A SHORT INTRODUCTION TO BANACH LATTICES AND

CHAPTER A SHORT INTRODUCTION TO BANACH LATTICES AND POSITIVE OPERATORS In tis capter we give a brief introduction to Banac lattices and positive operators. Most results of tis capter can be found, e.g., in [26], [] or [2].. BANACH LATTICES A non empty set M wit a relation is said to be an ordered set if te following conditions are satisfied. i) x x for every x M, ii) x y and y x implies x y, and iii) x y and y z implies x z. Let A be a subset of an ordered set M. Te element x M (resp. z M) is called an upper bound (lower bound resp.) of A if y x for all y A (resp. z y for all y A). Moreover, if tere is an upper bound (resp. lower bound) of A, ten A is said bounded from above (bounded from below resp.). If A is bounded from above and from below, ten A is called order bounded. Let x y M suc tat x y. We denote by x y : z M : x z y te order interval between x and y. It is obvious tat a subset A is order bounded if and only if it is contained in some order interval.

2 A sort introduction to Banac lattices and positive operators Definition.. A real vector space E wic is ordered by some order relation is called a vector lattice if any two elements x y E ave a least upper bound denoted by x y sup x y and a greatest lower bound denoted by x y inf x y and te following properties are satisfied. (L) x y implies x z y z for all x y z E, (L2) 0 x implies 0 tx for all x E and t. Let E be a vector lattice. We denote by E : x E : 0 x te positive cone of E. For x E let x : x 0 x : x 0 and x : x be te positive part, te negative part, and te absolute value of x, respectively. Two elements x y E are called ortogonal (or lattice disjoint) (denoted by x y) if x y 0. For a vector lattice E we ave te following properties (cf. [26, Proposition II..4, Corollary II.. and II..2] or [2, Teorem..]). Proposition..2 For all x y z E and a te following assertions are satisfied. (i) x y x y x y x y x y x y z x z y z and x y z x z y z (ii) x x x. (iii) x x x ax a x and x y x y. (iv) x x and te decomposition of x into te difference of two ortogonal positive elements in unique. (v) x y is equivalent to x y and y x. (vi) x y is equivalent to x y x y. In tis case we ave x y x y. (vii) x y z x z y z and x y z x z y z. x (viii) For all x y z E we ave x y z x z y z. (ix) x y x y x y and x y x z y z! " x z A norm on a vector lattice E is called a lattice norm if x y implies # x# $# y# for x y E y z. Definition..3 A Banac lattice is a real Banac space E endowed wit an ordering suc tat E is a vector lattice and te norm on E is a lattice norm.

. Banac lattices 3 For a Banac lattice E te following properties old (cf. [26, Proposition II.5.2] or [2, Proposition..6]). Proposition..4 Let E be a Banac lattice. Ten, (a) te lattice operations are continuous, (b) te positive cone E is closed, and (c) order intervals are closed and bounded. Sublattices, solids, bands and ideals A vector subspace F of a vector lattice E is a vector sublattice if and only if te following are satisfied. () x (2) x F for all x F, F or x F for all x F. A subset S of a vector lattice E is called solid if x S y x implies y S. Tus a norm on a vector lattice is a lattice norm if and only if its unit ball is solid. A solid linear subspace is called an ideal. Ideals are automatically vector sublattices since x y x y. One can see tat a subspace I of a Banac lattice E is an ideal if and only if x I implies x I and 0 y x I implies y I Consequently, a vector sublattice F is an ideal in E if x F and 0 y x imply y F. A subspace B E is a band in E if B is an ideal in E and sup M is contained in B wenever M is contained in B and as an upper bound (supremum) in E. Since te notion of sublattice, ideal, band are invariant under te formation of arbitrary intersections, tere exists, for any subset M of E, a uniquely determined smallest sublattice (ideal, band) of E containing M. Tis will be called te sublattice (ideal, band) generated by M. Next, we summarize all properties wic we will need in te sequel (cf. [2, Proposition..5,.2.3 and.2.5]). Proposition..5 If E is a Banac lattice, ten te following properties old. (i) If I I 2 are ideals of E, ten I I 2 is an ideal and if furtermore I and I 2 are closed, ten I I 2 is also a closed ideal. (ii) Te closure of every solid subset of E is solid. (iii) Te closure of every sublattice of E is a sublattice. (iv) Te closure of every ideal of E is an ideal. (v) Every band in E is closed.

4 A sort introduction to Banac lattices and positive operators (vi) For every non-empty subset A E, te ideal generated by A is given by I A n y y : n y x x r x x r A (vii) For every x E, te ideal generated by x is E x n x x : n Example..6. If E L p Ω µ p, were µ is σ-finite, ten te closed ideals in E are caracterized as follows: A subspace I of E is a closed ideal if and only if tere exists a measurable subset Y of Ω suc tat I ψ E : ψ x 0 a.e. x Y 2. If E C 0 X, were X is a locally compact topological space, ten a subspace J of E is a closed ideal if and only if tere is a closed subset A of X suc tat J ϕ E : ϕ x 0 for all x A Let E be a Banac lattice. If E e E olds for some e E, ten e is called an order unit. If E e E, ten e E is called a quasi interior point of E. It follows tat e is an order unit of E if and only if e is an interior point of E. Quasi interior points of te positive cone exist, for example, in every separable Banac lattice. Example..7. If E C K K compact, ten te function constant I K equal to is an order unit. In fact, for every f E, tere is n suc tat # f # n. Hence, f s ni K s for all s K. Tis implies f n I K I K. 2. If E L p µ wit σ-finite measure µ and p, ten te quasi interior points of E coincide wit te µ a.e. strictly positive functions, wile E does not contain any interior point. Spaces wit order continuous norm If te norm on E satisfies # x y# sup # x# # y# for x y E ten E is called an AM-space. Te above condition implies tat te dual norm satisfies Suc spaces are called AL-spaces. # x y # # x # # y # for x y Definition..8 Te norm of a Banac lattice E is called order continuous if every monotone order bounded sequence of E is convergent. E

. Banac lattices 5 One can prove te following result (cf. [2, Teorem 2.4.2]). Proposition..9 A Banac lattice E as order continuous norm if and only if every order interval of E is weakly compact. As a consequence one obtains te following examples. Example..0 Every reflexive Banac lattice and every L -space as order continuous norm. Te Banac space dual E of a Banac lattice E is a Banac lattice wit respect to te ordering defined by 0 x if and only if x x 0 for all x E A linear form x all 0 x (means 0 x and x 0). Te absolute value of x Han-Banac s teorem E is called strictly positive if x x 0 (notation: x 0) for x x sup y x : y x x E Te following results are consequences of te Han-Banac teorem. E being given by Proposition.. Let E be a Banac lattice. Ten 0 x is equivalent to x x 0 for all x E. Proposition..2 Let E be a Banac lattice. For eac 0 x E tere exists x E suc tat # x # and x x # x#. Proposition..3 In a Banac lattice E every weakly convergent increasing sequence x n is norm-convergent. Proof: Let A : n i a ix i : n a i 0 a! a n be te convex ull of x n : n. By te Han-Banac teorem, te norm-closure of A coincide wit te weak closure. Tis implies tat x A, were x : weak ε 0 tere exist y a x a n x n A a a n 0 a a n lim n x n. Tus, for suc tat # y x# ε. Since x k x, it follows tat # x x k # # x y# ε for all k n. Te following lemma will be useful in te proof of Proposition 2.5.3. Lemma..4 Let E be a totally ordered (tis means x E 0 x or x 0) real Banac lattice. Ten dime

6 A sort introduction to Banac lattices and positive operators Proof: Let e E and x E. We consider te closed subsets C : α : αe x and C : α : αe x of. It is obvious tat C C. Since is connected, it follows tat C C /0. Hence tere is α suc tat x αe. Complexification of real Banac lattices (cf. [26, II.]) It is often necessary to consider complex vector spaces (for instance in spectral teory). Terefore, we introduce te concept of a complex Banac lattice. Te complexification of a real Banac lattice E is te complex Banac space E wose elements are pairs x y E E, wit addition and scalar multiplication defined by x 0 y 0 x y : x 0 x y 0 y and a ib x y : ax by ay bx, and norm # x y # : # sup xsinθ 0 θ 2π ycosθ # One can sow tat te above supremum exists in E (cf. [26], p. 34). By identifying x 0 E wit x E, E is isometrically isomorpic to a real linear subspace of E. We write 0 x E if and only if x E. A complex Banac lattice is an ordered complex Banac space E tat arises as te complexification of a real Banac lattice E. Te underlying real Banac lattice E is called te real part of E and is uniquely determined as te closed linear span of all x E. Instead of te notation x y for elements of E, we usually write x iy. Te complex conjugate of an element z x iy E is te element z x iy. we use also te notation R z : x for z x iy E. Te modulus in E extends to E by x iy : sup xsinθ 0 θ 2π ycosθ All concepts first introduced for real Banac lattices ave a natural extension to complex Banac lattices. A complex Banac lattice as order continuous norm if its real part as..2 POSITIVE OPERATORS Tis section is concerned wit positive operators and teir properties. Let E F be two complex Banac lattices. A linear operator T from E into F is called positive (notation: T 0) if T E F, wic is equivalent to Tx T x for all x E Every positive linear operator T : E F is continuous (cf. [2, Proposition.3.5]). Furtermore, # T # sup # Tx# : x E # x# We denote by L E F te set of all positive linear operators from E into F. For positive operators one can prove te following properties.

.2 Positive operators 7 Proposition.2. Let T L E F (i) T x Tx and T x T x for all x E.. Ten te following properties old. (ii) If S L E F suc tat 0 S T (tis means tat 0 Sx Tx for all x E ), ten # S## T #. Let A D A be a linear operator on a Banac lattice E. It is a resolvent positive operator if tere is ω suc tat ω ρ A and 0 R λ A for all λ ω. A C 0 -semigroup on E is called positive if 0 T t for all t 0. Since R λ A 0 T t x e λt T t dt for λ lim n n t R n t A n x ω 0 A and for all x E and t 0 (cf. [2, Corollary 3.3.6]), it follows tat a C 0 -semigroup on a Banac lattice E is positive if and only if its generator is resolvent positive operator. For resolvent positive operators one as te following result (see [2, Teorem 3..8]). Teorem.2.2 Let E be a Banac lattice wit order continuous norm. If A is a resolvent positive operator, ten D A is an ideal in E. Proof: Since E is te complexification of a real Banac lattice E and R λ A E E λ ω we ave R z D A for z D A. Remark tat if I is a closed ideal of E, ten I ii is a closed ideal of E. Terefore we can suppose, witout loss of generality, tat E is a real Banac lattice. Moreover, we assume s A 0, by considering A ω instead of A oterwise. a) Let 0 y R 0 A x x E. We claim tat y D A. In fact, for λ 0 we ave 0 λr λ A y λr λ A R 0 A x R 0 A x R λ A x R 0 A x From Proposition..3 it follows tat 0 R 0 A x is weakly compact. Hence, tere is z E suc tat z weak lim λ λr λ A y. In particular, z D A weak (because D A D A ). Terefore, weak lim λ R 0 A y R λ A y weak lim λr λ A R 0 A y λ R 0 A z Since 0 R λ A y λr 0 A x, we ave R 0 A y R 0 A z and ence y z. b) Let y D A. Ten tere is y n D A suc tat lim n y n y. Moreover, tere exists x n E wit y n R 0 A x n and ten 0 y n R 0 A x n. Now a) implies tat y n D A and ence y D A.

8 A sort introduction to Banac lattices and positive operators c) Let 0 y x D A. Let x n x n D A. On te oter and, D A wit lim n x n x. From b) we ave y x n $ x n R 0 A Ax n R 0 A Ax n and a) implies tat y x n D A. Hence, y lim n y x n D A Positive operators on C K wit T I K I K are contraction operators (cf. [22, B.III. Lemma 2.]). Lemma.2.3 Suppose tat K is compact and T : C K C K is a linear operator satisfying T I K I K. Ten 0 T if and only if # T #. Proof: If 0 T, ten T f T f T # f # I K # f # I K Hence # T #. To prove te converse, we first observe tat I K f I K # f iri K # ρ r : r 2 for all r (.) Let f C K wit 0 f 2I K. Ten I K f I K I K. By (.) we ave # f I K iri K # ρ r for all r Since TI k I K and # T #, # T f I K iri K # ρ r for all r. So by (.) we obtain I K T f I K I K. Tis implies 0 T f 2I K. Lattice omomorpism and signum operators Let E F be two Banac lattices and T L E F. It is called lattice omomorpism if one of te following equivalent conditions is satisfied (cf. [2, Proposition.3.]). (a) T x y T x Ty and T x y T x Ty for all x y E. (b) T x T x x E (c) T x Tx 0 x E Te following result, due to Kakutani, sows tat for every e E te generated ideal satisfies E e C K for some compact K. Here, E e is equipped wit te norm # x# e : inf λ 0 : x λ e e x E e (cf. [2, Teorem 2..3]). Teorem.2.4 Let e E and take E e te ideal generated by e. Let B : x E e : e x and K ex B te set of all extreme points of B. Ten K is σ E E -compact and te mapping U e : E e x f x C K ; f x x x x x K, is an isometric lattice isomorpism.

#.2 Positive operators 9 If is a quasi interior point of E, ten E is a dense subspace of E isomorpic to a space C K. Consider te lattice isomorpism U from Kakutani s teorem. Let : U. Ten, U I K. Consider te operator S 0 : C K C K ; f sign f : f and put S : U S 0 U. Ten S is a linear mapping from E into itself satisfying (i) S, (ii) S x x for every x E, f (iii) S x 0 for every x E ortogonal to. Since (ii) implies te continuity of S for te norm induced by E and E E, S can be uniquely extended to E. Tis extension will be also denoted by S and is called signum operator wit respect to. We now give te following auxiliary result wic we need in Section 2.5. See [22, B.III. Lemma 2.3] for a similar result. Lemma.2.5 Let T R L E and assume tat is a quasi interior point of E. Suppose we ave R T, and Rx T x for all x E. Ten T S RS. Proof: It follows from Rx T x x E, tat T is a positive operator. Since T, E is T and R invariant. Consider te operators T : U TU R : U RU, and put : U. We ten ave R TI K I K R f T f for all f C K (.2) Define T : M RM, were M is te multiplication operator by on C K. By (.2) we ave T I K I K and T f M RM f RM f T M f T f (.3) for all f C K. Hence # T # # T # # T I K #. So by Lemma.2.3, T is a positive operator and (.3) implies tat 0 T T. Terefore, # T T # T T I K # 0. Since U U I K, it follows tat S 0 M. Tus, S U M U and T T implies tat T S RS.