Tensor products in Riesz space theory

Transcription

1 Tensor products in Riesz space theor Jan van Waaij Master thesis defended on Jul 16, 2013 Thesis advisors dr. O.W. van Gaans dr. M.F.E. de Jeu Mathematical Institute, Universit of Leiden

2 CONTENTS 2 Contents Contents 2 Preface 4 Acknowledgments 5 List of smbols 6 1 Riesz space theor Partiall ordered vector spaces Morphisms Order ideals Relativel uniform topolog Norms on partiall ordered vector spaces Order denseness Free spaces Free vector spaces and free Riesz spaces Free normed Riesz space and free Banach lattice Riesz completion The Riesz completion of a directed partiall ordered vector space The Riesz completion of a pre-riesz space Morphisms Riesz* bimorphisms 29 5 The Archimedean completion The Archimedean completion of a partiall ordered vector space The Archimedean completion of a Riesz space The Archimedean Riesz tensor product 39 7 The integrall closed Riesz* tensor product 43 8 The positive tensor product Tensor cones Construction of the positive tensor product via a free Riesz space Construction of the positive tensor product via the Archimedean Riesz tensor product 50 9 Eamples Tensor product of R m and R n Tensor product of polhedral cones

3 CONTENTS 3 10 Fremlin space tensor product Definition and properties Construction of Fremlin space tensor product Banach lattice tensor product Definitions and properties Construction of the normed Riesz space and Banach lattice tensor product via a free normed Riesz space Fremlin space tensor product and Banach lattice tensor product Open problems 62 References 63 Inde 64

4 PREFACE 4 Preface The aim of m master project is to stud several tensor products in Riesz space theor and in particular to give new constructions of tensor products of integrall closed directed partiall ordered vector spaces, also known as integrall closed pre-riesz spaces, and of Banach lattices without making use of the constructions b D.H. Fremlin [3, 4]. To do this, we made use of socalled free Riesz spaces and free Banach lattices. Our second aim was to find out if positive linear maps as universal mappings are reall so natural as the seems to be. The so-called Riesz* homomorphism and bimorphisms seems to be more natural as universal mappings for pre-riesz spaces, since Riesz* homomorphisms etends to Riesz homomorphism between the Riesz completions. Therefore we define the integrall closed Riesz* tensor product. Under some conditions, Riesz* bimorphisms etends to Riesz bimorphisms between the Riesz completions and we can prove that the integrall closed Riesz* tensor product actuall eists. In the cases where it eists, it is equal to the usual tensor product of integrall closed pre-riesz spaces; the positive tensor product. It turns out that the positive tensor product E F of integrall closed pre-riesz spaces E and F has the nice propert that the Riesz completion of E F is the Archimedean Riesz tensor product of the Riesz completions E r and F r. Moreover, : E F E F is a Riesz* bimorphism and the restriction of the Riesz bimorphism : E r F r E r F r where we view E as a subspace of E r and F as a subspace of F r. We give a positive answer to the open problem in [7]; the Archimedean tensor cone is equal to the Fremlin tensor cone. After all we consider the tensor product of partiall ordered vector spaces with Fremlin norm and norm closed cone and we calculate the positive tensor product of polhedral cones.

5 ACKNOWLEDGMENTS 5 Acknowledgments I would like to thank Marcel de Jeu and Onno van Gaans for their support and help during m master project and b writing this thesis.

6 LIST OF SYMBOLS 6 List of smbols A l A u [, ] A A A E + E FBLA FNRSA FVSA FRSA K F K I K T E F L M L M L M r A : R B R A set of lower bounds of a set A set of upper bounds of a set A order interval for non-empt set A we have a for all a A for non-empt set A we have a for all a A A and A is orthogonal to supremum of and infimum of and absolute value of positive elements of E the space of all order bounded linear functionals on E free Banach lattice over non-empt set A free normed Riesz space over non-empt set A free vector space over set A free Riesz space over set A Fremlin tensor cone integrall closed tensor cone projective tensor cone positive tensor product or vector space tensor product of E and F Riesz space tensor product of L and M Archimedean Riesz space tensor product of L and M Banach lattice tensor product of L and M restriction map for subset A B j A : R RA R RB j A fξ = fξ A, f R RA, ξ R B ω ξ : FRSA R, ξ R A ω ξ f = fξ

7 1 RIESZ SPACE THEORY 7 1 Riesz space theor Man vector spaces have a natural ordering. For eample R or R n but also the continuous functions on R, CR, with the point-wise ordering. In this section we treat the most basic theor about Riesz spaces and partiall ordered vector spaces. For the standard theor and notation we refer to [1, 2]. 1.1 Partiall ordered vector spaces Definition 1.1. A partiall ordered vector space is a real vector space E with a partiall ordering defined on it, such that for all,, z E and α R + i implies that + z + z, ii impies that α α. The ordering on a partiall ordered vector space is called a vector space ordering. An alternative notation for is. If and we also write < or >. We call an element E positive if 0 and negative if 0. Two elements and are comparable if or and we denote this b. We denote b FinE the set of all finite subsets of E. The trivial ordering on a vector space E, is defined b if and onl if = for all, E. The set of all positive elements of a partiall ordered vector space E is denoted b E +. It is called the positive cone. Definition 1.2. Let E be a partiall ordered vector space. Let A E be a set. We call A i bounded from below if there is an E such that a for all a A, and we write A or A. ii bounded from above if there is an E such that for all a A we have that a. In this case we write A or A. iii order bounded if it is both bounded from below and from above. iv solid if A = A [, ], or equivalentl, A is solid, if [, ] A for all A [5, Definition 351I]. A linear subspace of E that is solid is called an order ideal or solid subspace. We define i the set of all lower bounds of A to be A l = { E : for all a A, a}, ii the set of all upper bounds of A to be A u = { E : for all a A, a }. We write A ul for A u l, etc. Remark 1.3. i [, ] if and onl if. ii If E is a Riesz space and I E a subspace, then is I an order ideal if and onl if for all E and I, implies that I. iii Ever ideal of a Riesz space is a Riesz subspace. iv l = u = E. Proposition 1.4. Let E be a partiall ordered vector space. i If A E then A ulu = A u, ii If E 0 then E l = E u =.

8 1 RIESZ SPACE THEORY 8 Proof. Let E be a partiall ordered vector space. i Let A E be a subset. Let A u. Then A ul, so A ulu, and hence A u A ulu. Let A ulu. Let a A, then is a A u, so a A ul, and therefore a. Hence A u, so A ulu A u. We conclude that A u = A ulu. ii Suppose E 0. If E has the trivial ordering, then the statements are clear. Suppose E has a non-trivial ordering. Then there is an E with < 0. Suppose that E l. Let E l. Then < 0. But then 2 < and 2 E, so E l. Contradiction. So E l =. There is an E, > 0. Suppose that E u. Let E u. Then > 0. But then 2 > and 2 E, thus E u. Contradiction. So E u =. Definition 1.5. Let E be a partiall ordered vector space. Let A E be a subset. An element E is called the infimum of A provided is a lower bound of A and, for all A l. Likewise E is called the supremum of A if it is an upper bound of A and, for all A u. Definition 1.6. A partiall ordered vector space E is called 1. directed or generating if for all, E there is a z E such that z and z, 2. Archimedean if for all, E with the propert that n for all n Z, we have that = 0, 3. integrall closed if for all, E with the propert that n for all n N 0, one has 0, 4. a pre-riesz space if E is directed and for all X FinE\{ } and for all E with + X u X u we have that 0 [9, Definition 1.1viii], 5. a Riesz space if for all, E the supremum and infimum of and eists, 6. a σ-dedekind complete Riesz space if for ever countable bounded set X in E the supremum sup X and infimum inf X of X eists. Proposition 1.7. For a partiall ordered vector space E the following statements hold i E is directed if and onl if E = E + E +, ii if E is integrall closed, then E is Archimedean. The converse is not true in general. iii Ever Riesz space is pre-riesz and ever directed integrall closed partiall ordered vector space is pre-riesz[9, Theorem 1.7]. iv E is a Riesz space if and and onl if one of the following is satisfied a eists, for all, E, or, b eists, for all, E, or, c for all E, the absolute value = of eists, or, d for all E, the positive part of, + = 0 eists, or, e for all E, the negative part of, = 0 eists. Follows directl from [1, Theorem 1.7]. v Let I be an inde set and let for ever i I, E i be a partiall ordered vector space. The point wise ordering on E := i I E i is given b i i I i i I if and onl if i i, for all i I. E is directed, pre-riesz, Riesz or σ-dedekind complete, respectivel if and onl if for each i I, E i is directed, pre-riesz, Riesz or σ-dedekind complete, respectivel. vi For a Riesz space the notions Archimedean and integrall closed are equivalent.

9 1 RIESZ SPACE THEORY 9 vii A σ-dedekind complete Riesz space is Archimedean. Proof. Let E be a partiall ordered vector space. We will onl prove i, ii, vi and vii. The other statements are trivial or we refer to the literature. i Suppose E is directed. Let E. Then there is a E with 0 and. Thus 0 and =, so we have E = E + E +. On the other hand, suppose E = E + E +. Let, E. Then there are 1, 2, 1, 2 E + such that = 1 2 and = 1 2. Hence and So E is directed. ii Suppose E is integrall closed. Let, E with n for all n Z. Then n and n for all n N 0. Thus 0 and 0. So = 0. It follows that E is Archimedean. For a counter-eample consider R 2 with 1, 2 1, 2 if 1 < 1 and 2 < 2, or 1, 2 = 1, 2. Suppose n 1, 2 1, 2 for all n Z. Then n 1 1 and n 2 2, for all n Z. Thus 1, 2 = 0. So R 2, is Archimedean. Note that n 1, 0 1, 1, for all n N 0 and 1, 0 0. Therefore R 2, is not integrall closed. vi B statement ii we onl need to prove that an Archimedean Riesz space E is integrall closed. Let, E and suppose that n for all n N 0. Then n + for all n Z. Thus + = 0 and 0. It follows that E is integrall closed. vii Suppose E is a σ-dedekind complete Riesz space. Let, E and suppose that n, for all n N 0. Then { n : n N} =: A. Since A is bounded and E is a σ-dedekind complete Riesz space i = inf A eists. We will prove that i = 0. Suppose i 0. Since 0 A, we have that i > 0. We have ni for all n N 0. Thus B := {ni : n N 0 } is a bounded set. Let s = sup B. Then n + 1i s, for all n N 0. So ni s i for all n N 0. Since s is the supremum of B we have that s s i so i 0. Therefore i > 0 and i 0 and that is a contradiction. We conclude that i = 0, thus 0. It follows that E is Archimedean. Proposition 1.8. Let E be a partiall ordered vector space. Then the following statements are equivalent. i E is integrall closed, ii 0 = inf { n : n N}, for all E +. Proof. Let E be a partiall ordered vector space. Assume i. Let E +. Clearl, 0 is a lower bound of { 1 n : n N}. Let { 1 n : n N} l. Then, n, for all n N0. Since E is integrall closed, we have that 0. So the infimum of { n : n N} eists and is equal to 0. Assume ii. Let, E, and assume that n for all n N 0. So 0 and is a lower bound of { n : n N}. Hence inf { n : n N} = 0. We conclude that E is integrall closed. Cones Cones in a vector space are in a one-to-one correspondence to the possible vector space orderings. Definition 1.9. Let E be vector space. A subset K E is called a cone provided 1. α, + K for all, K and α R +, 2. K K = 0. If K satisfies i, then it is called a wedge.

10 1 RIESZ SPACE THEORY 10 Proposition Let E be a vector space. Let K be a cone in E. Define on E b if and onl if K. Then is a partiall ordered vector space. On the other hand if is a vector space ordering, then E + is a cone. The vector space ordering and the ordering induced from E + coincide. Proof. Let E be a vector space and K a cone in E. Let be defined through if and onl if K. Let,, z E and suppose that and z. Then z, K. Thus z + = z K, so z. If and, then, K, so K K = 0, thus =. For all E we have. Thus is a partiall ordering. It is straightforward that is a vector space ordering. On the other hand, suppose is a vector space ordering on E. Note that +, α E +, for all, E + and α R +. Suppose E + E +, then 0 and 0. Thus = 0. Therefore E + is a cone. The last statement is trivial. Remark If is a vector space ordering induced from a cone K, we denote E, sometimes b E, K. 1.2 Morphisms Definition Let E and F be partiall ordered vector spaces and let φ : E F be linear, then we call φ 1. order bounded if for ever order bounded set A E, the set φa is order bounded, 2. positive or increasing if for all, E with one has φ φ or equivalentl, φ 0 as soon as 0, 3. bipositive if 0, for all E, is equivalent to φ 0, 4. an order isomorphism if it is bipositive and surjective, 5. regular or a difference of positive linear maps if there are positive linear maps ψ, ω : E F such that φ = ψ ω. If E and F are directed, then we call φ 1. a Riesz* homomorphism if φ{, } ul φ{, } ul for all, E [9, Definition 5.1], 2. a Riesz homomorphism if φ{, } u l = φ{, } ul for all, E [9, Definition 2.1i], 3. a complete Riesz* homomorphism if for ever set X in E bounded from above, we have that φx ul φx ul [9, Definition 5.11], 4. a complete Riesz homomorphism if for ever set X in E with inf X = 0, also inf φx = 0[9, Definition 2.1ii]. Remark i Ever positive operator is regular and ever regular operator is order bounded. Not ever order bounded linear map is regular, for a countereample, see the Eample of Lotz in [1, Eample 1.16]. ii Ever bipositive linear map is injective. iii Ever Riesz homomorphism is a Riesz* homomorphism[9, Remark 5.2i]. iv If E and F are Riesz spaces, then ever Riesz* homomorphism φ : E F is also a Riesz homomorphism[9, Remark 5.2ii]. v The composition of two Riesz homomorphism is not necessaril a Riesz homomorphism[9, Remark 2.3].

11 1 RIESZ SPACE THEORY 11 vi There eists an eample of a Riesz homomorphism φ : E F and a directed subspace D E, such that φ D : D F is not a Riesz homomorphism[9, Remark 2.10]. vii The composition of two Riesz* homomorphisms is a Riesz* homomorphism[9, Remark 5.2iii]. viii If E and F are Riesz spaces, then our definition of a Riesz homomorphism coincide with the usual definition of a Riesz homomorphism[9, Remark 2.2]. i A complete Riesz homomorphism φ : E F is not necessaril a Riesz homomorphism [9, Remark 2.2], but if E is pre-riesz, then φ is also a Riesz homomorphism [9, Corollar 2.7]. Theorem Let E be a directed partiall ordered vector space. Let I be an inde set. Let F i be a directed partiall ordered vector space for ever i I. Let, for ever i I, φ i : E F i be a linear map. Let F = i I F i with the pointwise ordering. Define φ : E F b φ i = φ i. Then i φ is a Riesz* homomorphism if and onl if for each i I, φ i is a Riesz* homomorphism. ii φ is a Riesz homomorphism if and onl if for each i I, φ i is a Riesz homomorphism. Proof. For the proof of ii see also [9, Theorem 2.14]. Note that for, E φ{, } ul = i I φ i {, } ul, φ{, } u l = i I φ i {, } u l and φ{, } ul = i I φ i {, } ul. Hence the Theorem follows. Proposition Let L and M be Riesz spaces. If φ : L M is an injective Riesz homomorphism, then φ is bipositive. Proof. Let L and M be Riesz spaces. Let φ : L M be an injective Riesz homomorphism. Clearl, φ is positive. Let L and φ 0. Then φ = φ 0 = φ φ0 = φ 0. Since φ is injective, we have that = 0 0. Thus φ is bipositive. Not ever injective positive linear map is bipositive, for eample f : R, {0} R, R +, is an injective positive linear map that is not bipositive. 1.3 Order ideals Recall that a subspace I of a partiall ordered vector space E is called an ideal, if for all I the order interval [, ] is contained in I. We will give some results here about ideals and what the are useful for. Proposition Let E and F be partiall ordered vector spaces. Let φ : E F be a positive linear map. Then ker φ is an order ideal. Proof. Let E and F be partiall ordered vector spaces. Let φ : E F be a positive linear map. Let ker φ. For all [, ], we have 0 = φ φ φ = 0. Hence φ = 0, therefore ker φ. We conclude that ker φ is an order ideal.

12 1 RIESZ SPACE THEORY 12 Theorem Let E be a partiall ordered vector space and let J E be an order ideal. Denote an equivalence class + J of E b []. Define a partiall ordering on E/J b, [] [] if there is an z J such that + z. This turns E/J into a partiall ordered vector space. We have for all X, Y E/J, X Y if and onl if there are, E such that, [] = X and [] = Y. Moreover E/J + = {[] : E + }. Define q : E E/J b []. If E is directed, then q is a Riesz homomorphism. Proof. For the first statements, see [5, Proposition 351J]. Suppose E is directed. Let, E. Then q{, } u l = {[z] : z {, } u } l {[], []} ul = q{, } ul. Since q is in particular positive, the converse inclusion holds b [9, Lemma 2.4]. Thus q is a Riesz homomorphism. We call q the quotient Riesz homomorphism and E/J the quotient space. Proposition See [10, Corollar ] and [11, Theorem 62.3]. Let L be a normed Riesz space and I a norm closed order ideal of L. Then L/I is a normed Riesz space with quotient norm [] = inf{ : L, [] = []}. If L is a Banach lattice, then L/I is also a Banach lattice. 1.4 Relativel uniform topolog Suppose I is an ideal of a Riesz space L, then L/I is also a Riesz space. Unfortunatel, even in the case that L is Archimedean, L/I need not be Archimedean. A sufficient and necessar condition for L/I to be Archimedean is that I is relativel uniforml closed, which will be defined net. Definition Let E be a partiall ordered vector space. A sequence { n } n 1 converges relativel uniforml ru-converges to E, if there eist a u E + and a sequence of positive real numbers {ε n } n 1 with ε n 0 such that ε n u n ε n u, for all n N. A set A E is relativel uniforml closed ru-closed if for ever sequence { n } n 1 A that is relativel uniforml convergent to an L, we have that A. Clearl, the intersection of an arbitrar collection of relativel uniforml closed sets if relativel uniforml closed. We define the relativel uniforml closure ru-closure of a set A to be the intersection of all relativel uniforml closed sets B that contain A. This collection is not empt since it contains E. Remark i If E is a Riesz space, then n converges relativel uniforml to, if there is a u E + and a real valued sequence {ε n } n 0, ε n 0 such that n ε n u, for all n N. ii Suppose L is a non-archimedean Riesz space. So there are, L such that n, for all n Z, but 0. Note that = 0. For all n N we have that n and n, thus n, for all n N 0. We conclude that 0 1 n, for all n N, and that the constant sequence {0} n 1 converges to 0. We see that {0} is not ru-closed and that that the constant sequence {0} n=1 converges to at least two different elements, namel 0 and. Theorem Let E be a partiall ordered vector space. If E + is relativel uniforml closed, then E is integrall closed. Proof. Let E be a partiall ordered vector space with relativel uniforml closed positive cone E +. Let, E be such that n, for all n N 0. Then 0 and 0 1 n, for all n N. Hence 1 n 1 n = 1 n, for all n N. We see that the sequence { 1 n } n 1 E + converges relativel uniforml to. Since E + is relativel uniforml closed we have that E +, thus 0 and E is integrall closed. Lemma Let E be a partiall ordered vector space. Let A E be a non-empt relativel uniforml closed set. Then for all E we have that + A = { + : A} is relativel uniforml closed.

13 1 RIESZ SPACE THEORY 13 Proof. Let E be a partiall ordered vector space. Let A E be a non-empt relativel uniforml closed set. Let E. Suppose {+ n } n=1 is a sequence in +A that converges relative uniforml to. Then, there eists a real sequence {ε n } n=1 and u E + such that for all n N. Thus for all n N we have that ε n u + n ε n u, ε n u n ε n u. It follows that { n } n=1 is a sequence in A that ru-converges to. Since A is ru-closed, we have that A, thus + A. Hence + A is ru-closed. Theorem Let E be a partiall ordered vector space. Let I be an order ideal of E. Then E/I is Archimedean if and onl if I is relativel uniforml closed. Proof. Let E be a partiall ordered vector space and let I be an order ideal of E. Let q : E E/I be the canonical quotient map. Suppose E/I is Archimedean. Let { n } n 1 be a sequence in I that relative uniform converges to E. B definition, there is a sequence of positive real numbers {ε n } n 1 and a u E + such that ε n 0 and ε n u n ε n u, for all n N. Thus ε n qu q ε n qu, for all n N. For ever m N there is an n m N such that 1 m ε n m, so 1 m qu q 1 mqu, for all m N. It follows that mq qu, for all m N. Since E/I is Archimedean we have that q = 0. Thus I and I is relativel uniforml closed. On the other hand, suppose that I is relativel uniforml closed. Let, E be such that n[] [], for all n Z, then [] 0. Therefore we ma assume that 0. We have for all n N, Thus there is an n I such that 1 n [] [] 1 n []. 1 n + n 1 n, for all n N. Hence { + n } n=1 is a sequence in + I that converges relativel uniforml to 0. B Lemma 1.22 on the preceding page + I is ru-closed and hence 0 + I, so [] = 0. We conclude that E/I is Archimedean. Theorem Let E be a partiall ordered vector space. Then E is Archimedean if and onl if ever relativel uniforml convergent sequence in E has a unique limit. Proof. First assume that E = E/{0} is Archimedean. B Theorem 1.23, {0} is ru-closed. Let { n } n 0 be a sequence in E that converges relativel uniforml to and in E. B definition there are sequences of positive real numbers {ε n } n 1 and {δ n } n 1 and u, v E + with ε n 0 and δ n 0 and ε n u n ε n u and δ n v n δ n v, for all n N. Thus δ n v n δ n v, for all n N. It follows that for all n N. We conclude that ε n u δ n v ε n u + δ n v, ε n δ n u + v 0 ε n δ n u + v, for all n N. Note that ε n δ n 0 and that u + v 0 and hence {0} n 0 ru-converges to. Since {0} is ru-closed, we have that = 0, that is =. We conclude that { n } n=1 has a unique limit. On the other hand, suppose that E is not Archimedean. Then there are, E such that n, for all n Z, but 0. We have 1 n 0 1 n for all n N. Thus the constant sequence {} n=1 converges to 0, but clearl it also converges to 0. So {} n=1 has at least two different limits.

14 1 RIESZ SPACE THEORY 14 Theorem Let E and F be partiall ordered vector spaces with F Archimedean. Let φ : E F be positive. Then ker φ is ru-closed. Proof. Let E and F be partiall ordered vector spaces with F Archimedean. Let φ : E F be positive. Let { n } n=1 be a sequence in ker φ that is ru-convergent to E. So there eists a real sequence {ε n } n=1, ε n 0 and a u E +, such that for all n N. Thus ε n u n ε n u, ε n φu φ ε n φu, for all n N. For ever m N, there is an n m N, such that ε nm 1 n. Thus 1 n φu φ 1 n φu, for all n N. Furthermore φu 0, so nφ φu, for all n Z. Since F is Archimedean, we have that φ = 0, that is ker φ. We conclude that ker φ is ru-closed. The relativel uniform topolog Clearl, if Λ is an inde set and, for all λ Λ, A λ is a ru-closed subset of a partiall ordered vector space E, then λ Λ A λ is ru-closed too. Triviall, and E are ru-closed subsets of E. Lemma Let E be a partiall ordered vector space and A and B ru-closed subsets of E, then A B is ru-closed. Proof. Let E be a partiall ordered vector space and let A and B be ru-closed subsets of E. Let { n } n=1 be a sequence in A B that ru-converges to E. Then there is a subsequence { nk } k=1 of { n } n=1 such that nk C, for all k N, where C is either A or B. It follows directl from the definition of a ru-convergent sequence that { nk } k=1 ru-converges to. Since C is ru-closed, we have that C A B. So A B is ru-closed. Now, the following theorem follows immediatel. Theorem Let E be a partiall ordered vector space. The complements of ru-closed sets of E form a topolog; the relative uniform topolog or ru-topolog T ru. A set O E is open if and onl if E\O is ru-closed. 1.5 Norms on partiall ordered vector spaces In this subsection we consider norms on partiall ordered vector spaces that respect the order structure. Definition Let E be a partiall ordered vector space. A norm ρ on E is called a Fremlin norm provided that for all, E with we have that ρ ρ. Definition Let L be a Riesz space. A Riesz norm or lattice norm on L is a norm on L such that if then, for all, L. The pair L, is called a normed Riesz space. If the induced metric b on L is complete, then we call L, a Banach lattice. Proposition Let L be a Riesz space and let ρ be a norm on L. Then ρ is a Riesz norm if and onl if ρ is a Fremlin norm and ρ = ρ, for all L. Proof. Let L be a Riesz space and ρ a norm on L. Suppose ρ is a Riesz norm on L. Suppose, L and. Then ± thus. Hence ρ ρ. So ρ is a Fremlin norm. Clearl, ρ = ρ, for all L. On the other hand suppose that ρ is a Fremlin norm and ρ = ρ, for all L. Let, L and suppose. Then, and. Thus. Hence. Since ρ is a Fremlin norm, we have that ρ ρ. Note that ρ = ρ thus ρ ρ. We conclude that ρ is a Riesz norm.

15 1 RIESZ SPACE THEORY 15 It follows that ever Riesz norm is a Fremlin norm. A Fremlin norm on a Riesz space is not necessaril a Riesz norm as the following eample shows. The eample is from [6, Eample 1.8i]. The condition that ρ = ρ for all L is reall necessar. Eample Consider R 2 with the standard ordering. Then R 2 is a Riesz space. Define ρ 1, 2 = Let = 1, 2, = 1, 2 R 2 and suppose that. Then and and Thus 1 1, 2 2 and It follows that ρ ρ. Hence ρ is a Fremlin norm. Note that 1, 1 = 1, 1 = 1, 1, but ρ1, 1 = 4 2 = ρ1, 1. Thus b Proposition 1.30 on the preceding page ρ is not a Riesz norm. Proposition Let E be a partiall ordered vector space with Fremlin norm ρ. Then E is Archimedean. If E + is closed under ρ, then E is integrall closed. Proof. Let E be a partiall ordered vector space with Fremlin norm ρ. Let, E be such that n, for all n Z. Thus 1 n 1 n, for all n N. So ρ ρ 1 n = 1 nρ, for all n N. Hence ρ = 0 and therefore = 0. It follows that E is Archimedean. Suppose further that E + is ρ-norm closed. Let, E be such that n, for all n N 0. We have n, for all n N, which ρ-converges to, as n. Since E+ is ρ-closed, we have that E +, that is 0. We conclude that E is integrall closed. Definition A Fremlin space is a partiall ordered vector space E with Fremlin norm ρ such that E + is closed under ρ. Remark E need not be directed. An eample is R with the trivial ordering and Fremlin norm. Ever Fremlin space can be considered as a subspace of a Banach lattice. Theorem 1.35 Van Gaans. [6, Corollar 4.8]. The following statements are equivalent. i F is a Fremlin space. ii There eists a Banach lattice F and an isometric bipositive linear map ι : F F. Remark Eample [6, Eample 4.9] shows that the embedding of F in F is not necessaril a Riesz homomorphism. 1.6 Order denseness Order denseness is a ver important notion in the theor about general partiall ordered vector spaces, in particular in the theor about Riesz and Dedekind completions. We give also a proof that order denseness is transitive, because there is no proof in the literature, as far as we know. Definition Let E be a partiall ordered vector space. A subspace D E is called order dense in E if for all E we have that = inf E {d D : d} and = sup E {d D : d }. The following eample is illustrative. Eample Let D L 1 R be the space of all integrable simple functions. Then D is order dense in L 1 R. The following characterization of order denseness is useful. Proposition Let E be a partiall ordered vector space and let D E be a subspace. Then the following statements are equivalent. 1. D E is order dense, 2. = inf E {d : d, d D}, for all E,

16 1 RIESZ SPACE THEORY = sup E {d : d, d D}, for all E. Proof. Clearl, 1 2 and : Suppose that 2 holds. Let E and note that {d D : d } = {d D : d} = { d : d, d D} = {d D : d}. Since 2 holds, we have sup E {d D : d } = inf E {d D : d} = =. 3 1: Similar to the proof of 2 1. Proposition Let G be a partiall ordered vector space, let E G be an order dense subspace and let F G be a subspace such that E F G. Then F G is order dense. Proof. Let E, F and G be as in the proposition. Let G. Note that = inf G {e E : e}. Triviall, {f F : f}. Let G with {f F : f}. Then {e E : e}, since {e E : e} {f F : f}. Thus = inf G {e E : e}. It follows that = inf G {f F : f}. In m bachelor thesis [14, Lemmas and Theorem 1.27], I proved that order denseness is transitive. That is, if E F is order dense and F G is order dense, then E G is order dense. To do this we need three lemmas. Lemma Let E be a partiall ordered vector space and let D be an order dense subspace of E. Suppose that S D is a subset such that the infimum of S in the space D eists and is equal to s. Then the infimum of S eists in the space E and is equal to s. Proof. Let E be a partiall ordered vector space and let D E be an order dense subspace of E. Let S D be a set, such that the infimum s of S in D eists. Note that s S in E. Suppose that t E and t S. Suppose that d D with d S, then d s = inf D S. If d D with d t, then d S, so d s = inf D S. Thus {d D : d t} {d D : d s}. Since D E is order dense, we have that t = sup E {d D : d t} sup E {d D : d s} = s. It follows that s is the infimum of S in E. Lemma Let E be a partiall ordered vector space and let D be an order dense subspace of E. Suppose that, E are such that. Then there is a d D such that d and d. Proof. Let E be a partiall ordered vector space and let D E be an order dense subspace. Let, E be such that. We argue b contradiction. Suppose that there is no d D with d and d. Then for an d D with d one has d. So {d D : d} {d D : d}. Therefore, since D E is order dense, = inf E {d D : d} inf E {d D : d} =. That contradicts our assumption that. Lemma Let G be a partiall ordered vector space and let E and F be subspaces of G such that E F G, E F is order dense and F G is order dense. Then for ever z G there eist an E such that z. Proof. Let E, F and G be as in the lemma. Let z G. Since z = inf G { F : z } we have { F : z } =. Let 0 { F : z }. Since 0 = inf F { E : 0 } we have { E : 0 }. Let 0 { E : 0 }. Then 0 0 z. Theorem Let G be a partiall ordered vector space. Let E and F be subspaces of G such that E F G, E F is order dense and F G is order dense. Then E G is order dense. Proof. Let E, F and G be as in the theorem. We argue b contradiction. So suppose that E G is not order dense. Then there is a z G such that z is not the infimum of S = { E : z } in G. B Lemma 1.43 we have S. Note that z S. Thus there is a z G with z S 1

17 1 RIESZ SPACE THEORY 17 and z z. Since F G is order dense, we have b Lemma 1.42 on the preceding page a F such that and z 2 z. 3 Since E F is order dense and F G is order dense we have b Lemma 1.41 on the previous page that = inf F { E : } = inf G { E : }. Let E, then z, b 2, thus S. From 1 follows that z. Since z S { E : } and = inf G { E : } we have that z. This contradicts 3. We are forced to conclude that E G is order dense. Theorem 1.44 on the preceding page implies that for ever finite sequence of order dense subspaces D 1 D 2... D n, D n D 1 is order dense. The following eample shows that Theorem 1.44 on the previous page does not hold in greater generalit, when we have an infinite decreasing sequence of order dense subsets. Eample Let for all D N C[0, 1] be the subspace of continuous functions f : [0, 1] R with the propert that f0 = f k 2, N N0, k = 0, 1,..., 2 N. We will show that D N N+1 D N is order dense for all N. Consider D := N=1 D N. Then this is a subspace of D 0. Consider f D. For all N N 0 and all k {0, 1,..., 2 N } we have f0 = f k 2. Since f is continuous N and { k : N N 2 N 0, k {0, 1,..., 2 N }} [0, 1] is topologicall dense, we have that f is constant f0. It is clear that D D 0 is not order dense. Let N N 0 and let f D N. Since f is continuous and [0, 1] is compact M := sup [0,1] f <. Define a sequence of functions f n n=1 D N+1 through f if f n = { ma { ma 4n2 N+1 k f 2 N M k } + M, f 4n2N+1 2N+1 k if 2 N+1 k 2 N n2 N+1, k {0, 1,..., 2N+1 1} k 2 N n2 N+1 < < k N+1 1 4n2 N+1, k {0, 1,..., 2N+1 1} 4n2 N+1 M f k N+1 1 4n2 N n2 N+1 k f k + 1 } 2 N+1 1, f 4n2N+1 if k N+1 1 4n2 N+1 k N+1,, {0, 1,..., 2N+1 1} N+1 + Then f n D N+1, for all n N, and f f n. Letting n, we see that inf DN {f n : n N 0 } = f. So inf DN {g : f g, g D N+1 } = f. We conclude that D N+1 D N is order dense. Instead of looking at a decreasing sequence, we can also view an increasing sequence D 1 D 2..., such that D n D n+1 is order dense, for all n N. Let X = n N D n, then D 1 X is order dense. Theorem Let {D n } n=1 be a sequence of partiall ordered vector spaces such that D n D n+1 is a order dense subspace, for all n N. Define X := n=1 D n. Then X is a partiall ordered vector space and D 1 X is an order dense subspace. Proof. Let {D n } n=1 and X be as in the theorem. For,, z X there is an n N such that,, z D n. So all aioms of a vector space are satisfied. Define on X b, if and onl if there is an n N such that, D n and in D n. Note that this is well defined, since for m N with, D m we either have that D n D m or D n D m, so that in both cases in D m. Note that satisfied all aioms of a partiall ordered vector space. Thus X, is a well

18 2 FREE SPACES 18 defined partiall ordered vector space. Suppose X. Then there is an n N such that D n. Note that D 1 D n is order dense; appl Theorem 1.44 on page 16. So = inf Dn {d D 1 : d}. Note that {d D 1 : d} in X. Suppose that X and {d D 1 : d}. There is an m > n such that D m. Appl Lemma 1.41 on page 16, m n times to get that = inf Dn {d D 1 : d} = inf Dn+1 {d D 1 : d} =... = inf Dm {d D 1 : d}. Thus and hence = inf X {d D 1 : d}. Therefore D 1 X is order dense. Remark Note that Theorem 1.44 on page 16 is a special case of Theorem 1.46 on the previous page, let D 1 = E, D 2 = F and for n 3 let D n = G. Theorem Let A be a directed inde set and let {E α } α A be an increasing net of partiall ordered vector spaces, such that E α E β is order dense for all α, β A with α β. Let X = α A and define on X through if and onl if there is an α A such that, E α and in E α, for all, X. Then X is a well defined vector space and is a well defined vector space ordering on X and E α X is order dense, for all α A. Proof. Let A be a directed inde set and let {E α } α A be an increasing net of partiall ordered vector spaces, such that E α E β is order dense, for all α, β A with α β. Let X = α A and define on X through if and onl if there is an α A such that, E α and in E α, for all, X. Suppose, E α and in E α and, E β, then there is a γ {α, β}. So, E γ E α, so in E γ. Since E β E γ, we also have that in E β. Thus on X is well defined. Let,, z X. Since A is directed, there is an α A such that,, z E α. So all aioma s of a vector space are satisfied for X and is a vector space ordering on X. Let α A. We will now prove that E α is an order dense subspace of X. Let X. Since A is directed, there is a β α such that E β. Since E α E β is order dense, we have that = inf Eβ {d E α : d}. Let X, {d E α : d}. There is a γ β such that E γ. Since E β E γ is order dense, we have b Lemma 1.41 on page 16 that = inf Eγ {d E α : d}, so. Thus, = inf X {d E α : d} and E α X is order dense. Theorem If F is a partiall ordered vector space and E is an order dense subspace of F, then E is integrall closed if and onl if F is integrall closed. Proof. Let F be a partiall ordered vector space and let E be an order dense subspace of F. Clearl, if F is integrall closed, then E is integrall closed. On the other hand, suppose E is integrall closed. Let, F and suppose that n, for all n N 0. Let 0 E with 0 and let 0 E with 0. Then n 0 0, for all n N 0. Since E is integrall closed, we have that 0 0. Thus for all 0 E with 0, we have that 0 0. Since = sup F { 0 E : 0 }, we have that 0. We conclude that F is integrall closed. Remark Suppose E is an order dense subspace of a partiall ordered vector space F. An interesting question is wether E is Archimedean if and onl if F is Archimedean. We could not find a proof. E α E α 2 Free spaces The free vector space over a set A are all linear combinations of elements of A. Thus A is the vector space basis for the free vector space. Something similar can be done for Riesz spaces, normed Riesz

19 2 FREE SPACES 19 spaces and Banach lattices. We define and stud these objects here. We need the free spaces for the construction of various tensor products. Let B be a set and let A B. We define r A : R B R A to be the restriction map, thus for f R B we define r A f = f A. It is clear that r A is a surjective Riesz homomorphism. Sometimes we write ξ A = r A ξ for ξ R B. We define j A : R RA R RB b j A fξ = fξ A = fr A ξ = fξ A, where f R RA and ξ R B. Then j A is an injective Riesz homomorphism. From Proposition 1.15 on page 11 follows that j A is bipositive. 2.1 Free vector spaces and free Riesz spaces In this subsection we stud the free vector space and the free Riesz space and we will show that the free vector space is a subspace of the free Riesz space. Definition 2.1. A free vector space over a set A is a pair V, ι where ι : A V is a map and V is a real vector space, such that for ever real vector space W and for ever map φ : A W there is a unique linear map φ : V W such that φ = φ ι. φ V W ι φ A Lemma 2.2. Let A be a set. Suppose V, ι and W, j are free vector spaces over A. Then there is a unique bijective linear map φ : V W such that j = φ ι. Proof. Let A be a set. Suppose that V, ι and W, j are free vector spaces over A. B definition there are unique linear maps ι : W V and j : V W such that ι = ι j and j = j ι. Thus ι = ι j ι. Note that also the identit map id V on V is a linear map such that ι = id V ι. From the uniqueness statement it follows that id V = ι j. Similarl, we have that the identit map id W on W is equal to j ι. Define φ = j : V W. Then φ is the unique isomorphism ψ : V W such that j = ψ ι. V j W ι ι j A Definition 2.3. A free Riesz space over a set A is a pair L, ι where ι : A L is a map and L is a Riesz space such that for ever Riesz space M and ever map φ : A M there is a unique Riesz homomorphism φ : L M such that φ = φ ι. L φ M φ A The net lemma can be found in [13, Proposition 3.3]. ι Lemma 2.4. A free Riesz space is unique if it eists, in the following sense: let L, ι and M, j be two free Riesz spaces over a set A, then there is a unique surjective Riesz isomorphism T : L M such that T ι = j. In particular T is an order isomorphism.

20 2 FREE SPACES 20 Proof. Let A be a set and suppose that L, ι and M, j are two free Riesz spaces over A. There is a unique Riesz homomorphism j : L M such that j = j ι and a unique Riesz homomorphism ι : M L such that ι = ι j. Note that ι = ι j = ι j j and ι j : L L is a Riesz homomorphism. Note that also that the identit map id L on L is a Riesz homomorphism such that ι = id L ι. From the uniqueness statement follows that id L = ι j. Likewise is the identit map id M on M satisfies id M = j ι. Define T = j : L M, then T is an invertible Riesz homomorphism and T 1 = ι is a Riesz homomorphism and T ι = j ι = j. Moreover T is the unique Riesz homomorphism with this properties. In particular T is an order isomorphism. j L M ι ι j A Theorem 2.5. [13, Proposition 3.2]. generated as Riesz space b ιa. If L, ι is a free Riesz space over a set A, then L is Proof. Let L, ι be a free Riesz space over a set A. Let M be the Riesz subspace of L generated b ιa. Define the map φ : A M b φa = ιa. B definition, there is a Riesz homomorphism φ : L M such that φ = φ ι. Let j : M L be the inclusion map. Then j φ : L L satisfies j φ ι = ι. B definition, the identit map on L, id L is the unique Riesz homomorphism ψ : L L that satisfies ι = ψ ι. Thus j φ = id L. But that implies that j is surjective, so M = L. We conclude that L is generated as Riesz space b ιa. φ L M j ι φ A For ever set the the free Riesz space eists, see [13, Proposition 3.7]. The case A = is trivial. Theorem 2.6. Let A be a set. If A =, then 0, is the free Riesz space over A. If A then FRSA, ι is the free Riesz space over A, where FRSA is the Riesz subspace of R RA generated b elements ξ a R RA, defined b ξ a f = fa for a A and f R A, and where ι : A FRSA is defined b a ξ a. Moreover ι is injective and FRSA is Archimedean. Lemma 2.7. Let A be a set. Let ι be as in Theorem 2.6. Then ιa is a linearl independent set. Proof. The case A = is trivial, so suppose that A. Let a 1,..., a n A be finitel man mutuall different elements. Let λ 1,..., λ n R be such that n i=1 λ iξ ai = 0. Then, for all f R A we have that n n λ i ξ ai f = λ i fa i = 0. i=1 Define f k R A b fa k = 1 and fa = 0 for a A\{a k }, where k {1,..., n}. Then n n λ i ξ ai f k = λ i f k a i = λ k = 0. i=1 Thus λ k = 0, for k {1..., n}. It follows that ξ 1,..., ξ n are linearl independent. i=1 i=1

21 2 FREE SPACES 21 Theorem 2.8. Let A be a set. If A =, then 0, is the free vector space over A. If A, let ξ a be as in Theorem 2.6 on the previous page, where a A. Let FVSA = Span{ξ a : a A} FRSA R RA. Define ι : A FVSA b a ξ a, where a A. Then FVSA, ι is the free vector space over A. Proof. Let A be a set. The statement is trivial if A =. So suppose that A has at least one element. Let FVSA and ι be as in the statement of the theorem. B Lemma 2.7 on the preceding page {ξ a : a A} is a linearl independent set. Let W be an arbitrar vector space and let φ : A W be a map. Define a linear map φ : FVSA W on its basis elements ξ a b ξ a φa, where a A. Thus φ = φ ι. For an other linear map ψ : FVSA W that satisfies φ = ψ ι, we have that ψξ a = φa = φ ξ a. Since the ξ a generate FVSA as vector space, we have that ψ = φ thus φ is unique. We conclude that FVSA, ι is the free vector space over A. Remark 2.9. We view the free vector space as a subspace of the free Riesz space. Theorem Let B be a set and let A B be a subset. Let FRSB, ι B be the free Riesz space over B. Let FRSA be the Riesz subspace of FRSB generated b the elements ι B a, where a A, and let ι A = ι B A. Then FRSA, ι A is the free Riesz space over A. Proof. Let B be a set and let A B be a subset. The map j A : R RA R RB is an injective Riesz homomorphism, and hence j A FRSA FRSB is an injective Riesz homomorphism. Proposition Let A be a finite set. Let FRSA, ι be the free Riesz space over A. Then ιa is a strong order unit for FRSA. a A Proof. Let A be a finite set with free Riesz space FRSA, ι. Then the statement is clear from the fact that A is finite and that FRSA is generated b elements ιa. Proposition Let A be a non-empt set and let FA denote the set of all finite subsets of A. Then FRSA = FRSB, B FA where we view FRSB as a Riesz subspace of FRSA. Proof. Let A be a non-empt set with free Riesz space FRSA, ι. From Theorem 2.10 it is clear that FRSB FRSA. B FA On the other hand ever element FRSA is generated b finitel man elements ι a1,..., ι an, so FRS{a 1,..., a n } and hence FRSA = FRSB. B FA Proposition Let A be a set. 1. If ξ R A, then ω ξ : FRSA R defined b ω ξ f = fξ, where f FRSA, is a Riesz homomorphism. 2. If ω : FRSA R is a Riesz homomorphism, then there is a ξ R A such that ω = ω ξ. Here, we view FRSA as a Riesz subspace of R RA. Proof. The first statement is trivial. Suppose ω : FRSA R is a Riesz homomorphism. Define ξ R A b ξa = ωιa, for a A. Then for a A we have ωιa = ξa = ιaξ = ω ξ ιa. Thus ω and ω ξ coincide on the set {ιa : a A} that generates FRSA as Riesz space hence ω = ω ξ.

22 2 FREE SPACES Free normed Riesz space and free Banach lattice Something similar to free Riesz space eists for Banach lattices. It is eas to generalize the results of De Pagter and Wickstead for Banach lattices [13] to normed Riesz spaces. We give an overview of the main results. Definition Let A be a non-empt set and let X be a normed space. A map φ : A X is norm bounded, if there eists an M > 0 such that φa M, for all a A. We define the norm of a norm bounded map φ : A X to be φ = sup{ φa : a A}. Remark Note that is norm on the vector space of all norm bounded maps φ : A X. Definition Let A be a non-empt set. The free normed Riesz space free Banach lattice over A is a pair L, ι where L is a normed Riesz space Banach lattice and ι : A L is a bounded map such that for an normed Riesz space Banach lattice M and for an bounded map φ : A M there is a Riesz homomorphism φ : L M with the propert φ = φ. Moreover φ is the unique Riesz homomorphism ψ : L M that satisifies φ = ψ ι. L φ M φ A From the fact that ι = id L follows that ι = ι = 1. But even more is true. ι Lemma [13, Proposition 4.2]. Let A be a non-empt set, and suppose that L, ι is a free Banach lattice or free normed Riesz space over A. Then ιa = 1, for ever a A. Proof. Let A be a non-empt set, and suppose that L, ι is a free Banach lattice over A. Define j : A R b a 1, a A. Then j = j = 1. Further, 1 = ja = j ιa j ιa = ιa, thus ιa 1, for all a A. Since ι = 1 we have ιa 1, for all a A. Therefore ιa = 1, for all a A. The case that L, ι is a free normed Riesz space over A is proven similarl. L j R j A ι L ι L ι A ι The free Banach lattice or normed Riesz space is unique if it eists, see [13, Proposition 4.3]. Lemma Let A be a non-empt set, and suppose that L, ι and M, j are free Banach lattices over A free normed Riesz spaces over A. Then there eists a unique isometric order isomorphism φ : L M, such that φ ι = j. Proof. Let A be a non-empt set, and suppose that L, ι and M, j are free Banach lattices over A. Let j : L M be the unique Riesz homomorphism such that j = j ι. Then b Lemma 2.17, j = j = 1. Similarl, there is a unique Riesz homomorphism ι : M L of norm 1, such that ι = ι j. So ι j : L L is a Riesz homomorphism and ι = ι j = ι j ι. From the uniqueness statement in the theorem follows that ι j is the identit map on L. Similarl, j ι is the identit map on M. So ι is an isometric order isomorphism. The case that L, ι is a free normed Riesz space over A is proven similarl. j L M ι ι j A

23 2 FREE SPACES 23 For an ordered vector space E we denote b E the space of all order bounded linear functionals on E. According to [1, Theorem 1.18] E is equal to the space of regular linear functionals as soon as E is an Archimedean Riesz space. De Pagter and Wickstead define in [13] a lattice norm F on FRSA, that turns FRSA into the free normed Riesz space over A and its norm completion is the free Banach lattice over A. We will now review this construction. Definition [13, Definition 4.4] For a non-empt set A we define a map : FRSA [0, ] b φ = sup{ φ ιa : a A}. Let FRSA = {φ FRSA : φ < }. The following is clear from the definition. Lemma FRSA is a vector lattice ideal in FRSA. Lemma Let A be a non-empt set. Let ξ R A. Let ω ξ : FRSA R be defined through ω ξ = ξ, FRSA See Proposition 2.13 on page 21. Then ω ξ < if and onl if ξ is bounded. Or, equivalentl, ω ξ FRSA if and onl if ξ is bounded. Proof. Let ξ R A. Since ξa = ιaξ = ω ξ ιa = ω ξ ιa = ω ξ ιa, for all a A, it follows that ω ξ < if and onl if ξ is bounded. Lemma If A is a non-empt set, then is a Riesz norm on FRSA. Proof. Let A be a non-empt set. is clearl a Riesz seminorm. Suppose that φ = 0. Then φ ιa = 0, for all a A, thus φ ιa = 0, for all a A. Let FRSA. B Proposition 2.12 on page 21 there are finitel man a 1,..., a n A such that FRS{a 1,..., a n }. B Proposition 2.11 on page 21 e = n i=1 ιa is a strong order unit for FRS{a 1,..., a n }. Thus there is an λ R + such that λe. It follows that φ φ λ φ e = 0 and hence φ = 0. So φ = 0. We conclude that is a Riesz norm. Definition For a non-empt set A and for FRSA we define F = sup{φ : φ FRSA +, φ 1}. Theorem If A is a non-empt set, then F is a lattice norm on FRSA. Proof. From the definition it is clear that for, FRSA with we have that F F and that F is positive homogenious and subadditive. First we will show that for all FRSA we have that F <. Let FRSA. B Proposition 2.12 on page 21 there are finite a 1,..., a n A such that FRS{a 1,..., a n }. B Proposition 2.11 on page 21 e = n i=1 ιa i is a strong order unit of FRS{a 1,..., a n }. So there is a λ > 0 such that λe. So F λ e F. For φ FRSA +, φ 1, we have φe = n i=1 φ ιa i n. Thus e F n and hence F λn <. It onl remains to show that for all FRSA we have that F = 0 implies that = 0. Let FRSA with F = 0. Clearl, for all φ FRSA + we have φ = 0. In particular ω ξ = ξ = 0, for ever ξ R A. B Proposition 2.12 on page 21 there is a finite subset B of A such that FRSB. In particular for ever ξ R B R A we have that ξ = 0. But that means that = 0, since FRSB R RB. Theorem Let A be a non-empt set. Then FRSA, F, ι is the free normed Riesz space over A.

24 3 RIESZ COMPLETION 24 Proof. Let A be a non-empt set. Consider the free Riesz space over A, FRSA, ι. Let M be an arbitrar normed Riesz space and let φ : A M a bounded map that maps into M 1, the closed unit ball of M, and suppose that φ = 1. B the definition of the free Riesz space, there is a unique Riesz homomorphism φ : FRSA M such that φ = φ ι. It onl remains to prove that φ = φ. B Lemma 2.17 on page 22 we have for all a A, ιa = 1. So φ φ ιa = φa, thus φ φ. Suppose that φ > φ. So for some FRSA with = 1 is φ = φ > φ. B Hahn-Banach there eists a positive functional ψ on M of norm at most one and ψφ > φ. We have ψ φ = sup{ ψ φ ιa : a A} = sup{ ψ φ ιa : a A} = sup{ ψφa : a A} 1. Thus F ψφ > φ = 1. This is a contradiction. Therefore φ = φ. Suppose φ : FRSA M is a bounded map and φ > 0. Then φ = φ/ φ is of norm one and maps into M 1. It is clear that φ = φ / φ, so φ = φ. Suppose φ : FRSA M is zero. Then φ = 0, so φ = φ. This concludes our proof. Remark We denote the free normed Riesz space over a non-empt set A b FNRSA. For the net theorem see [1, Theorems 4.1 and 4.2]. Theorem Let E be a normed Riesz space and let E denote its norm dual space. Then E is a Banach lattice. Consider E as a subspace of its double dual E = E. Then the norm completion of E is a Banach lattice. Theorem Let A be a non-empt set. Let FBLA be the F -norm completion of FRSA. Then FBLA is a Banach lattice and FBLA, ι is the free Banach lattice over A. Proof. Let A be a non-empt set and let FBLA be the F -norm completion of FRSA. B Theorem 2.27, FBLA is a Banach lattice. Suppose M is Banach lattice and φ : A M a bounded map. B Theorem 2.25 on the previous page there is a Riesz homomorphism φ : FRSA M such that φ = φ ι and φ = φ, moreover φ is the unique Riesz homomorphism ψ that satisfies φ = ψ ι. The Riesz homomorphism φ etends b continuit to a Riesz homomorphism φ : FBLA M and we still have φ = φ ι and φ = φ. Suppose ψ : FBLA M is also a Riesz homomorphism that satisfies φ = ψ ι. Then ψ and φ coincide on FRSA. Since the continuous etension is unique, we have that ψ = φ. So FBLA, ι is the free Banach lattice over A. 3 Riesz completion Riesz spaces have man nice properties and there is an abundance of theor about them. That is the reason that one like, to embed partiall ordered vector spaces in a nice wa into Riesz spaces. It turns out that ever pre-riesz space can be bipositivel embedded into a Riesz space, its Riesz completion, such that the embedding is order dense and the Riesz completion is generated as a Riesz space b the pre-riesz space. Of course we could make use of the so-called Dedekind completion but then we should require that the space is integrall closed. Moreover, the Dedekind completion ma be too large. We give a short overview of the construction and theor of the Riesz completion. We follow the Ph.D. thesis of Van Haandel [9]. 3.1 The Riesz completion of a directed partiall ordered vector space We start with a definition, taken from [9, Definition 3.1]. Definition 3.1. Let E be a directed partiall ordered vector space. A pair L, φ, consisting of a Riesz space L and a Riesz homomorphism φ : E L, is called a Riesz completion of E if for