aa r X i v : . [ m a t h . F A ] J u l TRANSFORMATIONS OF MOMENT FUNCTIONALS
PHILIPP J. DI DIO
Abstract.
In measure theory several results are known how measure spacesare transformed into each other. But since moment functionals are repre-sented by a measure we investigate in this study the effects and implicationsof these measure transformations to moment funcationals. We gain charac-terizations of moments functionals. Among other things we show that for acompact and path connected set K ⊂ R n there exists a measurable func-tion g : K → [0 ,
1] such that any linear functional L : R [ x , . . . , x n ] → R isa K -moment functional if and only if it has a continuous extension to some L : R [ x , . . . , x n ]+ R [ g ] → R such that ˜ L : R [ t ] → R defined by ˜ L ( t d ) := L ( g d )for all d ∈ N is a [0 , f : [0 , → K independent on L such that the representing measure ˜ µ of ˜ L provides the representing measure˜ µ ◦ f − of L . We also show that every moment functional L : V → R isrepresented by λ ◦ f − for some measurable function f : [0 , → R n where λ is the Lebesgue on [0 , Contents
1. Introduction 12. Preliminaries: Measure Theory and the Lebesgue Integral 43. Transformations of linear Functionals: Basic Properties 74. Non-trivial Transformations of linear Functionals 95. Conclusion and Open Questions 20Appendix A. Daniell’s Representation Theorem 22References 251.
Introduction
Linear functionals L : V → K with K = R or C belong to the most importantstructures in mathematics, e.g. for separation arguments. If V is a vector space offunctions v : X → K then L is called a moment functional if it is represented by a(non-negative) measure µ on X : L ( v ) = Z X v ( x ) d µ ( x ) for all v ∈ V . If supp µ ⊆ K ⊆ X , then L is called a K -moment functional.Among the moment functionals the most important ones act on polynomials V = R [ x , . . . , x n ] on some K ⊆ R n , n ∈ N . Here, the name moment actuallycomes from. If K is closed then Haviland’s Theorem [Hav35, Hav36] states thata linear functional L : R [ x , . . . , x n ] → R is a K -moment functional if and only if Technische Universit¨at Berlin, Institut f¨ur Mathematik, Straße des 17. Juni 136,D-10623 Berlin, Germany
E-mail address : [email protected] .2010 Mathematics Subject Classification.
Key words and phrases. moment functional, representation, measure. L ( p ) ≥ p ∈ R [ x , . . . , x n ] with p ≥
0. On the other side p ≥ K if andonly if L ( p ) ≥ K -moment functionals L since every point evaluation is amoment functional. These are the two directions in the duality theorem and themany connections between the moment problem (deciding when a linear functionalis a moment functional) and non-negative polynomials (and therefore optimizationand many other applications) only start here. See e.g. [AK62], [Akh65], [dDS18],[CF00], [Fia16], [KN77], [Las15], [Lau09], [Mar08], [Sch17], and references withinfor more on the moment problem, the connection to non-negative polynomials, andapplications.Besides the one-point evaluation L ( f ) = f ( x ) the following is probably the sim-plest moment functional. Example 1.1.
Let λ be the Lebesgue measure on [0 ,
1] and let V = R [ t ]. Thenthe functional L Leb : R [ t ] → R with L Leb ( t d ) = Z t d d λ ( t ) = 1 d + 1 for all d ∈ N , (1)is the unique linear functional such that L ( t d ) = d +1 holds for all d ∈ N . ◦ Besides this the general [0 , Hausdorff Moment Problem 1.2 (see [Hau21] or [KN77, Thm. 1.1 and 1.2]) . (a) Let d ∈ N . The following are equivalent.(i) L : R [ x ] ≤ d → R is a [0 , -moment functional.(ii) L ( p ) ≥ holds for all p ∈ R [ x ] ≤ d such that p ≥ on [0 , .(b) The following are equivalent.(i) L : R [ x ] → R is a [0 , -moment functional.(ii) L ( p ) ≥ holds for all p ∈ R [ x ] such that p ≥ on [0 , . This problem is fully solved since by the univariate Positivstellensatz every poly-nomial p ∈ R [ x ] which is non-negative on [0 ,
1] has the form p ( x ) = p ( x ) + x · (1 − x ) · p ( x ) = q ( x ) + x · q ( x ) + (1 − x ) · q ( x )for some p i , q i ∈ P R [ x ] sums of squares. This also holds with the degree bounddeg p ≤ d .In higher dimensions the problem is not completely solved and several problemsappear, especially since in R n with n ≥ X , . . . , X n ) of pairwise commuting andsymmetric multiplication operators must have an extension to pairwise commutingand self-adjoint multiplication operators ( X , . . . , X n ).To understand moment functionals better and to simplify them we investigatein this article the possibility of transforming a linear (moment) functional into an-other linear (moment) functional based on several isomorphism and transformationresults between measure spaces. But before we give the formal definition of a trans-formation of a linear functional let us have a look at the following theorem to seewhat kind of results we are looking for. Theorem 1.3.
Let S be a Souslin set (e.g. a Borel set S ⊆ R n ), V be a vectorspace of real measurable functions v : S → R , and L : V → R be a linear functional.Then the following are equivalent:(i) L : V → R is a S -moment functional.(ii) There exists a measurable function f : [0 , → S such that L ( v ) = Z v ( f ( t )) d λ ( t ) (2) RANSFORMATIONS OF MOMENT FUNCTIONALS 3 for all v ∈ V where λ is the Lebesgue measure on [0 , , i.e., λ ◦ f − is arepresenting measure of L .Proof. (i) → (ii): Let µ be a representing measure of L . By Corollary 2.15 thereexists a measurable function f : [0 , → S such that µ = λ ◦ f − and hence L ( v ) = Z S v ( x ) d µ ( x ) = Z S v ( x ) d( λ ◦ f − )( x ) Lemma 2.1 = Z v ( f ( t )) d λ ( t ) ( ∗ )for all v ∈ V .(ii) → (i): λ ◦ f − is a representing measure of L by Lemma 2.1. (cid:3) Theorem 1.3 can be seen as a complete characterization of ( S -)moment function-als, i.e., every moment functional L : V → R has the form (2) for some f : [0 , → S .Additionally, Theorem 1.3 also shows that every moment functional L is representedby λ ◦ f − for a measurable function f : [0 , → R n .Hence, the aim of this paper is to characterize and represent moment functionalsin the form of (2) and especially to find additional properties of f : [0 , → S .The notation and result in Theorem 1.3 stimulate the notation of a transforma-tion of a linear (moment) functional. We introduce the following definitions. Definition 1.4.
Let X and Y be two Souslin spaces, U and V two vector spacesof real measurable functions on X resp. Y , and K : U → R and L : V → R betwo linear functionals. We say L (continuously) transforms into K , symbolized by L K resp. L c K , if there exists a Borel (resp. continuous) function f : X → Y such that
V ◦ f ⊆ U and L ( v ) = K ( v ◦ f ) for all v ∈ V .We say L strongly (and continuously) transforms into K , symbolized by L s K resp. L sc K , if there exists a surjective Borel (resp. surjective and continuous)function f : X ։ Y such that V ◦ f = U and L ( v ) = K ( v ◦ f ) for all v ∈ V .If in this definition of a transformation a function f : X → Y is fixed becauseit has special properties, then we denote that in the transformation by f . Ofcourse, we have the implications L s K ⇒ L K and L sc K ⇒ L c K ⇒ L K. With this definition Theorem 1.3 can be reformulated to the following statement.
Corollary 1.5. L : V → R is a moment functional iff L [ K : L ([0 , , λ ) → R ] . The paper is structured as follows. In Section 2 we will give the preliminaries onmeasure theory and integration. Since most of the measure theoretic terminologyand results in Section 2 (Souslin sets, Lebesgue–Rohlin spaces, isomorphisms be-tween measure spaces etc.) have to our knowledge never been used in connectionwith the moment problem before, we give the complete definitions, results, andimportant examples which are essential for this paper (but without proofs).In Section 3 we present basic properties of transformations (Definition 1.4). E.g.in Theorem 3.3 we show that if there exists a transformation L K and K is amoment functional, then also L is a moment functional. So, the transformation (literally and symbolically) aims at moment functionals K to determine whetheralready L was a moment functional.Section 4 contains then the main results where several non-trivial transforma-tions to [0 , I k -moment functionals are presented, I k finite union of compactintervals in R . We show, which might already be apparent from Theorem 1.3, thatthe structure of possible moment functionals K are quite simple. These are always TRANSFORMATIONS OF MOMENT FUNCTIONALS [0 , I k -moment functionals. However, this simplicity of K has the price that f : [0 , → S has little properties. In the worst case as in Theorem 1.3 we onlyhave that f is measurable. We therefore also present results where f is at least con-tinuous and can therefore approximated by polynomials on [0 ,
1] in the supremumnorm.In Section 5 we give the conclusions and open problems. Additionally, we giveand discuss several open questions, especially the restriction that f is a rational ora polynomial map.2. Preliminaries: Measure Theory and the Lebesgue Integral
We give here the measure theoretic results used in our paper. Of course, it ispossible to go directly to Section 3 and the main results in Section 4 and consultthis Section 2 if necessary while reading the results and proofs.In this article we follow the monographs [Fed69], [LL01], and [Bog07] for themeasure theory and Lebesgue integral. We denote by P ( X ) the power set of a set X , i.e., the set of all subsets of X . Let A ⊆ P ( X ) be a σ -algebra on a set X ,then we call ( X , A ) a measurable space . A function f : ( X , A ) → ( Y , B ) betweenmeasurable spaces is called measurable if f − ( B ) ∈ A for all B ∈ B holds.Given F ⊆ P ( X ), then by σ ( F ) we denote the σ -algebra generated by F , i.e.,the smallest σ -algebra containing F . The Borel σ -algebra B ( X ) of a topological(e.g. Hausdorff) space X is generated by all open sets in X .Given a measurable space ( X , A ), a measure µ on ( X , A ) is a countably additivefunction µ : A → [0 , ∞ ]. I.e., dissident from [Bog07] for us all measures are non-negative if not otherwise explicitly stated as signed . ( X , A , µ ) is called a measurespace . ( X , A , µ ) is called probability measure space if additionally µ ( X ) = 1. An atom δ x is a measure such that δ x ( A ) = ( x ∈ A x A .
The special (Carath´eodory) outer measures are used and treated in Appendix A.Let X be a topological (e.g. locally compact Hausdorff) space. A measure on( X , B ( X )) is called Borel measure . A
Radon measure µ is a measure over ( X , B ( X ))such that µ ( K ) < ∞ for all compact K ⊆ X and µ ( V ) = sup { µ ( K ) | K is compact, K ⊆ V } . By λ n we denote the n -dimensional Lebegue measure on ( R n , B ( R n )).Let ( X , A , µ ) be a measure space and f : X → [0 , ∞ ] be a non-negativemeasurable function. The Lebesgue integral is defined by Z X f ( x ) d µ ( x ) := Z ∞ µ ( f − (( t, ∞ ))) d t (3)since h ( t ) := µ ( f − (( t, ∞ ))) is non-increasing, i.e., Riemann integrable, with theRiemann integral R ∞ h ( t ) d t . f is called µ -integrable if (3) is finite. A generalmeasurable function f : X → [ −∞ , ∞ ] is called µ -integrable if f + := max( f,
0) and f − := − min( f,
0) are µ -integrable. The Lebesgue integral is then defined by Z X f ( x ) d µ ( x ) := Z X f + ( x ) d µ ( x ) − Z X f − ( x ) d µ ( x ) . We have the following transformation formula.
Lemma 2.1.
Let f : ( Y , B ) → ( R , B ( R )) and g : ( X , A ) → ( Y , B ) be measurablefunctions, µ a measure on ( X , A ) such that f ◦ g is µ -integrable. Then µ ◦ g − is ameasure on ( Y , B ) and f is µ ◦ g − -integrable with Z X ( f ◦ g )( x ) d µ ( x ) = Z Y f ( y ) d( µ ◦ g − )( y ) . (4) RANSFORMATIONS OF MOMENT FUNCTIONALS 5
Proof.
It is sufficient to show (4) for f ≥ Z X ( f ◦ g )( x ) d µ ( x ) = Z ∞ µ (( f ◦ g ) − (( t, ∞ ))) d t = Z ∞ µ ( g − ( f − (( t, ∞ )))) d t = Z ∞ ( µ ◦ g − )( f − (( t, ∞ ))) d t = Z Y f ( y ) d( µ ◦ g − )( y ) . (cid:3) We have the first result from measure theory. We apply it in Proposition 4.1.
Proposition 2.2 (see e.g. [Bog07, Prop. 9.1.11]) . Let µ be an atomless probabilitymeasure on a measurable space ( X , A ) . Then there exists an A -measurable function f : X → [0 , such that µ ◦ f − = λ is the Lebesgue measure on [0 , . The following is a central definition.
Definition 2.3 ([Bog07, Def. 6.6.1]) . A set in a Hausdorff space is called a
Souslinset if it is the image of a complete separable metric space under a continuousmapping. A
Souslin space is a Hausdorff space that is a Souslin set.The empty set is a Souslin set. Souslin sets are fully characterized.
Proposition 2.4 (see e.g. [Bog07, Prop. 6.6.3]) . Every non-empty Souslin set isthe image of [0 , \ Q under some continuous function and also the image of (0 , under some Borel mapping. More concrete examples which are important to us are the following.
Example 2.5.
The unit interval [0 , ⊂ R is of course a complete separable metricspace (with the usual distance metric d ( x, y ) := | x − y | ). The question which setsare the continuous images of [0 ,
1] is partially answered by space filling curves , seee.g. [Sag94, Ch. 5]. So the
Peano curves as continuous and surjective functions f : [0 , → [ a , b ] × · · · × [ a n , b n ]with n ∈ N and −∞ < a i < b i < ∞ for all i = 1 , . . . , n show that all hyper-rectangles are Souslin spaces/sets. Especially [0 ,
1] is a Souslin set/space.A full answer gives the following theorem.
Hahn–Mazurkiewicz’ Theorem 2.6 (see e.g. [Sag94, Thm. 6.8]) . A set K in a non-empty Hausdorff space is the continuous image of [0 , if andonly if it is compact, connected, and locally connected. So sets K ⊆ R n are continuous images of [0 ,
1] if and only if they are compact andpath-connected. Hahn–Mazurkiewicz also implies that PR n is a Souslin space. ◦ Lemma 2.7 (see e.g. [Bog07, Lem. 6.6.5, Thm. 6.6.6 and 6.7.3]) . (i) The image of a Souslin set under a continuous function to a Hausdorff spaceis a Souslin set.(ii) Every open or closed set of a Souslin space is Souslin.(iii) If A n are Souslin sets in X n for all n ∈ N then Q n ∈ N A n is a Souslin set in Q n ∈ N X n .(iv) If A n ⊆ X are Souslin sets in a Hausdorff space X , then T n ∈ N A n and S n ∈ N A n are Souslin sets.(v) Every Borel subset of a Souslin space is a Souslin space.(vi) Let A ⊆ X and B ⊆ Y be Souslin sets of Souslin spaces and f : X → Y be aBorel function. Then f ( A ) and f − ( B ) are Souslin sets.Remark . The reverse of Lemma 2.7(v) is in general not true. Not every Souslinset is Borel. In fact, every non-empty complete metric space without isolated pointscontains a non-Borel Souslin set, see e.g. [Bog07, Cor. 6.7.11].
TRANSFORMATIONS OF MOMENT FUNCTIONALS (vi) demonstrates the difference between Souslin sets and Borel sets (in R n ).While the continuous image of a Borel set is again a Borel set, this no longer holdsfor Borel functions. But as (vi) shows for the Souslin sets the preimage and imageunder measurable functions remain Souslin sets.From Example 2.5 and Lemma 2.7 we get the following additional explicitexamples of Souslin sets. Example 2.9. R n and every compact semi-algebraic set in R n (resp. PR n ) areSouslin sets. ◦ Definition 2.10.
Let ( X , A ) and ( Y , B ) be two measurable spaces. A measurablefunction i : ( X , A ) → ( Y , B ) is called an isomorphism and the two measurablespaces isomorphic if i is bijective, i ( A ) = B , and i − ( B ) = A .The reason why we work with Souslin spaces is revealed in the following theorem. Theorem 2.11 (see e.g. [Bog07, Thm. 6.7.4]) . Let X be a Souslin space. Thenthere exist a Souslin set S ⊆ [0 , and an isomorphism h : ( S, B ( S )) → ( X , B ( X )) . The existence of an isomorphism can be weakened. For Borel measurable func-tion f : X → Y between two Souslin spaces X and Y with f ( X ) = Y one alwaysfinds nice (i.e., Borel measurable) one-sided inverse functions. Jankoff ’s Theorem 2.12 (see e.g. [Bog07, Thm. 6.9.1 and 9.1.3]) . Let X and Y be two Souslin spaces and let f : X → Y be a surjective Borel mapping. Then thereexists a Borel measurable function g : Y → X such that f ( g ( y )) = y for all y ∈ Y . In other words, restricting f so some X ⊆ X makes ˜ f := f | X not only bijectivebut ˜ f and ˜ f − are measurable. We have Y g → X f → Y with f ◦ g = id Y , i.e., g is injective, f is surjective, and with X = im g := g ( Y ) we have ˜ f − = g . Definition 2.13 (see e.g. [Bog07, Def. 9.2.1]) . Let ( X , A , µ ) and ( Y , B , ν ) be twomeasure spaces with non-negative measures.i) A point isomorphism T : X → Y is a bijective mapping such that T ( A ) = B and µ ◦ T − = ν .ii) The spaces ( X , A , µ ) and ( Y , B , ν ) are called isomorphic mod0 if there existsets N ∈ A µ , M ∈ B ν with µ ( N ) = ν ( M ) = 0 and a point isomorphism T : X \ N → Y \ M that are equipped with the restriction of the measures µ and ν and the σ -algebras A µ and B ν .A point isomorphism T between ( X , A , µ ) and ( Y , B , ν ) is of course measurablesince ν ( B ) = ( µ ◦ T − )( B ) = µ ( T − ( B )) implies T − ( B ) ∈ A for all B ∈ B .Like Theorem 2.11 also the next result shows the importance of working onSouslin sets. Theorem 2.14 (see e.g. [Bog07, Thm. 9.2.2]) . Let ( X , A ) be a Souslin spacewith Borel probability measure µ . Then ( X , A , µ ) is isomorphic mod0 to the space ([0 , , B ([0 , , ν ) for some ν Borel probability measure. If µ is an atomless mea-sure, then one can take for ν the Lebesgue measure λ . Corollary 2.15 (see e.g. [Bog07, Rem. 9.7.4]) . Let µ be a probability measure on aSouslin space X . Then there exists a measurable function f : [0 , → X such that µ = λ ◦ f − where λ is the Lebesgue measure on [0 , . For both results note the difference to Proposition 2.2. In Proposition 2.2 wefind for any measurable space X and measure µ a map f : X → [0 ,
1] such that µ = λ ◦ f − . RANSFORMATIONS OF MOMENT FUNCTIONALS 7
But for Souslin spaces X in Corollary 2.15 we find a map f : [0 , → X such that λ = µ ◦ f − . Theorem 2.14 restricts f : [0 , → X to isomorphisms and hence not all measurescan be transformed into λ . Atoms in the measure µ prevent it from being isomorphicto λ . In fact, as explained in [Bog07, Rem. 9.7.4], Corollary 2.15 follows fromTheorem 2.14 by introducing atoms into f : [0 , → X by introducing constantfunctions into f .But Theorem 2.14 provides that if µ has atoms, it can still be isomorphic mod0be transformed into a measure ν on [0 , ν = λ .So is it possible to transform the non-atomic part of µ to λ and then add the atomsfrom µ to λ ? Yes, we can. This is done on the following spaces. Definition 2.16 (see e.g. [Bog07, Def. 9.4.6]) . A measure space ( X , A , µ ) is calleda Lebesgue–Rohlin space if it is isomorphic mod0 to some measure space ( Y , B , ν )with a countable basis with respect to which Y is complete. Example 2.17 (see e.g. [Bog07, Exm. 9.4.2]) . ( M, B ( M ) , µ ), where M is a Borelset of a complete separable metric space X and µ is a Borel measure on M , is aLebesgue–Rohlin space. Especially X = R n or PR n are complete metric spaces andtherefore any Borel measure on a Borel subset M ∈ B ( R n ) gives a Lebesgue–Rohlinspace. ◦ We can now transform any measure by an isomorphism mod0 to the Lebesguemeasure λ plus atoms. Theorem 2.18 (see e.g. [Bog07, Thm. 9.4.7]) . Let ( X , A , µ ) be a Lebesgue–Rohlinspace with a probability measure µ . Then it is isomorphic mod0 to the interval [0 , with the measure ν = cλ + P ∞ i =1 c n · δ /n , where c = 1 − P ∞ i =1 c i , µ ( a i ) = c i and { a i } ⊆ X is the family of all atoms of µ . So we can transform any measure to the Lebesgue measure λ on [0 ,
1] or to λ on[0 ,
1] plus atoms. But these transformations are performed mainly by measurablefunctions because the set X where the original measure lives it to large. If werestrict the space where the measure lives, we get better transformations, especiallycontinuous ones. Theorem 2.19 (see e.g. [Bog07, Thm. 9.7.1]) . Let K be a compact metric space thatis the image of [0 , under a continuous mapping ˜ f and let µ be a Borel probabilitymeasure on K such that supp µ = K . Then there exists a continuous and surjectivemapping f : [0 , → K such that µ = λ ◦ f − , λ is the Lebesgue measure on [0 , . We will apply Theorem 2.19 especially in connection with the Hahn–Mazurkiewicz’Theorem 2.6. The advantage is here that f on [0 ,
1] is continuous and can thereforebe approximated by polynomials up to any precision ε >
Transformations of linear Functionals: Basic Properties
For the transformation in Definition 1.4 we get the following technical result.
Lemma 3.1.
Let X , Y , and Z be Souslin spaces; U , V , and W be vector spacesof real measurable functions on X , Y , and Z respectively; and M : W → R , L : V → R , and K : U → R be linear functionals. The following hold:(i) M L and L K imply M K .(ii) M c L and L c K imply M c K .(iii) M s L and L s K imply M s K .(iv) M sc L and L sc K imply M sc K . TRANSFORMATIONS OF MOMENT FUNCTIONALS
Proof. (i): Since M L there exists a Borel function f : Y → Z such that
W ◦ f ⊆ V and M ( w ) = L ( w ◦ f ) for all w ∈ W . And since L K there existsa Borel function g : X → Y such that
V ◦ g ⊆ U and L ( v ) = K ( v ◦ g ) for all v ∈ V . Hence, h = f ◦ g : X → Z implies
W ◦ h = W ◦ f ◦ g ⊆ V ◦ g ⊆ U and M ( w ) = L ( w ◦ f ) = K ( w ◦ f ◦ g ) = K ( w ◦ h ) for all w ∈ W , i.e., M K .(ii)-(iv) follow in the same way as (i). (cid:3) Lemma 3.1 can be seen as shortening the sequence: M L K ⇒ M K. The next lemma shows, that a strong transformation L s K implies the reversetransformation K L . Lemma 3.2.
Let X and Y be Souslin sets, U and V vector spaces of real functionson X resp. Y , and L : V → R and K : U → R be linear functionals. Then L s K implies K L .Proof. Since L s K there exists a surjective Borel function f : X → Y such that L ( v ) = K ( v ◦ f ) and V ◦ f = U . Since f is surjective by Jankoff’s Theorem 2.12there exists a Borel function g : Y → X such that f ( g ( y )) = y for all y ∈ Y . Let u ∈ U = V ◦ f , then v in u = v ◦ f is unique since for v and v with that propertywe have v = v ◦ f ◦ g = u ◦ g = v ◦ f ◦ g = v . Hence,
U ◦ g = V and for all u ∈ U we have K ( u ) = K ( v ◦ f ) = L ( v ) = L ( v ◦ f ◦ g ) = L ( u ◦ g ) . (cid:3) While we have so far only transformed linear functionals, the importance of thetransformation is revealed in the following result. It shows that the property ofbeing a moment functional is preserved in one or both directions.
Theorem 3.3.
Let X and Y be Souslin sets, U and V vector spaces of real functionson X resp. Y , and L : V → R and K : U → R be linear functionals. If L K , then(i) K is a moment functionalimplies(ii) L is a moment functional.If L s K , then (i) ⇔ (ii).Proof. Since L K there exists a Borel function f : X → Y such that
V ◦ f ⊆ U and L ( v ) = K ( v ◦ f ) for all v ∈ V .(i) → (ii): Let K be a moment functional with representing measure ν on X , then L ( v ) = K ( v ◦ f ) = Z X ( v ◦ f )( x ) d ν ( x ) Lemma 2.1 = Z Y v ( y ) d( ν ◦ f − )( y ) , i.e., ν ◦ f − is a representing measure of L and hence L is a moment functional.(ii) → (i): When L s K , then Lemma 3.2 implies K L . (cid:3) The importance of the transformation and hence Theorem 3.3 can be seen in L L L L L L L K L . (5)If K is a moment functional, then all L , . . . , L are moment funtionals. Assumein (5) all transformations are strong transformations s . Then: If one L i or K is a moment functional, then all K, L , . . . , L are moment functionals. RANSFORMATIONS OF MOMENT FUNCTIONALS 9
Note, the transformation in Definition 1.4 also covers extensions and restric-tions of functionals. Let f = id X and let V be a vector space of measurable functionson X , V ⊆ V be a linear subspace, and L : V → R a linear functional. Then L | V id X L. Or if L i : V i → R are extensions of L , i.e., V ⊆ V ⊆ V ⊆ · · · ⊆ V k with L i = L i +1 | V i , then L id X L X L X . . . id X L k or short L L L . . . L k shows that if L k is a moment functional, then all L i and L are moment functionals.So far we introduced the transformation of a linear functional and gained basicproperties. But as seen from Theorem 1.3 and Corollary 1.5, there are non-trivialresults for the transformations. The next section is devoted to these non-trivialtransformation results.4. Non-trivial Transformations of linear Functionals
Let V be (finite or infinite dimensional) vector space of measurable functions ona Souslin space X . Then by Theorem 2.11 there exist a Souslin set S ⊆ [0 ,
1] andan isomorphism h : ( S, B ( S )) → ( X , B ( X )). This implies that ˜ L : ˜ V → R with˜ V := { f ◦ h | f ∈ V} and ˜ L ( g ) := L ( g ◦ h − ), g ∈ ˜ V , is a linear functional but nowthe functions ˜ V live on S ⊆ [0 , L is a moment functional if and onlyif ˜ L is a moment functional.For example, let L : R [ x , . . . , x n ] → R be a moment functional with X = R n .Then h = ( h , . . . , h n ) : S ⊆ [0 , → R n is an isomorphism between ( S, B ( S )) and( R n , B ( R n )) and ˜ L is a moment functional with ˜ L ( h α ) = L ( x α ).However, by Remark 2.8 S needs not to be a Borel set. So determining whether˜ L is a moment functional might be as hard as determining whether L is a mo-ment functional. Additionally, ˜ L now no longer lives on polynomials but evaluatesmeasurable functions h α = h α · · · h α n n with α = ( α , . . . , α n ) ∈ N n .Allowing general Borel measurable functions on measurable spaces instead ofisomorphisms we get Theorem 1.3 in the introduction. There we showed that anymoment functional can be expressed as integration with respect to the Lebesguemeasure λ on [0 , ,
1] evaluated on R [ t ]. Proposition 4.1.
Let V be a vector space of real measurable functions on a mea-surable space ( X , A ) such that there exists an element v ∈ V with ≤ v on X and let L : V → R be a moment functional which has an atomless representingmeasure. Then there exists a measurable function f : X → [0 , and an extension L : V + R [ f ] → R of L such that L ( f d ) = L (1) d +1 for all d ∈ N , i.e., ˜ L : R [ t ] → R with ˜ L ( t d ) := L ( f d ) for all d ∈ N is represented by L (1) · λ where λ is the Lebesguemeasure λ on [0 , .Proof. Let µ be a representing measure of L . By Proposition 2.2 there exists ameasurable f : R n → [0 ,
1] such that µ ◦ f − = λ on [0 , f is measurable, | f | ≤ R n , and L (1) < ∞ , all f d , d ∈ N , are µ -integrable: (cid:12)(cid:12)(cid:12)(cid:12)Z R n f d ( x ) d µ ( x ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ Z R n | f ( x ) | d d µ ( x ) ≤ Z R n µ ( x ) = L (1) . Define L : R [ f ] → R by L ( f d ) := R R n f d ( x ) d µ ( x ). Then L ( f d ) = Z R n f d ( x ) d µ ( x ) Lemma 2.1 = Z t d d( µ ◦ f − )( t ) = Z t d d λ ( t ) = L (1) d + 1is represented by L (1) · λ on [0 , (cid:3) Hence, for any moment functional with an atomless representing measure thereexists a function f (a direction) such that it acts on R [ f ] ∼ = R [ t ] as (1), i.e., theLebesgue measure on [0 , formulation with L Leb from Example 1.1 we canvisualize Proposition 4.1 as L : V → R id X L Leb : R [ t ] → R f L : V + R [ f ] → R . Note the reverse statement of Proposition 4.1. If a linear functional L can neverbe (continuously) extended to R [ f ] with L ( f d ) = L (1) d +1 for some measurable f , then L is not a moment functional with an atomless representing measure.Theorem 1.3 and Proposition 4.1 are very general. Especially Theorem 1.3 workson arbitrary Borel sets of R n (in fact on every Souslin space). For this generalitywe have to pay the price that f is in general only measurable. Additionally, sincewe always express L as integration with respect to λ on [0 , f dependson L . If we want additional properties for f to hold, especially continuity andindependence from L , then we need to restrict the functionals we want to transform.This can be achieved by restricting the investigation to K -moment functionals oncompact and path-connected sets K ⊂ R n . Then from the Hahn–Mazurkiewicz’Theorem 2.6 we get the existence of surjective and continuous functions f : [0 , → K . We find the following result. Theorem 4.2.
Let n ∈ N be a natural number, K ⊂ R n be a compact and path-connected set, and let V be a vector space of real measurable functions on ( K, B ( K )) .Then any surjective and continuous function f : [0 , → K induces for any linearfunctional L : V → R a strong and continuous transformation L : V → R sc : f ˜ L : V ◦ f → R , i.e., for any linear functional L : V → R the following are equivalent:(i) L : V → R is a K -moment functional.(ii) ˜ L : V ◦ f → R defined by ˜ L ( v ◦ f ) := L ( v ) is a [0 , -moment functional.If ˜ µ is a representing measure of ˜ L , then ˜ µ ◦ f − is a representing measure of L .There exists a measurable function g : K → [0 , such that f ( g ( x )) = x forall x ∈ K and if µ is a representing measure of L , then µ ◦ g − is a representingmeasure of ˜ L .Proof. Since K ⊂ R n is compact and path-connected, by the Hahn–Mazurkiewicz’Theorem 2.6 there exists a continuous and surjective function f : [0 , → K . ByExample 2.5 or Lemma 2.7 [0 ,
1] and K are Souslin spaces and f is Borel measur-able (since it is continuous). By Jankoff’s Theorem 2.12 there exists a measurablefunction g : K → [0 ,
1] such that f ( g ( x )) = x for all x ∈ K. ( ∗ ) RANSFORMATIONS OF MOMENT FUNCTIONALS 11 ( ∗ ) implies that ˜ L is well-defined by ˜ L ( v ◦ f ) = L ( v ). To show this, for ˜ v ∈ ˜ V let v , v ∈ V be such that v ◦ f = ˜ v = v ◦ f . But then g resp. ( ∗ ) implies v = v ◦ f ◦ g = ˜ v ◦ g = v ◦ f ◦ g = v , i.e., for any ˜ v ∈ V there is a unique v ∈ V with ˜ v = v ◦ f .(i) → (ii): Let L : V → R be a K -moment functional and µ be a representingmeasure of L , i.e., supp µ ⊆ K and L ( v ) = Z K v ( x ) d µ ( x ) for all v ∈ V . Then ˜ L ( v ◦ f ) = L ( v ) = Z K v ( x ) d µ ( x ) = Z K ( v ◦ f )( g ( x )) d µ ( x ) Lemma 2.1 = Z ( v ◦ f )( y ) d( µ ◦ g − )( y ) , i.e., µ ◦ g − is a representing measure of ˜ L and hence ˜ L is a [0 , → (i): Let ˜ µ be a representing measure of ˜ L : ˜ V → R . Then L ( v ) = ˜ L ( v ◦ f ) = Z ( v ◦ f )( y ) d˜ µ ( y ) Lemma 2.1 = Z K v ( x ) d(˜ µ ◦ f − )( x ) , i.e., ˜ µ ◦ f − is a representing measure of L with supp ˜ µ ◦ f − ⊆ K and L is thereforea K -moment sequence. (cid:3) In the previous result the functions f : [0 , → K and g : K → [0 ,
1] do notdepend on the functions V or the functional L : V → R . They depend only on K .We can therefore fix such functions f and g and investigate any L resp. ˜ L .If the continuous f can be chosen for each L , then in Theorem 4.2(ii) we caneven ensure that ˜ L is represented by the Lebesgue measure λ on [0 ,
1] if and onlyif L has a representing measure µ with supp µ = K , see Theorem 4.11 below.In Theorem 4.2 we required that K consists of one path-connected component.If K consists of more than one component, then we can glue the parts together. Corollary 4.3.
Let n ∈ N and K ⊂ R n be the union of k ∈ N ∪ {∞} compact,path-connected and pairwise disjoint sets K i ⊂ R n : K = S ki =1 K i . Let V be a vectorspace of real valued measurable functions on ( K, B ( K )) . There exists a continuoussurjective function f : k [ i =1 [2 i − , i − → K such that for any linear functional L : V → R the following are equivalent:(i) L : V → R is a K -moment functional.(ii) ˜ L : ˜ V → R on ˜ V := { v ◦ f | v ∈ V} and defined by ˜ L ( v ◦ f ) := L ( v ) is a S ki =1 [2 i − , i − -moment functional.Proof. It is sufficient to show the existence of the function f (and g ). The rest ofthe proof is verbatim the same as in the proof of Theorem 4.2.Since for each i = 1 , , . . . , k the set K i is compact and path-connected andthe translation of the unit interval [0 ,
1] to [2 i − , i −
1] is continuous, by theHahn–Mazurkiewicz’ Theorem 2.6 there exists a continuous and surjective f i :[2 i − , i − → K i . Define f : S ki =1 [2 i − , i − → K by f ( x ) := f i ( x ) if x ∈ [2 i − , i −
1] for an i ∈ { , , . . . , k } . Then f is continuous and surjective.For g : K → S ki =1 [2 i − , i −
1] we proceed in the same way. By Jankoff’sTheorem 2.12 for each f i : [2 i − , i − → K i there exists a measurable g i : K i → [2 i − , i − g as g ( x ) := g i ( x ) if x ∈ K i . (cid:3) Note, that when K consists of countably many compact and path-connectedcomponents ( k = ∞ ), then in Corollary 4.3 f is no longer supported on a bounded(and therefore compact) set: S ki =1 [2 i − , i − K is a compact andsemi-algebraic set, then K has only finitely many path-connected components.An advantage in Theorem 4.2 is that f = ( f , . . . , f n ) : [0 , → K ⊂ R n is continuous. Hence, all coordinate functions f i : [0 , → R are continuous.By the Stone–Weierstrass Theorem we can approximate each f i in the sup-normon [0 ,
1] by polynomials to any precision. f can therefore be approximated toany precision by a polynomial map. A representing measure ˜ µ of ˜ L provides therepresenting measure ˜ µ ◦ f − of L . An approximation f ε ∈ R [ x , . . . , x n ] n of f , i.e.,sup t ∈ [0 , k f ( t ) − f ε ( t ) k < ε with any (fixed) norm k · k on R n and ε >
0, providesan approximate representing measure ˜ µ ◦ f − ε of L .Let K ⊂ R n be a compact and path-connected set, V = R [ x , . . . , x n ], and L : V → R be a linear functional. Then the induced functional ˜ L : ˜ V → R on [0 , L ( p ◦ f ) := L ( p ). It depends on p ◦ f , i.e., f α = f α · · · f α n n , α =( α , . . . , α n ) ∈ N n . So as in Theorem 1.3 the algebraic structure of R [ x , . . . , x n ]remains but the domain K is pulled back to [0 ,
1] by the continuous f .That the algebraic structure remains also reveals one big difference between L and ˜ L . E.g. V = R [ x , . . . , x n ] separates points and is therefore dense in C ( K, R ).But f : [0 , → K is a space filling curve and therefore never injective (Netto’sTheorem). Hence, there are t , t ∈ [0 ,
1] with t = t and f ( t ) = f ( t ). Theset ˜ V := { p ◦ f | p ∈ V} therefore does not separate t from t and is by theStone–Weierstrass Theorem not dense in C ([0 , , R ). So the ˜ L in Theorem 4.2 andCorollary 4.3 can at this point not extended to the Hausdorff Moment Problem 1.2.In the next theorem we will identify each K -moment functional with a [0 , Theorem 4.4.
Let n ∈ N be a natural number and K ⊂ R n be a compact andpath-connected set. Then there exists a measurable function g : K → [0 , such that for all linear functionals L : V → R with ∈ V ⊆ C ( K, R ) the followingare equivalent:(i) L : V → R is a K -moment functional.(ii) L : V → R continuously extends to L : V + R [ g ] → R such that ˜ L : R [ t ] → R defined by ˜ L ( t d ) := L ( g d ) for all d ∈ N is a [0 , -moment functional, i.e., L : V → R id X ˜ L : R [ t ] → R g L : V + R [ g ] → R . (6) If µ is the representing measure of L , then µ ◦ g − represents ˜ L .Additionally, there exists a continuous and surjective function f : [0 , → K independent on L resp. ˜ L such that f ( g ( x )) = x for all x ∈ K and if ˜ µ is therepresenting measure of ˜ L , then ˜ µ ◦ f − is the representing measure of L .Proof. Since K is a compact and path-connected set, by the Hahn–Mazurkiewicz’Theorem 2.6 there exists a continuous and surjective function f : [0 , → K . ByLemma 2.7 [0 ,
1] and K are Souslin sets and hence by Jankoff’s Theorem 2.12 thereexists a measurable function g : K → [0 ,
1] such that f ( g ( x )) = x for all x ∈ K. ( ∗ ) If p i ∈ R [ t ] with p i ⇒ p ∈ C ([0 , , R ) and p ◦ g ∈ V then L ( p i ◦ g ) → L ( p ◦ g ). RANSFORMATIONS OF MOMENT FUNCTIONALS 13 (i) → (ii): Let L : V → R be a K -moment functional and µ be a representingmeasure of L with supp µ ⊆ K . g is measurable with | g | ≤ g d , d ∈ N , are µ -integrable by (cid:12)(cid:12)(cid:12)(cid:12)Z K g ( x ) d d µ ( x ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ Z K | g ( x ) | d d µ ( x ) ≤ Z K µ ( x ) = µ ( K ) = L (1) ($)and hence L extents to R [ g ]. Let p ∈ R [ t ], then˜ L ( p ) = L ( p ◦ g ) = Z K ( p ◦ g )( x ) d µ ( x ) Lemma 2.1 = Z p ( t ) d( µ ◦ g − )( t )and µ ◦ g − is a representing measure of ˜ L , i.e., ˜ L is a [0 , → (i): Let ˜ L : R [ t ] → R be a [0 , µ be its uniquerepresenting measure. Since by the Stone–Weierstrass Theorem R [ t ] is dense in C ([0 , , R ) the moment functional ˜ L extends uniquely to C ([0 , , R ). For simplic-ity we denote this extension also ˜ L : C ([0 , , R ) → R . Since f : [0 , → K iscontinuous we have v ◦ f ∈ C ([0 , , R ) for all v ∈ V . By ( ∗ ) we have v = v ◦ f ◦ g for all v ∈ V and hence L ( v ) = L ( v ◦ f ◦ g ) . (&)But since v ◦ f : [0 , → R is continuous and ˜ L : R [ t ] → R uniquely extends to C ([0 , , R ) we have L ( v ◦ f ◦ g ) = ˜ L ( v ◦ f ) . ( L ( v ) (&) = L ( v ◦ f ◦ g ) ( = ˜ L ( v ◦ f ) = Z ( v ◦ f )( t ) d˜ µ ( t ) Lem. 2.1 = Z K v ( x ) d(˜ µ ◦ f − )( x )for all v ∈ V , i.e., ˜ µ ◦ f − is a representing measure of L and L is therefore a K -moment functional. (cid:3) We see that all about L is already known if we know how it acts (via ˜ L ) onpowers of the fixed (and independent on L ) function g . ˜ L : R [ t ] → R is only aHausdorff moment problem and its representing measure ˜ µ provides a representingmeasure µ = ˜ µ ◦ f − via a fixed (and independent on L ) continuous function f . Remark . Note, that in Theorem 4.4 and therefore also in Corollary 4.8 thecondition 1 ∈ V can be weakened to:There shall exists a v ∈ V ⊆ C ( K, R ) such that v > K. By compactness of K and continuity of v this implies 1 ≤ c · v ∈ V for some c >
0, i.e., µ ( K ) < ∞ in ($). However, since we have to extend L : V → R to L : V + R [ g ] → R and 1 ∈ R [ g ] we can assume w.l.o.g. already 1 ∈ V . If 1
6∈ V and L can not be extended to 1, then L can definitely not be extended to R [ g ] and thestatements of Theorem 4.4 and Corollary 4.8 remain valid. ◦ Theorem 4.4 requires the existence of a continuous extension L : V + R [ g ] → R of L . Under the very mild condition 1 ∈ V (resp. v ∈ V with v > K by theprevious remark) extensions (not necessarily continuous) exist. Lemma 4.6.
Let g be as in Theorem 4.4 (resp. Corollary 4.8) and L : V → R be alinear functional on the vector space V with ∈ V ⊆ C ( K, I k ) and L (1) > . Thenthere exists an extension L : V + R [ g ] → R of L : V → R .Proof. Since g : K → I k ⊆ [0 ,
1] in Theorem 4.4 (resp. Corollary 4.8) we have | g | ≤
1. Hence, 1 ∈ V ∩ R [ g ] = ∅ and V + R [ g ] = V ⊕ ( R [ g ] \ V ), i.e., f = f + f ∈ V + R [ g ]with unique f ∈ V and f ∈ R [ g ] \ V . Define p : V + R [ g ] → R by p ( f ) := | L ( f ) | + L (1) · k f k ∞ for all f = f + f ∈ V + R [ g ], f ∈ V , and f ∈ R [ g ] \ V . Hence, L ( f ) ≤ p ( f ) forall f ∈ V . Then p ( f + g ) ≤ p ( f ) + p ( g ) and p ( α · f ) = α · p ( f )hold for all f, g ∈ V + R [ g ] and α ≥
0. By the Hahn–Banach Theorem there existsan extension L : V + R [ g ] → R of L . (cid:3) An extension L in Lemma 4.6 is in general not unique. If V is a point separatingalgebra on K and L is a K -moment functional, then the extension L is unique (andcontinuous), since then the representing measure µ of L is unique.For the extension L it is only necessary that 1 ∈ V to ensure | g | ≤ ∈ V . V ⊆ C ( K, I k ) continuous is actually not necessary and hence Lemma 4.6 can beeasily weakened.As in Theorem 4.2 also in Theorem 4.4 the functions f and g do not dependon L or ˜ L . They depend only on K . And as in Proposition 4.1 the functional ˜ L is defined in one “direction” R [ g ] ∼ = R [ t ] by ˜ L ( t d ) := L ( g d ). But now it no longerneeds to be L Leb as in Example 1.1.The problem of determining whether ˜ L : R [ t ] → R in Theorem 4.4(ii) is a [0 , g : K → [0 ,
1] toestablish the equivalence (i) ⇔ (ii) in Theorem 4.4 is a measurable function andnot a polynomial. Hence, L ( g d ) is not directly accessible unless of course d = 0.Fortunately, since K ⊂ R is compact, R [ x , . . . , x n ] is dense in C ( K, R ). Hence,for any given finite measure µ on K , i.e., µ ( K ) = L (1) < ∞ , we can approximate g by a polynomial g ε ∈ R [ x , . . . , x n ] in the L ( µ )-norm to any arbitrary precision. Theorem 4.7.
Let n ∈ N be a natural number, K ⊂ R n be a compact andpath-connected set, and let g : K → [0 , be from Theorem 4.4. Then for any ε > and K -moment functional L : R [ x , . . . , x n ] → R there exists a polynomial g ε ∈ R [ x , . . . , x n ] such that L ( | g ε − g | ) ≤ ε and | L ( g d ) − L ( g dε ) | ≤ d · L ( | g − g ε | ) ≤ d · ε hold for all d ∈ N . g ε can be chosen to be a square: g ε = p ε , p ε ∈ R [ x , . . . , x n ] .Proof. L is a K -moment functional and therefore has a unique representing measure µ with supp µ ⊆ K . g ≥ p : K → [0 ,
1] such that g = p . Since K is compact and µ ( K ) = L (1) < ∞ the polynomials R [ x , . . . , x n ] are dense in L ( K, µ ). By (cid:12)(cid:12)(cid:12)(cid:12)Z K p ( x ) d µ ( x ) (cid:12)(cid:12)(cid:12)(cid:12) ≤ Z K | p ( x ) | d µ ( x ) ≤ Z K µ ( x ) = L (1) < ∞ we have p ∈ L ( K, µ ) and therefore for any ε > p ε ∈ R [ x , . . . , x n ]such that p ε ≤ K and k p − p ε k L ( K,µ ) = Z K | p ( x ) − p ε ( x ) | d µ ( x ) ≤ ε. Set g ε := p ε . Then L ( | g − g ε | ) = Z K | g − g ε | d µ ( x ) = Z K | p ( x ) − p ε ( x ) | d µ ( x )= Z K | p − p ε | · | p + p ε | d µ ( x ) ≤ Z K | p ( x ) − p ε ( x ) | d µ ( x ) ≤ ε. (7) RANSFORMATIONS OF MOMENT FUNCTIONALS 15
For d = 0 we have g = g ε = 1, i.e., L ( g ) = L (1) = L ( g ε ), and for d = 1 we have | L ( g ) − L ( g ε ) | ≤ L ( | g − g ε | ) ≤ ε . So let d ≥
2. Then | L ( g d ) − L ( g dε ) | ≤ L ( | g d − g dε | ) = Z K | g ( x ) d − g ε ( x ) d | d µ ( x )= Z K | g ( x ) − g ε ( x ) | · (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) d − X i =0 g ( x ) i · g ε ( x ) d − − i (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) d µ ( x ) (8) ≤ d · Z K | g ( x ) − g ε ( x ) | d µ ( x ) ≤ d · ε. (cid:3) Note, the g ε not only depends on ε > L resp. its representingmeasure µ . Since g is measurable (but not necessarily continuous) it is not possibleto get sup x ∈ K | g ( x ) − g ε ( x ) | ≤ ε . So g ε depends on L . Otherwise assume we finda g ε ∈ R [ x , . . . , x n ] such that for any moment functional L (with L (1) = 1), i.e.,measure µ on K with µ ( K ) = 1, we have k g − g ε k L ( K,µ ) ≤ ε . Then for µ = δ x , x ∈ K , we get sup x ∈ K | g ( x ) − g ε ( x ) | = sup x ∈ K k g − g ε k L ( K,δ x ) ≤ ε, a contradiction. So the choice of g ε depends on L resp. µ .Additionally, note that in fact we can g ε not only chose to be a square, but infact any power: g ε = p kε for a fixed k ∈ N . Just replace p := √ g by p := k √ g in theproof since g ≥ K ⊂ R n to an at most countable union of pairwise disjoint, compact, and path-connected K i ’s. In Theorem 4.4 we required that K is a compact and path-connected set. Since we needed compactness of [0 ,
1] in Theorem 4.4 we can at leastextend Theorem 4.4 to a finite (disjoint) union of compact and path-connected sets.
Corollary 4.8.
Let k , n ∈ N be natural numbers and K ⊂ R n be the union offinitely many compact, path-connected, and pairwise disjoint sets K i : K = S ki =1 K i .Then there exists a measurable function g : K → I k := k [ i =1 (cid:20) i − k − , i − k − (cid:21) ⊂ [0 , such that for all linear functionals L : V → R with ∈ V ⊆ C ( K, R ) the followingare equivalent:(i) L : R [ x , . . . , x n ] → R is a K -moment functional.(ii) L : V → R continuously extends to L : V + R [ g ] → R such that ˜ L : R [ t ] → R defined by ˜ L ( t d ) := L ( g d ) for all d ∈ N is a [0 , -moment functional.Proof. For all i = 1 , . . . , k the sets K i and [ i − k − , i − k − ] are compact and path-connected and therefore by the Hahn–Mazurkiewicz’ Theorem 2.6 there exist con-tinuous and surjective functions f i : [ i − k − , i − k − ] → K i . By Lemma 2.7 all K i and [ i − k − , i − k − ] are Souslin sets and hence by Jankoff’s Theorem 2.12 there existmeasurable functions g i : K i → [ i − k − , i − k − ] such that f i ( g i ( x )) = x for all x ∈ K i , i = 1 , . . . , k . Define f : I k → K = k [ i =1 K i by f ( x ) = f i ( x ) for x ∈ K i and g : K = k [ i =1 K i → I k by g ( x ) = g i ( x ) for x ∈ (cid:20) i − k − , i − k − (cid:21) . Then f ( g ( x )) = x for all x ∈ K and I k ⊂ [0 , → (ii) and (ii) → (i) are verbatim the same as in the proof of Theorem 4.4. (cid:3) We are again facing the problem, that g is measurable but not necessarily apolynomial. But as in Theorem 4.7 we can approximate g by polynomials. Corollary 4.9.
Let n, k ∈ N be natural numbers, K ⊂ R n the union of finitelymany compact, path-connected, and pairwise disjoint sets K i , K = S ki =1 K i , and let g : K → I k be from Corollary 4.8. Then for any ε > and K -moment functional L : R [ x , . . . , x n ] → R there exists a polynomial g ε ∈ R [ x , . . . , x n ] such that L ( | g ε − g | ) ≤ ε and | L ( g d ) − L ( g dε ) | ≤ d · L ( | g − g ε | ) ≤ d · ε hold for all d ∈ N . g ε can be chosen to be a square: g ε = p ε , p ε ∈ R [ x , . . . , x n ] .Proof. Since I k ⊂ [0 ,
1] it is verbatim the same as the proof of Theorem 4.7. (cid:3)
Note, that in Theorem 4.7 and Corollary 4.9 we have | ˜ L ( t d ) | ≤ ˜ L (1) = L (1), i.e.,the error bounds ≤ d · ε exceed 2 · ˜ L (1) at some point and become unreasonable.We have seen in Theorem 4.2 resp. Corollary 4.3 that a linear functional L : V → R is a K -moment functional ( K is the countable union of compact and path-connected sets) if and only if it can be transformed by a continuous function f : I → K to a I -moment functional ( I is the countable union of intervals [ a i , b i ] ∈ R ).If we allow not only continuous functions f , then we can generalize this. Ifwe drop continuity of f but add bijectivity almost everywhere we find that anyfunctional on a Borel set of R n is a moment functional if and only if we can transformit into a moment functional with representing measure “Lebesgue measure on [0 , Theorem 4.10.
Let n ∈ N be a natural number, B ∈ B ( R n ) be a Borel set, and V be a vector space of real measurable functions on B with ∈ V . Then the followingare equivalent.(i) L : V → R is a B -moment functional.(ii) There exist Borel sets M ∈ B ( B ) and N ∈ B ([0 , and a bijective andmeasurable function (isomorphism) f : [0 , \ N → B \ M such that L ( v ) = Z v ( f ( t )) d ν ( t ) with ν = c · λ + X i ∈ N c i · δ /i (9) for all v ∈ V , where c , c i ≥ and c + P i ∈ N c i = L (1) , i.e., ν ◦ f − is arepresenting measure of L .Proof. (ii) → (i): Clear since ν ◦ f − is a representing measure of L .(i) → (ii): Let µ be a representing measure of L . Then ( B, B ( B ) , µ ) is by Exam-ple 2.17 a Lebesgue–Rohlin space and therefore by Theorem 2.18 isomorph mod0to ([0 , , B ([0 , , ν ) with ν as in (9), i.e., there exist Borel sets M ∈ B ( B ) and N ∈ B ([0 , f : [0 , \ N → B \ M suchthat ν = µ ◦ f and µ ( M ) = ν ( N ) = 0. Then by Lemma 2.1 for all v ∈ V we have L ( v ) = Z B v ( x ) d µ ( x ) = Z B \ M v ( f ◦ f − ) d µ ( x )= Z [0 , \ N v ( f ( t )) d( µ ◦ f )( t ) = Z v ( f ( t )) d ν ( t ) . (cid:3) If we drop bijectivity almost everywhere for f then we get Theorem 1.3, i.e., in(9) we can chose c = L (1) and c i = 0 for all i ∈ N .In Theorem 1.3 and Theorem 4.10 we can only ensure that f is measurable, butnot necessarily continuous or even a polynomial map. The reason is that we can RANSFORMATIONS OF MOMENT FUNCTIONALS 17 not control the support of a representing measure of L . In Theorem 4.2 we alreadyshowed that f can be chosen as continuous and surjective, independent on L . Butif we restrict the moment functionals resp. the support of a representing measureand chose f tailor made for each K -moment functional, then f can be chosen tobe continuous and surjective and the representing measure will be the Lebesguemeasure λ on [0 , Theorem 4.11.
Let n ∈ N , K ⊂ R n be a compact and path-connected set, V be avector space of real function on K , and L : V → R be a linear functional. Then thefollowing are equivalent:(i) L : V → R is a K -moment functional with representing measure µ such that supp µ = K .(ii) There exists a continuous and surjective function f : [0 , → K such that L ( v ) = Z v ( f ( t )) d λ ( t ) for all v ∈ V where λ is the Lebesgue measure on [0 , , i.e., L f L Leb : L ([0 , , λ ) → R . Proof. (i) → (ii): Let L : V → R be a K -moment functional and let µ be its uniquerepresenting measure with supp µ = K . Since K is a compact and path-connectedset, by the Hahn–Mazurkiewicz’ Theorem 2.6 there exists a continuous and sur-jective function ˜ f : [0 , → K . By Theorem 2.19 there exists a continuous andsurjective function f : [0 , → K such that µ = λ ◦ f − . For all v ∈ V we get L ( p ) = Z K p ( x ) d µ ( x ) = Z K p ( x ) d( λ ◦ f − )( x ) Lemma 2.1 = Z p ( f ( t )) d λ ( t ) . ( ∗ )(ii) → (i): By ( ∗ ) µ = λ ◦ f − is a representing measure of L , i.e., L is a K -moment functional. To show that supp µ = K holds, let U ⊆ K be open. Since f is continuous, f − ( U ) ⊆ [0 ,
1] is open and therefore µ ( U ) = λ ( f − ( U )) > (cid:3) So far we transformed moment functionals to [0 , R n -moment functionals can not be continuously transformedinto [0 , R n -moment functionals con-tinuously into [0 , ∞ )-moment functionals. We need the following. Lemma 4.12.
Let n ∈ N and ε > . Then there exists a continuous and surjectivefunction f ε : [0 , ∞ ) → R n with t − ε ≤ k f ε ( t ) k ≤ t + ε for all t ≥ and there exists a measurable function g ε : R n → [0 , ∞ ) such that f ε ( g ε ( x )) = x and k x k − ε ≤ g ε ( x ) ≤ k x k + ε for all x ∈ R n .Proof. Set A n := { x ∈ R n | ( n − · ε ≤ k x k ≤ n · ε } for all n ∈ N . Then all A n ’s are compact and path-connected and by the Hahn–Mazurkiewicz’ Theorem 2.6 there exist continuous and surjective functions f ε,n :[( n − · ε, n · ε ] → A n for all n ∈ N such that f ε,n ( n · ε ) = f ε,n +1 ( n · ε ), i.e., k f ε,n ( n · ε ) k = k f ε,n +1 ( n · ε ) k = n · ε for all n ∈ N . Since R n = S n ∈ N A n define f ε : [0 , ∞ ) → R n by f ε | [ n − ,n ] := f ε,n . Then for t ∈ [( n − · ε, n · ε ] we have t − ε ≤ ( n − · ε ≤ k f ε ( t ) k = k f ε,n ( t ) k ≤ n · ε ≤ t + ε. ( ∗ ) Since f : [0 , ∞ ) → R n is surjective and [0 , ∞ ) and R n are Souslin sets byLemma 2.7 then by Jankoff’s Theorem 2.12 there exists a g ε : R n → [0 , ∞ ) with f ε ( g ε ( x )) = x for all x ∈ R n . ( ∗ ) implies g ε ( x ) − ε ≤ k x k = k f ε ( g ε ( x )) k ≤ g ε ( x ) + ε and therefore k x k − ε ≤ g ε ( x ) ≤ k x k + ε for all x ∈ R n . (cid:3) Similar to Theorem 4.2 we then get the continuous transformation into [0 , ∞ )-moment functionals. Theorem 4.13.
Let n ∈ N , f : [0 , ∞ ) → R n be a continuous and surjectivefunction, and V be a vector space of measurable functions on R n . Then for alllinear functionals L : V → R the following are equivalent:(i) L : V → R is a moment functional.(ii) ˜ L : V ◦ f → R defined by ˜ L ( v ◦ f ) := L ( v ) is a [0 , ∞ ) -moment functional.I.e., L sc ˜ L . If ˜ µ is a representing measure of ˜ L , then ˜ µ ◦ f − . There exists afunction g : R n → [0 , ∞ ) such that f ( g ( x )) = x for all x ∈ R n and if µ is arepresenting measure of L , then µ ◦ g − is a representing measure of ˜ L .Proof. Since R n and [0 , ∞ ) are Souslin sets and f is surjective, by Jankoff’s The-orem 2.12 there exists a function g : R n → [0 , ∞ ) such that f ( g ( x )) = x for all x ∈ R n . It follows that ˜ L is well defined by ˜ L ( v ◦ f ) = L ( v ).(i) → (ii): Let µ be a representing measure of L , then˜ L ( v ◦ f ) = L ( v ) = Z R n v ( x ) d µ ( x ) = Z R n v ( f ( g ( x ))) d µ ( x ) Lemma 2.1 = Z ∞ ( v ◦ f )( t ) d( µ ◦ g − )( t ) , i.e., µ ◦ g − is a representing measure of ˜ L .(ii) → (i): Let ˜ µ be a representing measure of ˜ L , then L ( v ) = ˜ L ( v ◦ f ) = Z ∞ ( v ◦ f )( t ) d˜ µ ( t ) Lemma 2.1 = Z R n v ( x ) d(˜ µ ◦ f − )( x ) , i.e., ˜ µ ◦ f − is a representing measure of L . (cid:3) Remark . Similar to Theorem 4.4 we get that for any ε > g ε fromLemma 4.12(i) L : R [ x , . . . , x n ] → R is a moment functionalimplies that(ii) L : R [ x , . . . , x n ] → R continuously extends to L : R [ x , . . . , x n , g ] → R suchthat ˜ L : R [ t ] → R defined by ˜ L ( t d ) := L ( g d ) is a [0 , ∞ )-moment functional,i.e., L : R [ x , . . . , x n ] → R id X ˜ L : R [ t ] → R g L : R [ x , . . . , x n , g ] → R . That follows easily from the fact that 0 ≤ g ε ( x ) ≤ k x k + ε ≤ k x k + 1 + ε ∈ R [ x , . . . , x n ]. However, it is open whether the strong direction (ii) → (i) as inTheorem 4.4 holds in general. In Theorem 4.4 compactness of K implied that R [ x , . . . , x n ] is dense in C ( K, R ) and hence f could be approximated and the rep-resenting measure of L is unique. On R n both do not hold and hence (ii) → (i) canso far not be ensured in the same fashion as in Theorem 4.4. ◦ RANSFORMATIONS OF MOMENT FUNCTIONALS 19
At the end of this section we want to discuss two things that can easily be missed.The first is a crucial technical remark and the second is a historical one.For most transformations we required that f : X → Y is surjective to applyJankoff’s Theorem 2.12 to get a right-side inverse g : Y → X , i.e., f ( g ( y )) = y for all y ∈ Y . E.g. in Theorem 4.4 we used this g directly to embed a [0 , L of L . However, for any f : X → Y of course f : X → f ( Y ) is surjective. If f is continuous and X Borel, then f ( X ) remainseven a Borel set. Otherwise f ( X ) is at least a Souslin set.To demonstrate, that f : X → Y needs to be surjective and the restriction f : X → f ( X ) can not be used, let L : R [ x , . . . , x n ] → R be a linear functionalsuch that L ( p ) ≥ p ∈ R [ x , . . . , x n ]. Let f ∈ R [ x , . . . , x n ], then define˜ L : R [ t ] → R by ˜ L ( t d ) := L ( f d ) for all d ∈ N . We have ˜ L ( p ) = L (( p ◦ f ) ) ≥ p ∈ R [ t ], i.e., ˜ L is a Hamburger moment functional and there exists a measure ν on R such that ˜ L ( p ) = Z R p ( t ) d ν ( t ) for all p ∈ R [ t ] , i.e., L ( f d ) = ˜ L ( t d ) = Z R t d d ν ( t ) for all d ∈ N . (10)The important thing is, that (10) does not imply that there exists a µ such that L ( f d ) = R R n f d ( x ) d µ ( x ) for all d ∈ N . Jankoff’s Theorem 2.12 incorrectly appliedin (X) would suggest that there is a g such that f ( g ( t )) = t , i.e., Z R t d d ν ( t ) (X) = Z R f ( g ( t )) d d ν ( t ) = Z R n f ( x ) d d( ν ◦ g − )( x )and hence ν ◦ g − is a representing measure for L ( f d ). Therefore (X) would imply L ( f ) ≥ f ∈ R [ x , . . . , x n ] with f ≥ ν ◦ g − is non-negative.Havilands Theorem then shows that L is a moment functional. But for L we onlyhad L ( p ) ≥ p ∈ R [ x , . . . , x n ] and for n ≥ L ( p ) ≥ ν ⊆ f ( R n ) holds to apply Jankoff’sTheorem 2.12.For the historical remark, in this study we frequently encountered the case wherea linear functional L : V → R (or its transformation) lives on measurable functions V , i.e., we apparently face the problem that our functions v ∈ V live on a measuablespace ( X , A ). But a main tool in the moment problem is the Riesz RepresentationTheorem and it works with (compactly supported) continuous functions on locallycompact Hausdorff spaces. While the linear functional is extended to compactlysupported continuous functions via e.g. the Hahn–Banach Theorem, changing orextending a measurable space ( X , A ) to a topological space, especially to a locallycompact Hausdorff space, is in general not possible. Another important case wherewe rather work on a measurable space than a locally compact Hausdorff space isthe Richter Theorem. Richter’s Theorem 4.15 (see [Ric57, Satz 4]) . Let V be a finite-dimensionalvector space of measurable functions on a measurable space ( X , A ) . Then everymoment functional L : V → R has a finitely atomic representing measure d X i =1 c i · δ x i with c i > , x i ∈ X , and d = dim V . This theorem of Richter from 1957 was in the broader mathematical communitynot known (despite the fact that it is stated in this generality e.g. in [Kem68,Thm. 1] and more recently in [FP01, pp. 198–199]). Several attempts where madeto generalize a much weaker result from [Tch57]. Richter’s Theorem can also becalled
Richter–Rogosinski–Rosenbloom Theorem to account for all contributions[Ric57, Rog58, Ros52]. See [dDS18] for more on the early history of this theorem.We include Appendix A to avoid a similar confusion how to handle the repre-sentations of linear functionals of measurable functions which live not necessarilyon a locally compact Hausdorff space. This question was already fully answeredby P. J. Daniell in 1918 [Dan18]. To our knowledge this result does not appear inany standard functional analytic textbooks or works on the moment problem. Itis treated e.g. in [Bog07, Ch. 7.8] and [Fed69, Ch. 2.5]. Especially the approach in[Fed69] via the outer measure gives a simple proof of the general statement whichworks without any completions in the lattice of functions.
Definition 4.16.
Let X be a space. We call a set F of functions f : X → R a lattice ( of functions ) if the following holds:i) c · f ∈ F for all c ≥ f ∈ F ,ii) f + g ∈ F for all f, g ∈ F ,iii) inf( f, g ) ∈ F for all f, g ∈ F ,iv) inf( f, c ) ∈ F for all c ≥ f ∈ F , andv) g − f ∈ F for all f, g ∈ F with f ≤ g .Some authors require that a lattice of functions is a vector space. But for provingDaniell’s Representation Theorem 4.17 it is only necessary that a lattice is a cone. Daniell’s Representation Theorem 4.17 (P. J. Daniell 1918 [Dan18]) . Let F be a lattice of functions on a space X and let L : F → R be such thati) L ( f + g ) = L ( f ) + L ( g ) for all f, g ∈ F ,ii) L ( c · f ) = c · L ( f ) for all c ≥ and f ∈ F ,iii) L ( f ) ≤ L ( g ) for all f, g ∈ F with f ≤ g ,iv) L ( f n ) ր L ( g ) as n → ∞ for all g ∈ F and f n ∈ F with f n ր g .There exists a measure µ on ( X , A ) with A := σ ( { f − (( −∞ , a ]) | a ∈ R , f ∈ F} ) such that L ( f ) = Z X f ( x ) d µ ( x ) for all f ∈ F . The most impressive part is that the functional L : F → R lives only on a lattice F of functions f : X → R where X is a set without any structure. Daniell’sRepresentation Theorem 4.17 provides a representing measure µ including the σ -algebra A for the measurable space ( X , A ).The proof of Daniell’s Representation Theorem 4.17 we give in Appendix A istaken from [Fed69, Thm. 2.5.2] with alterations to fit in the non-outer-measureapproach.Riesz Representation Theorem follows directly from Daniell’s RepresentationTheorem 4.17. C ( X , R ), X a locally compact Hausdorff space, is a lattice offunctions, (i) and (ii) are the linearity of L , (iii) non-negativity of L , and thecontinuity condition (iv) of L follows easily from uniform convergence in C ( X , R ).5. Conclusion and Open Questions
We end with some conclusions, outlook, and some open questions which appearedduring our investigation.
RANSFORMATIONS OF MOMENT FUNCTIONALS 21
We gained in Section 3 basic properties of the transformation of linear func-tioals. Especially in Theorem 3.3 that a strong transformation L s K implies that L is a moment functional if and only if K is a moment functional. In Lemma 3.2we have seen that L s K implies the weaker statements L K and K L . Soit is natural to ask if the reverse holds. Open Problem 5.1.
Does L K and K L imply L s K ?Additionally, can the requirement of a strong transformation be weakened?While we have seen that surjectivity of f : X → Y is necessary and can in generalnot be omitted, it should be possible to weaken the condition that
V ◦ f = U from L : V → R and K : U → R . It is in fact only necessary that V and U (and therefore L and K ) can be extended to some V ⊇ V and
U ⊇ U such that
V ◦ f = U .In Proposition 4.1 we have seen that for a moment functional L with an atomlessrepresenting measure there exists an integrable function f such that L extended to L : V + R [ f ] → R which obeys L | R [ f ] = L Leb , i.e., L ( f d ) = L (1) d +1 for all d ∈ N .Because of the simplicity of L Leb in Example 1.1, are there other “directions”, i.e., f ’s, with similar properties? Open Problem 5.2.
Are there other “directions” f with L ( f d ) = L (1) d +1 or a similarbehavior?The importance of this question is again revealed in Theorem 4.4 where we havea similar structure in (6): L : V → R id X ˜ L : R [ t ] → R g L : V + R [ g ] → R . There exists a function g : K → [0 ,
1] such that: A linear functional L : V → R is a K -moment problem if and only if it continuously extends to some L : V + R [ g ] → R and ˜ L : R [ t ] → R defined by ˜ L ( t d ) := L ( g d ) for all d ∈ N is a [0 , L : R [ x , . . . , x n ] → R be a linear functional with L ( p ) ≥ p ∈ R [ x , . . . , x n ]. ( C [ x , . . . , x n ] , h · , · i ) with h p, q i := L ( p · q ) is a pre-Hilbert space viacomplexification of L by linearity (and removing the possible kernel of L ), and forall i = 1 , . . . , n the multiplication operators X i are defined by ( X i p )( x , . . . , x n ) := x i · p ( x , . . . , x n ) for all p ∈ C [ x , . . . , x n ]. ( X , . . . , X n ) is a tuple of commutingsymmetric operators on ( C [ x , . . . , x n ] , h · , · i ). Then L is a moment functional ifand only if ( X , . . . , X n ) extends to a tuple ( X , . . . , X n ) of communting self-adjointoperators on some Hilbert space H ⊃ ( C [ x , . . . , x n ] , h · , · i ).But extending L to R [ x , . . . , x n , g ] ⊇ R [ x , . . . , x n ] + R [ g ] gives L : R [ x , . . . , x n ] → R id X ˜ L : R [ t ] → R g L : R [ x , . . . , x n , g ] → R . By Theorem 4.4 it is sufficient to ensure that the multiplication operator G on C [ x , . . . , x n , g ], i.e., ( Gp )( x ) := g ( x ) · p ( x ), has a self-adjoint extension. So thetuple ( X , . . . , X n ) is replaced by G and the open question is loosely the following: Open Problem 5.3.
What is the functional analysis behind the g in Theorem 4.4?Note, that in the setting of Theorem 4.4 the multiplication operators are boundedsince K is compact. In the setup of K = R n , see Remark 4.14, we have in generalunbounded operators and only the easy direction (i) → (ii) was shown. It is open if(ii) → (i) also holds in the unbounded case. Open Problem 5.4.
Does (ii) → (i) in Remark 4.14 holds in general or is there acounter example?In Theorem 4.7 we have seen that this g in Theorem 4.4 can be approximated bypolynomials g ε ∈ R [ x , . . . , x n ]. So a natural question (especially in applications)is to ask the following: Open Problem 5.5.
How does deg g ε of g ε in Theorem 4.7 grow with ε → g in Theorem 4.4 is only a measurable function but not apolynomial even for V = R [ x , . . . , x n ] is a consequence of the reduction of thedimension. We reduce the dimension of K , in general dim K ≥
2, to 1, i.e., thedimension of [0 , f not necessarily needs to reducethe dimension of K .To remain in the algebraic setup we have to investigate transformations f oflinear functionals on R [ x , . . . , x n ] where f is a (bi)rational or polynomial function.Since a linear functional L is a moment functional if and only if L ( f ) ≥ f ≥ K , f ∈ R [ x , . . . , x n ], i.e., it has long been known that moment func-tionals are closely related to non-negative polynomials (Haviland Theorem), thesetransformations of moment functionals with (bi)rational or polynomial functionsmight give deeper insight into non-negative polynomials. Open Problem 5.6.
Do transformations f of moment functionals with poly-nomial or (bi)rational f give deeper insight into/characterizations of non-negativepolynomials? Appendix A. Daniell’s Representation Theorem
In this section we give a proof of Daniell’s Representation Theorem 4.17 from1918 [Dan18] in more recent mathematical notations following the proof in [Fed69,Thm. 2.5.2].
Definition A.1.
Let X be a set. A set function µ : P ( X ) → [0 , ∞ ] withi) µ ( ∅ ) = 0,ii) µ ( A ) ≤ µ ( B ) for all A ⊆ B ⊆ X ,iii) µ ( S ∞ i =1 A i ) ≤ P ∞ i =1 µ ( A i ) for all A i ∈ X is called a ( Carath´eodory) outer measure . Definition A.2.
For an outer measure µ on X a set A ⊆ X is called ( Carath´eodory ) µ -measurable if for every E ⊆ X we have µ ( E ) = µ ( E ∩ A ) + µ ( E \ A ). Remark
A.3 . Since by Definition A.1(iii) we always have µ ( E ) = µ (( E ∩ A ) ∪ ( E \ A )) ≤ µ ( E ∩ A ) + µ ( E \ A )it is sufficient for µ -measurability to test µ ( E ) ≥ µ ( E ∩ A ) + µ ( E \ A ) . (11)An outer measure is in fact a measure on all its measurable sets. Theorem A.4.
Let µ be an outer measure on a set X and A µ ⊆ P ( X ) be theset of all µ -measurable sets. Then A µ is a σ -algebra of X and µ is a measure on ( X , A µ ) .Proof. See e.g. [Bog07, Thm. 1.11.4(iii)]. (cid:3)
Outer measures give another characterization of measurable functions.
RANSFORMATIONS OF MOMENT FUNCTIONALS 23
Lemma A.5.
Let µ be an outer measure on X and f : X → [ −∞ , ∞ ] be a function.Then f is µ -measurable if and only if µ ( A ) ≥ µ ( { x ∈ X | f ( x ) ≤ a } ) + µ ( { x ∈ X | f ( x ) ≥ b } ) for all A ⊆ X and −∞ < a < b < ∞ .Proof. See e.g. [Fed69, 2.3.2(7) pp. 74/75]. (cid:3)
Definition A.6.
An outer measure µ is called regular if for each set A ⊆ X thereexists a µ -measurable set B ⊆ X with A ⊆ B and µ ( A ) = µ ( B ).Let f, g : ( X , A ) → R be two functions. Then we define inf( f, g ) by inf( f, g )( x ) :=inf( f ( x ) , g ( x )) for all x ∈ X and similarly sup( f, g ). Additionally, f ≤ g iff f ( x ) ≤ g ( x ) for all x ∈ X .Given a lattice F the following result [Fed69, 2.5.1, p. 91] shows that it inducesanother lattice F + by taking only the non-negative functions. Lemma A.7.
Let F be a non-empty lattice on a space X and set F + := F ∩ { f : X → R | f ≥ } . Theni) f + , f − , | f | ∈ F + for all f ∈ F andii) F + is a non-empty lattice on X .Proof. i): Since inf( f, ∈ F and inf( f, ≤ f we have f + = sup( f,
0) = f − inf( f, ∈ F + for all f ∈ F . Since f ≤ f + = sup( f, ∈ F we have f − = f + − f ∈F + for all f ∈ F . It follows that | f | = f + + f − ∈ F + for all f ∈ F .ii): Since F is non-empty there is a f ∈ F and by (ii) we have | f | ∈ F and hence | f | ∈ F + . F + is a lattice by directly checking the Definition 4.16. (cid:3) Note, that h n ր g means a sequence ( h n ) n ∈ N with h ≤ h ≤ ... ≤ g , i.e.,point-wise non-decreasing, with lim n →∞ h n ( x ) = g ( x ) for all x ∈ X . Equivalently, h n ց n →∞ h n ( x ) = 0 forall x ∈ X . Proof of Daniell’s Representation Theorem 4.17.
By assumption (iii) we have L ( f ) ≥ L (0 · f ) = 0 for all f ∈ F + .For any A ⊆ X we say a sequence ( f n ) n ∈ N suits A if and only if f n ∈ F + and f n ≤ f n +1 for all n ∈ N andlim n →∞ f n ( x ) ≥ x ∈ A. Note, that we can even assume equality by replacing the f n ’s by ˜ f n = inf( f n , ∈F + . Then we define µ ( A ) := inf n lim n →∞ L ( f n ) (cid:12)(cid:12)(cid:12) ( f n ) n ∈ N suits A o (12)which is ∞ if there is no sequence ( f n ) n ∈ N that suits A .We prove that µ is an outer measure, see Definition A.1. By assumption (iii) L ( f n ) is a non-negative increasing sequence and therefore lim n →∞ L ( f n ) exists andis in [0 , ∞ ] and therefore µ : P ( X ) → [0 , ∞ ]. For A = ∅ the zero sequence f n =0 ∈ F + is suited and therefore µ ( ∅ ) = 0. Let A ⊆ B ⊆ X , then a suited sequence( f n ) n ∈ N of B is also a suited sequence for A and therefore µ ( A ) ≤ µ ( B ). Let A i ⊆ X , i ∈ N , and set A := S ∞ i =1 A i . Any suited sequence for A is a suitedsequences for all A i . Assume there is an A i which has no suited sequence, then A has no suited sequence and µ ( A ) = ∞ ≤ P ∞ i =1 µ ( A i ) = ∞ . So assume all A i havesuited sequences, say ( f i,n ) n ∈ N suits A i , i ∈ N . Then f n := P ni =1 f i,n suits A and µ ( A ) ≤ lim n →∞ L ( f n ) = lim n →∞ n X i =1 L ( f i,n ) ≤ ∞ X i =1 lim m →∞ L ( f i,m ) . Taking the infimum on the right side for all A i ’s retains the inequality and gives µ ∞ [ i =1 A i ! = µ ( A ) ≤ ∞ X i =1 µ ( A i ) . Hence, all conditions in Definition A.1 are fulfilled and µ is an outer measure.Since µ is an outer measure on X by Theorem A.4 the set ˜ A of all µ -measurablesets of X is a σ -algebra and µ is a measure on ( X , ˜ A ).It remains to show that all f ∈ F are µ -measurable, µ is a measure on ( X , A )with A = σ ( { f − (( −∞ , a ]) | a ∈ R , f ∈ F} ), and L ( f ) = R X f ( x ) d µ ( x ) for all f ∈ F .Since f = f + − f − with f + , f − ∈ F + it is sufficient to show that every functionin F + is µ -measurable. So let f ∈ F + . To show that f is µ -measurable it issufficient to show that A := f − (( −∞ , a ]) = { x ∈ X | f ( x ) ≤ a } ∈ A for all a ∈ R ,i.e., A is µ -measurable by Definition A.2 resp. Remark A.3 if (11) holds for all E ⊆ X . From E \ A = E ∩ ( X \ A ) = E ∩ { x ∈ X | f ( x ) > a } we have to verify µ ( E ) ≥ µ ( E ∩ { x ∈ X | f ( x ) ≤ a } ) + µ ( E ∩ { x ∈ X | f ( x ) > a } )and by Lemma A.5 this is equivalent to µ ( E ) ≥ µ ( E ∩ { x ∈ X | f ( x ) ≤ a } | {z } =: E a ) + µ ( E ∩ { x ∈ X | f ( x ) ≥ b } | {z } =: E b ) ( ∗ )for all a < b . For a < µ ( E ) = ∞ ( ∗ ) is trivial, so assume a ≥ µ ( E ) < ∞ .Let ( g n ) n ∈ N be a sequence that suits E and set h := ( b − a ) − · [inf( f, b ) − inf( f, a )] ∈ F + and k n := inf( g n , h ) ∈ F + . Then we have 0 ≤ k n +1 − k n ≤ g n +1 − g n , h ( x ) = 1 for all x ∈ X with f ( x ) ≥ b, and h ( x ) = 0 for all x ∈ X with f ( x ) ≤ a. It follows that ( k n ) n ∈ N suits E b and ( g n − k n ) n ∈ N suits E a . Therefore,lim n →∞ L ( g n ) = lim n →∞ [ L ( g n − k n ) + L ( k n )] ≥ µ ( E a ) + µ ( E b )and taking the infimum on the left side retains the inequality and proves ( ∗ ). Hence,all f ∈ F + and therefore all f ∈ F are µ -measurable.Let us show that µ remains a measure on ( X , A ). Since all f ∈ F are µ - and A -measurable we have f − (( −∞ , a ]) ∈ ˜ A for all a ∈ R and f ∈ F . Therefore, A µ := σ ( { f − (( −∞ , a ]) | a ∈ R , f ∈ F} ) ⊆ ˜ A is a σ -algebra and we can restrict µ resp. ˜ A to A . µ is a measure on ( X , A ).We show that L ( f ) = R X f ( x ) d µ ( x ) holds for all f ∈ F + . Let f ∈ F + and set f t := inf( f, t )for t ≥
0. If ε > k ∈ N then0 ≤ f kε ( x ) − f ( k − ε ( x ) ≤ ε for all x ∈ X , RANSFORMATIONS OF MOMENT FUNCTIONALS 25 f kε ( x ) − f ( k − ε ( x ) = ε for all x ∈ X with f ( x ) ≥ kε, and f kε ( x ) − f ( k − ε ( x ) = 0 for all x ∈ X with f ( x ) ≤ ( k − ε. The constant sequence ( ε − · ( f kε − f ( k − ε )) n ∈ N suits { x ∈ X | f ( x ) ≥ kε } andconsequently L ( f kε − f ( k − ε ) ≥ ε · µ ( { x ∈ X | f ( x ) ≥ kε } ) ≥ Z X f ( k +1) ε ( x ) − f kε ( x ) d µ ( x ) ≥ ε · µ ( { x ∈ X | f ( x ) ≥ ( k + 1) ε } ) ≥ L ( f ( k +2) ε − f ( k +1) ε ) . Summing with respect to k from 1 to n we find L ( f nε ) ≥ Z X f ( n +1) ε ( x ) − f ε ( x ) d µ ( x ) ≥ L ( f ( n +2) ε − f ε )and since f nε ր f as n → ∞ we get from assumption (iv) for n → ∞ L ( f ) ≥ Z X f ( x ) − f ε ( x ) d µ ( x ) ≥ L ( f − f ε )which gives again from assumption (iv) for ε ց L ( f ) ≥ Z X f ( x ) d µ ( x ) ≥ L ( f ) . Hence, L ( f ) = R X f ( x ) d µ ( x ) for all f ∈ F + .Finally, for all f ∈ F we have f = f + − f − with f + , f − ∈ F + which implies Z X f ( x ) d µ ( x ) = Z X f + ( x ) d µ ( x ) − Z X f − ( x ) d µ ( x ) = L ( f + ) − L ( f − ) = L ( f )where the last equality follows from f + = f + f − and assumption (i). (cid:3) Remark
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