On the ϰ th root of a Stieltjes moment sequence
aa r X i v : . [ m a t h . F A ] O c t On the κ th root of a Stieltjes moment sequence Jan Stochel and Jerzy Bart lomiej Stochel
Abstract.
Stieltjes moment sequences { a n } ∞ n =0 whose κ th roots { κ √ a n } ∞ n =0 are Stieltjes moment sequences are studied ( κ is a fixed integer greater than orequal to 2). A formula connecting the closed supports of representing measuresof { a n } ∞ n =0 and { κ √ a n } ∞ n =0 is established. The relationship between the holesof the supports of these measures is investigated. The set of all pairs ( M, N )of positive integers for which there exists a Stieltjes moment sequence whosesquare root is a Stieltjes moment sequence and both of them have representingmeasures supported on subsets of (0 , ∞ ) of cardinality M and N , respectively,is described.
1. Introduction
In [
6, 7, 8 ] Horn laid the foundations of the theory of infinitely divisible ma-trices, kernels and positive definite sequences. Among other things, he describedStieltjes moment sequences { a n } ∞ n =0 whose all powers { a αn } ∞ n =0 with positive realexponents α are Stieltjes moment sequences (cf. [ , Theorem 2.9]). In a recentpaper [ ] we asked the question: for what positive integers M is it true that thesquare root of a Stieltjes moment sequence (cid:8) α ϑ n + . . . + α M ϑ nM (cid:9) ∞ n =0 is not a Stielt-jes moment sequence for all real numbers α , . . . , α M > < ϑ < . . . < ϑ M ?We will show that the answer to this question is in the affirmative for M ∈ { , } ,and in the negative for M / ∈ { , } (cf. Corollary 8.2).The present work is motivated by the aforementioned papers (including [
10, 2 ]).We will consider Stieltjes moment sequences { a n } ∞ n =0 whose κ th roots { κ √ a n } ∞ n =0 are Stieltjes moment sequences, where κ is a fixed integer greater than or equalto 2; by the Schur product theorem (see [ , p. 14] or [ , Theorem 7.5.3]) this isequivalent to considering the κ th powers of Stieltjes moment sequences. Underthe assumption that { a n } ∞ n =0 is determinate, we give a formula for the (closed)support of a representing measure µ of { a n } ∞ n =0 written in terms of the support ofa representing measure ν of { κ √ a n } ∞ n =0 (see Theorem 3.3). In Section 5 and 6 we Mathematics Subject Classification.
Primary 44A60, Secondary 28A99.
Key words and phrases.
Stieltjes moment sequence, κ th root of a Stieltjes moment sequence,representing measure, support of a measure, hole of a support.This work was supported by the MNiSzW grant No. NN201 546438 (2010-2013) as well asby the MNiSzW grant No. 10.42004 (the second author). The case of M = 2, which was answered in [ , Lemma 3.3], immediately implies thatRhoades’ square root problem (cf. [ , p. 296]) has a negative solution (see [ , Theorem 7] fora solution of this problem based on a particular choice of a Hausdorff moment sequence whoserepresenting measure is supported on a two point set). provide some solutions to the following problem: given a hole of the support of themeasure µ , determine the circumstances under which the support of ν has a holeand then localize it (see Theorems 5.3 and 6.3); the converse problem is studiedas well (see Theorem 5.1). Some solutions of this problem are written in terms ofthe parameters ι s and ι ∗ s that describe, in a sense, the geometry of the hole of thesupport of µ (see Theorem 6.3). Using the results of Sections 2-6, we constructa variety of examples illustrating the concepts of the paper (see Section 7). Inparticular, an example of a Stieltjes moment sequence which κ th root is a Stieltjesmoment sequence for κ = 2 ,
4, but not for κ = 3 is furnished (see Example 7.6).We conclude this paper with a description of the set of all pairs ( M, N ) of positiveintegers for which there exists a Stieltjes moment sequence whose square root is aStieltjes moment sequence and both of them have representing measures supportedon subsets of (0 , ∞ ) of cardinality M and N , respectively (see Theorem 8.1). Thereader is encouraged to refer to [
1, 3, 13, 14 ] for the mathematical details of thetheory of classical moment problems.
2. Preliminaries
From now on, the fields of real and complex numbers are denoted by R and C ,respectively, and Z stands for the set of all integers. Set R + = { x ∈ R : x > } , Z + = { n ∈ Z : n > } and N = { n ∈ Z : n > } . Given x ∈ R , we define ⌊ x ⌋ = max { n ∈ Z : n x } and ⌈ x ⌉ = min { n ∈ Z : x n } . The cardinality ofa set X is denoted by card ( X ). X κ stands for the κ -fold Cartesian product of X by itself. We write supp µ for the (closed) support of a regular positive Borelmeasure µ on a Hausdorff topological space. Given θ ∈ R + , we denote by δ θ theBorel probability measure on R + concentrated at { θ } .A sequence { a n } ∞ n =0 ⊆ R + is said to be a Stieltjes moment sequence if thereexists a positive Borel measure µ on R + such that a n = R R + x n d µ ( x ) for all n ∈ Z + ;such µ is called a representing measure of { a n } ∞ n =0 . If a Stieltjes moment sequencehas a unique representing measure, we call it determinate . Recall that (cf. [ ])each Stieltjes moment sequence which has a compactly supported rep-resenting measure is determinate.(2.1)It is well known that if ( X, Σ, µ ) is a measure space, f is a complex Σ -measurablefunction on X and R X | f | r d µ < ∞ for some r ∈ (0 , ∞ ), then the µ -essential supre-mum of | f | is equal to lim p →∞ (cid:16) R X | f | p d µ (cid:17) /p (cf. [ , p. 71]). This implies thatif { a n } ∞ n =0 is a Stieltjes moment sequence with a representingmeasure µ , then lim n →∞ a /nn = sup supp µ .(2.2)Given an integer κ >
2, we define the continuous mapping π κ : R κ + → R + by π κ ( x , . . . , x κ ) = x · · · x κ , x , . . . , x κ ∈ R + . (2.3) Lemma 2.1.
Let F be a subset of R + and κ be an integer greater than or equalto . If F is compact, then π κ ( F κ ) is compact. If / ∈ F and F is closed, then / ∈ π κ ( F κ ) and π κ ( F κ ) is closed. Proof of Lemma 2.1.
Clearly, it is enough to consider the case when 0 / ∈ F and F is closed. We claim that if { ( x ( n )1 , . . . , x ( n ) κ ) } ∞ n =1 ⊆ F κ and the sequence Such µ being finite is automatically regular (see e.g., [ , Theorem 2.18]). N THE κ TH ROOT OF A STIELTJES MOMENT SEQUENCE 3 { π κ ( x ( n )1 , . . . , x ( n ) κ ) } ∞ n =1 is convergent in R + , then there exists a subsequence of { ( x ( n )1 , . . . , x ( n ) κ ) } ∞ n =1 which is convergent in F κ .First note that the sequence { x ( n ) j } ∞ n =1 is bounded for each j ∈ { , . . . , κ } .Indeed, otherwise, by passing to a subsequence if necessary, we may assume thatfor some k ∈ { , . . . , κ } , x ( n ) k → ∞ as n → ∞ . Since F is closed and 0 / ∈ F , wededuce that inf n > x ( n ) j > j ∈ { , . . . , κ } . This implies that the sequence { π κ ( x ( n )1 , . . . , x ( n ) κ ) } ∞ n =1 is unbounded, which is a contradiction. Now using thecompactness argument, we establish our claim.By the assertion just proved, we see that 0 / ∈ π κ ( F κ ) and π κ ( F κ ) is closed. (cid:3) Note that Lemma 2.1 is no longer true if the assumption 0 / ∈ F is dropped.Indeed, if κ > F = { , , , . . . } ∪ { , , , . . . } , then F is closed, while π κ ( F κ ), being equal to the set of all nonnegative rational numbers, is not closed.
3. The relationship between supp µ and supp ν In the present paper we consider the following situation: { a n } ∞ n =0 is a Stieltjes moment sequence with a representing measure µ , κ isan integer greater than or equal to 2 and { κ √ a n } ∞ n =0 is a Stieltjes momentsequence with a representing measure ν .(3.1)We begin by proving the basic relation between µ and ν . Lemma 3.1. If (3.1) holds and { a n } ∞ n =0 is determinate, then µ ( E ) = ν ⊗ κ ( π − κ ( E )) for all Borel subsets E of R + , (3.2) where ν ⊗ κ is the κ -fold product of the measure ν by itself and π κ is as in (2.3) . Proof.
It follows from Fubini’s theorem and [ , Theorem C, p. 163] that a n = Z R κ + ( x · · · x κ ) n d ν ⊗ κ ( x , . . . , x κ ) = Z R + z n d (cid:0) ν ⊗ κ ◦ π − κ (cid:1) ( z ) , n ∈ Z + , where ν ⊗ κ ◦ π − κ is a positive Borel measure on R + given by the right hand sideof the equality in (3.2). Thus, by the determinacy of { a n } ∞ n =0 , the measures µ and ν ⊗ κ ◦ π − κ coincide. (cid:3) The following lemma which describes the support of the transport of a measureis surely folklore. For the reader’s convenience, we include its proof.
Lemma 3.2.
Let X and Y be Hausdorff topological spaces, ρ and σ be regularpositive Borel measures on X and Y , respectively, and φ : X → Y be a continuousmapping such that σ ( E ) = ρ ( φ − ( E )) for all Borel subsets E of Y . Then supp σ = φ (supp ρ ) . Proof.
We begin by showing that φ (supp ρ ) ⊆ supp σ . Take x ∈ supp ρ . Let V be an open neighbourhood of φ ( x ). By the continuity of φ , the set φ − ( V ) is anopen neighbourhood of x , and thus σ ( V ) = ρ ( φ − ( V )) >
0. Since V is an arbitraryopen neighbourhood of φ ( x ), we get φ ( x ) ∈ supp σ .It remains to show that Y \ φ (supp ρ ) ⊆ Y \ supp σ . Take y ∈ Y \ φ (supp ρ ).Then V := Y \ φ (supp ρ ) is an open neighbourhood of y such that V ∩ φ (supp ρ ) = ∅ .This implies that φ − ( V ) ∩ supp ρ = ∅ . Hence σ ( V ) = ρ ( φ − ( V )) = 0, which yields y ∈ Y \ supp σ . This completes the proof. (cid:3) J. STOCHEL AND J. B. STOCHEL
We are now ready to provide a formula connecting supp µ with supp ν . Theorem 3.3.
Suppose (3.1) holds and { a n } ∞ n =0 is determinate. Then supp µ = π κ (cid:0) (supp ν ) κ (cid:1) . (3.3) Moreover, if supp ν is compact or supp ν , then supp µ = π κ (cid:0) (supp ν ) κ (cid:1) . Proof.
Applying Lemmata 3.1 and 3.2 and the well known equality supp ν ⊗ κ =(supp ν ) κ , we get (3.3). The “moreover” part follows from (3.3) and Lemma 2.1. (cid:3) We will show in Example 7.1 that the closure sign in (3.3) cannot be omitted.
Corollary 3.4. If (3.1) holds and { a n } ∞ n =0 is determinate, then (i) supp ν ⊆ κ √ supp µ , (ii) card (supp ν ) card (supp µ ) , (iii) sup supp ν = κ √ sup supp µ . Proof.
Conditions (i) and (iii) follow from Theorem 3.3 (condition (iii) canalso be deduced from (2.2)). Condition (ii) is a direct consequence of (i). (cid:3)
As shown in Example 7.2, inclusion (i) in Corollary 3.4 may be proper. In turn,inequality (ii) in Corollary 3.4 may be strict (however, in view of Theorem 3.3, ifcard (supp µ ) = ℵ , then card (supp ν ) = card (supp µ )). In fact, it may happenthat supp ν is discreet and has only one accumulation point, while supp µ = R + (cf. Example 7.1).
4. Transforming holes of supp µ and supp ν Suppose (3.1) holds. In this and the subsequent two sections we will study therelationship between the following two situations:supp µ ⊆ [0 , ϑ ] ∪ [ ϑ , ϑ ] with ϑ , ϑ , ϑ ∈ R such that 0 ϑ < ϑ ϑ , supp ν ⊆ [0 , α ] ∪ [ β, γ ] with α, β, γ ∈ R such that 0 α < β γ. Hereafter we will consider a transformation ( ϑ , ϑ , ϑ ) → ( α, β, γ ) between tripletsof real numbers satisfying the inequalities 0 ϑ < ϑ ϑ and 0 α < β γ which is given by α = ϑ ϑ κ p ϑ , β = κ p ϑ and γ = κ p ϑ . (4.1)This transformation is well defined (because ( ϑ ϑ ) κ < ( ϑ ϑ ) κ ϑ ϑ ) and injective,but not surjective. A triplet ( α, β, γ ) with 0 α < β γ is the image of some( ϑ , ϑ , ϑ ) under this transformation if and only if αγ κ − < β κ . If this is the case,then ϑ = αγ κ − , ϑ = β κ and ϑ = γ κ . Given real numbers ϑ , ϑ , ϑ such that 0 ϑ < ϑ ϑ , we set α † = ϑ ϑ κ p ϑ , β † = κ p ϑ . (4.2)The quantities just defined will play an essential role in Theorem 5.3 below. Clearly,if α , β and γ are defined by (4.1), then0 α < β γ , α < α † β , 0 β † < β and αγ κ − < β κ .(4.3) N THE κ TH ROOT OF A STIELTJES MOMENT SEQUENCE 5
Note also that α † < β if and only if ϑ < ϑ .In general, there is no order relationship between α † and β † . Obviously we have α † < β † ⇐⇒ ϑ κ < ϑ ϑ κ − and α † = β † ⇐⇒ ϑ κ = ϑ ϑ κ − . The reader should be aware of the fact that the quantities α , β , γ , α † and β † dependnot only on ( ϑ , ϑ , ϑ ) but also on κ . Making explicit the dependence on κ , wecan formulate the following result. Proposition 4.1. If ϑ < ϑ ϑ < ∞ , then (i) lim κ →∞ α † ( κ ) = ϑ /ϑ , (ii) lim κ →∞ β † ( κ ) = 1 provided that ϑ > , (iii) if α † ( κ ) < β † ( κ ) for some κ > , then α † ( κ ′ ) < β † ( κ ′ ) for all κ ′ > κ .Moreover, if < ϑ < ϑ < ϑ < ∞ , then (iv) α † ( ι s ) < β † ( ι s ) , (v) the sequence { β † ( κ ) − α † ( κ ) } ∞ κ = ι s is strictly increasing provided that either ϑ or ϑ > and log ϑ ϑ ϑ log ϑ ,where ι s = ι s ( ϑ , ϑ , ϑ ) := 1 + (cid:22) log ( ϑ /ϑ )log ( ϑ /ϑ ) (cid:23) > . (4.4) Proof.
The properties (i)-(iv) are easily seen to be true.(v) Define the function f : [ ι s , ∞ ) → R + by f ( x ) = x √ ϑ − ϑ ϑ x √ ϑ for x ∈ [ ι s , ∞ ). It follows from (iv) that ϑ ϑ < ι s r ϑ ϑ x r ϑ ϑ , x ∈ [ ι s , ∞ ) . (4.5)Note that − x f ′ ( x ) = x p ϑ log ϑ − ϑ ϑ x p ϑ log ϑ , x ∈ [ ι s , ∞ ) . (4.6)If ϑ
1, then − x f ′ ( x ) (4.6) < (cid:16) x p ϑ − ϑ ϑ x p ϑ (cid:17) log ϑ , x ∈ [ ι s , ∞ ) . In turn, if ϑ > ϑ ϑ ϑ log ϑ , then − x f ′ ( x ) (4.6) (cid:16) x p ϑ − x p ϑ (cid:17) log ϑ < , x ∈ [ ι s , ∞ ) . Hence, in both cases, f ′ ( x ) > x ∈ [ ι s , ∞ ), which implies that f is strictlyincreasing. Since f ( κ ) = β † ( κ ) − α † ( κ ) for κ = 2 , , . . . , the proof is complete. (cid:3) Regarding Proposition 4.1 (v), we note that the sequence { β † ( κ ) − α † ( κ ) } ∞ κ = ι s may be strictly decreasing (e.g., consider ϑ = 2, ϑ = 5 / ϑ = 2 · ). J. STOCHEL AND J. B. STOCHEL
5. Relating holes of supp µ and supp ν Our next goal is to analyze the situation when the support of a representingmeasure ν of the κ th root of a Stieltjes moment sequence { a n } ∞ n =0 has a hole. Anatural question arises as to whether the support of a representing measure µ of { a n } ∞ n =0 has a hole and what is the relationship between these two holes. We alsostudy the reverse influence.If a hole of supp ν is properly suited to supp ν , then we can locate the corre-sponding hole of supp µ . Theorem 5.1. If (3.1) holds and ν (( α, β )) = 0 for some α, β ∈ R such that α < β γ := sup supp ν < ∞ and αγ κ − < β κ , then (i) µ (( ϑ , ϑ )) = 0 and ϑ = sup supp µ < ∞ , (ii) α ∈ supp ν if and only if ϑ ∈ supp µ , (iii) β ∈ supp ν if and only if ϑ ∈ supp µ ,where ϑ := αγ κ − , ϑ := β κ and ϑ := γ κ . Proof.
Employing (2.1) and (2.2), we deduce that the Stieltjes moment se-quences { a n } ∞ n =0 and { κ √ a n } ∞ n =0 are determinate, supp µ ⊆ [0 , γ κ ] and γ κ ∈ supp µ . It is also clear that supp ν ⊆ [0 , α ] ⊔ [ β, γ ] and γ ∈ supp ν .(i) Take x ∈ supp µ . By Theorem 3.3, there exist x , . . . , x κ ∈ supp ν such that x = x · · · x κ . If x i ∈ [ β, γ ] for every i ∈ { , . . . , κ } , then evidently x > β κ = ϑ .Otherwise, there exists i ∈ { , . . . , κ } such that x i ∈ [0 , α ], and thus x = x i · Y j = i x j αγ κ − = ϑ . Hence supp µ ⊆ [0 , ϑ ] ∪ [ ϑ , ϑ ], which gives (i).Now if α ∈ supp ν (respectively: β ∈ supp ν ), then the fact that γ ∈ supp ν enables us to infer from (3.3) that ϑ = αγ κ − ∈ supp µ (respectively: ϑ = β κ ∈ supp µ ). Therefore, it remains to prove the “if” parts of assertions (ii) and (iii).(ii) Assume that ϑ ∈ supp µ . Then, by Theorem 3.3, ϑ = x · · · x κ forsome x , . . . , x κ ∈ supp ν . Suppose that, contrary to our claim, α / ∈ supp ν . If x i ∈ [ β, γ ] for every i ∈ { , . . . , κ } , then clearly ϑ > β κ = ϑ , which contradictsthe assumption that αγ κ − < β κ . Otherwise, there exists i ∈ { , . . . , κ } suchthat x i ∈ [0 , α ]. As α / ∈ supp ν , we must have x i < α and so0 < αγ κ − = ϑ = x i · Y j = i x j < αγ κ − , which is a contradiction.(iii) Assume that ϑ = β κ ∈ supp µ . Suppose that, contrary to our claim, β supp ν . Since γ ∈ supp ν , we must have β < γ . Clearly, the set supp ν ∩ [ β, γ ]is compact and nonempty. Set β ′ = min(supp ν ∩ [ β, γ ]). Since β supp ν , we get0 α < β < β ′ γ and ν (( α, β ′ )) = 0. Note that αγ κ − < β κ < β ′ κ . (5.1)Hence, by (i), applied to the triplet ( α, β ′ , γ ), we have µ (( αγ κ − , β ′ κ )) = 0, whichtogether with (5.1) contradicts the fact that β κ ∈ supp µ . (cid:3) Regarding Theorem 5.1, one might expect that the idea of the proof of (iii)would apply to the proof of (ii) in the case of α >
0. However, it may happen thatthere is no point in supp ν lying on the left hand side of α (see (7.4) in Example N THE κ TH ROOT OF A STIELTJES MOMENT SEQUENCE 7 ν (( α, β )) = 0 (though their “onlyif” parts are always true). Remark 5.2.
Under the assumptions of Theorem 5.1, if α ′ := ( sup([0 , α ] ∩ sup ν ) when [0 , α ] ∩ sup ν = ∅ , β ′ := inf([ β, γ ] ∩ supp ν ) , then 0 α ′ α , β β ′ γ , ν (( α ′ , β ′ )) = 0 and α ′ γ κ − < β ′ κ , which means thatthe numbers α ′ , β ′ and γ satisfy the assumptions of Theorem 5.1.Our goal now is to look for holes of supp ν that may correspond to a given holeof supp µ . Theorem 5.3. If (3.1) holds and µ (( ϑ , ϑ )) = 0 for some ϑ , ϑ ∈ R such that ϑ < ϑ , then (i) ν (( β † , β )) = 0 provided that { a n } ∞ n =0 is determinate, (ii) ν (( α, α † )) = 0 provided that ϑ ϑ := sup supp µ < ∞ , (iii) ν (( α, β )) = 0 provided that ϑ ϑ := sup supp µ < ∞ and any of thefollowing two conditions holds :(iii-a) either β † < α † , or β † α † and κ > , or β † = α † and β ∈ supp ν , (iii-b) γβ α < α † and β ∈ supp ν ,where α = ϑ ϑ κ √ ϑ , β = κ √ ϑ , γ = κ √ ϑ , α † = ϑ ϑ κ √ ϑ and β † = κ √ ϑ . Remark 5.4.
Note that if κ is an integer greater than or equal to 2, 0 ϑ <ϑ ϑ := sup supp µ < ∞ and β † α † , then γβ α < α † . Indeed, since the case ϑ = ϑ is obvious, we can assume that ϑ < ϑ . Then ϑ ϑ (cid:16) ϑ ϑ (cid:17) κ < (cid:16) ϑ ϑ (cid:17) κ , which yields γβ α < α † . It may happen that condition (iii-b) of Theorem 5.3 issatisfied, while condition (iii-a) does not hold (cf. Example 7.2). Moreover, assertion(iii) is no longer true if we drop either the assumption that β ∈ supp ν (cf. Example7.3), or the assumption that γβ α < α † (cf. Example 7.4). Proof of Theorem 5.3. (i) Suppose that, contrary to our claim, there exists x ∈ supp ν such that β † < x < β . Then, by (3.3), we have x κ ∈ supp µ . This and ϑ < x κ < ϑ lead to the contradiction that µ (( ϑ , ϑ )) = 0.(ii) By (2.1), the Stieltjes moment sequence { a n } ∞ n =0 is determinate. Hence,by Corollary 3.4(iii), we have γ = κ √ ϑ ∈ supp ν . Suppose that, contrary to ourclaim, ν (( α, α † )) >
0. Then there exists x ∈ supp ν ∩ ( α, α † ). It follows from (3.3)that xγ κ − ∈ supp µ . This and the inequalities ϑ = αγ κ − < xγ κ − < α † γ κ − = ϑ lead to µ (( ϑ , ϑ )) >
0, which contradicts our assumption that µ (( ϑ , ϑ )) = 0.(iii-a) In view of (2.1), { a n } ∞ n =0 is determinate. Hence, by (i) and (ii), we have ν (( α, α † ) ∪ ( β † , β )) = 0. So if β † < α † , then ν (( α, β )) = 0. Assume now that J. STOCHEL AND J. B. STOCHEL β † = α † and κ >
3. It is enough to show that α † / ∈ supp ν . Suppose that, contraryto our claim, α † ∈ supp ν . By Corollary 3.4(iii), γ = κ √ ϑ ∈ supp ν . This and (3.3)imply that ( α † ) κ − γ ∈ supp µ . Since κ > α † = β † < β γ , we have ϑ = ( β † ) κ = ( α † ) κ < ( α † ) κ − γ < α † γ κ − = ϑ . Hence µ (( ϑ , ϑ )) >
0, which is a contradiction. Finally, the case of β † = α † and β ∈ supp ν follows from Remark 5.4 and (iii-b).(iii-b) Set α j = ( γβ ) j α and α † j = ( γβ ) j α † for j = 0 , . . . , κ −
1. Note that α j < α † j for j = 0 , . . . , κ −
1, and α = α < β = α † κ − . (5.2)We claim that ν (( α j , α † j )) = 0 , j = 0 , . . . , κ − . (5.3)Suppose that, contrary to our claim, ν (( α j , α † j )) > j ∈ { , . . . , κ − } .Then there exists x ∈ ( α j , α † j ) ∩ supp ν . By Corollary 3.4(iii), γ ∈ supp ν . Since β, γ ∈ supp ν , we infer from (3.3) that xβ j γ κ − − j ∈ supp µ . Noting that ϑ = αγ κ − = (cid:16) γβ (cid:17) j αβ j γ κ − − j = α j β j γ κ − − j < xβ j γ κ − − j < α † j β j γ κ − − j = (cid:16) γβ (cid:17) j α † β j γ κ − − j = α † γ κ − = ϑ , we get µ (( ϑ , ϑ )) >
0, in contradiction with our assumption that µ (( ϑ , ϑ )) = 0.The inequality γβ α < α † is easily seen to be equivalent to α j +1 < α † j , j = 0 , . . . , κ − . (5.4)We will show that ( α, β ) = κ − [ j =0 ( α j , α † j ) . (5.5)Indeed, assuming that ϑ < ϑ , we see that if x ∈ ( α, α κ − ], then x ∈ ( α j , α j +1 ]for some j ∈ { , . . . , κ − } . Hence, by (5.4), x ∈ ( α j , α † j ). Since, by (5.2),( α κ − , α † κ − ) = ( α κ − , β ), the equality (5.5) is proved.Combining (5.3) with (5.5), we conclude that ν (( α, β )) = 0. This completesthe proof. (cid:3) Regarding Theorem 5.3, we can write the following analogue of Remark 5.2.
Remark 5.5.
Suppose that (3.1) holds and ϑ , ϑ , ϑ are real numbers such that0 ϑ < ϑ ϑ = sup supp µ and µ (( ϑ , ϑ )) = 0. Set ϑ ′ = ϑ , ϑ ′ = inf([ ϑ , ϑ ] ∩ supp µ ) and ϑ ′ = ϑ . Then 0 ϑ ′ < ϑ ′ ϑ ′ = sup supp µ and µ (( ϑ ′ , ϑ ′ )) = 0.Let α, α † , β, β † , γ (respectively α ′ , α †′ , β ′ , β †′ , γ ′ ) be numbers attached to ϑ , ϑ , ϑ (respectively ϑ ′ , ϑ ′ , ϑ ′ ) via (4.1) and (4.2). Then α = α ′ , α † α †′ , β β ′ , β † = β †′ and γ = γ ′ . Hence, if β † < α † or β † α † , then β † < α †′ or β † α †′ respectively. This means that we can apply Theorem 5.3 to the new system ofnumbers ϑ , ϑ ′ , ϑ , α, α †′ , β ′ , β † , γ (note that if β ∈ supp ν , then by Corollary 3.4 (i)we have ϑ = β κ ∈ supp µ , and so ϑ ′ = ϑ ). The case of ϑ = ϑ is obvious. N THE κ TH ROOT OF A STIELTJES MOMENT SEQUENCE 9
Corollary 5.6.
Suppose (3.1) holds. Let ϑ , ϑ , ϑ be real numbers such that ϑ < ϑ ϑ and let α := ϑ ϑ κ √ ϑ , β := κ √ ϑ , γ := κ √ ϑ and β † := κ √ ϑ . Thenthe following assertions are valid. (i) If ϑ < ϑ , { a n } ∞ n =0 is determinate and µ (( ϑ , ϑ )) = µ (( ϑ , ϑ )) = 0 ,then ν (( β † , β )) = ν (( β, γ )) = 0 . (ii) If ϑ = ϑ , ϑ = sup supp µ and µ (( ϑ , ϑ )) = 0 , then ν (( α, β )) = 0 and β ∈ supp ν . (iii) If ϑ = ϑ , ϑ ∈ supp µ , ϑ = sup supp µ and µ (( ϑ , ϑ )) = 0 , then ν (( α, β )) = 0 and α, β ∈ supp ν . (iv) If ϑ = ϑ , then the following two conditions are equivalent :(a) ϑ ∈ supp µ , ϑ = sup supp µ and µ (( ϑ , ϑ )) = 0 , (b) α ∈ supp ν , β = sup supp ν and ν (( α, β )) = 0 . Proof. (i) Apply Theorem 5.3 (i) to intervals ( ϑ , ϑ ) and ( ϑ , ϑ ).(ii) and (iii) These two assertions can be deduced from assertion (ii) of Theorem5.3, (2.2) and assertions (ii) and (iii) of Theorem 5.1 (because β = α † ).(iv) Implication (a) ⇒ (b) can be inferred from (iii) and (2.2). The reverseimplication is an immediate consequence of Theorem 5.1. (cid:3) See Appendix for another proof of the implication (a) ⇒ (b) of Corollary 5.6 (iv).We conclude this section by discussing the case when the support of a repre-senting measure of a Stieltjes moment sequence is contained in a closed interval[ ϑ, ∞ ), ϑ >
0. Note that ϑ corresponds to ϑ in Theorem 5.1. Proposition 5.7.
Suppose that (3.1) holds and { a n } ∞ n =0 is determinate. Then thefollowing assertions are valid. (i) If ν ([0 , β )) = 0 for some real number β > , then µ ([0 , β κ )) = 0 . (ii) If µ ([0 , ϑ )) = 0 for some real number ϑ > , then ν ([0 , κ √ ϑ )) = 0 . (iii) If µ ([0 , ϑ )) = 0 for some real number ϑ > such that ϑ ∈ supp µ , then ν ([0 , κ √ ϑ )) = 0 and κ √ ϑ ∈ supp ν . Proof.
Assertions (i) and (ii) follow from Theorem 3.3.(iii) By (ii), we have ν ([0 , κ √ ϑ )) = 0, and thus 0 / ∈ supp ν . It follows fromTheorem 3.3 that ϑ = x · · · x κ for some x , . . . , x κ ∈ supp ν . Suppose that, con-trary to our claim, κ √ ϑ / ∈ supp ν . Then x j > κ √ ϑ for all j ∈ { , . . . , κ } , and thus ϑ = x · · · x κ > ϑ , which is a contradiction. (cid:3)
6. When is ( α, β ) a hole of supp ν ? Before stating a theorem which provides an answer to the above question, weintroduce a new parameter ι ∗ s and prove two technical lemmas about the parameters ι s (cf. (4.4)) and ι ∗ s . For real numbers ϑ , ϑ , ϑ such that 0 < ϑ < ϑ < ϑ , we set ι ∗ s = ι ∗ s ( ϑ , ϑ , ϑ ) := 1 + (cid:22) log ( ϑ /ϑ )log ( ϑ /ϑ ) (cid:23) > . (6.1)Note that the following equalities hold for all t ∈ (0 , ∞ ), ι s ( ϑ , ϑ , ϑ ) = ι s (cid:0) tϑ , tϑ , tϑ (cid:1) and ι ∗ s ( ϑ , ϑ , ϑ ) = ι ∗ s (cid:0) tϑ , tϑ , tϑ (cid:1) . Lemma 6.1. If ϑ , ϑ , ϑ ∈ R are such that < ϑ < ϑ < ϑ , then the followingassertions are valid :(i) ι s = 2 implies ι ∗ s > , (ii) ι s = 3 implies ι ∗ s ∈ { , } , (iii) ι s = 3 and ι ∗ s = 2 if and only if ϑ ϑ = ϑ , (iv) ι s > implies ι ∗ s = 1 , (v) ι ∗ s = 1 if and only if ϑ ϑ < ϑ , (vi) ι ∗ s = 1 implies ι s > , (vii) ι ∗ s = 2 implies ι s ∈ { , } , (viii) ι ∗ s > implies ι s = 2 .Moreover, for every integer p > , there exist real numbers ϑ , ϑ , ϑ such that < ϑ < ϑ < ϑ , ι s = 2 and ι ∗ s = p . Proof.
Suppose that ϑ , ϑ , ϑ ∈ R are such that 0 < ϑ < ϑ < ϑ . First weshow that there exists r ∈ [0 ,
1) such that ι s − r > ι ∗ s = 1 + (cid:22) ι s − r (cid:23) . (6.2)Indeed, by definition of ι s , there exists r ∈ [0 ,
1) such that ι s − r > log ( ϑ /ϑ )log ( ϑ /ϑ ) = ι s − r . This gives us ϑ ϑ = (cid:0) ϑ ϑ (cid:1) ι s − r . As a consequence, we have ϑ ϑ = (cid:16) ϑ ϑ (cid:17) ι s − r , (6.3)which yields (6.2).Conditions (i) to (viii) except for (iii) and (v) can be deduced from (6.2).Condition (iii) follows from (6.2) and (6.3). Clearly, ι ∗ s = 1 if and only if log ( ϑ /ϑ )log ( ϑ /ϑ ) <
1, or equivalently if and only if ϑ ϑ < ϑ , which gives (v).Now we justify the “moreover” part. Set ϑ = 1. For fixed ϑ ∈ (0 ,
1) and p ∈{ , , . . . } , we define ϑ = ϑ p − p +1 . It is a simple matter to verify that ϑ < ϑ < ϑ , ι s = 2 and ι ∗ s = p . This completes the proof. (cid:3) Lemma 6.2.
Let ϑ , ϑ , ϑ be real numbers such that < ϑ < ϑ < ϑ . Then thefollowing assertions hold :(i) if α † = β † and κ > , then ι ∗ s = 1 , (ii) if κ = 2 , then α † = β † if and only if ι s = 3 and ι ∗ s = 2 , (iii) if κ > ι ∗ s , then γβ α < α † ,where α , α † , β , β † and γ are as in (4.1) and (4.2) . Proof. (i) By (4.2), we have ϑ = ϑ κ r ϑ ϑ . (6.4)Since κ >
3, we get 1 − κ >
0, and thus ϑ − κ < ϑ − κ , which together with (6.4)implies that ϑ ϑ < (cid:16) ϑ κ q ϑ ϑ (cid:17) = ϑ . By Lemma 6.1(v), ι ∗ s = 1.(ii) By (4.2), α † = β † if and only if ϑ ϑ = ϑ . Applying Lemma 6.1(iii)completes the proof of (ii).(iii) Since κ > ι ∗ s , we see that κ > log ( ϑ /ϑ )log ( ϑ /ϑ ) . Hence ϑ ϑ > (cid:0) ϑ ϑ (cid:1) κ , and thus ϑ ϑ > (cid:0) ϑ ϑ (cid:1) κ , which implies that γβ α < α † . (cid:3) Theorem 6.3 below gives sufficient conditions for an interval ( α, β ) to be a holeof supp ν written in terms of the parameters ι s and ι ∗ s . N THE κ TH ROOT OF A STIELTJES MOMENT SEQUENCE 11
Theorem 6.3. If (3.1) holds, µ (( ϑ , ϑ )) = 0 for some ϑ , ϑ ∈ R such that < ϑ < ϑ < ϑ := sup supp µ < ∞ and any of the conditions (i) - (v) below issatisfied, then ν (( α, β )) = 0 . (i) κ > ι ∗ s and β ∈ supp ν . (ii) ι s > ι ∗ s and β ∈ supp ν . (iii) ι s > and β ∈ supp ν . (iv) ι s > . (v) ι ∗ s = 1 . ( α, β, ι s , ι ∗ s are as in (4.1) , (4.4) and (6.1) . ) Proof. (i) Apply Lemma 6.2(iii) and Theorem 5.3 (iii-b).(v) We deal first with the case in which ϑ ∈ supp µ . We will show that β ∈ supp ν . By (2.1), Corollary 3.4 (iii) and Theorem 3.3, there exist x , . . . , x κ ∈ supp ν such that ϑ = x · · · x κ . It follows from (4.2) and (4.3) that ϑ = α † γ κ − and 0 < α † , x , . . . , x κ γ . Hence x j ∈ [ α † , γ ] , j ∈ { , . . . , κ } . (6.5)Note that x = · · · = x κ . Indeed, otherwise x k < x l for some k = l . Since, byLemma 6.1(v), ϑ ϑ < ϑ , we have ϑ > x k · Y j / ∈{ k,l } x j = ϑ x k x l (6.5) > ϑ α † γ = ϑ ϑ > ϑ . (6.6)Applying Theorem 3.3 we see that x k · Q j / ∈{ k,l } x j ∈ supp µ , which together with(6.6) shows that µ (( ϑ , ϑ )) >
0. This leads to a contradiction. Since x = · · · = x κ ,we deduce that β = κ √ ϑ = x ∈ supp ν . Applying (i), we get ν (( α, β )) = 0.If ϑ / ∈ supp µ , then we argue as follows. Set ϑ ′ = inf([ ϑ , ϑ ] ∩ supp µ ) and β ′ = κ p ϑ ′ . Then β β ′ and µ (( ϑ , ϑ ′ )) = 0. If ϑ ′ = ϑ , then by Corollary 5.6(ii)we have ν (( α, β )) ν (( α, β ′ )) = 0. Otherwise ϑ ′ < ϑ . Since 1 ι ∗ s ( ϑ , ϑ ′ , ϑ ) ι ∗ s ( ϑ , ϑ , ϑ ) = 1 and ϑ ′ ∈ supp µ , we may apply the argument from the previousparagraph to ϑ , ϑ ′ , ϑ . What we obtain is ν (( α, β )) ν (( α, β ′ )) = 0.(iv) Apply (v) and Lemma 6.1(iv).(ii) If κ > ι ∗ s , then we may apply (i). Consider the case of κ < ι ∗ s . Then, by ι s > ι ∗ s , we have 2 κ < ι s . Thus, if ι s >
4, then we may apply (iv). It remainsto consider the case of ι s = 3. Then clearly κ = 2. It follows from ι s = 3 that log ( ϑ /ϑ )log ( ϑ /ϑ ) >
2, or equivalently that ϑ ϑ > ( ϑ ϑ ) , which is equivalent to β † α † .Applying Theorem 5.3 (iii-a) gives ν (( α, β )) = 0.(iii) Since, by parts (ii) and (iv) of Lemma 6.1, the inequality ι s > ι s > ι ∗ s , we can apply (ii). This completes the proof. (cid:3) Note that conditions (ii)-(v) of Theorem 6.3 impose no restriction on κ , andthat Theorem 6.3 is no longer true if any of the assumptions (i)-(v) is dropped (seeExample 7.3 for the discussion concerning the assumption β ∈ supp ν , and Example7.4 for the discussion concerning the remaining assumptions). Corollary 6.4.
Let { a n } ∞ n =0 be a Stieltjes moment sequence with a representingmeasure µ such that < ϑ := sup supp µ < ∞ . Suppose µ (( ϑ , ϑ )) = 0 for some ϑ , ϑ ∈ R such that < ϑ < ϑ < ϑ . Assume that the set J := n κ ∈ Z + : κ > and (cid:8) κ √ a n (cid:9) ∞ n =0 is a Stieltjes moment sequence o (6.7) is nonempty. If ι ∗ s = 1 , then the following conditions are equivalent :(i) ϑ ∈ supp µ , (ii) β ( κ ) ∈ supp ν κ for some κ ∈ J , (iii) β ( κ ) ∈ supp ν κ for every κ ∈ J ,where β ( κ ) = κ √ ϑ and ν κ is a representing measure of { κ √ a n } ∞ n =0 . Proof.
By (2.2), we have γ ( κ ) := κ √ ϑ = sup supp ν κ for every κ ∈ J .(i) ⇒ (iii) It follows form Theorem 6.3 (v) that ν κ (( α ( κ ) , β ( κ ))) = 0 for every κ ∈ J , where α ( κ ) = ϑ ϑ κ √ ϑ . Hence, by Theorem 5.1 (iii), β ( κ ) ∈ supp ν κ forevery κ ∈ J .(iii) ⇒ (ii) Evident.(ii) ⇒ (i) Applying (3.3), we see that ϑ = β ( κ ) κ ∈ supp µ . (cid:3) It is worth mentioning that implication (i) ⇒ (iii) of Corollary 6.4 is no longertrue if we drop the assumption that ι ∗ s = 1 (cf. Example 7.3), though the reverseimplication is always true. What is more, the set J defined in (6.7) may not be aset of consecutive integers (cf. Example 7.6). As shown in Example 7.3, the set J may contain only one point. It may also happen that J = { , , , . . . } (cf. [ ]).
7. Examples
In this section we gather examples that illustrate the delicate nature of resultsappearing in Sections 3, 5 and 6. In what follows we adhere to the notation in(4.1), (4.2), (4.4) and (6.1). If { a n } ∞ n =0 and { κ √ a n } ∞ n =0 are determinate Stieltjesmoment sequences, then their representing measures will be denoted by µ and ν respectively.We begin by showing that the closure sign in (3.3) cannot be omitted. Example 7.1.
Let κ = 2. For τ ∈ { , } , we set √ a n = ∞ X j =2 j ( j + τ ) n + ∞ X j =2 j n e j , n ∈ Z + . (7.1)It is clear that {√ a n } ∞ n =0 and { a n } ∞ n =0 are Stieltjes moment sequences, and that ν := P ∞ j =2 − j δ j + τ + P ∞ j =2 e − j δ j is a representing measure of {√ a n } ∞ n =0 . Hencesupp ν = { } ∪ n . . . ,
14 + τ ,
13 + τ ,
12 + τ o ∪ { , , , . . . } . (7.2)Now we show that { a n } ∞ n =0 is a determinate Stieltjes moment sequence. For n > f n : R + → R + by f n ( x ) = x n e − x for x ∈ R + . It is easilyseen that f n is strictly increasing on the interval [0 , √ n ] and strictly decreasing onthe interval [ √ n, ∞ ). This implies that √ a n (7.1) ∞ X j =2 f n ( j ) ⌊√ n ⌋− X j =1 f n ( j ) + ∞ X j = ⌊√ n ⌋ +2 f n ( j ) + f n ( ⌊√ n ⌋ ) + f n ( ⌊√ n ⌋ + 1) Z ∞ f n ( x ) d x + 2 f n ( √ n ) , n > . N THE κ TH ROOT OF A STIELTJES MOMENT SEQUENCE 13
By a suitable change of variables, we have Z ∞ f n ( x ) d x = 12 Z ∞ x n − e − x d x Z x n − e − x d x + Z ∞ x n − e − x d x Z ∞ x n e − x d x Z ∞ x n e − x d x = 1 + n ! , n > . It is also clear that f n ( √ n ) n n for all n >
1. Putting all these together yields √ a n n n for all n >
4, which implies that P ∞ n =1 a − / n n = ∞ . Hence, by theCarleman criterion (cf. [ , Theorem 1.10]), the Stieltjes moment sequence { a n } ∞ n =0 is determinate.We first consider the case of τ = 0. By (7.2), the set π κ (cid:0) (supp ν ) κ (cid:1) coincideswith the set of all nonnegative rational numbers, and so, in view of Theorem 3.3,we have supp µ = R + .Now suppose that τ = / . It follows from (7.2) that 1 / ∈ π κ (cid:0) (supp ν ) κ (cid:1) . Since1 = lim j →∞ jj + , we infer from (3.3) that 1 ∈ supp µ .As above we verify that the Stieltjes moment sequence { a n } ∞ n =0 given by √ a n = ∞ X j =2 j ( j + τ ) n + ∞ X j =2 j n e j , n ∈ Z + (cid:16) τ ∈ n , o(cid:17) , satisfies the inequality √ a n n n for all n >
2, which implies that P ∞ n =1 a − / n n = ∞ . Hence, by the Carleman criterion (cf. [ , Theorem 1.11]), the Stieltjes momentsequence { a n } ∞ n =0 is determinate (in fact, it is determinate as a Hamburger momentsequence, cf. [ , Corollary 4.5]). Choosing τ ∈ { , } , we obtain new examples ofStieltjes moment sequences with the required properties.The next example is related to Corollary 3.4 and Theorems 5.1 and 5.3. Example 7.2.
Set κ = 3. Let ϑ , ϑ , ϑ be real numbers such that0 < ϑ < ϑ < ϑ , ϑ ϑ < r ϑ ϑ and r ϑ ϑ < r ϑ ϑ (e.g., ϑ = √ , ϑ = 1 and ϑ = 2). Then 0 < α < α † < β † < β < γ and α < α β < α γ < αβ < αβγ< αγ = ϑ < ϑ = β < β γ < βγ < γ = ϑ . (7.3)Set a n = ( α n + β n + γ n ) for n ∈ Z + . Clearly { a n } ∞ n =0 and { √ a n } ∞ n =0 are determi-nate Stieltjes moment sequences. We easily verify that the terms of the sequence(7.3) form the support of µ and that supp ν = { α, β, γ } . Thus µ (( ϑ , ϑ )) = 0, { ϑ , ϑ } ⊆ supp µ and µ ([0 , ϑ )) >
0, though ν (( α, β )) = 0, { α, β } ⊆ supp ν and ν ([0 , α )) = 0.(7.4)Since card (supp µ ) = 10, we see that supp ν κ √ supp µ . Moreover, if we replacethe inequality q ϑ ϑ < q ϑ ϑ by the stronger one q ϑ ϑ < q ϑ ϑ (which is still satisfiedby ϑ = √ , ϑ = 1 and ϑ = 2), then γβ α < α † and β ∈ supp ν , though β † > α † .Example 7.3 below shows that the assumption that one or two endpoints of theinterval ( α, β ) belong to supp ν is essential for Theorems 5.1, 5.3 and 6.3 as well as for Corollary 6.4. Moreover, in this example, the set J defined in (6.7) consistsonly of one point 2. Example 7.3.
Let κ = 2, ϑ = 1, ϑ = √ a and ϑ = a with a ∈ (1 , ∞ ). Then α = √ a , α † = β † = 1, β = √ a and γ = √ a . Set a n = (( α † ) n + γ n ) = ϑ n + 2 ϑ n + ϑ n , n ∈ Z + . (7.5)Clearly { a n } ∞ n =0 and {√ a n } ∞ n =0 are determinate Stieltjes moment sequences. Itfollows from (7.5) that supp µ = { ϑ , ϑ , ϑ } and supp ν = { α † , γ } . Hence γβ α < α † , ν (( α, β )) > α, β supp ν . Moreover, we have ι s = 3 > ι ∗ s .We will show that { κ √ a n } ∞ n =0 is not a Stieltjes moment sequence for everyinteger κ >
3. Suppose that, contrary to our claim, { κ √ a n } ∞ n =0 is a Stieltjesmoment sequence for some integer κ >
3. Denote by ν κ the representing mea-sure of { κ √ a n } ∞ n =0 . By (2.1) and Corollary 3.4, card (supp ν κ ) < ∞ and κ √ ϑ =sup supp ν κ . Hence, in view of Theorem 3.3, supp µ = π κ (cid:0) (supp ν κ ) κ (cid:1) . If supp ν κ = { κ √ ϑ } , then supp µ = { ϑ } , a contradiction. Otherwise, by (3.3), there exists x ∈ supp ν κ ∩ (0 , κ √ ϑ ). Then, by (3.3) again, x (cid:0) κ √ ϑ (cid:1) κ , x (cid:0) κ √ ϑ (cid:1) κ − , . . . , x κ (cid:0) κ √ ϑ (cid:1) is a strictly decreasing sequence of κ + 1 elements of supp µ . Since κ + 1 >
4, wearrive at a contradiction with card (supp µ ) = 3.The subsequent example is related to Theorems 5.3 and 6.3. Example 7.4.
Let κ = 2. Set ϑ = , ϑ = 1 and ϑ = 9. Then α = , α † = , β † = √ , β = 1 and γ = 3. Set a n = ( α n + ( α † ) n + β n + γ n ) for n ∈ Z + . TheStieltjes moment sequences { a n } ∞ n =0 and {√ a n } ∞ n =0 are determinate. It is easilyseen that supp µ = { , , , , , , , , } , µ (( ϑ , ϑ )) = 0, { ϑ , ϑ , ϑ } ⊆ supp µ , ν (( α, β )) > { α, β } ⊆ supp ν . Moreover, we have α † < β † , γβ α > α † , ι s = 2 < ι ∗ s . (7.6)Now we show that Theorems 5.3 (iii) and 6.3 do not cover all possible situationsin which supports of representing measures of the κ th roots of Stieltjes momentsequences may be involved. Example 7.5.
Let κ = 2. Set ϑ = 1, ϑ = a and ϑ = a with a ∈ (1 , ∞ ).Then α = a , α † = a , β † = 1, β = a and γ = a . Set a n = ( α n + β n + γ n ) for n ∈ Z + . Clearly { a n } ∞ n =0 and {√ a n } ∞ n =0 are determinate Stieltjes momentsequences. We verify directly that supp µ = { a , a , , a , a , a } , µ (( ϑ , ϑ )) = 0, { ϑ , ϑ , ϑ } ⊆ supp µ , ν (( α, β )) = 0 and { α, β } ⊆ supp ν . Moreover, (7.6) issatisfied.We conclude this section with an example of a Stieltjes moment sequence whose κ th root is a Stieltjes moment sequence for κ = 2 ,
4, but not for κ = 3 (consultCorollary 6.4). Example 7.6.
Fix a real number a >
1. Set ˆ α = a − , ˆ β = 1, ˆ γ = a and a n = (ˆ α n + ˆ β n + ˆ γ n ) , n ∈ Z + . It is clear that { a n } ∞ n =0 , {√ a n } ∞ n =0 and { √ a n } ∞ n =0 are determinate Stieltjes mo-ment sequences whose representing measures have finite supports. We claim that { √ a n } ∞ n =0 is not a Stieltjes moment sequence. N THE κ TH ROOT OF A STIELTJES MOMENT SEQUENCE 15
For this, denote by µ and ν the representing measures of Stieltjes momentsequences { a n } ∞ n =0 and { √ a n } ∞ n =0 , respectively. By Theorem 3.3, we havesupp µ = π (cid:0) (supp ν ) (cid:1) = π (cid:0) { ˆ α, ˆ β, ˆ γ } (cid:1) = { a j − i : i, j ∈ Z + , i + j } . This implies that card (supp µ ) = 15 . (7.7)Set ϑ = ˆ α ˆ γ κ − , ϑ = ˆ β κ and ϑ = ˆ γ κ with κ = 4. Clearly ϑ = a − , ϑ = 1 and ϑ = a . Since supp ν = { ˆ α, ˆ β, ˆ γ } , ν (cid:0) (ˆ α, ˆ β ) (cid:1) = 0 and ˆ α ˆ γ < ˆ β , we infer fromTheorem 5.1 that µ (cid:0) ( ϑ , ϑ ) (cid:1) = 0, ϑ = sup supp µ and ϑ , ϑ , ϑ ∈ supp µ .Suppose that, contrary to our claim, { √ a n } ∞ n =0 is a Stieltjes moment sequence.Let ν be its representing measure. In view of Corollary 3.4, supp ν is finite. Hence,by Theorem 3.3, we have supp µ = π (cid:0) (supp ν ) (cid:1) . (7.8)Let α , β , γ , α † and β † be as in (4.1) and (4.2) with κ = 3. Then α = a − , β = 1, γ = a , α † = β † = a − and αγ < αγ = ϑ < . (7.9)By Theorem 5.3 (iii-a), ν (cid:0) ( α, β ) (cid:1) = 0. It follows from Corollary 3.4 (iii) that γ = sup supp ν . Applying Theorem 5.1, we deduce that α, β, γ ∈ supp ν . Ifcard (supp ν ) = 3, then supp ν = { α, β, γ } , and by (7.8) we have card (supp µ ) =10, which contradicts (7.7). The remaining possibility is that card (supp ν ) > x ∈ supp ν ∩ [(0 , α ) ∪ ( β, γ )]. We will show that card (supp µ ) >
16, again contradicting (7.7).Consider first the case of x ∈ (0 , α ). If α = xγ , then x = a − and the sequence { x , x α, xα , α , x , xα, xαγ, α γ, x, xγ, xγ , αγ , , γ, γ , γ } ⊆ supp µ is strictly increasing. If α < xγ , then (7.9) implies that the sequence { x , x α, x , xα, x γ, xαγ, x, α, xγ, αγ, xγ , αγ , , γ, γ , γ } ⊆ supp µ is strictly increasing as well. Finally, if α > xγ , then, by (7.9) again, the sequence { ξ n } n =1 := { x , x α, x , x γ, xα, xαγ, x, xγ, xγ , αγ, αγ , , γ, γ , γ } ⊆ supp µ is strictly increasing and ξ < α < ξ . If α = ξ , then evidently card (supp µ ) > x = a − , and thus ξ < α < ξ , which yields card (supp µ ) > x ∈ ( β, γ ). Then by (7.9) the sequence { α , α , xα , α γ, α, xα, x α, xαγ, αγ , , x, x , x , x γ, xγ , γ } ⊆ supp µ is strictly increasing. This completes the proof of our claim. We use (7.8) to justify that the terms of the sequences being considered are in supp µ .
8. Square roots
In this section, we concentrate on square roots of Stieltjes moment sequenceswhich have representing measures supported in finite sets. For
M, N ∈ N , we definethe following classes of Stieltjes moment sequences: • S M stands for the set of all Stieltjes moment sequences having represent-ing measures µ such that supp µ ⊆ (0 , ∞ ) and card (supp µ ) = M , • S / M stands for the set of all sequences { a n } ∞ n =0 ∈ S M such that {√ a n } ∞ n =0 is a Stieltjes moment sequence, • S / M,N stands for the set of all { a n } ∞ n =0 ∈ S M such that {√ a n } ∞ n =0 ∈ S N .By (2.1), any member of S M is determinate. It follows from Theorem 3.3 that S / M = ∞ [ N =1 S / M,N . (8.1)We now describe all pairs ( M, N ) for which the classes S / M,N are nonempty.
Theorem 8.1.
Let
M, N ∈ N . Then the following conditions are equivalent :(i) S / M,N = ∅ , (ii) 2 N − M (cid:0) N +12 (cid:1) , (iii) n − ( M ) N n + ( M ) ,where n − ( M ) = (cid:6) √ M +1 − (cid:7) and n + ( M ) = (cid:4) M +12 (cid:5) . Moreover, the following holds :(a) { n − ( M ) } ∞ M =1 and { n + ( M ) } ∞ M =1 are monotonically increasing sequences, (b) for each k ∈ N , exactly k terms of { n − ( M ) } ∞ M =1 are equal to k , (c) for each k ∈ N , exactly terms of { n + ( M ) } ∞ M =1 are equal to k . Proof. (i) ⇒ (ii) Take { a n } ∞ n =0 ∈ S / M,N . Let µ and ν be as in (3.1) with κ = 2.Then supp ν = { ξ , . . . , ξ N } , where 0 < ξ < . . . < ξ N . First note that 2 N − M .Indeed, this can be inferred from (3.3) and the following inequalities ξ < ξ ξ < ξ < . . . < ξ N − < ξ N − ξ N < ξ N . Set Ω N = (cid:8) ( k, l ) ∈ J N × J N : k l (cid:9) with J N := { , . . . , N } . By (3.3) the mapping Ω N ∋ ( i, j ) ξ i ξ j ∈ supp µ is surjective, and thus M card ( Ω N ) = (cid:0) N +12 (cid:1) .(ii) ⇒ (i) We shall prove that for every N ∈ N and for every M ∈ N such that2 N − M (cid:0) N +12 (cid:1) there exists a sequence ξ < . . . < ξ N of positive realnumbers such that card (cid:0)(cid:8) ξ i ξ j : i, j ∈ J N (cid:9)(cid:1) = M . Once this is done, we see that (cid:8) ( P Nj =1 ξ nj ) (cid:9) ∞ n =0 ∈ S / M,N .We proceed by induction on N . The cases of N = 1 and N = 2 are easily seento hold. Suppose that our claim is valid for a fixed integer N which is greater thanor equal to 2. Take M ∈ N such that2( N + 1) − M (cid:18) N + 22 (cid:19) . (8.2)First we consider the case when M N. (8.3) For simplicity, we write (cid:8) ξ i ξ j : i, j ∈ J N (cid:9) in place of (cid:8) x ∈ R | ∃ i, j ∈ J N : x = ξ i ξ j (cid:9) . N THE κ TH ROOT OF A STIELTJES MOMENT SEQUENCE 17
It is clear that k := (cid:0) N +22 (cid:1) − N + 1) > N > k = M − N + 1).It follows from (8.2) that − k k . From (8.3) we infer that k + 3 N + 1. Fix ξ ∈ (0 , ∞ ) and t ∈ (1 , ∞ ), and set ξ j = ξ t k + j for j = 2 , . . . , N + 1. It is easily seenthat the sets Ξ N +1 := { ξ i ξ j : i, j ∈ J N +1 } and { ξ } ∪{ ξ t i } k + N +1 i = k +2 ∪{ ξ t i } k +2( N +1) i =2 k +4 coincide. Thus, since k + 3 N + 1, we can arrange the elements of the set Ξ N +1 as follows ξ < ξ ξ < . . . < ξ ξ k +3 < ξ < ξ ξ < ξ < . . . < ξ N < ξ N ξ N +1 < ξ N +1 . Hence card (Ξ N +1 ) = M .Now we consider the remaining possibility, namely that M > N. (8.4)Set M ′ = M − ( N + 1). Then by (8.2) we have2 N − N − ( N + 1) (8.4) < M ′ (8.2) (cid:18) N + 22 (cid:19) − ( N + 1) = (cid:18) N + 12 (cid:19) . By the induction hypothesis applied to M ′ , there exists a sequence ξ < . . . < ξ N +1 of positive real numbers such that card (cid:0)(cid:8) ξ i ξ j : i, j = 2 , . . . , N + 1 (cid:9)(cid:1) = M ′ . Thenthere exists ξ ∈ (0 , ξ ) such that ξ < ξ ξ < . . . < ξ ξ N +1 < ξ . Since ξ ξ i ξ j for all i, j = 2 , . . . , N + 1, we conclude that card (cid:0)(cid:8) ξ i ξ j : i, j = 1 , . . . , N + 1 (cid:9)(cid:1) = M ′ + N + 1 = M . This completes the induction argument. Hence (i) is valid.It is a matter of routine to show that the conditions (ii) and (iii) are equivalent.The assertions (a) and (c) are easily seen to hold, so we only explicitly prove (b).Set M k = (cid:0) k +12 (cid:1) for k ∈ N . Then M = 1 and M k +1 − M k = k + 1 for k ∈ N . Notethat n − (1) = 1 and √ M k + 1 −
12 = (2 k + 1) −
12 = k, k ∈ N . This implies that for every k ∈ N and for every M ∈ N such that M k + 1 M M k +1 , n − ( M ) = k + 1. This completes the proof. (cid:3) Using assertions (a), (b) and (c) of Theorem 8.1, one can easily specify suc-cessive terms of the sequences { n − ( M ) } ∞ M =1 and { n + ( M ) } ∞ M =1 . Below, we list thefirst fifteen terms of each of these sequences. M · · · n − ( M ) 1 · · · n + ( M ) 1 · · · Table 1
Applying (8.1) and Theorem 8.1 (see also Table 1), we get the following corollary.
Corollary 8.2. If M ∈ { , } , then S / M = ∅ . If M ∈ N \ { , } , then the set A M := { N ∈ N : n − ( M ) N n + ( M ) } is nonempty, S / M,N = ∅ for every N ∈ A M and S / M,N = ∅ for every N ∈ N \ A M . It is worth mentioning that the Stieltjes moment sequences { a n } ∞ n =0 constructedin Examples 7.3, 7.5 and 7.4 belong to the classes S / , , S / , and S / , , respectively.According to Corollary 8.2, the square root of a Stieltjes moment sequence whose representing measure is concentrated on either a two point or a four point subsetof (0 , ∞ ) is never a Stieltjes moment sequence. Hence, the following holds. Corollary 8.3 ([ , Lemma 3.3]) . If α , α , ϑ and ϑ are positive real numbers,then the sequence (cid:8)p α ϑ n + α ϑ n (cid:9) ∞ n =0 is a Stieltjes moment sequence if and onlyif ϑ = ϑ . Let { a n } ∞ n =0 be a Stieltjes moment sequence whose representing measure is con-centrated on a three point subset of (0 , ∞ ). Then there exit α , α , α , ϑ , ϑ , ϑ ∈ (0 , ∞ ) such that ϑ < ϑ < ϑ and a n = α ϑ n + α ϑ n + α ϑ n for all n ∈ Z + . Hence,in view of (8.1) and Corollary 8.2, the sequence {√ a n } ∞ n =0 is a Stieltjes momentsequence if and only if ϑ = ϑ ϑ and α = 4 α α .
9. Appendix
Here we present a proof of the implication (a) ⇒ (b) of Corollary 5.6 (iv) whichis independent of Theorems 5.1 and 5.3. First, we state an auxiliary result. Lemma 9.1. If { a n } ∞ n =0 is a Stieltjes moment sequence with a representing measure µ supported in [0 , , then lim k →∞ a k = µ ( { } ) and { a n − lim k →∞ a k } ∞ n =0 is aStieltjes moment sequence. Proof.
By Lebesgue’s monotone convergence theorem, lim k →∞ a k = µ ( { } ),and thus { a n − lim k →∞ a k } ∞ n =0 is a Stieltjes moment sequence with a representingmeasure µ | [0 , . (cid:3) Proof of implication (a) ⇒ (b) of Corollary 5.6 (iv) . According to (2.1)and (2.2), the Stieltjes moment sequences { a n } ∞ n =0 and { κ √ a n } ∞ n =0 are determinate.Without loss of generality we can assume that ϑ = 1 (consider { ϑ − n a n } ∞ n =0 insteadof { a n } ∞ n =0 ). We infer from (2.2) applied to { a n } ∞ n =0 thatlim n →∞ ( κ √ a n ) /n = 1 . (9.1)By (2.2), supp ν ⊆ [0 , n →∞ κ √ a n = κ p µ ( { } ).Now, by Lemma 9.1 applied to { κ √ a n } ∞ n =0 , we see that { κ √ a n − κ p µ ( { } ) } ∞ n =0 isa Stieltjes moment sequence. According to our assumption, { a n − µ ( { } ) } ∞ n =0 isa Stieltjes moment sequence with the representing measure µ = µ | [0 ,ϑ ] . Since ϑ ∈ supp µ , we get ϑ = sup supp µ , which together with (2.2) leads tolim n →∞ (cid:0) a n − µ ( { } ) (cid:1) /n = ϑ . (9.2)By the mean value theorem, we have (recall that a n − µ ( { } ) > n > κ √ a n − κ p µ ( { } ) = a n − µ ( { } ) κ τ κ − κ n , n = 0 , , , . . . , where τ n is a real number such that µ ( { } ) τ n a n . This implies that a n − µ ( { } ) κ a κ − κ n κ √ a n − κ p µ ( { } ) a n − µ ( { } ) κ µ ( { } ) κ − κ , n = 0 , , , . . . (9.3)The conditions (9.1), (9.2) and (9.3) combined givelim n →∞ (cid:0) κ √ a n − κ p µ ( { } ) (cid:1) /n = ϑ . (9.4) N THE κ TH ROOT OF A STIELTJES MOMENT SEQUENCE 19
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Instytut Matematyki, Uniwersytet Jagiello´nski, ul. Lojasiewicza 6, PL-30348 Kra-k´ow, Poland
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Faculty of Applied Mathematics, AGH University of Science and Technology, Al.Mickiewicza 30, PL-30059 Krak´ow, Poland
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