The Kadison-Singer and Paulsen Problems in Finite Frame Theory

20 Peter G. CasazzaThis problem appears to be quite innocent, but it is surprisingly difficult. Itis immediate that the Feichtinger Conjecture implies the Sundberg Problem.Theorem 12. A positive solution to the Feichtinger Conjecture implies apositive solution to the Sundberg Problem.Proof. If {ϕ i } ∞ i=1 is a unit norm Bessel sequence then by FC, we can partitionthe natural numbers into a finite number of sets {A j } r j=1 so that {ϕ i} i∈Aj isa Riesz sequence for all j = 1, 2, . . . , r. For each j = 1, 2, . . . , r choose i j ∈ A j .Then neither ϕ ij nor {ϕ i } i∈Aj\{i j} can span the space.1.4 The Paulsen ProblemThe Paulsen Problem has been intractable for over a dozen years despitereceiving quite a bit of attention. In this section we look at the current stateof the art on this problem. First we need two definitions.Definition 4. A frame {ϕ i } M i=1 for HN with frame operator S is said to beɛ-nearly equal norm if(1 − ɛ) N M ≤ ‖ϕ i‖ 2 ≤ (1 + ɛ) N , for all i = 1, 2, . . . , M,Mand it is ɛ-nearly Parseval ifWe also need:(1 − ɛ)Id ≤ S ≤ (1 + ɛ)Id.Definition 5. Given frames Φ = {ϕ i } M i=1 and Ψ = {ψ i} M i=1 for HN , we definethe distance between them byd(Φ, Ψ) =M∑‖ϕ i − ψ i ‖ 2 .i=1The above is not exactly a distance function since we have not taken thesquare root of the right-hand-side of the equality. But since this formulationis standard, we will use it. We can now state the Paulsen Problem.Problem 2 (Paulsen Problem). How close is an ɛ-nearly equal norm andɛ-nearly Parseval frame to an equal norm Parseval frame?The importance of the Paulsen Problem is that we have algorithms forconstructing frames which are equal norm and nearly Parseval. The questionis, if we work with these frames, are we sure that we are working with a framewhich is close to some equal norm Parseval frame? We are looking for the