Constructive proof of the Carpenter's Theorem
CCONSTRUCTIVE PROOF OF THE CARPENTER’S THEOREM
MARCIN BOWNIK AND JOHN JASPER
Abstract.
We give a constructive proof of Carpenter’s Theorem due to Kadison [14, 15].Unlike the original proof our approach also yields the real case of this theorem. Kadison’s theorem
In [14] and [15] Kadison gave a complete characterization of the diagonals of orthogonalprojections on a Hilbert space H . Theorem 1.1 (Kadison) . Let { d i } i ∈ I be a sequence in [0 , . Define a = (cid:88) d i < / d i and b = (cid:88) d i ≥ / (1 − d i ) . There exists a projection P with diagonal { d i } if and only if one of the following holds (i) a, b < ∞ and a − b ∈ Z , (ii) a = ∞ or b = ∞ . The goal of this paper is to give a constructive proof of the sufficiency direction of Kadison’stheorem. Kadison [14, 15] referred to the necessity part of Theorem 1.1 as the PythagoreanTheorem and the sufficiency as Carpenter’s Theorem. Arveson [3] gave a necessary conditionon the diagonals of a certain class of normal operators with finite spectrum. When specializedto the case of two point spectrum Arveson’s theorem yields the Pythagorean Theorem, i.e.,the necessity of (i) or (ii) in Theorem 1.1. Whereas Kadison’s original proof is a beautifuldirect argument, Arveson’s proof uses the Fredholm Index Theory.In contrast, until very recently there were no proofs of Carpenter’s Theorem other thanthe original one by Kadison, although its extension for II factors was studied by Argeramiand Massey [2]. A notable exception is a recent paper by Argerami [1] about which webecame aware only after completing this work. In this paper we give an alternative proof ofCarpenter’s Theorem which has two main advantages over the original. First, the originalproof does not yield the real case, which ours does. Second, our proof is constructive in thesense that it gives a concrete algorithmic process for finding the desired projection. This isdistinct from Kadison’s original proof, which is mostly existential.The paper is organized as follows. In Section 2 we state preliminary results such as finiterank Horn’s theorem. These results are then used in Section 3 to show the sufficiency of(i) in Theorem 1.1. The key role in the proof is played by a lemma from [8] which enables Date : October 15, 2018.2000
Mathematics Subject Classification.
Primary: 42C15, 47B15, Secondary: 46C05.
Key words and phrases. diagonals of self-adjoint operators, the Schur-Horn theorem, the Pythagoreantheorem, the Carpenter theorem, spectral theory.This work was partially supported by a grant from the Simons Foundation ( a r X i v : . [ m a t h . F A ] S e p odifications of diagonal sequences into more favorable configurations. Section 4 containsthe proof of sufficiency of (ii) in Theorem 1.1. To this end we introduce an algorithmicprocedure for constructing a projection with prescribed diagonal which is reminiscent ofthe spectral tetris construction introduced by Casazza et al. [10] in their study of tightfusion frames. Finally, in Section 5 we formulate an open problem of characterizing spectralfunctions of shift-invariant spaces in L ( R d ), introduced by the first author and Rzeszotnikin [9], which was a motivating force behind this paper.2. Preliminary results
The main goal of this section is to give a constructive proof of Horn’s Theorem [18,Theorem 9.B.2], which is the sufficiency part of the Schur-Horn Theorem [13, 21]. Wepresent this proof both for the sake of self-sufficiency of part (i) of Carpenter’s Theorem andalso to cover the more general case of finite rank operators on an infinite dimensional Hilbertspace, see also [4, 16, 17]. Moreover, we also give an argument reducing Theorem 1.1 to thecountable case.
Theorem 2.1 (Horn’s Theorem) . Let { λ i } Ni =1 be a positive nonincreasing sequence, and let { d i } Mi =1 be a nonnegative nonincreasing sequence, where M ∈ N ∪ {∞} and M ≥ N . If n (cid:88) i =1 d i ≤ n (cid:88) i =1 λ i for all n ≤ N, M (cid:88) i =1 d i = N (cid:88) i =1 λ i , (2.1) then there is a positive rank N operator S on a real M -dimensional Hilbert space H withpositive eigenvalues { λ i } Ni =1 and diagonal { d i } Mi =1 . We need a basic lemma.
Lemma 2.2.
Let M ∈ N ∪ {∞} . If { d i } Mi =1 is a nonzero nonnegative sequence with M (cid:88) i =1 d i = λ < ∞ , then there is a positive rank operator S on an M -dimensional Hilbert space H with eigen-value λ and diagonal { d i } .Proof. Let { e i } Mi =1 be an orthonormal basis for the Hilbert space H . Set v = M (cid:88) i =1 (cid:112) d i e i , and define S : H → H by Sf = (cid:104) f, v (cid:105) v for each f ∈ H . Clearly S is rank 1, and since (cid:107) v (cid:107) = λ the vector v is an eigenvector with eigenvalue λ . Finally, it is simple to check that S has the desired diagonal. (cid:3) roof of Theorem 2.1. The proof proceeds by induction on N . The base case N = 1 followsfrom Lemma 2.2. Suppose that Theorem 2.1 holds for ranks up to N −
1. Define m = max (cid:26) m : M (cid:88) i = m d i ≥ λ N (cid:27) and(2.2) η = (cid:18) M (cid:88) i = m d i (cid:19) − λ N = N − (cid:88) i =1 λ i − m − (cid:88) i =1 d i . Note that the maximality of m implies that m ≥ N . For each n ≤ N define δ n = n (cid:88) i =1 ( λ i − d i ) ≥ . For a certain value 0 ≤ ∆ ≤ η , which will be specified later, define the sequence(2.3) (cid:101) d i = d + ∆ i = 1 d m − ∆ i = m d i i (cid:54) = 1 , m . From the maximality of m we have (cid:101) d m = d m − ∆ ≥ d m − η = λ N − M (cid:88) i = m +1 d i > . This shows that { (cid:101) d i } is a nonnegative sequence. However, note that this sequence mightmight fail to be nonincreasing at the position i = m , which requires extra care in ourconsiderations.Our next goal is to construct an operator (cid:101) S with positive eigenvalues { λ i } Ni =1 , diagonal { (cid:101) d i } Mi =1 with respect to the orthonormal basis { e i } Mi =1 , and the property that (cid:104) (cid:101) Se , e m (cid:105) = 0.The argument splits into two cases. Case 1:
Assume there exists n ≤ min { N, m − } such that δ n < η . Fix n ≤ min { N, m − } such that δ n ≤ δ n for all n ≤ min { N, m − } . Define { (cid:101) d i } as in (2.3) with ∆ = δ n .Note that(2.4) M (cid:88) i = n +1 (cid:101) d i = − δ n + M (cid:88) i = n +1 d i = M (cid:88) i =1 d i − n (cid:88) i =1 λ i = N (cid:88) i =1 λ i − n (cid:88) i =1 λ i = N (cid:88) i = n +1 λ i . Since m > n and (cid:101) d m >
0, from (2.4) we see that n < N .For n ≤ n n (cid:88) i =1 (cid:101) d i = δ n + n (cid:88) i =1 d i ≤ δ n + n (cid:88) i =1 d i = n (cid:88) i =1 λ i with equality when n = n . Since n < N , by the inductive hypothesis there is a positiverank n operator (cid:101) S with eigenvalues { λ i } n i =1 and diagonal { (cid:101) d i } n i =1 with respect to the basis { e i } n i =1 . bserve that the subsequence { (cid:101) d i } N − i = n +1 coincides with { d i } N − i = n +1 since N − < m .Thus, for any n + 1 ≤ n ≤ N − n (cid:88) i = n +1 (cid:101) d i = n (cid:88) i = n +1 d i ≤ δ n − δ n + n (cid:88) i = n +1 d i = n (cid:88) i = n +1 λ i . Moreover, by (2.4) we have N (cid:88) i = n +1 (cid:101) d i ≤ M (cid:88) i = n +1 (cid:101) d i = N (cid:88) i = n +1 λ i . Thus, { λ i } Ni = n +1 and the nonincreasing rearrangement of { (cid:101) d i } Mi = n +1 satisfy the inductivehypothesis (2.1). That is, there is a positive rank N − n operator (cid:101) S with eigenvalues { λ i } Ni = n +1 and diagonal { (cid:101) d i } Mi = n +1 with respect to the basis { e i } Mi = n +1 . Thus, the operator (cid:101) S = (cid:101) S ⊕ (cid:101) S has the desired properties. Indeed, the property that (cid:104) (cid:101) Se , e m (cid:105) = 0 followsimmediately from the definition of (cid:101) S and the fact that n < m . Case 2:
Assume η ≤ δ n for all n ≤ min { N, m − } . Define { (cid:101) d i } as in (2.3) with ∆ = η .For n ≤ N − n (cid:88) i =1 (cid:101) d i = η + n (cid:88) i =1 d i ≤ δ n + n (cid:88) i =1 d i = n (cid:88) i =1 λ i . We also have by (2.2) m − (cid:88) i =1 (cid:101) d i = η + m − (cid:88) i =1 d i = N − (cid:88) i =1 λ i . By the inductive hypothesis there is a positive rank N − (cid:101) S with diagonal { (cid:101) d i } m − i =1 and positive eigenvalues { λ i } N − i =1 . Using the equality in (2.1) we have M (cid:88) i = m (cid:101) d i = − η + M (cid:88) i = m d i = M (cid:88) i =1 d i − N − (cid:88) i =1 λ i = λ N . By Lemma 2.2 there is a positive rank 1 operator (cid:101) S with diagonal { (cid:101) d i } Mi = m and eigenvalue λ N . Thus, the operator (cid:101) S = (cid:101) S ⊕ (cid:101) S has the desired properties.Combining the above two cases shows that the desired operator (cid:101) S exists. Let α ∈ [0 , α ( d + ∆) + (1 − α )( d m − ∆) = d . Define the unitary operator U on theorthonormal basis { e i } Mi =1 by U ( e i ) = √ αe − √ − αe m i = 1 , √ − αe + √ αe m i = m ,e i otherwise . A simple calculation shows that S = U ∗ (cid:101) SU has diagonal { d i } Mi =1 in the basis { e i } Mi =1 . Thiscompletes the proof of Theorem 2.1. (cid:3) he following “moving toward 0-1” lemma first appeared in [8]. Its proof is constructive asit consists a finite number of “convex moves” as at the end of the previous proof. Moreover,from the proof in [8] it follows that Lemma 2.3 holds for real Hilbert spaces as well ascomplex. Lemma 2.3.
Let { d i } i ∈ I be a sequence in [0 , . Let I , I ⊂ I be two disjoint finite subsetssuch that max { d i : i ∈ I } ≤ min { d i : i ∈ I } . Let η ≥ and η ≤ min (cid:26) (cid:88) i ∈ I d i , (cid:88) i ∈ I (1 − d i ) (cid:27) . (i) There exists a sequence { (cid:101) d i } i ∈ I in [0 , satisfying (cid:101) d i = d i for i ∈ I \ ( I ∪ I ) , (2.5) (cid:101) d i ≤ d i i ∈ I , and (cid:101) d i ≥ d i , i ∈ I , (2.6) η + (cid:88) i ∈ I (cid:101) d i = (cid:88) i ∈ I d i and η + (cid:88) i ∈ I (1 − (cid:101) d i ) = (cid:88) i ∈ I (1 − d i ) . (2.7) (ii) For any self-adjoint operator (cid:101) E on H with diagonal { (cid:101) d i } i ∈ I , there exists an operator E on H unitarily equivalent to (cid:101) E with diagonal { d i } i ∈ I . We end this section by remarking that the indexing set I in Theorem 1.1 need not becountable. In [15] the possibility that I is an uncountable set is addressed in all but themost difficult case where { d i } and { − d i } are nonsummable [15, Theorem 15]. However,the case when I is uncountable is a simple extension of the countable case, as we explainbelow. Proof of reduction of Theorem 1.1 to countable case.
First, we consider a projection P withdiagonal { d i } i ∈ I with respect to some orthonormal basis { e i } i ∈ I of a Hilbert space H . If a or b is infinite then there is nothing to show, so we may assume a, b < ∞ . Set J = { i ∈ I : d i =0 } ∪ { i ∈ I : d i = 1 } , and let P (cid:48) be the restriction of P to the subspace H (cid:48) = span { e i } i ∈ I \ J .Since e i is an eigenvector for each i ∈ J , H (cid:48) is an invariant subspace P (cid:48) ( H (cid:48) ) ⊂ H (cid:48) . Hence, P (cid:48) is a projection with diagonal { d i } i ∈ I \ J . The assumption that a, b < ∞ implies I \ J is atmost countable. Thus, the countable case of Theorem 1.1 applied to the operator P (cid:48) yields a − b ∈ Z . This shows that (ii) is necessary.To show that (i) or (ii) is sufficient, we claim that it is enough to assume that all of d i ’sare in (0 , P with only these d i ’s, then we take I to be theidentity and the zero operator on Hilbert spaces with dimensions equal to the cardinalitiesof the sets { i ∈ I : d i = 1 } and { i ∈ I : d i = 0 } , respectively. Then, P ⊕ I ⊕ has diagonal { d i } . Since a and b do not change when we restrict to (0 , { d i } i ∈ I has uncountably many terms and is contained in (0 , n ∈ N such that J = { i ∈ I : 1 /n < d i < − /n } has the same cardinality as I . Thus, we can partition I into a collection of countable infinite sets { I k } k ∈ K such that I k ∩ J is infinite for each k ∈ K .Each sequence { d i } i ∈ I k contains infinitely many terms bounded away from 0 and 1, thus (ii)holds. Again, by the countable case of Theorem 1.1, for each k ∈ K there is a projection P k with diagonal { d i } i ∈ I k . Thus, (cid:76) k ∈ K P k is a projection with diagonal { d i } i ∈ I . (cid:3) . Carpenter’s Theorem part i
The goal of this section is to give a proof of the sufficiency of (i) in Theorem 1.1. As acorollary of Theorem 2.1 we have the summable version of the Carpenter’s Theorem.
Theorem 3.1.
Let M ∈ N ∪ {∞} , and let { d i } Mi =1 be a sequence in [0 , . If (cid:80) Mi =1 d i ∈ N ,then there is a projection P with diagonal { d i } .Proof. Let { d (cid:48) i } M (cid:48) i =1 be the terms of { d i } in (0 , N = (cid:80) Mi =1 d i , and define λ i = 1 for i = 1 , . . . , N . Since d (cid:48) i ≤ i we have(3.1) n (cid:88) i =1 d (cid:48) i ≤ n (cid:88) i =1 λ i for n = 1 , , . . . , N. We also have M (cid:48) (cid:88) i =1 d (cid:48) i = N = N (cid:88) i =1 λ i . By Theorem 2.1 there is a rank N self-adjoint operator P (cid:48) with positive eigenvalues { λ i } Ni =1 and diagonal { d (cid:48) i } M (cid:48) i =1 . Since λ i = 1 for each i , the operator P (cid:48) is a projection. Let be thezero operator on a Hilbert space with dimension equal to |{ i : d i = 0 }| . The operator P (cid:48) ⊕ is a projection with diagonal { d i } Mi =1 . (cid:3) Corollary 3.2.
Let M ∈ N ∪ {∞} and { d i } Mi =1 be a sequence in [0 , . If (cid:80) Mi =1 (1 − d i ) ∈ N ,then there is a projection P with diagonal { d i } .Proof. This follows immediately from the observation that a projection P has diagonal { d i } if and only if I − P is a projection with diagonal { − d i } . (cid:3) Finally, we can handle the general case (i) of the Carpenter’s Theorem.
Theorem 3.3.
Let { d i } i ∈ I be a sequence in [0 , . If (3.2) a = (cid:88) d i < / d i < ∞ , b = (cid:88) d i ≥ / (1 − d i ) < ∞ , and a − b ∈ Z , then there exists a projection P with diagonal { d i } .Proof. First, note that if { d i } or { − d i } is summable, then by (3.2) its sum is in N . Thus, wecan appeal to Theorem 3.1 or Corollary 3.2, resp., to obtain the desired projection. Hence,we may assume both 0 and 1 are limit points of the sequence { d i } .Next, we claim that it is enough to prove the theorem under the assumption that d i ∈ (0 , i . Indeed, if P is a projection with diagonal { d i } d i ∈ (0 , , I is the identity operator ona space of dimension |{ i : d i = 1 }| , and is the zero operator on a space of dimension |{ i : d i = 0 }| , then P ⊕ I ⊕ is a projection with diagonal { d i } i ∈ I .Define J = { i ∈ I : d i < / } and J = { i ∈ I : d i ≥ / } . Choose i ∈ J such that d i ≤ d i for all i ∈ J . Choose J (cid:48) ⊆ J such that J \ J (cid:48) is finite and (cid:88) i ∈ J (cid:48) d i < − d i . et i ∈ J be such that d i > d i and d i + (cid:88) i ∈ J (cid:48) d i ≥ . Set(3.3) η = (cid:88) i ∈ J (cid:48) d i − (1 − d i ) < (cid:88) i ∈ J (cid:48) d i < − d i . Let I ⊂ J (cid:48) be a finite set such that(3.4) (cid:88) i ∈ I d i > η . By (3.3) and (3.4), we can apply Lemma 2.3 to finite subsets I and I = { i } to obtain asequence { (cid:101) d i } i ∈ I coinciding with { d i } i ∈ I outside of I ∪ I and such that (cid:88) i ∈ I (cid:101) d i = (cid:88) i ∈ I d i − η and 1 − (cid:101) d i = 1 − d i − η . Note that (cid:88) i ∈ J (cid:48) ∪{ i } (cid:101) d i = d i + (cid:88) i ∈ J (cid:48) \ I d i + (cid:88) i ∈ I (cid:101) d i = d i + (cid:88) i ∈ J (cid:48) \ I d i + (cid:88) i ∈ I d i − η = 1 . Thus, by Theorem 3.1 there is a projection P with diagonal { (cid:101) d i } i ∈ J (cid:48) ∪{ i } . Next, we notethat (cid:88) i ∈ I \ ( J (cid:48) ∪{ i } ) (1 − (cid:101) d i ) = (cid:88) i ∈ J \ J (cid:48) (1 − (cid:101) d i ) + (cid:88) i ∈ J \{ i } (1 − (cid:101) d i )= | J \ J (cid:48) | − (cid:88) i ∈ J \ J (cid:48) d i + (cid:88) i ∈ J \{ i } (1 − d i ) − η = | J \ J (cid:48) | − (cid:88) i ∈ J d i + (cid:88) i ∈ J (1 − d i ) = | J \ J (cid:48) | − a + b ∈ N . By Corollary 3.2 there is a projection P with diagonal { (cid:101) d i } i ∈ I \ ( J (cid:48) ∪{ i } ) .The projection P ⊕ P has diagonal { (cid:101) d i } i ∈ I . By Lemma 2.3 (ii) there is an operator P with diagonal { d i } i ∈ I which is unitarily equivalent to P ⊕ P . Thus, P is the requiredprojection. (cid:3) In [14, Remark 8] Kadison asked whether it is possible to construct projections withspecified diagonal so that all its entries are real and nonnegative. While the answer ispositive for rank one, in general it is negative for higher rank projections.
Example . Consider any sequence { d i } i =1 of numbers in (0 ,
1) such that d + d + d = 2.By Theorem 3.1 there exists a projection P on R with such diagonal. However, some entriesof P must be negative. Indeed, I − P is rank one projection. Hence, ( I − P ) x = (cid:104) x, v (cid:105) v forsome unit vector v = ( v , v , v ) ∈ R . That is, ( i, j ) entry of I − P equals v i v j . In particular,( v i ) = 1 − d i > i . This implies that for some i (cid:54) = j , the off-diagonal entry ( i, j )of I − P must be positive. Consequently, ( i, j ) entry of P is negative. . The algorithm and Carpenter’s Theorem part ii
In this section we introduce an algorithmic technique for finding a projection with pre-scribed diagonal. The main result of this section is Theorem 4.3. Given a non-summablesequence { d i } with all terms in [0 , / / , { d i } . Applying this result countablymany times allows us to deal with all possible diagonal sequences in part (ii) of Carpenter’sTheorem.The procedure of Theorem 4.3 is reminiscent to spectral tetris construction of tight framesintroduced by Casazza et al. in [10], and further investigated in [11]. In fact, the infinitematrix constructed in the proof of Theorem 4.3 consists of column vectors forming a Parsevalframe with squared norms prescribed by the sequence { d i } . However, our construction wasdiscovered independently with a totally different aim than that of [10]. Lemma 4.1.
Let σ, d , d ∈ [0 , . If max { d , d } ≤ σ and σ ≤ d + d , then there exists anumber a ∈ [0 , such that the matrix (4.1) (cid:20) a σ − ad − a d − σ + a (cid:21) has entries in [0 , and (4.2) a ( d − a ) = ( σ − a )( d − σ + a ) . Moreover, if d + d < σ , then a is unique and given by (4.3) a = σ ( σ − d )2 σ − d − d . Proof.
First, assume max { d , d } ≤ σ and σ ≤ d + d . If d = d = σ then any a ∈ [0 , σ ]will satisfy (4.2) and the matrix (4.1) will have entries in [0,1]. Thus, we may additionallyassume d + d < σ , and hence σ >
0. Since the quadratic terms in (4.2) cancel out, theequation is linear and the unique solution is given by (4.3). It remains to show that theentries of the matrix in (4.1) are in [0 , a ≥
0. Next, we calculate(4.4) σ − a = σ (cid:18) − σ − d σ − d − d (cid:19) = σ ( σ − d )2 σ − d − d , which implies that σ − a ≥
0. Since σ ≤ a, σ − a ∈ [0 , d + d ∈ [ σ, σ ) we have( d − a ) + ( d − σ + a ) = d + d − σ ∈ [0 , σ ) . If one of d − a or d − σ + a is negative, then the other must be positive. From (4.2) wesee that a = σ − a = 0. This contradicts the assumption that σ >
0. Thus, both d − a and d − σ + a are nonnegative. (cid:3) Lemma 4.2.
Let { d i } i ∈ N be a sequence such that d ∈ [0 , , d i ∈ [0 , ] for i ≥ and (cid:80) ∞ i =1 d i = ∞ . There is a bijection π : N → N such that for each n ∈ N we have (4.5) d π ( k n − ≥ d π ( k n ) where k n := min (cid:40) k ∈ N : k (cid:88) i =1 d π ( i ) ≥ n (cid:41) . roof. For n ∈ N define(4.6) m n := min (cid:40) k ∈ N : k (cid:88) i =1 d i ≥ n (cid:41) . Define a bijection π n : { m n − + 1 , . . . , m n } → { m n − + 1 , . . . , m n } such that { d π ( i ) } m n i = m n − +1 isin nonincreasing order with the convention that m = 0. Finally, define a bijection π : N → N by π ( i ) = π n ( i ) if m n − < i ≤ m n , n ∈ N . We claim that(4.7) m n − + 2 ≤ k n ≤ m n for all n ∈ N . Indeed, by the minimality of m n − we have for n ≥ m n − +1 (cid:88) i =1 d π ( i ) = m n − (cid:88) i =1 d i + d π ( m n − +1) < ( n − /
2) + 1 / n. The above holds also holds trivially for n = 1. Thus, k n > m n − + 1 for all n ∈ N . On theother hand, we have m n (cid:88) i =1 d π ( i ) = m n (cid:88) i =1 d i ≥ n. This yields k n ≤ m n and, thus, (4.7) is shown. By (4.7) we have m n − +1 ≤ k n − < k n ≤ m n .Since { d π ( i ) } m n i = m n − +1 is nonincreasing, this yields (4.5). (cid:3) Theorem 4.3.
Let { d i } i ∈ I be a sequence such that d i ∈ [0 , for some i ∈ I , d i ∈ [0 , ] forall i (cid:54) = i , and (cid:80) i ∈ I d i = ∞ . There exists an orthogonal projection P with diagonal { d i } i ∈ I .Proof. Since I is a countable set and (cid:80) i ∈ I d i = ∞ we may assume without loss of generalitythat I = N and i = 1. By Lemma 4.2 there is a bijection π : N → N such that (4.5) holds.For each n ∈ N set(4.8) σ n = n − k n − (cid:88) i =1 d π ( i ) . From the definition of k n we see that(4.9) σ n = n − k n (cid:88) i =1 d π ( i ) + d π ( k n − + d π ( k n ) ≤ d π ( k n − + d π ( k n ) . From the minimality of k n and (4.5) we see that σ n = n − k n − (cid:88) i =1 d π ( i ) + d π ( k n − ≥ d π ( k n − ≥ d π ( k n ) , which implies that(4.10) σ n ≥ max { d π ( k n − , d π ( k n ) } . By Lemma 4.1 for each n there exists a n ∈ [0 ,
1] such that the matrix (cid:20) a n σ n − a n d π ( k n − − a n d π ( k n ) − σ n + a n (cid:21) as non-negative entries and(4.11) a n ( d π ( k n − − a n ) = ( σ n − a n )( d π ( k n ) − σ n + a n ) . Let { e i } i ∈ N be an orthonormal basis for a Hilbert space H . Set v = k − (cid:88) i =1 d / π ( i ) e i + a / e k − − ( σ − a ) / e k , and for n ≥ v n = ( d π ( k n − − − a n − ) / e k n − − + ( d π ( k n − ) − σ n − + a n − ) / e k n − + k n − (cid:88) i = k n − +1 d / π ( i ) e i + a / n e k n − − ( σ n − a n ) / e k n . We can visualize { v n } n ∈ N as row vectors expanded in the orthonormal basis { e i } i ∈ I by thefollowing infinite matrix. v v v · · · = √ d · · · · √ a −√ σ − a √ d · − a √ d · − σ + a √ d · · · · √ a −√ σ − a √ d · − a √ d · − σ + a · · ·· · · In the above matrix empty spaces represents 0 and d · is an abbreviation for d π ( i ) in i thcolumn.We claim that { v n } n ∈ N is an orthonormal set in H . Indeed, by (4.8) we have for n ≥ (cid:107) v n (cid:107) = d π ( k n − − − a n − + d π ( k n − ) − σ n − + a n − + k n − (cid:88) i = k n − +1 d π ( i ) + a n + σ n − a n = k n − (cid:88) i = k n − − d π ( i ) + σ n − σ n − = k n − (cid:88) i = k n − − d π ( i ) + (cid:32) n − k n − (cid:88) i =1 d π ( i ) (cid:33) − (cid:32) n − − k n − − (cid:88) i =1 d π ( i ) (cid:33) = 1 . A similar calculation yields (cid:107) v (cid:107) = 1. This means that rows of our infinite matrix haveeach norm 1. Moreover, they are mutually orthogonal since any two vectors v n and v m havedisjoint supports unless they are consecutive: v n and v n +1 . However, in the latter case theorthogonality is a consequence of (4.11).Define the orthogonal projection P by P v = (cid:88) n ∈ N (cid:104) v, v n (cid:105) v n , v ∈ H . It is easy to check that the i th column of our infinite matrix has norm equal to (cid:112) d π ( i ) . Inother words, for each i ∈ N we have (cid:104) P e i , e i (cid:105) = || P e i || = (cid:88) n ∈ N |(cid:104) e i , v n (cid:105)| = d π ( i ) . This completes the proof of Theorem 4.3. (cid:3) e are now ready to prove Carpenter’s Theorem under assumption (ii). Theorem 4.4. If { d i } i ∈ I is a sequence in [0 , such that (4.12) a = (cid:88) d i < / d i = ∞ or b = (cid:88) d i ≥ / (1 − d i ) = ∞ , then there is a projection P with diagonal { d i } .Proof. Set I = { i : d i ≤ / } and I = { i : d i > / } . Our hypothesis (4.12) implies that(4.13) a (cid:48) = (cid:88) i ∈ I d i = ∞ or b = ∞ . Case 1.
Assume that a (cid:48) = ∞ . We can partition I into countably many sets { J n } n ∈ N suchthat each J n contains at most one element in I and (cid:88) i ∈ J n d i = ∞ for all n ∈ N . This is possible since I satisfies (4.13). By Theorem 4.3, for each n ∈ N there is a projection P n with diagonal { d i } i ∈ J n . Thus, the projection P = (cid:77) n ∈ N P n has the desired diagonal { d i } i ∈ I . This completes the proof of Case 1. Case 2.
Assume that b = ∞ . Note that b = (cid:88) − d i ≤ / (1 − d i ) . Thus, by Case 1 there is a projection P (cid:48) with diagonal { − d i } . Hence, P = I − P (cid:48) is aprojection with diagonal { d i } . (cid:3) A selector problem
Kadison’s Theorem 1.1 is closely connected with an open problem of characterizing allspectral functions of shift-invariant spaces. Shift-invariant (SI) spaces are closed subspaces of L ( R d ) that are invariant under all shifts, i.e., integer translations. That is, a closed subspace V ⊂ L ( R d ) is SI if T k ( V ) = V for all k ∈ Z d , where T k f ( x ) = f ( x − k ) is the translationoperator. The theory of shift-invariant spaces plays an important role in many areas, mostnotably in the theory of wavelets, spline systems, Gabor systems, and approximation theory[5, 6, 7, 19, 20]. The study of analogous spaces for L ( T , H ) with values in a separableHilbert space H in terms of the range function, often called doubly-invariant spaces, is quiteclassical and goes back to Helson [12].In the context of SI spaces a range function is any mapping J : T d → { closed subspaces of (cid:96) ( Z d ) } , where T d = R d / Z d is identified with its fundamental domain [ − / , / d . We say that J is measurable if the associated orthogonal projections P J ( ξ ) : (cid:96) ( Z d ) → J ( ξ ) are operatormeasurable, i.e., ξ (cid:55)→ P J ( ξ ) v is measurable for any v ∈ (cid:96) ( Z d ). We follow the convention hich identifies range functions if they are equal a.e. A fundamental result due to Helson[12, Theorem 8, p. 59] gives one-to-one correspondence between SI spaces V and measurablerange functions J , see also [7, Proposition 1.5]. Among several equivalent ways of introducingthe spectral function of a SI space the most relevant definition uses a range function. Definition 5.1.
The spectral function of a SI space V is a measurable mapping σ V : R d → [0 ,
1] given by(5.1) σ V ( ξ + k ) = || P J ( ξ ) e k || = (cid:104) P J ( ξ ) e k , e k (cid:105) for ξ ∈ T d , k ∈ Z d , where { e k } k ∈ Z d denotes the standard basis of (cid:96) ( Z d ) and T d = [ − / , / d . In other words, { σ V ( ξ + k ) } k ∈ Z d is a diagonal of a projection P J ( ξ ).Note that σ V ( ξ ) is well defined for a.e. ξ ∈ R d , since { k + T d : k ∈ Z d } is a partition of R d . As an immediate consequence of Theorem 1.1 we have the following result. Theorem 5.2.
Suppose that V ⊂ L ( R d ) is a SI space. Let σ = σ V : R d → [0 , be itsspectral function. For ξ ∈ T d define a ( ξ ) = (cid:88) k ∈ Z d , σ ( ξ + k ) < / σ ( ξ + k ) and b ( ξ ) = (cid:88) k ∈ Z d , σ ( ξ + k ) ≥ / (1 − σ ( ξ + k )) . Then, for a.e. ξ ∈ R d we either have (i) a ( ξ ) , b ( ξ ) < ∞ and a ( ξ ) − b ( ξ ) ∈ Z , or (ii) a ( ξ ) = ∞ or b ( ξ ) = ∞ . It is an open problem whether the converse to Theorem 5.2 holds.
Problem . Suppose that a measurable function σ : R d → [0 ,
1] satisfies either (i) or (ii) fora.e. ξ ∈ R d . Does there exists a SI space V ⊂ L ( R d ) such that its spectral function σ V = σ ?The sufficiency part of Theorem 1.1, i.e., Carpenter’s Theorem, suggests a positive answerto this problem. That is, for a.e. ξ it yields a projection P J ( ξ ) whose diagonal satisfies(5.1). However, it does not guarantee a priori that the corresponding range function J ismeasurable. This naturally leads to the following selector problem. Problem . Let X be a finite (or σ -finite) measure space and let I be a countable index set.Let σ : X × I → [0 ,
1] be a measurable function. For ξ ∈ X define a ( ξ ) = (cid:88) i ∈ I, σ ( ξ,i ) < / σ ( ξ, i ) and b ( ξ ) = (cid:88) i ∈ I, σ ( ξ,i ) ≥ / (1 − σ ( ξ, i )) . Suppose that for a.e. ξ ∈ X we either have(i) a ( ξ ) , b ( ξ ) < ∞ and a ( ξ ) − b ( ξ ) ∈ Z , or(ii) a ( ξ ) = ∞ or b ( ξ ) = ∞ .Does there exists a measurable range function J : X → { closed subspaces of (cid:96) ( I ) } such thatthe corresponding orthogonal projections P J ( ξ ) have diagonal { σ ( ξ, i ) } i ∈ I for a.e. ξ ∈ X ?In other words, Problem 2 asks whether it is possible to find a measurable selector ofprojections in Theorem 1.1. The constructive proof of Carpenter’s Theorem given in thispaper might be a first step toward resolving this problem. However, Problem 2 remainsopen. eferences [1] M. Argerami, Majorisation and Kadison’s Carpenter’s Theorem , preprint arXiv:1304.1232 .[2] M. Argerami, P. Massey,
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