The Distributions of Hilbert Space Frame Vectors and Frame Coefficients

arXiv:1504.03722v1 [math.FA] 14 Apr 2015

THE DISTRIBUTIONS OF HILBERT SPACE FRAME VECTORS AND FRAME COEFFICIENTS KEVIN BREWSTER, PETER G. CASAZZA, ERIC PINKHAM AND LINDSEY WOODLAND Abstract. The most fundamental notion for Hilbert space frames is the sequence of frame coefficients for a vector x in the space. Yet, we know little about the distribution of these coefficient sequences. In this paper, we make the first detailed study of the distribution of the frame coefficients for vectors in a Hilbert space, as well as the related notion of the square sums of the distances of vectors in the space from the frame vectors. We will give some surprisingly exact calculations for special frames such as unit norm tight frames and equiangular frames. This is a study which should have been done 20 years ago.

1. Introduction The most fundamental notion for a Hilbert space frame Φ = {ϕi }N i=1 for HM is the sequence of frame coefficients for a vector x, namely {hx, ϕi i}N i=1 . Yet, we know little about the distribution of these coefficients, even for very specific frames. In this paper, we will make the first detailed study of the distribution of the frame coefficients for general frames and then strengthen the results for several special classes of frames including unit norm tight frames and equiangular frames. For this we use the concept of majorization. We will also study the distributions of products of frame coefficients for different vectors x and y. We will then make a detailed study of the square sums of the distances from a vector x to the frame vectors and discover that in quite general cases, these sums are nearly equal for all vectors. 2. Preliminaries While the following work is largely self contained, an interested reader can find a more detailed introduction to frame theory in [2, 5, 6]. Given M ∈ N, HM represents a (real or complex) Hilbert space of (finite) dimension M . Definition 2.1. A family of vectors {ϕi }N i=1 in an M -dimensional Hilbert space HM is a frame if there are constants 0 < A ≤ B < ∞ so that for all 1991 Mathematics Subject Classification. 42C15. Key words and phrases. Frame Coefficients, Tight Frames, Equiangluar Tight Frames. The authors were supported by NSF DMS 1307685; NSF ATD 1321779; and AFOSR DGE51: FA9550-11-1-0245. 1

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BREWSTER, CASAZZA, PINKHAM AND WOODLAND

x ∈ HM , Akxk2 ≤

N X i=1

|hx, ϕi i|2 ≤ Bkxk2 ,

where A and B are lower and upper frame bounds, respectively. The largest A and smallest B satisfying these inequalities are called the optimal frame bounds. (i) If A = B is possible, then {ϕi }N i=1 is a A-tight frame. If A = B = 1 is possible, then {ϕi }N is a Parseval frame. i=1 N (ii) If kϕi k = 1 for all i ∈ [N ] then {ϕi }i=1 is an unit norm frame. (iii) {hx, ϕi i}N i=1 are called the frame coefficients of the vector x ∈ HM with respect to frame {ϕi }N i=1 . (iv) If the frame is unit norm and there is a constant d so that |hϕi , ϕj i| = d for all 1 ≤ i 6= j ≤ N , then {ϕi }N i=1 is an equiangular frame. (v) The frame operator S of the frame is given by N X hx, ϕi iϕi . Sx = i=1

The frame operator is a particularly important object in the study of frame theory. Here we present an important result result about this operator. The proof can be found in [2]. Theorem 2.2. Let {ϕi }N i=1 be a frame for HM with frame bounds A and B and frame operator S. S is positive, invertible, and self adjoint. Moreover, the optimal frame bounds of {ϕi }N i=1 are given by A = λmin (S) and B = λmax (S) (the maximum and minimum eigenvalues of S). Now we make a simple observation about the frame bound of a unit norm tight frame. Proposition 2.3. If {ϕi }N i=1 is a unit norm tight frame for HM then the frame bound will be A = N/M . 3. Estimating the number of non-zero frame coefficients In this section we will provide estimates on the number of indices for which the frame coefficients are non-zero. We will be looking for vectors producing the minimal number of non-zero frame coefficients, since there is always a dense set of vectors which have non-zero inner products with all ⊥ the frame vectors. For example, if {ϕi }N i=1 is a frame for HM , then ϕi is a hyperplane for every i ∈ [N ]. If we choose ⊥ x∈ / ∪N i=1 ϕi ,

we have that |{i ∈ [N ] : hx, ϕi i 6= 0}| = N.

THE DISTRIBUTION OF FRAME VECTORS AND COEFFICIENTS

3

In general, given a vector x there may be only a few indices for which hx, ϕi i = 6 0. For example, given K copies of an orthonormal basis {ei }M i=1 for HM , say {eij }1≤i≤M, 1≤j≤K , and choosing x = e1 gives hx, eij i = 0,

∀ i ∈ {2, . . . , M }, j ∈ {1, . . . , K}.

That is, we have only K non-zero coefficients out of a total of KM . In other words, there are KM/M non-zero coefficients. This is minimal with respect to this property as we demonstrate shortly. Theorem 3.1. Let Φ = {ϕi }N i=1 be a frame in HM with frame bounds A, B 2 and set D := max{kϕi k : i ∈ [N ]}. For any unit norm x ∈ HM define Jx := {i ∈ [N ] : hx, ϕi i 6= 0}. Then

A . D So if Φ is a unit norm frame, then |Jx | ≥ A and if it is a unit norm tight frame then |Jx | ≥ N/M . Moreover, if we have equality in (1) then the sub-collection of frame vectors {ϕi : i ∈ Jx } spans a one-dimensional space. |Jx | ≥

(1)

Proof. Pick a unit norm x ∈ HM and set Jx := {1 ≤ i ≤ N : hx, ϕi i = 6 0}. Then A = Akxk2 ≤ (2)

≤

X

i∈Jx

N X i=1

|hx, ϕi i|2 =

kxk2 kϕi k2 =

X

i∈Jx

X

i∈Jx

|hx, ϕi i|2

kϕi k2 ≤ D|Jx |.

Hence, A/D is a lower bound for |Jx | independent of each unit norm x ∈ HM . It follows that (1) holds true. Concerning the moreover part, if we have equality in (2), then |hx, ϕi i|2 = kϕi k2 , for all i ∈ Jx . It follows that ϕi = ci x, for some |ci | = 1, and all i ∈ Jx .

Note that the example preceding the theorem satisfies the minimal number of nonzero frame coefficients in the theorem. We may be interested in not only knowing when coefficients are nonzero, but when they are bounded away from zero as well. We provide lower bounds in the following theorem which can be viewed as a generalization of the previous. Theorem 3.2. Let C ∈ (0, 1), let Φ = {ϕi }N i=1 be a frame for HM with frame bounds A, B and set D := max{kϕi k2 : i ∈ [N ]}. For a unit norm x ∈ HM , define Kx := {i ∈ [N ] : |hx, ϕi i|2 > (CA)/N }. We have (3)

|Kx | ≥ (1 − C)

A . D

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In particular, if Φ is a unit norm tight frame, then |Kx | ≥ (1 − C) Proof. For unit norm x ∈ HM , we have X CA c |hx, ϕi i|2 ≤ |Kx | and N c

N . M X

i∈Kx

i∈Kx

|hx, ϕi i|2 ≤ D|Kx |.

Hence, 2

A = Akxk ≤

N X i=1

|hx, ϕi i|2 =

X

i∈Kx

|hx, ϕi i|2 +

X

i∈Kxc

|hx, ϕi i|2

CA c CA |Kx | = D|Kx | + (N − |Kx |) N N CA |Kx | + CA. = D− N It follows that CA A(1 − C) ≤ D − |Kx |. N By the definition of Kx and the Cauchy-Schwarz inequality, we may deduce that D > (CA)/N . Consequently ≤ D|Kx | +

|Kx | ≥ (1 − C)

(4)

A A ≥ (1 − C) . D − (CA)/N D

Since the lower bound in (4) is independent of the unit norm x ∈ HM , the validity of (3) follows. 4. The Distribution of the Frame Coefficients In this section we will classify the distribution of the frame coefficients using majorization. Majorization has an interesting theory in itself of which we will need only a small portion. The interested reader should refer to [3] for a detailed study of majorization inequalities and their applications. N Definition 4.1. Let a = (ai )N i=1 and b = (bi )i=1 be vectors with nonnegative, non-increasing components. (i) We say a majorizes b and write a ≻ b if k X i=1

ai ≥

k X i=1

bi , ∀ k ∈ {1, 2, . . . , N }

and

N X

ai =

i=1

(ii) We say a weakly majorizes b and write a ≻W b if k X i=1

ai ≥

k X i=1

bi , ∀ k ∈ {1, 2, . . . , N }.

N X i=1

bi .

THE DISTRIBUTION OF FRAME VECTORS AND COEFFICIENTS

5

If the two vectors do not have the same number of terms we agree to pad the shorter vector with zeros so that they have the same length. It will be important for our work to understand what happens when partial sums of the majorization vectors are equal. N Proposition 4.2. Let a := (ai )N i=1 ∈ R be a non-negative, non-increasing sequence of numbers, let b := (A, A, . . . , A) ∈ RN where A > 0, and assume m P a ≻W b. If there exists m ∈ {1, . . . , N } so that ai = mA, then ai = A i=1

for every i ∈ {m + 1, . . . , N }.

Proof. Assume there exists m ∈ {1, . . . , N } so that

m P

ai = mA. Then

i=1

am ≤ A and, hence, ai ≤ A for every i ∈ {m + 1, . . . , N }. Assume for the moment that ai < A for some i ∈ {m + 1, . . . , N }, then NA =

N X i=1

A≤

N X

ai =

m X

ai +

ai

i=m+1

i=1

i=1

N X

< mA + (N − m)A = N A,

which is a contradiction. Thus, ai = A for every i ∈ {m + 1, . . . , N }.

The following proposition establishes a relationship between the largest eigenvalue of the frame operator and the number of frame coefficients of modulus one. Proposition 4.3. Let {ϕi }N i=1 be a unit norm frame in HM whose frame operator has eigenvalues λ1 ≥ · · · ≥ λM and let x ∈ HM be unit norm. Define N 2 N the vector a := (ai )N i=1 ∈ R to be (|hx, ϕi i| )i=1 arranged in non-increasing order. Then ai = 1 for at most ⌊λ1 ⌋ indices i (where ⌊ · ⌋ represents the least integer function). Proof. By definition,

N P

i=1

|hx, ϕi i|2 ≤ λ1 kxk2 = λ1 . Suppose for the moment

that ai = 1 for every i ∈ {1, . . . , ⌊λ1 ⌋ + 1}, then λ1 ≥

N X i=1

|hx, ϕi i|2 =

= (⌊λ1 ⌋ + 1) + which is a contradiction.

⌊λ1 ⌋+1

X i=1

N X

i=⌊λ1 ⌋+2

|hx, ϕi i|2 +

N X

i=⌊λ1 ⌋+2

|hx, ϕi i|2

|hx, ϕi i|2 > λ1 ,

We are ready to give one of the main majorization results for frame coefficients of unit norm vectors.

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Theorem 4.4. Let {ϕi }N i=1 be a frame for HM with frame bounds A ≤ B and let A A A b := , ,..., ∈ RN . N N N N to be (|hx, ϕ i|2 )N For any unit norm x ∈ HM , define a := (ai )N i i=1 ∈ R i=1 arranged in non-increasing order. Then a ≻W b. In particular, a1 ≥ A/N . m P Moreover, if there exists an m ∈ {1, . . . N } such that ai = m(A/N ) i=1

then ai = A/N for all i ∈ {m + 1, . . . , N }.

Proof. We proceed by way of contradiction. If there is an m ∈ {1, . . . , N } so that m m X X A A =m , ai < N N i=1

i=1

then am < A/N . Furthermore, it follows that

ai ≤ am , ∀ i ∈ {m, . . . , N },

and we may write for any unit vector x ∈ HM A = Akxk2 ≤

N X i=1

ai =

m X i=1

ai +

N X

ai

i=m+1

A A A + (N − m)am < m + (N − m) N N N = A,