The Feichtinger Problem

Now we show that the solutions to Problems 5.2.1, 5.2.2 and 5.2.3 are negative by providing counterexamples for each of them. We answer negatively Problems 5.2.2 and 5.2.3 by constructing a counterexample for the complex Hilbert space H, in Proposition 5.2.5. This example is then modified to generate an example when the Hilbert space H is over the real field, in Proposition 5.2.7. From there, we answer the Feichtinger original problem in Theorem 5.2.8.

Proposition 5.2.5. Let H=ℓ²({1,2, ...}), and choose p > 1. Let {w_ℓ}^∞ℓ=1 denote the standard orthonormal basis of H, i.e., w_ℓ = δ_ℓ. Then H¹ = ℓ¹({1,2, ...}). For each n≥1, let {e_n,k}ⁿk=0⁻¹ be the Fourier ONB of Cⁿ defined by

e_n,k= 1

√n

e^−2πikln n−1 l=0 = 1

√n

1, e^−2πikn , e^−4πikn , ..., e^−2πik(n

−1) n

T

, and consider the n×n matrix T_n given by

T_n=

n−1

X

k=0

λ_n,k(e_n,k⊗e_n,k) = 1 n³

n−1

X

k=0

1 + k

n^p 2

(e_n,k⊗e_n,k)∈C^n×n

where λ_n,k = ¹

n³

1 + _n^kp

2

. We define an infinite block-diagonal matrix T by T = T₁⊕T₂⊕...⊕T_n⊕...Then,T is a positive semi-definite trace-class operator of Type B but not of Type A with respect to the orthonormal basis {w_ℓ}.

Proof. By construction, the blocks T_n that make up T are pairwise orthogonal.

Furthermore, for eachn≥1, the spectrum of T_nconsists of simple eigenvectors e_n,k with corresponding eigenvalues λ_n,k for k = 0, . . . , n−1. Consequently, for each n≥1, and each k∈ {0, . . . , n−1},e_n,k generates a one-dimensional eigenspace of T corresponding to the eigenvalue λ_n,k. It is clear that T is positive semi-definite.

Since,ke_n,kk2 = 1 and T =⊕^∞n=1 n−1

X

k=0

λ_n,k(e_n,k⊗e_n,k),we see that

kTkop≤ X∞ n=1

n−1

X

k=0

1 n³

1 + k

n^p 2

ke_n,k⊗e_n,kkop

= X∞ n=1

n−1

X

k=0

1 n³

1 + k n^p

2

ke_n,kk²= X∞ n=1

n−1

X

k=0

1 n³

1 + k n^p

2

<∞.

5.2. The Feichtinger Problem 109

Furthermore,

kTk_I1 = trace(T) = X∞ n=1

n−1

X

k=0

1 n³

1 + k n^p

2

= X∞ n=1

1 n³

n+n(n−1)

n^p +n(n−1)(2n−1) 6n^2p

< ∞, since p >1.

HenceT is a well-defined trace-class operator on H. We now show thatT is of Type B. To this end we observe that for eachn≥1,ⁿP⁻¹

k=0

e_n,k⊗e_n,k =I_n, whereI_ndenotes the identity of ordern. Then

T_n = 1 n³

nX−1 k=0

1 + k

n^p 2

(e_n,k⊗e_n,k)

= 1

n³

nX−1 k=0

(e_n,k⊗e_n,k) + 1 n^3+p

n−1

X

k=0

2k+k² n^p

(e_n,k⊗e_n,k)

= 1

n³I_n+ 1 n^3+p

n−1

X

k=0

2k+ k² n^p

(e_n,k⊗e_n,k).

ThusT can be written as T = ⊕n≥1T_n=⊕n≥1

1

n³I_n+ 1 n^3+p

nX−1 k=0

2k+k² n^p

(e_n,k⊗e_n,k)

= ⊕n≥1

1 n³I_n

+⊕n≥1

1 n^3+p

nX−1 k=0

2k+k² n^p

(e_n,k⊗e_n,k)

= ⊕n≥1

1 n³

Xn k=1

(wn(n−1)

2 +k⊗wn(n−1) 2 +k) +⊕n≥1

1 n^3+p

n−1

X

k=0

2k+k² n^p

(e_n,k⊗e_n,k).

Then we have

wn(n−1)

2 +k

= 1, e_n,k=√ n, and

X

n≥1

1 n³ ·

Xn k=1

1 +X

n≥1

1 n^3+p

n−1

X

k=0

2k+ k² n^p

·(√ n)²

= X

n≥1

1 n² +X

n≥1

1 n^2+p

2·n(n−1)

2 + 1

n^p ·n(n−1)(2n−1) 6

= X

n≥1

1

n² +n−1

n^1+p +(n−1)(2n−1) 6n^2+2p

<∞, for any p >1.

Hence, T is of Type B with respect to {w_ℓ}ℓ≥1. We now show that T is not of Type A with respect to {w_ℓ}ℓ. The key point is that T has only one-dimensional eigenspaces, so

X∞ n=1

nX−1 k=0

λ_n,k(e_n,k⊗e_n,k) = 1 n³

nX−1 k=0

1 + k

n^p 2

(e_n,k⊗e_n,k)

is the unique decomposition of T using sum of rank one projections generated by orthogonal eigenfunctions ofT. Note that e_n,k= √¹

n ·n=√ n, and

λ_n,ke_n,k= 1 n³

1 + k

n^p 2

·√

n <∞. However,

X∞ n=1

n−1

X

k=0

λ_n,ke_n,k² = X∞ n=1

1 n²

n−1

X

k=0

1 + k n^p

2

= X∞ n=1

1 n²

n+ n(n−1)

n^p +n(n−1)(2n−1) 6n^2p

≥ X∞ n=1

1 n =∞.

We can modify the counter-example in Proposition 5.2.5 to deal with the case of a real Hilbert space H. This amount to use a real-valued ONB for Rⁿ instead of the ONB{e_n,k}ⁿk=0⁻¹.

For a fix n≥1, let S_n:= n

2πk

n : 0≤k≤n−1o

. It is easy to see that for any

5.2. The Feichtinger Problem 111

0≤l≤n−1, we have S_n=

2πkl

n (mod 2π) : 0≤k≤n−1

=

−2πk

n (mod 2π) : 0≤k≤n−1

.

Now for eachn≥1, let{h_n,k}ⁿk=0⁻¹ denote the Hartley ONB basis forRⁿ (see [139]), where

h_n,k = 1

√n

cos 2πkl

n

+ sin 2πkl

n

n−1 l=0

= r2

n

cos 2πkl

n − π 4

n−1 l=0

.

Lemma 5.2.6. For a fixed n≥1 and any 0≤k≤n−1 we have h_n,k=

n−1

X

l=0

cos

2πkl n

+ sin

2πkl n

≥ n

√2.

Proof. Let E=P

x∈Sn|cosx+ sinx|.Then

2E = X

x∈S_n

|cosx+ sinx|+ X

−x∈S_n

|cosx+ sinx|

= √

2

n−1

X

k=0

cos

2πk n −π

4

+√ 2

n−1

X

k=0

cos

2πk n +π

4

= √

2

n−1

X

k=0

cos 2πk

n −π 4

+ sin

2πk n −π

4

. (5.9)

Now for anyx∈R,

(|sinx|+|cosx|)² =|sinx|²+|cosx|²+ 2|sinxcosx|= 1 +|sin 2x| ≥1,

⇒ |sinx|+|cosx| ≥1.

Therefore by (5.9) we get 2E ≥√ 2

nP−1 k=0

1 =√

2n⇒E≥ ^√n₂.

Proposition 5.2.7. Let H=ℓ²({1,2, ...}), and choose p > 1. Let {w_ℓ}^∞ℓ=1 denote the standard orthonormal basis of H, i.e., w_ℓ = δ_ℓ. For each n ≥ 1 let T_n denote

the n×n matrix given by

T_n=A_n

n−1

X

k=0

1 + k

n^p 2

(h_n,k⊗h_n,k)∈R^n×n,

where A_n =n⁻³. We define an infinite block-diagonal matrix T by T =T₁⊕T₂⊕ ...⊕T_n⊕... Then, T is a positive semi-definite trace-class operator of Type B but not of Type A with respect to the orthonormal basis {w_ℓ}ℓ≥1.

Proof. The proof is identical to that of Proposition 5.2.5. But for the sake of com- pleteness we provide the details.

By construction, the blocks T_n that make up T are pairwise orthogonal. Fur- thermore, for each n ≥ 1, the spectrum of T_n consists of simple eigenvectors h_n,k with corresponding eigenvalues λ_n,k = ¹

n³(1 +_n^kp)² fork= 0, . . . , n−1. Thus T is positive semi-definite. Consequently, for each n ≥ 1, and each k ∈ {0, . . . , n−1}, h_n,k generates a one-dimensional eigenspace of T corresponding to the eigenvalue λ_n,k. Since,kh_n,kk2 = 1 and T =⊕^∞n=1

Pn−1

k=0λ_n,k(e_n,k⊗e_n,k),we see that kTk_I1 = trace(T) =

X∞ n=1

n−1

X

k=0

A_n

1 + k n^p

2

= X∞ n=1

A_n

n+n(n−1)

n^p +n(n−1)(2n−1) 6n^2p

.

The choice A_n= 1/n³ guarantees that the series is convergent in trace-class norm i.e., kTk_I1 <∞. Thus T is a positive semi-definite trace-class operator onH.

We now show that T is not of Type A. The key point is that T has only one-dimensional eigenspaces, so

X∞ n=1

n−1

X

k=0

g_n,k⊗g_n,k,whereg_n,k=√ A_n

1 + ^k

n^p

h_n,k is the unique decomposition of T using sum of rank one projections generated by orthogonal eigenfunctions ofT. Then

g_n,k = p A_n

1 + k

n^p h_n,k

= p

A_n

1 + k n^p

· 1

√n

n−1

X

l=0

cos

2πkl n

+ sin

2πkl n

5.2. The Feichtinger Problem 113

≥ p A_n

1 + k

n^p

· 1

√n

√n 2 =

s n 2 ·A_n

1 + k

n^p 2

and X∞ n=1

nX−1 k=0

g_n,k² ≥ X∞ n=1

n 2 ·A_n

nX−1 k=0

1 + k

n^p 2

= X∞ n=1

n 2 ·A_n

n+n(n−1)

n^p +n(n−1)(2n−1) 6n^2p

≥ 1 2

X∞ n=1

n²A_n = X∞ n=1

1 2n =∞. So T is not of Type A.

We now show that T is of TypeB. Let λ_k =√ A_n

1 + ^k

n^p

and let U be the unitary operator (matrix) given byU = [h_n,0 h_n,1... h_n,n−1].It follows that

T_n = [λ₀h_n,0 λ₁h_n,1 ... λ_n−1h_n,n−1] ·(U^∗U)· [λ₀h^∗_n,0 λ₁h^∗_n,1 ... λ_n−1h^∗_n,n−1]

= [λ₀h_n,0 λ₁h_n,1 ... λ_n−1h_n,n−1]U^∗· [λ₀h_n,0 λ₁h_n,1 ... λ_n−1h_n,n−1]U^∗_∗

= G_n·G^∗_n,

whereG_n= [λ₀h_n,0 λ₁h_n,1 ... λ_n−1h_n,n−1]U^∗. Now we compute G_n G_n = [λ₀h_n,0 λ₁h_n,1 ... λ_n−1h_n,n−1]·[h^∗_n,0h^∗_n,1... h^∗_n,n−1]

= λ₀h_n,0h^∗_n,0+λ₁h_n,1h^∗_n,1+...+λ_n−1h_n,n−1h^∗_n,n−1

= p

A_n+λ₀−p A_n

h_n,0h^∗_n,0+p

A_n+λ₁−p A_n

h_n,1h^∗_n,1+...

+p

A_n+λ_n−1−p A_n

h_n,n−1h^∗_n,n−1

= p

A_n h_n,0h^∗_n,0+h_n,1h^∗_n,1+...+h_n,n−1h^∗_n,n−1 +

λ₀−p A_n

h_n,0h^∗_n,0 +

λ₁−p A_n

h_n,1h^∗_n,1+...+

λ_n−1−p A_n

h_n,n−1h^∗_n,n−1

= p

A_nI_n+ (λ₁−λ₀)h_n,1h^∗_n,1+...+ (λ_n−1−λ₀)h_n,n−1h^∗_n,n−1, whereI_nis the identity matrix of order n and √

A_n=λ₀. Let Λ_n=λ_n−λ₀ then

T_n = G_n·G^∗_n

= p

A_nI_n+ Λ₁h_n,1h^∗_n,1+...+ Λ_n−1h_n,n−1h^∗_n,n−1 p ·

A_nI_n+ Λ₁h_n,1h^∗_n,1+...+ Λ_n−1h_n,n−1h^∗_n,n−1

= A_nI_n+ 2Λ₁p

A_nh_n,1h^∗_n,1+ 2Λ₂p

A_nh_n,2h^∗_n,2+...+ 2Λ_n−1p

A_nh_n,n−1h^∗_n,n−1 +Λ²₁h_n,1h^∗_n,1+ Λ²₂h_n,2h^∗_n,2+...+ Λ²_n−1h_n,n−1h^∗_n,n−1

= A_nI_n+ 2Λ₁p

A_n+ Λ²₁

h_n,1h^∗_n,1+...+

2Λ_n−1p

A_n+ Λ²_n−1

h_n,n−1h^∗_n,n−1

= A_n δ₀δ₀^∗+δ₁δ^∗₁+..+δ_n−1δ_n^∗₋₁ +

2Λ₁p

A_n+ Λ²₁

h_n,1h^∗_n,1+...

+

2Λ_n−1p

A_n+ Λ²_n−1

h_n,n−1h^∗_n,n−1

= f_n,0f_n,0^∗ +...+f_n,n−1f_n,n^∗ ₋₁+f_n,nf_n,n^∗ +...+f_n,2n−2f_n,2n^∗ ₋₂, wheref_n,k=√

A_nδ_k, for 0≤k≤n−1 andf_n,k= q

2Λ_k−n+1√

A_n+ Λ²_k−n+1h_n,k−n+1, forn≤k≤2n−2. Now we compute |||·|||-norm of f_n,k

f_n,k =







√A_n|||δ_k|||, 0≤k≤n−1 q

2Λ_k−n+1√

A_n+ Λ²_k−n+1h_n,k−n+1, n≤k≤2n−2

=







√A_n, 0≤k≤n−1

q

2Λ_k−n+1√

A_n+ Λ²_k−n+1h_n,k−n+1, n≤k≤2n−2.

Since Λ_k=λ_k−λ₀ =√ A_n·

1 + ^k

n^p

−√

A_n=√

A_n·_n^kp, then q

2Λ_k−n+1p

A_n+ Λ²_k−n+1 = sp

A_n·k−n+ 1 n^p

2p

A_n+p

A_n·k−n+ 1 n^p

= p

A_n· s

k−n+ 1 n^p

2 + k−n+ 1 n^p

≤ p A_n·

r3n n^p. Also h_n,k−n+1≤√

2n. Therefore,

5.2. The Feichtinger Problem 115 q

2Λ_k−n+1p

A_n+ Λ²_k−n+1h_n,k−n+1

≤ p A_n·

r3n n^p ·√

2n=p 6A_n·

s n² n^p =

r 6

n^1+p, sinceA_n= 1 n³. Now

X∞ n=1

2nX−2 k=0

f_n,k² ≤ X∞ n=1

(2n−1)· 6

n^1+p ≤12 X∞ n=1

1

n^p <∞, sincep >1,

and T =⊕^∞n=1

²ⁿX⁻²

k=0

f_n,k⊗f_n,k

.So T is of TypeB.

We can now give an answer to Feichtinger’s question, i.e., Problem 5.2.2.

Theorem 5.2.8. Suppose that{w_n}n≥1 is a Wilson basis forL²(R)withg∈M¹(R).

Let p ≥ 2, and for each n ≥ 1 set λ_n,k = ¹

n³(1 + _n^kp)². For fix n ≥ 1 and any 0≤k≤n−1, let h_n,k ∈L²(R) where

h_n,k = 1

√n

n−1

X

l=0

cos

2πkl n

+ sin

2πkl n

wn(n−1) 2 +l+1.

Let T be an operator define by T = X∞ n=1

n−1

X

k=0

λ_n,kh_n,k⊗h_n,k.The following holds:

(i) {h_n,k : 0≤k≤n−1, n≥1} is an orthonormal basis for L²(R).

(ii) T is a positive semi-definite trace-class operator on L²(R) that provides a counter-example to Problem 5.2.2.

Proof. (i) It is easy to see that for each n ≥ 1, {h_n,k}ⁿk=0⁻¹ is an orthogonal set in L²(R). Also, hh_n,k, h_n^′_,k^′i = 0, for n 6= n^′. They are normalized in L²(R), since hw_n, w_mi=δ_n,m.

(ii) It is also easy to see that T is a well defined operator on L²(R). In fact, the series defining T converges in the operator norm. Furthermore, since kh_n,k⊗

h_n,kk_I1 = 1, it follows that

kTk_I1 = X∞ n=1

n−1

X

k=0

λ_n,k= X∞ n=1

1 n³

n−1

X

k=0

1 + k

n^p 2

= X∞ n=1

1 n³

n+n(n−1)

n^p +n(n−1)(2n−1) 6n^2p

<∞.

Consequently, T is a trace-class operator. By Lemma 5.2.6, kh_n,kk_M¹ =

X∞ m=1

|hh_n,k, w_mi|

= 1

√n X∞ m=1

*_n−1 X

l=0

cos

2πkl n

+ sin

2πkl n

wn(n−1)

2 +l, w_m+

= 1

√n

n−1

X

l=0

cos

2πkl n

+ sin

2πkl n

≥ 1

√n · n

√2 = rn

2. Also each term

λ_n,kkh_n,kk_M¹ = 1

n³(1 + k n^p)²· 1

√n

n−1

X

l=0

cos

2πkl n

+ sin

2πkl n

≤ 1

n³(1 + k

n^p)²·2√

n <∞, but

X∞ n=1

n−1

X

k=0

λ_n,kkh_n,kk²_M¹ ≥ X∞ n=1

1 2n²

n−1

X

k=0

(1 + k n^p)²

= X∞ n=1

1 2n²

n+ n(n−1)

n^p +n(n−1)(2n−1) 6n^2p

≥ X∞ n=1

1 2n =∞.

bac

Chapter 6

Balian-Low Type Theorems on L ² ( C )

In this chapter, we prove the Balian-Low type theorem (BLT) on L²(C) using the operators Z and ¯Z i.e., we prove that kZgk2 and kZg¯ k2 cannot both be si- multaneously finite if the twisted Gabor frame generated by g ∈ L²(C) forms an orthonormal basis or an exact frame for L²(C). The operators

Z = d dz +1

2z¯ and Z¯= d d¯z −1

2z are associated with the special Hermite operator

L=−∆_z+ 1

4|z|²−i

x d dy −y d

dx

onC, where ∆_z is the standard Laplacian onCandz=x+iy. Also we present the amalgam version of BLT using Weyl transform and illustrate the distinction between BLT and amalgam BLT by examples. We introduce the twisted Zak transform and using it we establish several versions of the Balian-Low type theorems on L²(C).