# THE KADISON-SINGER PROBLEM IN MATHEMATICS AND ENGINEERING: A DETAILED ACCOUNT

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1 THE ADISON-SINGER PROBLEM IN MATHEMATICS AND ENGINEERING: A DETAILED ACCOUNT PETER G. CASAZZA, MATTHEW FICUS, JANET C. TREMAIN, ERIC WEBER Abstract. We will show that the famous, intractible 1959 adison-singer problem in C -algebras is equivalent to fundamental unsolved problems in a dozen areas of research in pure mathematics, applied mathematics and Engineering. This gives all these areas common ground on which to interact as well as explaining why each of these areas has volumes of literature on their respective problems without a satisfactory resolution. In each of these areas we will reduce the problem to the minimum which needs to be proved to solve their version of adison-singer. In some areas we will prove what we believe will be the strongest results ever available in the case that adison-singer fails. Finally, we will give some directions for constructing a counter-example to adison-singer. 1. Introduction The famous 1959 adison-singer Problem [61] has defied the best efforts of some of the most talented mathematicians of our time. adison-singer Problem (S). Does every pure state on the (abelian) von Neumann algebra D of bounded diagonal operators on l 2 have a unique extension to a (pure) state on B(l 2 ), the von Neumann algebra of all bounded linear operators on the Hilbert space l 2? A state of a von Neumann algebra R is a linear functional f on R for which f(i) = 1 and f(t ) 0 whenever T 0 (whenever T is a positive operator). The set of states of R is a convex subset of the dual space of R which is compact in the ω -topology. By the rein-milman theorem, this convex set is the closed convex hull of its extreme points. The extremal elements in the space of states are called the pure states (of R). This problem arose from the very productive collaboration of adison and Singer in the 1950 s which culminated in their seminal work on triangular operator algebras. During this collaboration, they often discussed the fundamental work of Dirac [38] on Quantum Mechanics. In particular, they kept returning to one part of 1991 Mathematics Subject Classification. Primary: 42A05,42A10,42A16,43A50,46B03,46B07,46L05, 46L30. The first and second authors were supported by NSF DMS , the last author was supported by NSF DMS

2 2 P.G. CASAZZA, M. FICUS, J.C. TREMAIN, E. WEBER Dirac s work because it seemed to be problematic. Dirac wanted to find a representation (an orthonormal basis) for a compatible family of observables (a commutative family of self-adjoint operators). On pages of [38] Dirac states: To introduce a representation in practice (i) We look for observables which we would like to have diagonal either because we are interested in their probabilities or for reasons of mathematical simplicity; (ii) We must see that they all commute a necessary condition since diagonal matrices always commute; (iii) We then see that they form a complete commuting set, and if not we add some more commuting observables to make them into a complete commuting set; (iv) We set up an orthogonal representation with this commuting set diagonal. The representation is then completely determined... by the observables that are diagonal... The emphasis above was added. Dirac then talks about finding a basis that diagonalizes a self-adjoint operator, which is troublesome since there are perfectly respectable self-adjoint operators which do not have a single eigenvector. Still, there is a spectral resolution of such operators. Dirac addresses this problem on pages of [38]: We have not yet considered the lengths of the basic vectors. With an orthonormal representation, the natural thing to do is to normalize the basic vectors, rather than leave their lengths arbitrary, and so introduce a further stage of simplification into the representation. However, it is possible to normalize them only if the parameters are continuous variables that can take on all values in a range, the basic vectors are eigenvectors of some observable belonging to eigenvalues in a range and are of infinite length... In the case of D, the representation is {e i } i I, the orthonormal basis of l 2. But what happens if our observables have ranges (intervals) in their spectra? This led Dirac to introduce his famous δ-function vectors of infinite length. From a mathematical point of view, this is problematic. What we need is to replace the vectors e i by some mathematical object that is essentially the same as the vector, when there is one, but gives us something precise and usable when there is only a δ-function. This leads to the pure states of

4 4 P.G. CASAZZA, M. FICUS, J.C. TREMAIN, E. WEBER fundamental principle for showing that very general problems are equivalent to S. In Section 3 we introduce our universal language of frame theory and introduce our second fundamental principle for reducing problems to S. In Section 4, we will show that S is equivalent to a fundamental result concerning inner products. This formulation of the problem has the advantage that it can be understood by a student one week into their first course in Hilbert spaces. In Section 5 we show that S is equivalent to the Bourgain-Tzafriri Conjecture (and in fact, a significantly weaker form of the conjecture is equivalent to S). This also shows that the Feichtinger Conjecture is equivalent to S. In Section 6, we show that a fundamental problem in harmonic analysis is equivalent to S. We also classify the uniform paving conjecture and the uniform Feichtinger Conjecture. As a consequence we will discover a surprising new identity in the area. In Section 7, we show that the Feichtinger Conjecture for frames of translates is equivalent to one of the fundamental unsolved problems in harmonic analysis. In Section 8, we look at how S arises naturally in various problems in signal-processing, internet coding, coding theory and more. Notation for statements of problems: Problem A (or Conjecture A) implies Problem B (or Conjecture B) means that a positive solution to the former implies a positive solution to the latter. They are equivalent if they imply each other. Notation for Hilbert spaces: Throughout, l 2 (I) will denote a finite or infinite dimensional complex Hilbert space with a fixed orthonormal basis {e i } i I. If I is infinite we let l 2 = l 2 (I), and if I = n write l 2 (I) = l n 2 with fixed orthonormal basis {e i } n i=1. For any Hilbert space H we let B(H) denote the family of bounded linear operators on H. An n-dimensional subspace of l 2 (I) will be denoted H n. For an operator T on any one of our Hilbert spaces, its matrix representation with respect to our fixed orthonormal basis is the collection ( T e i, e j ) i,j I. If J I, the diagonal projection Q J is the matrix whose entries are all zero except for the (i, i) entries for i J which are all one. For a matrix A = (a ij ) i,j I let δ(a) = max i I a ii. A universal language: We are going to show that the adison-singer problem is equivalent to fundamental unsolved problems in a dozen different areas of research in both mathematics and engineering. But each of these areas is overrun with technical jargon which makes it difficult or even impossible for those outside the field to understand results inside the field. What we need is a universal language for interactions between a broad spectrum of research. For our universal language, we have chosen the language of Hilbert space frame theory (See Section 3) because it is simply stated and easily understood while being fundamental enough to quickly pass quite technical results between very

6 6 P.G. CASAZZA, M. FICUS, J.C. TREMAIN, E. WEBER (2) If A j = {m 1 < m 2 < }, then for all natural numbers l we have {m 1, m 2,, m l } A km l j. Proof: For each natural number n, 1 is in one of the sets {A n j } r j=1. Hence, there are natural numbers n 1 1 < n 1 2 < and a 1 j r so that 1 A n1 i j for all i N. Now, for every natural number n 1 i, 2 is in one of the sets {A n1 i j }r j=1. Hence, there is a subsequence {n 2 i } of {n 1 i } and a 1 j r so that 2 A n2 i j, for all i N. Continuing by induction, for all l N we get a subsequence {n l+1 i } of {n l i} and a 1 j r so that l + 1 A nl+1 j j, for all i N. Letting k i = n i i for all i N gives the conclusion of the proposition. Theorem 2.3. The Paving Conjecture is equivalent to the Paving Conjecture for operators on l 2. Proof: Assume PC holds for operators on l n 2. Let T = (a ij ) i,j=1 be a bounded linear operator on l 2. Fix ɛ > 0. By our assumption, for every natural number n there is a partition {A n j } r j=1 of {1, 2,, n} so that if T n = (a ij ) n i,j=1 then for all j = 1, 2,, r Q A n j T n Q A n j ɛ 2 T n ɛ T. 2 Let {A j } r j=1 be the partition of N given in Proposition 2.2. Fix 1 j r, let A j = {m 1 < m 2 < }, and for all l N let Q l = Q Il where I l = {m 1, m 2,, m l }. Fix f l 2 (N). For all large l N we have: Q Aj T Q Aj (f) 2 Q l Q Aj T Q Aj Q l (f) = 2 Q l Q km l A j 2 Q km l A j T kml Q km l A j Q l (f) T kml Q km l A j Q l (f) 2 ɛ T f = ɛ T f. 2 Hence, Q Aj T Q Aj ɛ T. Conversely, assume PC holds for operators on l 2. We assume that PC fails for operators on l n 2 and get a contradiction. If (1) fails, a little thought will yield that there must be an ɛ > 0, a partition {I n } n=1 of N into finite subsets, operators T n : l 2 (I n ) l 2 (I n ) with T n = 1 and for every partition {A n j } n j=1 of I n there exists a 1 j n so that Let T = Q A n j T n Q A n j ɛ. ( ) T n : l 2 (I n ) n=1 n=1 l 2 ( ) l 2 (I n ) n=1 l 2.

7 THE ADISON-SINGER PROBLEM IN MATHEMATICS AND ENGINEERING 7 Then, T = sup n T n = 1. By (2), there is a partition {A j } r j=1 of N so that for all j = 1, 2,, r Q Aj T Q Aj ɛ. For every n N and every j = 1, 2,, r let A n j = A j I n. Then, {A n j } r j=1 is a partition of I n. Hence, for every j = 1, 2,, r we have Q A n j T n Q A n j = Q A n j T Q A n j Q Aj T Q Aj ɛ. If n r, this contradicts our assumption about T n. It is known [12] that the class of operators satisfying PC (the pavable operators) is a closed subspace of B(l 2 ). The only large classes of operators which have been shown to be pavable are diagonally dominant matrices [10, 11, 12, 53], matrices with all entries real and positive [56] and Toeplitz operators over Riemann integrable functions (See also [57] and Section 6). Also, in [13] there is an analysis of the paving problem for certain Schatten C p - norms. We strongly recommend that everyone read the argument of Berman, Halpern, aftal and Weiss [12] showing that matrices with positive entries satisfy PC. This argument is a fundamental principle concerning decompositions of matrices which has applications across the board here, you will see it used in the proof of Theorem 8.16, and it was vaguely used in producing a generalization of the Rado-Horn Theorem [29] (See Theorem 8.3). We next note that in order to verify PC, it suffices to show that PC holds for any one of your favorite classes of operators. Theorem 2.4. The Paving Conjecture has a positive solution if any one of the following classes satisfies the Paving Conjecture: (1) Unitary operators. (2) Orthogonal projections. (3) Positive operators. (4) Self-adjoint operators. (5) Gram matrices ( f i, f j ) i,j I where T : l 2 (I) l 2 (I) is a bounded linear operator, and T e i = f i, T e i = 1 for all i I. (6) Invertible operators (or invertible operators with zero diagonal). Proof: (1): This is immediate from the fact that every bounded operator is a multiple of a sum of three unitary operators [23]. (2): This follows from the Spectral Theorem (or see Fundamental Principle II: Theorem 3.5). (3), (4): Since (3) or (4) immediately implies (2). (5): We will show that (5) implies a positive solution to the Bourgain- Tzafriri Conjecture (See Section 5) and hence to PC by Theorem 5.1. Given T : l 2 (I) l 2 (I) with T e i = 1 for all i I, let G = ( T e i, T e j ) i,j I. By

8 8 P.G. CASAZZA, M. FICUS, J.C. TREMAIN, E. WEBER (5), there is a partition {A j } r j=1 of I which paves the Gram operator. Hence, for all j = 1, 2,, r and all f = i A j a i e i we have i A j a i T e i 2 = i A j a i T e i, k A j a k T e k = i A j a i 2 T e i 2 + i k A j a i a k T e i, T e j = i A j a i 2 + Q Aj (G D(G))Q Aj f, f i A j a i 2 Q Aj (G D(G))Q Aj Q Aj f 2 i A j a i 2 ɛ i A j a i 2 = (1 ɛ) i A j a i 2. Hence, the Bourgain-Tzafriri Conjecture holds (See section 5). Now we need to jump ahead to Theorem 5.3 to see that the proof of BT implies S is done from the definition and does not need any theorems developed between here and there. (6): Given an operator T, ( T +1)I +T is invertible and if it is pavable then so is T. For the second statement, given an operator T, let S = T +( T 2 +2)U where U = (b ij ) i,j I is the unitary matrix given by the bilateral shift on N (the wrap-around shift on l n 2 if I = n). Then, S D(S) is invertible and has zero diagonal. By (6), for 0 < ɛ < 1 there is a partition {A j } r j=1 of I so that for all j = 1, 2,, r we have Q Aj (S D(S))Q Aj ɛ. Note that for any i I, if i A j then i + 1 / A j, since otherwise: (Q Aj (S D(S))Q Aj e i+1 )(i) = T e i, T e i+1 + ( T 2 + 2) 1. Hence, Q Aj (S D(S))Q Aj 1, which contradicts our paving of S D(S). It follows that Q Aj (S D(S))Q Aj = Q Aj (T D(T ))Q Aj, for all j = 1, 2,, r. So, our paving of S also paves T. Akemann and Anderson [1] showed that the following conjecture implies S. Conjecture 2.5. There exists 0 < ɛ, δ < 1 with the following property: for any orthogonal projection P on l n 2 with δ(p ) δ, there is a diagonal projection Q such that QP Q 1 ɛ and (I Q)P (I Q) 1 ɛ.

9 THE ADISON-SINGER PROBLEM IN MATHEMATICS AND ENGINEERING 9 It is important that ɛ, δ are independent of n in Conjecture 2.5. It is unknown if S implies Conjecture 2.5. Weaver [82] showed that a conjectured strengthening of Conjecture 2.5 fails. Recently, Weaver [81] provided important insight into S by showing that a slight weakening of Conjecture 2.5 will produce a conjecture equivalent to S. This is our first Fundamental Principle. Conjecture 2.6 (Fundamental Principle I: Weaver). There exist universal constants 0 < δ, ɛ < 1 and r N so that for all n and all orthogonal projections P on l n 2 with δ(p ) δ, there is a paving {A j } r j=1 of {1, 2,, n} so that Q Aj P Q Aj 1 ɛ, for all j = 1, 2,, r. This needs some explanation since there is nothing in [81] which looks anything like Conjecture 2.6. In [81], Weaver introduces what he calls Conjecture S r (See Section 3). A careful examination of the proof of Theorem 1 of [81] reveals that Weaver shows Conjecture S r implies Conjecture 2.6 which in turn implies S which (after the theorem is proved) is equivalent to S r. We will see in Section 3 (Conjecture 3.10, Theorem 3.11) that we may assume P e i = P e j for all i, j = 1, 2,, n in Conjecture 2.6 with a small restriction on the ɛ > Frame Theory: The Universal Language A family of vectors {f i } i I in a Hilbert space H is a Riesz basic sequence if there are constants A, B > 0 so that for all scalars {a i } i I we have: A i I a i 2 i I a i f i 2 B i I a i 2. We call A, B the lower and upper Riesz basis bounds for {f i } i I. If the Riesz basic sequence {f i } i I spans H we call it a Riesz basis for H. So {f i } i I is a Riesz basis for H means there is an orthonormal basis {e i } i I so that the operator T (e i ) = f i is invertible. In particular, each Riesz basis is bounded. That is, 0 < inf i I f i sup i I f i <. Hilbert space frames were introduced by Duffin and Schaeffer [42] to address some very deep problems in nonharmonic Fourier series (see [83]). A family {f i } i I of elements of a (finite or infinite dimensional) Hilbert space H is called a frame for H if there are constants 0 < A B < (called the lower and upper frame bounds, respectively) so that for all f H (3.1) A f 2 i I f, f i 2 B f 2. If we only have the right hand inequality in Equation 3.1 we call {f i } i I a Bessel sequence with Bessel bound B. If A = B, we call this an A-tight frame and if A = B = 1, it is called a Parseval frame. If all the frame

10 10 P.G. CASAZZA, M. FICUS, J.C. TREMAIN, E. WEBER elements have the same norm, this is an equal norm frame and if the frame elements are of unit norm, it is a unit norm frame. It is immediate that f i 2 B. If also inf f i > 0, {f i } i I is a bounded frame. The numbers { f, f i } i I are the frame coefficients of the vector f H. If {f i } i I is a Bessel sequence, the synthesis operator for {f i } i I is the bounded linear operator T : l 2 (I) H given by T (e i ) = f i for all i I. The analysis operator for {f i } i I is T and satisfies: T (f) = i I f, f i e i. In particular, T f 2 = i I f, f i 2, for all f H, and hence the smallest Bessel bound for {f i } i I equals T 2. Comparing this to Equation 3.1 we have: Theorem 3.1. Let H be a Hilbert space and T : l 2 (I) H, T e i = f i be a bounded linear operator. The following are equivalent: (1) {f i } i I is a frame for H. (2) The operator T is bounded, linear, and onto. (3) The operator T is an (possibly into) isomorphism. Moreover, if {f i } i I is a Riesz basis, then the Riesz basis bounds are A, B where A, B are the frame bounds for {f i } i I. It follows that a Bessel sequence is a Riesz basic sequence if and only if T is onto. The frame operator for the frame is the positive, self-adjoint invertible operator S = T T : H H. That is, ( ) Sf = T T f = T f, f i e i = f, f i T e i = f, f i f i. i I i I i I In particular, Sf, f = f, f i 2. i I It follows that {f i } i I is a frame with frame bounds A, B if and only if A I S B I. So {f i } i I is a Parseval frame if and only if S = I. Reconstruction of vectors in H is achieved via the formula: f = SS 1 f = i I S 1 f, f i f i = i I f, S 1 f i f i = f, f i S 1 f i i I = f, S 1/2 f i S 1/2 f i. i I

11 THE ADISON-SINGER PROBLEM IN MATHEMATICS AND ENGINEERING 11 It follows that {S 1/2 f i } i I is a Parseval frame equivalent to {f i } i I. Two sequences {f i } i I and {g i } i I in a Hilbert space are equivalent if there is an invertible operator T between their spans with T f i = g i for all i I. We now show that there is a simple way to tell when two frame sequences are equivalent. Proposition 3.2. Let {f i } i I, {g i } i I be frames for a Hilbert space H with analysis operators T 1 and T 2, respectively. The following are equivalent: (1) The frames {f i } i I and {g i } i I are equivalent. (2) ker T 1 = ker T 2. Proof: (1) (2): If Lf i = g i is an isomorphism, then Lf i = LT 1 e i = g i = T 2 e i quickly implies our statement about kernels. (2) (1): Since T i (ker Ti ) is an isomorphism for i = 1, 2, if the kernels are equal, then ( ) 1 T 2 T1 (ker T2 ) fi = g i is an isomorphism. In the finite dimensional case, if {g j } n j=1 is an orthonormal basis of l n 2 consisting of eigenvectors for S with respective eigenvalues {λ j } n j=1, then for every 1 j n, i I f i, g j 2 = λ j. In particular, i I f i 2 = trace S (= n if {f i } i I is a Parseval frame). An important result is Theorem 3.3. If {f i } i I is a frame for H with frame bounds A, B and P is any orthogonal projection on H, then {P f i } i I is a frame for P H with frame bounds A, B. Proof: For any f P H, f, P f i 2 = P f, f i 2 = f, f i 2. i I i I i I A fundamental result in frame theory was proved independently by Naimark and Han/Larson [35, 54]. For completeness we include its simple proof. Theorem 3.4. A family {f i } i I is a Parseval frame for a Hilbert space H if and only if there is a containing Hilbert space H l 2 (I) with an orthonormal basis {e i } i I so that the orthogonal projection P of l 2 (I) onto H satisfies P (e i ) = f i for all i I. Proof: The only if part is Theorem 3.3. For the if part, if {f i } i I is a Parseval frame, then the synthesis operator T : l 2 (I) H is a partial isometry. So T is an isometry and we can associate H with T H. Now, for all i I and all g = T f T H we have T f, P e i = T f, e i = f, T e i = f, f i = T f, T f i.

12 12 P.G. CASAZZA, M. FICUS, J.C. TREMAIN, E. WEBER It follows that P e i = T f i for all i I. Now we can establish our Fundamental Principle II which basically states that any bounded operator on a finite dimensional Hilbert space is really just a multiple of a piece of a projection from a larger space. Theorem 3.5 (Fundamental Principle II). Let H n be an n-dimensional Hilbert space with orthonormal basis {g i } n i=1. If T : H n H n is any bounded linear operator with T = 1, then there is a containing Hilbert space H n l M 2 (M=2n-1) with an orthonormal basis {e i } M i=1 so that the orthogonal projection P from l M 2 onto H n satisfies: P e i = T g i, for all i = 1, 2,, n. Proof: Let S be the frame operator for the Bessel sequence {f i } n i=1 = {T g i } n i=1 having eigenvectors {x i } n i=1 with respective eigenvalues {λ i } n i=1 where 1 = λ 1 λ 2 λ n. For i = 2, 3,, n let h i = (1 λ i )x i. Then, {f i } n i=1 {h i } n i=2 is a Parseval frame for H since for every f H we have n n n f, f i 2 + f, h i 2 = Sf, f + (1 λ i ) f, x i 2 i=1 i=2 = = i=2 n λ i f, x i 2 + i=1 n (1 λ i ) f, x i 2 i=2 n f, x i 2 = f 2. i=1 Now, by Theorem 3.4, there is a containing Hilbert space l 2n 1 2 with an orthonormal basis {e i } 2n 1 i=1 so that the orthogonal projection P satisfies: P e i = T g i for i = 1, 2,, n and P e i = h i for i = n + 1,, 2n 1. For an introduction to frame theory we refer the reader to Christensen [35]. Weaver [81] established an important relationship between frames and S by showing that the following conjecture is equivalent to S. Conjecture 3.6. There are universal constants B 4 and ɛ > B and an r N so that the following holds: Whenever {f i } M i=1 is a unit norm B-tight frame for l n 2, there exists a partition {A j } r j=1 of {1, 2,, M} so that for all j = 1, 2,, r and all f l n 2 we have (3.2) f, f i 2 (B ɛ) f 2. i A j In his work on time-frequency analysis, Feichtinger [32] noted that all of the Gabor frames he was using (see Section 7) had the property that they could be divided into a finite number of subsets which were Riesz basic sequences. This led to the conjecture:

13 THE ADISON-SINGER PROBLEM IN MATHEMATICS AND ENGINEERING 13 Feichtinger Conjecture (FC). Every bounded frame (or equivalently, every unit norm frame) is a finite union of Riesz basic sequences. There is a significant body of work on this conjecture [10, 11, 32, 53]. Yet, it remains open even for Gabor frames. In [25] it was shown that FC is equivalent to the weak BT, and hence is implied by S (See Section 5). In [31] it was shown that FC is equivalent to S (See Theorem 5.3). In fact, we now know that S is equivalent to the weak Feichtinger Conjecture: Every unit norm Bessel sequence is a finite union of Riesz basic sequences (See Section 5). In [30] it was shown that FC is equivalent to the following conjecture. Conjecture 3.7. Every bounded Bessel sequence is a finite union of frame sequences. Let us mention two more useful equivalent formulations of S due to Weaver [81]. Conjecture 3.8 (S r ). There is a natural number r and universal constants B and ɛ > 0 so that the following holds. Let {f i } M i=1 be elements of l n 2 with f i 1 for i = 1, 2,, M and suppose for every f l n 2, M (3.3) f, f i 2 B f 2. i=1 Then, there is a partition {A j } r j=1 of {1, 2,, n} so that for all f l n 2 and all j = 1, 2,, r, f, f i 2 (B ɛ) f 2. i A j Weaver [81] also shows that Conjecture S r is equivalent to PC if we assume equality in Equation 3.4 for all f l n 2. Weaver further shows that Conjecture 3.8 is equivalent to S even if we strengthen its assumptions so as to require that the vectors {f i } M i=1 are of equal norm and that equality holds in 3.4, but at great cost to our ɛ > 0. Conjecture 3.9 (S r ). There exists universal constants B 4 and ɛ > B so that the following holds. Let {f i } M i=1 be elements of l n 2 with f i 1 for i = 1, 2,, M and suppose for every f l n 2, M (3.4) f, f i 2 = B f 2. i=1 Then, there is a partition {A j } r j=1 of {1, 2,, n} so that for all f l n 2 and all j = 1, 2,, r, f, f i 2 (B ɛ) f 2. i A j

14 14 P.G. CASAZZA, M. FICUS, J.C. TREMAIN, E. WEBER We now strengthen the assumptions in Fundamental Principle I, Conjecture 2.6. Conjecture There exist universal constants 0 < δ, δ ɛ < 1 and r N so that for all n and all orthogonal projections P on l n 2 with δ(p ) δ and P e i = P e j for all i, j = 1, 2,, n, there is a paving {A j } r j=1 of {1, 2,, n} so that Q Aj P Q Aj 1 ɛ, for all j = 1, 2,, r. Using Conjecture 3.9 we can see that S is equivalent to Conjecture Theorem S is equivalent to Conjecture Proof: It is clear that Conjecture 2.6 (which is equivalent to S) implies Conjecture So we assume that Conjecture 3.10 holds and we will show that Conjecture 3.9 holds. Let {f i } M i=1 be elements of H n with f i = 1 for i = 1, 2,, M and suppose for every f H n, M (3.5) f, f i 2 = B f 2, i=1 where 1 δ. It follows from Equation 3.5 that { 1 B B f i } M i=1 is an equal norm Parseval frame and so there is a larger Hilbert space l M 2 and a projection P : l M 2 H n so that P e i = f i for all i = 1, 2,, M. Now, P e i 2 = P e i, e i = 1 δ for all i = 1, 2,, M. So by Conjecture 3.10, there B is a paving {A j } r j=1 of {1, 2,, M} so that Q Aj P Q Aj 1 ɛ, for all j = 1, 2,, r. Now, for all 1 j r and all f l n 2 we have: Q Aj P f 2 = M Q Aj P f, e i 2 = i=1 = 1 f, f i 2 B i A j Q Aj P 2 f 2 It follows that for all f H n we have M f, P Q Aj e i 2 i=1 = Q Aj P Q Aj f 2 (1 ɛ) f 2. i A j f, f i 2 (B ɛb) f 2. Since ɛb > B, we have verified Conjecture 3.9. We give one final formulation of S in Hilbert space frame theory. Theorem The following are equivalent: (1) The Paving Conjecture.

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