FRAMES IN FINITE-DIMENSIONAL INNER PRODUCT SPACES IN FINITE-DIMENSIONAL INNER PRODUCT SPACES ... 1 Frames in Finite-dimensional Inner Product Spaces 4 ... A frame for a vector space

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  • 1

    VIETNAM NATIONAL UNIVERSITY

    UNIVERSITY OF SCIENCE

    FACULTY OF MATHEMATICS, MECHANICS AND INFORMATICS

    Hoang Dinh Linh

    FRAMES IN FINITE-DIMENSIONAL INNERPRODUCT SPACES

    Undergraduate Thesis

    Undergraduate Advanced Program in Mathematics

    Thesis advisor: PhD. Dang Anh Tuan

    Hanoi - 2013

  • Contents

    Contents 1

    Introduction 2

    1 Frames in Finite-dimensional Inner Product Spaces 41.1 Some basic facts about frames . . . . . . . . . . . . . . . . . . . . . . 41.2 Frames in Cn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201.3 The discrete Fourier transform . . . . . . . . . . . . . . . . . . . . . 291.4 Pseudo-inverses and the singular value decomposition . . . . . . . . 331.5 Finite-dimensional function spaces . . . . . . . . . . . . . . . . . . . 41

    Conclusion 51

    References 52

    1

  • Introduction

    One of the important concepts in the study of vector spaces is the concepts of abasis for the vectors spaces, which allows every vector to be uniquely representedas a linear combination of the basis elements. However, the linear independenceproperty for a basis is restrictive; sometimes it is impossible to find vector whichboth fulfill the basis requirements and also satisfy external condition demandedby applied problems. For such purpose, we need to look for more flexible type ofspanning sets.

    Frames are such tools which provide these alternatives. They not only havegreat variety for use in applications, but also have a rich theory from a pureanalysis point of view. A frame for a vector space equipped with an inner prod-uct also allows each element in the space to be written as a linear combination ofthe elements in the frame, but linear independence between the frame elementsis not required. Intuitively, one can think about a frame as a basis to whichone has added more elements. The theory for frames and bases has developedrapidly in recent years because of its role as a mathematical tool in signal andimage processing.

    Lets say you want to send a signal across some kind of communication sys-tem, perhaps by talking on wireless phone or sending a photo to your friend overthe internet. We think that signal as a vector in a vector space. The way it gettransmitted is as a sequence of coefficients which represent the signal in termof a spanning set. If that spanning set is an orthonormal basis, then comput-ing those coefficients just involves finding some inner product of vectors, whicha computer can accomplish very quickly. As a result, there is not a significanttime delay in sending your voice or the photograph. This is a good feature fora communication system to have, so orthonormal bases are used a lot in suchsituation.

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  • CONTENTS

    Orthogonality is a very restrictive property, though. What if one of the coeffi-cients representing a vector gets lost in transmission? That piece of informationcannot be reconstructed. It is lost. Perhaps, wed like our system to have someredundancy, so that if one piece gets lost, the information can be pieced togetherfrom what does get through. This is where frames come in.

    By using a frame instead of an orthonormal basis, we do give up the unique-ness of coefficients and orthogonality of the vectors. However, these propertiesare supperfluous. If you are sending your side of a phone conversation or a photo,what matter is quickly computing a working set of expansion coefficients, notwhether those coefficients are unique. In fact, in some setting the linear inde-pendence and orthogonality restrictions inhibit the use of orthonormal bases.Frame can be constructed with a wider variety of characteristics, and can thusbe tailored to match the needs of a particular system.

    This thesis will introduce the concept of a frame for a finite-dimensionalHilbert space. We begin with the characteristics of frames. The first section dis-cusses some basic facts about frames, giving a standard definition of a frame.Then proceed in the latter to understood a litter bit about what frames are?,and how to construct a frames in finite-dimensional spaces. From that, thinkingabout the connection between frames in finite-dimensional vector spaces and theinfinite-dimensional constructions.

    Most of contents of the thesis come from the book "An Introduction toFrames and Riesz Bases", written by Prof. Ole Christensen. Moreover, I haveused some results of Prof. Ingrid Daubechies in [2] and some results of Prof.Nguyn Hu Vit Hng in [3].

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  • CHAPTER 1

    Frames in Finite-dimensional Inner

    Product Spaces

    1.1 Some basic facts about frames

    Let V be a finite-dimensional vector space, equipped with an inner product., .

    Definition 1.1 [1] A countable family of elements { fk}kI in V is a frame for Vif there exist constants A, B > 0 such that

    A f 2 kI| f , fk|2 B f 2 , f V. (1.1)

    A , B are called frame bounds, and they are not unique! Indeed, we canchoose A = A2 , B

    = B + 1 as other frame bounds.

    The frame are normalized if fk = 1 , k I.

    In the finite-dimensional space, { fk}kI can be having infinitely many ele-ments by adding infinitely many zero elements to the given frame.

    Now, we will only consider finite families { fk}kI , I finite. Then, the upperframe condition is automatically satisfied by Cauchy-Schwartzs inequality

    m

    k=1| f , fk|2

    m

    k=1 fk2 f 2 , f V. (1.2)

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  • 1.1. Some basic facts about frames

    For all f V, we are easily to prove that

    | f , fk|2 fk2 f 2 , k {1, 2, 3, ..., m}. (1.3)

    Indeed, for each k {1, 2, ..., m} ; f , fk V.If f = 0 then

    | f , fk| = |0, fk| = 0 = f 2 fk2.

    So, the result holds automatically.Now assume f 6= 0, for any C we have:

    0 fk + f 2 = fk + f , fk + f . (1.4)

    Expanding the right side

    0 fk, fk+ fk, f + f , fk+ ||2 f , f = fk2 + fk, f + f , fk+ ||2 f 2.

    Now select

    = fk, f f 2 .

    Substituting this into preceding expression yields

    0 fk2 2| f , fk|2 f 2 +

    | f , fk|2 f 4 f

    2

    = fk2 | f , fk|2 f 2 .

    which yields

    | f , fk|2 f 2 fk2.

    Thus, we can choose the upper frame bound B =m

    k=1 fk2.

    Recall the Lemma about a vector spaces decomposition.

    Definition 1.2 Let W is a subspace of a finite-dimensional vector space V. Then

    W = { V| W, i.e.,, = 0, W} (1.5)

    is said to be orthogonal complement of W in V.

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  • 1.1. Some basic facts about frames

    We have the following lemma:

    Lemma 1.3 [3] Let W be a subspace of finite-dimensional vector space V. Then(W) = W , and V can be decomposed as V = W W.

    PROOF Let (e1, e2, ..., em) be an orthogonal basis of W, and extend it to be a basis(e1, e2, ..., em, m+1, ..., n) of V. Applying Schmidts orthogonalization process tothis basis. Then we obtain an orthogonal basis (e1, e2, ..., em, em+1, ..., en) for V.The vectors em+1, ..., en are orthogonal to each element in (e1, e2, ..., em) , then theyare orthogonal to W. So, em+1, ..., en W.Let W then , = 0, W.So, (W). Therefore, W (W).In addition, if is an arbitrary vector in (W) V then

    = a1e1 + a2e2 + ... + anen.

    Since em+1, ..., en W then

    0 = , ej = a1e1, ej+ a2e2, ej+ ... + anen, ej == ajej, ej j = m + 1, n.

    Then, am+1 = am+2 = ... = an = 0.So, represents linearly in (e1, ..., em).Therefore, W. Thus, (W) W.Hence, (W) = W.Finally, we will prove that W W = {0}.Indeed, if W W then 2 = , = 0. Thus = 0.In conclusion,

    V = span(e1, e2, ..., em) span(em+1, ..., en) = W W

    .

    In order for the lower condition to be satisfied, if and only if span{ fk}mf=1 = V.Then we have the following theorem:

    Theorem 1.4 [1] A family of elements { fk}mk=1 in V is a frame for V if and onlyif span{ fk}mk=1 = V.

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  • 1.1. Some basic facts about frames

    PROOF :If span{ fk}mf=1 = V then we consider the following mapping:

    :V R

    f 7m

    k=1| f , fk|2.

    Firstly, we will prove that there exist A, B > 0 such that

    A f 2 kI| f , fk|2 B f 2 , f V, f = 1. (1.6)

    Take B =m

    k=1| fk|2 then (1.6) holds automatically by (1.2). So, we only need to

    show the existence of A.For any f , g V, we have:

    |( f ) (g)| = |m

    k=1

    (| f , fk|2 |g, fk|2)|

    = |m

    k=1

    (| f , fk| |g, fk|)(| f , fk|+ |g, fk|)|

    m

    k=1| f g, fk|( f fk+ g fk)

    m

    k=1 f g( f + g) fk2

    = f g( f + g)m

    k=1 fk2.

    Then, ( f ) tends to (g) as f tends to g. So, is continuous.Moreover, the set { f V| f = 1} ,the unit sphere in finite-dimensional spaceV, is compact. By using Weierstrasss theorem, we know that ( f ) has a mini-mum on the unit sphere.Choose A = min

    f =1( f ), we will prove that ( f ) > 0 when f = 1, then A > 0.

    Suppose the contrary that ( f ) = 0 for some f V, f = 1. We have:

    ( f ) =m

    k=1| f , fk|2 = 0.

    It implies that

    | f , fk| = 0, k = {1, 2, ..., n}.

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  • 1.1. Some basic facts about frames

    Besides, f V = span{ fk}mk=1 then {k}mk=1 : f =

    m

    k=1k fk and

    1 = f 2

    = f , f = f ,m

    k=1

    k fk

    =m

    k=1

    k f , fk = 0.

    It is impossible!Thus, ( f ) > 0 , f V when f = 1. Then for A > 0, we have

    m

    k=1| f , fk|2 A = A f 2 , f V, f = 1.

    So the lower bound could be chosen as A = min f =1

    ( f ) > 0.

    Indeed, if f = 0 then

    0 =m

    k=1| f , fk|2 = A f 2

    If f 6= 0, we can choose f = f f then f = 1.

    Apply the above result then

    m

    k=1| f , fk|2 A

    equivalent to

    m

    k=1| f f , fk|

    2 A.

    Hence,

    m

    k=1| f , fk|2 A f 2 , f V.

    Conv