The multidimensional truncated Moment Problem: Carathéodory Numbers
aa r X i v : . [ m a t h . F A ] A p r THE MULTIDIMENSIONAL TRUNCATED MOMENT PROBLEM:CARATH´EODORY NUMBERS
PHILIPP J. DI DIO AND KONRAD SCHM ¨UDGEN
Abstract.
Let A be a finite-dimensional subspace of C ( X ; R ), where X isa locally compact Hausdorff space, and A = { f , . . . , f m } a basis of A . Asequence s = ( s j ) mj =1 is called a moment sequence if s j = R f j ( x ) dµ ( x ), j = 1 , . . . , m , for some positive Radon measure µ on X . Each moment se-quence s has a finitely atomic representing measure µ . The smallest possiblenumber of atoms is called the Carath´eodory number C A ( s ). The largest num-ber C A ( s ) among all moment sequences s is the Carath´eodory number C A . Inthis paper the Carath´eodory numbers C A ( s ) and C A are studied. In the caseof differentiable functions methods from differential geometry are used. Themain emphasis is on real polynomials. For a large class of spaces of polynomi-als in one variable the number C A is determined. In the multivariate case weobtain some lower bounds and we use results on zeros of positive polynomialsto derive upper bounds for the Carath´eodory numbers. AMS Subject Classification (2000) . 44A60, 14P10.
Key words: truncated moment problem, Carath´eodory number, convex cone,positive polynomials 1.
Introduction
The present paper continues the study of the truncated moment problem beganin our previous papers [Sch15] and [dDS]. Here we investigate the Carath´eodorynumber of moment sequences and moment cones.Throughout this paper, we assume that X is a locally compact topologicalHausdorff space, A is a finite-dimensional real linear subspace of C ( X ; R ) and A = { f , . . . , f m } is a fixed basis of the vector space A .Let s = ( s j ) mj =1 be a real sequence and let L s be the linear functional on A definedby L s ( f j ) = s j , j = 1 , . . . , m . We say that s is a moment sequence , equivalently, L s is a moment functional on A , if there exists a (positive) Radon measure µ on X such that f j is µ -integrable and s j = Z f j ( x ) dµ ( x ) for j = 1 , . . . , m, equivalently, L s ( f ) = Z X f ( x ) dµ ( x ) for f ∈ A . Such a measure µ is called a representing measure of s resp. L s . The Richter–Tchakaloff Theorem (see Proposition 1 below) implies that each moment sequencehas a k -atomic representing measure, where k ≤ m = dim A . The smallest number k is called the Carath´eodory number C A ( s ) and the smallest number K such thateach moment sequence s has a k -atomic representing measure with k ≤ K is the Carath´eodory number C A .Let L s be a moment functional. Determining a k -atomic representing measure ν for L s is closely related to the problem of finding quadrature or cubature formulas in numerical integration, see for instance [DR84], [SW97]. The Carath´eodory number C A ( s ) corresponds then to the smallest possible number of nodes.A large part of our considerations is developed in this general setup. Neverthelesswe are mainly interested in the case when A consists of real polynomials and X isa closed subset of R n or of the projective real space P ( R n ). In this case momentsequences are usually called truncated moment sequences in the literature.This paper is organized as follows. In Section 2, we define and investigateCarath´eodory numbers and the cone S A of moment sequences in the case when A ⊆ C ( X , R ). In Section 3, we assume that the functions of A are differentiable andapply differential geometric methods to study the moment cone and Carath´eodorynumbers. Important technical tools are the total derivative DS k, A ( C, X ) associatedwith a k -atomic measure µ = P ki =1 c i δ x i and the smallest number N A of atoms suchthat DS k, A ( C, X ) has full rank m = | A | . This number N A is a lower bound of theCarath´eodory number C A . The remaining four sections are concerned with polynomials. Section 4 dealswith polynomials in one variable. For A = { , x, . . . , x m } it is a classical fact that C A = (cid:6) m (cid:7) . We investigate a set A and its homogenization B with gaps, that is, A = { , x d , ..., x d m } and B = { y d , x d y d − d , ..., x d m − y d − d m − , x d } , where 0 = d < ... < d m = 2 d . Our main result (Theorem 45) gives sufficientconditions for the validity of the formula C A = C B = (cid:6) m (cid:7) .Sections 5–7 are devoted to the multivariate case. Except from a few simplecases the Carath´eodory number C A is unknown for polynomials in several variables.In Section 5 we give a new lower bound of C A and relate the number N A to theAlexander–Hirschowitz Theorem. Another group of main results of this paper isobtained in Section 6. Here we use known results on zeros of non-negative poly-nomials to derive upper bounds for Carath´eodory numbers (Theorems 57, 59, and62). Section 7 deals with signed Carath´eodory numbers and the real Waring rank.The multidimensional truncated moment problem was first studied in the The-sis of J. Matzke [Mat92] and independently by R. Curto and L. Fialkow [CF96a],[CF96b]. It is an active research topic, see e.g. [Ric57], [Kem68], [Rez92], [Sch15],[Lau09], [FN10], [CF13], [Fiaa], [Fiab], [dDS]. Carath´eodory numbers of multivari-ate polynomials have been investigated by C. Riener and M. Schweighofer [RS].Carath´eodory numbers of general convex cones are studied in [Tun01].For r ∈ R let ⌈ r ⌉ denote the smallest integer larger or equal to r .2. Carath´eodory Numbers: Continuous Functions
Let δ x be the delta measure at x ∈ R n , that is, δ x ( M ) = 1 if x ∈ M and δ x ( M ) = 0 if x / ∈ M . By a signed k -atomic measure µ we mean a signed measure µ = P kj =1 c j δ x j , where x , . . . , x k are pairwise different points of R n and c , . . . , c k are nonzero real numbers. If all numbers c , . . . , c k are positive, then µ is a positivemeasure and is called simply a k -atomic measure . The points x j are called theatoms of µ . The zero measure is considered as 0-atomic measure.A crucial result for our considerations is the Richter–Tchakaloff Theorem provedin [Ric57]. In the present context it can be stated as follows.
Proposition 1.
Each truncated moment sequence s of A has a k -atomic represent-ing measure with k ≤ m = | A | . Definition 2.
The moment cone S A ≡ S ( A , X ) is the set of all truncated X -momentsequences. Obviously, S A is a convex cone in R m . Since the functions f , . . . , f m form avector space basis of A , it follows easily that R m = S A − S A . HE MULTIDIMENSIONAL TRUNCATED MOMENT PROBLEM: CARATH´EODORY NUMBERS3
Definition 3.
The
Carath´eodory number C A ( s ) ≡ C A , X ( s ) of s ∈ S ( A , X ) is thesmallest k such that s has a k -atomic representing measure with all atoms in X .The Carath´eodory number C A ≡ C A , X of the moment cone S ( A , X ) is the smallestnumber C A such that each moment sequence s ∈ S ( A , X ) has a k -atomic representingmeasure with all atoms in X and k ≤ C A . Definition 4.
The signed Carath´eodory number C A , ± ( s ) ≡ C A , X , ± ( s ) of s ∈ R m isthe smallest number k such that s has a signed k -atomic representing measure withall atoms in X . The signed Carath´eodory number C A , ± ≡ C A , X , ± is the smallestnumber C A , ± such that every sequence s has a signed k -atomic representing measurewith all atoms in X and k ≤ C A , ± . Since R m = S A − S A as noted above, Proposition 1 implies each vector s ′ ∈ R m has a signed k -atomic representing measure, where k ≤ m , and we have C A ( s ) ≤ C A ≤ m for s ∈ S A and C ± ( s ′ ) ≤ C A , ± ≤ m for s ′ ∈ R m . (1) Remark 5.
The above definitions of moment sequences, moment cones and Cara-th´eodory numbers make sense for Borel functions rather than continuous functions.For instance, let x , . . . , x m be pairwise different points of R n and let A be theset of characteristic functions of the points x j . Then it is easily verified that theCarath´eodory number C A is equal to m = | A | . Definition 6.
The moment curve of A in R m is defined by (2) s A : X → R m , x s A ( x ) := f ( x ) ... f m ( x ) and we set (3) S k, A : ( R ≥ ) k × X k → R m , ( C, X ) S k, A ( C, X ) := k X i =1 c i · s A ( x i ) , where C = ( c , ..., c k ) , X = ( x , ..., x k ) . Clearly, s A ( x ) is the moment sequence of the delta measure δ x and S k, A ( C, X )is the moment sequence with representing measure µ = P ki =1 c i δ x i :(4) S k, A ( C, X ) = k X i =1 c i s A ( x i ) = (cid:18) Z X f j ( x ) dµ ( x ) (cid:19) mj =1 . By Proposition 1, each moment sequence s ∈ S A is of the form S m, A ( C, X ) for some(
C, X ) ∈ ( R ≥ ) m × X m . Further, let us introduce a convenient notation:(5) Pos( A , X ) ≡ Pos( X ) := { f ∈ A : f ( x ) ≥ x ∈ X } . The following proposition restates a known result (see e.g. Lemma 3 and Propo-sition 27(i) in [dDS]).
Proposition 7.
Suppose that s ∈ S A is a boundary point of S A . Then there exists p ∈ Pos( A , K ) , p = 0 , such that L s ( p ) = 0 and each representing measure of s issupported on the set of zeros Z ( p ) of p . The next proposition is a crucial technical ingredient of many proofs given below.The following condition is used at several places of this paper:(6)
There exists e ∈ A such that e ( x ) ≥ for x ∈ X . PHILIPP J. DI DIO AND KONRAD SCHM¨UDGEN
Proposition 8.
Let s ∈ S A and x ∈ X . Suppose that condition (6) is satisfied.Define (7) c s ( x ) := sup { c ∈ R : ( s − c · s A ( x )) ∈ S A } . Then c s ( x ) ≤ e ( x ) − L s ( e ) and ( s − c s ( x ) s A ( x )) ∈ ∂ S A . If K is compact, then the supremum in (7) is attained, the moment cone S A isclosed in R m , and we have (8) C A ≤ max {C A ( s ) : s ∈ ∂ S A } + 1 . Proof.
Let c ∈ R . If ( s − cs A ( x )) ∈ S A , then L s − cl x is a moment functional on A andtherefore ( L s − cl x )( e ) ≥
0, so that c ≤ e ( x ) − L s ( e ). Hence c s ( x ) ≤ e ( x ) − L s ( e ) . The definition of c s ( x ) implies that s − c s ( x ) s A ( x ) belongs to the boundary of S A .Since X is compact, it was shown in [FN10] that the moment cone S A is closedin R m . We choose a sequence ( c n ) n ∈ N such that s − c n s A ( x ) ∈ S A for all n andlim n c n = c s ( x ). Then s − c n s A ( x ) → s − c s ( x ) s A ( x ). Since S A is closed, we have( s − c s ( x ) s A ( x )) ∈ S A , that is, the supremum (7) is attained.Note that ( s − c s ( x ) s A ( x )) ∈ ∂ S A ∩S A . Obviously, C A ( s ) ≤ C A ( s − c s ( x ) s A ( x ))+1.This implies the inequality (8). (cid:3) The following example shows that the number c s ( x ) is not equal to(9) c s ( x ) := sup { c ∈ R : ( s − c · s A ( x )) ∈ S A } . However, if s ∈ int S A , then c s ( x ) = c s ( x ) by Proposition 10(vi) below. Example 9.
Set X = [ − , π ] , f ( x ) := 1 , f ( x ) := ( x ∈ [ − , x x ∈ (0 , π ] , f ( x ) := ( x + 1 x ∈ [ − , x x ∈ (0 , π ] , and g i = f i | [ − ,π ) for i = 1 , , . Set A = { f , f , f } and B = { g , g , g } . Then S A is closed, but S B is not closed. In fact, S B = S A . Let s = s A ( −
1) = (1 , , T , Then s ′ = s − s A (0) / / , , − / T = s A ( π ) / ∈ ∂ S A = ∂ S B , but s A ( π )
6∈ S A . Thus c s (0) = 0 and c s (0) = 1 / . Recall from [Sch15] the maximal mass function ρ L ( x ) of a moment functional L :(10) ρ L ( x ) := sup { µ ( { x } ) : µ is a representing measure of L } , x ∈ X . Proposition 10.
Suppose that condition (6) holds and retain the notation fromProposition 8.(i) s − c · s A ( x )
6∈ S A for all c > c s ( x ) .(ii) If s ∈ int S A , then s − c · s A ( x ) ∈ int S A for all c < c s ( x ) .(iii) The map int S A ∋ s c s ( x ) ∈ R is concave and continuous for all x ∈ X .(iv) The map X ∋ x c s ( x ) ∈ R is continuous for all s ∈ int S A .(v) c s ( x ) = ρ L s ( x ) .(vi) If s ∈ int S A , then c s ( x ) = c s ( x ) .Proof. (i) is clear from the definition (7).(ii): Since s is an inner point, there exists ε > B ε ( s ) ⊂ int S A . Fromthe convexity of S A it follows that B cs ( x ) − ccs ( x ) ε ( s − c · s A ( x )) ⊂ int S A ∀ c < c s ( x ) . (iii): Let s, t ∈ S A and λ ∈ (0 , c, c ′ ∈ R such that c < c s ( x ) and c ′ < c t ( x ). Then s − cs A ( x ) and t − c ′ s A ( x ) are in S A . Since S A is convex, we have λ [ s − cs A ( x )] + (1 − λ )[ t − c ′ s A ( x )]= [ λs + (1 − λ ) t ] − [ λc + (1 − λ ) c ′ ] s A ( x ) ∈ S A , HE MULTIDIMENSIONAL TRUNCATED MOMENT PROBLEM: CARATH´EODORY NUMBERS5 i.e., λc + (1 − λ ) c ′ ≤ c λs +(1 − λ ) t ( x ). Taking the suprema over c and c ′ it follows that λc s ( x ) + (1 − λ ) c t ( x ) ≤ c λs +(1 − λ ) t ( x ). Hence s c s ( x ) is a concave function andtherefore continuous on int S A by [Sch14, Thm. 1.5.3].(iv): Let x ∈ X . Let K be a compact neighborhood of x and ( x i ) i ∈ I a net in K such that lim i ∈ I x i = x . Since K is compact, we have e ( y ) ≥ δ > k s A ( y ) k ≥ δ for y ∈ K . Hence c s ( y ) is bounded on K , say by k , by Proposition 8. Since s A ( y )is continuous, there exist M > k c s ( y ) s A ( y ) k ≤ M on K . Further, from(i) and (ii) it follows that ∂ S A ∩ ( s + R · s A ( y )) = { s − c s ( y ) s A ( y ) } for y ∈ K .Define s ′ y := s − c s ( y ) s A ( y ). Then s ′ y ∈ B M ( s ) ∩ ∂ S A for all y ∈ K . Since ∂ S A is closed and B M ( s ) is compact, B M ( s ) ∩ ∂ S A is also compact. Therefore,( s ′ x i ) i ∈ I ⊆ B M ( s ) ∩ ∂ S A has an accumulation point, say a . Since ∂ S A is closed, a ∈ ∂ S A . Since c s ( x i ) is bounded by k and s A is continuous, |h v, s ′ x i − s i| = |h v, − c s ( x i ) s A ( x i ) i| ≤ k · |h v, s A ( x i ) i| → k · |h v, s A ( x ) i| = 0for all v ⊥ s A ( x ), i.e., a − s ∈ [ − k, k ] · s A ( x ), so that a ∈ s + [ − k, k ] · s A ( x ). Then a ∈ ∂ S A ∩ ( s + [ − k, k ] · s A ( x )) ⊆ ∂ S A ∩ ( s + R · s A ( x )) = { s ′ x } , so ( s ′ x i ) i ∈ I has a unique accumulation point s ′ x . Thus lim i ∈ I s ′ x i = s ′ x . This provesthat the map y s ′ y is continuous at x . Therefore, k s − s ′ y k · k s A ( y ) k − = k c s ( y ) s A ( y ) k · k s A ( y ) k − = k c s ( y ) k = c s ( y )is continuous at x . Since x ∈ X was arbitrary, x c s ( x ) is continuous on X .(v): Let c ∈ R be such that ˜ s := s − c · s A ( x ) ∈ S A . Then L s = L ˜ s + c · δ x . Hencethere is a representing measure µ of s such that c ≤ µ ( { x } ) ≤ ρ L s ( x ). Taking thesupremum over c yields c s ( x ) ≤ ρ L s ( x ).Assume that c s ( x ) < ρ L s ( x ). By the definition of ρ L s ( x ), there exist a c ∈ ( c s ( x ) , ρ L s ( x )) and a representing measure µ of s such that µ ( { x } ) = c. Then˜ µ := µ − c · δ x is a positive Radon measure representing ˜ s = s − c · s A ( x ). But ˜ s
6∈ S A by (i), a contradiction. This proves that c s ( x ) < ρ L s ( x ) . Thus, c s ( x ) = ρ L s ( x ).(vi): Since s ∈ int S A , it follows from (i) and (ii) that ∂ S A ∩ ( s + R · s A ( x )) = { s ′ x = s − c s ( x ) s A ( x ) } . Both numbers s − c s ( x ) s A ( x ) and s − c s ( x ) s A ( x ) belong to the set on left hand sideset. Hence they are equal and therefore c s ( x ) = c s ( x ). (cid:3) From Proposition 10(iii) we easily derive that the supremum in (10) is attainedif X is compact. This was proved in [Sch15, Prop. 6] by using the weak topologyon the set of representing measures and the Portmanteau Theorem.The following example shows that (iv) is false in general if s ∈ ∂ S A . Example 11.
Let { x , ..., x } be the zero set of the Robinson polynomial, A thehomogeneous polynomials of degree 6 on P ( R ) , and s := P i =1 s A ( x i ) . By Theorem18 and Example 18 in [dDS] , s is determinate. Therefore, ρ s ( x ) = c s ( x ) = ( for x ∈ { x , ..., x } , else. If K is not compact, then the supremum in (7) is not attained in general. Thisis shown by the following simple example. Example 12.
Let X = R , A = { , x, x } . Set s = (1 , , T = ( s A ( −
1) + s A (1)) .Clearly, s A (0) = (1 , , T . Then c s (0) = 1 , but s ′ = s − c s (0) s A (0) = (0 , , T isnot in S A . The following theorem improves the first equality in (1) and Proposition 1.
PHILIPP J. DI DIO AND KONRAD SCHM¨UDGEN
Theorem 13.
Suppose that condition (6) holds. If m ≥ and X has at most m − path-connected components, then C A ≤ m − .Proof. Obviously, the Carath´eodory number C A depends only on the linear span A ,but not on the particular basis A of A = Lin A . Hence we can assume without lossof generality that e = f m . Since e ( x ) > X by assumption, b j := f j e − ∈ C ( X )for j = 1 , . . . , m . Set B = { b , . . . , b m } . Let s be a moment sequence of B . First we prove that s has a finitely atomicrepresenting measure of a most m − s m = 1. By Proposition 1, s has a k -atomic measure µ = P kj =1 c j δ x j , where k ≤ m and x j ∈ X and c j > j . If k < m , we are done, so we can assume that k = m . Since X consists of at most m − x i , say x and x , are in the same component, say X , of X .Then there is a connecting path γ : [0 , → X such that γ (0) = x and γ (1) = x .For t ∈ [0 ,
1] we denote by ∆ t the simplex in R m − × { } spanned by the points s B ( x ) , s B ( γ ( t )) , s B ( x ) , . . . , s B ( x m ) . Since s m = 1, we have P mj =1 c j = 1. Hence s := ( s , . . . , s m ) belongs to the convex hull of s B ( x ) , s B ( x ) , s B ( x ) , . . . , s B ( x m ),that is, s is in the simplex ∆ . By decreasing t to 0 it follows from the continuity of b i that there exists a t ∈ [0 ,
1] such that s belongs to the boundary of the simplex∆ t . Then s is a convex combination of at most m − k -representing measure ˜ µ of s with k ≤ m − A has a k -atomic representingmeasure with k ≤ m −
1. This in turn implies the assertion C A ≤ m −
1. Let s ′ bea moment sequence of A and let µ ′ be a finitely atomic representing measure of s ′ .Let s be the moment sequence of B given by the measure e ( x ) dµ . As shown in thepreceding paragraph, s has a k -atomic representing measure ν , where k ≤ m − e ( x ) − dν is a k -atomic representing measure of s ′ . (cid:3) Corollary 14.
Let A = { p ∈ R [ x , . . . , x n ] : deg( p ) ≤ d } and X = R n . Then C A ≤ | A | − (cid:18) n + dn (cid:19) − . We give two somewhat pathological examples. Example 15 shows that the as-sertion of Theorem 13 is not true if the assumption on the function e ( x ) is omitted. Example 15.
Set ϕ ( x ) := x for x ∈ [0 , , − x + 2 for x ∈ (1 , , elsewhere. ϕ ( x ) := ϕ ( x ) , ϕ ( x ) := ϕ ( x − , ϕ ( x ) := ϕ ( x − . Then A := { ϕ , ϕ , ϕ } ⊂ C ( R ) . Using the moment sequence s = (1 , , we find that C A = 3 . Example 16 gives a three-dimensional moment cone with C A = 1. A slightmodification of this idea yields for m ∈ N an m -dimensional space A such that C A = 1. Example 16.
Let x L and y L be the coordinate functions of a space filling curve [Sag94, Ch. 5] , i.e., x L , y L : [0 , → [0 , are continuous, nowhere differentiable onthe Cantor set C , differentiable on [0 , \ C , and the curve ( x L , y L ) : [0 , → [0 , is surjective. Set A := { x L , y L , } and X = [0 , . Then ( ∗ ) s A ([0 , , × { } and the moment cone S A = { ( x, y, z ) : z ≥ , ≤ x ≤ z, ≤ y ≤ z } is full-dimensional. Clearly, ( ∗ ) implies that C A = 1 . HE MULTIDIMENSIONAL TRUNCATED MOMENT PROBLEM: CARATH´EODORY NUMBERS7
Remark 17.
In this paper the vector space A is finite-dimensional. However thedefinitions of the moment cone and the Carath´eodory number can be extended toinfinite-dimensional vector spaces A . The following example shows that even in thiscase it is possible that C A = 1 . Let A = { ϕ n } n ∈ N be the coordinate functions of the ℵ -dimensional Sch¨onberg space filling curve [Sag94, Ch. 7] , i.e., ϕ n is continuousand nowhere differentiable on [0 , for all n , and set ϕ = 1 . Then ( ∗ ) ( ϕ n ) n ∈ N : [0 , → { } × [0 , N is surjective. The moment cone S A = { ( x n ) n ∈ N : 0 ≤ x n ≤ x } is full dimensional,closed, and C A = 1 from ( ∗ ) . Theorem 18.
Let p ∈ A and x , . . . , x k ∈ X , k ∈ N . Suppose that p ( x ) ≥ for x ∈ X , Z ( p ) = { x , ..., x k } and the set { s A ( x i ) : i = 1 , ..., k } is linearly independent.Then C A ≥ k .Proof. Let s = P ki =1 s A ( x i ). Clearly, L s ( p ) = 0 and hence supp µ ⊆ Z ( p ) = { x , . . . , x k } for any representing measure µ of s by Proposition 7. Assume thereis an at most ( k − µ . Without loss of generality weassume that x / ∈ supp µ , so µ is of the form µ = P ki =2 c i δ x i , c i ≥
0. Then0 = s − s = k X i =1 s A ( x i ) − k X i =2 c i s A ( x i ) ⇒ s A ( x ) = k X i =2 ( c i − s A ( x i ) . Since the set { s A ( x i ) : i = 1 , ..., k } is linear independent, this is a contradiction.Therefore, k = C A ( s ) ≤ C A . (cid:3) Applications of the previous theorem will be given in Examples 31 and 63.Deeper results on the connections between the Carath´eodory number and the zerosof positive polynomials are treated in Section 6.We derive some useful facts which will be used several times. We investigatesome properties of the set S k := range S k, A of moment sequences which are given by measures of at most k atoms. Lemma 19.
For fixed k ∈ N the following are equivalent:(i) S k is convex, or equivalently, S k + S k ⊆ S k .(ii) S k = S k +1 .(iii) k ≥ C A .Proof. (i) ⇒ (ii): Let s = (1 − λ ) s + λs A ( x ) ∈ S k +1 with s , s A ( x ) ∈ S k . Since S k is convex , s ∈ S k . Hence S k = S k +1 .(ii) ⇒ (iii): Let s = s + λ s A ( x ) + ... + λ l s A ( x l ) ∈ S k + l be an arbitrary momentsequence. Set s i := s + λ s A ( x ) + ... + λ l s A ( x i ). Then s = s + λ s A ( x ) ∈ S k +1 = S k ⇒ s = s + λ s A ( x ) ∈ S k +1 = S k ... ⇒ s = s l − + λ l s A ( x l ) S k +1 = S k Thus C A ≤ k .(iii) ⇒ (ii): Since C A ≤ k , we have S C A ⊆ S k ⊆ S C A . Here the last inclusion followsfrom the mimimality of C A . Hence, S k = S C A is convex. (cid:3) An immediate consequence of the preceding lemma are the following inclusions: { } = S $ S $ ... $ S C A = S C A + j , j ∈ N . (11) Proposition 20. (i) C A = min { k : S k is convex } = min { k : S k = S k +1 } . PHILIPP J. DI DIO AND KONRAD SCHM¨UDGEN (ii) For each k = 0 , , ..., C A there is a moment sequence s such that C A ( s ) = k .Proof. (i) follows at once from the minimality of C A in Lemma 19.(ii): By (11), we have S k − $ S k for k = 0 , ..., C A , where we set S − := ∅ .Therefore, S k \ S k − = ∅ . (cid:3) Proposition 21.
Suppose that condition (6) is satisfied.(i) The cone S is pointed, that is, S ∩ ( −S ) = { } . (ii) If S is closed, then S k is closed for all k .(iii) If the set X is compact, then S k is closed for all k .Proof. (i): Suppose that s, − s ∈ S . Using that e ( x ) > X we conclude that L s ( e ) ≥ L − s ( e ) = − L s ( e ) ≥
0, so L s ( e ) = 0 and therefore s = 0.(ii): The proof follows by induction. Assume S and S k is closed for some k . Weshow that S k +1 is also closed.Let ( s n ) n ∈ N be a sequence of S k +1 such that s n → s ∈ S k +1 . We can write s n = α n x n + β n y n such that x n ∈ S k , y n ∈ S , α n , β n ∈ [0 , + ∞ ), and k x n k = k y n k = 1for all n . Since S k and S are closed, the sets S k ∩ B (0) and S ∩ B (0) are bothcompact. Hence we can find a subsequence ( n i ) such that x n i → x ∈ S k ∩ B (0) , and y n i → y ∈ S ∩ B (0). Let us assume for a moment that the sequences ( α n i )and ( β n i ) are bounded. There is a subsequence n i j such that α n ij → α ∈ [0 , + ∞ )and β n ij → β ∈ [0 , + ∞ ). Then s n ij → s = αx + βy ∈ S k +1 . Thus, S k +1 is closed.We show that the sequence ( β n i ) is unbounded if ( α n i ) is unbounded. Takingthe standard scalar product h · , ·i in R m , we can uniquely write y n = y ⊥ n + y k n with x n k y k n , x n ⊥ y ⊥ n . Then k s n i k = k α n i x n i + β n i y k n i k = k α n i x n i + β n i y k n i k + β n i k y ⊥ n i k ≥ β n i k y ⊥ n i k . Since ( s n i ) converges, the sequence ( k s n i k ) is bounded by some k . Thus, k ≥ k α n i x n i + β n i y k n i k ≥ | α n i k x n i k − β n i k y k n i k| = | α n i − β n i k y k n i k| and if ( α n i ) is unbounded, so ( β n i ) is unbounded. The same reasoning shows that( α n i ) is unbounded if ( β n i ) is unbounded.If the sequence ( β n i ) is unbounded, ( y ⊥ n i ) converges to 0 and hence y = − x .Since S is pointed by (i), this implies x = y = 0, a contradiction to k x k = k y k = 1.This completes the proof.(iii): By (ii) it suffices to prove that S is closed. Clearly, condition (6) impliesthat s A ( x ) = 0 for all x ∈ X . Since A ⊆ C ( X , R ) and X is compact, we have k s A k − s A ∈ C ( X , S m − ) ( S m − denotes the unit sphere in R m ) and range k s A k − s A is closed. Hence, S ≡ R ≥ · range k s A k − s A is closed. (cid:3) More on the moment cone can be found in Proposition 30.3.
Caratheodory Numbers: Differentiable Functions
In the rest of this paper we assume that X = R n or P ( R n ) and A is a finite-dimensional linear subspace of C r ( R n ; R ), r ∈ N .Clearly, S k, A in Definition 6 is a C r -map of R k ≥ × R kn into R m . Let DS k, A denote its total derivative. We can write DS k, A = ( ∂ c S k, A , ∂ x (1)1 S k, A , ..., ∂ x ( n )1 S k, A , ∂ c S k, A , ..., ∂ x ( n ) k S k, A )= ( s A ( x ) , c ∂ s A | x = x , ..., c ∂ n s A | x = x , s A ( x ) , ..., c k ∂ n s A | x = x k ) . (12)The following number is crucial in what follows. HE MULTIDIMENSIONAL TRUNCATED MOMENT PROBLEM: CARATH´EODORY NUMBERS9
Definition 22. (13) N A := min { k ∈ N : rank DS k, A = m } , i.e., N A is the smallest number k of atoms such that DS k, A has full rank m = | A | . A lower bound for N A is given by the following proposition. Proposition 23.
We have l | A | n +1 m ≤ N A . If all functions f i are homogeneous ofthe same degree, then l | A | n m ≤ N A .Proof. Since DS k, A has | A | rows and each atom contributes n + 1 columns, we needat least k ≥ | A | n +1 atoms for full rank. Thus, N A ≥ l | A | n +1 m . If all functions f i are homogeneous of degree r , then f i ( λx ) = λ r f i ( x ) and so δ λx = λ r δ x . Hence DS , A has rank at most d and kernel dimension at least 1.Therefore, at least k ≥ | A | n atoms are needed, so that N A ≥ l | A | n m . (cid:3) Example 24.
Let ϕ ∈ C ∞ ( R ) , ϕ = 0 , and supp ( ϕ ) ⊆ (0 , . Set ϕ i ( x ) := ϕ ( x − i + 1) for i = 1 , ..., m ∈ N and A := { ϕ , ..., ϕ m } . Then ∂s A ( x ) = ϕ ′ i ( x ) e i for x ∈ ( i − , i ) or otherwise. Then N A = C A = m. Theorem 25.
Suppose that A ⊆ C ( R n , R ) . Then (14) C A , ± ≤ N A . Set C = (1 , ..., ∈ R N A . There exists X ∈ R N A · n and an open neighborhood U of ( C, X ) such that for every ε > there are ( C ε , X ε ) ∈ U and λ ε ∈ R such that (15) s = λ ε ( S A , N A ( C, X ) − S A , N A ( C ε , X ε )) . Proof.
It clearly suffices to prove the second part of the theorem. The first assertionfollows then from the second.Since DS N A has full rank, there is a ( C, X ) ∈ R N A > × R N A n such that DS N A ( C, X )has full rank. Since scaling the columns of DS N A ( C, X ) does not change the rank,we can assume without loss of generality that C = (1 , ..., U of ( C, X ) such that( ∗ ) S N A , A ( C, X ) ∈ int S N A , A ( U ) . Let s ∈ R m . By ( ∗ ) there is a ( C ε , X ε ) ∈ U such that S A , N A ( C, X ) − S A , N A ( C ε , X ε )is a multiple of s , i.e., (15) holds for some λ ε ∈ R . (cid:3) Definition 26.
Let n, k ∈ N with k ≥ N A . A k -atomic measure ( C, X ) on R n is called regular (for S k, A ) iff DS k, A ( C, X ) has full rank. Otherwise the measure ( C, X ) is called singular (for S k, A ) .A real sequence s = ( s α ) α ∈ A is called regular iff S − k, A ( s ) is empty (that is, s is not a moment sequence) or consists solely of regular measures. Otherwise, s iscalled singular . Theorem 27.
Suppose that A ⊂ C r ( R n ; R ) and r > N A · ( n + 1) − m . Then (16) N A ≤ C A . Further, the set of moment sequences s which can be represented by less than N A atoms has | A | -dimensional Lebesgue measure zero in R m .Proof. By Proposition 23 we have r > N A · ( n + 1) − m ≥
0, so that r ≥ N A atoms aresingular values. Hence the second assertion follows from Sard’s Theorem [Sar42].To prove (16) assume to the contrary that C A < N A . Then every momentsequence in the moment cone is singular. This is a contradiction to Sard’s Theoremsince the moment cone has non-empty interior. (cid:3) Remark 28.
Theorem 27 also holds for the signed Carath´eodory number with ver-batim the same proof. With Theorem 25 we get (17) N A ≤ C A , ± ≤ N A for A ⊂ C r ( R n ; R ) and r > N A · ( n + 1) − m . Without these conditions the lowerbound needs not to hold, neither for C A nor for C A , ± , see [Fed69, pp. 317–318] . Proposition 29.
Suppose that A ⊆ C ( R n , R ) and { x ∈ R n : s A ( x ) = 0 } isbounded. Let γ ∈ C ( R n , R ) be such that γ ( x ) ≥ and lim | x |→∞ f i ( x ) γ ( x ) = 0 for i = 1 , . . . , m and let s be a moment sequence of A . Set (18) Γ l,c ( s ) := S − A , C A ( s )+ l ( s ) ∩ ( ( C, X ) ∈ R C A ( s )+ l ≥ × R n · ( C A ( s )+ l ) (cid:12)(cid:12)(cid:12)(cid:12)(cid:12) X i c i γ ( x i ) ≤ c ) with l ∈ N and c ≥ . Then:(i) Γ l,c ( s ) is closed for all l ∈ N and c ≥ .(ii) Γ ,c ( s ) is compact for all c ≥ .If, in addition, s is regular, then:(iii) ∃ c ≥ l,c ( s ) unbounded ⇔ l ≥ .(iv) Γ l,c ( s ) compact ∀ c ≥ ⇔ l = 0 .Proof. (i): If f ∈ C ( R n , R m ) and K ⊆ R m is closed, then f − ( K ) is closed by thecontinuity of f . Since both intersecting sets in (18) are of the form f − ( K ), theyare closed and so is their intersection.(ii): Suppose Γ ,c is non-empty. Since Γ ,c is closed by (i), it suffices to showthat it is bounded. Assume to the contrary that it is unbounded and let ( C ( i ) , X ( i ) )be a sequence such that lim i →∞ k ( C ( i ) , X ( i ) ) k = ∞ . Since0 ≤ c ( i ) j ≤ c ( i ) j γ ( x j ) ≤ X l c ( i ) l γ ( x l ) ≤ c the sequence ( C ( i ) ) is bounded. The sequence ( X ( i ) ) is unbounded. After renum-bering and passing to subsequences we can assume that c ( i ) j → c ∗ j for all j , x ( i ) j → x ∗ j for j = 1 , ..., k and k x ( i ) j k → ∞ for j = k + 1 , ..., C A ( s ) as i → ∞ . Since s = S A , C A (( C ( i ) , X ( i ) )) = k X j =1 c ( i ) j s A ( x ( i ) j ) + C A X i = k +1 c ( i ) j s A ( x ( i ) j )for all i , it follows that s = lim i →∞ k X j =1 c ( i ) j s A ( x ( i ) j ) + lim i →∞ C A X j = k +1 c ( i ) j s A ( x ( i ) j )= k X j =1 c ∗ j s A ( x ∗ j ) + lim i →∞ C A X j = k +1 s A ( x ( i ) j ) γ ( x ( i ) j ) | {z } → · c ( i ) j γ ( x ( i ) j ) | {z } ≤ c | {z } → = k X j =1 c ∗ j s A ( x ∗ j ) . Therefore, µ ∗ = (( c ∗ , ..., c ∗ k ) , ( x ∗ , ..., x ∗ k )) is a k -atomic representing measure of s with k ≤ C A ( s ). By the minimality of C A ( s ), k = C A ( s ). Hence all sequence ( x ( i ) j )are bounded which is a contradiction. Thus Γ ,c ( s ) is bounded.It is clear that (iii) and (iv) are equivalent. Thus it suffices to prove (iii). HE MULTIDIMENSIONAL TRUNCATED MOMENT PROBLEM: CARATH´EODORY NUMBERS11 (iii) “ ⇒ ”: By (ii), if Γ l,c ( s ) is unbounded, we find a k -atomic representing mea-sure with k < C A ( s ) + l , i.e., l ≥ ⇐ ”: We will show that there is a c > x ∈ R n there isa representing measure µ in Γ ,c ( s ) which has x as an atom. This will prove thatΓ ,c , hence also Γ l,c , is unbounded for l ≥ µ = ( C , X ) = (( c , , ..., c , C A ( s ) ) , ( x , , ..., x , C A ( s ) )) be a representing mea-sure of s . Set c := Z γ dµ + 1 . Since s is regular, all representing measures have full rank. Hence there existvariables y , ..., y m from c , ..., c C A ( s ) , x , , ..., x C A ( s ) ,n such that D y S ( µ ) is a squarematrix with full rank. Then F (( C, X ) , t ) = S C A ( s ) , A (( C, X )) − s + t · s A ( x )is a C -function such that F (( C , X ) ,
0) = 0 and D y F (( C , X )) = D y S ( µ ) isbijective. Thus, F fulfills all assumptions of the implicit function theorem, hencethere are an ε > C –function ( C ( t ) , X ( t )) such that F (( C ( t ) , X ( t )) , t ) = 0for all t ∈ ( − ε, ε ). Since c i, >
0, there is t ∈ (0 , ε ) such that c i ( t ) > i , so µ ( t ) = C A ( s ) X i =1 c i ( t ) δ x i ( t ) + t δ x with Z γ dµ ( t ) ≤ c is a ( C A ( s ) + 1)-atomic representing measure of s which has x as an atom. (cid:3) (iii) and (iv) no longer hold if s is singular. E.g., let s be moment sequence ofthe measure µ = P i =1 δ x i where x i are the ten zeros of the Robinson polynomial,then S − k, A ( s ) ⊆ [0 , k × { x , ..., x } k is compact for all k ≥ S k andthe Carath´eodory number. Recall that B ρ ( t ) is the ball with center t and radius ρ in R m . Proposition 30. (i) Suppose that S k − is closed for some k , N A ≤ k ≤ C A .Then there exist a moment sequence s and an ε > such that C A ( t ) = k for all t ∈ B ε ( s ) .(ii) s ∈ int S C A if and only if there exists ( C, X ) such that S A ( C, X ) = s and rank DS A ( C, X ) = | A | .(iii) s ∈ ∂ S C A if and only if rank DS A ( C, X ) < | A | for all ( C, X ) such that S A ( C, X ) = s .(iv) Suppose that N A < C A and S k is closed for all k = N A , ..., C A . Then for eachsuch k there exists s ∈ int S C A such that all k -atomic representing measuresof s are singular, but s has a regular representing measure with at least k +1 atoms.(v) Suppose that N A < C A , S k is closed for k = C A − , C A and S C A = R | A | . If R | A | \ S C A − is path-connected, there exists s ∈ ∂ S C A such that C A ( s ) = C A .Proof. (i): Fix such a number k and assume (int S k ) \ S k − = ∅ . Then we have S k − ⊇ int S k ⊇ int S k − . Taking the closure, S k − = S k − ⊇ S k ⊇ S k − = S k − ,so that S k ⊆ S k − which contradicts (11). Thus, (int S k ) \ S k − = ∅ .(ii): “ ⇐ ”: Let ( C, X ) be a full rank measure of s . Then a neighborhood U of( C, X ) is mapped onto a neighborhood of s , that is, s is an inner point.“ ⇒ ”: Let s be an inner point. Choose ν such that S A ( ν ) has full rank. Since s is an inner point, there exists ε > s ′ := s − ε · S A ( ν ) is also an innerpoint. In particular, s ′ is a moment sequence. Let µ ′ be a representing measureof s ′ . Then µ = µ ′ + ε · ν is a representing measure of s and has full rank, sincealready DS A ( ν ) has full rank. (iii) follows from (ii).(iv): Let s ∈ int S k ⊆ int S C A . By (i), there exists t ∈ (int S C A ) \ S k for all k = N A , ..., C A −
1. Then [ s, t ] := { λs + (1 − λ ) t | λ ∈ [0 , } ⊆ int S C A by theconvexity of S C A . Therefore, since s ∈ int S k but s int S k , we haveint S C A ∩ ∂ S k ⊇ [ s, t ] ∩ ∂ S k = ∅ . Hence there exists s ∈ int S C A ∩ ∂ S k . Then all k -atomic representing measures of s are singular. Otherwise, a full rank k -atomic measures implies that s is an innerpoint of S k . But, by (iv), s has a regular l -atomic measure with l > k .(v): Let s ∈ int S C A \ S C A − by (i) and t ∈ R | A | \ S C A . Since s, t ∈ R | A | \ S C A − ,they are path-connected, so there exists a continuous path γ : [0 , → R | A | \ S C A − with γ (0) = s and γ (1) = t . But since s = γ (0) ∈ int S C A , t = γ (1) int S C A ,and γ ([0 , ⊆ R | A | \ S C A − , we have γ ([0 , ∩ ( ∂ S C A \ S C A − ) = ∅ . Therefore, ∂ S C A \ S C A − = ∅ . (cid:3) In Sections 4 and 6 we derive upper bounds of C A by using Proposition 8 and theinequality (8). As (v) implies, this inequality can be strict, since the Carath´eodorynumber C A can be attained at a boundary point, see the following example. Example 31.
The (homogeneous) Motzkin polynomial M ( x, y, z ) = z + x y + x y − x y z has the projective roots Z ( M ) = { (1 , , , (1 , , − , (1 , − , , (1 , − , − , (1 , , , (0 , , } . We consider the truncated moment problem on the projective space P ( R ) for B := { z + x y + x y − x y z , x , y , z , x y, x z, x yz } . Clearly, M ∈ lin B . Since M is non-negative and has a discrete set of roots, s = P ξ ∈Z ( M ) s B ( ξ ) is a boundary point of the closed moment cone. The matrix ( s B ( ξ )) ξ ∈Z ( M ) = − − − − − − has rank , i.e., the set { s B ( ξ ) } ξ ∈Z ( M ) is linearly independent. Hence C B ( s ) = 6 byTheorem 18 and C B ≤ | B | − by Theorem 13. Thus, C B = C B ( s ) = 6 , that is,the Carath´eodory number is attained at the boundary moment sequence s . Next we derive an upper bound for the Carath´eodory number in terms of zerosof positive elements of A . For the rest of this section we assume that X is a closedsubset of R n or P ( R n ) and A ⊆ C ( X , R ). By the latter we mean that there existsan open subset U of R n or P ( R n ) such that X ⊆ U and A ⊆ C ( U , R ) . Definition 32.
Let M A be the largest number k obeying the following property: ( ∗ ) k : There exist f ∈ A and x , . . . , x k ∈ Z ( f ) such that f ( x ) ≥ on X and { s A ( x i ) } i =1 ,...,k is linearly independent ( DS k, A ((1 , ..., , ( x , ..., x k )) does not havefull rank). From the definition it is clear that M A is the largest dimension an exposed faceof S A . Proposition 33.
For each s ∈ ∂ S A ∩ S A we have C A ( s ) ≤ M A . HE MULTIDIMENSIONAL TRUNCATED MOMENT PROBLEM: CARATH´EODORY NUMBERS13
Proof.
In this proof we abbreviate N := C A ( s ). Let µ = P Ni =1 c i δ x i be an N -atomicrepresenting measure of s . Since s ∈ ∂ S A ∩ S A , there exists f ∈ A , f = 0 , suchthat f ( x ) ≥ X and L s ( f ) = 0. From the latter it follows that supp µ ⊆ Z ( f )and hence x , . . . , x N ∈ Z ( f ) . Further, by Proposition 30(iii), s ∈ ∂ S A ∩ S A impliesthat DS N, A ( C, X ) does not have full rank | A | . Since c i > i , we haverank DS N, A ( C, X ) = rank DS N, A ((1 , . . . , , X ) . Finally, by Theorem 18 the set { s A ( x i ) } i =1 ,...,N is linearly independent. Thus, property ( ∗ ) N in Definition 32 holds,so that C A ( s ) = N ≤ M A . (cid:3) Theorem 34.
Suppose that X is a compact subset of R n or P ( R n ) , condition (6)is satisfied, and A ⊆ C ( X , R ) . Then C A ≤ M A + 1 . Proof.
The assumptions of this theorem ensure that Proposition 8 applies. Hencethe assertion follows by combining Proposition 33 with the inequality (8). (cid:3) Carath´eodory Numbers: One-dimensional Monomial Case
For the one-dimensional truncated moment problem the number N A can be cal-culated from the formula for the Vandermonde determinant. Lemma 35.
Let A := { , x, ..., x n } , where n ∈ N .(i) If n = 2 k − , k ∈ N , then (19) det DS k, A = c · · · c k · Y ≤ i 1. The even case n = 2 k is derived in a similar manner.We have ∂ c i S k ( C, X ) = s A ( x i ) and ∂ x i S k ( C, X ) = c i s ′ A ( x i ). Therefore,det DS k, A = c · · · c k det( s A ( x ) , s ′ A ( x ) , ..., s A ( x k ) , s ′ A ( x k ))(21)and we computedet( s A ( x ) , s ′ A ( x ) , ..., s A ( x k ) , s ′ A ( x k ))= lim h → ... lim h k → det (cid:18) s A ( x ) , s A ( x + h ) − s A ( x ) h , ..., s A ( x k + h k ) − s A ( x k ) h k (cid:19) = lim h → ... lim h k → det( s A ( x ) , s A ( x + h ) , ..., s A ( x k + h k )) h · · · h k = lim h → ... lim h k → Q ki =1 (cid:16) h i Q kj = i +1 ( x j + h j − x i )( x j − x i )( x j − x i − h i )( x j + h j − x i − h i ) (cid:17) h · · · h k = Y ≤ i H. Richter [Ric57] has shown that for the one-dimensional truncatedmoment problem A = { , x, ..., x d } the Carath´eodory number is C A = (cid:6) d +12 (cid:7) . Thisresult will also follow from Theorem 45 below. If we take this equality for grantedand combine it with Lemma 35(iii), then we obtain N A = (cid:22) d (cid:23) + 1 = (cid:24) d + 12 (cid:25) = C A . Now we turn to the general case and assume that A = { x d , ..., x d m } , where 0 ≤ d < d < ... < d m , d , . . . , d m ∈ N . (22)Then we compute f A ( x ,..., x m ) := det( s A ( x ) , ..., s A ( x m )) = | ( x d j i ) i,j =1 ,...,m | = (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) x d x d · · · x d m x d x d · · · x d m ... ... . . . ... x d m x d m · · · x d m m (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) = ( x · · · x m ) d · (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) · · · x d − d x d − d · · · x d − d m ... ... . . . ... x d m − d x d m − d · · · x d m − d m (cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12)(cid:12) . From the latter equation it follows that each linear polynomial x j − x i , j = i , dividesthe polynomial f A . Hence there exists a polynomial p A such that f A ( x , ..., x m ) = | ( x d j i ) i,j =1 ,...,m | = ( x · · · x m ) d Y ≤ i Assume that A is as in (22) and p A is defined by (23). Set q A ( x , ..., x k ) := p A ( x , x , ..., x k , x k ) if m = 2 k is even and q A ,i ( x , ..., x k ) := p A ( x , x , ..., x i − , x i − , x i , x i +1 , x i +1 , ..., x k , x k ) for all i = 1 , ..., k if m = 2 k − is odd. Lemma 39. (i) If m is even then q A is symmetric.(ii) If m is odd then q A ,i ( x , ..., x k ) = q A ,k ( x , ..., x i − , x i +1 , ..., x k , x i ) for all i = 1 , ..., k .Proof. (i): Since the Schur polynomial p A is symmetric, so is q A .(ii): We derive q A ,i ( x , ..., x k ) = p A ( x , x , ..., x i − , x i − , x i , x i +1 , x i +1 , ..., x k , x k )= p A ( x , x , ..., x i − , x i − , x i +1 , x i +1 , ..., x k , x k , x i )= q A ,k ( x , ..., x i − , x i +1 , ..., x k , x i ) . (cid:3) In the odd case it suffices to prove formula (27) below. All other determinantsare then obtained by interchanging variables and Lemma 39(ii). Lemma 40. Suppose that A is of the form (22).(i) If m = 2 k is even, then det DS k, A ( c , ..., c k , x , ..., x k ) = c · · · c k · ( x · · · x k ) d Y ≤ i In the case B = { xy , x y , x y } we have det(( s B ( x i , y i )) i =1 ) = x y x y x y · Y ≤ i For even m = 2 k we define (29) q B ( x , y , . . . , x k , y k ) := ( y · · · y k ) d m − d − m )+3 q A (cid:18) x y , · · · , x k y k (cid:19) . HE MULTIDIMENSIONAL TRUNCATED MOMENT PROBLEM: CARATH´EODORY NUMBERS17 For odd m = 2 k − we set (30) q B ,k ( x , y , ..., x k , y k ) := ( y · · · y k − ) d m − d − m +6 y d m − d − m +1 k q A ,k (cid:18) x y , . . . , x k y k (cid:19) . Lemma 43. Let B be of the form (28).(i) If m = 2 k is even, then det D c,x S k, B ( c , ..., c k , x , y , ..., x k , y k )= c · · · c k · ( x · · · x k ) d · Y ≤ i Let B = { xy , x y , x y } . Then we have det( s B ( x , y ) , ∂ x s B ( x , y ) , s B ( x , y ))= 3 x y x y ( x y − x y ) ( x y + x x y y + x y ) . (2) B = { xy , x y , x } . Then det( s B ( x , y ) , ∂ x s B ( x , y ) , s B ( x , y ))= x x y ( x y − x y ) (4 x y + 3 x x y y + 2 x x y y + x y ) . The following theorem is the main result of this section. It gives sufficient con-ditions for the validity of formula (32) concerning the Carath´eodory number C B . Theorem 45. Let m, d , d , . . . , d m , d ∈ N be such that d < ... < d m = 2 d ,put A = { , x d , ..., x d m } , B = { y d , x d y d − d , ..., x d m − y d − d m − , x d } , and Z := Z ( q A ) if m is even or Z := Z ( q A , ) ∩ ... ∩ Z ( q A ,k ) if m = 2 k − is odd, where q A and q A ,j are as in Definition 38. Suppose that (31) ( x , ..., x k ) ∈ Z ⇒ ∃ i = j : x i = x j . HE MULTIDIMENSIONAL TRUNCATED MOMENT PROBLEM: CARATH´EODORY NUMBERS19 Then C A = C B = N A = N B = l m m . (32) Proof. Recall that S A and S B denote the moment cones of A and B , respectively.We set ∂ ∗ S A := ∂ S A ∩ S A .By Lemmas 40(iii) and 43(iv) we have N A = N B = (cid:6) m (cid:7) . Further, N A ≤ C A and N A ≤ C B by Theorem 27. Therefore, it suffices to show that C A ≤ N A and C B ≤ N A .First we prove that C B ≤ N A . Let s ∈ S B . Since X = P ( R ) is compact and condition (6) is satisfied (with e ( x, y ) := x d + y d ∈ B ), Proposition 8 applies with x = (1 , . Hence the supremum c s (1 , 0) := sup { c ∈ R : s − cs B (1 , ∈ S B } is attained and s ′ := s − c s (1 , s B (1 , ∈ ∂ S B . By Proposition 30(iii) all representing measures ( C ′ , X ′ ) of s ′ are singular.They do not contain (1 , 0) as an atom. (Indeed, otherwise c s (1 , 0) could be increasedwhich contradicts to the maximality of c s (1 , B arehomogeneous, we can assume without loss of generality that X ′ i = ( x ′ i , 1) with x ′ i pairwise different, say x ′ < x ′ < ... < x ′ l , and s ′ ∈ ∂ ∗ S A , i.e., s ′ is a boundarymoment sequence of S A . But from (31) and Lemma 43,(i) and (ii), it follows that l < N A , that is, C B ( s ) ≤ l + 1 ≤ N B = N A . This completes the proof of theinequality C B ≤ N A .Next we show that C A ≤ N A . If s ∈ ∂ ∗ S A , then C A ( s ) < N A by the preceding proof. Now let s ∈ int S A = int S B .Then C B ( s ) ≤ N A by the preceding paragraph and it suffices to show that s hasan at most N A -atomic representing measure which does not have an atom at (1 , ε > B ε ( s ) ⊆ int S A .Let c t ( x ) be defined by (7). Since t L t is a continuous map of R m → A ∗ , t L t ( e ) is continuous. Hence, c t ( x ) ≤ e ( x ) − L t ( e ) (by Proposition 8) is boundedfrom above on B ε ( s ). Then the supremum C of c t (1 , 0) on B ε ( s ) is finite. Let T := [ c ∈ [0 ,C +1] B ε ( s − c · s B (1 , ε -tube around the line γ := s − [0 , C +1] · s B (1 , T = T ∪ T ∪ T with T := T ∩ ∂ S B , T := T ∩ int S B , and T := T \ ( T ∪ T ), i.e., T is the part inside S B , T is the part outside S B , and T is the boundary part of S B in T . Since S B is closedand convex, T is closed and every path in T starting in T and ending in T containsat least one point in T . By construction, t ′ := t − c t (1 , s B (1 , ∈ T for all t ∈ T and no representing measure of t ′ contains (1 , 0) as an atom, i.e., T ⊂ ∂ ∗ S A . Then γ = s − [0 , · ( C + 1) s B (1 , ⊂ T and s ∈ γ ∩ T and s − ( C + 1) s B (1 , ∈ γ ∩ T ,so that s ′ = s − c s (1 , s B (1 , ∈ T . Since s B is continuous and C < ∞ there is a δ > k ( s − ( C + 1) s B (1 , − ( s − ( C + 1) s B (1 , δ )) k = ( C + 1) k s B (1 , − s B (1 , δ ) k < ε. Thus, s − ( C + 1) s B (1 , δ ) ∈ T . Then γ δ := s − [0 , · ( C + 1) s B (1 , δ ) ⊂ T and s ∈ γ δ ∩ T and s − ( C + 1) s B (1 , δ ) ∈ γ δ ∩ T , i.e., s ′ δ = s − c s (1 , δ ) s B (1 , δ ) ∈ T ⊂ ∂ ∗ S A . Summarizing, s = s ′ δ + c s (1 , δ ) s B (1 , δ ) and s ′ δ has a k -atomic representing measure( k < N A ) which has no atom at (1 , s has an l -atomic presentingmeasure ( l ≤ N A ) which has no atom at (1 , C A ( s ) ≤ N A . (cid:3) We illustrate the preceding by the following examples. Example 46. Let A = { , x , x , x , x } and B = { y , x y , x y , x y, x } , that is, m = 5 . Then we have (33) det( s A ( x ) , s ′ A ( x ) , s A ( y ) , s ′ A ( y ) , s A ( z )) = ( x − y ) ( x − z ) ( y − z ) · f ( x, y, z ) , where f ( x, y, z ) := xy ( x y + 4 x y + xy + 2 x z + 10 x yz + 10 xy z + 2 y z + 4 x z + 7 xyz + 4 y z ) . (34) This implies N B = N A = 3 as also proved in Lemma 35 and 43. Hence C A ≥ .From the Richter–Tchakaloff Theorem (Proposition 1) we find C A ≤ m = 5 , whileTheorem 13 gives a better bound C A ≤ m − .To apply Theorem 45 we have to check that the assumptions are satisfied. Clearly, d = 0 and d m = 6 is even. It remains to show that (31) is true. By symmetry itsuffices to verify (31) for f ( x, y, z ) := f ( x, y, z ) , f ( x, y, z ) := f ( y, z, x ) , and f ( x, y, z ) := f ( z, x, y ) . Set Z := Z ( f ) ∩Z ( f ) ∩Z ( f ) and let X = ( x, y, z ) ∈ Z . If X = 0 , then (31) holds.Now let X = 0 . Since f is homogeneous, we can scale X such that x + y + z = 1 .Then we derive (for instance, by using spherical coordinates) (35) Z ( f ) ∩ Z ( f ) ∩ Z ( f ) = { ( ± , , , (0 , ± , , (0 , , ± } , so (31) is fulfilled. Therefore, by Theorem 45 we have C A = 3 . A nice application of the preceding example is the following corollary. Corollary 47. Let p ( x ) = a + bx + cx + dx + ex be a non-negative polynomialwhich is not the zero polynomial. Then p has at most distinct real zeros.Proof. Assume to the contrary that p has three distinct zeros, say x, y, z . Let s bethe moment sequence of the measure µ = δ x + δ y + δ z . Then L s ( p ) = 0, so s is aboundary point of the moment cone. But from (35) it follows that the determinant(33) is non-zero, so s is an inner point, a contradiction. (cid:3) In the following example the assumption (31) of Theorem 45 is not satisfied andthe assertion (32) does not hold. Example 48. Let A = { , x, x , x } and B = { y , xy , x y , x } . From Theorem13, C B ≤ m − , while Theorem 27 and det( s A ( x ) , s ′ A ( x ) , s A ( y ) , s ′ A ( y )) = 2( y − x ) ( x + y )(2 x + xy + 2 y ) yield N B ≤ C A , so that C B ∈ { , } . We prove that C B = 3 . Let ν := ( δ − + δ − + δ + δ ) . Then s = ( s , s , s , s ) T = ( s A ( − 2) + s A ( − 1) + s A (1) + s A (2)) / , , . , . T . By some straightforward computations it can be shown that s has no k -atomic rep-resenting measure with k ≤ . Therefore, since C B ∈ { , } , we have C B = 3 = (cid:6) (cid:7) .Note that N B = 2 = (cid:6) (cid:7) . Thus, the equality (32) fails. Carath´eorody Numbers: Multidimensional Monomial Case Definition 49. For n, d ∈ N set A n,d := { x α : α ∈ N n , | α | ≤ d } , (36) B n,d := { x α : α ∈ N n , | α | = d } . (37)Note that | A n,d | = (cid:0) n + dd (cid:1) and | B n,d | = (cid:0) n + d − d (cid:1) . Throughout this section, we assume the following: For the polynomials A n,d :=Lin A n,d we consider the truncated moment problem on X = R n , while for thehomogeneous polynomials B n,d := Lin B n,d the moment problem is treated on thereal projective space X := P ( R n − ). Let S n − be the unit sphere in R n , n ≥ , and S n − the set of points x ∈ S n − for which the first non-vanishing coordinateis positive. We consider S n − as a realization of the projective space P ( R n − ). HE MULTIDIMENSIONAL TRUNCATED MOMENT PROBLEM: CARATH´EODORY NUMBERS21 The following simple fact is often used without mention: A polynomial of B n, d is non-negative on S n − , equivalently on P ( R n − ), if and only if it is on R n − .The following example shows how differential geometric methods can be used forthe truncated moment problem. Example 50. Let n = d = k = 2 , x α = ( x (1) ) α ( x (2) ) α and A , = { x α : α ∈ N , | α | ≤ } = { x α : α = (0 , , (0 , , (0 , , (2 , , (1 , , (0 , } . Then DS , A ( C, X ) = x (1)1 c x (1)2 c x (2)1 c x (2)2 c ( x (1)1 ) c x (1)1 x (1)2 ) c x (1)2 x (1)1 x (2)1 c x (2)1 c x (1)1 x (1)2 x (2)2 c x (2)2 c x (1)2 ( x (2)1 ) c x (2)1 ( x (2)2 ) c x (2)2 , where C = ( c , c ) and X = ( x , x ) , x i = ( x (1) i , x (2) i ) . From this we find that ker DS , A ( C, X ) = R · v ( C, X ) with v ( C, X ) := − c − ( x (1)1 − x (1)2 ) c − ( x (2)1 − x (2)2 )2 c − ( x (1)1 − x (1)2 ) c − ( x (2)1 − x (2)2 ) . Hence rank DS , A , = 5 at each point ( x , x ) , x = x , so the local rank theorem ofdifferential geometry applies. Fix ( C, X ) as above. The local rank theorem [Hil03,Proposition 1, p. 309] implies that there is a one-parameter family ( C ( t ) , X ( t )) which has the same moments as ( C, X ) satisfying the differential equations ˙ γ ( t ) = v ( C ( t ) , X ( t )) with initial condition ( C (0) , X (0)) = ( C, X ) . This system is ˙ c = − c = 2 c · ˙ x (1)1 = x (1)1 − x (1)2 c · ˙ x (1)2 = x (1)1 − x (1)2 c · ˙ x (2)1 = x (2)1 − x (2)2 c · ˙ x (2)2 = x (2)1 − x (2)2 and its solution is given by c ( t ) = c , − t x ( i )1 ( t ) = γ ,i + γ ,i c , + c , · s c , + 2 tc , − tc ( t ) = c , + 2 t x ( i )2 ( t ) = γ ,i − γ ,i c , + c , · s c , − tc , + 2 t with t ∈ ( − c , , c , ) . Here C = ( c , , c , ) and X = (( γ , , γ , ) , ( γ , , γ , )) are theinitial values at t = 0 . It should be noted that the corresponding moment sequenceis indeterminate, but it is a boundary point of the moment cone. Recall that N A ≤ C A by Theorem 27. There are various other lower boundsfor Carath´eodory numbers in the literature, see e.g. [DR84, p. 366]. In the case A , k − , M. M¨oller [M¨ol76] obtained the lower boundM¨o(2 , k − 1) := (cid:18) k + 12 (cid:19) + (cid:22) k (cid:23) . The following result improves M¨oller’s lower bound. Proposition 51. (38) M¨o(2 , k − ≤ (cid:24) | A , k − | (cid:25) ≤ N A , k − ≤ C A , k − for k ∈ N . For k ≥ we have (39) (cid:24) | A , k − | (cid:25) − M¨o(2 , k − ≥ ( k − − . Proof. The second inequality of (38) has been stated in Proposition 23. It reaminsto prove the first inequality of (38). In the cases k = 1 , , k ≥ (cid:24) (cid:18) k + 12 (cid:19)(cid:25) − (cid:18)(cid:18) k + 12 (cid:19) + (cid:22) k (cid:23)(cid:19) ≥ (cid:18) k + 12 (cid:19) − (cid:18)(cid:18) k + 12 (cid:19) + k (cid:19) = (2 k + 1) k − ( k + 1) k − k k − − . (cid:3) Before we turn to our next result we restate the Alexander–Hirschowitz Theorem[AH95]. We denote by V n,d,r the vector space of polynomials in n variables of degreeat most d having singularities at r general points in R n . Proposition 52. The subspace V n,d,r has the expected codimension min (cid:18) r ( n + 1) , (cid:18) n + dd (cid:19)(cid:19) except for the following cases:(i) d = 2 ; ≤ r ≤ n , codim V n, ,r = r ( n + 1) − r ( r − / ;(ii) d = 3 ; n = 4 , r = 7 , dim V , , = 1 ;(iii) d = 4 ; ( n, r ) = (2 , , (3 , , (4 , , dim V n, ,r = 1 . Theorem 53. We have (40) N A n,d = (cid:24) n + 1 (cid:18) n + dn (cid:19)(cid:25) , except for the following cases(i) d = 2 : N A n, = n + 1 .(ii) n = 4 , d = 3 : N A , = 8 .(iii) n = 2 , d = 4 : N A , = 6 .(iv) n = 3 , d = 4 : N A , = 10 .(v) n = d = 4 : N A , = 15 .Proof. From the corresponding definitions of V n,d,r and DS k, A n,d we obtaincodim V n,d,r = | A n,d | − dim V n,d,r = rank DS k, A n,d ((1 , ..., , X ) . Therefore, apart from exceptional cases, (40) follows at once from the Alexander–Hirschowitz Theorem. Next we treat the exceptions.(i): Note that N A n, ≥ ⌈ n/ ⌉ + 1. Since for all k satisfying ⌈ n/ ⌉ + 1 ≤ k ≤ n the matrix DS k, A n, ((1 , ..., , X ) has not the expected full rank for any X , the first k with full rank is k = n + 1.(ii): Since N A , ≥ DS , A , ((1 , ..., , X ) has not the expected full rankfor any X , N A , = 8.(iii): We have N A , ≥ DS , A , ((1 , ..., , X ) has not the expected full rankfor any X . Hence N A , = 6.(iv): Then N A , ≥ DS , A , ((1 , ..., , X ) has not the expected full rankfor any X . Thus N A , = 10. HE MULTIDIMENSIONAL TRUNCATED MOMENT PROBLEM: CARATH´EODORY NUMBERS23 (v): Then N A , ≥ 14 and DS , A , ((1 , ..., , X ) has not the expected full rankfor any X . Therefore, N A , = 15. (cid:3) For the homogeneous case we have Corollary 54. N B n +1 ,d = N A n,d .Proof. Let X = ( X , ..., X k ) ∈ R nk , k = N A n,d , be such that DS k, A n,d (1 , X ) hasfull rank. Then DS k, B n +1 ,d (1 , Y ) with Y = (( X , , ..., ( X k , N A n,d = k ≥ N B n +1 ,d .On the other hand, let Y = ( Y , ..., Y k ) ∈ R ( n +1) k , k = N B n +1 ,d , be such that DS k, B n +1 ,d (1 , Y ) has full rank. We can assume that all ( n + 1)-th coordinates of Y i are non-zero by the continuity of the determinant and therefore they can be chosento be 1, since we are in P ( R n ). The column ∂ n +1 s B n +1 ,d ( Y i ) depends linearly on s B n +1 ,d ( Y i ) and ∂ j s B n +1 ,d ( Y i ), j = 1 , ..., n . Therefore, omitting this column doesnot change the rank. Hence DS k, A n,d (1 , X ) with Y i = ( X i , 1) has full rank, that is, N A n,d ≤ k = N B n +1 ,d . (cid:3) Carath´eodory Numbers and Zeros of positive Polynomials For f ∈ B , d , Z P ( f ) denotes the projective zero set of f . Set(41) α (2 d ) := 32 d ( d − 1) + 1 . In this section we use the following proposition of Choi, Lam, and Reznick[CLR80]. Proposition 55. Let f ∈ B , d . Suppose that f ∈ Pos( R ) and |Z P ( f ) | > α (2 d ) . Then |Z P ( f ) | is infinite and there are polynomials p ∈ B , d , q ∈ B ,d suchthat f = pq , where d + d = d , p ∈ Pos( R ) , |Z P ( p ) | < ∞ , q is indefinite,and |Z P ( q ) | is infinite. (It is possible that p is a positive real constant; in this case d = 0 and we set B , := R .) The main aim of this section is to derive upper bounds for the Carath´eodorynumber C B n, d , n = 3. The first approach (Theorem 57) applies also to cases with n > d ∈ N let β (2 d ) denote the maximum of |Z P ( f ) | , where f ∈ B , d , f ∈ Pos( R ) and Z P ( f ) is finite. By the Choi–Lam–Reznick Theorem (Proposition55), β ( d ) ≤ α ( d ) for d ∈ N . We abbreviate C d := C B , d . Theorem 56. C d ≤ max k =0 ,...,d (cid:26)(cid:18) d + 22 (cid:19) − (cid:18) d + 2 − k (cid:19) + β (2( d − k )) (cid:27) + 1 . (42) Proof. Let s ∈ S . Since the projective space P ( R n − ) is compact and condition(6) holds with e := x d + x d + x d , it follows from Proposition 8 that C d ≤ max s ∈ ∂ S C d ( s ) + 1. Therefore, it is sufficient to show( ∗ ) C d ( s ) ≤ max k =0 ,...,d (cid:26)(cid:18) d + 22 (cid:19) − (cid:18) d + 2 − k (cid:19) + β (2( d − k )) (cid:27) for all s ∈ ∂ S .Let s = P li =1 c i s B , d ( x i ) be an l -atomic representating measure of s ∈ ∂ S . Since s ∈ ∂ S , there exists a polynomial p ∈ B , d , p = 0, such that p ( x ) ≥ P ( R )and L s ( p ) = 0. Then supp µ ⊆ Z ( p ), that is, x , . . . , x l ∈ Z ( p ).We can assume without loss of generality that the set { s B , d ( x i ) } i =1 ,...,l is lin-early independent. Indeed, assume that these vectors are linearly dependent and let P li =1 d i s B , d ( x i ) = 0 be a non-trivial linear combination. Since all c i > 0, thereexists ε > c i + εd i ≥ i and c j + εd j = 0 for one j . Hence µ ′ = P li =1 ( c i + εd i ) · s B , d ( x i ) is a ( l − s .The polynomial p ∈ B , d is non-negative on P ( R ), hence on R , so the Choi–Lam–Reznick Theorem (Proposition 55) applies. There are two cases:a) |Z ( p ) | ≤ β (2 d ).b) p = h q , where k := deg( h ) ≥ |Z ( p ) | ≤ β (2 d ) by the definition of β (2 d ) and therefore C d ( s ) ≤ β (2 d ). This is the case d = k in ( ∗ ).Now we turn to case b). Then k = 1 , . . . , d − . Let D ( k ) denote the largest l for which there exist y , . . . , y l ∈ Z ( h ) such that the vector s B , d ( y ) , ...., s B , d ( y l )are linearly independent. Then, by the paragraph before last, we have C d ( s ) ≤ D ( k ) + β (2( d − k )) . (43)Let y , ..., y l ∈ Z ( h ). We define M ( y , ..., y l ) := s B , d ( y ) T ... s B , d ( y l ) T and h α := x α h for α ∈ N , | α | = 2 d − k . Let ˜ h α be the coefficient vector of h α , thatis, h α ( · ) = h ˜ h α , s B , d ( · ) i . Since s B , d ( y i ) T · ˜ h α = h ˜ h α , s B , d ( y i ) i = y αi h ( y i ) = 0,we have ˜ h α ∈ ker M ( y , ..., y l ). Clearly, the vectors ˜ h α are linearly independent.Therefore, using (43) we derive C d ( s ) ≤ D ( k ) + β (2( d − k )) ≤ max rank M ( y , ..., y l ) + β (2( d − k )) ≤ | B , d | − | B , d − k | + β (2( d − k ))= (cid:18) d + 22 (cid:19) − (cid:18) d + 2 − k (cid:19) + β (2( d − k ))which is the k -th term in ( ∗ ).Summarizing, we have k = d in case a) and k = 1 , ..., d − ∗ ) for arbitrary s ∈ ∂ S which completes the proof. (cid:3) As far as the authors know, the numbers β (2 d ) are not yet known for d ≥ 4, butwe have β (2 d ) ≤ α (2 d ) by Proposition 55. Theorem 57. For d ∈ N we have C d ≤ α (2( d + 1)) = 32 d ( d + 1) + 1 . (44) Proof. Since β (2 d ) ≤ α (2 d ) = d ( d − 1) + 1 and ( d − k )( k + 3) − ≥ d ∈ N and k = 0 , ..., d − 1, we have for (42)32 d ( d + 1) = (cid:18) d + 22 (cid:19) − (cid:18) d + 22 (cid:19) = (cid:18) d + 22 (cid:19) − (cid:18) d − k + 22 (cid:19) + α ( d − k ) + ( d − k )( k + 3) − ≥ (cid:18) d + 22 (cid:19) − (cid:18) d − k + 22 (cid:19) + α ( d − k ) . Inserting the latter into (42) we obtain the assertion. (cid:3) In Table 1 we collect some numerical cases of Carath´eodory bounds.The next proposition is also due to Choi–Lam–Reznick [CLR80]. We will use itto derive a bound for the Carath´eodory number C B , . HE MULTIDIMENSIONAL TRUNCATED MOMENT PROBLEM: CARATH´EODORY NUMBERS25 Lower Upper Bounds for C d from known2 d Bounds N B , d Prop. 1 Thm. 13 Thm. 57 Thm. 62 C d Table 1. Bounds on the Carath´eodory numbers C d for d =1 , ..., , , , 500 from Proposition 1 and Theorems 13, 57, 62. Proposition 58. If p ∈ B , and |Z P ( p ) | > , then p is a sum of at most sixsquares of quadratics. Theorem 59. C B , ≤ .Proof. Let s be a boundary moment sequence. Then there exists p ∈ B , , p =0 , such that p ∈ Pos( R ) and L s ( p ) = 0. By Proposition 58, |Z ( p ) | ≤ 11 orwe have p = f + ... + f for some f , ..., f ∈ B , . In the following proof wegive an upper bound on the maximal number l of linearly independent vectors s B , ( x ) , ..., s B , ( x l ) with x i ∈ Z ( p ) . By Theorem 18, this number l is an upperbound of C B , ( s ). We proceed in a similar manner as in the proof of Theorem 57.By Proposition 58 we have two cases:a) |Z ( p ) | ≤ p = f + ... + f k , k ≤ l ≤ |Z ( p ) | ≤ Z ( f + ... + f k ) ⊆ Z ( f ) = Z ( f ) . Hence itsuffices to determine the maximal number l for a single square p = f , where f ∈ B , , f = 0. Let x , ..., x l ∈ Z ( f ) be such that the set { s B , ( x i ) } i =1 ,...,l islinearly independent. Define M ( x , ..., x l ) := s B , ( x ) T ... s B , ( x l ) T ,f α := x α f for α ∈ N , | α | = 2, and ˜ f α by f α ( · ) = h ˜ f α , s B , ( · ) i . Then we have ˜ f α ∈ ker M ( x , ..., x l ), since s B , ( x i ) T · ˜ f α = h ˜ f α , s B , ( x i ) i = x αi f ( x i ) = 0. The vectors˜ f α are linearly independent. Therefore, dim ker M ( x , ..., x l ) ≥ f α = | B , | and l = rank M ( x , ..., x l ) ≤ | B , | − | B , | = (cid:18) (cid:19) − (cid:18) (cid:19) = 25 . This proves that the moment sequence s can be represented by at most 25 atoms.Summarizing both cases, we have shown that each s ∈ ∂ S has a k -atomic repre-senting measure with k ≤ 25. Therefore, by Proposition 8, C B , ≤ 25 + 1 = 26. (cid:3) Proposition 1 yields C B , ≤ 35, while Theorem 13 gives C B , ≤ 34. Combiningthe upper bound of Theorem 59 with the lower bound from Theorem 27 we get(45) N B , = N A , = 10 ≤ C B , ≤ . Now we give another approach to obtain estimates of the Carath´eodory number C B , d from above. It is based on Bezout’s Theorem.Let f ∈ B ,d and f ∈ B ,d . For each point t ∈ Z P ( f ) ∩Z P ( f ) the intersectionmultiplicity I t ( f , f ) ∈ N of the projective curves f = 0 and f = 0 at t is definedin [Wal78, III, Section 2.2]. We do not restate the precise definition here. In whatfollows we use only the fact that I t ( f , f ) ≥ t is a singular point of one of thecurves f = 0 or f = 0.We use the following version of Bezout’s Theorem . The symbol | Z | denotes thenumber of points of a set Z . Lemma 60. If f ∈ B ,d and f ∈ B ,d are relatively prime in R [ x , x , x ] , then X t ∈Z P ( f ) ∩Z P ( f ) I t ( f , f ) ≤ d d . Proof. See e.g. [Wal78, p. 59]. (cid:3) Lemma 61. Let s be a moment sequence for B , d . Suppose p ∈ B ,k is irreduciblein R [ x , x , x ] , k ≤ d , and L s ( p ( x + x + x ) d − k ) = 0 . Then C d ( s ) ≤ dk + 1 . Proof. Consider the moment cone ˜ S := S ( B , d , Z ( p )). Then ˜ S is an exposed faceof the moment cone S = S ( B , d , P ( R )) and s ∈ ˜ S . By Proposition 21, S is closedand so is ˜ S . Clearly, each point of ˜ S is the limit of relative inner points of ˜ S .Therefore, since the sets ˜ S k are closed by Proposition 21, it is sufficient to provethe assertion for all relatively inner points of the cone ˜ S .Let s be a relatively inner point of ˜ S and x ∈ Z ( p ). Setting e := x d + x d + x d ,condition (6) holds. Since Z ( p ) is compact, Proposition 8 applies, so the supremum c s ( x ) := sup { c : s − c · s B , d ( x ) ∈ ˜ S} is attained and s ′ := s − c s ( x ) · s B , d ( x ) ∈ ∂ ˜ S .Thus there exists a supporting hyperplane of the cone ˜ S at s ′ . Hence there existsa polynomial q ∈ B , d such that L s ′ ( q ) = 0, L s ( q ) > 0, and q ≥ Z ( p ).From L s ′ ( q ) = L s ( q ) − c s ( x ) q ( x ) = 0 it follows that q ( x ) = 0. (Indeed, otherwise L s ( q ) = 0, so s would be a boundary point of ˜ S , a contradiction.) Since p ( x ) = 0and q ( x ) = 0, the irreducible polynomial p is not a factor of q , so p and q arerelatively prime and Bezout’s Theorem applies.Since q ( y ) ≥ Z ( p ), for each intersection point of q and p has the intersectionmultiplicity of at least 2. Therefore, by Lemma 60,2 |Z ( q ) ∩ Z ( p ) | ≤ deg( q ) deg( p ) = 2 dk. (46)Since each representing measure of s ′ is supported on Z ( p ) ∩ Z ( q ), (46) implies that C d ( s ′ ) ≤ dk . Hence C d ( s ) ≤ C d ( s ′ ) + 1 ≤ dk + 1. (cid:3) Our main result in this section is the following theorem. Theorem 62. C d ≤ α (2 d ) + 1 = d ( d − 1) + 2 for d ∈ N , d ≥ .Proof. Let us consider the moment cone S := S ( B , d , P ( R )). We proceed in asimilar manner as in the proof of Lemma 61. By Proposition 21, the sets S k areclosed. Hence it suffices to prove the inequality C d ( s ) ≤ α (2 d ) + 1 for all relativelyinner points of the cone S .Let s be an inner point of S and x ∈ P ( R ). By Proposition 8, the supremum c s ( x ) := sup { c : s − c · s B , d ( x ) ∈ S} is attained and s ′ := s − c s ( x ) · s B , d ( x ) ∈ ∂ S .Then there exists a supporting hyperplane of S at s , hence there is a polynomial HE MULTIDIMENSIONAL TRUNCATED MOMENT PROBLEM: CARATH´EODORY NUMBERS27 f ∈ B , d such that L s ′ ( f ) = 0 and f ≥ P ( R ). We apply Proposition 55to f . Then, we can write f = p · q · · · q r ( r ≤ d ), where p ∈ Pos( P ( R )), all q i are indefinite and irreducible in R [ x , x , x ], Z ( p ) < ∞ and all |Z ( q i ) | are infinite.Since Z ( f ) = Z ( p ) ∪ Z ( q ) ∪ · · · ∪ Z ( q r )we find a disjoint decomposition Z ∪ Z ∪ · · · ∪ Z r of Z ( f ) with Z ⊆ Z ( p ) and Z i ⊆ Z ( q i ). Let µ ′ = P mj =1 c j δ x j be a representing measure of s ′ and set s := X x j ∈ Z c j s B , d ( x j ) and s i := X x j ∈ Z i c j s B , d ( x j ) . Clearly, s ′ = s + s + · · · + s r . Setting d i = deg( q i ) and 2 k = deg( p ), we have d = k + d + · · · + d r and r ≤ d − k . Using Proposition 55 and Lemma 61 we derive C B , d ( s ′ ) ≤ C B , d ( s ) + C B , d ( s ) + · · · + C B , d ( s r ) ≤ α (2 k ) + ( d · d + 1) + · · · + ( d · d r + 1) = α (2 k ) + d ( d − k ) + r ≤ α (2 k ) + ( d + 1)( d − k ) = α (2 d ) − ( α (2 d ) − α (2 k ) − ( d + 1)( d − k )) | {z } = ( d − k )( d +3 k − ≥ ∀ d ≥ , k =0 ,...,d ≤ α (2 d ) . Therefore, C B , d ( s ) ≤ C B , d ( s ′ ) + 1 ≤ α (2 d ) + 1 = d ( d − 1) + 2 for all d ≥ (cid:3) Example 63 ( d = 5) . W. R. Harris [Har99] discovered a polynomial h ∈ B , thatis nonnegative on P ( R ) with projective zero set Z P ( h ) = { (1 , , ∗ , (1 , , √ ∗ , (1 , , / ∗ } , where ( a, b, c ) ∗ denotes all permutations of ( a, b, c ) including sign changes. Hence h has exactly projective zeros z i , i = 1 , . . . , . A computer calculation shows thatthe matrix ( s B , ( z )) z ∈Z P ( h ) has rank , i.e., the set { s B , ( z i ) : i = 1 , . . . , } islinearly independent. Therefore, C B , ≥ by Theorem 18. Further, we compute N B , = 15 and have C B , ≤ α (10) + 1 = 37 by Theorem 62. Summarizing, N B , = 15 < ≤ C B , ≤ . (47)The following corollary reformulates Theorem 18 in the present context. Corollary 64. Let d ∈ N and p ∈ B , d . Suppose that p ∈ Pos( R ) , |Z ( p ) | = β (2 d ) , and the set { s B , d ( z ) : z ∈ Z ( p ) } is linearly independent. Then β (2 d ) ≤ C B , d . It seems natural to ask whether or not the assumption on the linear independenceof the set { s B , d ( z ) : z ∈ Z ( p ) } in Corollary 64 can be omitted. This leads to the Question: Suppose p ∈ B , d , p ∈ Pos( R ) , and |Z ( p ) | < ∞ (or |Z ( p ) | = β (2 d ) ).Is the set { s B , d ( z ) : z ∈ Z ( p ) } linearly independent? Note that for the Robinson polynomial R ∈ B , the answer is “Yes”.Recall that β (2 d ) ≤ α (2 d ) by the Choi–Lam–Reznick Theorem (Proposition 55).It seems likely to conjecture that(48) Conjecture : β (2 d ) ≤ C B , d ≤ β (2 d ) + 1 for d ≥ . The Robinson polynomial has 10 projective zeros, so that α (6) = β (6) = 10 . There-fore, since C B , = 11 as shown in [Kun14], this conjecture is true for d = 3. Asnoted above, the Harris polynomial R ∈ B , has 30 projective zeros. Hence30 ≤ β (10) ≤ α (10) = 31.From the proof of Theorem 62 it follows that (48) holds if β ( d ) + ( d ′ + 1)( d ′ − d ) ≤ β ( d ′ ) for d ′ ∈ N , d ∈ N , d < d ′ , ( d ′ , d ) = (3 , . Carath´eodory Numbers and Real Waring Rank In Definition 4 we introduced the signed Carath´eodory number C A , ± . In thissection we connect it to the real Waring rank w ( n, d ), that is, to the smallestnumber w ( n, d ) such that each f ∈ B n, d can be written as real linear combination(49) f ( x ) = k X i =1 c i ( x · λ i ) d of 2 d -powers of linear forms x · λ i = λ i, x + · · · + λ i,n x n , where k ≤ w ( n, d ), c i ∈ R , λ i ∈ R n .Let us recall some basics on the apolar scalar product [ · , · ], see e.g. [Rez92]. For α = ( α , ..., α n ) ∈ N n with | α | := α + · · · + α n = 2 d we set γ α := (2 d )! α ! ··· α n ! . Let p, q ∈ B n, d . We write p ( x ) = P α γ α a α x α and q ( x ) = P α γ α b α x α and define[ p, q ] := X α γ α a α b α . Then ( B n, d , [ · , · ]) becomes a finite-dimensional real Hilbert space. Setting f λ ( x ) :=( λ · x ) d , we obtain(50) [ p, f λ ] = X α γ α a α λ α = p ( λ ) . Let f be of the form (49). Then, for p ∈ B n, d it follows from (50) that(51) L f ( p ) := [ f, p ] = hX i c i f λ i , p i = k X i =1 c i p ( λ i ) , that is, the linear functional L f on B n, d is the integral with respect to the signedmeasure µ := P ki =1 c i δ λ i . Conversely, each signed atomic measure yields a function f of the form (49) such that (51) holds. By the Riesz Theorem all linear functionalson B n, d are of the form L f , where f is as in (49). Theorem 65. (i) w ( n, d ) = C B n, d , ± .(ii) N B n, d ≤ w ( n, d ) ≤ N B n, d .(iii) Set N := N B n, d . Then there exists λ = ( λ , ..., λ N ) ∈ R N · n such that forall ε > and p ∈ B n, d we have p ( x ) = c · N B n, d X i =1 (cid:2) ( λ i · x ) d − c i ( λ εi · x ) d (cid:3) for some λ ε = ( λ ε , ..., λ εN ) with k λ − λ ε k < ε , | − c i | < ε , c ∈ R .(iv) The set of vectors λ as in (iii) is open and dense in R N B n, d · n .Proof. (i) is clear from the preceding considerations on the apolar scalar product.Remark 28 and (i) imply (ii), while (iii) follows from Theorem 25 combined with(i). (iv) is a consequence of Sard’s Theorem as in Theorem 27. (cid:3) With Theorem 53 the upper bound in (ii) was already obtained in [BT15, Cor.9]. Acknowledgment The authors are grateful to G. Blekherman, M. Schweighofer, and C. Riener forvaluable discussions on the subject of this paper. K.S. thanks also J. St¨uckrad tohelpful discussions. P.dD. was supported by the Deutsche Forschungsgemeinschaft(SCHM1009/6-1). 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Universit¨at Leipzig, Mathematisches Institut, Augustusplatz 10/11, D-04109 Leipzig,Germany Max Planck Institute for Mathematics in the Sciences, Inselstraße 22, D-04103Leipzig, Germany E-mail address : [email protected] Universit¨at Leipzig, Mathematisches Institut, Augustusplatz 10/11, D-04109 Leipzig,Germany E-mail address ::