuous, and order preserving.Then there exists some v ∈ ˙ X + with B ( v ) = r + ( B ) v if r + ( B ) > andone of the two following conditions is satisfied in addition:(a) H is Lipschitz continuous on X + , k H ( x ) − H ( y ) k ≤ Λ k x − y k for all x, y ∈ X + , with Λ < r + ( B ) .(b) X + is normal, H is cone-additive ( H ( x + y ) = H ( x ) + H ( y ) for all x, y ∈ X + ) , and r + ( H ) < r + ( B ) . The following observation is worth mentioning [21, Thm.9.3] [30, Thm.2.2].
Proposition 1.11.
Let B : X + → X + be positively linear , i.e., B ( αx + y ) = αB ( x ) + B ( y ) for all x, y ∈ X + and α ≥ . If r > , w ∈ ˙ X + , m ∈ N and B m ( w ) = r m w , then B ( v ) = rv for some v ∈ ˙ X + . Simply set v = m − X j =10 r − j B j ( w ) . (1.22)It seems to be an open problem (with a beer barrel scent [28, 41]) whetherany of the results in Theorem 1.9 holds if B or some higher power of B iscompact rather than B itself. For additional conditions to make such a resulthold, see [1, Thm.7.3], conditions (i) and (iii), and [37, Sec.7].The construction in (1.22) can be modified to yield the following result. Proposition 1.12.
Let X + be a sup-semilattice and B : X + → X + be homo-geneous and order-preserving. If r > , w ∈ ˙ X + , m ∈ N and B m ( w ) ≥ r m w ,then B ( v ) ≥ rv for some v ∈ ˙ X + . This time, choose v as in the proof of [1, Thm.5.1], v = m − sup j =0 r − j B j ( w ) . (1.23)12otice that the number r in Proposition 1.12 satisfies r ≤ [ B ] v ≤ cw ( B ).This observation provides conditions for equality to hold in Theorem 1.8.In order not to burden our representation with technical language or a list ofvarious special cases and to include possible further developments, we makethe following definition. Definition 1.13.
A homogeneous bounded map B : X + → X + has the KR property (the Krein-Rutman property) if there is some v ∈ ˙ X + with B ( v ) = r + ( B ) v whenever r + ( B ) > B has the lower KR property if there is some v ∈ ˙ X + with B ( v ) ≥ r + ( B ) v whenever r + ( B ) > B : X + → X + is called power-compact if B m is compact for some m ∈ N .We emphasize the lower KR property in addition to the KR propertybecause of our interest in population dynamics. If B is the homogeneousapproximation of an nonlinear map F at 0, the condition r + ( B ) < F whether or not r + ( B ) is an eigenvalueassociated with a positive eigenvector. To prove persistence of the dynamicalsystem, it is very helpful to have a positive lower eigenvector B ( v ) ≥ rv with r > Theorem 1.14.
Let B be bounded, homogeneous, order-preserving and B m have the lower KR property for some m ∈ N . Then r cw ( B ) = r ♯o ( B ) = r ♯ + ( B ) = r o ( B ) = r + ( B ) . If X + is a sub-semilattice or m = 1 , then also cw ( B ) = r + ( B ) and B hasthe lower KR property. The significance of cw ( B ) = r + ( B ) is that the lower Collatz-Wielandtnumbers [ B ] x provide lower estimates for r + ( B ) that get arbitrarily sharp bychoosing x ∈ ˙ X + in the right way. See Section 14 for some crude attemptsin that direction for a rank-structured population model with mating.13 roof. The equalities follow from Theorem 1.8 if r + ( B ) = 0. So we can as-sume that r + ( B ) >
0. If B has the lower KR property, it follows immediatelyfrom the definitions that r + ( B ) ≤ cw ( B ) and equality follows from Theorem1.8. If B m has the lower KR property,( r + ( B ) m = r + ( B m ) ≤ cw ( B m ) ≤ r cw ( B m ) ≤ ( r cw ( B )) m , and the equalities follow again from Theorem 1.8. If r = r + ( B ) >
0, thereexists some w ∈ ˙ X + with B m ( w ) ≥ r m w . By Proposition 1.12, B ( v ) ≥ rv for some v ∈ ˙ X + and r ≥ cw ( B ).The following monotone dependence of the various spectral radii on themap can be shown without assuming normality of the cone. Theorem 1.15.
Let
A, B : X + → X + be bounded and homogeneous. Assumethat A ( x ) ≤ B ( x ) for all x ∈ X + .If A is order-preserving and has the lower KR property, then r + ( A ) = cw ( A ) ≤ cw ( B ) . For x ∈ ˙ X + , the upper Collatz-Wielandt numbers of x [16] is defined as k B k x = inf { λ ≥ B ( x ) ≤ λx } , with the convention that inf ∅ = ∞ , and a local upper Collatz-Wielandt radius η u ( B ) at u ∈ ˙ X + , η u ( B ) = inf n ∈ N k B n k /nu . Collatz-Wielandt numbers [16], without this name, became more widelyknown when Wielandt used them for a new proof of the Perron-Frobeniustheorem [43]. We will use them closer to Collatz’ original purpose, namelyto prove inclusion theorems (Einschließungss¨atze) for r + ( B ) which generalizethose in [10, 11].If the cone X + is solid, with nonempty interior ˘ X + , one can define an upper Collatz-Wielandt bound (cf. [1, Sec.7]) by CW ( B ) = inf x ∈ ˘ X + k B k x . These new numbers, bound, and radius relate to the former bounds and radiiin the following way and show that upper Collatz-Wielandt numbers taken14t interior points provide upper estimates of the cone spectral radius. Forthis expos´e, we illustrate our results for the special case of a solid cone.
Theorem 1.16.
Let X + be solid with nonempty interior ˘ X + and B : X + → X + be homogeneous, bounded and order-preserving. Then, for all u ∈ ˘ X + , γ B ( u ) = lim n →∞ k B n ( u ) k /n , η u ( B ) = lim n →∞ k B n k /nu and cw ( B ) ≤ r cw ≤ r ♯ + ( B ) ≤ η u ( B ) ≤ ( γ u ( B ) CW ( B ) . We obtain some equalities in the following cases. • If X + is normal, r + ( B ) = γ u ( B ) = η u ( B ) ≤ CW ( B ) for all u ∈ ˘ X + . • If B m has the lower KR property for some m ∈ N , r cw ( B ) = r + ( B ) = γ u ( B ) = η u ( B ) ≤ CW ( B ) for all u ∈ ˘ X + . • If B has the lower KR property, cw ( B ) = r cw ( B ) = r + ( B ) = γ u ( B ) = η u ( B ) ≤ CW ( B ) for all u ∈ ˘ X + . Notice that the equality r + ( B ) = γ u ( B ) = lim n →∞ k B n ( u ) k /n meansthat the cone spectral radius can be determined by following the growthfactors of an arbitrarily chosen u ∈ ˘ X + .The estimates from above by upper Collatz-Wielandt numbers k B k u with u ∈ ˘ X + can be arbitrarily sharp by appropriate choice of u if the map iscompact. The next result should be compared to [1, Thm.7.3]. Notice that CW ( B ) here is cw ( B ) in [1, Thm.7.3]. Use of the companion half-normmakes it possible to drop the normality of the cone. Theorem 1.17.
Let X + be solid and complete. Let B : X + → X + and B = K + A where K : X + → X + is homogeneous, continuous, order preserving andcompact and A : X → X is linear, positive, and bounded and r ( A ) < r + ( B ) .Then cw ( B ) = r cw ( B ) = r + ( B ) = γ B ( u ) = η u ( B ) = CW ( B ) for all u ∈ ˘ X + .If r = CW ( B ) > , then there exists some v ∈ ˙ X + such that B ( v ) = rv . In Section 12, we prove more general versions of these theorems replacingsolidity of the cone by uniform order-boundedness of the map.15sing the monotone companion half-norm, we show for certain classes ofhomogeneous maps B for which it is not clear whether they have the lowerKR property that there is some v ∈ ˙ X + such that Bv ≥ rv for r = η u ( B )provided that r > We start with some more cone properties.The cone X + is called regular if any decreasing sequence in X + converges. X + is called fully regular if any increasing bounded sequence in X + con-verges.The norm of X is called additive on X + if k x + z k = k x k + k z k for all x, z ∈ X + . If the norm is additive, then the cone X + is normal. The following result is well-known [20, Sec.1.2].
Theorem 2.1.
The following three properties are equivalent:(i) X + is normal: There exists some δ > such that k x + z k ≥ δ whenever x ∈ X + , z ∈ X + and k x k = 1 = k z k .(ii) The norm is semi-monotonic: There exists some M ≥ such that k x k ≤ M k x + z k for all x, z ∈ X + .(iii) There exists some ˜ M ≥ such that k x k ≤ ˜ M k y k whenever x ∈ X , y ∈ X + , and − y ≤ x ≤ y . Remark 2.2. If X + were just a wedge, property (iii) would be rewritten asThere exists some ˜ M ≥ k x k ≤ ˜ M k y k whenever x ∈ X , y ∈ X + , and y + x ∈ X + , y − x ∈ X + .Notice that this property implies that X + is cone: If x ∈ X + and − x ∈ X + ,then 0 + x ∈ X + and 0 − x ∈ X + and (iii) implies k x k ≤ ˜ M k k = 0.16 efinition 2.3. An element u ∈ X + is called a normal point if the set {k x k ; x ∈ X + , x ≤ u } is a bounded subset of R .The following is proved in [21, Thm.4.1]. Theorem 2.4.
Let X be an ordered normed vector space with cone X + . If X + is a normal cone, then all elements of X + are normal points and all sets {k x k ∈ X ; u ≤ x ≤ v } with u, v ∈ X are bounded subsets of R .If all elements of X + are normal points and X + is complete or fullyregular, then X + is a normal cone.Proof. The first statement is obvious. Assume that X + is not a normal conebut complete or fully regular. Then there exist sequences ( x n ) and ( y n ) in X + such that x n ≤ y n and k x n k ≥ n k y n k for all n ∈ N . Set v n = y n n k y n k and u n = x n n k y n k . Then u n ≤ v n and k v n k ≤ − n and k u n k ≥ n . Since X + iscomplete or fully regular, the series v = P ∞ n =1 v n converges in X + and u n ≤ v for all n ∈ N . So v is not a normal point.Completeness or full regularity of X + are necessary for the normality of X + as shown by the forthcoming Example 2.11.Connections between completeness, normality, regularity and full regu-larity of cones are spelt out in the next result. Theorem 2.5.
Let X be an ordered normed vector space with cone X + .(a) If X + is complete and regular, then X + is normal.(b) If X + is complete and fully regular, then X + is normal.(c) If X + is normal and fully regular, then X + is regular.(d) If X + is complete and fully regular, then X + is regular.Proof. Notice that the proofs in [20, 1.5.2] and [20, 1.5.3] only need com-pleteness of X + and not of X . Theorem 2.6.
Let X be an ordered normed vector space with cone X + . If X + is complete with additive norm, then X + is fully regular. roof. Let ( x n ) be an increasing sequence in X + such that there is some c > k x n k ≤ c for all n ∈ N . Define y n = x n +1 − x n . Then y n ∈ X + and, for m ≥ j , P mk = j y n = x m +1 − x j . Since the norm is additiveon X + , m X k =1 k y n k = (cid:13)(cid:13)(cid:13) m X k =1 y n (cid:13)(cid:13)(cid:13) = k x m +1 − x k ≤ c, m ∈ N . This implies that ( x n ) is a Cauchy sequence in the complete cone X + andthus converges.The standard cones of nonnegative functions of the Banach spaces L p (Ω),1 ≤ p < ∞ , are regular and completely regular, while the cones of BC (Ω),the Banach space of bounded continuous functions, and of L ∞ (Ω) are neitherregular nor completely regular. All these cones are normal. Forthcomingexamples will present a cone that is regular, but neither completely regular,normal, nor complete. Recall the Banach sequence spaces ℓ ∞ , c, c of bounded sequences, convergingsequences and sequences converging to 0 with the supremum norm and thespace ℓ of summable sequences with the sum-norm.The subsequent example for an ordered Banach space whose cone is notnormal follows a suggestion by Wolfgang Arendt.A sequence ( x n ) in R N is called of bounded variation if the following seriesconverges | x | + ∞ X n =1 | x n +1 − x n | =: k ( x n ) k bv . (2.1)The sequences of bounded variation form a vector space, bv , over R with k · k bv being a norm called the variation-norm [14, IV.2.8]. Notice ℓ ⊆ bv ⊆ c, (cid:26) k x k ∞ ≤ k x k bv , x ∈ bv, k x k bv ≤ k x k , x ∈ ℓ . (2.2) Lemma 2.7. bv with the variation-norm is a Banach space. ℓ ∞ is complete under thesup-norm.Notice that bv contains all constant sequences and k ( x n ) k bv = k x k = k ( x n ) k ∞ if ( x n ) is a constant sequence. Actually, all monotone boundednonnegative sequences are of bounded variation. Lemma 2.8. If ( x n ) is a nonnegative bounded increasing sequence, then ( x n ) is of bounded variation and k ( x n ) k bv = k ( x n ) k ∞ = lim n →∞ x n .If ( x n ) is a nonnegative decreasing sequence, then ( x n ) is of bounded vari-ation and k ( x n ) k bv = 2 x − lim n →∞ x n . There are several cones we can consider in bv . The one we are goingto consider here is the cone of nonnegative sequences of bounded variation, bv + . Others are the cone of nonnegative increasing sequences and the coneof nonnegative decreasing sequences. Proposition 2.9. bv + is generating, solid, but not normal (and not regularand not fully regular). X is a lattice and the lattice operations are continuous:If x = ( x j ) ∈ bv , then | x | = ( | x j | ) ∈ bv and (cid:13)(cid:13) | x | (cid:13)(cid:13) bv ≤ k x k bv with strictinequality being possible. Notice that every monotone nonnegative sequence that is bounded awayfrom zero is the interior of bv + as can be seen from (2.2). The space of Lips-chitz continuous functions with Lipschitz norm is an example of an orderednormed vector space and lattice where the cone of nonnegative functions isnot normal and the lattice operations are not continuous [26, p.535].That bv + is not normal follows from the following result that characterizesthe normal points. Since bv + is complete, bv + is then also not regular andnot fully regular (Theorem 2.5). Theorem 2.10. x = ( x n ) ∈ bv + is a normal point if and only if x ∈ ℓ . If x ∈ ℓ , then k x k − (1 / x ≤ sup {k v k bv ; v ∈ bv + , x − v ∈ bv + } ≤ k x k . Proof.
One half of the statement is easy to see. Recall (2.2). For the otherpart, assume that x = ( x n ) ∈ bv + is a normal element. Let y be the se-quence ( x , , x , , x , . . . ). Then 0 ≤ y ≤ x and y ∈ bv and k y k bv = | x | + 2 P ∞ j =1 | x j +1 | . Further let z be the sequence (0 , x , , x , , . . . ). Then0 ≤ z ≤ x and z ∈ bv and k z k bv = 2 P ∞ j =1 k x j k . Hence x ∈ ℓ and2 k x k − x = k y k bv + k z k bv ≤ {k v k bv ; 0 ≤ v ≤ x } . xample 2.11 (Wolfgang Arendt) . The cone ℓ in ℓ with the variationnorm is not normal though all points in ℓ are normal elements. It is regular,but not fully regular. Proof.
Let x m be the sequence where x mj = 0 for j > m , x mj = 0 for allodd indices and x mj = 1 otherwise. Let u m be the sequence where u mj = 0for j > m and u mj = 1 otherwise. Then k u m k bv = 2 and k x m k bv = 2 m . Forall m ∈ N , x m ≤ u m but k x m k bv = m k u m k bv . So ℓ is not normal under thevariation norm. Theorem 2.4 implies that ℓ is not fully regular under thevariation norm. Let ( x n ) be a decreasing sequence in ℓ . Since ℓ is regularunder the sum-norm, there exists some x ∈ ℓ with k x n − x k →
0. By (2.2), k x n − x k bv → The following convergence principle, which has been distilled from the proofof [1, Thm.5.2], will be applied several times.
Proposition 2.12.
Let X be an ordered normed vector space. Let S ⊆ X N bea set of sequences with terms in X with the property that with ( x n ) ∈ S also allsubsequences ( x n j ) ∈ S . Assume that every increasing (decreasing) sequence ( x n ) ∈ S has a convergent subsequence. The every increasing (decreasing)sequence ( x n ) ∈ S converges.Proof. Let ( x n ) ∈ S be increasing (the decreasing case is similar). Thenthere exist a subsequence ( x n j ) and some x ∈ X such that x n j → x . Supposethat ( x n ) does not converge to x . Then there exist some ǫ > k i ) ∈ N such that k x k i − x k ≥ ǫ for all i ∈ N . Byassumption, ( x k i ) ∈ S ; further it is an increasing sequence. So, after choosinga subsequence, x k i → y for some y = x . Fix n j . If i is sufficiently large, k i ≥ n j and x n j ≤ x k i . Taking the limit i → ∞ yields x n j ≤ y because thecone is closed. Now we let j → ∞ and obtain x ≤ y . By symmetry, y ≤ x ,a contradiction. Let B : X + → X + be homogeneous and x ∈ X + . We prove (1.14), γ ( x, B m ) ≤ ( γ ( x, B )) m , m ∈ N , (3.1)20ith equality holding if B is bounded.From the properties of the limit superior, γ ( x, B m ) = lim sup k →∞ k ( B m ) k ( x ) k /k = lim j →∞ sup k ≥ j k B mk ( x ) k m/ ( mk ) ≤ lim j →∞ sup n ≥ mj ( k ( B n ( x ) k /n ) m = (cid:16) lim sup n →∞ k ( B n ( x ) k /n (cid:17) m = ( γ ( x, B )) m . To prove the opposite inequality if B is bounded, suppose that γ ( x, B m ) < ( γ ( x, B )) m . Then there exists some s ∈ (0 ,
1) such that γ ( x, B m ) < s m ( γ ( x, B )) m . By definition of γ ( x, B m ), there exists some N ∈ N such that k B mn ( x ) k / ( mn ) < sγ ( x, B ) , n ≥ N. (3.2)Choose a sequence ( k j ) such that k j → ∞ and k B k j ( x ) k /k j → γ ( x, B ).Then there exist sequences n j and p j such that n j → ∞ and 0 ≤ p j < m and k j = mn j + p j . By (1.9), k B k j ( x ) k /k j ≤ k B p j k /k j + k B mn j ( x ) k /k j . By the properties of the limit superior, γ ( x, B ) ≤ lim sup j →∞ ( k B p j k /k j + ) lim sup j →∞ ( k B mn j ( x ) k / ( mn j ) ) ( k j − p j ) /k j . By (3.2) and p j /k j → γ ( x, B ) ≤ lim sup j →∞ ( sγ ( x, B )) ( k j − p j ) /k j ≤ lim sup j →∞ ( sγ ( x, B )) − ( p j /k j ) = sγ ( x, B ) , a contradiction. Every ordered normed vector space carries an order-preserving half-normwhich we call the (monotone) companion half-norm (called the canonicalhalf-norm in [2]). 21 roposition 4.1 (cf. [2], [21, (4.2)], [9, L.4.1]) . We define the (monotone)companion half-norm ψ : X → R + by ψ ( x ) = inf {k x + z k ; z ∈ X + } = d ( x, − X + ) (4.1)= inf {k y k ; x ≤ y ∈ X } , x ∈ X. (4.2) Then the following hold:(a) ψ is positively homogenous and order-preserving on X .(b) ψ is subadditive on X ( ψ ( x + y ) ≤ ψ ( x ) + ψ ( y ) , x, y ∈ X ), | ψ ( x ) − ψ ( y ) | ≤ k x − y k , x, y ∈ X. (c) For x ∈ X , ψ ( x ) = 0 if and only if x ∈ − X + . In particular ψ is strictlypositive: ψ ( x ) > for all x ∈ ˙ X + .(d) X + is normal if and only if there exists some δ > such that δ k x k ≤ ψ ( x ) for all x ∈ X + .(e) If the original norm k · k is order-preserving on X + , then k x k = ψ ( x ) for all x ∈ X + . Here d ( x, − X + ) denotes the distance of x from − X + . Proof.
The functional ψ inherits positive homogeneity from the norm. That ψ is order-preserving is immediate from (4.2). For all x ∈ X , x ≤ x and so k x k ≥ ψ ( x ).Most of the other properties follow from (4.1) and the assumption that X + is a cone.Since ψ is subadditive, | ψ ( x ) − ψ ( y ) | ≤ ψ ( x − y ) ≤ k x − y k . If − x ∈ X + , then x ≤ ψ ( x ) ≤ k k = 0.Assume that x ∈ X and ψ ( x ) = 0. By definition, there exists a sequence( y n ) in X with k y n k → y n ≥ x for all n ∈ N . Then y n − x ∈ X + . Since X + is closed, − x = lim n →∞ ( y n − x ) ∈ X + .The strict positivity of ψ follows from X + ∩ ( − X + ) = { } .22d) Assume that X + is normal. By Theorem 2.1 (ii), there exists some c > k x k ≤ c k y k whenever x, y ∈ X + and x ≤ y . Hence k x k ≤ cψ ( x ) for all x ∈ X + . Set δ = 1 /c .The converse follows from ψ being order-preserving and ψ ( x ) ≤ k x k forall x ∈ X + .One readily checks that the monotone companion half-norm inherits equiv-alence of norms. Remark 4.2. If k · k ∼ is a norm on X that is equivalent to k x k , i.e., thereexists some c > /c ) k x k ≤ k x k ∼ ≤ c k x k for all x ∈ X , thenthen respective monotone companion half-norms satisfy (1 /c ) ψ ( x ) ≤ ˜ ψ ( x ) ≤ cψ ( x ) for all x ∈ X .The functional ψ induces a monotone norm on X which is equivalent tothe original norm if and only if X + is a normal cone. Theorem 4.3 (cf. [21, Thm.4.4]) . Define ♯x♯ = max { ψ ( x ) , ψ ( − x ) } , x ∈ X, with ψ from Proposition 4.1. Then ♯ · ♯ is a norm on X with the followingproperties. • ♯x♯ ≤ k x k for all x ∈ X , ♯x♯ = ψ ( x ) if x ∈ X + , and ♯x♯ = ψ ( − x ) if − x ∈ X + . • ♯ · ♯ is order-preserving on X + : ♯x♯ ≤ ♯y♯ for all x, y ∈ X + with x ≤ y .Moreover, for all x, y, z ∈ X with x ≤ y ≤ z , ♯y♯ ≤ max { ♯x♯, ♯z♯ } . (4.3) • X + is a normal cone if and only if the norm ♯ · ♯ is equivalent to theoriginal norm. • If the original norm k · k is order-preserving on X + , then k x k = ♯x♯ forall x ∈ X + ∪ ( − X + ) .Proof. It is easy to see from the properties of ψ that ♯αx♯ = | α | ♯x♯ forall α ∈ R , x ∈ X , and that ♯ · ♯ is subadditive. Now ψ ( x ) ≤ k x k and ψ ( − x ) ≤ k − x k = k x k and so ♯x♯ ≤ k x k .23o prove (4.3), let x, y, z ∈ X and x ≤ y ≤ z . Then y ≤ z and − y ≤ − x .Since ψ is order-preserving on X , ψ ( y ) ≤ ψ ( z ) ≤ ♯z♯, ψ ( − y ) ≤ ψ ( − x ) ≤ ♯x♯. (4.3) now follows from the definition of ψ as do most of the remaining asser-tions.That k · k and ♯ · ♯ are equivalent norms if the cone is normal is shown in[21, Thm.4.4]. The converse follows from Theorem 2.1 (ii). Definition 4.4.
The norm ♯ · ♯ is called the (monotone) companion norm onthe ordered normed vector space X . Corollary 4.5.
Let X + be a normal cone. Then there exists some c ≥ such that k y k ≤ c max {k x k , k z k} for all x, y, z ∈ X with x ≤ y ≤ z .Proof. Let ♯ · ♯ be the monotone companion norm from Theorem 4.3, whichis equivalent to the original norm because X + is normal. Choose c ≥ ♯x♯ ≤ k x k ≤ c♯x♯ for all x ∈ X . Let x ≤ y ≤ z . By Theorem 4.3, k y k ≤ c♯y♯ ≤ c max { ♯x♯, ♯z♯ } ≤ c max {k x k , k z k} . Corollary 4.6 (Squeezing theorem [21, Thm.4.3]) . Let X + be a normal cone.Let y ∈ X and ( x n ) , ( y n ) , ( z n ) be sequences in X with x n ≤ y n ≤ z n for all n ∈ N and x n → y and z n → y . Then y n → y .Proof. Notice that x n − y ≤ y n − y ≤ z n − y . By Corollary 4.5, with some c ≥ n , k y n − y k ≤ c max {k z n − y k , k x n − y k} → . Example 4.7. (a) Let X = R with the absolute value. Then the monotonecompanion norm is also the absolute value and the companion half-norm isthe positive part, ψ ( x ) = max { x, } for all x ∈ R .(b) Let X = R with the maximum norm and X + = R . Then themonotone companion norm is also the maximum norm.(c) Let X = R with k · k being either the Euclidean norm or the sumnorm and X + = R . Then ♯x♯ = k x k if x ∈ X + ∪ ( − X + ) while ♯x♯ is themaximum norm of x otherwise. 24 xample 4.8. Let X be a normed vector lattice [34, II.5]. Since x ≤ x + , ψ ( x ) ≤ k x + k . Let x ≤ y . Then x + ≤ y + , and k x + k ≤ k y + k ≤ k y k . Hence k x + k ≤ ψ ( x ). In combination, ψ ( x ) = k x + k , x ∈ X. Further ψ ( − x ) = k ( − x ) + k = k x − k . So ♯x♯ = max {k x + k , k x − k} , x ∈ X, and ♯x♯ = k x k for all x ∈ X + ∪ ( − X + ). If X is an abstract M space [34, II.7], ♯x♯ = k x + ∨ x − k = (cid:13)(cid:13) | x | (cid:13)(cid:13) = k x k . We turn to ordered Banach space the cone of which may be not normal.
Example 4.9.
Let X ⊆ ˜ X where X, ˜ X are ordered normed vector spaceswith norms k · k and k · k ∼ and k x k ≥ k x k ∼ for all x ∈ X . Let ψ, ˜ ψ bethe respective monotone companion half-norms. Then ψ ( x ) ≥ ˜ ψ ( x ) for all x ∈ X .Assume that ( ˜ X, k · k ∼ ) is an abstract M space. By the previous example, ♯x♯ ≥ k x k ∼ , x ∈ X. Now assume that there exists some u ∈ X with k u k ≤ | x | ≤ k x k ∼ u forall x ∈ X . Then ψ ( ± x ) ≤ k x k ∼ and ♯x♯ ≤ k x k ∼ . In combination, ♯x♯ = k x k ∼ , x ∈ X. Example 4.10 (Space of sequences of bounded variation) . To determine themonotone companion (half-) norm in a concrete case where the cone is notnormal we revisit the space bv of sequences of bounded variation with thevariation-norm.Recall that bv ⊆ ℓ ∞ and k x k bv ≥ k x k ∞ for all x ∈ bv . ℓ ∞ is an abstractM space under the sup-norm. We apply the previous example with u beingthe sequence all the terms of which are 1. Then k u k bv = 1 and | x | ≤ k x k ∞ u .We obtain that ♯x♯ = k x k ∞ for all x ∈ bv .Notice that bv is not complete under the monotone companion norm. Bythe open mapping theorem, an ordered Banach space is complete under themonotone companion norm if and only if the cone is normal.25 Order-preserving maps and the companionnorm
In the following, let X and Y be ordered normed vector spaces. We use thesame symbols k · k , ψ and ♯ · ♯ for the norms and the monotone companion(half-) norms on X and Y . Theorem 5.1.
Let B : X + → Y + be bounded, homogeneous, and order-preserving. Then ψ ( B ( x )) ≤ k B k + ψ ( x ) for all x ∈ X + . In particular, B isbounded with respect to the monotone companion norms and ♯B♯ + ≤ k B k + .Proof. Since B is order-preserving, for x ∈ X + , { y ∈ Y ; B ( x ) ≤ y } ⊇ { B ( z ); x ≤ z ∈ X + } . By definition of ψ , ψ ( B ( x )) ≤ inf {k B ( z ) k ; z ∈ X + , x ≤ z } ≤ inf {k B k + k z k ; x ≤ z ∈ X } = k B k + ψ ( x ) . Theorem 5.2.
Let B : X → Y be bounded, linear and positive. Then B isbounded with respect to the monotone companion norm, ψ ( Bx ) ≤ k B k ψ ( x ) for all x ∈ X , and ♯B♯ ≤ k B k .Proof. By a similar proof as for Theorem 5.1, since B is order-preserving, ψ ( Bx ) ≤ k B k ψ ( x ) for all x ∈ X . Then ψ ( − Bx ) = ψ ( B ( − x )) ≤ k B k ψ ( − x ) . The assertion now follows from ♯x♯ = max { ψ ( x ) , ψ ( − x ) } .Various concepts of continuity are preserved if one switches from theoriginal norm to the monotone companion norm. We only look at the mostusual concept here. Proposition 5.3.
Let B : X → Y be order-preserving and continuous at x with respect to the original norms. Then B is continuous at x with respectto the monotone companion norms.Proof. Let x ∈ X and B be continuous at x . Let y ∈ X as well.For any z ≥ y − x , z ∈ X , we have B ( y ) − B ( x ) = B ( x + y − x ) − B ( x ) ≤ B ( x + z ) − B ( x ) .
26y definition of the monotone companion half-norm, ψ ( B ( y ) − B ( x )) ≤ k B ( x + z ) − B ( x ) k , y − x ≤ z ∈ X. Let ǫ >
0. Then there exists some δ + > k B ( x + z ) − B ( x ) k < ǫ for all z ∈ X , k z k < δ + . Now let y ∈ X and ψ ( y − x ) < δ + . Then there existssome z ∈ X such that y − x ≤ z and k z k < δ + . Then k B ( x + z ) − B ( x ) k < ǫ and ψ ( B ( y ) − B ( x )) < ǫ .Also, for any z ∈ X with x − y ≤ z , we have B ( x ) − B ( y ) = B ( x ) − B ( x − ( x − y )) ≤ B ( x ) − B ( x − z ) . By definition of the monotone companion half-norm ψ ( B ( x ) − B ( y )) ≤ k B ( x ) − B ( x − z ) k , z ∈ X, x − y ≤ z. Let ǫ >
0. Then there exists some δ − > k B ( x ) − B ( x − z ) k < ǫ for all z ∈ X , k z k < δ − . Now let y ∈ Y and ψ ( x − y ) < δ − . Then thereexists some z ∈ X such that x − y ≤ z and k z k = k − z k < δ − . Thus ψ ( B ( x ) − B ( y )) < ǫ .Set δ = min { δ + , δ − } and ♯y − x♯ < δ . Then ψ ( ± ( y − x )) < δ ± and ψ ( ± ( B ( y ) − B ( x )) < ǫ . This implies ♯B ( y ) − B ( x ) ♯ < ǫ . Proposition 5.4.
Let φ : X + → R + be homogeneous.(a) If φ is bounded with respect to the monotone companion norm, then itis bounded with respect to the original norm and k φ k + ≤ ♯φ♯ + .(b) If φ is order-preserving and bounded with respect to the original norm,then it is bounded with respect to the monotone companion norm and k φ k + = ♯φ♯ + . Further φ ( x ) ≤ ψ ( x ) k φ k + for all x ∈ X + .Proof. (a) Let x ∈ X + . Since φ is bounded with respect to the monotonecompanion norm, φ ( x ) ≤ ♯φ♯ + ♯ x♯ ≤ ♯φ♯ + k x k . (b) This follows from part (a) and Theorem 5.1. Proposition 5.5.
Let φ : X → R be linear.(a) If φ is bounded with respect to the monotone companion norm, then itis bounded with respect to the original norm and k φ k ≤ ♯φ♯ . b) If φ is positive and bounded with respect to the original norm, then it isbounded with respect to the monotone companion norm and k φ k = ♯φ♯ and φ ( x ) ≤ k φ k ψ ( x ) for all x ∈ X .Proof. (a) The proof is similar to the one for Proposition 5.4.(b) follows from part (a) and Theorem 5.2 and the fact that the monotonehalf-norm on R is given by the positive part.Let X ∗ + be the dual wedge of positive linear functionals on X that arebounded with respect to the original norm k · k . By Proposition 5.5, X ∗ + isalso the dual wedge of linear functions that are bounded with respect to themonotone companion norm ♯ · ♯ and the norms induced on X ∗ are the sameon X ∗ + . Proposition 5.6.
Let x ∈ X . Then there exists x ∗ ∈ X ∗ + such that x ∗ x = ψ ( x ) and k x ∗ k = ♯x ∗ ♯ ≤ . If x ∈ X + , k x ∗ k = 1 can be achieved.Actually ψ ( x ) = max { x ∗ x ; x ∗ ∈ X ∗ + , k x ∗ k ≤ } , x ∈ X,ψ ( x ) = max { x ∗ x ; x ∗ ∈ X ∗ + , k x ∗ k = 1 } , x ∈ X + . (5.1) Further [21, (4.3)], for all x ∈ X , ♯x♯ = max {| x ∗ x | ; x ∈ X ∗ + , k x ∗ k = 1 } = max {| x ∗ x | ; x ∈ X ∗ + , k x ∗ k ≤ } . (5.2) Proof.
Recall that X ∗ + and its norm do not depend on whether we consider X with the original norm or its monotone companion norm (Proposition 5.5).Let u ∈ X . By [44, IV.6], we find x ∗ ∈ X ∗ with x ∗ u = ψ ( u ) and − ψ ( − x ) ≤ x ∗ x ≤ ψ ( x ) , x ∈ X, and so | x ∗ x | ≤ ♯x♯ ≤ k x k , x ∈ X. Since ψ ( − x ) = 0 for all x ∈ X + , x ∗ ∈ X ∗ + . By Proposition 5.5, ♯x ∗ ♯ = k x ∗ k ≤
1. Now let x ∈ X + . Then x ∗ x = ψ ( u ) = ♯u♯ and so ♯x ∗ ♯ = 1.Notice that we have proved ≤ in (5.1).Suppose that ψ ( u ) > ψ ( − u ). Then there exists x ∗ ∈ X ∗ + with ♯x ∗ ♯ ≤ x ∗ u = ψ ( u ) = ♯u♯ . Hence ♯x ∗ ♯ = 1.28f ψ ( u ) < ψ ( − u ), there exists x ∗ ∈ X ∗ + with ♯x ∗ ♯ ≤ x ∗ ( − u ) = ♯u♯ = | x ∗ u | . Again ♯x ∗ ♯ = 1.If ψ ( u ) = ψ ( − u ) >
0, then u, − u X + and we can make the sameconclusion.It remains the case ψ ( u ) = ψ ( − u ) = 0. But then u = 0, and all equalitieshold trivially.So the ≤ inequalities hold in (5.2).The ≥ inequalities follow from Proposition 5.5 (b). Corollary 5.7. X + is closed with respect to the monotone companion norm.Proof. Let ( x n ) be a sequence in X + , x ∈ X and ♯x n − x♯ →
0. Suppose that x / ∈ X + . By a theorem of Mazur [44, IV.6.Thm.3’], there exists a boundedlinear φ : X → R such that φ ( x ) < − φ ( y ) ≥ − y ∈ X + .Let z ∈ X + , n ∈ N . Then nz ∈ X + and φ ( nz ) ≥ −
1. So φ ( z ) ≥ − /n .We let n → ∞ and obtain φ ( z ) ≥
0. By Proposition 5.5, φ is continuouswith respect to the monotone companion norm. So 0 ≤ φ ( x n ) → φ ( x ) and φ ( x ) ≥
0, a contradiction.
We consider integral inequalities of the following kind on an interval [0 , b ],0 < b < ∞ , u ( t ) ≥ Z t K ( t, s ) u ( s ) ds, t ∈ [0 , b ] . (6.1)Here u : [0 , b ) → X is a continuous function, K ( t, s ), 0 ≤ s ≤ t ≤ b , arebounded linear positive operators such that, for each x ∈ X , K ( t, s ) x is acontinuous function of ( t, s ), 0 ≤ s ≤ t ≤ b . Theorem 6.1.
Let X be an ordered Banach space. Let u : [0 , b ] → X be acontinuous solution of the inequality (6.1). Then u ( t ) ∈ X + for all t ∈ [0 , b ] .Proof. We define v : [0 , b ) → X by v ( t ) = − u ( t ). Then v ( t ) ≤ Z t K ( t, s ) v ( s ) ds, t ∈ [0 , b ] . (6.2)29ince the monotone companion half-norm ψ is order-preserving, ψ ( v ( t )) ≤ ψ (cid:16) Z t K ( t, s ) v ( s ) ds (cid:17) , t ∈ [0 , b ] . Since ψ is convex and homogeneous, ψ ( v ( t )) ≤ Z t ψ ( K ( t, s ) v ( s )) ds, t ∈ [0 , b ] . (6.3)By Theorem 5.2, ψ ( K ( t, s ) u ( s )) ≤ k K ( t, s ) k ψ ( s ) , ≤ s ≤ t ≤ b. By the uniform boundedness theorem, there exists some c ≥ k K ( t, s ) k ≤ c whenever 0 ≤ s ≤ t ≤ b . So ψ ( K ( t, s ) v ( s )) ≤ cψ ( v ( s )) , ≤ s ≤ t ≤ b. (6.4)We substitute the last inequality into (6.3), ψ ( v ( t )) ≤ c Z t ψ ( v ( s )) ds, t ∈ [0 , b ] . Let λ >
0. Then e − λt ψ ( v ( t )) ≤ c Z t e − λ ( t − s ) e − λs ψ ( v ( s )) ds, t ∈ [0 , b ] . Define α ( λ ) = sup ≤ t ≤ b e − λt ψ ( v ( t )). Then e − λt ψ ( v ( t )) ≤ c Z t e − λ ( t − s ) α ( λ ) ds ≤ c α ( λ ) λ , ≤ t ≤ b, and α ( λ ) ≤ c α ( λ ) λ . Choosing λ > α ( λ ) ≤ α ( λ ) = 0.This implies ψ ( v ( t )) = 0 for all t ∈ [0 , b ]. By Proposition 4.1, v ( t ) ∈ − X + and so u ( t ) = − v ( t ) ∈ X + for all t ∈ [0 , b ].30 Towards a compact attractor of boundedsets
The monotone companion half-norm can also be useful if the cone X + isnormal. The following result generalizes [35, Prop.7.2]. Proposition 7.1.
Let X + be a normal cone and F : X + → X + . Assumethat there exist some linear, bounded, positive map A : X → X , r ( A ) < ,and real numbers R, c > such thatfor any x ∈ X + with k x k ≤ R there exists some y ∈ X with k y k ≤ c and F ( x ) ≤ A ( x ) + y. (7.1) Then, after introducing an equivalent norm, there exists some ˜ R > suchthat F ( X + ∩ ¯ B s ) ⊂ ¯ B s , s ≥ ˜ R, for all closed balls ¯ B s with radius s ≥ ˜ R and the origin as center. Further,there exists some ˆ R > such that lim sup n →∞ k F n ( x ) k ≤ ˆ R, x ∈ X + . Proof.
Since X + is normal, the original norm is equivalent to its monotonecompanion norm ♯ · ♯ . Let ψ be the monotone companion half-norm. Then(7.1) can be rewritten as ψ ( F ( x ) − A ( x )) ≤ c, x ∈ X + , k x k ≥ R. (7.2)In a next step we can switch to an equivalent monotone norm, denoted by k · k ∼ , such that the associated operator norm of A satisfies k A k ∼ < k x k ∼ = m X k =0 r − m ♯A k ( x ) ♯, x ∈ X, where r + ( A ) < r < m ∈ N is chosen large enough such that k A m +1 k 1. Let ˜ ψ be the monotone companionhalf-norm of k · k ∼ . Since k · k ∼ is order-preserving on X + , k x k ∼ = ˜ ψ ( x ) for x ∈ X + .The inequality (7.2) remains valid for k · k ∼ and ˜ ψ after possibly changing R and c . 31o, without loss of generality, we can assume that k · k is order-preservingon X + , k A k < 1, and the monotone companion half-norm of k · k satisfies(7.2).Since ψ is monotone and subadditive on X , for x ∈ X + with k x k ≥ R , k F ( x ) k = ψ ( F ( x )) ≤ ψ ( F ( x ) − Ax ) + ψ ( Ax ) ≤ c + k Ax k ≤ c + k A k k x k . (7.3)Choose some ξ ∈ ( k A k , R > R . Then, if x ∈ X + and k x k ≥ ˘ R , k F ( x ) k ≤ ξ k x k + c + ( k A k − ξ ) k x k ≤ ξ k x k + c + ( k A k − ξ ) ˘ R. Choosing ˘ R > R large enough, we can achieve that c + ( k A k − ξ ) ˘ R ≤ k F ( x ) k ≤ ξ k x k , k x k ≥ ˘ R. (7.4)Set ˜ R = max { ˘ R, c + k A k ˘ R } . By (7.3) and (7.4), k F ( x ) k ≤ max { ˜ R, ξ k x k} , x ∈ X + . For any s ≥ ˜ R , k x k ≤ s implies k F ( x ) k ≤ s . By (7.4) k F n ( x ) k ≤ ξ n k x k as long as k F n ( x ) k ≥ ˜ R ≥ ˇ R . Once k F m ( x ) k ≤ ˜ R for some m ∈ N , then k F n ( x ) k ≤ ˜ R for all n ≥ m . Proof of Theorem 1.3. Let k · k ∼ be an equivalent norm such that there issome ˜ R > F ( X + ∩ ¯ B s ) ⊆ ( ¯ B s ) for all s ≥ ˜ R where ¯ B s is the closedball with radius s and the origin as center taken with respect to k · k ∼ . Let B be a bounded subset of X + with respect to the original norm. Then B ⊆ ¯ B s for some s ≥ ˜ R . By Proposition 7.1, F n ( B ) ⊆ ¯ B s for all n ∈ N . Noticethat ˜ B = B s is bounded with respect to the original norm. The statementconcerning the compact attractor K follows from [35, Thm.2.33]. Definition 8.1. Let x ∈ X and u ∈ X + . Then x is called u - bounded if thereexists some c > − cu ≤ x ≤ cu . If x is u -bounded, we define k x k u = inf { c > − cu ≤ x ≤ cu } . (8.1)32he set of u -bounded elements in X is denoted by X u . If x, u ∈ X + and x isnot u -bounded, we define k x k u = ∞ . Two elements v and u in X + are called comparable if v is u -bounded and u is v -bounded, i.e., if there exist ǫ, c > ǫu ≤ v ≤ cu . Comparabilityis an equivalence relation for elements of X + , and we write u ∼ v if u and v are comparable. Notice that X u = X v if and only if u ∼ v .If X is a space of real-valued functions on a set Ω, k x k u = sup n | x ( ξ ) | u ( ξ ) ; ξ ∈ Ω , u ( ξ ) > o . Since the cone X + is closed, − k x k u u ≤ x ≤ k x k u u, x ∈ X u . (8.2) X u is a linear subspace of X , k · k u is a norm on X u , and X u , under thisnorm, is an ordered normed vector space with cone X + ∩ X u which is normal,generating, and has nonempty interior. Lemma 8.2. Let u ∈ ˙ X + . Then the following hold:(a) ♯x♯ ≤ k x k u ♯ u♯ = k x k u max { d ( u, − X + ) , d ( u, X + ) } , x ∈ X u . (b) u is a normal point of X + if and only if there exists some c ≥ suchthat k x k ≤ c k x k u for all x ∈ X u ∩ X + .(c) If X + is solid and u in the interior of X + , then X = X u , d ( u, X \ X + ) > , and k x k ≥ k x k u d ( u, X \ X + ) , x ∈ X. (d) In turn, if X u = X and there exists some ǫ > such that k x k ≥ ǫ k x k u for all x ∈ X , then u is an interior point of X + .(e) If X + is solid and u is both an interior and a normal point of X + , then k · k and k · k u are equivalent. roof. (a) By (8.2), if x ∈ X u , x, − x ≤ k x k u u. Since the companion functional is order-preserving on X , ψ ( x ) ≤ k x k u ψ ( u ) , ψ ( − x ) ≤ k x k u ψ ( u )and so ♯x♯ ≤ k x k u ♯u♯ .(b) Let u be normal point of X + . Then there exists some c > k y k ≤ c for all y ∈ X + with y ≤ u . For x ∈ X + ∩ X u , x ≤ k x k u u . If x = 0in addition, k x k − u x ≤ u . Hence (cid:13)(cid:13) k x k − u x (cid:13)(cid:13) ≤ c .The other direction is obvious.l(c) Let u be an interior element of X + . Then d ( u, X \ X + ) > 0. For any δ ∈ (0 , d ( u, X \ X + )), we have u ± δ k x k x ∈ X + for all x ∈ ˙ X . So ± x ≤ k x k δ u and so x ∈ X u and k x k u ≤ k x k δ . Since this holds for any δ ∈ (0 , d ( u, X \ X + ),it also holds for δ = d ( u, X \ X + ).(d) Assume that X u = X and there exists some ǫ > k x k ≥ ǫ k x k u for all x ∈ X . This means that ± x ≤ k x k u u ≤ (1 /ǫ ) k x k u. Hence u ± ǫ k x k x ∈ X + for all x ∈ ˙ X . This implies that u is an interior pointof X + .(e) follows from combining (b) and (c).If X + is normal, by Theorem 2.1, there exists some M ≥ k x k ≤ M k x k u k u k , x ∈ X u . (8.3)If X + is a normal and complete cone of X , then X + ∩ X u is a complete subsetof X u with the metric induced by the norm k · k u . For more information see[20, 1.3] [6, I.4], [21, 1.4].For u ∈ X + , one can also consider the functionals( x/u ) ⋄ = inf { α ∈ R ; x ≤ αu } ( x/u ) ⋄ = sup { β ∈ R ; βu ≤ x } (cid:27) x ∈ X, with the convention that inf( ∅ ) = ∞ and sup( ∅ ) = −∞ . If X is a space ofreal-valued functions on a set Ω,( x/u ) ⋄ = sup (cid:8) x ( ξ ) u ( ξ ) ; ξ ∈ Ω , u ( ξ ) > (cid:9) ( x/u ) ⋄ = inf (cid:8) x ( ξ ) u ( ξ ) ; ξ ∈ Ω , u ( ξ ) > (cid:9) ) x ∈ X. x ∈ X + , k x k u = ( x/u ) ⋄ . Since we will use this functional for x ∈ X + only, we willstick with the notation k x k u . Again for x ∈ X + , ( x/u ) ⋄ is a nonnegative realnumber, and we will use the leaner notation[ x ] u = sup { β ≥ βu ≤ x } , x, u ∈ X + . (8.4)Since the cone X + is closed, x ≥ [ x ] u u, x, u ∈ X + . (8.5)Further [ x ] u is the largest number for which this inequality holds. Lemma 8.3. Let u ∈ ˙ X + . Then the functional φ = [ · ] u : X + → R + ishomogeneous, order-preserving and concave. It is bounded with respect to theoriginal norm on X and also to the monotone companion norm, [ x ] u ≤ ♯x♯♯u♯ ≤ k x k ♯u♯ , x ∈ X + , and k φ k + = ♯φ♯ + ≤ ♯ u♯ .φ is upper semicontinuous with respect to the original norm and (cid:12)(cid:12) [ y ] u − [ x ] u (cid:12)(cid:12) ≤ k y − x k u , y, x ∈ X u ∩ X + . Recall that ♯u♯ = d ( u, − X + ). Proof. We apply the monotone companion norm to (8.5), ♯ x♯ ≥ [ x ] u ♯ u♯, x ∈ X + . The equality k φ k + = ♯φ♯ + follows from Proposition 5.4. The other propertiesare readily derived from the definitions.See [22] for an in-depth treatment of this functional.35 Lower Collatz-Wielandt numbers, bounds,and radius Let B : X + → X + be homogeneous and order-preserving. We do not assumethat B is continuous at 0; so concepts like the cone spectral radius [24, 26,27, 30] may not apply. For x ∈ ˙ X + , the lower Collatz-Wielandt number of x is defined as [16] [ B ] x = [ Bx ] x = sup { λ ≥ Bx ≥ λx } . (9.1)By (8.5), Bx ≥ [ B ] x x, x ∈ ˙ X + . (9.2)The following result is immediate. Lemma 9.1. Let B, C : X + → X + be homogeneous, bounded, and order-preserving. Let x ∈ ˙ X + . Then [ CB ] x ≥ [ C ] x [ B ] x . This implies b n + m ≥ b n b m , b n = [ B n ] x , n, m ∈ N . (9.3)The lower local Collatz-Wielandt radius of x is defined as η x ( B ) = sup n ∈ N [ B n ] /nx . (9.4)This implies η x ( B n ) ≤ ( η x ( B )) n , n ∈ N . (9.5)The lower Collatz-Wielandt bound is defined as cw ( B ) = sup x ∈ ˙ X + [ B ] x , (9.6)and the Collatz-Wielandt radius of B is defined as r cw ( B ) = sup x ∈ ˙ X + η x ( B ) . (9.7)36 .1 Companion spectral radii If B : X + → X + is homogeneous and bounded with respect to the origi-nal norm, then, by Proposition 5.1, B is also bounded with respect to themonotone companion norm ♯ · ♯ and ♯B♯ + ≤ k B k + . So, we can define thecompanion cone spectral radius , the companion growth bounds , and the com-panion orbital spectral radius by r ♯ + ( B ) = inf n ∈ N ♯B n ♯ /n + = lim n →∞ ♯B n ♯ /n + (9.8)and r ♯o ( B ) = sup x ∈ X + γ ♯B ( x ) , γ ♯B ( x ) = lim sup n →∞ ♯B n x♯ /n . (9.9)Since ♯ x♯ ≤ k x k for all x ∈ X , we have the estimates r ♯o ( B ) ≤ r ♯ + ( B ) ≤ r + ( B ) , r ♯o ( B ) ≤ r o ( B ) ≤ r + ( B ) . (9.10)If the cone X + is normal, the companion norm is equivalent to the originalnorm and the respective spectral radii equal their companion counterparts.The following proposition implies Theorem 1.8. Proposition 9.2. Let B be bounded, homogeneous and order-preserving.Then, for all x ∈ ˙ X + , [ B ] x ≤ η x ( B ) ≤ γ ♯B ( x ) ≤ γ B ( x ) . Further cw ( B ) ≤ ♯B♯ ≤ k B k and cw ( B ) ≤ r cw ( B ) ≤ r ♯o ( B ) ≤ (cid:26) r ♯ + ( B ) r o ( B ) (cid:27) ≤ r + ( B ) . Proof. Let x ∈ ˙ X B . The first inequality follows from (1.18). Since the coneis closed, B ( x ) ≥ [ B ] x x . By induction B n ( x ) ≥ [ B ] nx x .We apply the monotone companion half-norm ψ , which is homogeneous,and obtain ψ ( B n ( x )) ≥ [ B ] nx ψ ( x ). Since ψ ( x ) > γ x ( B ) ≥ γ ♯x ( B ) ≥ [ B ] x .Since, for all n ∈ N , B n is homogeneous and order-preserving,[ B n ] x ≤ γ ♯x ( B ) ≤ γ ♯x ( B ) n . The last equality follows (3.1). Since this holds for all n ∈ N , by (1.18) η x ( B ) ≤ γ ♯x ( B ). The last inequality follows from (9.10).The remaining assertions follow directly from the definitions.37he following criteria for the positivity of the lower Collatz-Wielandtbound and the Collatz-Wielandt radius are obvious from their definitions. Lemma 9.3. Let B : X + → X + be homogeneous and order-preserving.Then cw ( B ) > if and only if there exist ǫ > and x ∈ ˙ X + such that B ( x ) ≥ ǫx .Further r cw ( B ) > if and only if there exists ǫ > , m ∈ N , and x ∈ ˙ X + such that B m ( x ) ≥ ǫx . Theorem 9.4. Let X be an ordered normed vector with complete cone X + and B : X + → X + be homogeneous, continuous and order preserving. Then r ♯ + ( B ) ≤ r o ( B ) .Proof. Let r > r o ( B ). It is sufficient to show that r ♯ + ( B ) < r . By definitionof r o ( B ), γ B ( x ) < r for all x ∈ X + . Let m ∈ N . For all x ∈ X + , by definitionof γ B ( x ), there exists some n ∈ N , n ≥ m , such that k B n ( x ) k ≤ r n . In otherwords, X + ⊆ [ n ≥ m C n , C n = { x ∈ X + ; k B n ( x ) k ≤ r n } . Since B is continuous, the sets C n are closed with respect to k · k . Since X + is complete with respect to k · k , by the Baire category theorem, there existssome n ∈ N , n ≥ m , such that C n contains a subset of X + that is relativelyopen with respect to k · k . So there exist some ǫ > x ∈ X + such that k B n ( x + ǫz ) k ≤ r n whenever z ∈ X , k z k ≤ 2, and x + ǫz ∈ X + . Let z ∈ X + , k z k ≤ 2. Since x ∈ X + and B is order preserving, B n ( ǫz ) ≤ B n ( x + ǫz ).Since the companion norm is monotone and B is homogeneous, ǫ♯B n z♯ ≤ ♯B n ( x + ǫz ) ♯ ≤ k B n ( x + ǫz ) k ≤ r n , z ∈ X + , k z k ≤ . Now let y ∈ X + with ♯y♯ = ψ ( y ) ≤ 1. By definition of the companion half-norm, there exists some z ∈ X with z ≥ y and k z k ≤ 2. Then z ∈ X + and,since B is order preserving, B n ( y ) ≤ B n ( z ). Since the companion norm ismonotone, ♯B n ( y ) ♯ ≤ ♯B n ( z ) ♯ ≤ r n /ǫ. Since this holds for all y ∈ X + with ♯y♯ ≤ ♯B n ♯ + ≤ r n /ǫ . Since n ≥ m ,inf n ≥ m ♯B n ♯ /n + ≤ r inf n ≥ m ǫ − /n . Thus r ♯ + ( B ) = lim n →∞ ♯B n ♯ /n + ≤ r . 38e conclude this section with a conditions under which the lower KRproperty implies the KR property (see Definition 1.13). The following defi-nition is similar to the one in [6, III.2.1]. Definition 9.5. Let θ : X + → R + be order-preserving and homogeneous.An order-preserving map B : X + → X + is called strictly θ -increasing if forany x, y ∈ X + with x ≤ y and θ ( x ) < θ ( y ) there exists some ǫ > m ∈ N such that B m ( y ) ≥ (1 + ǫ ) B m ( x ). B is called strictly increasing if B is strictly θ -increasing where θ is therestriction of the norm to X + . Theorem 9.6. Let θ : X + → X + be order-preserving and homogeneousand B : X + → X + be continuous, homogeneous, and strictly θ -increasing.Assume that there is some p ∈ N such that ( B p ( x n )) n ∈ N has a convergentsubsequence for any increasing sequence ( x n ) in X + where { θ ( x n ); n ∈ N } is bounded. Then B has the KR property whenever it has the lower KRproperty.Proof. Since B is homogeneous, we can assume that r + ( B ) = 1 and thatthere exists some x ∈ ˙ X + such that B ( x ) ≥ x . Then the sequence ( x n )defined by x n = B n ( x ) is increasing. We claim that ( θ ( x n )) is bounded. Ifnot, then there exists some n ∈ N with θ ( x n − ) < θ ( x n ) where x = x . Since B is strictly θ -increasing, there exists some ǫ > m ∈ N suchthat B m ( x n ) ≥ (1 + ǫ ) B m ( x n − ). By definition of ( x n ), B ( y ) ≥ (1 + ǫ ) y for y = B m ( x n − ) ≥ x . Since y ∈ ˙ X + , r + ( B ) ≥ cw ( B ) ≥ ǫ , a contradiction.Choose p ∈ N according to the assumption of the theorem. We ap-ply the convergence principle in Proposition 2.12. Let S be the set of se-quences ( B p ( y n )) where ( y n ) is a increasing sequence in X + such that ( θ ( y n ))is bounded. Then S has the property required in Proposition 2.12 and soevery increasing sequence in S converges.Since ( x n ) is increasing and bounded, ( B p ( x n )) ∈ S converges with limit v . Since B p ( x n ) = x n + p , x n → v . Since x n +1 = B ( x n ) and B is continuous, B ( v ) = v .We mention some interesting properties of strictly θ -increasing maps. Proposition 9.7. Let θ : X + → R + be order-preserving and homogeneousand B : X + → X + be homogeneous and strictly θ -increasing. Let r, s > and v, w ∈ X + with θ ( v ) > and θ ( w ) > . a) If B ( v ) ≥ rv and B ( w ) ≤ sw and v is w -bounded, then r ≤ s and r = s implies w ≥ θ ( w ) θ ( v ) v .(b) If B ( v ) = rv and B ( w ) = sw and v and w are comparable, then r = s and w = θ ( w ) θ ( v ) v .Proof. We first assume that θ ( v ) = 1 = θ ( w ).(a) Since v is w -bounded, w ≥ [ w ] v v with [ w ] v > θ ( w ) = θ ([ w ] v v ).Then 1 = [ w ] v and w ≥ v . So rv ≤ B ( v ) ≤ B ( w ) ≤ sw . We apply θ andobtain r ≤ s .Case 2: θ ( w ) > θ ([ w ] v v )Since B is strictly θ -increasing, there exists some δ > sw ≥ B ( w ) ≥ (1 + ǫ ) B ([ w ] v v ) = (1 + ǫ )[ w ] v B ( v ) ≥ (1 + ǫ )[ w ] v rv, which implies that [ w ] v ≥ (1 + ǫ )( r/s )[ w ] v and so r < s .In either case r ≤ s . If r = s , the second case cannot occur and the firstcase holds where w ≥ v .(b) From part (a), by symmetry, r = s and w ≥ v . Again, by symmetry, w = v .If just θ ( v ) > θ ( w ) > 0, we set ˜ v = θ ( v ) v and ˜ w = θ ( w ) w . Then θ (˜ v ) = 1 = θ ( ˜ w ) and B ˜ v ≥ r ˜ v and B ˜ w ≤ s ˜ w .We apply the previous considerations to ˜ v and ˜ w and obtain the results. 10 Order-bounded maps The following terminology has been adapted from various works by Kras-nosel’skii [20, Sec.2.1.1] and coworkers [21, Sec.9.4] though it has been mod-ified. Definition 10.1. Let B : X + → X + , u ∈ X + . B is called pointwise u -bounded if, for any x ∈ X + , there exist some n ∈ N and γ > B n ( x ) ≤ γu . The point u is called a pointwise order bound of B . B is called uniformly u -bounded if there exists some c > B ( x ) ≤ c k x k u for all x ∈ X + . The element u is called a uniform order bound of B . 40 is called uniformly order-bounded if it is uniformly u -bounded for some u ∈ X + . B is called pointwise order-bounded if it is pointwise u -bounded forsome u ∈ X + .If B : X + → X + is bounded and X + is solid, then B is uniformly u -bounded for every interior point u of X + .Uniform order boundedness is preserved if the original norm k · k is re-placed by its monotone companion norm ♯ · ♯ . Proposition 10.2. Let B : X + → X + be order-preserving. Let u ∈ X + and B be uniformly u -bounded. Then B is also uniformly u -bounded with respectto the monotone companion norm.Proof. Let x ∈ X + . By (4.3), for each n ∈ N there exists some y n ∈ X + suchthat x ≤ y n and ♯x♯ ≤ k y n k ≤ ♯x♯ + (1 /n ). Since B is order-preserving anduniformly u -bounded, B ( x ) ≤ B ( y n ) ≤ c k y n k u, n ∈ N . We take the limit as n → ∞ and obtain B ( x ) ≤ c♯x♯u . Proposition 10.3. Let X + be a complete cone, u ∈ X + , and B : X + → X + be continuous, order-preserving and homogeneous. Then the following hold.(a) B is uniformly u -bounded if for any x ∈ X + there exists some c = c x ≥ such that B ( x ) ≤ cu .(b) If B be pointwise u -bounded, then some power of B is uniformly u -bounded.Proof. We prove (b); the proof of (a) is similar. Define M n,k = { x ∈ X + ; B n ( x ) ≤ ku } , n, k ∈ N . Since B is continuous and X + is closed, each set M n,k is closed. Since B is assumed to be pointwise u -bounded, X + = S k,n ∈ N M n,k . Since X + isa complete metric space, by the Baire category theorem, there exists some n, k ∈ N such that M n,k contains a relatively open subset of X + : There existssome y ∈ X + and ǫ > y + ǫz ∈ M n,k whenever z ∈ X , k z k ≤ y + ǫz ∈ X + . Now let z ∈ X + and k z k ≤ 1. Since B is order-preservingand y + ǫz ∈ X + , B n ( ǫz ) ≤ B n ( y + ǫz ) ≤ ku . Since B is homogeneous, forall x ∈ ˙ X + , B n ( x ) = k x k ǫ B n (cid:16) ǫ k x k x (cid:17) ≤ kǫ k x k u. In order to be able to compare the companion spectral radius to the upperCollatz-Wielandt bound which will be defined later, some results on theupper semicontinuity of the cone spectral radius are useful. For more resultsof that kind see [25].In the following, let X be an ordered normed vector space with cone X + . Lemma 11.1. Let B be bounded and homogeneous, x ∈ X + , and B be contin-uous at B n ( x ) for all n ∈ N . Let ( B k ) be a sequence of bounded homogeneousmaps such that k B k − B k + → as k → ∞ and ( x k ) be a sequence in X + such that x k → x . Then, for all n ∈ N , B nk ( x k ) → B n ( x ) as k → ∞ .Proof. For k ∈ N , k B k ( x k ) − B ( x ) k ≤ k B k ( x k ) − B ( x k ) k + k B ( x k ) − B ( x ) k≤k B k − B k + k x k k + k B ( x k ) − B ( x ) k k →∞ −→ . This provides the basis step for an induction proof. The induction stepfollows in the same way. Theorem 11.2. Let u ∈ X + , u = 0 . Let B : X + → X + be homogeneous andbounded and B be continuous at B n ( u ) for all n ∈ N . Let ( B k ) be a sequenceof bounded, homogeneous, order preserving maps such that k B k − B k + → as k → ∞ . Assume that there exist m ∈ N and c ≥ such that B mk ( x ) ≤ c k x k u for all k ∈ N and all x ∈ X + . Then lim sup k →∞ r ♯ + ( B k ) ≤ γ ♯B ( u ) ≤ r ♯o ( B ) ≤ r ♯ + ( B ) . Proof. We can also assume that k u k = 1. Choose m ∈ N and c ≥ B mk ( x ) ≤ c♯x♯u, k ∈ N , x ∈ X + . Since B k is order-preserving and homogeneous, B n + mk ( x ) ≤ c♯x♯B nk ( u ) , k, n ∈ N , x ∈ X + . 42e apply the monotone companion norm, ♯B n + mk x♯ ≤ ♯x♯ ♯B nk ( u ) ♯, k, n ∈ N , x ∈ X + . So ♯B n + mk ♯ + ≤ ♯B nk ( u ) ♯, n ∈ N . Let r > γ ♯B ( u ). Then there exists some N ∈ N such that ♯B n ( u ) ♯ < r n , n > N. Let n ∈ N , n > N . Since B nk ( u ) k →∞ −→ B n ( u ) by Lemma 11.1, there existssome k n ∈ N such that ♯B nk ( u ) ♯ < r n , k ≥ k n . We combine the inequalities. ♯B n + mk ♯ + ≤ r n , k ≥ k n . Then r ♯ + ( B k ) ≤ ♯B n + mk ♯ / ( n + m )+ ≤ r n/ ( n + m ) , k ≥ k n . So, for any n > N , lim sup k →∞ r ♯ + ( B k ) ≤ r n/ ( n + m ) . We take the limit as n → ∞ and obtainlim sup k →∞ r + ( B k ) ≤ r. Since this holds for any r > γ ♯B ( u ), the assertion follows. 12 Upper Collatz-Wielandt numbers The upper Collatz-Wielandt number of B at x ∈ ˙ X + is defined as k B k x = k B ( x ) k x = inf { r ≥ B ( x ) ≤ rx } . (12.1)where k B k x = ∞ if B ( x ) is not x -bounded. Collatz-Wielandt numbers [16], without this name, became more widelyknown when Wielandt used them to give a new proof of the Perron-Frobeniustheorem [43]. We will use them closer to Collatz’ original purpose, namelyto prove inclusion theorems (Einschließungss¨atze) for r + ( B ) which generalizethose in [10, 11]. 43 The upper local Collatz-Wielandt radius of x ∈ X + is defined as η x ( B ) = inf n ∈ N k B n k /nx . (12.2) Lemma 12.1. Let B, C : X + → X + be homogeneous and order-preserving.Let x ∈ ˙ X + , and B ( x ) and C ( x ) be x -bounded. Then C ( B ( x )) is x -boundedand k CB k x ≤ k C k x k B k x .Proof. By (8.2), C ( x ) ≤ k C k x x and B ( x ) ≤ k B k x x and CB ( x ) ≤ k B k x C ( x ) ≤k B k x k C k x x ; so k CB k x ≤ k B k x k C k x . Lemma 12.2. Let u ∈ ˙ X + .(a) Then η u ( B m ) ≥ ( η u ( B )) m for all m ∈ N .(b) If there exists some k ∈ N such that B m u is u -bounded for all m ≥ k ,then η u ( B ) = lim n →∞ k B n k /nu < ∞ and η u ( B m ) = ( η u ( B )) m for all m ∈ N .Proof. (a) Let u ∈ ˙ X + , m ∈ N . Then η u ( B m ) = inf n ∈ N k B mn k /nu = ( inf n ∈ N k B mn k / ( mn ) u ) m ≥ (inf k ∈ N k B k k /ku ) m . (b) Now let k ∈ N such that B m ( u ) is u -bounded for all m ≥ k . ByLemma 12.1, c n + m ≤ c n c m , n, m ≥ k, c n = k B n k u . Let r be an arbitrary number such that η u ( B ) = inf n ∈ N k B n k /nu < r . Thenthere exists some m ∈ N such that c /mm = k B m k /mu ≤ r . So B m ( u ) ≤ r m u .By applying B m as often as necessary, we can assume that m ≥ k .Any number n ∈ N with n ≥ m has a unique representation n = pm + q with p ∈ N and m ≤ q < m . Then, for n ≥ m , c n ≤ c pm c q ≤ r mp c q . If c q = 0, then both the limit inferior and the limit are zero and equal. Sowe can assume that c q = 0. We have c /nn ≤ r pm/n c /nq . n → ∞ , pm/n → n →∞ c /nn ≤ r . Since r was any numberlarger than the infimum , the limes superior and inferior coincide and thelimit exists and equals the infimum.The second equality in (b) follows from the fact that every subsequenceof a convergent sequence converges to the same limit. Remark 12.3. If B ( u ) is u -bounded, k B k u is the cone norm of B in thespace X u with u -norm. Then η u ( B ) is the asymptotic growth factor of u taken in X u . Theorem 12.4. Let B : X + → X + be homogeneous, bounded, and order-preserving. Let u ∈ X + , α ∈ R + and k ∈ N such that B k ( u ) ≤ α k u .Let u be a normal point of X + or B be power-compact. Then γ B ( u ) ≤ α .Proof. Since γ ( u, B k ) = ( γ ( u, B )) k by (1.14), we can assume that k = 1.Since B is homogenous, it is enough to show that B ( u ) ≤ u implies that γ B ( u ) = γ ( u, B ) ≤ B ( u ) ≤ u . Then B n ( u ) ≤ u for all n ∈ N .We first assume that u is a normal point of X + . By Definition 2.3, thereexists some c > k B n ( u ) k ≤ c for all n ∈ N . This implies γ B ( u ) ≤ B ℓ is compact for some ℓ ∈ N and that γ B ( u ) > a n ) with a n = k B n ( u ) k (12.3)is unbounded. By a lemma by Bonsall [8], there exists a subsequence a n j such that a n j → ∞ , j → ∞ , a k ≤ a n j , k = 1 , . . . , n j , j ∈ N . Set v j = a nj B n j ( u ). Then v j = B ℓ ( w j ) , w j = 1 k B n j ( u ) k B n j − ℓ ( u ) . Now k w j k ≤ a n j − ℓ a n j ≤ . v j ) converges to some v ∈ X + , k v k = 1.Since B n ( u ) ≤ u for all n ∈ N , v j ≤ a n j u. Since a n j → ∞ and X + is closed, we have v ≤ 0, a contradiction. Corollary 12.5. Let B : X + → X + be homogeneous, bounded, and order-preserving. Let u ∈ ˙ X + and B k ( u ) be u -bounded for all but finitely many k ∈ N . Assume that u is a normal point of X + or that B is power-compact.Then γ B ( u ) ≤ η u ( B ) .Proof. By definition, B k ( u ) ≤ k B k k u u for all k ∈ N with k B k k u < ∞ . ByTheorem 12.4, γ B ( u ) ≤ k B k k u for all k ∈ N . So γ B ( u ) ≤ η u ( B ). Theorem 12.6. Let B : X + → X + be homogeneous, bounded, and order-preserving. Let u ∈ X + , α ∈ R + and k ∈ N such that B k ( u ) ≤ α k u .Let u be a normal point of X + Then r o ( B ) ≤ α if B is pointwise u -bounded, and r + ( B ) ≤ α if some power of B is uniformly u -bounded,Proof. As in the proof of Theorem 12.4, we can reduce the proof to theimplication B ( u ) ≤ u = ⇒ r o ( B ) ≤ . Assume that B ( u ) ≤ u . Let x ∈ X + . Since B is pointwise u -bounded,there exists some m ∈ N and c > B m ( x ) ≤ cu . For all n ∈ N , B m + n ( x ) ≤ cB n ( u ) ≤ cu . Since u is a normal point of X + , by Definition2.3, there exists some ˜ c > k B m + n ( x ) k ≤ c ˜ c for all n ∈ N . Thisimplies γ B ( x ) ≤ 1. Since x ∈ X + has been arbitrary, r o ( B ) ≤ B m is uniformly u -bounded, we can replace c by c k x k and we obtain r + ( B ) ≤ Corollary 12.7. Let B : X + → X + be homogeneous, bounded, and order-preserving.Let u be a normal point of X + . Then r o ( B ) ≤ η u ( B ) if B is pointwise u -bounded, and r + ( B ) ≤ η u ( B ) if some power of B is uniformly u -bounded. x ∈ X + for which the sequence ♯B n x♯ /n is bounded, we extendthe definition of the companion growth bound of the B -orbit of x by γ ♯B ( x ) := lim sup n →∞ ♯B n ( x ) ♯ /n (12.4)and set it equal to infinity otherwise. We extend the definition of the orbitalcompanion spectral radius of B by r ♯o ( B ) := sup x ∈ X + γ ♯B ( x ) . (12.5) Theorem 12.8. Let B : X + → X + be homogeneous, bounded, and order-preserving, u ∈ X + . Let B be pointwise u -bounded and assume that there issome ℓ ∈ N such that B n ( u ) is u -bounded for all n ≥ ℓ .Then η x ( B ) ≤ γ ♯B ( x ) ≤ γ ♯B ( u ) ≤ η u ( B ) for all x ∈ X + and cw ( B ) ≤ r cw ( B ) ≤ r ♯o ( B ) = γ ♯B ( u ) ≤ η u ( B ) . If B m has the lower KR property for some m ∈ N , γ B ( u ) ≤ r + ( B ) ≤ η u ( B ) .Proof. Let x ∈ X + . We can assume x = 0. Since B is pointwise u -bounded,there exists some k = k ( x ) ∈ N and some c = c ( x ) > B k ( x ) ≤ cu .Since B is order-preserving and homogeneous, B n ( x ) ≤ cB n − k ( u ) for all n > k . This implies B n ( x ) ≤ cB n − k ( u ) ≤ c k B n − k k u u. We apply the monotone companion norm, ♯B n ( x ) ♯ ≤ c♯B n − k ( u ) ♯ ≤ c k B n − k k u ♯u♯. So ♯B n ( x ) ♯ /n ≤ c /n ♯B n − k ( u ) ♯ /n ≤ k B n − k k /nu (cid:0) c♯u♯ (cid:1) /n . We take the limit as n → ∞ , use Lemma 12.2 (b), recall Proposition 9.2 andobtain the first inequality. The second then follows by taking the supremumover x ∈ X + and recalling cw ( B ) ≤ r cw ( B ) from Proposition 9.2.47ssume that B is bounded and B m has the lower KR property for some m ∈ N . We can assume that r = r + ( B ) > 0. Then B m ( v ) ≥ r m v with r = r + ( B ) and some v ∈ ˙ X + . Since B is pointwise u -bounded, there existssome c > k ∈ N (which depend on v ) such that B k ( v ) ≤ cu . So r m + k v = B m + k ( v ) ≤ cB m ( u ). For all n ∈ N , r m + k + n ( v ) ≤ cB m + n ( u ). Then r k + j k v k u ≤ c k B j k u , j ∈ N , j ≥ m. So r ≤ ( c/r k k v k u ) /j k B j k /ju , j ∈ N , j ≥ m. Taking the limit as j → ∞ yields the desired result. Theorem 12.9. Let B : X + → X + be homogeneous, bounded and order-preserving, u ∈ X + . Let some power of B be uniformly u -bounded.Then cw ( B ) ≤ r cw ( B ) ≤ r ♯o ( B ) = r ♯ + ( B ) = γ ♯B ( u ) = η u ( B ) ≤ γ B ( u ) ≤ r + ( B ) . Under additional assumptions, the following hold: • If u is a normal point in X + , then η u ( B ) = r o ( B ) = r + ( B ) = γ B ( u ) =lim n →∞ k B n ( u ) k /n . • If some power of B has the lower KR property, then r cw ( B ) = η u ( B ) = γ B ( u ) = r o ( B ) = r + ( B ) . • If B has the lower KR property, then cw ( B ) = r cw ( B ) = η u ( B ) = γ B ( u ) = r o ( B ) = r + ( B ) .Proof. Let k ∈ N such that B k is uniformly u -bounded. By Proposition 10.2, B k is also uniformly u -bounded with respect to the monotone companionnorm. Then there exists some c > B k ( x ) ≤ c♯x♯u for all x ∈ X + .For all n ∈ N , since B is order-preserving and homogeneous, B k + n ( x ) = B k ( B n ( x )) ≤ ♯B n ( x ) ♯cu,B k + n ( x ) = B n ( B k ( x )) ≤ c♯x♯B n ( u ) . By definition of upper Collatz-Wielandt numbers, k B k + n k u ≤ ♯B n ( u ) ♯c, n ∈ N . 48y (12.2), ( η u ( B )) ( k + n ) /n ≤ ♯B n ( u ) ♯ /n c /n , n ∈ N . We take the limit as n → ∞ , η u ( B ) ≤ lim inf n →∞ ♯B n ( u ) ♯ /n ≤ lim inf n →∞ k B n ( u ) k /n . (12.6)This implies η u ( B ) ≤ γ ♯B ( u ). The other inequality follows from Theorem12.8.Let u be a normal point. By Corollary 12.7, η u ( B ) ≥ r + ( B ) ≥ r o ( B ) ≥ γ B ( u ) = lim sup n →∞ k B n ( u ) k /n . Together with (12.6), this implies equalities.Since the companion norm is order-preserving, ♯B k + n ( x ) ♯ ≤ c♯x♯ ♯B n ( u ) ♯ and ♯B k + n ♯ ≤ c♯B n ( u ) ♯, n ∈ N . Since B is bounded, r ♯ + ( B ) ≤ γ ♯B ( u ).The other statements now follow from the previous theorems.In view of estimating the cone spectral radius from above the followingobservation may be of interest. Corollary 12.10. Let B : X + → X + be a homogeneous, bounded, order-preserving map. Assume that X + is normal and complete or some power of B has the lower KR property.Then r + ( B ) is a lower bound for all upper Collatz-Wielandt numbers k B k u where u ∈ ˙ X + , B ( u ) is u -bounded and B is pointwise u -bounded.Proof. Combine the previous theorems with (12.2) and recall that r ♯o ( B ) = r + ( B ) if X + is normal and complete.49 Let u ∈ ˙ B + and B ( u ) be u -bounded. Then B ( x ) is x -bounded for any u -comparable x ∈ X + .So we define the upper Collatz-Wielandt bound with respect to u by CW u ( B ) = inf {k B k x ; x ∈ X + , x ∼ u } . (12.7)If x ∈ X + and x ∼ u , η x ( B ) = η u ( B ). Since η x ( B ) ≤ k B k x , η u ( B ) ≤ CW u ( B ) . (12.8)We have the following inequalities from Theorem 12.8 and Theorem 12.9. Theorem 12.11. Let B be homogeneous and order-preserving. Let u ∈ ˙ X + , B ( u ) be u -bounded and B be pointwise u -bounded. Then cw ( B ) ≤ r cw ( B ) ≤ η u ( B ) ≤ CW u ( B ) . Lower KR property of the map turns some of the inequalities in equalities(Theorem 12.9). Theorem 12.12. Let B be homogeneous, bounded, and order-preserving andsome power of B have the lower KR property. Let u ∈ ˙ X + and some powerof B be uniformly u -bounded. Then r cw ( B ) = r o ( B ) = r + ( B ) = η u ( B ) ≤ CW u ( B ) . If the cone X + is normal, the cone and orbital spectral radius are increasingfunctions of the homogeneous bounded order-preserving maps (cf. [1, L.6.5]).Collatz-Wielandt numbers, bounds, and radii and the companion radii areincreasing functions of the map even if the cone is not normal. Theorem 12.13. Let A, B : X + → X + be bounded and homogeneous. As-sume that A ( x ) ≤ B ( x ) for all x ∈ X + and that A or B are order-preserving.Then cw ( A ) ≤ cw ( B ) , r cw ( A ) ≤ r cw ( B ) , r ♯ + ( A ) ≤ r ♯ + ( B ) r ♯o ( A ) ≤ r ♯o ( B ) .Further, for all x ∈ X + , k A k x ≤ k B k x and η x ( A ) ≤ η x ( B ) .If u ∈ ˙ X + and A ( u ) and B ( u ) are u -bounded, then CW u ( A ) ≤ CW u ( B ) .If X + is a normal cone, then also r + ( A ) ≤ r + ( B ) and r o ( A ) ≤ r o ( B ) . roof. We claim that A n ( x ) ≤ B n ( x ) for all x ∈ X + and all n ∈ N . For n = 1, this holds by assumption. Now let n ∈ N and assume the statementholds for n . If A is order-preserving, then, for all x ∈ X + , since B n ( x ) ∈ X + , A n +1 ( x ) = A ( A n ( x )) ≤ A ( B n ( x )) ≤ B ( B n ( x )) = B n +1 ( x ) . If B is order-preserving, then, for all x ∈ X + , since A n ( x ) ∈ X + , A n +1 ( x ) = A ( A n ( x )) ≤ B ( A n ( x )) ≤ B ( B n ( x )) = B n +1 ( x ) . Since the companion norm is order-preserving, ♯A n ( x ) ♯ ≤ ♯B n ( x ) ♯ for all x ∈ X + , n ∈ N . Further ♯A n ♯ + ≤ ♯B n ♯ + for all n ∈ N and r ♯ + ( A ) ≤ r ♯ + ( B ).Further γ ♯A ( x ) ≤ γ ♯B ( x ) and so r ♯o ( A ) ≤ r ♯o ( B ).If X + is normal, the respective order radii taken with the original normcoincide with those taken with the companion norm.As for the Collatz-Wielandt radius, B n x ≥ A n x ≥ [ A n ] x x. By (9.1), [ B n ] x ≥ [ A n ] x , and the claim follows from (9.7) and (9.4). Theproofs for k · k x , η x , and CW u are similar. Proof of Theorem 1.15. By Theorem 12.13, cw ( A ) ≤ cw ( B ). The assertionnow follows from Theorem 1.14. Conditions which make CW u ( B ) an eigenvalue of B with positive eigenvectorand imply equality between all these numbers including CW u ( B ) can befound in [1, Thm.7.3]. Using the companion half-norm ψ , one can dropthat the cone is normal and complete provided that the map is compact.Solidity of the cone can be replaced by the weaker assumption that the mapis uniformly u -bounded. Theorem 12.14. Let B : X + → X + be continuous, compact, homogeneous,and order-preserving. Let u ∈ ˙ X + and B be uniformly u -bounded.Then cw ( B ) = r cw ( B ) = r + ( B ) = η u ( B ) = CW u ( B ) .If r = CW u ( B ) > , then there exists some v ∈ ˙ X u such that B ( v ) = rv . emark 12.15. If X + is complete, we also obtain this result if we replacecompactness of B by the assumptions in Theorem 1.10 with part (a) or byassumption (ii) in [1, Thm.7.3].More generally, the following holds. Theorem 12.16. Let B : X + → X + be homogeneous and order-preserving.Let u ∈ ˙ X + and B be uniformly u -bounded and continuous at B n ( u ) for all n ∈ N . Assume there is some ǫ > such that, for all ǫ ∈ (0 , ǫ ) , the maps B ǫ , B ǫ ( x ) = B ( x ) + ǫψ ( x ) u , have eigenvectors B ǫ ( v ǫ ) = λ ǫ v ǫ with v ǫ ∈ ˙ X + and λ ǫ > .Then r + ( B ) ≥ γ u ( B ) ≥ CW u ( B ) = η u ( B ) with equality holding every-where if u is a normal point of X + or some power of B has the lower KRproperty.Further, if B has the KR property and CW u ( B ) > , there exists some v ∈ ˙ X + such that B ( v ) = CW u ( B ) v .Proof. Choose a sequence ( ǫ n ) in (0 , ǫ ) with ǫ n → 0. Set B n = B ǫ n . Themaps B n inherit uniform u -boundedness from B .By assumption, there exist v n ∈ ˙ X + and r n > B ( v n ) + ǫ n ψ ( v n ) u = r n v n . Since B is uniformly u -bounded and ψ ( v n ) > v n is u -comparable. By (12.8), r n ≥ CW u ( B n ). Also r n ≤ cw ( B n ) by (1.20).By Theorem 12.11 and Theorem 12.9, η u ( B n ) = r ♯ + ( B n ) = CW u ( B n ) for all n ∈ N . Further, CW u ( B n ) ≥ CW u ( B ).Suppose that r ♯ + ( B ) < CW u ( B ). Since ǫ n → k B n − B k + → 0. ByTheorem 11.2, r ♯ + ( B n ) < CW u ( B ) for large n , a contradiction.So r ♯ + ( B ) ≥ CW u ( B ). By Theorem 12.9, also η u ( B ) = r ♯ + ( B ). Since η u ( B ) ≤ CW u ( B ), we have η u ( B ) = CW u ( B ). The other inequalities followfrom Theorem 12.9.If some power of B has the lower KR property, equality holds by Theorem12.12.If u is a normal point of X + , equality follows from Theorem 12.9.Assume that B has the KR property and CW u ( B ) > 0. Then r + ( B ) = CW u ( B ) > v ∈ ˙ X + such that B ( v ) = r + ( B ) v .The equality r + ( B ) = CW u ( B ) guarantees that, at least in theory, onecan get arbitrarily sharp estimates of r + ( B ) from above in terms of up-per Collatz-Wielandt numbers k B k x by choosing an appropriate x ∈ ˙ X + for which B ( x ) is x -bounded and B pointwise x -bounded (Corollary 12.10).52rude attempts in this direction are made for the rank-structured discretepopulation model with mating in Section 14.The idea of perturbing the map B as above or in a similar way is quiteold; see [31, Satz 3.1] and [40, Thm.3.6]. Theorem 12.17. Let X + be complete. Assume that B = K + A where K : X + → X + is compact, homogeneous, continuous and order preservingand A : X → X is linear, positive and bounded and r ( A ) < r + ( B ) . Let u ∈ ˙ X + and B be uniformly u -bounded.Then cw ( B ) = r cw ( B ) = r + ( B ) = η u ( B ) = CW u ( B ) .If r = CW u ( B ) > , then there exists some v ∈ ˙ X u such that B ( v ) = rv .Proof. For ǫ ∈ [0 , B ǫ : X + → X + by B ǫ ( x ) = B ( x ) + ǫψ ( x ) u where ψ is the companion half-norm. Then B ǫ = K ǫ + A with K ǫ ( x ) = K ( x )+ ǫψ ( x ) u . Then K ǫ is compact, continuous, order-preserving and homogeneous.Since A is linear and bounded, B nǫ = K n,ǫ + A n with compact, continuous,homogeneous, order-preserving maps K n,ǫ .If n is chosen large enough, k A n k < r + ( B n ). By Theorem 1.10 (a), somepower of B has the KR property. Since B ( x ) ≤ B ǫ ( x ) for all x ∈ X + , r + ( B ) = cw ( B ) ≤ cw ( B ǫ ) ≤ r + ( B ǫ ) by Theorem 1.15.So r ( A ) < r + ( B ) ≤ r + ( B ǫ ). By Theorem 1.10 (a), for large enough n ,there exist eigenvectors v ǫ ∈ ˙ X + such that B nǫ ( v ǫ ) = r nǫ v ǫ with r ǫ = r + ( B ǫ ).It is easy to see that, for ǫ ∈ (0 , B ǫ is strictly ψ -increasing and that B nǫ inherits this property. Set w ǫ = B ǫ ( v ). Then B nǫ ( w ǫ ) = r nǫ w ǫ . Since B ǫ isuniformly u -bounded and ψ ( v ǫ ) > w ǫ is u -comparable. Since B kǫ ( v ǫ ) ∈ ˙ X + for all k ∈ N , B nǫ ( v ǫ ) is u -comparable and so v ǫ and w ǫ are comparable.By Lemma 9.7, B ǫ ( v ǫ ) = α ǫ v ǫ for some α ǫ > r ǫ .Since v ǫ is u -comparable r ǫ = cw ( B ǫ ) = r cw ( B ǫ ) = r + ( B ǫ ) = CW u ( B ǫ )for all ǫ ∈ (0 , CW u ( B ) = r + ( B ) = η u ( B ) = γ u ( B ).Now choose a decreasing sequence ( ǫ k ) in (0 , 1] with ǫ n → 0. Then r k = r + ( B ǫ k ) form a decreasing sequence with r k ≥ CW u ( B ). Suppose CW u ( B ) > 0. Set v k = v ǫ k . We can assume that k v k k = 1 for all k ∈ N . Then r k v k = K ( v k ) + A ( v k ) + ǫ k ψ ( v k ) u. Let r = lim k →∞ r k . Since K is compact,( r − A ) v k = ( r − r k ) v k + K ( v k ) + ǫ k ψ ( v k ) u k → ∞ after choosing a subsequence. Since r ≥ CW u ( B ) = r + ( B ) > br + ( A ), ( r − A ) − = P ∞ j =1 (1 /r ) j +1 A k exists as a continuous additivehomogeneous map and acts as the inverse of r − A . This implies that v k → v for some v ∈ X + , k v k = 1. Since B is continuous, rv = B ( v ) which impliesthat r = r + ( B ). Then r ≤ cw ( B ) and equality follows. If CW u ( B ) = 0,equality holds anyway. 13 Monotonically compact order-boundedmaps on semilattices As before, let X be an ordered normed vector space with cone X + . Definition 13.1. Let u ∈ ˙ X + . B : X + → X + ∩ X u is called antitonically u -compact if ( B ( x n )) has a convergent subsequence for each decreasing sequence( x n ) in X + for which there is some c > x n ≤ cu for all n ∈ N . B is called monotonically u -compact if ( B ( x n )) has a convergent subse-quence for each monotone sequence ( x n ) in X + for which there is some c > x n ≤ cu for all n ∈ N .If X + is regular (Section 2), then every order-preserving homogeneous B : X + → X + is monotonically u -compact for any u ∈ X + . Definition 13.2. Let u ∈ ˙ X + , B : X + → X + and B ( u ) be u -bounded. B iscalled antitonically continuous at x ∈ X + if, for for any decreasing sequence( x k ) in X + with x ≤ x k ≤ cu for all k ∈ N (with some c > k ) and k x k − x k → 0, we have ψ ( B ( x k ) − B ( x )) → B is called monotonically continuous at x ∈ X + with for every monotonesequence ( x k ) in X u with x k ≤ cu for all k ∈ N (with some c > k ) and k x k − x k → 0, we have ♯B ( x k ) − B ( x ) ♯ → x ∧ y = inf { x, y } exists for all x, y ∈ X + ) and the upper local Collatz-Wieland spectral radiusat u ∈ X + , η u ( B ) = inf n ∈ N k B n k /nu . Theorem 13.3. Let X + be a minihedral cone and B : X + → X + be order-preserving and homogeneous. Let u ∈ ˙ X + , and B ( u ) be u -bounded. Assume hat B is antitonically u -compact and antitonically continuous. Finally as-sume that η u ( B ) > and k B ( y n ) k u → for any decreasing sequence ( y n ) in X u ∩ X + with k y n k → .Then there exists some x ∈ ˙ X + , such that B ( x ) ≥ η u ( B ) x and η u ( B ) = r ♯o ( B ) = r cw ( B ) = cw ( B ) . The first part of the proof has been adapted from [21, L.9.5] where B is assumed to be a linear operator on the ordered Banach space X and thecone X + to be normal. Use of the monotone companion metric allows todrop normality as assumption. However, without u being a normal point ofthe cone, compactness of B may not imply monotonic compactness. Proof of Theorem 13.3. Let u ∈ X + such that B ( u ) is u -bounded. Since B is homogeneous, we can assume that η u ( B ) = 1. Otherwise, we consider η u ( B ) B . We define x = u, x k = y k ∧ u, y k = B ( x k − ) + 2 − k u, k ∈ N . (13.1)Then x k ≤ u = x for all n ∈ N . By induction, since B is order-preserving, x k +1 ≤ x k for all k ∈ N . We apply the convergence principle in Proposition2.12 with S being the set of sequences ( B ( v n )) with ( v n ) being decreasing and v ≤ cu for some c > S has the properties requested in Proposition 2.12.Since B is antitonically u -compact, every sequence in S has a convergentsubsequence. So every sequence in S converges.Since ( B ( x n )) ∈ S , there exists some z ∈ X + such that ( B ( x k )) convergesto z as k → ∞ and B ( x k ) ≥ z for all k ∈ N . By (13.1), y k → z . Further z ≤ B ( x k − ) ≤ y k ≤ B ( u ) + 2 − k u. By (13.1), x k = y k ∧ u ≥ z ∧ u =: x. Notice that y k ∧ u + z − y k ≤ z ∧ u = x. So 0 ≤ x k − x ≤ y k − z, k ∈ N . (13.2)Further ♯B ( x k ) − B ( x ) ♯ = ψ ( B ( x k ) − B ( x )) ≤ ψ ( B ( y k − z + x ) − B ( x )) ≤k B ( y k − z + x ) − B ( x ) k . x ≤ y k − z + x ≤ B ( u ) + u + x ≤ ( c k u k + 2) u . Recall that ( y k − z + x )converges x with respect to the the original norm.Since B is antitonically continuous, ( B ( x k )) converges to B ( x ) with re-spect to the monotone companion norm. Since B ( x k ) → z with respect tothe monotone companion norm, we have B ( x ) = z ≥ x .Moreover, x = z ∧ u = B ( x ) ∧ u. It remains to show that x = 0. Suppose that x = 0. Then z = B ( x ) =0. Recall that x k = y k ∧ u and k y k k → 0. Since B is order-preserving, k B ( x k ) k u ≤ k B ( y k ) k u → m ∈ N such that B ( x k − ) + 2 − k u ≤ u for all k ≥ m .Hence x k = B ( x k − ) + 2 − k u = y k , k ≥ m. In particular, 2 m x m ≥ u and x k ≥ B ( x k − ) for all k ≥ m . Since B isorder-preserving and homogeneous,2 m x m + n ≥ B n (2 m x m ) ≥ B n ( u )and 2 m B ( x m + n ) ≥ B n +1 ( u ) . Now 2 m B ( x m + n ) ≤ (1 / u for sufficiently large n . This shows that, forsome n ∈ N , B n +1 ( u ) ≤ (1 / u and k B n +1 k u ≤ / 2. By Lemma 12.2, η u ( B ) = inf n ∈ N k B n k /nu < 1, a contradiction.This shows that x = 0 and B ( x ) ≥ x . Then B n ( x ) ≥ x for all n ∈ N .By (9.1), [ B n ] x ≥ cw ( B ) ≥ η u ( B ) ≥ r ♯ + ( B ) ≥ r cw ( B ) ≥ cw ( B ). Theorem 13.4. Let the cone X + be a lattice and B : X + → X + be order-preserving and homogeneous. Further let u ∈ ˙ X + and some power of B bemonotonically u -compact and continuous and some power of B be uniformly u -bounded. Finally assume that r = η u ( B ) > .Then there exists some x ∈ ˙ X + such that B ( x ) ≥ rx and η u ( B ) = r ♯o ( B ) = r cw ( B ) = cw ( B ) .Proof. Replacing B by η u ( B ) B , we can assume that η u ( B ) = 1. Let B m bemonotonically u -compact and B ℓ be uniformly u -bounded. Set p = m + ℓ .Then B p is monotonically u -compact and uniformly u -bounded and continu-ous. By Theorem 13.3, there exists some w ∈ ˙ X + such that B p ( w ) ≥ w . ByProposition 1.12, there exists some v ∈ ˙ X + such that B ( v ) ≥ v .56ecall the definition of B being strictly increasing in Definition 9.5. Theorem 13.5. Let X + be a lattice and B : X + → X + be monotonically con-tinuous, strictly increasing, and homogeneous. Further assume that a powerof B is monotonically u -compact and some power is uniformly u -bounded forsome u ∈ ˙ X + . Finally assume that r = η u ( B ) > .Then there exists some x ∈ ˙ X + such that B ( x ) = rx .Proof. We can assume that η u ( B ) = 1. By Theorem 13.3, there exists some x ∈ X + , k x k = 1 such that B ( x ) ≥ x . Then the sequence ( x n ) n ∈ Z + in X + defined by x n = B n ( x ) is increasing. The same proof as for Theorem 9.6implies that {k x n k ; n ∈ N } is a bounded set in R .Set x = x . Then x n = B ( x n − ) for n ∈ N . Since some power of B is uniformly u -bounded, there exists some c ≥ m ∈ N such that x n ≤ c k x n − m k u = cu for n ≥ m . Since some power of B is monotonically u -compact and ( x n ) = ( B n ( x )) is increasing, a similar application of theconvergence principle in Proposition 2.12 provides that x n → y for some y ∈ X + with k y k = 1. Since B is monotonically continuous, ψ ( B ( y ) − B ( x n )) → 0. So ψ ( B ( y ) − x n +1 ) → 0. Then ♯y − B ( y ) ♯ ≤ ♯y − x n +1 ♯ + ♯x n +1 − B ( y ) ♯ = ψ ( y − x n +1 ) + ψ ( B ( y ) − x n +1 ) → 0. Thus y = B ( y ). Proposition 13.6. Let u ∈ ˙ X + and B : X + → X + ∩ X u be homogeneousand order-preserving. Further let B be monotonically u -compact, uniformly u -bounded and monotonically continuous.Let ǫ > and ψ the companion half-norm. Set B ǫ ( x ) = B ( x ) + ǫψ ( x ) u .Then there exists some v ∈ ˙ X + such that B ǫ ( v ) = r ǫ v with r ǫ = η u ( B ǫ ) = CW u ( B ǫ ) = r cw ( B ǫ ) = cw ( B ǫ ) > .Proof. One readily checks that B ǫ satisfies the assumptions of Theorem 13.3and r ǫ = η u ( B ǫ ) ≥ ǫ > r ǫ = 1. By Theorem 13.3, there exists some w ∈ X + , ψ ( w ) = 1, such that B ǫ w ≥ w . Let w n = B nǫ ( w ) for n ∈ Z + . Then ( w n ) isan increasing sequence in X + ∩ X u . We claim that ψ ( w n ) = 1 for all n ∈ N .Suppose not. Then there exists some n ∈ N such that ψ ( w n ) > ψ ( w n − ).Then there exists some δ > B ǫ ( w n ) ≥ B ǫ ( w n − ) + δu. Since B ǫ ( w n − ) ∈ X u , there exists some ǫ > B ǫ ( w n ) ≥ (1 + ǫ ) B ǫ ( w n − ) . η u ( B ǫ ) ≥ ǫ , a contradiction. Since B is uniformly u -bounded, it is also uniformly u -bounded with respect to the companionhalf-norm ψ by Proposition 10.2. So there exists some c > w n = B ( w n − ) + ǫu ≤ cu . Since ( w n ) is increasing and B ǫ is monotonically u -compact, w n +1 = B ǫ ( w n ) → v for some v ∈ X + , ψ ( v ) = 1. Since B ǫ ismonotonically continuous, v = B ǫ ( v ) and 1 ≤ cw ( B ǫ ) ≤ r cw ( B ǫ ) ≤ η u ( B ǫ ).Since v ≥ ǫu and B is uniformly u -bounded, v is u -comparable. This implies CW u ( B ǫ ) ≤ η u ( B ). Since CW u ( B ǫ ) ≥ η u ( B ǫ ), we have equality. Theorem 13.7. Let u ∈ ˙ X + and B : X + → X + be homogeneous, order-preserving and continuous. Assume that B is uniformly u -bounded monoton-ically u -compact.Then r + ( B ) ≥ CW u ( B ) = η u ( B ) with equality holding if u is a normalpoint of X + or a power of B has the lower KR property. If r := CW u ( B ) > ,there exists some v ∈ ˙ X + such that B ( v ) ≥ rv .Proof. By Theorem 12.9, r + ( B ) ≥ γ B ( u ) ≥ η u ( B ) with equality holding if u is a normal point of X + or some power of B has the lower KR property. By(12.8), η u ( B ) ≤ CW u ( B ).So, if CW u ( B ) = 0, the assertion holds, and we can assume that CW u ( B ) > ǫ n ) in (0 , 1) with ǫ n → 0. Let B n : X + → X + be given by B n ( x ) = B ( x ) + ǫ n ψ ( x ) u . We combine Proposition 13.6 andTheorem 12.16 and obtain η u ( B ) = CW u ( B ) ≤ r + ( B ).If u is a normal point of X + or some power of B has the lower KRproperty, η u ( B ) = r + ( B ) which implies CW u ( B ) = r + ( B ).Assume that r := CW u ( B ) > 0. Then η u ( B ) = r > v ∈ ˙ X + with B ( v ) ≥ rv by Theorem 13.3. 14 Application to a rank-structured popula-tion model with mating Let X ⊆ R N be an ordered normed vector space with cone X + = X ∩ R N + .Assume that the norm has the property that x j ≤ k x k for all x = ( x j ) ∈ X + and all j ∈ N . This implies that X ⊆ ℓ ∞ and k x k ∞ ≤ k x k for all x ∈ X .58efine a map B : X + → R N + , B ( x ) = ( B j ( x )), by B ( x ) = q x + ∞ X j,k =1 β jk min { x j , x k } B j ( x ) = max { p j − x j − , q j x j } , j ≥ x = ( x j ) ∈ X + . (14.1)Here p j , q j ≥ β j,k ≥ j, k ∈ N .The dynamical system ( B n ) n ∈ N can be interpreted as the dynamics of arank-structured population and, in a way, is a discrete version (in a dou-ble sense) of the two-sex models with continuous age-structure in [17, 18]. B ( x ) is the number of newborn individuals who all have the lowest rank 1.Procreation is assumed to require some mating. Mating is assumed to berank-selective and is described by taking the minimum of individuals in tworanks. The numbers β jk represent the fertility of a pair when the female hasrank j and the male has rank k where a 1:1 sex ratio is assumed at each rank.The maps B j , j ≥ 2, describe how individuals survive and move upwards inthe ranks from year to year where one cannot move by more than one rankwithin a year. If the population size is modeled in number of individuals onewould assume that p j , q j ≤ 1, but if it is modeled in biomass, then such anassumption would not be made. However, an individuals at rank j is at least j − p j → q j → B j ( x ) ≤ ( p j − + q j ) k x k for j ≥ B is u -bounded with respect to u = ( u j ) with u = 1 , u j = p j − + q j , j ≥ , (14.2)provided that u ∈ X .If we choose X = ℓ , it is sufficient to assume thatsup k ∈ N ∞ X j =1 β jk < ∞ or sup j ∈ N ∞ X k =1 β jk < ∞ (14.3)If X = ℓ ∞ , it is sufficient to assume that ∞ X j,k =1 β jk < ∞ . (14.4)It does not appear that (14.4) can be improved for X = c or X = bv .59 is compact on X and uniformly u -bounded if u ∈ ℓ p , X = ℓ p , p ∈ [1 , ∞ ) u ∈ c , X = c , c, ℓ ∞ ( p j ) , ( q j ) ∈ bv ∩ c , X = bv, bv ∩ c . Notice that, if x n = ( x nj ) is a bounded sequence in X , then it is a boundedsequence in ℓ ∞ and, after a diagonalization procedure, has a subsequence( y m ) such that ( y mj ) m ∈ N converges for each j ∈ N .Apparently, X = ℓ ∞ , c, c require the weakest assumption in terms of u .It is not clear, however, whether an eigenvector in any of these three spaceswould also be in bv if ( q j ) , ( p j ) ∈ bv ∩ c . If u ∈ ℓ , then u is also in all theother spaces we have considered, and cw ( B ) = r cw ( B ) = η u ( B ) = CW u ( B )in all spaces with the radii and bounds not depending on the space. If X isnormal, then also r + ( B ) = cw ( B ) does not depend on the space. For X = bv ,we have r + ( B ) = cw ( B ) as well provided that ( p j ) , ( q j ) ∈ bv ∩ c such that B is compact also in bv or in bv ∩ c .To derive estimates for the cone spectral radius of B , one runs into al-gebraic difficulties very soon except when attempting cw ( B ). Let x n be thesequence where the first n terms are 1 and all others zero. Then,[ B ] x = q + β , [ B ] x m = min n q + m X j,k =1 β j,k , m inf j =2 max { p j − , q j } o , m ≥ . Let e m be the sequence where the m th term is one and all other terms zero.Then [ B ] e = q + β , [ B ] e m = q m , m ≥ . Since cw ( B ) is an upper bound for all these numbers, we obtain the followingtwo conditions for cw ( B ) > q + β > q j > j ∈ N , j > j, k ∈ N and m = max { j, k } , we have β jk > p i > i = 1 , . . . , m − r + ( B ) > B ( x ) = ∞ X j,k =1 β jk min { x j , x k } B j ( x ) = p j − x j − , j ≥ x = ( x j ) ∈ X + . (14.5)If r = r + ( B ) = cw ( B ) > 0, then there exists some x ∈ ˙ X + with(1 /r ) B ( x ) = x . (1 /r ) B corresponds to the map where all parameters aredivided by r . This division does not affect our assumptions and conditions.So we can assume that B ( x ) = x , x ∈ ˙ X + . By (14.5), x j = p j − x j − , j ≥ . By recursion, x j = j − Y i =1 p j x , j ≥ . Since x = P ∞ j,k =1 β jk min { x j , x k } , failure of condition 2 implies x = 0 whichimplies x = 0. This contradiction tells us that r + ( B ) = 0 if both condition 1and 2 fail.In order to find estimates of r + ( B ) from above, we can use Corollary12.10.Take the sequence e where all terms are 1 for X = ℓ ∞ or X = bv providedthat P ∞ j,k =1 β jk < ∞ . Then r + ( B ) ≤ η e ( B ) ≤ k B ( e ) k e = max n q + ∞ X j,k =1 β jk , sup m ≥ max { p m − , q m } o . We obtain this estimate also for X = c , bv ∩ c , ℓ provided that there exist θ ∈ (0 , 1) and c > { p m − , q m } ≤ cθ m for all m ≥ 2. Then B is uniformly w -bounded for w = ( θ n ) and r + ( B ) ≤ k B k w = max n q + ∞ X j,k =1 β jk θ j + k − , sup m ≥ max { p m − /θ, q m } o . This estimate also holds for all ˜ θ ∈ ( θ, θ → α > c > { p m − , q m } ≤ m − α for all m ≥ 2. Then B is uniformly w -bounded for w = n − α and r + ( B ) ≤ k B k w = max n q + ∞ X j,k =1 β jk j − α k − α , sup m ≥ max { p m − ( m/m − α , q m } o . This estimate also holds for all ˜ α ∈ (1 , α ), so we can take the limit α → β jk , p j and q j arefunctions of x ∈ X + and let F ( x ) be B ( x ) with this functional dependence.We assume that p j − ( x ) + q j ( x ) → j ∈ N as k x k → ∞ ,further P ∞ j,k =1 β jk ( x ) → k x k → ∞ . Let ǫ > c > p j − ( x ) + q j ( x ) ≤ ǫ ∞ X j,k =1 β jk ( x ) ≤ ǫ k x k ≥ c. Choose A : X → X , A ( x ) = ( A j ( x )), A ( x ) = ǫx + ∞ X j,k =1 ˜ β jk x j A j ( x ) = ǫ ( x j − + x j ) , j ≥ x = ( x j ) ∈ X + . (14.6)Here ˜ β jk = sup k x k≥ c β jk ( x ). Then B ( x ) ≤ A ( x ) whenever k x k ≥ c . Choose ǫ < / X = ℓ ∞ and ǫ < / X = ℓ . Then k A k < F ispoint-dissipative by Theorem 1.3.To make F compact in X = ℓ , we assume that for any c > u ∈ ℓ such that sup k ≥ P ∞ j =1 β jk ( x ) ≤ u and p j − ( x ) + q j ( x ) ≤ u j for j ≥ x ∈ X + with k x k ≤ c . Under these assumptions, the discretesemiflow ( F n ) has a compact attractor of bounded sets. Acknowledgement. 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