aa r X i v : . [ m a t h . F A ] D ec Yosida duality
Bas Westerbaan
Radboud Universiteit Nijmegen [email protected]
January 26, 2018
Abstract
In this note we prove Yosida duality — that is: the category of compactHausdorff spaces with continuous maps is dually equivalent to the categoryof uniformly complete Archimedean Riesz spaces with distinguished unitsand unit-preserving Riesz homomorphisms between them.
For a compact Hausdorff space X , write C ( X ) for the vector space of real-valued continuous functions on it. The space C ( X ) contains sufficient additionalstructure to reconstruct X from it. In fact, there are several ways to do it.One particularly well-known method is using Gel’fand-duality: it states thatany commutative unital C ∗ -algebra is isomorphic to the algebra of complex val-ued continuous functions on some up-to-isomorphism unique compact Haus-dorff spaces. The categorical corollary is that the category of unital C ∗ -algebraswith unital ∗ -homomorphisms is dually equivalent to that of compact Hausdorffspaces.The proof of Gel’fand’s Theorem is rather involved — in this note we willpresent a lesser known method due to Yosida[Yos42] which will lead to a differentduality. Gel’fand uses the product on C ( X ) to reconstruct X . Yosida insteaduses the lattice order. At the end of this note, we will have shown that thecategory of compact Hausdorff spaces is dually equivalent to the category ofuniformly complete Archimedean unitary Riesz spaces.The present proof is based on a lecture series given by A. van Rooij in 2011of which notes can be found here[vR11]. The present text takes a short path toYosida duality; for a broad treatment of Riesz spaces, see e.g. [vR11, DJR77,Zaa12]. We will first define Riesz spaces and derive their elementary theory.
Definition 1.
1. A real ordered vector space E is a real vector spacewith a partial order ≤ such that for all x, y ∈ E with x ≤ y we have(a) x + a ≤ y + a for any a ∈ E and1b) λ · x ≤ λ · y for all scalars λ ≥ Riesz space is a real ordered vector space that is a lattice (i.e. all finiteinfima and suprema exist.)3. A linear map ϕ : E → E ′ between Riesz spaces is called a Riesz homo-morphism if ϕ preserves finite infima and suprema. Example 2.
1. For any topological space X , the vector space C ( X ) of con-tinuous real-valued functions on X is a Riesz space.2. Let ≤ lex be the lexicographic order on R — that is( x, y ) ≤ lex ( x ′ , y ′ ) ⇐⇒ x < x ′ or ( x = x ′ and y ≤ y ′ ) . Then R ordered by ≤ lex is a Riesz space.In a Riesz space the lattice operations interact nicely with other structure: Lemma 3.
For a Riesz space E with elements x, y, a ∈ E it holds1. λ ( x ∨ y ) = ( λx ) ∨ ( λy ) for scalars λ ≥ − ( x ∨ y ) = ( − x ) ∧ ( − y ) ( x + a ) ∨ ( y + a ) = ( x ∨ y ) + a x ∨ y + x ∧ y = x + y ( x ∧ y ) ∨ a = ( x ∨ a ) ∧ ( y ∨ a ) Proof.
The maps x λx and x x + a are order isomorphisms (if λ >
0) andthese preserve suprema, hence 1 and 3. Negation is an order anti-isomorphismand so 2. Point 4 follows from the previous two:( x ∨ y ) − x − y = ( x − x − y ) ∨ ( y − x − y ) = ( − y ) ∨ ( − x ) = − ( x ∧ y ) . In every lattice ( x ∧ y ) ∨ a ≤ ( x ∨ a ) ∧ ( y ∨ a ). For the other inequality, note( x ∨ a ) ∧ ( y ∨ a ) ≤ x ∨ a = x + a − x ∧ a ≤ x + a − x ∧ y ∧ a and so ( x ∨ a ) ∧ ( y ∨ a )+ x ∧ y ∧ a − a ≤ x . Similarly ( x ∨ a ) ∧ ( y ∨ a )+ x ∧ y ∧ a − a ≤ y .And so ( x ∨ a ) ∧ ( y ∨ a ) + x ∧ y ∧ a − a ≤ x ∧ y .( x ∨ a ) ∧ ( y ∨ a ) ≤ x ∧ y + a − x ∧ y ∧ a = ( x ∧ y ) ∨ a. The lattice-order on a Riesz space gives quite some structure.
Definition 4.
For any Riesz space E , define1. the absolute value | x | := x ∨ ( − x ) for x, y ∈ E ;2. the positive cone E + := { e ∈ E ; e ≥ } ;3. orthogonality x ⊥ y : ⇔ | x | ∧ | y | = 0 for x, y ∈ E and
4. the positive x + := x ∨ negative parts x − := ( − x ) ∨ C ( X ) the absolute value is point-wise; C ( X ) + are the positivevalued functions and f ⊥ g iff f and g have disjoint support. Lemma 5.
For any Riesz space E with x, y ∈ E we have1. x = x + − x − | x | = x + + x − | x | ≥ | x | = x + ∨ x − x + ⊥ x − | λx | = | λ || x | | x | = 0 = ⇒ x = 0 ( x + y ) + ≤ x + + y + | x + y | ≤ | x | + | y | x ∨ y = x + y + (cid:12)(cid:12) x − y (cid:12)(cid:12) Proof. As x = x + 0 = x ∨ x ∧ x ∨ − ( − x ) ∨ x + − x − we have 1. So x + + x − = 2 x + − x = 2( x ∨ − x = (2 x ) ∨ − x = (2 x − x ) ∨ ( − x ) = | x | hence 2. Clearly 0 ≤ ∨ x = x + and similarly 0 ≤ x − . So by the previous point0 ≤ x + + x − = | x | hence 3. So | x | = | x |∨ x ∨ ( − x ) ∨ x ∨ ∨ (( − x ) ∨
0) = x + ∨ x − , which is point 4. Also x + + x − = x + ∨ x − + x + ∧ x − thus we musthave x + ∧ x − = 0 viz point 5. For λ ≥ | λ || x | = λ ( x ∨ ( − x )) =( λx ) ∨ ( − λx ) = | λx | and as | x | = | − x | we establish 6. Note −| x | ≤ x ≤ | x | andso if | x | = 0 we have x = 0 i.e. point 7. Always x ≤ x + . So x + y ≤ x + + y + .Hence ( x + y ) + ≤ ( x + + y + ) + = x + + y + so 8. Similarly ( x + y ) − ≤ x − + x − .So 9 follows: | x + y | ≤ | x | + | y | . From Lemma 3 point 3 it follows x + y + | x − y | = x + y + ( x − y ) ∨ ( y − x )= ( x + y + x − y ) ∨ ( x + y + y − x ) = 2( x ∨ y ) , which (together with pt. 2) shows pt. 10 and completes the proof. Lemma 6 (Riesz Decomposition Lemma) . Let E be any Riesz space. Forany positive x, a, b ∈ E + with x ≤ a + b there is a decomposition x = a ′ + b ′ with ≤ a ′ ≤ a and ≤ b ′ ≤ b .Proof. Define a ′ := x ∧ a and b ′ := x − x ∧ a . Clearly x = a ′ + b ′ and a ′ ≤ a . Wehave to show b ′ ≤ b . Note b ′ = x − x ∧ a = x ∨ a − a . By assumption x ≤ a + b and so x ∨ a ≤ ( a + b ) ∨ a = a + b hence b ′ ≡ x ∨ a − a ≤ b .3 Riesz subspaces and ideals
To construct Riesz homomorphisms into R we study Riesz ideals. Definition 7.
Let E be a Riesz space. A subset D ⊆ E is called a Riesz ideal if it is the kernel of some Riesz homomorphism. The ideal D is said to be proper when D = E . The set D ⊆ E is called a Riesz subspace if x ∨ y, x ∧ y ∈ D whenever x, y ∈ D . Lemma 8.
A linear subspace D ⊆ E is a Riesz subspace if and only if x + ∈ D whenever x ∈ D .Proof. If D is a Riesz subspace, then clearly x + = x ∨ ∈ D whenever x ∈ D .The converse follows from x ∨ y = x + y + (cid:12)(cid:12) x − y (cid:12)(cid:12) and | x | = x + + ( − x ) + . Lemma 9.
Given a Riesz space E and a D ⊆ E , then TFAE:1. D is a Riesz ideal.2. D is a linear subspace and for all a ∈ D and x ∈ E :if | x | ≤ | a | , then x ∈ D .3. D is a Riesz subspace and for all a ∈ D + and x ∈ E + :if x ≤ a , then x ∈ D .Proof. ⇒
2. Assume D is a Riesz ideal, say it is the kernel of a Rieszhomomorphism f . Clearly D is a linear subspace. Assume | x | ≤ | a | forsome x ∈ E and a ∈ D . Note Riesz homomorphisms preserve absolute valuesas | x | ≡ x ∨ ( − x ) and so | f ( x ) | = f ( | x | ) ≤ f ( | a | ) = | f ( a ) | = 0 hence f ( x ) = 0.Thus x ∈ D , as desired.2 ⇒
3. Assume D is a linear subspace for which x ∈ D whenever | x | ≤ | a | for some a ∈ D . We only need to show D is a Riesz subspace. Assume a ∈ D .We have to show a + ∈ D . Recall a + ≤ a + + a − = | a | and so a + ∈ D , as desired.3 ⇒
1. Consider the linear quotient q : E → E/ D . We will turn E/ D into aRiesz space. We want to use q ( E + ) as positive cone for a partial order on E/ D ;that is we define q ( x ) ≤ q ( y ) ⇐⇒ q ( y ) − q ( x ) ∈ q ( E + ). This turns E/ D into a ordered vector space if q ( E + ) is closed under addition, scalar multipli-cation by positive reals and if we have q ( a ) , − q ( a ) ∈ q ( E + ) ⇒ q ( a ) = q (0).Clearly q ( E + ) is closed under addition and scalar multiplication by positivereals. Note q ( a ) ∈ q ( E + ) iff there is d ∈ D with d ≤ a . To show the last condi-tion, assume q ( a ) , − q ( a ) ∈ q ( E + ). Then d ≤ a and d ′ ≤ − a for some d, d ′ ∈ D .Hence 0 ≤ a − d ≤ − d ′ − d . By assumption on D we find a − d ∈ D . Thus d ∈ D and q ( a ) = q (0). Indeed q ( E + ) is a positive cone and we can order E/ D usingit. Automatically q is order preserving.To show E/ D is a Riesz space assume q ( x ) , q ( y ) ∈ E/ D . Clearly q ( x ∧ y ) ≤ q ( x ) and q ( x ∧ y ) ≤ q ( y ). Assume there is a q ( a ) ∈ E/ D with q ( a ) ≤ ( x ) and q ( a ) ≤ q ( y ). Then x − c ≥ d and y − c ≥ d ′ for some d, d ′ ∈ D .Consequently x ∧ y − c = ( x − c ) ∧ ( y − c ) ≥ d ∧ d ′ . As d ∧ d ′ ∈ D wesee q ( c ) ≤ q ( x ∧ y ). We have shown E/ D has pairwise infima. As x
7→ − x is anorder anti-isomorpism we see E/ D has pairwise suprema as well. We also saw q is a Riesz homomorphism and by construction D is its kernel. Lemma 10.
For any a ∈ E the set a ⊥ := { x ; x ∈ E ; x ⊥ a } is a Riesz ideal.Proof. We will first prove that a ⊥ is a linear subspace. Let x ∈ a ⊥ and λ ∈ R .Assume | λ | ≤
1. Then 0 ≤ | λx | ∧ | a | = | λ || x | ∧ | a | ≤ | x | ∧ | a | = 0 and so λx ⊥ a .In the other case | λ | ≥ n for some n ∈ N with n = 0 and so | λn | ≤
1. By theprevious | λn | ∧ | a | = 0. Hence 0 = | λ | ∧ ( n | a | ) ≥ | λ | ∧ | a | , as desired.Let x, y ∈ a ⊥ . By distributivity ( | x | ∨ | y | ) ∧ | a | = ( | x | ∧ | a | ) ∨ ( | y | ∧ | a | ) = 0so | x | ∨ | y | ⊥ a . By the previous 2( | x | ∨ | y | ) ⊥ a . Thus | x + y | ∧ | a | ≤ ( | x | + | y | ) ∧ | a | ≤ (2( | x | ∨ | y | )) ∧ | a | = 0 . Thus a ⊥ is a linear subspace. To show a ⊥ is a Riesz ideal, we will demonstratecondition 3 from Lemma 9. Assume 0 ≤ x ≤ y ∈ a ⊥ . Then 0 ≤ x ∧ | a | ≤ y ∧ | a | = 0 and so x ∈ a ⊥ — it remains to be shown that a ⊥ is a Riesz subspace.It is sufficient to show x + ∈ a ⊥ whenever x ∈ a ⊥ . Indeed from0 = | x | ∧ | a | = ( x + ∨ x − ) ∧ | a | = ( x + ∧ | a | ) ∨ ( x − ∧ | a | )it follows x + ⊥ a as desired. Definition 11.
1. An element u ∈ E + is called an (order) unit if for ev-ery x ∈ E there exists a n ∈ N such that | x | < n · u . A Riesz space iscalled unitary if it has a unit.2. An element ε ∈ E is called infinitesimal if there is a b ∈ E such that n · ε ≤ b for all n ∈ Z . A Riesz space is called Archimedean if its onlyinfinitesimal is 0.
Definition 12.
Given an Riesz space E with unit u ∈ E write k x k u := inf { λ ; λ ∈ R + ; | x | ≤ λu } . Lemma 13.
Let E be a Riesz space with units e, u .1. k k u is a seminorm.2. If k x k u = 0 , then x is infinitesimal.3. If E is Archimedean, then k k u is a norm and | x | ≤ k x k u u .4. k k e and k k u are equivalent. roof. First we’ll show k k u is a seminorm. Let ε > x ∈ E we have | x | ≤ ( k x k u + ε ) u . So | a + b | ≤ | a | + | b | ≤ ( k a k u + k b k u + 2 ε ) u .Hence k a + b k u ≤ k a k u + k b k u + 2 ε . Thus, as ε > k k u .Now we’ll show the absolute homogeneity. Let µ > | µa | = | µ || a | and similar reasoning as before we see k µa k u ≤ | µ |k a k u . To prove theother inequality first note that for any k a k u ≥ ε > | a | (cid:2) ( k a k u − ε ) u .Thus | µa | = | µ || a | (cid:2) ( | µ |k a k u − | µ | ε ) u . As we can make | µ | ε arbitrarily small wesee k µa k u ≥ | µ |k a k u . This gives absolute homogeneity except for the case µ = 0,which is trivially true. Finally, by definition k k u is positive, so it’s a seminorm.Assume k x k u = 0. Then | x | ≤ n u for any n ∈ N with n >
0. Hence n | x | ≤ u Now, for any m ∈ Z we have mx ≤ | mx | = | m || x | ≤ u and so x is infinitesimal.If E is Archimedean then directly by the previous we see k k u is a norm.Now we show | a | ≤ k a k u u . Suppose n ∈ N with n = 0. Then | a | ≤ ( k a k u + n ) u = k a k u u + n u . By taking the infimum over n ∈ N > , we see it sufficesto show inf { n u ; n ∈ N > } = 0. Clearly 0 is a lower-bound. Assume b ∈ E is another lower-bound, i.e. b ≤ n u for all n ∈ N > . Then b + ≤ ( n u ) + = n u so nb + ≤ n u and − nb + ≤ ≤ n u for all n ∈ N . Thus b + is infinitesimal:so b + = 0 and b ≤
0, as desired.Finally we show point 4. As for any ε > | x | ≤ ( k x k u + ε ) u weget k x k e ≤ ( k x k u + ε ) k u k e and so k x k e ≤ k x k u k u k e . Similarly k x k u ≤ k x k e k e k u and so k e k u k x k u ≤ k x k e ≤ k u k e k x k u , which shows k k u and k k e are equivalentnorms. Lemma 14.
Any non-zero Riesz homomorphism f : E → E ′ between unitalArchimedean Riesz spaces is continuous with respect to the norms induced byany unit.Proof. Pick any unit u ∈ E and u ′ ∈ E ′ . As all unit norms are equivalent it issufficient to show f is continuous with respect to k k u and k k u ′ . If k f ( u ) k u ′ =0, then f ( u ) = 0 and so f = 0 as −k x k u u ≤ x ≤ k x k u u for any x ∈ E .Thus k f ( u ) k u ′ >
0. Let ε > k x − y k u ≤ δ for some δ > | x − y | ≤ δu and so | f ( x ) − f ( y ) | = f ( | x − y | ) ≤ δf ( u ) = δ | f ( u ) | ≤ δ k f ( u ) k u ′ u ′ . Thus it is sufficient to choose δ ≤ ε k f ( u ) k u ′ . Definition 15.
An Archimedean unitary Riesz space is called uniformly com-plete if k k u is complete for some unit u ∈ E .As order-unit norms are equivalent we know that if E is complete withrespect to one unit, then it is complete with respect to all units. Definition 16.
We write
CAURiesz for the category of uniformly completeArchimedean Riesz spaces with distinguished unit as objects and Riesz homo-morphisms that preserve the selected unit as arrows between.6
Maximal Riesz ideals
The real numbers form a totally ordered Archimedean Riesz space and is, in fact,the only one. This Proposition will play the same rˆole as the Banach–MazurTheorem in the development of Gel’fand Duality.
Proposition 17. If E is a Riesz space with more than one element, then TFAE.1. E ∼ = R
2. The only ideals of E are { } and E .3. E is totally ordered and Archimedean.Proof. ⇒
2. Direct.2 ⇒
3. First we will prove E is totally ordered. Assume x, y ∈ E . Then J = { v ∈ E ; v ⊥ ( x − y ) + } is a Riesz ideal by Lemma 10. If J = { } thensince ( x − y ) − ⊥ ( x − y ) + , we know ( x − y ) − = 0 and thus 0 ≤ x − y . Hence x ≥ y .In the other case: if J = E , then ( x − y ) + ⊥ ( x − y ) + and thus ( x − y ) + = 0.Hence x − y = − ( x − y ) − ≤ x ≤ y .Now we will prove E is Archimedean. Assume ε is an infinitesimal of E . Wewill show ε = 0. Let b ∈ E such that n | ε | ≤ b for all n ∈ N . Note J := { x ∈ E ; n | x | ≤ b for all n ∈ N } is a Riesz ideal. If J = { } we are done. In the othercase J = E . Then b ∈ J and n | b | ≤ b ≤ n | b | for all n = 0. Thus 2 | b | = b = | b | so | ε | ≤ b = 0 and we are done.3 ⇒
1. Pick any a >
0. The map λ λa is a linear order isomorphismof R onto a 1-dimensional subspace of E . We are done if this subspace is E itself. Let b ∈ E + . Define r := sup { r ; r ∈ R + ; rb ≤ a } . As E is Archimedean,there is a n with nb (cid:2) a . Thus nb ≥ a and so r < ∞ . For any ε > r + ε ) b > a as E is totally ordered. Taking infimum over ε > a = rb thus indeed E is 1-dimensional.An important corollary is the following: Proposition 18.
Given an Archimedean Riesz space E with unit e . For ev-ery a ∈ E , a = 0 , there is a Riesz homomorphism ϕ : E → R such that ϕ ( e ) = 1 and ϕ ( a ) = 0 .Proof. It is sufficient to find such ϕ for | a | as 0 = ϕ ( | a | ) = | ϕ ( a ) | implies ϕ ( a ) =0. Thus assume a >
0. As E is Archimedean, there is an n ∈ N with na (cid:2) e .That is na − e (cid:2)
0. But na − e ≤ ( na − e ) + , so ( na − e ) + = 0.Consider the proper Riesz ideal J := { x ; x ⊥ ( na − e ) + } . Find using Zorn’sLemma a maximal proper Riesz ideal J ⊇ J . Then E/ J contains preciselytwo ideals and must be isomorphic to R . Define ϕ as the composition ofthis isomorphism with the quotient-map E → E/ J ∼ = R . Note ϕ ( e ) > ϕ = 0. Define ϕ := ϕ ϕ ( e ) . Clearly ϕ ( e ) = 1.Note ( na − e ) − ∈ J and so 0 = ϕ (( na − e ) − ) = ( nϕ ( a ) − ϕ ( e )) − whichimplies nϕ ( a ) − ϕ ( e ) = ( nϕ ( a ) − ϕ ( a )) + ≥
0. Thus nϕ ( a ) ≥ ϕ ( e ) > ϕ ( a ) = 0. 7 Function spaces
Example 19.
Let X be a compact Hausdorff space. Then C ( X ) with theobvious pointwise operations and order, is a Riesz space. Since X is compact,all these functions are bounded, ( x ) := 1 is a unit and k k is a supnorm forwhich C ( X ) is complete. It is also obvious that the zero function is the onlyinfinitesimal. Thus C ( X ) is Archimedean. Thus C ( X ) with unit is an objectof CAURiesz .Let X be a compact Hausdorff space and x ∈ X . Then point evaluation δ x : C ( X ) → R defined by δ x ( f ) = f ( x ) is a Riesz homomorphism into R with δ x ( ) = 1. Every unit-preserving Riesz homomorphism C ( X ) → R is ofthis form: Proposition 20.
Let X be a compact Hausdorff space. For any Riesz homo-morphism ϕ : C ( X ) → R with ϕ ( ) = 1 there is a point x ∈ X with ϕ = δ x .Proof. For any f ∈ C ( X ) define ˜ f := | f − ϕ ( f ) | . If we can find an x ∈ X with x / ∈ supp ˜ f := { x ; ˜ f ( x ) = 0 } for all f ∈ C ( X ), then we are doneas ˜ f ( x ) = 0 implies ϕ ( f ) = f ( x ) = δ x ( f ). Note that by definition ˜ f ≥ ϕ ( ˜ f ) = 0. For a single f ∈ C ( X ) there must be an x ∈ X with x / ∈ supp ˜ f (i.e. ˜ f ( x ) = 0) for otherwise ˜ f > f > ε forsome ε > ϕ ( ˜ f ) > ε quod non. Now, let f , . . . , f n ∈ C ( X )be given. Write g := ˜ f ∨ . . . ∨ ˜ f n . Note ϕ ( g ) = 0 and so there must also bean x ∈ X with g ( x ) = 0. As supp g = supp ˜ f ∪ . . . ∪ supp ˜ f n we see x / ∈ supp ˜ f i for every 1 ≤ i ≤ n . Thus no finite subset of the family of open sets A := { supp ˜ f ; f ∈ C ( X ) } can cover X . Thus by contraposition of compactness wesee that ∪A cannot cover X . Hence there is an x ∈ X with x / ∈ supp ˜ f forany f ∈ C ( X ) as desired.This Proposition is a cornerstone of the duality. Before we will look intothat, we will proof a variant of the Stone-Weierstrass theorem in the languageof Riesz spaces. Theorem 21 (Stone-Weierstrass) . For any compact Hausdorff space X andRiesz subspace D of C ( X ) such that:1. D is unital — that is: ∈ D D separates the points — that is: for all x, y ∈ X with x = y there isan f ∈ D with f ( x ) = f ( y ) .Then: D is k k -dense in C ( X ) .Proof. Let g ∈ C ( X ) and ε > g ∈ D with k g − g k ≤ ε . By approximating g + and − g − separately we see we may assume withoutloss of generality g ≥
0. We will approximate g progressively in three steps.Let y, z ∈ X be any points with y = z . By assumption there is an f ∈ D with f ( z ) = f ( y ). Define f y := g ( z ) f ( z ) − f ( y ) ( f − f ( y ) ). Note f y ( y ) = 0, f y ( z ) = g ( z ) and still f y ∈ D . 8rite U f y := { x ; x ∈ X ; f y ( x ) < g ( x ) + ε } . Clearly both z, y ∈ U f y .Thus the U f y are an open cover of X . By compactness we can find n ∈ N and y , . . . , y n such that U y ∪ · · · ∪ U y n = X . Define f ′ z := f y ∧ · · · ∧ f y n . Themap f ′ z ∈ D is a better approximation: f ′ z ( z ) = g ( z ) and f ′ z ≤ g + ε .Write V f ′ z := { x ; g ( x ) − ε < f ′ z ( x ) } , which are also open sets with z ∈ V f ′ z . Bycompactness we can find m ∈ N and z , . . . , z m with V f ′ z ∪· · · ∪ V f ′ zm = X . Nowdefine g := f ′ z ∨ · · · ∨ f ′ z m . Note g − ε ≤ g ≤ g + ε and so k g − g k ≤ ε . We can turn a compact Hausdorff space X into an Archimedean uniformlycomplete Riesz space C ( X ) — how do we go the other way? Definition 22.
Let E be an Archimedean Riesz space with a unit u . The spectrum of E is defined to be the setΦ( E ) := { ϕ ; ϕ : E → R Riesz homomorphism with ϕ ( u ) = 1 } . which is a topological space with induced topology of the inclusion Φ( E ) ⊆ R E where we take the product topology on the latter.At first glance, the spectrum seems to depend on the choice of unit. However: Claim 23.
Let E be an Archimedean Riesz space with units e, u ∈ E . We willwrite Φ e ( E ) and Φ u ( E ) for the spectra respective to e and u .Then: Φ e ( E ) ∼ = Φ u ( E ) via ϕ ( x ϕ ( x ) ϕ ( u ) ) . Lemma 24.
The spectrum Φ( E ) is a compact Hausdorff space.Proof. As for any x ∈ E it holds −k x k u ≤ ϕ ( x ) ≤ k x k u we have the followinginclusions of topological spacesΦ( E ) ⊆ Y x ∈ E [ −k x k u , k x k u ] ⊆ R E . The middle one is compact by Tychonoff’s Theorem theorem. Thus if we canshow Φ( E ) is closed in R E , we know Φ( E ) is compact and Hausdorff.Let ( ϕ α ) α in Φ( E ) be a net converging to some ϕ in R E . As addition, infi-mum, supremum and scalar multiplication are continuous the map ϕ must be aRiesz homorphism as well. Trivially ϕ ( u ) = lim α ϕ α ( u ) = 1. Thus indeed Φ( E )is closed in R E as desired. We are almost ready to state and prove Yosida duality. First a notationalmatter: recall the objects of
CAURiesz are pairs (
E, u ) of a uniformly completeArchimedean Riesz space E together with a distinguished unit u ∈ E . Forbrevity we’ll simply write E and denote the implicit distinguished unit by u E .9 heorem 25 (Yosida) . C and Φ are categorically dual:1. The map X C ( X ) extends to a functor CHsdf → CAURiesz ∂ taking as unit on C ( X ) and sending a continuous map f : X → Y to the Rieszhomomorphism C ( f ) : C ( Y ) → C ( X ) given by C ( f )( g ) = g ◦ f .2. The map E Φ( E ) extends to a functor CAURiesz → CHsdf ∂ by map-ping a Riesz homomorphism f : E → E ′ to Φ( f ) : Φ( E ′ ) → Φ( E ) givenby Φ( f )( ϕ ) = ϕ ◦ f .3. We have C Φ ∼ = id (every uniformly complete Archimedean Riesz spaceis naturally isomorphic to a continuous function space) and Φ C ∼ = id (every compact Hausdorff space is naturally isomorphic to the spectrumof an Archimedean Riesz space) and so CHsdf and
CAURiesz are duallyequivalent.Proof.
We first show 1 and 2.1. We already saw C ( X ) is in CAURiesz (Example 19 and that Φ( E ) isin CHsdf (Lemma 24). Maps on arrows defined by pre-composition (suchas C and Φ) are automatically functorial.2. Clearly C ( f )( g ) ≡ g ◦ f is continuous as it is the composition of twocontinuous maps. To finish 1, we still have to check C ( f ) is a Rieszhomomorphism. It is additive: C ( f )( g + g ′ )( x ) = ( g + g ′ ) ◦ f ( x ) = g ( f ( x ))+ g ′ ( f ( x )) = C ( f )( g )( x ) + C ( f )( g )( x ). In a similar fashion one checks C ( f )preserves scalar multiplication, ∨ and ∧ . Unit-preservation is different,but also simple: C ( f )( )( x ) = ◦ f ( x ) = 1 = ( x ).3. To wrap up 2 we need to check Φ( f ) is a continuous map for all f : E → E ′ .Let ( ϕ α ) α be a net in Φ( E ′ ) converging to ϕ . In particular we have forany x ∈ E that ϕ α ( f ( x )) → ϕ ( f ( x )). Hence Φ( f )( ϕ α ) ≡ ϕ α ◦ f → ϕ ◦ f ≡ Φ( f )( ϕ ) in Φ( E ), as desired.Now we turn to the proof of C Φ ∼ = id. Let E be in CAURiesz .4. First we define what will turn out to be an isomorphism. For x ∈ E ,define ˆ x (initially in R Φ( E ) ) by ˆ x ( ϕ ) := ϕ ( x ). To prove ˆ x is continuous,assume ϕ α → ϕ is a converging net in Φ( E ). Then ˆ x ( ϕ α ) = ϕ α ( x ) → ϕ ( x ) = ˆ x ( ϕ ), thus ˆ x ∈ C Φ( E ). For the moment, write ˆ E := { ˆ x ; x ∈ E } .5. Note that (ˆ x + ˆ y )( ϕ ) = ϕ ( x + y ) = ϕ ( x ) + ϕ ( y ) = ˆ x ( ϕ ) + ˆ y ( ϕ ) andthus ˆ x + ˆ y = [ x + y ∈ ˆ E . The same reasoning applies to · , ∧ and ∨ .Thus x ˆ x is a Riesz-homomorphism and ˆ E , being its image, is a Rieszsubspace of C Φ( E ).6. Suppose x ∈ E and x = 0. Then by Proposition 18, there is a ϕ ∈ Φ( E )such that ϕ ( x ) = 0. Thus ˆ x = 0. Thus x ˆ x is injective. Conse-quently E ∼ = ˆ E . 10. Given ϕ, ψ ∈ Φ( E ) with ϕ = ψ . By definition there is an x ∈ E suchthat ϕ ( x ) = ψ ( x ). Together with ˆ u E ( ϕ ) = ϕ ( u E ) = , we concludewith the theorem of Stone-Weierstrass (see Thm. 21) that ˆ E is k k -densein C Φ( E ). Since E ∼ = ˆ E is uniformly complete (i.e. k k -complete) wefind C Φ( E ) ∼ = E .8. We finish with naturality. Assume f : E → E ′ in CAURiesz , x ∈ E and ϕ ∈ Φ( E ′ ). Expanding definitions we find C Φ( f )(ˆ x )( ϕ ) = ˆ x (Φ( f )( ϕ )) = ˆ x ( ϕ ◦ f ) = ϕ ◦ f ( x ) = d f ( x )( ϕ ). Thus C Φ ∼ = id.Now we prove the other direction: Φ C ∼ = id. Let X ∈ CHsdf be given.9. Recall δ x ∈ Φ C ( X ) defined by δ x ( f ) ≡ f ( x ). Write ∆( X ) := { δ x ; x ∈ X } .10. Suppose x α → x in X . Then for f ∈ C ( X ) we have δ x α ( f ) = f ( x α ) → f ( x ) = δ x ( f ). Thus x δ x is continuous.11. Let x, y ∈ X with x = y be given. By Urysohn’s Lemma there is an f ∈ C ( X ) such that f ( x ) = f ( y ). Hence δ x = δ y . Thus x δ x is injective.12. Assume ϕ ∈ Φ C ( X ). By Proposition 20 there is an x ∈ X such that ϕ = δ x . Thus the map x δ x is surjective. Together with the previous twopoints we see x δ x is a continuous bijection between compact Hausdorffspaces and thus a homeomorphism.13. The demonstration of naturality is the same as in the previous Proposition.Thus Φ C ∼ = id as desired. Acknowledgements
Robert Furber kindly encouraged me to publish these notes. I am grateful toAbraham Westerbaan for his suggestions.
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