The Kadison-Singer and Paulsen Problems in Finite Frame Theory

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8 Peter G. CasazzaSimilarly, ‖ ∑ i∈A ja i T e i ‖ 2 ≤ (1 + ɛ) ∑ i∈A j|a i | 2 .(2) ⇒ (3): This is obvious.(3) ⇒ (1): Let T ∈ B(l N 2 ) with T e i = ϕ i and ‖ϕ i ‖ = 1 for all 1 ≤ i ≤ N.By Theorem 1 part 6, it suffices to show that the Gram operator G of {ϕ i } N i=1is pavable. Fix 0 < δ < 1 and let ɛ > 0. Let ψ i = √ 1 − δ 2 ϕ i ⊕ δe i ∈ l N 2 ⊕ l N 2 .Then ‖ψ i ‖ = 1 for all 1 ≤ i ≤ N and for all scalars {a i } N i=1N∑N∑N∑δ |a i | 2 ≤ ‖ a i ψ i ‖ 2 = (1 − δ 2 )‖ a i T e i ‖ 2 + δ 2i=1i=1i=1≤ [ (1 − δ 2 )‖T ‖ 2 + δ 2] ∑N |a i | 2 .i=1N ∑i=1|a i | 2So {ψ i } N i=1 is a unit norm Riesz basic sequence and 〈ψ i, ψ k 〉 = (1−δ 2 )〈ϕ i , ϕ k 〉for all 1 ≤ i ≠ k ≤ N. By the R ɛ -Conjecture, there is a partition {A j } r j=1 of{1, 2, . . . , N} so that for all j = 1, 2, . . . , r and all x = ∑ i∈A ja i e i ,(1 − ɛ) ∑i∈A j|a i | 2 ≤ ‖ ∑a i ψ i ‖ 2 = 〈 ∑a i ψ i , ∑a k ψ k 〉i∈A j i∈A j k∈A j= ∑|a i | 2 ‖ψ i ‖ 2 +∑a i a k 〈ψ i , ψ k 〉i∈A j i≠k∈A j= ∑∑|a i | 2 + (1 − δ 2 ) a i a k 〈ϕ i , ϕ k 〉i∈A j i≠k∈A j= ∑|a i | 2 + (1 − δ 2 )〈Q Aj (G − D(G))Q Aj x, x〉i∈A j≤ (1 + ɛ) ∑|a i | 2 .i∈A jSubtracting ∑ i∈A j|a i | 2 through the inequality yields,That is,−ɛ ∑i∈A j|a i | 2 ≤ (1 − δ 2 )〈Q Aj (G − D(G))Q Aj x, x〉 ≤ ɛ ∑(1 − δ 2 )|〈Q Aj (G − D(G))Q Aj x, x〉| ≤ ɛ‖x‖ 2 .i∈A j|a i | 2 .Since Q Aj (G−D(G))Q Aj is a self-adjoint operator, we have (1−δ 2 )‖Q Aj (G−D(G))Q Aj ‖ ≤ ɛ, i.e., (1 − δ 2 )G (and hence G) is pavable.Remark 1. The proof of (3) ⇒ (1) of Theorem 2 illustrates a standard methodfor turning conjectures about unit norm Riesz basic sequences {ψ i } i∈I intoconjectures about unit norm familes {ϕ i } i∈I with T ∈ B(l 2 (I)) and T e i = ϕ i .Namely, given {ϕ i } i∈I and 0 < δ < 1 let ψ i = √ 1 − δ 2 f i ⊕δe i ∈ l 2 (I)⊕l 2 (I).
1 The Kadison-Singer and Paulsen Problems in Finite Frame Theory 9Then {ψ i } i∈I is a unit norm Riesz basic sequence and for δ small enough, ψ iis close enough to ϕ i to pass inequalities from {ψ i } i∈I to {ϕ i } i∈I .The R ɛ -Conjecture is different from all other conjectures in this chapterin that it does not hold for equivalent norms on the Hilbert space in general.For example, if we renorm l 2 by: |{a i }| = ‖{a i }‖ l2 + sup i |a i | then the R ɛ -Conjecture fails for this equivalent norm. To see this, we proceed by way ofcontradiction and assume there is an 0 < ɛ < 1 and an r = r(ɛ, 2) satisfyingthe R ɛ -Conjecture. Let {e i } 2Ni=1 be the unit vectors for l2N2 and let x i =e 2i+e √2+1 2i+1for 1 ≤ i ≤ N. This is now a unit norm Riesz basic sequence withupper Riesz bound 2. Assume we partition {1, 2, . . . , 2N} into sets {A j } r j=1 .Then for some 1 ≤ k ≤ r we have |A k | ≥ N r . Let A ⊂ A k with |A| = N randa i = √ 1Nfor i ∈ A. Then| ∑ ( )1 √2 ra i x i | = √ + √i∈A 2 + 1 NSince the norm above is bounded away from one for large N, we cannot satisfythe requirements of the R ɛ -Conjecture. It follows that a positive solution toKS would imply a fundamental new result concerning “inner products”, notjust norms.Another important equivalent form of PC comes from [26]. This is, onits face, a significant weakening of the R ɛ -Conjecture while it still remainsequivalent to PC.Conjecture 4. There exists a constant A > 0 and a natural number r so thatfor all natural numbers N and all T : l N 2 → l N 2 with ‖T e i ‖ = 1 for alli = 1, 2, . . . , N and ‖T ‖ ≤ 2, there is a partition {A j } r j=1 of {1, 2, . . . , N} sothat for all j = 1, 2, . . . , r and all scalars {a i } i∈Aj we have‖ ∑i∈A ja i T e i ‖ 2 ≥ A ∑Theorem 3. Conjecture 4 is equivalent to PC.i∈A j|a i | 2 .Proof. Since PC is equivalent to the R ɛ -Conjecture, which in turn impliesConjecture 4, we just need to show that Conjecture 4 implies Conjecture1. So choose r, A satisfying Conjecture 4. Fix 0 < δ ≤ 3 4and let P be anorthogonal projection on l N 2 with δ(P ) ≤ δ Now, 〈P e i , e i 〉 = ‖P e i ‖ 2 ≤ δimplies ‖(I − P )e i ‖ 2 ≥ 1 − δ ≥ 1 4 . Define T : lN 2 → l N 2 by T e i =For any scalars {a i } N i=1 we have(I−P )ei‖(I−P )e i‖ .
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