Abstract

Almost orthogonal frames have been introduced and studied. It has been proved that a bounded almost orthogonal frame satisfies Feichtinger conjecture. Also, we prove that a bounded almost orthogonal frame contains a Riesz basis.

1. Introduction

Frames were formally introduced in 1952 by Duffin and Schaeffer [1]. In 1985, frames were resurfaced in the book by Young [2]. The theory of frames began to be more widely studied only after the landmark paper of Daubechies et al. [3] in 1986. For an introduction to frames, one may refer to [4–6].

Feichtinger in his work on time frequency analysis noted that all Gabor frames (which he was using for his work) had the property that they could be divided into a finite number of subsets which were Riesz basis sequences. This observation led to the following conjecture, called the Feichtinger conjecture “Every bounded frame can be written as a finite union of Riesz basic sequences.”

Feichtinger conjecture is connected to the famous Kadison-Singer conjecture. It was shown in [7] that Kadison-Singer conjecture implies Feichtinger conjecture. For literature related to Feichtinger conjecture, one may refer to [7, 8].

In the present paper, we introduce and study almost orthogonal frames in Hilbert spaces and prove that a bounded almost orthogonal frame satisfies Feichtinger conjecture. Also, we prove that a bounded almost orthogonal frame contains a Riesz basis.

2. Preliminaries

Throughout the paper, 𝐻 will denote an infinite-dimensional Hilbert space, {𝑛𝑘} an infinite-increasing sequence in ℕ, [𝑥𝑛] the closed linear span of {𝑥𝑛}, and for any set 𝐷, |𝐷| will denote cardinality of 𝐷.

Definition 2.1. A sequence {𝑥𝑛} in a Hilbert space 𝐻 is said to be a frame for 𝐻 if there exist constants 𝐴 and 𝐵 with 0<𝐴≤𝐵<∞ such that 𝐴‖𝑥‖2≤𝑛||⟨𝑥,𝑥𝑛⟩||2≤𝐵‖𝑥‖2,𝑥∈𝐻.(2.1)
The positive constants 𝐴 and 𝐵, respectively, are called lower and upper frame bounds for the frame {𝑥𝑛}. The inequality (2.1) is called the frame inequality for the frame {𝑥𝑛}.
A frame {𝑥𝑛} in 𝐻 is called tight if it is possible to choose 𝐴,𝐵 satisfying inequality (2.1) with 𝐴=𝐵 as frame bounds and is called normalized tight if 𝐴=𝐵=1. A frame {𝑥𝑛} in 𝐻 is called exact if removal of any 𝑥𝑛 renders the collection {𝑥𝑛} no longer a frame for 𝐻. A sequence {𝑥𝑛}∈𝐻 is called a Bessel sequence if it satisfies upper frame inequality in (2.1).

Definition 2.2. A sequence {𝑥𝑛} in 𝐻 is called a Riesz basic sequence if there exist positive constants 𝐴 and 𝐵 such that for all finite sequence of scalars {𝛼𝑘}, we have 𝐴𝑘||𝛼𝑘||2≤‖‖‖‖𝑘𝛼𝑘𝑥𝑘‖‖‖‖2≤𝐵𝑘||𝛼𝑘||2.(2.2)
In case, the Riesz basic sequence {𝑥𝑛} is complete in 𝐻, it is called a Riesz basis for 𝐻.

Definition 2.3. A sequence {𝑦𝑛} in a Hilbert space 𝐻 is said to be a block sequence with respect to a given sequence {𝑥𝑛} in 𝐻, if it is of the form 𝑦𝑛=𝑖∈𝐷𝑛𝛼𝑖𝑥𝑖≠0,𝑛∈ℕ,(2.3) where 𝐷𝑛's are finite subsets of ℕ with 𝐷𝑛∩𝐷𝑚=∅, 𝑛≠𝑚, ⋃𝑛∈ℕ𝐷𝑛=ℕ and 𝛼𝑖's are any scalars.

It has been observed in [9] that a block sequence with respect to a frame in a Hilbert space may not be a frame for 𝐻. Also, a block sequence with respect to a sequence in 𝐻 which is not even a frame for 𝐻 may be a frame for 𝐻.

3. Main Results

We begin with a sufficient condition for a bounded frame to satisfy the Feichtinger conjecture.

Theorem 3.1. Let {𝑥𝑛} be a bounded frame for 𝐻. If there exists a sequence of finite subsets {𝐷𝑛}𝑛∈ℕ of ℕ with 𝐷𝑖∩𝐷𝑗=∅, for all 𝑖≠𝑗, ⋃∞𝑖=1𝐷𝑖=ℕ and sup𝑛{|𝐷𝑛|}<∞ such that ⨁𝐻=𝑛∈ℕ𝑉𝑛, where 𝑉𝑛=[𝑥𝑛]𝑖∈𝐷𝑛, then {𝑥𝑛} can be decomposed into a finite union of a Riesz basic sequences.

Proof. Suppose the problem has an affirmative answer. Let {𝐷𝑛}𝑛∈ℕ be sequence of finite subsets of ℕ with 𝐷𝑛∩𝐷𝑛=∅, 𝑛≠𝑚 and ⋃𝑛∈ℕ𝐷𝑛=ℕ such that ⨁𝐻=𝑛∈ℕ𝑉𝑛, where 𝑉𝑛=[𝑥𝑖]𝑖∈𝐷𝑛 and {𝑥𝑛} is a bounded frame for 𝐻. Let {𝐺𝑛} be a sequence of sets given by 𝐺𝑛=𝑥𝑖𝑖∈𝐷𝑛,∀𝑛∈ℕ.(3.1) Now, for each 𝑗∈ℕ, choose a sequence {𝑦𝑗𝑖}𝑖∈ℕ such that 𝑦𝑗𝑖=𝑗thelementof𝐺𝑖,if𝐺𝑖contains𝑗thelement,∅,otherwise.(3.2) Then, for each 𝑗∈ℕ, {𝑦𝑗𝑖}𝑖∈ℕ is a sequence of orthogonal vectors which are norm bounded. So, {𝑦𝑗𝑖}𝑖∈ℕ is a Riesz basic sequence for 𝐻, for each 𝑗∈ℕ. Also, note that 𝑥𝑛=𝑗𝑦𝑗𝑖.(3.3) Since 𝐷𝑛's are finite, 𝑗 varies on a finite set. Hence {𝑥𝑛} is decomposed into finite number of Riesz basic sequences.

We will now introduce a concept which is more general than orthogonal frame and call it almost orthogonal frame. We give the following definition of almost orthogonal frame.

Definition 3.2. A frame {𝑥𝑛} in a Hilbert space 𝐻 is called an almost orthogonal frame of order 𝐾(𝐾∈ℕ) if 𝐾 is the smallest natural number for which there exists a permutation {𝜎𝑛} of ℕ such that 𝑥𝜎𝑛,𝑥𝜎𝑚=0,∀𝜎𝑛,𝜎𝑚||𝜎suchthat𝑛−𝜎𝑚||≥𝐾.(3.4)

Note 1. We use ⟨𝑥𝑛⟩ instead of ⟨𝑥𝜎𝑛⟩ for convenience.

Example 3.3. (I) An orthogonal basis is an almost orthogonal frame of order 1.
(II) {𝑒1,𝑒1,𝑒2,𝑒2,…,𝑒𝑛,𝑒𝑛,…} is an almost orthogonal frame of order 2.
(III) {𝑒1,𝑒2/√2,𝑒2/√2,𝑒3/√3,𝑒3/√3,𝑒3/√3,…} is not an almost orthogonal frame of any order.
(IV) {𝑒1,𝑒1+𝑒2,𝑒2+𝑒3,𝑒3,𝑒3,𝑒3+𝑒4,…} is an almost orthogonal frame of order 3.
(V) {𝑒1,𝑒2+𝑒1/4,𝑒3+𝑒1/8,…} is not an almost orthogonal frame of any order.
(VI) {𝑒𝑖+(1/𝑖)𝑒𝑖+1}∞𝑖=1 is an almost orthogonal frame of order 2.
(VII) {𝑒1,(1/2)𝑒2,(1−(1/22))1/2𝑒2,(1/3)𝑒3,(1−(1/32))1/2𝑒3,…}={(1/𝑛)𝑒𝑛}∪{(1−(1/22))1/2𝑒𝑛} is a tight frame with 𝐴=𝐵=1, which is almost orthogonal of order 2 and is not bounded below.

Observations
(I)A bounded frame may or may not be an almost orthogonal frame. (See Example I and Example V.) (II)An almost orthogonal frame of some finite (≠1) order may or may not be a Riesz basis. (See Example II and Example V.) (III)A Riesz basis may or may not be an almost orthogonal frame. (See Example I and Example V.)

Theorem 3.4. A bounded almost orthogonal frame satisfies Feichtinger conjecture.

Proof. Let {𝑥𝑛} be a bounded almost orthogonal frame of order 𝐾. Define a sequence {𝐺𝑛} of subspaces as follows: 𝐺1=𝑥1,𝑥2,…,𝑥𝐾,𝐺2=𝑥𝐾+1,…,𝑥2𝐾,⋮𝐺𝑛=𝑥(𝑛−1)𝐾+1,𝑥(𝑛−1)𝐾+2,…,𝑥𝑛𝐾,𝑛∈ℕ.(3.5) Now, since {𝑥𝑛} is an almost orthogonal frame of degree 𝐾. This gives ⟨𝑥𝑛,𝑥𝑚⟩=0∀𝑛,𝑚∈ℕsuchthat|𝑛−𝑚|≥𝐾.(3.6) Let 𝑥∈𝐺𝑛 and 𝑦∈𝐺𝑛+2, for any 𝑛∈ℕ. Then 𝑥=𝑛𝐾(𝑛−1)𝐾+1𝛼𝑖𝑥𝑖,𝑦=(𝑛+2)𝐾(𝑛+1)𝐾+1𝛽𝑗𝑥𝑗.(3.7) Therefore, we have ⟨𝑥,𝑦⟩=𝑛𝐾(𝑛−1)𝐾+1𝛼𝑖𝑥𝑖,(𝑛+2)𝐾(𝑛+1)𝐾+1𝛽𝑗𝑥𝑗=𝑛𝐾(𝑛−1)𝐾+1𝛼𝑖𝑥𝑖,(𝑛+2)𝐾(𝑛+1)𝐾+1𝛽𝑗𝑥𝑗=𝑛𝐾(𝑛−1)𝐾+1𝛼𝑖(𝑛+2)𝐾(𝑛+1)𝐾+1𝛽𝑗𝑥𝑖,𝑥𝑗=0.⇒𝐺𝑛∩𝐺𝑛+2=𝜙,∀𝑛∈ℕ,⇒span𝐺𝑛,𝐺𝑛+2=𝐺𝑛⊕𝐺𝑛+2⇒∀𝑛,𝐺span1,𝐺3,𝐺5,…=𝐺1⊕𝐺3⊕𝐺5⊕⋯=𝑛∈ℕ𝐺2𝑛−1=𝐻1.(3.8) Also, we have 𝐺span2,𝐺4,𝐺6=,…𝑛∈ℕ𝐺2𝑛=𝐻2.(3.9) So, by Theorem 3.1𝑥1,𝑥2,…,𝑥𝐾,𝑥2𝐾+1,𝑥2𝐾+2,…,𝑥3𝐾,…(3.10) can be written as finite union of Riesz basic sequences.
Similarly, using Theorem 3.1, 𝑥𝐾+1,𝑥𝐾+2,…,𝑥2𝐾,𝑥3𝐾+1,…,𝑥4𝐾,…(3.11) can be written as finite union of Riesz basic sequences.
Hence, {𝑥𝑛} can be written as finite union of Riesz basic sequences.

Remark 3.5. Almost orthogonal frames produce fusion frames (nonorthogonal) and fusion frame systems. Indeed, let {𝑥𝑛} be an almost orthogonal frame of order 𝐾. Proceeding as in Theorem 3.4, we get a sequence of subspaces {𝐺n} satisfying 𝐺span1,𝐺3,𝐺5,…=𝐺1⊕𝐺3⊕𝐺5⊕⋯=𝐻1,𝐺span2,𝐺4,𝐺6,…=𝐺2⊕𝐺4⊕𝐺6⊕⋯=𝐻2.(3.12) Now, define a sequence of projections {𝑣𝑖} (𝑣𝑖∶𝐻→𝐺𝑖). Then, we can easily verify that {𝑣2𝑖−1,𝐺2𝑖−1}𝑖∈ℕ is a fusion frame for 𝐻1 and {𝑣2𝑖,𝐺2𝑖}𝑖∈ℕ is a fusion frame for 𝐻2. So, {𝑣𝑖,𝐺𝑖}𝑖∈ℕ is a fusion frame for 𝐻.

Finally, we prove that for any bounded almost orthogonal frame, there exists a block sequence with respect to the almost orthogonal frame such that the block sequence is a Riesz basis. More precisely, we have the following.

Theorem 3.6. A bounded almost orthogonal frame contains a Riesz basis.

Proof. Let {𝑥𝑛} be an almost orthogonal frame of order 𝐾. Consider {𝑥1,𝑥2,…,𝑥𝐾,𝑥𝐾+1,…,𝑥2𝐾,𝑥2𝐾+1,…}. Then, following the steps in Theorem 3.4, we get a sequence of subspaces {𝐺𝑛} which are finite dimensional. So, we can extract a Riesz basis for 𝐺𝑛 out of {𝑥(𝑛−1)𝐾+1,𝑥(𝑛−1)𝐾+2,…,𝑥𝑛𝐾} and let it be {𝑥𝑛𝑖}. Then ⋃𝑛∈ℕ{𝑥𝑖2𝑛−1} is a Riesz basis for 𝐻1 and ⋃𝑛∈ℕ{𝑥𝑖2𝑛} is a Riesz basis for 𝐻2, where 𝐻1 and 𝐻2 are as in Theorem 3.4. Write 𝐹𝑛=𝐺𝑛∩𝐺𝑛+1 for all 𝑛∈ℕ, then, for each 𝑛∈ℕ, 𝐹𝑛 is a finite-dimensional subspace of 𝐺𝑛. Let {𝑥𝑖𝑛} be an extracted Riesz basis for 𝐹𝑛 which is extracted from {𝑥(𝑛−1)𝐾+1,𝑥(𝑛−1)𝐾+2,…,𝑥𝑛𝐾} or {𝑥𝑛𝐾+1,…,𝑥(𝑛+1)𝐾}. Then, ⋃𝑛{𝑥𝑛𝑖⋃}∼𝑛{𝑥𝑖𝑛} is the desired Riesz basis for 𝐻.

Acknowledgment

The authors thank the anonymous referees for their useful and valuable suggestions which greatly helped to improve this paper.