chapter 1 the kadison-singer and paulsen problems in finite frame … · 2017-03-27 · the...

Chapter 1

The Kadison-Singer and PaulsenProblems in Finite Frame Theory

Peter G. Casazza

Abstract We now know that some of the basic open problems in frame theoryare equivalent to fundamental open problems in a dozen areas of research inboth pure and applied mathematics, engineering and more. This includesthe 1959 Kadison-Singer Problem in C∗-algebras, the Paving Conjecture inoperator theory, the Bourgain-Tzafriri Conjecture in Banach space theory,the Feichtinger Conjecture and the Rε-Conjecture in frame theory and muchmore. In this chapter we will show these equivalences and many more. Wewill also consider a slight weakening of the Kadison-Singer Problem calledthe Sundberg Problem. Then we will look at the recent advances on anotherdeep problem in frame theory called the Paulsen Problem. In particular, wewill see that this problem is also equivalent to a fundamental open problemin operator theory. Namely, if a projection on a finite dimensional Hilbertspace has a nearly constant diagonal, how close is it to a constant diagonalprojection?

Key words: Kadison-Singer Problem, Paving Conjecture, State, Riesz-able, Discrete Fourier Transform, Rε-Conjecture, Feichtinger Conjecture,Bourgain-Tzafriri Conjecture, Restricted Invertibility Principle, SundbergProblem, Paulsen Problem, Principle Angles, Chordal Distance

1.1 Introduction

Finite frame theory is beginning to have an important impact on some of thedeepest problems in both pure and applied mathematics. In this chapter wewill look at two cases where finite frame theory is having a serious impact:

Peter G. CasazzaUniversity of Missouri, Mathematics Department, Columbia, MO 65211, USA, e-mail:[email protected]

1

2 Peter G. Casazza

The Kadison-Singer Problem and the Paulsen Problem. We warn the readerthat because we are restricting ourselves to finite dimensional Hilbert spaces,a significant body of literature on the infinite dimensional versions of theseproblems does not appear here.

1.2 The Kadison-Singer Problem

For over 50 years the Kadison-Singer Problem [36] has defied the best effortsof some of the most talented mathematicians of our time.

Kadison-Singer Problem 1 (KS) Does every pure state on the (abelian)von Neumann algebra D of bounded diagonal operators on `2, the Hilbert spaceof square summable sequences on the integers, have a unique extension to a(pure) state on B(`2), i.e., the von Neumann algebra of all bounded linearoperators on the Hilbert space `2?

A state of a von Neumann algebraR is a linear functional f onR for whichf(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 iscompact in the ω∗-topology. By the Krein-Milman theorem, this convex setis the closed convex hull of its extreme points. The extremal elements in thespace of states are called the pure states (of R).

This problem arose from the very productive collaboration of Kadison andSinger in the 1950s when they were studying Dirac’s Quantum Mechanicsbook [30] which culminated in their seminal work on triangular operatoralgebras.

It is now known that the 1959 Kadison-Singer Problem is equivalent tofundamental unsolved problems in a dozen areas of research in pure mathe-matics, applied mathematics and engineering (See [1, 2, 3, 4, 17, 26, 27, 37]and their references). We will not develop this topic in detail here since it isfundamentally an infinite dimensional problem and we are concentrating onfinite dimensional frame theory. In this chapter we will look at a number ofthese finite dimensional problems which are equivalent to KS and which areimpacted by finite frame theory. Most people today seem to agree with theoriginal statement of Kadison and Singer [36] that KS will have a negativeanswer and so all the equivalent forms will have negative answers also.

1.2.1 The Paving Conjecture

A significant advance on KS was made by Anderson [2] in 1979 when hereformulated KS into what is now known as the Paving Conjecture (Seealso [3, 4]). Lemma 5 of [36] shows a connection between KS and Paving. For

1 The Kadison-Singer and Paulsen Problems in Finite Frame Theory 3

notation, if J ⊂ {1, 2, . . . , n}, the diagonal projection QJ is the matrixwhose entries are all zero except for the (i, i) entries for i ∈ J which are allone. For a matrix A = (aij)

Ni,j=1 let δ(A) = max1≤i≤N |aii|.

Definition 1. An operator T ∈ B(`N2 ) is said to have an (r, ε)-paving if thereis a partition {Aj}rj=1 of {1, 2, . . . , N} so that

‖QAjTQAj‖ ≤ ε‖T‖.

Paving Conjecture 1 (PC) For every 0 < ε < 1, there is a natural num-ber r so that for every natural number N and every linear operator T on lN2whose matrix has zero diagonal, T has an (r, ε)-paving.

It is important that r not depend on N in PC. We will say that an arbitraryoperator T satisfies PC if T −D(T ) satisfies PC where D(T ) is the diagonalof T .

The only large classes of operators which have been shown to be pavableare “diagonally dominant” matrices [6, 7, 11, 31], `1-localized operators [24],matrices with all entries real and positive [32], matrices with small coefficientsin comparison with the dimension [13] (See [40] for a paving into blocks ofconstant size), and Toeplitz operators over Riemann integrable functions (Seealso [33]). Also, in [9] there is an analysis of the paving problem for certainSchatten Cp-norms.

Theorem 1. The Paving Conjecture has a positive solution if any one of thefollowing classes satisfies the Paving Conjecture:

1. Unitary operators. [27]

2. Orthogonal projections. [27]

3. Orthogonal projections with constant diagonal 1/2. [18]

4. Positive operators. [27]

5. Self-adjoint operators. [27]

6. Gram matrices (〈ϕi, ϕj〉)i,j∈I where T : `2(I)→ `2(I) is a bounded linearoperator, and Tei = ϕi, ‖Tei‖ = 1 for all i ∈ I. [27]

7. Invertible operators (or invertible operators with zero diagonal). [27]

8. Triangular operators [38]

Recently, Weaver [43] provided important insight into KS by giving anequivalent problem to PC in terms of projections.

Conjecture 1 (Weaver). There exist universal constants 0 < δ, ε < 1 andr ∈ N so that for all N and all orthogonal projections P on `N2 with δ(P ) ≤ δ,there is a paving {Aj}rj=1 of {1, 2, . . . , N} so that ‖QAjPQAj‖ ≤ 1 − ε, forall j = 1, 2, . . . , r.

This needs some explanation since there is nothing in [43] that looks any-thing like Conjecture 1. Weaver observes that the fact that Conjecture 1

4 Peter G. Casazza

implies PC follows by a minor modification of Propositions 7.6 and 7.7 of[1]. Then he introduces what he calls “Conjecture KSr” (See Conjecture 8).A careful examination of the proof of Theorem 1 of [43] reveals that Weavershows Conjecture KSr implies Conjecture 1 which in turn implies KS which(after the theorem is proved) is equivalent to KSr.

In [18] it was shown that PC fails for r = 2, even for projections withconstant diagonal 1/2. Recently [21] there appeared a frame theoretic con-crete construction of non-2-pavable projections. If this construction can begeneralized, we would have a counterexample to PC and KS. We now lookat the construction from [21].

Definition 2. A family of vectors {ϕi}Mi=1 for an N -dimensional Hilbertspace HN is (δ, r)-Rieszable if there is a partition {Aj}rj=1 of {1, 2, . . . ,M}so that for all j = 1, 2, . . . , r and all scalars {ai}i∈Aj we have

‖∑i∈Aj

aiϕi‖2 ≥ δ∑i∈Aj

|ai|2.

A projection P on HN is (δ, r)-Rieszable if {Pei}Ni=1 is (δ, r)-Rieszable.

We now have:

Proposition 1. Let P be an orthogonal projection on HN . The following areequivalent:

(1) The vectors {Pei}Ni=1 are (δ, r)-Rieszable.(2) There is a partition {Aj}rj=1 of {1, 2, . . . , N} so that for all j =

1, 2, . . . , r and all scalars {ai}i∈Aj we have

‖∑i∈Aj

ai(I − P )ei‖2 ≤ (1− δ)∑i∈Aj

|ai|2.

(3) The matrix of I − P is (δ, r)-pavable.

Proof. (1)⇔ (2): For any scalars {ai}i∈AJ we have∑i∈Aj

|ai|2 = ‖∑i∈Aj

aiPei‖2 + ‖∑i∈Aj

ai(I − P )ei‖2.

Hence,

‖∑i∈Aj

ai(I−P )ei‖2 ≤ (1−δ)∑i∈Aj

|ai|2 if and only if ‖∑i∈Aj

aiPei‖2 ≥ δ∑i∈Aj

|ai|2.

(2) ⇔ (3): Given any partition {Aj}rj=1, any 1 ≤ j ≤ r and any x =∑i∈Aj aiei we have


〈(I − P )x, x〉 = ‖(I − P )x‖2 = ‖∑i∈Aj

ai(I − P )ei‖2

≤ (1− δ)∑i∈Aj

|ai|2 = 〈(1− δ)x, x〉,

if and only if I − P ≤ (1− δ)I.

Given N ∈ N, let ω = exp( 2πiN ), we define the discrete Fourier transform

(DFT) matrix in CN by

DN =

√1

N

(ωjk)N−1j,k=0

.

The main point of these DN matrices is that they are unitary matrices

for which the modulus of all the entries are equal to√

1N . The following is a

simple observation.

Proposition 2. Let A = (aij)Ni,j=1 be a matrix with orthogonal rows and

satisfying |aij |2 = a for all i, j. If we multiply the jth-row of A by a constantCj to get a new matrix B, then:

(1) The rows of B are orthogonal.(2) The square sums of the entries of any column of B all equal

a

N∑j=1

C2j .

(3) The square sum of the entries of the jth row of B equals aC2j .

To construct our example, we start with a 2N × 2N DFT and multiply

the first N − 1 rows by√

2 and the remaining rows by√

2N+1 to get a new

matrix B1. Next, we take a second 2N × 2N DFT matrix and multiply the

first N − 1 rows by 0 and the remaining rows by√

2N2N+1 to get a matrix B2.

We then put the matrices B1, B2 side by side to get a N × 2N matrix B ofthe form

B =(N-1) Rows

√2 0

(N+1) Rows√

2N+1

√2NN+1

This matrix has 2N rows and 4N columns. Now we show that this matrixgives the required example.

Proposition 3. The matrix B satisfies:(1) The columns are orthogonal and the square sum of the coefficients of

every column equals 2.

6 Peter G. Casazza

(2) The square sum of the coefficients of every row equals 1.The row vectors of the matrix B are not (δ, 2)-Rieszable, for any δ inde-

pendent of N .

Proof. A direct calculation yields (1) and (2).We will now show that the column vectors of B are not uniformly two

Rieszable independent of N . So let {A1, A2} be a partition of {1, 2, . . . , 4N}.Without loss of generality, we may assume that |A1 ∩ {1, 2, . . . , 2N}| ≥ N .Let the column vectors of the matrix B be {ϕi}4Ni=1 as elements of C2N . LetPN−1 be the orthogonal projection of C2N onto the first N − 1 coordinates.Since |A1| ≥ N , there are scalars {ai}i∈A1

so that∑i∈A1

|ai|2 = 1 and

PN−1

(∑i∈A1

aiϕi

)= 0.

Also, let {ψj}2Nj=1 be the orthonormal basis consisting of the original columnsof the DFT2N . We now have:

‖∑i∈A1

aiϕi‖2 = ‖(I − PN−1)(∑i∈A1

aiϕi)‖2

=2

N + 1‖(I − PN−1)(

∑i∈A1

aiψi)‖2

≤ 2

N + 1‖∑i∈A1

aiψi‖2

=2

N + 1

∑i∈A1

|ai|2

=2

N + 1.

Letting N →∞, this class of matrices is not (δ, 2)-Rieszable, and hence not(δ, 2)-pavable for any δ > 0.

If this argument could be generalized to yield non-(δ, 3)-Rieszable (Pavable)matrices, then such an argument will lead to a complete counterexample toPC.

1.2.2 The Rε-Conjecture

In this section we will define the Rε-Conjecture and show it is equivalent tothe Paving Conjecture.

Definition 3. A family of vectors {ϕi}Mi=1 is an ε-Riesz basic sequence for0 < ε < 1 if for all scalars {ai}Mi=1 we have


(1− ε)M∑i=1

|ai|2 ≤ ‖M∑i=1

aiϕi‖2 ≤ (1 + ε)

M∑i=1

|ai|2.

A natural question is whether we can improve the Riesz basis bounds fora unit norm Riesz basic sequence by partitioning the sequence into subsets.

Conjecture 2 (Rε-Conjecture). For every ε > 0, every unit norm Riesz basicsequence is a finite union of ε-Riesz basic sequences.

This conjecture was first stated by Casazza and Vershynin and was firststudied in [16] where it was shown that PC implies the conjecture. Oneadvantage of the Rε-Conjecture is that it can be shown to students at thebeginning of a course in Hilbert spaces.

The Rε-Conjecture has a natural finite dimensional form.

Conjecture 3. For every ε > 0 and every T ∈ B(`N2 ) with ‖Tei‖ = 1 for i =1, 2, . . . , N there is an r = r(ε, ‖T‖) and a partition {Aj}rj=1 of {1, 2, . . . , N}so that for all j = 1, 2, . . . , r and all scalars {ai}i∈Aj we have

(1− ε)∑i∈Aj

|ai|2 ≤ ‖∑i∈Aj

aiTei‖2 ≤ (1 + ε)∑i∈Aj

|ai|2.

Now we show that the Rε-Conjecture is equivalent to PC.

Theorem 2. The following are equivalent:(1) The Paving Conjecture.(2) For 0 < ε < 1, there is an r = r(ε, B) so that for every N ∈ N, if

T : `N2 → `N2 is a bounded linear operator with ‖T‖ ≤ B and ‖Tei‖ = 1 forall i = 1, 2, . . . , N , then there is a partition {Aj}rj=1 of {1, 2, . . . , N} so thatfor each 1 ≤ j ≤ r, {Tei}i∈Aj is an ε-Riesz basic sequence.

(3) The Rε-Conjecture.

Proof. (1)⇒ (2): Fix 0 < ε < 1. Given T as in (2), let S = T ∗T . Since S hasones on its diagonal, by the Paving Conjecture there is a r = r(ε, ‖T‖) and apartition {Aj}rj=1 of {1, 2, . . . , N} so that for every j = 1, 2, . . . , r we have

‖QAj (I − S)QAj‖ ≤ δ‖I − S‖

where δ = ε‖S‖+1 . Now, for all x =

∑Ni=1 aiei and all j = 1, 2, . . . , r we have

‖∑i∈Aj

aiTei‖2 = ‖TQAjx‖2 = 〈TQAjx, TQAjx〉 = 〈T ∗TQAjx,QAjx〉

= 〈QAjx,QAjx〉 − 〈QAj (I − S)QAjx,QAjx〉≥ ‖QAjx‖2 − δ‖I − S‖‖QAjx‖2

≥ (1− ε)‖QAjx‖2 = (1− ε)∑i∈Aj

|ai|2.

8 Peter G. Casazza

Similarly, ‖∑i∈Aj aiTei‖

2 ≤ (1 + ε)∑i∈Aj |ai|

2.

(2)⇒ (3): This is obvious.(3)⇒ (1): Let T ∈ B(`N2 ) with Tei = ϕi and ‖ϕi‖ = 1 for all 1 ≤ i ≤ N .

By Theorem 1 part 6, it suffices to show that the Gram operator G of {ϕi}Ni=1

is pavable. Fix 0 < δ < 1 and let ε > 0. Let ψi =√

1− δ2ϕi ⊕ δei ∈ `N2 ⊕ `N2 .Then ‖ψi‖ = 1 for all 1 ≤ i ≤ N and for all scalars {ai}Ni=1

δ

N∑i=1

|ai|2 ≤ ‖N∑i=1

aiψi‖2 = (1− δ2)‖N∑i=1

aiTei‖2 + δ2N∑i=1

|ai|2

≤[(1− δ2)‖T‖2 + δ2

] N∑i=1

|ai|2.

So {ψi}Ni=1 is a unit norm Riesz basic sequence and 〈ψi, ψk〉 = (1−δ2)〈ϕi, ϕk〉for all 1 ≤ i 6= k ≤ N . By the Rε-Conjecture, there is a partition {Aj}rj=1 of{1, 2, . . . , N} so that for all j = 1, 2, . . . , r and all x =

∑i∈Aj aiei,

(1− ε)∑i∈Aj

|ai|2 ≤ ‖∑i∈Aj

aiψi‖2 = 〈∑i∈Aj

aiψi,∑k∈Aj

akψk〉

=∑i∈Aj

|ai|2‖ψi‖2 +∑

i 6=k∈Aj

aiak〈ψi, ψk〉

=∑i∈Aj

|ai|2 + (1− δ2)∑

i 6=k∈Aj

aiak〈ϕi, ϕk〉

=∑i∈Aj

|ai|2 + (1− δ2)〈QAj (G−D(G))QAjx, x〉

≤ (1 + ε)∑i∈Aj

|ai|2.

Subtracting∑i∈Aj |ai|

2 through the inequality yields,

−ε∑i∈Aj

|ai|2 ≤ (1− δ2)〈QAj (G−D(G))QAjx, x〉 ≤ ε∑i∈Aj

|ai|2.

That is,(1− δ2)|〈QAj (G−D(G))QAjx, x〉| ≤ ε‖x‖2.

Since QAj (G−D(G))QAj is a self-adjoint operator, we have (1−δ2)‖QAj (G−D(G))QAj‖ ≤ ε, i.e., (1− δ2)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 {ψi}i∈I intoconjectures about unit norm familes {ϕi}i∈I with T ∈ B(`2(I)) and Tei = ϕi.Namely, given {ϕi}i∈I and 0 < δ < 1 let ψi =

√1− δ2fi⊕δei ∈ `2(I)⊕`2(I).


Then {ψi}i∈I is a unit norm Riesz basic sequence and for δ small enough, ψiis close enough to ϕi to pass inequalities from {ψi}i∈I to {ϕi}i∈I .

The Rε-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 `2 by: |{ai}| = ‖{ai}‖`2 + supi |ai| then the Rε-Conjecture fails for this equivalent norm. To see this, we proceed by way ofcontradiction and assume there is an 0 < ε < 1 and an r = r(ε, 2) satisfyingthe Rε-Conjecture. Let {ei}2Ni=1 be the unit vectors for `2N2 and let xi =e2i+e2i+1√

2+1for 1 ≤ i ≤ N . This is now a unit norm Riesz basic sequence with

upper Riesz bound 2. Assume we partition {1, 2, . . . , 2N} into sets {Aj}rj=1.

Then for some 1 ≤ k ≤ r we have |Ak| ≥ Nr . Let A ⊂ Ak with |A| = N

r andai = 1√

Nfor i ∈ A. Then

|∑i∈A

aixi| =1√

2 + 1

(√2 +

r√N

)Since the norm above is bounded away from one for large N , we cannot satisfythe requirements of the Rε-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 Rε-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 : `N2 → `N2 with ‖Tei‖ = 1 for alli = 1, 2, . . . , N and ‖T‖ ≤ 2, there is a partition {Aj}rj=1 of {1, 2, . . . , N} sothat for all j = 1, 2, . . . , r and all scalars {ai}i∈Aj we have

‖∑i∈Aj

aiTei‖2 ≥ A∑i∈Aj

|ai|2.

Theorem 3. Conjecture 4 is equivalent to PC.

Proof. Since PC is equivalent to the Rε-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

4 and let P be anorthogonal projection on `N2 with δ(P ) ≤ δ Now, 〈Pei, ei〉 = ‖Pei‖2 ≤ δ

implies ‖(I − P )ei‖2 ≥ 1 − δ ≥ 14 . Define T : `N2 → `N2 by Tei = (I−P )ei

‖(I−P )ei‖ .

For any scalars {ai}Ni=1 we have

10 Peter G. Casazza

‖N∑i=1

aiTei‖2 =

∥∥∥∥∥N∑i=1

ai‖(I − P )ei‖

(I − P )ei

∥∥∥∥∥2

≤N∑i=1

∣∣∣∣ ai‖(I − P )ei‖

∣∣∣∣2

≤ 4

N∑i=1

|ai|2.

So ‖Tei‖ = 1 and ‖T‖ ≤ 2. By Conjecture 4, there is a partition {Aj}rj=1 of{1, 2, . . . , N} so that for all j = 1, 2, . . . , r and all scalars {ai}i∈Aj we have

‖∑i∈AJ


|ai|2.

Hence, ∥∥∥∥∥∥∑i∈Aj

ai(I − P )ei

∥∥∥∥∥∥2

=

∥∥∥∥∥∥∑i∈Aj

ai‖(I − P )ei‖Tei

∥∥∥∥∥∥2

≥ A∑i∈Aj

|ai|2‖(I − P )ei‖2

≥ A

4

∑i∈Aj

|ai|2.

It follows that for all scalars {ai}i∈Aj ,∑i∈Aj

|ai|2 = ‖∑i∈Aj

aiPei‖2 + ‖∑i∈Aj

ai(I − P )ei‖2

≥ ‖∑i∈Aj

aiPei‖2 +A

4

∑i∈Aj

|ai|2.

Now, for all x =∑Ni=1 aiei we have

‖PQAjx‖2 = ‖∑i∈Aj

aiPei‖2 ≤ (1− A

4)∑i∈Aj

|ai|2.

Thus,

‖QAjPQAj‖ = ‖PQAj‖2 ≤ 1− A

4.

So Conjecture 1 holds.Weaver [43] established an important relationship between frames and PC

by showing that the following conjecture is equivalent to PC.


Conjecture 5. There are universal constants B ≥ 4 and α >√B and an

r ∈ N so that the following holds. Whenever {ϕi}Mi=1 is a unit norm B-tightframe for `N2 , there exists a partition {Aj}rj=1 of {1, 2, . . . ,M} so that for all

j = 1, 2, . . . , r and all x ∈ `N2 we have∑i∈Aj

|〈x, ϕi〉|2 ≤ (B − α)‖x‖2. (1.1)

Using Conjecture 5 we can show that the following conjecture is equivalentto PC:

Conjecture 6. There is a universal constant 1 ≤ D so that for all T ∈ B(`N2 )with ‖Tei‖ = 1 for all i = 1, 2, . . . , N , there is an r = r(‖T‖) and a partition{Aj}rj=1 of {1, 2, . . . , N} so that for all j = 1, 2, . . . , r and all scalars {ai}i∈Aj

‖∑i∈Aj

aiTei‖2 ≤ D∑i∈Aj

|ai|2.

Theorem 4. Conjecture 6 is equivalent to PC.

Proof: Since Conjecture 3 clearly implies Conjecture 6, we just need toshow that Conjecture 6 implies Conjecture 5. So, choose D as in Conjecture6 and choose B ≥ 4 and α >

√B so that D ≤ B − α. Let {ϕi}Mi=1 be a

unit norm B tight frame for `N2 . If Tei = ϕi is the synthesis operator for thisframe, then ‖T‖2 = ‖T ∗‖2 = B. So by Conjecture 6, there is an r = r(‖B‖)and a partition {Aj}rj=1 of {1, 2, . . . ,M} so that for all j = 1, 2, . . . , r and allscalars {ai}i∈Aj

‖∑i∈Aj

aiTei‖2 = ‖∑i∈Aj

aiϕi‖2 ≤ D∑i∈Aj

|ai|2 ≤ (B − α)∑i∈Aj

|ai|2.

So ‖TQAj‖2 ≤ B − α and for all x ∈ `N2∑i∈Aj

|〈x, ϕi〉|2 = ‖(QAjT )∗x‖2 ≤ ‖TQAj‖2‖x‖2 ≤ (B − α)‖x‖2.

This verifies that Conjecture 5 holds and so KS holds. utRemark 1 and Conjecture 6 show that we only need any universal upper

bound in the Rε-Conjecture to hold for KS.

1.2.3 The Feichtinger Conjecture

While working in time-frequency analysis, Feichtinger [16] observed that allof the Gabor frames he was using had the property that they could be divided

12 Peter G. Casazza

into a finite number of subsets which were Riesz basic sequences. This led tothe conjecture:

Feichtinger Conjecture 1 (FC) Every bounded frame (or equivalently,every unit norm frame) is a finite union of Riesz basic sequences.

The finite dimensional form of FC looks like:

Conjecture 7 (Finite Dimensional Feichtinger Conjecture). For every B,C >0, there is a natural number r = r(B,C) and a constant A = A(B,C) > 0so that whenever {ϕi}Ni=1 is a frame for HN with upper frame bound B and‖ϕi‖ ≥ C for all i = 1, 2, . . . , N , then {1, 2, . . . , N} can be partitioned intosubsets {Aj}rj=1 so that for each 1 ≤ j ≤ r, {ϕi}i∈Aj is a Riesz basic sequencewith lower Riesz basis bound A and upper Riesz basis bound B.

There is a significant body of work on this conjecture [6, 7, 16, 31], yet, itremains open even for Gabor frames.

We now check that the Feichtinger Conjecture is equivalent to PC.

Theorem 5. The following are equivalent:(1) The Paving Conjecture.(2) The Feichtinger Conjecture.

Proof. (1) ⇒ (2): Part (2) of Theorem 2 is equivalent to PC and clearlyimplies FC.

(2)⇒ (1): We will observe that FC implies Conjecture 4 which is equiva-lent to PC by Theorem 3. In Conjecture 4, {Tei}Ni=1 is a frame for its spanwith upper frame bound 2. It is now immediate that the Finite DimensionalFeichtinger Conjecture implies Conjecture 4.

Another equivalent formulation of KS due to Weaver [43].

Conjecture 8 (KSr). There are universal constants B and ε > 0 so that the fol-lowing holds. Let {ϕi}Mi=1 be elements of `N2 with ‖ϕi‖ ≤ 1 for i = 1, 2, . . . ,Mand suppose for every x ∈ `N2 ,

M∑i=1

|〈x, ϕi〉|2 ≤ B‖x‖2. (1.2)

Then, there is a partition {Aj}rj=1 of {1, 2, . . . ,M} so that for all x ∈ `N2 andall j = 1, 2, . . . , r, ∑

i∈Aj

|〈x, ϕi〉|2 ≤ (B − ε)‖x‖2.

Theorem 6. The following are equivalent:(1) The Paving Conjecture.(2) Conjecture KSr holds for some r ≥ 2.


Proof. Assume Conjecture KSr is true for some fixed r,B, ε. We will showthat Conjecture 1 is true. Let P be an orthogonal projection on HM withδ(P ) ≤ 1

B . If P has rank N, then its range is a N-dimensional subspace W of

HM . Define ϕi =√B · Pei ∈W for all 1 ≤ i ≤M . We check

‖ϕi‖2 = B · ‖Pei‖2 = B〈Pei, ei〉 ≤ Bδ(P ) ≤ 1, for all i = 1, 2, . . . ,M.

Now, if x ∈W is any unit vector then

M∑i=1

|〈x, ϕi〉|2 =

M∑i=1

|〈x,√BPei〉|2 = B ·

M∑i=1

|〈x, ei〉|2 = B.

By Conjecture KSr, there is a partition {Aj}rj=1 of {1, 2, . . . ,M} satisfyingfor all 1 ≤ j ≤ r and all unit vectors x ∈W ,∑

i∈Aj

|〈x, ϕi〉|2 ≤ B − ε.

Then∑rj=1QAj = Id, and for any unit vector x ∈W we have

‖QAjPx‖2 =

M∑i=1

|〈QAjPx, ei〉|2 =

M∑i=1

|〈x, PQjei〉|2

=1

B

∑i∈Aj

|〈x, ϕi〉|2 ≤ε

B.

Thus Conjecture 1 holds and so PC holds.Conversely, assume KSr fails for all r. Fix B = r ≥ 2 and let {ϕi}Mi=1

in HN be a counterexample with ε = 1. Let ψi = ϕi√B

and note that

‖ψiψiT ‖ = ‖ψi‖2 ≤ 1B , for all i = 1, 2, . . . ,M and

∑Mi=1 ψiψi

T ≤ Id. Then

Id −∑Mi=1 ψiψ

Ti is a positive finite rank operator, so we can find positive

rank one operators ψiψTi for M + 1 ≤ i ≤ K such that ‖ψiψTi ‖ ≤ 1

B for all

1 ≤ i ≤ K and∑Ki=1 ψiψ

Ti = Id.

Let T be the analysis operator for {ψi}Ki=1, which is an isometry and if Pis the orthogonal projection of HK with range T (HN ), then Pei = Tψi forall i = 1, 2, . . . ,K. Let D be the diagonal matrix with the same diagonal asP . Then

‖D‖ = max1≤i≤K

‖ψi‖2 ≤1

B.

Let {Qj}rj=1 be any K×K diagonal projections which sum to the identity.Define a partition {Aj}rj=1 of {1, 2, . . . ,K by letting Aj be the diagonal of Qj .

By our choice of {ϕi}Mi=1, there exists 1 ≤ j ≤ r and x ∈ HN with ‖x‖ = 1and

14 Peter G. Casazza∑i∈Aj∩{1,2,...,M}

|〈x, ϕi〉|2 > B − 1.

Hence, ∑i∈Aj

|〈x, ψi〉|2 > 1− 1

B.

It follows that for all j,

‖QjP (Tx)‖2 ≥O∑i=1

|〈QjP (Tx), ei〉|2 =∑i∈Aj

|〈Tx, ei〉|2

=∑i∈Aj

|〈x, ψi〉|2 > 1− 1

B.

Thus, ‖QjPQj‖ = ‖QjP‖2 > 1 − 1B . Now, the matrix A = P −D has zero

diagonal and satisfies ‖A‖ ≤ 1 + 1B and the above shows that for any K ×K

diagonal projections {Qj}rj=1 with∑rj=1Qj = Id we have

‖QjAQj‖ ≥ ‖QjPQj‖ − ‖QjDQj‖ ≥ 1− 2

B, for some j .

Finally, as B = r → ∞, we obtain a sequence of examples which negate thePaving Conjecture.

Weaver [43] also shows that Conjecture KSr is equivalent to PC if weassume equality in Equation 1.2 for all x ∈ `M2 . Weaver further shows thatConjecture KSr is equivalent to PC even if we strengthen its assumptions soas to require that the vectors {ϕi}Mi=1 are of equal norm and that equalityholds in 1.2, but at great cost to our ε > 0.

Conjecture 9 (KSr′). There exists universal constants B ≥ 4 and ε >

√B so

that the following holds. Let {ϕi}Mi=1 be elements of `N2 with ‖ϕi‖ = 1 fori = 1, 2, . . . ,M and suppose for every x ∈ `N2 ,

M∑i=1

|〈x, ϕi〉|2 = B‖x‖2. (1.3)

Then, there is a partition {Aj}rj=1 of {1, 2, . . . ,M} so that for all x ∈ `M2 andall j = 1, 2, . . . , r, ∑

i∈Aj

|〈x, ϕi〉|2 ≤ (B − ε)‖x‖2.

We introduce one more conjecture.

Conjecture 10. There exist universal constants 0 < δ,√δ ≤ ε < 1 and r ∈ N

so that for all N and all orthogonal projections P on `N2 with δ(P ) ≤ δ


and ‖Pei‖ = ‖Pej‖ for all i, j = 1, 2, . . . , N , there is a paving {Aj}rj=1 of{1, 2, . . . , N} so that ‖QAjPQAj‖ ≤ 1− ε, for all j = 1, 2, . . . , r.

Using Conjecture 9 we can see that PC is equivalent to Conjecture 10.

Theorem 7. PC is equivalent to Conjecture 10.

Proof: It is clear that Conjecture 1 (which is equivalent to (PC)) impliesConjecture 10. So we assume that Conjecture 10 holds and we will showthat Conjecture 9 holds. Let {ϕi}Mi=1 be elements of HN with ‖ϕi‖ = 1 fori = 1, 2, . . . ,M and suppose for every x ∈ HN ,

M∑i=1

|〈x, ϕi〉|2 = B‖x‖2, (1.4)

where 1B ≤ δ. It follows from Equation 1.4 that { 1√

Bϕi}Mi=1 is an equal norm

Parseval frame and so by Naimark’s Theorem, we may assume there is alarger Hilbert space `M2 and a projection P : `M2 → HN so that Pei = ϕi forall i = 1, 2, . . . ,M . Now ‖Pei‖2 = 〈Pei, ei〉 = 1

B ≤ δ for all i = 1, 2, . . . , N .So by Conjecture 10, there is a paving {Aj}rj=1 of {1, 2, . . . ,M} so that‖QAjPQAj‖ ≤ 1 − ε, for all j = 1, 2, . . . , r. Now for all 1 ≤ j ≤ r and allx ∈ `N2 we have:

‖QAjPx‖2 =

M∑i=1

|〈QAjPx, ei〉|2

=

M∑i=1

|〈x, PQAjei〉|2

=1

B

∑i∈Aj

|〈x, ϕi〉|2

≤ ‖QAjP‖2‖x‖2

= ‖QAjPQAj‖‖x‖2

≤ (1− ε)‖x‖2.

It follows that for all x ∈ `N2 we have∑i∈Aj

|〈xϕi〉|2 ≤ (B − εB)‖x‖2.

Since εB >√B, we have verified Conjecture 9. ut

16 Peter G. Casazza

1.2.4 The Bourgain-Tzafriri Conjecture

We start with a fundamental theorem of Bourgain and Tzafriri called theRestricted Invertibility Principle. This theorem led to the (strong and weak)Bourgain-Tzafriri Conjectures. We will see that these conjectures are equiv-alent to PC.

In 1987, Bourgain and Tzafriri [12] proved a fundamental result in Banachspace theory known as the Restricted Invertibility Principle.

Theorem 8 (Bourgain-Tzafriri). There is a universal constants 0 < c < 1so that whenever T : `N2 → `N2 is a linear operator for which ‖Tei‖ = 1,for 1 ≤ i ≤ N , then there exists a subset σ ⊂ {1, 2, . . . , N} of cardinality|σ| ≥ cN/‖T‖2 so that for all choices of scalars {aj}j∈σ,

‖∑j∈σ

ajTej‖2 ≥ c∑j∈σ|aj |2.

A close examination of the proof of the theorem [12] yields that c is onthe order of 10−72. The proof of the theorem uses probabilistic and functionanalytic techniques, is non-trivial and is non-constructive. A significant break-through occurred recently when Spielman and Srivastava [39] presented analgorithm for proving the Restricted-Invertibility Theorem. Moreover, theirproof gives the best possible constants in the theorem.

Theorem 9 (Restricted Invertibility Theorem: Spielman-Srivastava

Form). Assume {vi}Mi=1 are vectors in `N2 with A =∑Mi=1 viv

Ti = I and

0 < ε < 1. If L : `N2 → `N2 is a linear operator, then there is a subset J ⊂{1, 2, . . . ,M} of size |J | ≥ ε2 ‖L‖

2F

‖L‖2 for which {Lvi}i∈J is linearly independent

and

λmin

(∑i∈J

Lvi(Lvi)T

)>

(1− ε)2‖L‖FM

,

where ‖L‖F is the Frobenious norm of L and λmin is the smallest eigenvalueof the operator computed on span {vi}i∈J .

This generalized form of the Restricted-Invertibility Theorem was intro-duced by Vershynin [41] where he studied the contact points of convex bodiesusing John’s decompositions of the identity. The corresponding theorem forinfinite dimensional Hilbert spaces is still open. But this case requires the setJ to be large with respect to the Beurling density [6, 7]. Special cases of thisproblem were solved in [24, 41].

The inequality in the Restricted Invertibility Theorem is referred to as alower `2-bound. It is known [28, 41] that there is a corresponding close to oneupper `2-bound which can be achieved in the theorem. Recently, a two-sidedSpielman-Srivastava algorithm was given [25] which gives the best two sidedlower and upper bounds in BT.


Corollary 1. For T : `N2 → `N2 and an orthonormal basis {ei}Ni=1 with‖Tei‖2 = 1, i = 1, . . . , N , and for any α ∈ (0, 1) there exists a subsetJα ⊆ {1, 2, . . . , N}, satisfying

1. |Jα| ≥⌊α2

2

N

‖T‖2

⌋,

and

2. The condition number of T |span {ej : j∈Jα} is bounded above by (1− α)−4 .

The corresponding theorem for infinite dimensional Hilbert spaces is stillopen. Special cases of this problem were solved in [24, 41].

Theorem 8 gave rise to a problem in the area which has received a greatdeal of attention [13, 26, 27].

Bourgain-Tzafriri Conjecture 1 (BT) There is a universal constant A >0 so that for every B > 1 there is a natural number r = r(B) satisfying: Forany natural number N , if T : `N2 → `N2 is a linear operator with ‖T‖ ≤ Band ‖Tei‖ = 1 for all i = 1, 2, . . . , N , then there is a partition {Aj}rj=1 of{1, 2, . . . , N} so that for all j = 1, 2, . . . , r and all choices of scalars {ai}i∈Ajwe have:

‖∑i∈Aj


|ai|2.

Sometimes BT is called strong BT since there is a weakening called weakBT. In weak BT we allow A to depend upon the norm of the operator T . Asignificant amount of effort over the years was invested in trying to show thatstrong and weak BT are equivalent. Casazza and Tremain finally proved thisequivalence [26]. We will not have to do any work here since we developed allof the needed results in earlier sections.

Theorem 10. The following are equivalent:(1) The Paving Conjecture.(2) The Bourgain-Tzafriri Conjecture.(3) The (weak) Bourgain-Tzafriri Conjecture.

Proof. (1) ⇒ (2) ⇒ (3): The Paving Conjecture is equivalent to the Rε-Conjecture which clearly implies the Bourgain-Tzafriri Conjecture and thisimmediately implies the (weak) Bourgain-Tzafriri Conjecture.

(3) ⇒ (1): The weak Bourgain-Tzafriri Conjecture immediately impliesConjecture 4 which is equivalent to the Paving Conjecture.

1.2.5 Partitioning Frames into Frame Subsets

A natural and frequently occurring problem in frame theory is to partitiona frame into subsets each of which has good frame bounds. This innocent

18 Peter G. Casazza

looking question turns out to be much deeper than it looks, and as we willnow see, it is equivalent to PC.

Conjecture 11. There exists an ε > 0 so that for large K, for all N and allequal norm Parseval frames {ϕi}KNi=1 for `N2 , there is a J ⊂ {1, 2, . . . ,KN} sothat both {ϕi}i∈J and {ϕi}i∈Jc have lower frame bounds which are greaterthan ε.

The ideal situation would be for Conjecture 11 to hold for all K ≥ 2. Inorder for {ϕi}i∈J and {ϕi}i∈Jc to both be frames for `N2 , they at least haveto span `N2 . So the first question is whether we can partition our frame intospanning sets. This follows from a generalization of the Rado-Horn theorem.See the chapter on independence and spanning properties of frames.

Proposition 4. Every equal norm Parseval frame {ϕi}KN+Li=1 , 0 ≤ L < N

for `N2 can be partitioned into K linearly independent spanning sets plus alinearly independent set of L elements.

The natural question is whether we can do such a partition so that eachof the subsets has good frame bounds, i.e., a universal lower frame bound forall subsets. Before addressing this question, we state another conjecture.

Conjecture 12. There exists ε > 0 and a natural number r so that for allN , all large K and all equal norm Parseval frames {ϕi}KNi=1 in `N2 there is apartition {Aj}rj=1 of {1, 2, . . . ,KN} so that for all j = 1, 2, . . . , r the Besselbound of {ϕi}i∈Aj is ≤ 1− ε.

We will now establish a relationship between our conjectures and PC.

Theorem 11. (1) Conjecture 11 implies Conjecture 12.(2) Conjecture 12 is equivalent to PC.

Proof. (1): Fix ε > 0, r,K as in Conjecture 11. Let {ϕi}KNi=1 be an equal normParseval frame for an N -dimensional Hilbert space HN . By Naimark’s Theo-rem we may assume there is an orthogonal projection P on `KN2 with Pei = ϕifor all i = 1, 2, . . . ,KN . By Conjecture 11, there is a J ⊂ {1, 2, . . . ,KN} sothat {Pei}i∈J and {Pei}i∈Jc both have a lower frame bound of ε > 0. Hence,for x ∈ HM = P (`KN2 ),

‖x‖2 =

KN∑i=1

|〈x, Pei〉|2 =∑i∈J|〈x, Pei〉|2 +

∑i∈Jc|〈x, Pei〉|2

≥∑i∈J|〈x, Pei〉|2 + ε‖x‖2.

That is,∑i∈J |〈x, Pei〉|2 ≤ (1−ε)‖x‖2. So the upper frame bound of {Pei}i∈J

(which is the norm of the analysis operator (PQJ)∗ for this frame) is ≤ 1− ε.Since PQJ is the synthesis operator for this frame, we have that ‖QJPQJ‖ =


‖PQJ‖2 = ‖(PQJ)∗‖2 ≤ 1− ε. Similarly, ‖QJcPQJc‖ ≤ 1− ε. So Conjecture12 holds for r = 2.

(2): We will show that Conjecture 12 implies Conjecture 5. Choose ε andr satisfying Conjecture 12 for all large K. In particular, choose any K with1√K< α. Let {ϕi}Mi=1 be a unit norm K-tight frame for an N -dimensional

Hilbert space HN . Then M =∑Mi=1 ‖ϕi‖2 = KN . Since { 1√

Kϕi}Mi=1 is an

equal norm Parseval frame, by Naimark’s Theorem we may assume there isan orthogonal projection P on `M2 with Pei = 1√

Kϕi, for i = 1, 2, . . . ,M . By

Conjecture 12 there is a partition {Aj}rj=1 of {1, 2, . . . ,M} so that the Bessel

bound ‖(PQAj )∗‖2 for each family {ϕi}i∈Aj is ≤ 1− ε. So for j = 1, 2, . . . , rand any x ∈ `N2 we have∑i∈Aj

|〈x, 1√Kϕi〉|2 =

∑i∈Aj

|〈x, PQAjei〉|2 =∑i∈Aj

|〈QAjPx, ei〉|2 ≤ ‖QAjPx‖2

≤ ‖QAjP‖2‖x‖2 = ‖(PQAj )∗‖2‖x‖2 ≤ (1− ε)‖x‖2.

Hence, ∑i∈Aj

|〈x, ϕi〉|2 ≤ K(1− ε)‖x‖2 = (K −Kε)‖x‖2.

Since Kε >√K, we have verified Conjecture 5.

For the converse, choose r, δ, ε satisfying Conjecture 1. If {ϕi}KNi=1 is anequal norm Parseval frame for an N -dimensional Hilbert space HN with 1

K ≤δ, by Naimark’s Theorem we may assume we have an orthogonal projectionP on `KN2 with Pei = ϕi for i = 1, 2, . . . ,KN . Since δ(P ) = ‖ϕi‖2 ≤ 1

K ≤ δ,by Conjecture 1 there is a partition {Aj}rj=1 of {1, 2, . . . ,KN} so that for allj = 1, 2, . . . , r,

‖QAjPQAj‖ = ‖PQAj‖2 = ‖(PQAj )∗‖2 ≤ 1− ε.

Since ‖(PQAj )∗‖2 is the Bessel bound for {Pei}i∈Aj = {ϕi}i∈Aj , we havethat Conjecture 12 holds.

1.3 The Sundberg Problem

Recently, an apparent weakening of the Kadison-Singer Problem has arisen.In his work on interpolation in complex function theory, Sundberg noticedthe following problem. Although this is an infinite dimensional problem, westate it here because of its connections to the Kadison-Singer Problem.

Problem 1 (Sundberg Problem). If {ϕ}∞i=1 is a unit norm Bessel se-quence, can we partition {ϕi}∞i=1 into a finite number of spanning sets?

20 Peter G. Casazza

This 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 {ϕi}∞i=1 is a unit norm Bessel sequence then by FC, we can partitionthe natural numbers into a finite number of sets {Aj}rj=1 so that {ϕi}i∈Aj isa Riesz sequence for all j = 1, 2, . . . , r. For each j = 1, 2, . . . , r choose ij ∈ Aj .Then neither ϕij nor {ϕi}i∈Aj\{ij} can span the space.

1.4 The Paulsen Problem

The 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 {ϕi}Mi=1 for HN with frame operator S is said to beε-nearly equal norm if

(1− ε)NM≤ ‖ϕi‖2 ≤ (1 + ε)

N

M, for all i = 1, 2, . . . ,M,

and it is ε-nearly Parseval if

(1− ε)Id ≤ S ≤ (1 + ε)Id.

We also need:

Definition 5. Given frames Φ = {ϕi}Mi=1 and Ψ = {ψi}Mi=1 for HN , we definethe distance between them by

d(Φ, Ψ) =

M∑i=1

‖ϕi − ψi‖2.

The 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 ε-nearly equal norm andε-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


function f(ε,N,M) so that every ε-nearly equal norm and ε-nearly Parsevalframe Φ = {ϕi}Mi=1 satisfies

d(Φ, Ψ) ≤ f(ε,N,M),

for some equal norm Parseval frame Ψ . A simple compactness argument dueto Hadwin (See [10]) shows that such a function must exist.

Lemma 1. The function f(ε,N,M) exists.

Proof. We will proceed by way of contradiction. If this fails, then there is an0 < ε so that for every δ = 1

n , there is a frame {ϕni }Mi=1 with frame bounds1− 1

n , 1 + 1n and satisfying(

1− 1

n

)N

M≤ ‖ϕni ‖ ≤

(1 +

1

n

)N

M,

while Φn = {ϕni }Mi=1 is a distance greater than ε from any equal norm Parsevalframe. By compactness and switching to a subsequence, we may assume that

limn→∞

ϕni = ϕi, exists for all i = 1, 2, . . . ,M.

But now Φ = {ϕi}Mi=1 is an equal norm Parseval frame contradicting the factthat

d(Φn, Φ) ≥ ε > 0, for all n = 1, 2, . . . ,

for any equal norm Parseval frame.

The problem with this argument is that it does not give any quantitativeestimate on the parameters. We do not have a good idea of what form thefunction f(ε,N,M) must have. We do not even know if M needs to be inthe function or if it is independent of the number of frame vectors. Thefollowing example shows that the Paulsen function is certainly a function ofthe dimension of the space.

Lemma 2. The Paulsen function satisfies

f(ε,N,M) ≥ ε2N.

Proof. Fix an ε > 0 and an orthonormal basis {ej}Nj=1 for HN . We define a

frame {ϕi}2Ni=1 by

ϕi =

{1−ε√

2ei if 1 ≤ i ≤ N

1+ε√2ei−N if N + 1 ≤ i ≤ 2N

By the definition, {ϕi}Mi=1 is ε-nearly equal norm. Also, for any x ∈ HN wehave

22 Peter G. Casazza

2N∑i=1

|〈x, ϕi〉|2 =(1− ε)2

2

N∑i=1

|〈x, ei〉|2 +(1 + ε)2

2

N∑i=1

|〈x, ei〉|2 = (1 + ε2)‖x‖2.

So {ϕi}2Ni=1 is a (1 + ε2) tight frame and hence an ε-nearly Parseval frame.The closest equal norm frame to {ϕi}2Ni=1 is { ei√

2}Ni=1 ∪ { ei√2

}Ni=1. Also,

N∑i=1

‖ ei√2−ϕi‖2+

2N∑i=N+1

‖ei−N√2−ϕi‖2 =

N∑i=1

‖ ε√2ei‖2+‖

N∑i=1

‖ ε√2ei‖2 = ε2N.

The main difficulty with solving the Paulsen Problem is that finding aclose equal-norm frame to a given frame involves finding a close frame whichsatisfies a geometric condition while finding a close Parseval frame to a givenframe involves satisfying a (algebraic) spectral condition. At this time, welack techniques for combining these two conditions. However, each of themindividually has a known solution. That is, we do know the closest equalnorm frame to a given frame [15] and we do know the closest Parseval frameto a given frame [5, 10, 15, 22, 35].

Lemma 3. If {ϕi}Mi=1 is an ε-nearly equal norm frame in HN , then the clos-est equal norm frame to {ϕi}Mi=1 is

ψi = aϕi‖ϕi‖

, for i = 1, 2, . . . ,M,

where

a =

∑Mi=1 ‖ϕi‖M

.

It is well known that for a frame {ϕi}Mi=1 for HN with frame operator S,the closest Parseval frame to {ϕi}Mi=1 is {S−1/2ϕi}Mi=1 [5, 10, 15, 22, 35]. Wewill give the version from [10] here.

Proposition 5. Let {ϕi}Mi=1 be a frame for a N -dimensional Hilbert spaceHN , with frame operator S = T ∗T . Then {S−1/2ϕi}Mi=1 is the closest Parsevalframe to {ϕi}Mi=1. Moreover, if {ϕi}Mi=1 is an ε-nearly Parseval frame then

M∑i=1

‖S−1/2ϕi − ϕi‖2 ≤ N(2− ε− 2√

1− ε) ≤ Nε2/4.

Proof. We first check that {S−1/2ϕi}Mi=1 is the closest Parseval frame to{ϕi}Mi=1.

The squared `2-distance between {ϕi}Mi=1 and any Parseval frame {ψj}nj=1

can be expressed in terms of their analysis operators T and T1 as

‖F − G‖2 = Tr[(T − T1)(T − T1)∗]

= Tr[TT ∗] + Tr[T1T∗1 ]− 2<Tr[TT ∗1 ]


Choosing a Parseval frame {ψi}Mi=1 is equivalent to choosing the isometry T1.To minimize the distance over all choices of T1, consider the polar decompo-sition T = UP , where P is positive and U is an isometry. In fact, S = T ∗Timplies P = S1/2, which means its eigenvalues are bounded away from zero.

Since P is positive and bounded away from zero, the term Tr[TT ∗1 ] =Tr[UPT ∗1 ] = Tr[T ∗1UP ] is an inner product between T1 and U . Its magnitudeis bounded by the Cauchy Schwarz inequality, and thus it has a maximal realpart if T1 = U which implies T ∗1U = I. In this case, T = T1P = T1S

1/2,or equivalently T ∗1 = S−1/2T ∗, and we conclude ψi = S−1/2ϕi for all i =1, 2, . . .M .

After choosing T1 = TS−1/2, the `2-distance is expressed in terms of theeigenvalues {λj}Nj=1 of S = T ∗T by

‖F − G‖2 = Tr[S] + Tr[I]− 2Tr[S1/2]

=

N∑j=1

λj +N − 2

N∑j=1

√λj .

If 1− ε ≤ λj ≤ 1 + ε for all j = 1, 2, . . . N , calculus shows that the maximumof λj − 2

√λj is achieved when λj = 1− ε.

Consequently,

‖F − G‖2 ≤ 2N −Nε− 2N√

1− ε.

Estimating√

1− ε by the first three terms in its decreasing power series givesthe inequality ‖F − G‖2 ≤ Nε2/4.

It can be shown that the estimate above is exact and so we have separateverification that the closeness function is a function of N .

There is a simple algorithm for turning any frame into an equal normframe with the same frame operator due to Holmes and Paulsen [34].

Proposition 6. There is an algorithm for turning any frame into an equalnorm frame with the same frame operator.

Proof. Let {ϕi}Mi=1 be a frame for HN with frame operator S and analysisoperator T . Then

M∑i=1

‖ϕi‖2 = Tr S.

Let λ = Tr SM . If ‖ϕi‖2 = λ, for all m = 1, 2, . . . ,M , then we are done.

Otherwise, there exists 1 ≤ i 6= j ≤ M with ‖ϕi‖2 > λ > ‖ϕj‖2. For any θ,replace the vectors ϕi, ϕj by the vectors

ψi = (cosθ)ϕi − (sinθ)ϕj , ψj = (sinθ)ϕi + (cosθ)ϕj , ψk = ϕk, for k 6= i, j.

Now, the analysis operator for {ψi}Mi=1 is T1 = UT for a unitary operator Uon `N2 given by the Givens’ rotation. Hence, T ∗1 T1 = T ∗U∗UT = T ∗T = S; so

24 Peter G. Casazza

the frame operator is unchanged for any value of θ. Now choose the θ yielding‖ψi‖2 = λ. Repeating this process at most M − 1 times yields an equal normframe with the same frame operator as {ϕi}Mi=1.

Using a Parseval frame in Proposition 6, we do get to an equal normParseval frame. The problem, again, is that we do not have any quantitativemeasure of how close these two Parseval frames are.

There is an obvious approach towards solving the Paulsen Problem. Givenan ε-nearly equal norm ε-nearly Parseval frame {ϕi}Mi=1 for HN with frameoperator S, we can switch to the closest Parseval frame {S−1/2ϕi}Mi=1. Thenswitch to the closest equal norm frame to {S−1/2ϕi}Mi=1, call it {ψi}Mi=1 with

frame operator S1. Now switch to {S−1/21 ψi}Mi=1 and again switch to the clos-est equal norm frame and continue. Unfortunately, even if we could show thatthis process converges and we could check the distance traveled through thisprocess, we would still not have an answer to the Paulsen Problem becausethis process does not have to converge to an equal norm Parseval frame. Inparticular, there is a fixed point of this process which is not an equal normParseval frame.

Example 1. Let {ei}Ni=1 be an orthonormal basis for `N2 and let {ϕi}N+1i=1 be

a equiangular unit norm tight frame for `N2 . Then {ei ⊕ 0}Ni=1 ∪ {0⊕ϕi}N+1i=1

in `N2 ⊕ `N2 is a ε = 1N -nearly equal norm and 1

N - nearly Parseval frame with

frame operator, say S. A direct calculation shows that taking S−1/2 of theframe vectors and switching to the closest equal norm frame leaves the frameunchanged.

The Paulsen Problem has proven to be intractable for over 12 years. Re-cently, two partial solutions to the problem were given in [10, 20] each havingsome advantages. Since each of these papers is technical, we will only outlinethe ideas here.

In [10], a new technique is introduced. This is a system of vector valuedODE’s which starts with a given Parseval frame and has the property that allframes in the flow are still Parseval while approaching an equal norm Parsevalframe. They then bound the arc length of the system of ODE’s by the frameenergy. Finally, giving an exponential bound on the frame energy, they havea quantitative estimate for the distance between the initial, ε-nearly equal-norm and ε-nearly Parseval frame F and the equal-norm Parseval frame G.For the method to work, they must assume that the dimension N of theHilbert space and the number M of frame vectors are relatively prime. Theauthors show that in practice, this is not a serious restriction. The main resultof [10] is

Theorem 13. Let N,M ∈ N be relatively prime, let 0 < ε < 12 , and assume

Φ = {ϕi}Mi=1 is an ε-nearly equal-norm and ε-nearly Parseval frame for areal or complex Hilbert space of dimension N . Then there is an equal-normParseval frame Ψ = {ψi}Mi=1 such that


‖Φ− Ψ‖ ≤ 29

8N2M(M − 1)8ε.

In [20], the authors present a new iterative algorithm—gradient descent ofthe frame potential—for increasing the degree of tightness of any finite unitnorm frame. The algorithm itself is trivial to implement, and it preservescertain group structures present in the initial frame. In the special case wherethe number of frame elements is relatively prime to the dimension of theunderlying space, they show that this algorithm converges to a unit normtight frame at a linear rate, provided the initial unit norm frame is alreadysufficiently close to being tight. The main difference between this approachand the approach in [10] is that in [10], the authors start with a nearly equalnorm Parseval frame and improve its closeness to an equal norm frame whilemaintaining Parseval, and in [20] the authors start with an equal norm nearlyParseval frame and give an algorithm for improving its algebraic propertieswhile changing its transform as little as possible. The main result from [20]is:

Theorem 14. Suppose M and N are relatively prime. Pick t ∈ (0, 12M ), and

let Φ0 = {ϕi}Mi=1 be a unit norm frame with analysis operator T0 satisfying‖T ∗0 T0−MN I‖

2HS ≤ 2

N3 . Now, iterate the gradient descent of the frame potentialmethod to obtain Φk. Then, Φ∞ := limk Φk exists and is a unit norm tightframe satisfying

‖Φ∞ − Φ0‖HS ≤ 4N20M8.5

1−2Mt

∥∥T ∗0 T0 − MN I∥∥HS.

In [10], the authors showed there is a connection between the PaulsenProblem and a fundamental open problem in operator theory.

Problem 3 (Projection Problem). Let HN be an N -dimensional Hilbertspace with orthonormal basis {ei}Ni=1. Find the function g(ε,N,M) satisfyingthe following. If P is a projection of rank M on HN satisfying

(1− ε)MN≤ ‖Pei‖2 ≤ (1 + ε)

M

N, for all i = 1, 2, . . . , N,

then there is a projection Q with ‖Qei‖2 = MN for all i = 1, 2, . . . , N satisfying

N∑i=1

‖Pei −Qei‖2 ≤ g(ε,N,M).

In [14], it was shown that the Paulsen problem is equivalent to the Pro-jection Problem and that their closeness functions are within a factor of 2 ofone another. The proof of this result gives several exact connections betweenthe distance between frames and the distance between the ranges of theiranalysis operators.

26 Peter G. Casazza

Theorem 15. Let Φ = {ϕi}i∈I , Ψ = {ψi}i∈I be Parseval frames for a Hilbertspace H with analysis operators T1, T2 respectively. If

d(Φ, Ψ) =∑i∈I‖ϕi − ψi‖2 < ε,

thend(T1(Φ), T2(Ψ)) =

∑i∈I‖T1ϕi − T2ψi‖2 < 4ε.

Proof. Note that for all j ∈ I,

T1ϕj =∑i∈I〈ϕj , ϕi〉ei, and T2ψj =

∑i∈I〈ψj , ψi〉ei.

Hence,

‖T1ϕj − T2ψj‖2 =∑i∈I|〈ϕj , ϕi〉 − 〈ψj , ψi〉|2

=∑i∈I|〈ϕj , ϕi − ψi〉+ 〈ϕj − ψj , ψi〉|2

≤ 2∑i∈I|〈ϕj , ϕi − ψi〉|2 + 2

∑i∈I|〈ϕj − ψj , ψi〉|2.

Summing over j and using the fact that our frames Φ and Ψ are Parsevalgives∑j∈I‖T1ϕj − T2ψj‖2 ≤ 2

∑j∈I

∑i∈I|〈ϕj , ϕi − ψi〉|2 + 2

∑j∈I

∑i∈I|〈ϕj − ψj , ψi〉|2

= 2∑i∈I

∑j∈I|〈ϕj , ϕi − ψi〉|2 + 2

∑j∈I‖ϕj − ψj‖2

= 2∑i∈I‖ϕi − ψi‖2 + 2

∑j∈I‖ϕj − ψj‖2

= 4∑j∈I‖ϕj − ψj‖2.

Next, we want to relate the chordal distance between two subspaces to thedistance between their orthogonal projections. First we need to define thedistance between projections.

Definition 6. If P,Q are projections on HN , we define the distance betweenthem by

d(P,Q) =

M∑i=1

‖Pei −Qei‖2,

where {ei}Ni=1 is an orthonormal basis for HN .


The chordal distance between subspaces of a Hilbert space was defined by[29] and has been shown to have many uses over the years.

Definition 7. Given M -dimensional subspaces W1,W2 of a Hilbert space,define the M -tuple (σ1, σ2, . . . , σM ) as follows:

σ1 = max{〈x, y〉 : x ∈ SpW1, y ∈ SpW2

} = 〈x1, y1〉,

where SpW is the unit sphere of the subspace W . For 2 ≤ i ≤M ,

σi = max{〈x, y〉 : ‖x‖ = ‖y‖ = 1, 〈xj , x〉 = 0 = 〈yj , y〉, for 1 ≤ j ≤ i− 1},

whereσi = 〈xi, yi〉.

The M -tuple (θ1, θ2, . . . , θM ) with θi = cos−1(σi) is called the principleangles between W1,W2. The chordal distance between W1,W2 is given by

d2c(W1,W2) =

M∑i=1

sin2θi.

So by the definition, there exists orthonormal bases {aj}Mj=1, {bj}Mj=1 forW1,W2 respectively satisfying

‖aj − bj‖ = 2sin

(θ

2

), for all j = 1, 2, . . . ,M.

It follows that for 0 ≤ θ ≤ π2 ,

sin2θ ≤ 4sin2(θ

2

)= ‖aj − bj‖2 ≤ 4sin2θ, for all j = 1, 2, . . . ,M.

Hence,

d2c(W1,W2) ≤M∑j=1

‖aj − bj‖2 ≤ 4d2c(W1,W2). (1.5)

We also need the following result [29].

Lemma 4. If HN is an N -dimensional Hilbert space and P,Q are rank Morthogonal projections onto subspaces W1,W2 respectively, then the chordaldistance dc(W1,W2) between the subspaces satisfies

d2c(W1,W2) = M − Tr PQ.

Next we give a precise connection between chordal distance for subspacesand the distance between the projections onto these subspaces. This resultcan be found in [29] in the language of Hilbert-Schmidt norms.

28 Peter G. Casazza

Proposition 7. Let HM be an M -dimensional Hilbert space with orthonor-mal basis {ei}Mi=1. Let P,Q be the orthogonal projections of HM onto N -dimensional subspaces W1,W2 respectively. Then the chordal distance betweenW1,W2 satisfies

d2c(W1,W2) =1

2

M∑i=1

‖Pei −Qei‖2.

In particular, there are orthonormal bases {ei}Ni=1 for W1 and {ei}Ni=1 for W2

satisfying

1

2

M∑i=1

‖Pei −Qei‖2 ≤N∑i=1

‖ei − ei‖2 ≤ 2

N∑i=1

‖Pei −Qei‖2.

Proof. We compute:

M∑i=1

‖Pei −Qei‖2 =

M∑i=1

〈Pei −Qei, P ei −Qei〉

=

M∑i=1

‖Pei‖2 +

M∑i=1

‖Qei‖2 − 2

M∑i=1

〈Pei, Qei〉

= 2N − 2

M∑i=1

〈PQei, ei〉

= 2N − 2Tr PQ

= 2N − 2[N − d2c(W1,W2)]

= 2d2c(W1,W2).

This combined with Equation 1.5 completes the proof.

The next problem to be addressed is to connect the distance betweenprojections and the distance between the corresponding ranges of analysisoperators for Parseval frames.

Theorem 16. Let P,Q be projections of rank N on HM and let {ei}Mi=1 bethe coordinate basis of HM . Further, assume that there is a Parseval frame{ϕi}Mi=1 for HN with analysis operator T satisfying Tϕi = Pei for all i =1, 2, . . . ,M . If

M∑i=1

‖Pei −Qei‖2 < ε,

then there is a Parseval frame {ψi}Mi=1 for HM with analysis operator T1satisfying

T1ψi = Qei, for all i = 1, 2, . . . ,M,

and


M∑i=1

‖ϕi − ψi‖2 < 2ε.

Moreover, if {Qei}Mi=1 is equal norm, then {ψi}Mi=1 may be chosen to be equalnorm.

Proof. By Proposition 7, there are orthonormal bases {aj}Mj=1 and {bj}Mj=1

for W1 = T (HN ), W2 = T1(HN ) respectively satisfying

M∑j=1

‖aj − bj‖2 < 2ε.

Let A,B be the N×M matrices whose jth columns are aj , bj respectively. Letaij , bij be the (i, j) entry of A,B respectively. Finally, let {ϕ′i}Mi=1, {ψ′i}Mi=1 bethe ith rows of A,B respectively. Then we have

M∑i=1

‖ϕ′i − ψ′i‖2 =

M∑i=1

N∑j=1

|aij − bij |2

=

N∑j=1

M∑i=1

|aij − bij |2

=

N∑i=1

‖aj − bj‖2

≤ 2ε.

Since the columns of A form an orthonormal basis for W1, we know that{ϕ′i}Mi=1 is a Parseval frame which is isomorphic to {ϕi}Mi=1. Thus there isa unitary operator U : HM → HM with Uϕ′i = ϕi. Now let {ψi}Mi=1 ={Uψ′i}Mi=1. Then

M∑i=1

‖ϕi − Uψ′i‖2 =

M∑i=1

‖U(ϕ′i)− U(ψ′i)‖2 =

M∑i=1

‖ϕ′i − ψ′i‖2 ≤ 2ε.

Finally, if T1 is the analysis operator for the Parseval frame {ψi}Mi=1, thenT1 is a isometry and since T1ψi = Qei, for all i = 1, 2, . . . , N , if Qei is equalnorm, so is {T1ψi}Mi=1 and hence so is {ψi}Ni=1.

Theorem 17. If g(ε,N,M) is the function for the Paulsen Problem andf(ε,N,M) is the function for the Projection Problem, then

f(ε,N,M) ≤ 4g(ε,N,M) ≤ 8f(ε,N,M).

Proof. First, assume that the Projection Problem holds with function f(ε,N,M).Let {ϕi}Mi=1 be a Parseval frame for HN satisfying

30 Peter G. Casazza

(1− ε)NM≤ ‖ϕi‖2 ≤ (1 + ε)

N

M.

Let T be the analysis operator of {ϕi}Mi=1 and let P be the projection of HMonto range T . So, Tϕi = Pei for all i = 1, 2, . . . ,M . By our assumption thatthe Projection Problem holds, there is a projection Q on HM with constantdiagonal so that

M∑i=1

‖Pei −Qei‖2 ≤ f(ε,N,M).

By Theorem 16, there is a a Parseval frame {ψi}Mi=1 for HN with analysisoperator T1 so that T1ψi = Qei and

M∑i=1

‖ϕi − ψi‖2 ≤ 2f(ε,N,M).

Since T1 is a isometry and {T1ψi}Mi=1 is equal norm, it follows that {ψi}Mi=1

is an equal norm Parseval frame satisfying the Paulsen problem.Conversely, assume the Parseval Paulsen problem has a positive solution

with function g(ε,N,M). Let P be an orthogonal projection onHM satisfying

(1− ε)NM≤ ‖Pei‖2 ≤ (1 + ε)

N

M.

Then {Pei}Mi=1 is a ε-nearly equal norm Parseval frame for HN and by thePaulsen problem, there is an equal norm Parseval frame {ψi}Mi=1 so that

M∑i=1

‖ϕi − ψi‖2 < g(ε,N,M).

Let T1 be the analysis operator of {ψi}Mi=1. Letting Q be the projection ontothe range of T1, we have that Qei = T1ψi, for all i = 1, 2, . . . ,M . By Theorem15, we have that

M∑i=1

‖Pei − T1ψi‖2 =

N∑i=1

‖Pei −Qei‖2 ≤ 4g(ε,N,M).

Since T1 is a isometry and {ψi}Mi=1 is equal norm, it follows that Q is aconstant diagonal projection.

In [14] there are several generalizations of the Paulsen and ProjectionProblems.


1.5 Final Comments

We have concentrated here on some problems in pure mathematics which havefinite dimensional formulations. There are many other infinite dimensionalversions of these problems [26, 27] in sampling theory, harmonic analysis andmore which we have not covered.

Because of the long history of these problems and their connections to somany areas of mathematics, we are naturally led to consider the decidabilityof KS. Since we have finite dimensional versions of the problem, it can bereformulated in the language of pure number theory, and hence it has aproperty logicians call absolutness. As a practical matter, the general feelingis that this means it is very unlikely to be undecidable.

Acknowledgements

The author acknowledges support from NSF DMS 1008183, NSF ATD1042701and AFOSR DGE51: FA9550-11-1-0245.

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