# On Almost Orthogonal Frames

Authors: Virender; A. Zothansanga; S. K. Kaushik
Journal: International Journal of Mathematics and Mathematical Sciences
 (2012)

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## 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.


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## Body

## 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
(2.1)A‖x‖2≤∑n‍|〈x,xn〉|2≤B‖x‖2,x∈H.
The positive constantsA 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
(2.2)A∑k‍|αk|2≤‖∑k‍αkxk‖2≤B∑k‍|αk|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
(2.3)yn=∑i∈Dn‍αixi≠0,n∈N,
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=1∞Di=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
(3.1)Gn={xi}i∈Dn,∀n∈N.
Now, for each j∈ℕ, choose a sequence {yij}i∈ℕ such that
(3.2)yij={jthelementofGi,ifGicontainsjthelement,∅,otherwise.
Then, for each j∈ℕ, {yij}i∈ℕ is a sequence of orthogonal vectors which are norm bounded. So, {yij}i∈ℕ is a Riesz basic sequence for H, for each j∈ℕ. Also, note that
(3.3){xn}=⋃j‍{yij}.
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 thanorthogonal 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
(3.4)〈xσn,xσm〉=0,∀σn,σmsuchthat|σn-σm|≥K.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:
(3.5)G1=[x1,x2,…,xK],G2=[xK+1,…,x2K],G2=⋮Gn=[x(n-1)K+1,x(n-1)K+2,…,xnK],n∈N.
Now, since {xn} is an almost orthogonal frame of degree K. This gives
(3.6)〈xn,xm〉=0∀n,m∈Nsuchthat|n-m|≥K.
Let x∈Gn and y∈Gn+2, for any n∈ℕ. Then
(3.7)x=∑(n-1)K+1nK‍αixi,y=∑(n+1)K+1(n+2)K‍βjxj.
Therefore, we have
(3.8)〈x,y〉=〈∑(n-1)K+1nK‍αixi,∑(n+1)K+1(n+2)K‍βjxj〉=∑(n-1)K+1nK‍αi〈xi,∑(n+1)K+1(n+2)K‍βjxj〉=∑(n-1)K+1nK‍αi(∑(n+1)K+1(n+2)K‍β¯j〈xi,xj〉)=0.⇒Gn∩Gn+2=ϕ,∀n∈N,⇒span⁡{Gn,Gn+2}¯=Gn⊕Gn+2∀n,⇒span⁡{G1,G3,G5,…}¯=G1⊕G3⊕G5⊕⋯=⨁n∈NG2n-1=H1.
Also, we have
(3.9)span⁡{G2,G4,G6,…}¯=⨁n∈NG2n=H2.
So, by Theorem 3.1(3.10){x1,x2,…,xK,x2K+1,x2K+2,…,x3K,…}
can be written as finite union of Riesz basic sequences.
Similarly, using Theorem3.1,
(3.11){xK+1,xK+2,…,x2K,x3K+1,…,x4K,…}
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
(3.12)span⁡{G1,G3,G5,…}¯=G1⊕G3⊕G5⊕⋯=H1,span⁡{G2,G4,G6,…}¯=G2⊕G4⊕G6⊕⋯=H2.
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 {xin}. 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{xin}~⋃n{xin′} is the desired Riesz basis for H.



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*Source: 920607-2012-06-25.xml*