The Kadison-Singer and Paulsen Problems in Finite Frame Theory

30 Peter G. Casazza(1 − ɛ) N M ≤ ‖ϕ i‖ 2 ≤ (1 + ɛ) N M .Let T be the analysis operator of {ϕ i } M i=1 and let P be the projection of H Monto range T . So, T ϕ i = P e i for all i = 1, 2, . . . , M. By our assumption thatthe Projection Problem holds, there is a projection Q on H M with constantdiagonal so thatM∑‖P e i − Qe i ‖ 2 ≤ f(ɛ, N, M).i=1By Theorem 16, there is a a Parseval frame {ψ i } M i=1 for HN with analysisoperator T 1 so that T 1 ψ i = Qe i andM∑‖ϕ i − ψ i ‖ 2 ≤ 2f(ɛ, N, M).i=1Since T 1 is a isometry and {T 1 ψ i } M i=1 is equal norm, it follows that {ψ i} M i=1is an equal norm Parseval frame satisfying the Paulsen problem.Conversely, assume the Parseval Paulsen problem has a positive solutionwith function g(ɛ, N, M). Let P be an orthogonal projection on H M satisfying(1 − ɛ) N M ≤ ‖P e i‖ 2 ≤ (1 + ɛ) N M .Then {P e i } M i=1 is a ɛ-nearly equal norm Parseval frame for HN and by thePaulsen problem, there is an equal norm Parseval frame {ψ i } M i=1 so thatM∑‖ϕ i − ψ i ‖ 2 < g(ɛ, N, M).i=1Let T 1 be the analysis operator of {ψ i } M i=1 . Letting Q be the projection ontothe range of T 1 , we have that Qe i = T 1 ψ i , for all i = 1, 2, . . . , M. By Theorem15, we have thatM∑N∑‖P e i − T 1 ψ i ‖ 2 = ‖P e i − Qe i ‖ 2 ≤ 4g(ɛ, N, M).i=1i=1Since T 1 is a isometry and {ψ i } M i=1constant diagonal projection.is equal norm, it follows that Q is aIn [14] there are several generalizations of the Paulsen and ProjectionProblems.