A Study On Some Function Spaces And

A Study On Some Function Spaces And Approximation Of Continuous Functions Through Polynomials

M.Sc. Thesis This Thesis is Submitted to the Department of Mathematics, University of Rajshahi, in Partial Fulfillment of the Requirements for the Degree of Master of Science in Mathematics (Stream-A) Submitted By Examination Roll: 10084031 Registration No: 1952 Session: 2013-2014 Year of the Examination-2014 Department of Mathematics University of Rajshahi Rajshahi-6205 Bangladesh June, 2016

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I would like to dedicate this thesis to my loving parents and my honorable supervisor.

Acknowledgements

First of all, I am very grateful to almighty Allah who has created me and has given me the opportunity and patience to do the thesis work. Then, I would like to express my sincere gratitude to my supervisor Mr. Md. Masum Murshed, Assistant Professor, Department of Mathematics, University of Rajshahi for the continuous support of my M.Sc. thesis, for his patience, motivation, enthusiasm, and immense knowledge. His guidance helped me in all the time of my thesis and writing of this thesis. I could not imagine to complete this thesis without his valuable guidance. Besides my supervisor, I would like to express my sincere gratitude to Professor Dr. Gour Chandra Paul, who provided me an opportunity to join their team as a member and inspired me to take thesis in my M.Sc. program. Without his valuable suggestion, help and inspiration it would not be possible to complete this thesis work properly. My sincere thanks also goes to Mr. Md. Rabiul Haque Assistant Professor of the same Department for his encouragement, suggestion and valuable guidance throughout my thesis work. Without his precious support it would not be possible to conduct this work. I would like to convey my gratitude to honorable Chairman, Prof. Dr. M. Wazed Ali Pramanik, Department of Mathematics, University of Rajshahi, Rajshahi-6205, for providing the Departmental facilities during my thesis. Also, I would like to express my special gratitude to Mr. A.H.M. Rashedunnabi and Mr. Md. Jakir Hossain Assistant Professor of the same Department for their inspiration and valuable suggestion. Also, I would like to express my heartiest thank to Professor Dr. Akhil Chandra Paul, UGC Professor Dr. Subrata Majumdar, Professor Dr. Assaduzaman, Professor Dr. Nasima Akhter, Professor Dr. Quazi Selina Sultana, Professor Dr. M. Ashabul Hoque, Professor Dr. M. Abdul Latif, Professor Dr. MD. Altab Hossain, Professor Dr. M. Sazuwar Raihan, Professor Dr. Jogendra Nath Pramanik, Associate Professor Dr. Md. Sahadat Hossain, Associate Professor Dr. Md. Ilias Hossain , Associate Professor Dr. M. Harunor Rashid and all of the teachers of the same department for delivering their suggestions and help during class lecture and in the time of my thesis.

vi I would like to thank all of my classmates and friends for the stimulating discussions, for the sleepless nights we were working together, and for all the fun we have had in the last five years. My special thanks also goes to Salma Nazmeen , Md. Morshed Bin Shiraj, Md.Nur Hossain, Md.Dalim Haque, Nazrina Parvin and Md. Mashiur Rahman for their limitless help throughout my Graduate and M.Sc. program. Here, I would like to say my gratitude thanks to my friends and senior Md. Rakibul Hasan (Kanon), Bisshojeet kumer, Md. Rejaul karim, Md. Faruk Hossain, Md. Babul Akter, Md. Amjad Hossin and all other who always loves me. Last but not the least, I would like to thank my family, my parents and my wife for their sacrifice and supporting me spiritually throughout writing this thesis and my life.

Abstract

In this study, we have reviewed some function spaces and have approximated some continuous functions by polynomials. Weierstrass approximation Theorem guaranteed that every continuous function can be approximated by polynomials. We have verified this Theorem for some continuous functions using Bernstein polynomials through computer programming. We have shown that C[a, b], the set of all continuous functions on [a, b] form a vector space. We have also shown that polynomials are continuous functions and the set of all polynomials up to degree n forms an inner product space. Also, we have shown that polynomials space is not only a vector space but also richer than it. We have approximated some known continuous functions through piecewise cubic polynomials. As tide at a point on the sea in the mean sea level is a continuous function with respect to time, we have approximated the tide at some points on the sea on the mean sea level for certain period from some hourly observed data collected from the Bangladesh Inland Water Transport Authority (BIWTA) by piecewise cubic polynomials. Finally, we have approximated the Bathymetric data for the Bay of Bengal region covering the area between 150 N and 230 N Latitudes and 850 E and 950 E Longitudes through inverse distance weighted interpolation from a figure with some doted depth contours. To ensure the depth values at every consecutive column in the matrix of depths, the doted depth contours are made continuous using the cubic spline interpolation. From the figure with continuous depth contours, a matrix of depth values is constructed by a MATLAB routine. Then inverse distance weighted interpolation is used to generate depth values at all other elements of the matrix. Such a type of depth matrix of the Bay of Bengal region can be used to compile the bathymetric data of that region, appropriately.

Contents List of Figures

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List of Tables

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1 1 1 2 2 2 3 3 4 4 5 6 6 7 7 8 8 8 9 9 9 10 10 11

Preliminaries 1.1 Preliminary Concepts of Algebra and Linear Algebra 1.1.1 Groupoid . . . . . . . . . . . . . . . . . . . 1.1.2 Semigroup . . . . . . . . . . . . . . . . . . 1.1.3 Monoid . . . . . . . . . . . . . . . . . . . . 1.1.4 Group . . . . . . . . . . . . . . . . . . . . . 1.1.5 Abelian Group . . . . . . . . . . . . . . . . 1.1.6 Ring . . . . . . . . . . . . . . . . . . . . . . 1.1.7 Field [9] . . . . . . . . . . . . . . . . . . . . 1.1.8 Vector Space [3] . . . . . . . . . . . . . . . 1.1.9 Linear functional . . . . . . . . . . . . . . . 1.1.10 Subspace [9] . . . . . . . . . . . . . . . . . 1.1.11 Linear combinations of a set of vector [9] . . 1.1.12 Linear Dependence and Independence [9] . . 1.1.13 Basis and dimension of a vector space . . . . 1.2 Preliminary Concepts of Real Analysis . . . . . . . . 1.2.1 Metric Space . . . . . . . . . . . . . . . . . 1.2.2 Sequence . . . . . . . . . . . . . . . . . . . 1.2.3 Convergent Sequence . . . . . . . . . . . . . 1.2.4 Bounded Sequence . . . . . . . . . . . . . . 1.2.5 Subsequence . . . . . . . . . . . . . . . . . 1.2.6 Cauchy Sequence . . . . . . . . . . . . . . 1.2.7 Complete Metric Space . . . . . . . . . . . . 1.2.8 Neighborhood of a point . . . . . . . . . . .

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1.2.9 Interior point . . . . . . . . . . . . . . . . 1.2.10 Open set . . . . . . . . . . . . . . . . . . 1.2.11 Limit point . . . . . . . . . . . . . . . . . 1.2.12 Closed set . . . . . . . . . . . . . . . . . . 1.2.13 Perfect Set . . . . . . . . . . . . . . . . . 1.2.14 Closure of a set . . . . . . . . . . . . . . . 1.2.15 Interior of a Set . . . . . . . . . . . . . . . 1.2.16 Continuous function . . . . . . . . . . . . 1.2.17 Uniformly Continuous function . . . . . . 1.2.18 Ordered set . . . . . . . . . . . . . . . . . 1.2.19 Bounded Above Subset of an ordered set S: 1.2.20 Bounded Below Subset of an ordered set S 1.2.21 Least upper bound or the Supremum . . . . 1.2.22 Greatest lower bound or the infimum . . . 1.2.23 Greatest lower bound property . . . . . . . 1.2.24 Least upper bound property . . . . . . . . Preliminary Concepts of Functional Analysis . . . 1.3.1 Dot Product in Rn . . . . . . . . . . . . . 1.3.2 Inner Product Space [10] . . . . . . . . . . 1.3.3 Normed Linear Space [10] . . . . . . . . . 1.3.4 Angle Between two vectors . . . . . . . . 1.3.5 Polarization Identity . . . . . . . . . . . . 1.3.6 Norm induced from inner product . . . . . 1.3.7 Quotient Normed Linear Space [2] . . . . . 1.3.8 Hilbert Space [10] . . . . . . . . . . . . . 1.3.9 Orthogonal Vectors [10] . . . . . . . . . . 1.3.10 Orthogonal complement of a Vector . . . . 1.3.11 Grahm-Schmidt orthogonalization process . 1.3.12 Best Approximation . . . . . . . . . . . .

Spaces of continuous functions 2.1 Spaces of Continuous Functions [7] . . . . 2.1.1 Lipschitz Continuity [7] . . . . . . 2.1.2 Contraction Mapping [5] . . . . . . 2.1.3 Equicontinuity [7] . . . . . . . . . 2.1.4 Pointwise Continuous Function [7] 2.1.5 Pointwise Convergence [7] . . . . .

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2.1.6 Uniform Convergence [7] 2.1.7 The spaces C(X,Y) . . . . The Space C(Y) [1] . . . . . . . . 2.2.1 The Space C[a,b] . . . . . 2.2.2 Support of a function . . . 2.2.3 The Space CC (X) . . . . 2.2.4 The Space C0 (X) . . . . .

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Measure Spaces and L p Spaces 3.1 Some basic concepts of Measure Theory . . . . . . . . . . . . . . . . . 3.1.1 Set Function . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.2 Extended Real Number System . . . . . . . . . . . . . . . . . 3.1.3 Intervals in R . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.4 Length of Intervals . . . . . . . . . . . . . . . . . . . . . . . . 3.1.5 Length of Sets . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.6 Monotone Class . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.7 Measurable Space and Measurable Set . . . . . . . . . . . . . . 3.1.8 Borel Measurable Set and Boerel Measure . . . . . . . . . . . . 3.1.9 Measurable Function . . . . . . . . . . . . . . . . . . . . . . . 3.1.10 Some properties of Measurable Function [? ] . . . . . . . . . . 3.1.11 Measure of a Set . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.12 Measure and Measure Space . . . . . . . . . . . . . . . . . . . 3.1.13 Simple Function . . . . . . . . . . . . . . . . . . . . . . . . . 3.1.14 Outer Measure And Properties of Outer Measure . . . . . . . . 3.1.15 Construction of Lebesgue Measure . . . . . . . . . . . . . . . 3.1.16 Some Properties of Lebesgue Measure . . . . . . . . . . . . . . 3.1.17 Existence and Construction of Nonmeasurable Set . . . . . . . 3.2 Construction and some properties of Riemann Integration . . . . . . . . 3.2.1 Riemann Integration . . . . . . . . . . . . . . . . . . . . . . . 3.2.2 Theorems on Riemann Integrals [19] . . . . . . . . . . . . . . 3.2.3 Limitation of Riemann Integration . . . . . . . . . . . . . . . . 3.2.4 Construction of Lebesgue Integration and some of its properties 3.2.5 Lebesgue Integral . . . . . . . . . . . . . . . . . . . . . . . . 3.2.6 Almost everywhere . . . . . . . . . . . . . . . . . . . . . . . 3.3 L p Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.1 Essential Supremum . . . . . . . . . . . . . . . . . . . . . . . 3.3.2 Convex Function . . . . . . . . . . . . . . . . . . . . . . . . .

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Contents 3.3.3

Conjugate Exponents . . . . . . . . . . . . . . . . . . . . . . . .

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Polynomial Space 4.1 Polynomial . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Polynomial Function . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.1 Classification of polynomials: . . . . . . . . . . . . . . . . . . . . 4.3 Polynomial Space . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4 Relation of Polynomial Space with Linear Algebra . . . . . . . . . . . . . 4.4.1 Polynomial Space is Richer than General vector Space . . . . . . . 4.4.2 Derivative of a polynomial . . . . . . . . . . . . . . . . . . . . . . 4.4.3 Inner product on Polynomial Space . . . . . . . . . . . . . . . . . 4.4.4 Norm in polynomial space . . . . . . . . . . . . . . . . . . . . . . 4.5 Bases for Polynomials Space . . . . . . . . . . . . . . . . . . . . . . . . . 4.5.1 Some Basis for the polynomial space P3 . . . . . . . . . . . . . . . 4.5.2 WEIERSTRASS APPROXIMATION THEOREM [21] . . . . . . . 4.6 Approximation of continuous functions using piecewise cubic polynomials 4.6.1 Cubic Spline Method . . . . . . . . . . . . . . . . . . . . . . . .

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Approximation of Continuous Functions by Polynomials 5.1 Approximation of Continuous Functions . . . . . . . . . . 5.2 Bernstein polynomials . . . . . . . . . . . . . . . . . . . 5.2.1 Recursive Definition of Bernstein Polynomials [20] 5.2.2 Some Properties of Bernstein Basis Polynomials . 5.2.3 Taylor Series vs Weierstrass Approximation . . . . 5.3 Approximation Of Some continuous functions . . . . . . . 5.3.1 Conclusion and Result discussion . . . . . . . . .

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Approximation of tidal data and bottom topography for the Bay of Bengal region 101 6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 101 6.2 Necessary background . . . . . . . . . . . . . . . . . . . . . . . . . . . . 106 6.2.1 Cubic Spline Methods . . . . . . . . . . . . . . . . . . . . . . . . 106 6.2.2 Inverse distance weighted interpolation (IDWI) . . . . . . . . . . . 110 6.3 Methodology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 111 6.4 Discussion of results and conclusion . . . . . . . . . . . . . . . . . . . . . 112

Bibliography

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List of Figures 4.1

Convex Linear Combination of Four Vectors . . . . . . . . . . . . . . . . .

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5.1 5.2 5.3 5.4 5.5 5.6 5.7 5.8

Bernstein Polynomial Basis of degree 1 . . . . . . . . . . . . . Bernstein Polynomial Basis of degree 2 . . . . . . . . . . . . . Bernstein Polynomial Basis of degree 3 . . . . . . . . . . . . . Approximation of linear function . . . . . . . . . . . . . . . . . Approximation of quadratic function by Bernstein Polynomial . Approximation of cubic function by Bernstein Polynomial . . . Approximation of sin(x) by Bernstein Polynomial . . . . . . . . Approximation of exponential function by Bernstein Polynomial

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6.1

Approximated water levels with respect to the mean sea level due to tide from 02.00 UTC of 28th April to 02.00 UTC of 30th April at 3 coastal stations; (a) Hiron Point, (b) Charchanga (Hatiya) and (c) Chittagong using cubic spline interpolation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102 Approximated water levels with respect to the mean sea level due to tide from 20.00 UTC of 23rd May to 20.00 UTC of 25th May at 3 coastal stations; (a) Hiron Point, (b) Charchanga (Hatiya) and (c) Chittagong using cubic spline interpolation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 103 Bathymetric figure (After Johns et al. [8]) from which water depth data for that region of CMS, which is outside FMS region are compiled . . . . . . . 104 Bathymetric figure covering the area between 150 N and 230 N Latitudes and 850 E and 950 E Longitudes taken from Fig. 6.3 . . . . . . . . . . . . . . . 105 A doted depth contour taken from Fig. 6.4 . . . . . . . . . . . . . . . . . . 106 A continuous depth contour made from Fig. 6.5 using cubic spline interpolation109 All continuous depth contours made from Fig. 6.4 using cubic spline interpolation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 110

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6.3 6.4 6.5 6.6 6.7

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List of Figures 6.8 6.9

Water and land area of Fig. 6.4, where blue colour represents water and white colour represents land . . . . . . . . . . . . . . . . . . . . . . . . . 111 A contour map of our interpolated bathymetry using inverse distance weighted interpolation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

List of Tables 5.1 5.2 5.3 5.4

Estimated root mean square errors for a quadratic function . . Estimated root mean square errors for a Cubic function . . . . Estimated root mean square errors for approximation of sin(x) Estimated root mean square errors for a exponential function .

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Chapter 1 Preliminaries Summary In this chapter, we have discussed about some basic concepts of Linear Algebra, Real Analysis and Functional Analysis. Especially, we have reviewed the prerequisite concepts of inner product and normed linear spaces. Also we have discussed about some important Theorems of Hilbert Space that are used in numerical analysis and approximation theory.

1.1 1.1.1

Preliminary Concepts of Algebra and Linear Algebra Groupoid

A non empty set together with a binary operation is said to be a groupoid. That is, if G is a non empty set and () is a binary operation on G then the pair (G , ) is said to be a groupoid. Example 1.1. Let us consider the set of all integers Z. Then (Z, −) is a groupoid, where (−) is a binary operation on Z. Example 1.2. Let us consider the set of all real numbers R. Then (R, +) is a groupoid, where (+) is a binary operation on R. Example 1.3. Let us consider the set of all positive integers N. Then (N, .) is a groupoid, where (.) is a binary operation called multiplication operation on N. Example 1.4. Let us consider the set of all rational numbers Q. Then (Q, +) is a groupoid, where (+) is a binary operation on Q.

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1.1.2

Preliminaries

Semigroup

A non empty set with an associative binary operation is said to be a semigroup. That is, a semigroup is an ordered pair (G, ∗), where G is a non empty set and (∗) is an associative binary operation on G. Example 1.5. A singleton set {a} with the operation

a ∗ a = a, is a semigroup.

Example 1.6. The set of all positive integers N with the binary operation addition is a semigroup i.e., the pair (N,+) is a semigroup. Example 1.7. The set of all rational numbers Q with the binary operation addition, substraction and multiplication form semigroups i.e., the pair (Q,+), (Q,-) and (Q,) are all semigroups under the respective operations. Example 1.8. The set of all real numbers R with the binary operation addition, substraction and multiplication form semigroups i.e., the pair (R,+), (R,-) and (R,) are all semigroups under the respective operations.

1.1.3

Monoid

A monoid is a set S, which is closed under an associative binary operation and has an identity element I in S such that for all a ∈ S, I.a = a.I = a. In other words, a semigroup with an identity element is called a monoid. A monoid must contain at least one element. Example 1.9. The set of all positive integers N forms a commutative monoid under multiplication, where the identity element is 1. A submonoid of N under addition is called a numerical monoid. Example 1.10. Given a set A, all subsets of A form a commutative monoid under intersection operation, where identity element is A itself). Example 1.11. Given a set A, all subsets of A form a commutative monoid under union operation, where identity element is the empty set.

1.1.4

Group

A group is an ordered pair (G, ∗), where G is a non empty set and (∗) is a binary operation on G such that the following properties are satisfied : (G1) For all x, y, z ∈ G, x ∗ (y ∗ z) = (x ∗ y) ∗ z. (G2) For all x ∈ G ∃ e ∈ G such that x ∗ e = x = e ∗ x here e is called the identity

1.1 Preliminary Concepts of Algebra and Linear Algebra

3

element in G. (G3) For every x ∈ G ∃ y ∈ G such that x ∗ y = e = y ∗ x here y is called the inverse of x. Example 1.12. The set of all integers Z is a group under the binary operation addition i.e., (Z,+) is a group. Example 1.13. The set of all real numbers R without zero forms a group under the binary operation multiplication i.e., (R-{0},.) is a multiplicative group. Example 1.14. (Q-{0},.) is a multiplicative group.

1.1.5

Abelian Group

A commutative group is called an Abelian group. If a group is commutative under the binary operation addition, then it is called additive Abelian group. Examples of Abelian group Examples 1.12, 1.13 and 1.14 are the examples of Abelian groups under the respective operations.

1.1.6

Ring

An algebraic system (R, +, ) consisting of a non empty set R with two binary operations (+) addition and () multiplication is said to be a ring, if the following properties are satisfied: (R1) (R,+) is an Abelian group. (R2) (R, ) is a semigroup. (R3) R satisfies distributive laws, (i) x  (y + z) = x  y + x  z for all x, y, z ∈ R. (ii) (x + y)  z = x  z + y  z for all x, y, z ∈ R. Example 1.15. The set of integers forms a ring under the operation addition and multiplication, i.e., (Z, +, ) is a ring. ( ! ) a 0 Example 1.16. Let R = M : M = where a, b ∈ R that is R is the set of all 0 b 2 × 2 diagonal matrices then (R, +, ) is a commutative ring with identity element.

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1.1.7

Preliminaries

Field [9]

A non empty set K together with two binary operations addition and multiplication is said to be a field, if it satisfies the following axioms (A), (M) and (D): (A1) if x ∈ K and y ∈ K then x + y is in K. (A2) Addition is commutative, that is x + y = y + x for all x, y ∈ K. (A3) Addition is associative, that is x + (y + z) = (x + y) + z for all x, y, z ∈ K. (A4) K contains an element 0 ∈ K such that x + 0 = x = 0 + x for all x in K. (A5) To every x ∈ K, there exists −x ∈ K such that x + (−x) = 0 = (x) + x (M1) if x ∈ K and y ∈ K then x  y is in K. (M2) multiplication is commutative that is x  y = y  x for all x, y ∈ K. (M3) multiplication is associative that is x  (y  z) = (x  y)  z for all x, y, z ∈ K. (M4) K contains an element 1 ∈ K such that 1  x = x for all x in K. (M5) To every x ∈ K with x ̸= 0 there exists 1x ∈ K such that x  1x = 1 (D) The distributive laws: (D1) x  (y + z) = x  y + x  z for all x, y, z ∈ K. (D2) (x + y)  z = x  z + y  z for all x, y, z ∈ K.

Example 1.17. R, Q, C are fields but N, Z are not fields.

1.1.8

Vector Space [3]

A non-empty set V with two binary operations addition and scalar multiplication is called a vector space over a field K, if the followings axioms (A) and (M) are satisfied: (A1) x + y = y + x , for all x, y ∈ V . (A2) x + (y + z) = (x + y) + z , for all x, y, z ∈ V . (A3) V contains an element 0 ∈ V such that x + 0 = x = 0 + x , for all x in V. (A4) To every x ∈ V there exists −x ∈ V such that x + (−x) = 0 = (−x) + x. (M1) k(x + y) = kx + ky for all k ∈ K and x, y ∈ V . (M2) (k1 + k2 )x = k1 x + k2 x for all k1 , k2 ∈ K and x ∈ V . (M3) (k1 k2 )x = k1 (k2 x) for all k1 , k2 ∈ K and x ∈ V . (M4) 1.x = x = x.1 for 1 ∈ K and ∀ x ∈ V . Note: A vector space over real or complex field is said to be a linear space. Example 1.18. Rn is vector space over the field R.

1.1 Preliminary Concepts of Algebra and Linear Algebra

5

Example 1.19. The set of all polynomial of degree less or equal to n is a vector space over the field R, where n is a positive integers. Example 1.20. The set of all continuous real valued function on an interval [a,b] is a vector space over the field R.

1.1.9

Linear functional

Let V be a vector space over the field K then a mapping φ : V → K is called a linear functional, if for every u, v ∈ V and every a, b ∈ K, φ (au + bv) = aφ (u) + bφ (v). In other words, a linear functional on a vector space V over a field is linear mapping from V into K. Example 1.21. The linear transformation defined by the Riemann integral, Z b

I( f ) =

f (x) dx a

is a linear functional from the vector space C[a,b] of all continuous functions on the interval [a,b] to the real numbers. The linearity of I follows from the standard facts about the integral: I(α f + β g) = ab (α f (x) + β g(x)) dx R R = ab α f (x) dx + ab β g(x) dx R R = α ab f (x) dx + β ab g(x) dx = αI( f ) + β I(g) R

Example 1.22. Suppose that vectors in the real coordinate space Rn are represented as   x1  ..  column vectors ⃗x =  . . Then any linear functional can be written in these coordinates as xn a sum of the form: f (⃗x) = a1 x1 + · · · + an xn . This is just the matrix product of the row vector (a1 , a2 , ..., an ) and the column vector :  ⃗x  x1 . f (⃗x) = [a1 . . . an ] .. . xn

6

1.1.10

Preliminaries

Subspace [9]

Let V is a vector space over the field K and W be a subset of V then, W is called a subspace of V, if W is itself a vector space over the same field K with respect to the same operations of vector addition and scalar multiplication. Theorem 1.1. If W ⊂ V is a subspace of V iff, (i) 0 ∈ W and (ii) au + bv ∈ W for a, b ∈ K and u, v ∈ V . Example 1.23. Any line through (0,0) is a subspace of R2 over R. Example 1.24. If F(X) = { f | f : X → R, the set o f all real then the set of all continious function on X is a subspace of F(X).

valued

f unction

on

Example 1.25. Let V be any vector space. Then the set {0} consisting the zero vector alone, and the entire space V are subspaces of V.

1.1.11

Linear combinations of a set of vector [9]

Suppose that V be a vector space over the field K and let v1 , v2 , v3 , ......, vn ∈ V then a1 v1 + a2 v2 + a3 v3 + ...... + an vn is called a linear combinations of v1 , v2 , v3 , ......, vn where a1 , a2 , a3 , ......, an ∈ K. The set of all such linear combinations is denoted by, span{v1 , v2 , v3 , ......, vn } called the linear span of V. Theorem 1.2. Let E be a subset of a linear space V. (i) Then span(E) is a subspace of V, containing E. (ii) If W is a subspace of V containing E, then span(E) ⊆ W . Example 1.26. Let e1 , e2 , e3 ∈ R3 , where e1 = (1, 0, 0), e2 = (0, 1, 0) and e3 = (0, 0, 1). Then any vectors (a, b, c) ∈ R3 can be written as a linear combination of the vectors e1 , e2 , e3 i.e., (a, b, c) = a(1, 0, 0) + b(0, 1, 0) + c(0, 0, 1) = ae1 + be2 + ce3 Example 1.27. The polynomials 1, x, x2 , ....... span the linear space P(X) of all polynomials in x, i.e., P(X) = span{1, x, x2 , .......}.

X},

1.1 Preliminary Concepts of Algebra and Linear Algebra

1.1.12

7

Linear Dependence and Independence [9]

Suppose that V is a vector space over the field K. Then the set of vectors v1 , v2 , v3 , ......, vn ∈ V are said to be linearly dependent, if there exist scalars a1 , a2 , a3 , ......, an ∈ K, not all of them zero such that a1 v1 + a2 v2 + a3 v3 + ...... + an vn = 0 and if all the scalars are zero then the set of vectors are said to be linearly independent. Example 1.28. The vectors u = (2, −2, 0), v = (2, 6, −2) and w = (10, 6, −4) are linearly dependent since 3u + 2v − w = 0 Example 1.29. The vectors u = (6, 2, 3, 4), v = (0, 5, −3, 1) and w = (0, 0, 7, −2) are linearly independent since xu + yv + zw = 0 ⇒ x = 0, y = 0, z = 0.

1.1.13

Basis and dimension of a vector space

A set S = {e1 , e2 , e3 , ..., en } of vectors is said to be a basis of a linear space X, if the followings two conditions hold : (i) e1 , e2 , e3 , ..., en are linearly independent, (ii) e1 , e2 , e3 , ..., en span X. A linear space X is said to be a finite dimensional space, if there exists a finite basis for X. That is, X is said to be n-dimensional, if X has a basis with n elements. A linear space which is not of finite dimensional space is said to be infinite dimensional space. Theorem 1.3. If a liner space has a finite spanning set, then it has a finite basis. Theorem 1.4. Let Y be a subspace of an n-dimensional linear space X. Then dim Y ≤ n. In particular if dim Y = n, then Y = X. Example 1.30. Let F be a field and V n is a vector space consisting of n-tuples of elements of F. The vectors e1 = (1, 0, 0, ..., 0) e2 = (0, 1, 0, ..., 0) e3 = (0, 0, 1, ..., 0) ................... en = (0, 0, 0, ..., 1) form a basis, called the usual basis of V n . Thus V n has dimension n.

8

Preliminaries

Example 1.31. Let P be a vector space of polynomials of degree less or equal to n over the field R. Then the set 1, x, x2 , ..., xn is linearly independent and generates W. Thus it is a basis of W and hence dimension of W is n+1. Theorem 1.5. Let V be a finite dimensional vector space . Then every basis of V has the same number of elements. Theorem 1.6. Let V be a vector space of dimension n over a field K. Then, (i) Any set of n+1 or more vectors is linearly dependent. (ii) Any linearly independent set is a part of a basis, that can be extended to a basis. (iii) A linearly independent set with n elements is a basis. Theorem 1.7. Let U and V be two finite dimensional subspaces of a vector space V. Then dim(U +W ) = dimU + dimW − dim (U

1.2 1.2.1

T

W)

Preliminary Concepts of Real Analysis Metric Space

A non empty set X together with a function d:X×X → [0, ∞) is said to be a metric space, if the function d satisfies the following conditions: (d1) d(x, x) ≥ 0 ∀x ∈ X and d(x, y) = 0 if and only if x = y for allx, y ∈ X. (d2) d(x, y) = d(y, x) for all x, y ∈ X. (d3) d(x, z) ≤ d(x, y) + d(y, z) for all x, y, z ∈ X. Then the function d is called a metric on X and the pair (X,d) is called a metric space. Example 1.32. The set of real numbers R is a metric space under the metric d, where d is defined by, d(x, y) = |x − y| for all x, y ∈ R.

1.2.2

Sequence

A sequence is a function whose domain is the set of natural numbers N. A sequence can be thought of as a list of elements with a particular order. There are a number of ways to denote a sequence some of which are more useful for specific types of sequences. We can denote a sequence as {xn }n∈N . Another, way to specify a sequence is to list the elements. For example, the infinite sequence of positive odd integers can be written as {1, 3, 5, 7, ...}.

1.2 Preliminary Concepts of Real Analysis Example 1.33. Let xn = n ∈ N.

1 n+1 ,

9

then {xn } is a decreasing sequence of real numbers, where

Example 1.34. Let xn = n(n + 1), then {xn } is an increasing sequence of real numbers, where n ∈ N. Example 1.35. Let an = xn , then {an } is a sequence of polynomials, where n ∈ N.

1.2.3

Convergent Sequence

A sequence {xn }n∈N is said to be convergent at a point x, if for every ε > 0 there exists a natural number N (depending on ε ) such that | x − xn |< ε for all n ≥ N. Then x is called the limit of the sequence and we can write, lim xn = x or xn → x.

n→∞

If a sequence is not convergent then the sequence is a divergent sequence. Example 1.36. Let xn = 1 lim xn = lim ( ) = 0. n→∞ n→∞ n

1 n

Example 1.37. Let xn =

n(n+1) n+2 ,

, then {xn } is a convergent sequence which converse to 0 since

then {xn } is not a convergent sequence. n

Example 1.38. Let sn = 1 + (−1) n , then {sn } is a convergent sequence.

1.2.4

Bounded Sequence

A sequence {xn } is said to be bounded if there exists a M ∈ R such that |xn | ≤ M for all n ∈ N. Note : If a sequence is convergent then the sequence is a bounded sequence.

1.2.5

Subsequence

A subsequence is a sequence that can be derived from another sequence by deleting some elements without changing the order of the remaining elements. A subsequence {sn } of the sequence {tn } defined by, tk = snk , where n1 < n2 < n3 < ... is an increasing sequence of indices. If {sni } converges then {tn } converges. Example 1.39. Let xn = converges to 0.

1 n+1 ,

then {xn } is a subsequence of the sequence { 1n } which

10

Preliminaries

Theorem 1.8. If limit of a sequence exists then this limit must be unique. Proof : Suppose that lim an = a and lim an = b. Given any ε > o there exists N1 , N2 ∈ N n→∞

n→∞

(dependimg on ε) such that |an − a| < ε2 and |an − b| < Choose N = max(N1 , N2 ) < n. Consequently,

ε 2

whenever N1 and N2 < n.

|a − b| = |a − an + an − b| ≤ |a − an | + |an − b| < ε2 + ε2 = ε. But, a and be are fixed real numbers, and ε is an arbitrary positive real number. Therefore, above inequality says that |a − b| is less than every positive real number. But by definition, |a − b| ≥ 0. Thus, |a − b| = 0 which implies a = b. Hence if the limit of a sequence exists then the limit must be unique.

1.2.6

Cauchy Sequence

A sequence {xn }n∈N is said to be a Cauchy Sequence if for every ε > 0 there exists a natural number N (depending on ε ) such that | xm − xn |< ε, for all m,n≥ N. Example 1.40. Let an =

n+3 2n+1 ,

then the sequence {an } is a Cauchy sequence.

Example 1.41. Let an = (−1)n , then the sequence {an } is not a Cauchy sequence. n

Example 1.42. Let sn = 1 + (−1) n , then {sn } is a convergent sequence which converges to 0.

1.2.7

Complete Metric Space

A metric space X is said to be a complete metric space if every Cauchy sequence in X converges in X. Theorem 1.9. Every convergent sequence is a Cauchy sequence but the converse is not true in general. Theorem 1.10. [22] (a) In any metric space X, every convergent sequence is a Cauchy sequence. (b) If X is a compact metric space and if {sn } is a Cauchy sequence in X, then {sn } converges to some point of X. (c) In Rn , every Cauchy sequence converges.

1.2 Preliminary Concepts of Real Analysis

11

Theorem 1.11. [22] The subsequential limit of a sequence {sn } in a metric space X form a closed subset of X. Theorem 1.12. [22] Suppose that {xn } and {yn } are complex sequences , and lim xn = x, lim yn = y. n→∞ n→∞ Then (a) lim (xn + yn ) = x + y. n→∞

(b) lim cxn = cx and lim (c + xn ) = c + x for any complex number c. n→∞

n→∞

(c) lim (xn yn ) = xy. n→∞ 1 1 = (d) lim n→∞ xn x Theorem 1.13. [22] Every Cauchy sequence of real numbers converges. Theorem 1.14. ( Bolzano Weierstrass Theorem) [22] Every bounded sequence has a convergent subsequence.

1.2.8

Neighborhood of a point

Let X be a metric space with the metric d. A neighborhood of a point p ∈ X , is a set Nr (p) consisting all point q ∈ X such that d(p, q) < r. That is, Nr (p) = {q ∈ X : d(p, q) < r}. Example 1.43. Let E = [−1, 10] , then the neighborhood of any point p ∈ E is Nr (p) = {q ∈ X : d(p, q) < r } where r is the radius of neighborhood. If we choose p = 1 and r = 5 then we have the neighborhood of the point 1 with radius 5 is N5 (1) = {q ∈ X : d(1, q) < 5}.

1.2.9

Interior point

Let X be a metric space and E be a non empty subset of X. Then a point p ∈ X , is said to be an interior point of E, if there exist a neighborhood Nε (p) of p contained in E. That is, if there exist a neighborhood Nε (p) of p such that Nε (p) ⊂ E. Example 1.44. Suppose that E = {x ∈ R : |x| < 1} then every point of E is an interior point of E.

1.2.10

Open set

Let X be a metric space and E be a non empty subset of X. Then E is said to be an open set, if every point of E is an interior point of E. That is, E is said to be open if for every point p ∈ E there exist a neighborhood Nε (p) of p such that Nε (p) ⊂ E.

12

Preliminaries

Example 1.45. Every open interval of R is an open set with the usual metric. Example 1.46. Suppose that E = {x ∈ R : |y| < r, x > 0} is an open set since every point of this set is an interior point of the set E. Example 1.47. Suppose that A = {y ∈ Z : |y| < r, r is any positive real number} is an open set. Theorem 1.15. A set E is said to be open if and only if E c is closed. Theorem 1.16. The intersection of any finite number of open sets is open. Theorem 1.17. The intersection of an infinite number of open sets need not be open. 1 For example consider the open set Gn = ( −1 n , n ). Here 1 G1 = ( −1 1 , 1) 1 G2 = ( −1 2 , 2) 1 G3 = ( −1 3 , 3) ········· 1 Gn = ( −1 n , n)

∩∞ i=1 Gi = {0} , which is not open. Therefore, the intersection of an infinite collection of open set need not be open.

1.2.11

Limit point

Let X be a metric space and E be a non empty subset of X. Then, a point p ∈ X is called a limit point of E, if every neighborhood of p contains a point of E other than p.

1.2.12

Closed set

A subset E of a matric space X is said to be closed, if E contains all of its limit points. Example 1.48. Every closed interval is a closed set in R with the usual metric. Example 1.49. Suppose that B = {y ∈ R : |y| ≤ r, r is any positive real number} , then B is closed set. Example 1.50. Suppose that C = {y ∈ Z : |y| ≤ r, r is any positive real number} , then C is a closed set.

1.2 Preliminary Concepts of Real Analysis

13

Theorem 1.18. A set E is closed if and only if E c is open. Theorem 1.19. (i) The intersection of any collection of closed sets is closed. Theorem 1.20. The union of any finite collection of closed sets is closed. Note: (i) The union of an infinite collection of closed sets need not be closed. For example, consider the closed set Fn = [1 + n1 , 3] Here F1 = [2, 3] F2 = [ 32 , 3] F3 = [ 43 , 3] ········· Fn = [1 + 1n , 3] S∞ i=1 Fi = (1, 3] which is not closed.

1.2.13

Perfect Set

A set subset E of a metric space X is said to be a perfect set, if every point of E is a limit point of E.

1.2.14

Closure of a set

The closure of a set E is the smallest closed superset containing E. That is closure of E is the intersection of all closed supersets of E. S

Example 1.51. Let E = [0, 3] [4, 7] then closure of E is [0,7]. Theorem 1.21. Closure of a set is closed. Theorem 1.22. The closure of a set is the union of the set E and all of its limit points.

1.2.15

Interior of a Set

The set of all interior points of a set A is called the interior of A, denoted by Int(A) or A◦ . Theorem 1.23. The interior of a set E is the union of all open subset of E. S

Example 1.52. Let E = [0, 3] (3, 5) then the interior of E is (0,5). Theorem 1.24. Interior of a set is open.

14

1.2.16

Preliminaries

Continuous function

Let X and Y be two metric spaces and let f : X → Y be a function. Then, the function f is said to be continuous at a point x = a if for every ε > 0 there exists a δ > 0 such that, d( f (x), f (a)) < ε whenever d(x, a) < δ . In otherwards, a function f is said to be continuous at a point x = a if lim f (x) = f (a). x→a A function f is called continuous, if f is continuous at every point of X . Example 1.53. The function f (x) = x is a continuous function, where f is a function from R to R. Example 1.54. The function f (x) = x3 + x is a continuous function, where f is a function from R to R.

1.2.17

Uniformly Continuous function

Let X and Y be two metric spaces and let f : X → Y . Then, the function f is said to be uniformly continuous if for every ε > 0 there exists a δ > 0 such that, dY ( f (p), f (q)) < ε whenever dX (p, q) < δ for all p, q ∈ X. Example 1.55. The function defined by fy (x) = ⟨x, y⟩ is a uniformly continuous function, where ⟨., .⟩ is known the inner product.

1.2.18

Ordered set

Let S be a set, then an order is a relation on S denoted by < with the followings properties: (i) if x ∈ S and x ∈ S, then only one of the statements x < y, x = y, y < x is true. (ii) if x, y, z ∈ S such that x < y and y < z then x < z An ordered set is a set S in which an order is defined. Example 1.56. The set of real numbers R is an ordered set with the usual order 0 be given then for any x1 , x2 ∈ H | fy (x1 ) − fy (x2 )| = |⟨x1 , y⟩ − ⟨x2 , y⟩|

1.3 Preliminary Concepts of Functional Analysis

23

= |⟨x1 − x2 , y⟩| ≤ ∥x1 − x2 ∥ ∥y∥ using Cauchy Schwarz inequality. < ε. ε Now Choose δ = ∥y∥ for y ̸= 0 and choose any δ > 0 for y = 0. Then we have, | fy (x1 ) − fy (x2 )| < ε whenever ∥x1 − x2 ∥ < δ . Hence fy (x) is uniformly continuous on H. Therefor fy (x) is continuous on H. Now from the triangle inequality we know that,

∥x1 + x2 ∥ ≤ ∥x1 ∥ + ∥x2 ∥

(1.1)

Replacing x1 by x1 − x2 in equation (1) we have, ∥x1 − x2 + (x2 )∥ ≤ ∥x1 − x2 ∥ + ∥x2 ∥ ⇒ ∥x2 ∥ ≤ ∥x1 − x2 ∥ + ∥x2 ∥ ⇒ ∥x1 ∥ − ∥x2 ∥ ≤ ∥x1 − x2 ∥

(1.2)

Again interchanging x1 and x2 in equation (2),we have ∥x2 ∥ − ∥x1 ∥ ≤ ∥x1 − x2 ∥

− (∥x1 ∥ − ∥x2 ∥) ≤ ∥x1 − x2 ∥ From equations (1.2) and (1.3) we have, |(∥x1 ∥ − ∥x2 ∥)| ≤ ∥x1 − x2 ∥ Hence ∥.∥ is a continuous function.

1.3.10

Orthogonal complement of a Vector

The orthogonal complement of a vector x ∈ H is given by, x⊥ = {y ∈ H : ⟨x, y⟩ = 0}

Theorem 1.31. If x ∈ H then x⊥ is a subspace of H. Proof: Since o ∈ x⊥ then x⊥ is non empty. Now, let x, y x⊥ and a,b are scalars then for z ∈ H

(1.3)

24

Preliminaries

we have ⟨x, z⟩ = 0 and ⟨y, z⟩ = 0. Then ⟨ax + by, z⟩ = ⟨ax, z⟩ + ⟨by, z⟩ ⇒ ⟨ax + by, z⟩ = a⟨x, z⟩ + b⟨y, z⟩ ⇒ ⟨ax + by, z⟩ = a.0 + b.0 ∴ ⟨ax + by, z⟩ = 0. Hence ax + by ∈ x⊥ and therefore x⊥ is a subspace of H. Definition 1.2. Let E ⊂ H and E is nonempty then the orthogonal complement of E is defined by, E ⊥ = {y ∈ H : ⟨x, y⟩} = 0 ∀x ∈ E Theorem 1.32. Let E be a subset of H then E ⊥ closed subspace of H. Proof: Since 0 ∈ E then E is non empty. Now, let x, y E ⊥ and a,b are scalars then for all z ∈ E we have ⟨x, z⟩ = 0 and ⟨y, z⟩ = 0. Then ⟨ax + by, z⟩ = ⟨ax, z⟩ + ⟨by, z⟩ ⇒ ⟨ax + by, z⟩ = a⟨x, z⟩ + b⟨y, z⟩ ⇒ ⟨ax + by, z⟩ = a.0 + b.0 ∴ ⟨ax + by, z⟩ = 0 Hence ax + by ∈ E ⊥ and therefore E ⊥ is a subspace of H. Since for a fixed x ∈ E the map fx : H → C defined by fx (y) = y⟨, x⟩ is continuous and {0} is a closed subset of C then Ker fx = fx−1 (0) = x⊥ . Hence x⊥ is a closed subspace of H. Since \ E⊥ = x⊥ , then E ⊥ is closed. x∈E

Hence, if E be a subset of H then E ⊥ closed subspace of H. Theorem 1.33. Every non empty closed convex set E ⊂ H contains a unique element of smallest norm. Proof: Let d = in f {∥x∥ : x ∈ E},then ∃ a sequence xn ∈ E such that ∥xn ∥ → d as n → ∞. Since E is convex therefore, 12 (xn + xm ) ∈ E ∴ ∥ 12 (xn + xm )∥2 ≥ d 2 ⇒ ∥(xn + xm )∥2 ≥ 4d 2 By parallelogram law, we have, ⇒ ∥(xn + xm )∥2 + ∥(xn − xm )∥2 = 2(∥xn ∥2 + ∥xm ∥2 ) ⇒ ∥(xn − xm )∥2 = 2(∥xn ∥2 + ∥xm ∥2 ) − ∥(xn + xm )∥2 ⇒ ∥(xn − xm )∥2 ≥ 2(∥xn ∥2 + ∥xm ∥2 ) − 4d 2 ∴ ∥(xn − xm )∥2 → 0 as m, n → ∞.

25

1.3 Preliminary Concepts of Functional Analysis

Hence {xn } is a Cauchy sequence in E. Since E is closed then {xn } converges in E. Hence, there exists x ∈ E such that xn → x. Since, the map x → ∥x∥ is continuous and ∥xn ∥ → ∥x∥ then ∥x∥ = d. For uniqueness let y ∈ E such that ∥y∥ = d. Since, E is closed and convex then 12 (x + y) ∈ E. Therefore∥(x + y)∥2 ≥ 4d 2 By parallelogram law we have, ∥(x − y)∥2 = 2(∥x∥2 + ∥y∥2 ) − ∥(x + y)∥2 ⇒ ∥(x − y)∥2 = 2(d 2 + d 2 ) − ∥(x + y)∥2 ⇒ ∥(x − y)∥2 ≤ 4d 2 − 4d 2 ⇒ ∥(x − y)∥2 ≤ 0 ∴ ∥(x − y)∥2 = 0 and hence x = y Hence for every non empty closed convex set E ⊂ H contains a unique element of smallest norm. Definition 1.3. A set {uα }α∈A of vectors in H is said to be an orthonormal set if, ( 0 i f α ̸= β ⟨uα , uβ ⟩ = 1 if α = β Theorem (Projection Theorem): If M is a closed subspace of H then, H =M

L

M⊥.

Proof: Let x ∈ H then the set, x+M = x+y : y ∈ M is closed and convex. Hence x + M contains a unique elements of smallest norm. Let this elements be z. Then z ∈ x + M ⇒ z − x ∈ M ⇒ x − z ∈ M. Now, ∥z∥ ≤ ∥z − αy∥........(1) for all y ∈ M and for all scalar α. Put w = z − x. Then w ∈ M and x = z + w. We now show that z ∈ M ⊥ . From (1) we have for all y ∈ M and for all scalar α ∥z − αy∥2 − ∥z∥2 ≥ 0 ⇒ ⟨z − αy, z − αy⟩ − ⟨z, z⟩ ≥ 0 ⇒ ⟨z, z⟩ − α⟨y, z⟩ − α⟨z, y⟩ + αα⟨y, y⟩ − ⟨z, z⟩ ≥ 0 ⇒ −α⟨y, z⟩ − α⟨z, y⟩ + αα⟨y, y⟩ ≥ 0.........(2).

26

Preliminaries

The inequality (2) holds for all y ∈ M and for any scalar α. Thus for any y ∈ M if we put ⟨z,y⟩ α = ∥y∥ 2 in (2) we have, ⟨z,y⟩ ⟨z,y⟩ ⟨z,y⟩ ⟨z,y⟩ ⇒ − ∥y∥ 2 ⟨y, z⟩ − ∥y∥2 ⟨z, y⟩ + ∥y∥2 ∥y∥2 ⟨y, y⟩ ≥ 0 2

⇒ − |⟨z,y⟩| ≥0 ∥y∥2 ⇒ −|⟨z, y⟩|2 ≥ 0 ⇒ ⟨z, y⟩ = 0 for all y ∈ M. ⇒ z ∈ M. Hence x = z + w ∈ M + M ⊥ . T Now we show that M M ⊥ = 0. T Let z1 ∈ M M ⊥ ∴ z1 ∈ Mand z1 ∈ M ⊥ . ⇒ ⟨z1 , z1 ⟩ = 0. ⇒ z1 = 0. T Thus M M ⊥ = 0. L Hence H = M M ⊥ . ⊥

Corollary 1.1. For any subset M of H, M ⊂ M ⊥ . ⊥

Corollary 1.2. If M is any closed subset of H, then M ⊥ = M. Theorem 1.34. Let M be a closed subspace H then there exists a unique pair of maps, P : H → M and Q : H → M ⊥ such that, (1) P + Q = 1H . These mapping has the following further properties: (2) P(x) = x for all x ∈ M and P(x) = 0 for all x ∈ M ⊥ . (3) ∥x∥2 = ∥P(x)∥2 + ∥Q(x)∥2 . (4) ∥x − P(x)∥ = in f {∥x − y∥ : y ∈ M} if x ∈ H. (5) P and Q are linear mappings. Definition 1.4. P and Q are respectively called the Projections of x on M and M ⊥ From the above theorem , we see that among the points of M, P(x) has the smallest distance from x. We call, ∥x − P(x)∥ as the distance of x from M. If P2 = P, then P is a projection mapping. Definition 1.5. A set E ⊂ H is said to be a set if for any x, y ∈ E, αx + β y ∈ E, where α +β = 1 Theorem 1.35. Representations of Continuous Linear Functionals ( Riesz representation theorem) : Let f be a continuous linear functional on the Hilbert space H then there is a unique y ∈ H such that f (x) = ⟨x, y⟩ for all x ∈ H .

1.3 Preliminary Concepts of Functional Analysis

27

Proof : If f (x) = 0 for all x ∈ H, then we have y = 0. Otherwise, define M = {x : f (x) = 0}. Then clearly, M is a subspace of H. Since f is continuous and {0} is a closed subset of H, M is closed. Again f (x) ̸= 0 for some x ∈ H, there exists an x ∈ H such that x ∈ / M. Thus, M ⊥ does not consists of 0 alone. Hence, there exists z ∈ M ⊥ with ∥z∥ = 1.

Put u = f (x)z − f (z)x. Since f (u) = f (x) f (z) − f (z) f (x) = 0 then u ∈ M. Thus ⟨u, z⟩ = 0. Hence ⟨ f (x)z − f (z)x, z⟩ = 0 ⇒ f (x)⟨z, z⟩ − f (z)⟨x, z⟩ = 0 ⇒ f (x) − f (z)⟨x, z⟩ = 0 ⇒ f (x) = f (z)⟨x, z⟩ ⇒ f (x) = ⟨x, f (z)z⟩ Put y = f (z)z. Then we have f (x) = ⟨x, y⟩ for all x ∈ H. To prove the uniqueness of the y suppose there are y1 , y2 ∈ H such that ⟨x, y1 ⟩ = ⟨x, y2 ⟩ for all x ∈ H. Then ⟨x, y1 ⟩ − ⟨x, y2 ⟩ = 0 ⇒ ⟨x, y1 − y2 ⟩ = 0 for x ∈ H ⇒ y1 − y2 = 0 ∴ y1 = y2 . This complete the proof of the theorem. Theorem 1.36. An orthonormal set of vectors in a Hilbert space is linearly independent. Proof :Let A be an orthonormal set in H . Assume A is finite and A = {u1 , u2 , ..., un }. Consider the equation, n

∑ ciui = 0 where c′is are scalars.

i=1

Taking inner product on both sides with u j , (1 ≤ j ≤ n) we have, n

⟨ ∑ ci ui , u j ⟩ = 0 i=1

28

Preliminaries n

⇒ ∑ ci ⟨ui , u j ⟩ = 0 i=1 n

⇒ ∑ ci δi, j = 0 i=1

⇒ c j = 0. Hence any arbitrary set A (finite or infinite ) is said to be linearly independent if every non-empty finite subset of A is linearly independent. Therefore it follows from our assertion is also valid for the case when A is infinite. Definition 1.6. Let {uα }α∈A be an orthonormal set from H. Let x ∈ H and consider the function xˆ : A → C defined by, x(α) ˆ = ⟨x, uα ⟩. Then, the numbers x(α) ˆ are called the Fourier coefficients of x with respect to the orthonormal set {uα }α∈A . Theorem 1.37. If {u1 , u2 , ..., un } be an orthonormal set in H and x ∈ H, then n

ˆ i is the projection of x on the subspace generated by {u1 , u2 , ..., un }. ∑ x(α)u

i=1

Theorem 1.38. If {u1 , u2 , ..., un } be an orthonormal set in H then for any x ∈ H, n

(a)

∑ |⟨x, ui⟩|2 ≤ ∥x∥2 (Bessel ′sInequality).

i=1

n

(b) x − ∑ ⟨x, ui ⟩ui ⊥ u j (J = 1, 2, 3, ..., n) i=1

The inequality (a) can be given, The sum of the squares of the components of a vector in various perpendicular direction does not exceed the square of the length of the vector itself. The relation (b) implies, If we subtract from a vector its components in several perpendicular directions, then the resultant has no component left in any of these directions. Hence, the resultant vector is perpendicular to each of these perpendicular directions. Bessel Inequality for n = 1 is essentially the Cauchy-Schwarz Inequality. Definition 1.7. Let E be an orthonormal set in an Hilbert space H. Then E is said to be an orthonormal basis of H if it is a maximal orthonormal set in H. Theorem 1.39. Let E be an orthonormal set in an Hilbert space H. Then E is an orthonormal basis if and only if E ⊥ = {0}.

1.3.11

Grahm-Schmidt orthogonalization process

Advantage of orthonormal sets over arbitrary linearly independent sets:

1.3 Preliminary Concepts of Functional Analysis

29

(i) x = ⟨x, u1 ⟩u1 + ⟨x, u2 ⟩u2 + ⟨x, u3 ⟩u3 + ... + ⟨x, un ⟩un the detemination of unknown coefficients for a linear combination is very simple. n

(ii) ∥x∥2 = ∑ |⟨x, ui ⟩|2 the determination of norm of a vector by inner product. i=1

Keeping in view the usefulness and convenience of orthonormal sequences over the linearly independent sequences , one is interested to generate orthonormal sequences from the linearly independent sequences . This is done by a constructive procedure, known as the Grahm (1883)Schmidt(1907) process. Let {xn } be a (finite or countably infinite) linearly independent set of vectors in inner product space X. The problem is to convert this set into an orthonormal set {un } such that span{u1 , u2 , ..., un } = span{x1 , x2 , ..., xn }, for each n. Step 1: Normalize x1 , which is necessarily non zero,so as to obtain u1 as u1 =

x1 ∥x1 ∥

Step 2: Write x2 = ⟨x2 , u1 ⟩u1 + v2 v2 = x2 − ⟨x2 , u1 ⟩u1 . Clearly v2 ̸= 0, since {xi } is linearly independent. Also, v2 ⊥ u1 , since ⟨v2 , u1 ⟩ = ⟨x2 − ⟨x2 , u1 ⟩u1 , u1 ⟩, = ⟨x2 , u1 ⟩ − ⟨x2 , u1 ⟩⟨u1 , u1 ⟩ =0 2 ,u1 ⟩u1 Now we can take u2 by normalizing v2 namely u2 = ∥vv22 ∥ that is u2 = x2 −⟨x . ∥v2 ∥ Now, since ∥u2 ∥ = 1 and ⟨u1 , u2 ⟩ = 0, it is clear that {u1 , u2 } is a orthonormal set in X. Further, u1 is a linear combination of x1 and x2 and x2 is a linear combination of u1 and u2 , we have span{u1 , u2 } = span{x1 , x2 }. Step 3: Similarly, the vector v3 is given by v3 = x3 − ⟨x3 , u1 ⟩u1 − ⟨x3 , u2 ⟩u2 is non zero, and v3 ⊥u1 , v3 ⊥u2 . Therefore, u3 =

v3 ∥v3 ∥

=

x3 −⟨x3 ,u1 ⟩u1 −⟨x3 ,u2 ⟩u2 ∥x3 −⟨x3 ,u1 ⟩u1 −⟨x3 ,u2 ⟩u2 ∥ .

obviously, {u1, u2, u3} is an orthonormal set in X and it can be easily verified that span{u1, u2, u3} = span{x1, x2, x3}.

30

Preliminaries

Proceeding in this way, we can easily covert {xn } to an orthonormal set of vectors {un } such that span{u1 , u2 , ..., un } = span{x1 , x2 , ..., xn }, for each n.

1.3.12

Best Approximation

Definition 1.8. Let X be a normed linear space and X0 be a subspace of X. For x ∈ X, an element x0 ∈ X0 is said to be a best approximation of x, if ∥x − x0 ∥ = dist(x, X0 ). A subspace X0 is said to have the best approximation if every x ∈ X has a best approximation in X0 . Definition 1.9. Let X be an inner product space and E be a subset of X. An element y ∈ E is said to be a best approximation to x from y if ∥x − y∥ ≤ ∥x − z∥ for all z ∈ E Theorem 1.40. Let Y be a subspace of an inner product space X and x ∈ X. Then, (a) an element y ∈ Y is a best approximation to x iff x − y ⊥ Y . (b) if a best approximation to x from Y exists then it is unique. Applications of Hilbert space Many of the applications of Hilbert spaces exploit the fact that Hilbert spaces support generalizations of simple geometric concepts like projection and change of basis from their usual finite dimensional setting. In particular, the spectral theory of continuous self-adjoint linear operators on a Hilbert space generalizes the usual spectral decomposition of a matrix, and this often plays a major role in applications of the theory to other areas of mathematics and physics.

Chapter 2 Spaces of continuous functions In this chapter, we have studied some continuous function spaces. First of all we have proved some theorems on continuous functions then, we have shown that the set of all continuous functions is a vector space. Lastly, we have reviewed some other continuous function spaces, their properties and some theorems related to them. Definition 2.1. Let X and Y be two non empty sets then a function is a correspondence (mapping or relation) from X to Y by which each element of X correspond to a unique element of Y. Then the set X is known as domain of the function and the set Y is called the codomain of the function. That is, a mapping f : X → Y is called a function on X if every element of X correspond to a unique element of Y. Example 2.1. Let f : R → R is a mapping defined by f (x) = 3x + 5 then the mapping f is a function.

Function Space Let X be a non empty set and let K be any arbitrary field. Consider the set F(X) of all functions from X into K. Since X is non empty hence F(X) is non empty. The sum of two functions f , g ∈ F(x) is the function f + g ∈ F(X) defined by, ( f + g)(x) = f (x) + g(x) for all x ∈ X, and the product of a scalar k ∈ K and a function f ∈ F(X) is the function k f ∈ F(X) defined by, (k f )(x) = k f (x) for all x ∈ X.

32

Spaces of continuous functions

Then F(X) with the above operations is a linear space over K. The zero vector in F(X) is the zero function 0 which maps each x ∈ X into 0 ∈ K , that is, 0(x) = 0 for all x ∈ X. Also, for any function f ∈ F(X), the function − f defined by, (− f )(x) = − f (x) for all x ∈ X is the additive inverse of the function f .

2.1

Spaces of Continuous Functions [7]

Let F(X) be the set of all continuous real valued function defined on a metric space (X,d). Then F(X) is a linear space and subspace of F(X). So, the linear space F(X) can be defined by, F(X) = { f : X → R| f is continuous}. Theorem 2.1. Every constant function is continuous i.e. f(x)=c is continuous. Proof of theorem 2.1 : Let f (x) = c0 be a constant function. Now, we show that f is continuous. Suppose that ε > 0 be given. Then, | f (x) − f (a)| = |c0 − c0 | = 0 < ε whenever |x − a| < δ for δ > 0 choosing δ = 1 This shows that lim f (x) = f (a). Hence by definition 1.2.16 f is continuous. x→a

Theorem 2.2. Every identity function is continuous i.e. f(x)=x is continuous. Proof of theorem 2.2 : Let f (x) = x be an identity function. Now, we show that f is continuous. Suppose that, ε > 0 be given . Then

| f (x) − f (a)| = |x − a| < ε taking δ = ε Hence | f (x) − f (a)| < ε whenever |x − a| < δ . That is, lim f (x) = f (a). Therefore, f is x→a continuous.

2.1 Spaces of Continuous Functions [7]

33

Theorem 2.3. Addition of two continuous functions is continuous. Proof of theorem 2.3: Now, let f and g be two continuous function at a point x = a then clearly we have lim f (x) = f (a) and lim g(x) = g(a). x→a

x→a

Then, let ε > 0 be given since f and g are continuous at x = a then there exist δ f > 0 and δg > 0 such that | f (x) − f (a)| < ε2 and |g(x) − g(a)| < ε2 whenever |x − a| < δ f and |x − a| < δg respectively. Now, let δ = min(δ f , δg ). Thus, whenever |x − a| < δ then, |( f (x) + g(x)) − ( f (a) + g(a))| = |( f (x) − f (a)) + (g(x) − g(a))| Now, using Triangle inequality, |( f (x) + g(x)) − ( f (a) + g(a))| ≤ |( f (x) − f (a))| + |(g(x) − g(a))| = ε2 + ε2 =ε ∴ lim ( f (x) + g(x)) = lim ( f (x)) + lim (g(x)) = ( f (a) + g(a)) = ( f + g)(a). Hence (f+g)(x) x→a x→a x→a is continuous. Therefore, sum of two continuous function is again a continuous function. Theorem 2.4. Product of two continuous functions is continuous. Proof of theorem 2.4: Suppose that f and g be two continuous function then by definition 1.2.16 lim f (x) = f (a) and lim g(x) = g(a). x→a

x→a

Then, let ε > 0 be given since f and g are continuous at x = a then there exist δ f > 0 and δg > 0 ε such that | f (x) − f (a)| < 2(1+|g(a)|) and |g(x) − g(a)| < 2(1+|εf (a)|) whenever |x − a| < δ f and |x − a| < δg respectively. Now, let δ = min(δ f , δg ). So, whenever |x − a| < δ then, |( f (x)g(x)) − ( f (a)g(a))| = |g(x)( f (x) − f (a)) + f (a)(g(x) − g(a))| Now, using Triangle inequality, |( f (x)g(x)) − ( f (a)g(a))| ≤ |g(x)||( f (x) − f (a))| + | f (a)||(g(x) − g(a))| ε < |g(x)| 2(1+|g(a)|) + | f (a)| 2(1+|εf (a)|) ε < (1 + |g(a)|) 2(1+|g(a)|) + (1 + |g(a) 2(1+|εf (a)|) < ε2 + ε2 =ε Since, |g(x)| = |g(x) − g(a) + g(a)| ≤ |g(x) − g(a)| + |g(a)| < |1 + g(a)| Similarly, | f (a)| < (1 + | f (a)|. Now, lim ( f (x)g(x)) = lim ( f (x)). lim (g(x)) = ( f (a).g(a)) = ( f g)(a) which shows that prodx→a x→a x→a uct of two continuous function is continuous function. Problem 2.1. Prove that F(X) is a vector space. Proof : To prove F(X) is a vector space we show that F(X) satisfies all the properties of vector space of definition 1.1.8. First of all, suppose that f , g ∈ F(X) then,

34

Spaces of continuous functions ( f + g)(x) = f (x) + g(x) = g(x) + f (x) = (g + f )(x).

Hence, f + g = g + f for all f , g ∈ F(X). Therefore F(X) satisfies (A1). Again, suppose that f , g, w ∈ F(X) then, (( f + g) + w)(x) = ( f + g)(x) + w(x) = f (x) + g(x) + w(x) = f (x) + (g(x) + w(x)) = f (x) + (g + w)(x) = ( f + (g + w))(x). Therefore, ( f + g) + w = f + (g + w) for all f , g, w ∈ F(X). Thus, F(X) satisfies (A2). Now, F(X) contains an element f (x) = 0 called zero function in F(X) such that, ( f + g)(x) = f (x) + g(x) = 0 + g(x) = g(x) That is, ( f + g) = g. Again, (g + f )(x) = g(x) + f (x) = g(x) + 0 = g(x). Hence, ( f + g) = g = g + f for all g ∈ F(X). Hence, F(X) satisfies (A3). Now, for every f ∈ F(X) there exist − f ∈ F(X) such that, ( f + (− f ))(x) = f (x) + (− f )(x) = f (x) − f (x) =0 Thus, f + (− f ) = 0. Similarly, (− f ) + f = 0. Thus , F(X) satisfies (A4). For, (M1),(M2), (M3) and (M4) let α1 , α2 ∈ R and f , g ∈ F(X) then,

(α1 ( f + g))(x) = (α1 f )(x) + (α1 g)(x) = α1 f (x) + α1 g(x)

2.1 Spaces of Continuous Functions [7]

35

Thus, (α1 ( f + g)) = α1 f + α1 g for all α1 ∈ R and f , g ∈ F(X). Again, ((α1 + α2 ) f )(x) = (α1 + α2 ) f (x) = α1 f (x) + α2 f (x). Therefore, (α1 α2 ) f = α1 (α2 f ) for all α1 , α2 ∈ R and f ∈ F(X). Also, for 1 ∈ R and f ∈ F(X) we have 1. f = f . Thus, F(X) satisfies (M1,(M2), (M3) and (M4). Hence F(X) is a vector space over the field R. Similarly, we can show that F(X) is a vector space over the field C.

2.1.1

Lipschitz Continuity [7]

Let X and Y be two metric spaces. Then, a function f : X → Y is said to be Lipschitz continuous with Lipschitz constant K if dY ( f (x), f (y)) ≤ KdX (x, y) for all x, y ∈ X.

2.1.2

Contraction Mapping [5]

A contraction mapping on a metric space (X,d) is a function f from X to itself, with the property that there is some nonnegative real numbers 0 ≤ k < 1 such that for all x and y in X, d( f (x), f (y)) ≤ k d(x, y). The smallest such value of k is called the Lipschitz constant of f . Contraction maps are sometimes known as Lipschitzian maps. If the above condition is instead satisfied for k = 1, then the mapping is said to be a non-expansive map. More generally, the idea of a contraction mapping can be defined for maps between metric spaces. Thus, if (X, dX ) and (Y, dY ) are two metric spaces, and f : X → Y , then there is a constant k < 1 such that dY ( f (x), f (y)) ≤ k dX (x, y) for all x and y in X.

2.1.3

Equicontinuity [7]

Let (X, dX ) and (Y, dY ) be metric spaces, and let F be a collection of functions f : X → Y . We say that F is equicontinuous if for all ε > 0 , there is a δ > 0 such that for all f ∈ F and all x, y ∈ X with dX (x, y) < δ , we have dY ( f (x), f (y)) < ε. Note that in the case, the same δ should not only hold at all points x, y ∈ X, but also for all functions f ∈ F

36

Spaces of continuous functions

Example 2.2. Let F be the set of all contractions f : X → X. Then F is equicontinuous, since we can choose δ = ε. To see this, just note that if dX (x, y) < δ = ε, then dX ( f (x), f (y)) = dX (x, y) < ε for all x, y ∈ X and all f ∈ F

2.1.4

Pointwise Continuous Function [7]

A function which is continuous at all points in X, but not uniformly continuous, is often called pointwise continuous. Example 2.3. The function f : R → R defined by f (x) = x2 is pointwise continuous, but not uniformly continuous. Note : If the underlying space X is compact, pointwise continuity and uniform continuity is the same. This means that a continuous function defined on a closed and bounded subset of Rn is always uniformly continuous.

2.1.5

Pointwise Convergence [7]

Let (X, dX ) and (Y, dY ) be two metric space, and let { fn } be a sequence of functions fn : X → Y . We say that { fn } converges pointwise to a function f : X → Y if fn (x) → f (x) for all x ∈ X. This means that for each x and each ε > 0, there is an N ∈ N such that dY ( fn (x), f (x)) < ε when n ≥ N.

2.1.6

Uniform Convergence [7]

Let (X, dX ) and (Y, dY ) be two metric space, and let { fn } be a sequence of functions fn : X → Y . We say that { fn } converges pointwise to a function f : X → Y if fn (x) → f (x) for all x ∈ X. This means that for each x and each ε > 0, there is an N ∈ N such that if n ≥ N , then dY ( fn (x), f (x)) < ε . It seems to be that the two definitions 2.1.5 and 2.1.6 may seem confusingly similar, but the difference is that in the last one, the same N should work simultaneously for all x, while in the first we can adapt N to each individual x. Hence uniform convergence implies pointwise convergence, but a sequence may converge pointwise but not uniformly. Before we look at an example, it will be useful to reformulate the definition of uniform convergence. Proposition 2.1. Let (X, dX ) and (Y, dY ) be two metric space, and let { fn } be a sequence of functions fn : X → Y . Then for any function f : X → Y the following are equivalent. (i) { fn } converge uniformly to f. (ii) sup {dY ( fn (x), f (x))| x ∈ X} → o as n → ∞.

37

2.1 Spaces of Continuous Functions [7]

Hence uniform convergence means that the maximal distance between f and fn goes to zero Proof : (i) ⇒ (ii) Let { fn } converges uniformly to f.For any ε > 0, we can find an N ∈ N such that dY ( fn (x), f (x)) < ε for all x ∈ X and n ≥ N. This means that sup {dY ( fn (x), f (x))|x ∈ X} ≤ ε for all n ≥ N and since ε is arbitrary then {dY ( fn (x), f (x))|x ∈ X} → 0 as n → ∞. (ii) ⇒ (i) Assume that {dY ( fn (x), f (x))|x ∈ X} → 0 as n → ∞. Given ε > 0 there exists N ∈ N such that sup{dY ( fn (x), f (x))|x ∈ X} < ε for n ≥ N. But then we have dY ( fn (x), f (x)) < ε for all x ∈ X and n ≥ N, which means that fn is uniformly continuous. Here is an example which shows clearly the distinction between pointwise and uniform convergence: Example 2.4. Let fn : [0, 1] → R be the function in Figure 1. It is constant zero except on the interval [0, 1n ] where it looks like a tent of height 1.

If we insist, the function is defined by,   2nx if  fn (x) = −2nx + 2 i f   0 if

1 0 ≤ x < 2n 1 1 2n ≤ x < n 1 n ≤x≤1

The sequence { fn } converges pointwise to 0, because at every point x ∈ [0, 1] the value of { fn }(x) eventually becomes 0 (for x = 0, the value is always 0, and for x > 0 the tent will eventually pass to the left of x.) However, since the maximum value of all { fn } is 1,

38

Spaces of continuous functions

sup{dY ( fn (x), f (x))|x ∈ [0, 1]} = 1 for all n, and hence { fn } does not converge uniformly to 0. Proposition 2.2. Let (X, dX ) and (Y, dY ) be two metric spaces, and assume that { fn } is a sequence of continuous functions fn : X → Y converging uniformly to a function f. Then f is continuous. Proof : Let a ∈ X. Given any ε > 0 we must find a δ > 0 such that dY ( f (x), f (a)) < ε, whenever dX (x, a) < δ . Since fn converges uniformly to f then there exists an N ∈ N when n ≥ N, dY ( f (x), fn (x)) < ε3 for all x ∈ X. Since fN is continuous at a then dY ( fN (x), fN (a)) < ε3 whenever dX (x, a) < δ . If dX (x, a) < δ then, dY ( f (x), f (a)) ≤ dY ( f (x), fN (a)) + dY ( fN (x), fN (a)) + dY ( fN (x), f (a)) < ε3 + ε3 + ε3 = ε. Thus, f is continuous. The next example shows that this is not always the case when we are dealing with pointwise convergence. Example 2.5. Let fn : R → R be the function in the figure 2.

It is defined by,

39

2.1 Spaces of Continuous Functions [7]    −1 i f fn (x) = nx i f   1 if

x ≤ −1 n −1 n ≤x< 1 n ≤x

1 n

The sequence { fn } converges pointwise to the function, f defined by,    −1 i f x < 0 f (x) = 0 if x = 0   1 if x > 0 Though all the functions { fn } are continuous, the limit function f is not. Lemma 2.1. Let (X, dX ) and (Y, dY ) are metric spaces, and assume that X is compact. If f , g : X → Y are continuous functions, then ρ( f , g) = sup {dY ( f (x), g(x))| x ∈ X}. is finite, and there is a point x ∈ X such that dY ( f (x), g(x)) = ρ( f , g)

2.1.7

The spaces C(X,Y)

If (X, dX ) and (Y, dY ) are metric spaces, consider C(X,Y ) = { f : X → Y | f is continuous and X is compact} be the collection of all continuous functions from X to Y . In this section we shall prove that C(X,Y ) is a metric space i.e., we can defined a metric on C(X,Y ). The basic idea is to measure the distance between two functions by looking at the point defined by, ρ( f , g) = sup {dY ( f (x), g(x))| x ∈ X}. Our first task is to show that ρ is a metric on C(X,Y ). By lemma 2.1, ρ( f , g) is always finite and now we have only to show that ρ satisfies the three properties of a metric on C(X,Y ). The first two are obvious, and we concentrate on the triangle inequality: Assume that f , g, h are three functions in C(X,Y ), we have to show that ρ( f , g) = ρ( f , h) + ρ(h, g). According to lemma 2.1, there is a point x ∈ X such that ρ( f , g) = dY ( f (x), g(x)) ⇒ ρ( f , g) ≤ dY ( f (x), h(x)) + dY (h(x), g(x)) ⇒ ρ( f , g) ≤ sup{dY ( f (x), h(x)) : x ∈ X} + sup{dY (h(x), g(x)) : x ∈ X} ∴ ρ( f , g) ≤ ρ( f , h) + ρ(h, g). Hence C(X,Y) is a metric space.

40

2.2

Spaces of continuous functions

The Space C(Y) [1]

Let Y be a metric space. The set of all bounded continuous K valued functions on Y is denoted by C(Y) i.e., C(Y ) = { f : Y → K| f is bounded continuous K valued f unction}. This is a linear space with respect to pointwise addition and scalar multiplication of functions. Also, ∥ f ∥∞ = sup{| f (x)| : x ∈ Y and f ∈ C(Y )} defines a norm called supnorm on C(Y).

2.2.1

The Space C[a,b]

The collection of all continuous functions defined on an interval [a,b] is denoted by C([a,b]). Then C[a,b] is a vector space over the field R. Also, C[a,b] is an inner product space and a normed linear space.

2.2.2

Support of a function

Suppose that f : M → R is a real-valued function whose domain is an arbitrary set M. The support of f, written supp(f), is the set of points in X where f is non-zero i.e.,

supp( f ) = {x ∈ X | f (x) ̸= 0}. If the supp(f) is a compact subset of M then the function is said to be function with compact support.

2.2.3

The Space CC (X)

Let X be a metric space then the set of all continuous complex functions on X whose support is compact, is denoted by Cc(X).

2.2.4

The Space C0 (X)

A complex function f defined on a locally compact Hausdorff space X is said to vanish at infinity if for every ε > 0 there exists a compact set K ⊂ X such that | f (x)| < ε for all x ∈ / K.

41

2.2 The Space C(Y) [1]

Then C0 (X) is the class of all continuous complex function f on X which vanish at infinity. Theorem 2.5. CC (X) is dense in C0 (X) Proof : If we define a metric d on in C0 (X) by, d( f , g) = ∥ f − g∥. Then in C0 (X) is a metric space. Given ε > 0, there exists a compact set K so that | f (x)| < ε outside K. Then, by Urysohn′ s lemma we have a function g ∈ CC (X) such that 0 ≤ g ≤ 1 and g(x) = 1 on K. Put h = f g, then h ∈ CC (X) and ∥ f (x) − h(x)∥ = sup | f (x) − h(x)| x∈X

= sup | f (x) − f (x)g(x)| x∈X

= sup | f (x)(1 − g(x))| x∈X

confirms that ∥ f (x) − h(x)∥ < ε. Hence CC (X) is dense in C0 (X). Theorem 2.6. C0 (X) is a complete metric space. Proof : If we define a metric d on in C0 (X) by, d( f , g) = ∥ f − g∥. Then in C0 (X) is a metric space. Let { fn } be a Cauchy sequence in C0 (X) i.e., assume that { fn } converges uniformly. Then, its pointwise limit function is continuous. Given ε > 0, there exists an n such that | fn − f | < ε2 and there is a compact K so that | fn (x)| < ε2 outside K. Hence | f (x)| = | f (x) − fn (x) + fn (x)| ≤ | f (x) − fn (x)| + | fn (x)| = ε2 + ε2 =ε which confirms that f vanishes at infinity. Thus, C0 (X) is complete. Theorem 2.7. If X be a locally compact Hausdorff space then C0 (X) is the completion of CC (X), under the metric defined by the supremum norm, ∥ f ∥ = sup | f (x)| x∈X

42

Spaces of continuous functions For the proof of the theorem 2.2.4 see the proof of theorems 2.5 and 2.6

Definition 2.2. Let T be the set of positive integers with discrete metric. Then C(T) is the set of all bounded sequences {xn } in K. The space is denoted by l ∞ and the norm is given by, ∥x∥∞ = supn≥n |xn |, x ∈ l ∞ . The subspace C0 (T ) becomes,

c0 = {x ∈ l ∞ : lim }xn = 0 n→∞

, and the subspace CC (T ) becomes, c00 = {x ∈ l ∞ : xn = 0 f or all but f inite number values o f n}.

Chapter 3 Measure Spaces and L p Spaces In this chapter, we have studied about measure spaces and L p spaces. At first, we have reviewed prerequisite concepts of measure theory. Then, we have studied about measure space, measurable functions, Lebesgue measure and some of their properties. Also, we have studied the construction, some properties and limitations of the Riemann integration. After that, we have studied the construction and advantages of Lebesgue integration and properties of Lebesgue measurable functions. Lastly, we have studied L p spaces some theorems related to them.

3.1 3.1.1

Some basic concepts of Measure Theory Set Function

A function is said to be a set function if its domain is the collection of sets. That is, a function S f : X → R {−∞, +∞} is said to be a set function if X is the collection of sets.

3.1.2

Extended Real Number System

Let R be the set of real number. Then, the set of extended real numbers is denoted by, R∗ and S is defined by R∗ = R {−∞, +∞}, where two new symbols −∞ and +∞ are known as minus infinity and plus infinity, respectively. We known that R is an ordered set. Now, the order relation on R∗ for any x, y ∈ R and x < y is −∞ < x < y < +∞. Again we have the Addition and Multiplication operation are defined as follows: (i) (−∞) + x = x + (−∞) = −∞ for all x ∈ R. (ii) (+∞) + x = x + (+∞) = +∞ for all x ∈ R. (iii) (+∞) − x = x − (+∞) = +∞ for all x ∈ R.

Measure Spaces and L p Spaces

44

(iv) (−∞) − x = x − (−∞) = −∞ for all x ∈ R. (v) (+∞).x = x.(+∞) = +∞ for all x ∈ R and x > 0. (vi) (−∞).x = x.(−∞) = (−∞) for all x ∈ R and x > 0. (vii) (+∞).x = x.(+∞) = (−∞) for all x ∈ R and x < 0. (viii) (−∞).x = x.(−∞) = (+∞) for all x ∈ R and x < 0. And (−∞) + (+∞), (−∞) − (+∞) are not defined. Then, the set R∗ with these operations is called an extended real number system.

3.1.3

Intervals in R

There are two types of intervals, one is finite or bounded interval and another one is infinite or unbounded interval. The interval of the types a < x < b , a ≤ x < b , a ≤ x ≤ b , a < x ≤ b are known as bounded or finite intervals where a, b, x ∈ R. The interval of the types −∞ < x < b , −∞ ≤ x < ∞ , −∞ ≤ x ≤ +∞ , a < x ≤ +∞ are known as unbounded or infinite intervals where a, b, x ∈ R.

3.1.4

Length of Intervals

The length of all finite intervals of the form a < x < b , a ≤ x < b , a ≤ x ≤ b , a < x ≤ b is defined by b-a. That is, ℓ((a, b)) = ℓ([a, b)) = ℓ((a, b]) = ℓ([a, b]) = b − a. Similarly, the length of all infinite intervals is defined by ∞. That is, ℓ((−∞, a)) = ℓ((a, ∞)) = ℓ([−∞, a)) = ℓ([a, ∞]) = ℓ((−∞, ∞)) = ∞. In the case of, a = b of finite intervals a ≤ x ≤ b has length zero. Hence length is an example of a set function which associates an extended real number to each set in some collection of sets. In the case of length, the domain is the collection of all intervals. The set function ℓ clearly satisfies the following properties : (i) ℓ(I) ≥ 0 , where I is an interval. S (ii) If {Ii } is a countable collection of mutually disjoint intervals such that i Ii is an intervals then, S ℓ( i Ii ) = ∑∞ i=1 ℓ(Ii ). (iii) ℓ(I + x) = ℓ(I) , where I is an interval and x ∈ R.

Example 3.1. ℓ((3, 7)) = 7 − 3 = 4 and ℓ((a, ∞)) = ∞. Example 3.2. I1 = [1, 3), I2 = [3, 6), I3 = [6, 8] then,

45

3.1 Some basic concepts of Measure Theory S

S

S

Ii = I1 I2 I3 S S = [1, 3) [3, 6) [6, 8]. = [1, 8]. S

Hence all Ii are mutually disjoint and their union is again an interval. So, ℓ( Ii ) = ℓ(I1 ) + ℓ(I2 ) + ℓ(I3 ) = 2 + 3 + 2 = 7 = ℓ([1, 8])

3.1.5

Length of Sets

We know that any open set can be expressed as a countable union of mutually disjoint open S intervals. That is, any open set O can be written as O = i Ii where each Ii are mutually S disjoint open intervals. Then, the length of open set O is defined by, ℓ(O) = ℓ( i Ii ) = ∑∞ i=1 ℓ(Ii ). Now, suppose that F is a closed set contained in an open interval (a, b), then F c = (a, b) − F c here the complement of F is taken with respect to (a,b). Then, ℓ(F ) = b−a−ℓ(F) therefore, the length of closed set F is defined by ℓ(F) = b − a − ℓ(F c ). So far we have extended the concept of length to open and closed sets. Since the class of these sets are too restricted so we would like to extend the concept of length to a wider class of sets in R. If possible, to all subsets of R. In this regard, we imagine a set function m which assigns to each set E in R to a nonnegative extended real number m(E) called the measure of E (an extension of the notion of length function) satisfying the following properties : (i) m(E) is defined for all subset E of R. (ii) m(I) = ℓ(I) for an interval I. S (iii) m( i Ei ) = ∑∞ i=1 ℓ(Ei ) for any sequence {Ei } of disjoint set in R. (iv) m(E + x) = m(E) , where x is any fixed number. But unfortunately, it is not possible to construct such a set function which satisfies all the properties (i) to (iv). Now, losing the first property that is, property (i) is replaced by m(E) is not defined for all E ∈ P(R) and weaken (iii) by finite additivity and countably subadditivity S S that is , m( i Ein ) = ∑ni=1 ℓ(Ei ) and m( i Ei∞ ) ≤ ∑∞ i=1 ℓ(Ei ). Thus if we wish to keep all the condition (i) to (iv) we must conclude that all subset of R have not a measure. Similarly, if we want that all set have a measure then we must sacrifice one or more of the properties from (i) to (iv). S

S

Example 3.3. Consider the open set G = (1, 3) (4, 7) (7, 9) and a closed set F = S [1, 3] [4, 7]. Then using the definition of length of open sets ℓ(G) = ℓ((1, 3)) + ℓ((4, 7)) + ℓ((7, 9)) since (1, 3), (4, 7), (7, 9) are mutually disjoint intervals. Hence, ℓ(G) = 3 − 1 + 7−4+9−7 = 7

Measure Spaces and L p Spaces

46

Since, F ⊂ [1, 7] then by using the definition of length of closed set ℓ(F) = 7 − 1 − ℓ(F c ) where F c is the complement of F relative to [1, 7]. Now, ([1, 3])c = [1, 7] − [1, 3] = (3, 7] similarly ([4, 7])c = [1, 7] − [4, 7] = [1, 4). We know by De’Morgan’s law that, F c = T T ([1, 3])c ([4, 7])c = (3, 7] [1, 4) = (3, 4). Therefore ℓ(F c ) = 1. Thus ℓ(F) = 7 − 1 − 1 = 5. Semi Algebra Let X be a non empty set and C be the collection of subsets of X then C is said to be a semi algebra on the set X if the following hold: (i) X, φ ∈ M T (ii) for all A, B ∈ C A B ∈ C S (iii) if A ∈ C then Ac = ni=1 Ai , where Ac is the compliment of A and Ai ’s are mutually disjoint members of C. Example 3.4. The collection of half-open intervals [a,b) in R is a Semi Algebra.

3.1.6

Monotone Class

Let X be a non empty set and N be the collection of subsets of X then N is said to be a monotone class if the following hold: S (i) if A1 ⊆ A2 ⊆ A3 ...... then ∞ i=1 Ai ∈ N. T∞ (ii) if A1 ⊇ A2 ⊇ A3 ...... then i=1 Ai ∈ N.

3.1.7

Measurable Space and Measurable Set

Let X be a non empty set and M be the collection of subsets of X then M is called an Algebra on the set X if the following hold: (i) X ∈ M (ii) i f A ∈ M the Ac ∈ M where Ac is the complement of A relative to X. S (iii) if A, B ∈ M then A B ∈ M. S An algebra M is said to be a σ −algebra on X, if for i = 1, 2, 3, ............ ∞ i=1 Ai ∈ M , whenever Ai ∈ M for each i . Then the set X together with a σ −algebra M is called a measurable space and the member of M are called measurable sets. Example 3.5. Let X be a non empty set then consider M = P(X) then M is a σ −algebra on X and each member of M is a measurable set.

3.1 Some basic concepts of Measure Theory

47

Example 3.6. Let X be a non empty set and E ⊂ X and M = {X, 0, / E, E c } where E c is the complement of E then M is a σ −algebra on X and each member of M is a measurable set. Example 3.7. Let X be a non empty set and M = {X, 0} / then M is a σ −algebra on X.

3.1.8

Borel Measurable Set and Boerel Measure

If F is any collection of subsets of X then there exists a smallest σ −algebra B in X such that F ⊂ B. If X is a topological space then there exists a smallest σ −algebra B in X such that every open set in X belongs to B. The σ −algebra B on X is called Borel σ −algebra and members of B are called Borel sets or Borel measurable sets. Then, a countably additive non-negative set function µ : B → [0, ∞] defined on a Borel σ −algebra is called Borel measure on X.

3.1.9

Measurable Function

If X is a measurable space with a σ −algebra M and Y is a topological space with a topology T on Y then the mapping f : X → Y is called a measurable function, if for every open set V in Y, f −1 (V ) is measurable in X.

3.1.10

Some properties of Measurable Function [? ]

Proposition 3.1. Let f be an extended real-valued function whose domain is measurable.Then the following statements are equivalent: (i) E( f > α) = x ∈ E : f (x) > α is measurable for all α ∈ R. (ii) E( f ≥ α) = x ∈ E : f (x) ≥ α is measurable for all α ∈ R. (iii) E( f < α) = x ∈ E : f (x) < α is measurable for all α ∈ R. (iv) E( f ≤ α) = x ∈ E : f (x) ≤ α is measurable for all α ∈ R. These statements imply, (v) E( f = α) = x ∈ E : f (x) = α is measurable for all α ∈ R. Proof : (i) ⇔ (iv), since E( f > α) = {x ∈ E : f (x) > α}, and a set is measurable if and only if its compliment is measurable. Similarly, (ii) ⇔ (iii), since E( f ≥ α) = {x ∈ E : f (x) ≥ α}=complement of E( f < α) = {x ∈ E : f (x) < α}. ∞ \ 1 Now, (i) ⇔ (ii), since E( f ≥ α) = {x ∈ E : f (x) ≥ α} = {x ∈ E : f (x) > α − }, and n n=1 the intersection of a sequence of measurable set is measurable.

48

Measure Spaces and L p Spaces

Similarly, (ii) ⇔ (i), since E( f > α) = {x ∈ E : f (x) > α} =

∞ [

1 {x ∈ E : f (x) > α + }, n n=1

and the union of a sequence of measurable set is measurable. this shows that the first four statements are equivalent. I f α is any real number, S E( f = α) = {x ∈ E : F(x) = α} = {x ∈ E : F(x) ≥ α} {x ∈ E : F(x) ≤ α}, and so (ii) and (iv) implies (v). Definition 3.1. An extended real-valued function f defined on a measurable set E is said to be measurable if the set E( f > α) is measurable for every real α. The measure of the set E( f > α) may be finite or infinite. As α varies, the behavior of the set E( f > α) describes how the values of the function f are distributed. Intuitively, it is obvious that the smoother the function f is, the smaller the variety of the sets will be. For instance, if E = R and f is continuous on R, the set E( f > α) is always open. Problem 3.1. Show that the constant function with a measurable domain is measurable. Solution : Let f : E(measurable) → R∗ be constant function defined by f (x) = k, where k is a constant. Then, for any real α, ( E if α < k E( f > α) = 0/ i f α ≤ k. This implies that, E( f > α) is measurable since both E and 0/ are measurable. Example 3.8. Let f : [0, 1] → R be defined by,  1   x if 0 < x < 1 f (x) = 5 if x=0   7 if x=1 then the function  f is measurable. Now, we show that f is measurable. Since  0/ if α ≤1    1   ( α , 1) if 1 < α < 5  S 1 E( f ≤ α) = ( 5 , 1) {0} i f α =5  S  1  ( , 1) {0} i f 5 < α < 7  α    ( 1 , 1) S{0, 1} i f α ≥7 α This implies that, E( f ≤ α) is a measurable set for all α since, each of the set in right side is measurable. Hence f is measurable. Problem 3.2. Let f be real valued function defined on a set E = [0, 1] by

49

3.1 Some basic concepts of Measure Theory ( f (x) =

0 i f x is an irrational in [0, 1] 1 if x is a rational in [0, 1]

Investigate whether f is measurable or not? Solution : Let α be any real number,   if α ≥1  T0/ E( f > α) = {x ∈ E : f (x) > α} = Q [0, 1] i f 0 ≤ α < 1   0, 1 if α α) = {x ∈ E : f (x) > α}. Clearly, E( f > α) ⊂ E and m(E) = 0. Thus E( f > α) is measurable. Hence f is measurable.

Problem 3.4. Show that a constant function with a measurable domain is measurable. Solution : Let f : E → R be a function defined by f (x) = k for all x ∈ E where k is a constant. For any real number α we have, ( E( f > α) =

E if α < k 0/ i f α ≥ k

This implies that the set E( f > α) is measurable since both E and φ are measurable.

Theorem 3.1. If f is a measurable function defined on E, then the set E( f = α) is measurable for each extended real number α. Proof : For a real number α we have, E( f = α) = {x ∈ E : f (x) = α} T = {x ∈ E : f (x) ≥ α} {x ∈ E : f (x) ≤ α} Since f is measurable then E( f = α) is measurable for each α. For α = ∞, we have

Measure Spaces and L p Spaces

50 ∞ \

E( f = ∞) = {x ∈ E : f (x) = ∞} =

{x ∈ E : f (x) ≥ n}.

n=1

Again for α = −∞, we have

E( f = −∞) = {x ∈ E : f (x) = −∞} =

∞ \

{x ∈ E : f (x) ≤ −n}. Hence the theorem is proved.

n=1

The converse of the Theorem need not be true. Theorem 3.2. (a) If f is a measurable function on the set E and E1 ⊂ E is a measurable set, then f is measurable on E1 . (b) If f is a measurable function on each of the sets in a countable collection {Ei } of n

disjoint measurable sets, then f is measurable on

∑ Ei.

i=1

(c) If f and g are measurable functions on a common domain E, then the set A( f , g) = {x ∈ E : f (x) < g(x)} is measurable. (d) Let f be a function defined on a measurable set E. Then f is measurable if and only if, for any open set G in R, the inverse image f −1 (G) is a measurable set.

Theorem 3.3. (Monotone Convergence Theorem)[23] Given a measure space (X, A, µ). let ( fn : n ∈ N) be an increasing sequence of nonnegative extended real-valued A-measurable Z Z function on a set D ∈ A and let f = lim fn on D. Then lim n→∞

n→∞ D

fn dµ =

f dµ. D

Theorem 3.4. (Fatou’s Lemma)[23] Given a measure space (X, A, µ). Then for every sequence ( fn : n ∈ N) of nonnegative extended real-valued A-measurable functions on a set D ∈ A, we have Z Z lim inf fn dµ ≤ lim inf fn dµ (3.1) D n→∞

n→∞

D

In particular if f = lim fn exists a.e. on D, then n→∞

Z

f dµ ≤ lim inf n→∞

D

Z D

(3.2)

fn dµ

Proof: By definition, lim inf fn = lim inf fk . Since (infk≥n fk : n ∈ N) is an increasing n→∞

n→∞ k≥n

sequence of nonnegative extended real-valued A-measurable functions on D, Theorem (3.3) implies that Z Z Z Z Z lim inf fn dµ =

D n→∞

lim inf fk dµ = lim

D n→∞ k≥n

inf fk dµ = lim inf

n→∞ D k≥nR

n→∞

inf fk dµ ≤ lim inf

D k≥n

n→∞

D

fn dµ,

where the third equality is from the fact that ( D infk≥n fk dµ : n ∈ N) is an increasing sequence in R so that its limit exists and the limit is equal to the limit inferior and the last

3.1 Some basic concepts of Measure Theory

51

inequality is from the fact that infk≥n fk ≤ fn . This proves (3.1). If f = lim fn exists a.e. on n→∞ D, then f = lim inf fn a.e. on D and thus (3.2) follows from (3.1) by Definition (??). n→∞

3.1.11

Measure of a Set

The measure of a set is a non-negative countably additive extended real valued set function i.e., a measure µ is a function from a class of set into the set of extended real numbers.

3.1.12

Measure and Measure Space

Let X be measurable space with a σ −algebra M on X. Then X together with a function µ : M → [0, ∞] called measure on M is said to be a measure space.

3.1.13

Simple Function

A measurable real-valued function ψ defined on a set E is said to be simple provided it takes only a finite number of real values. Let ψ be a simple measurable function with its canonical representation given by ψ(x) = ∑ni=1 αi χEi where the set Ei = {x|ψ(x) = αi } are disjoint and measurable and αi are distinct non-zero numbers.

3.1.14

Outer Measure And Properties of Outer Measure

Let E ⊆ R be arbitrary set. The outer measure or exterior measure of the set E is denoted by S∞ µ ∗ (E) and defined by µ ∗ (E) = in f ZE where ZE = {∑∞ n=1 ℓ(In ) : E ⊆ n=1 In } where In are intervals. The outer measure µ ∗ is a set function from the power set P(R) into the set of all non-negative extended real numbers. Hence the outer measure of a set E is the infimum of the lengths of all possible covers of E. Since The set ZA is bounded from below by 0 so the infimum always exists for any subset of R but it is not bounded above. We can deduce something about the set ZA straight away : it is an interval of the form [x, ∞] or (x, ∞] where x is the infimum of ZA . To see that ZA is an interval of one of these forms, suppose that r is the length of one of the covers of A. Therefore r ∈ ZA . There may be lengths of covers of A that are smaller than r (if r is not the infimum of ZA ), but it is definitely true that every number from r to ∞ is the length of a possible cover of A. This is because we can always increase the total length of any given cover by any number, simply by increasing the widths of the constituent covering intervals. Therefore if r ∈ ZA then it must also be true that every number bigger than r is in ZA , so [r, ∞] ⊆ ZA . If x is the infimum

52

Measure Spaces and L p Spaces

of ZA , it follows that ZA is either the interval [x, ∞] (if x ∈ ZA ) or the interval (x, ∞] (if x in not in ZA ) because every number bigger than x must be in ZA and there is no cover of A with a length smaller than x. Therefore ZA must be an interval as claimed. Thus the outer measure of any set E ⊆ R has the following properties: (i) µ ∗ (E) is defined for all subset E of R. (ii) µ ∗ (I) = ℓ(I) for an interval I. S S (iii) µ ∗ ( ni=1 Ei ) = ∑ni=1 ℓ(Ei ) for any sequence {Ei } of disjoint set in R and µ ∗ ( i Ei∞ ) ≤ ∑∞ i=1 ℓ(Ei ). (iv) µ ∗ (E + x) = µ ∗ (E) where x is any fixed number. Theorem 3.5. If A ⊆ R is a null set then µ ∗ (A) = 0. Proof : Since the empty set 0/ is contained in every non empty set that is every non empty set will cover 0. / So, let 0/ ⊂ (a, a) where a ∈ R. Since µ ∗ ((a, a)) = 0 then µ ∗ (0) / ≤ 0 by ∗ using the monotonicity property of outer measure. Hence µ (0) / = 0. Theorem 3.6. If A ⊆ B then µ ∗ (A) ≤ µ ∗ (B). Proof : Suppose that A ⊆ B that is every cover of B must be a cover of A so r ∈ ZB implies r ∈ ZA . It follows that ZA ⊆ ZB . Now, let xA = in f ZA and xB = in f ZB . Since, ZA is an interval of the form [xA , ∞] or (xA , ∞] and ZB is an interval of the form [xB , ∞] or (xB , ∞]. Then the only way that ZA ⊆ ZB is xA ≤ xB which is the same as saying µ ∗ (A) ≤ µ ∗ (B). Theorem 3.7. [17] The outer measure of an interval equals its length. Proof: First, consider an unbounded interval I . Since it cannot be covered by a system of intervals with finite total length then we have µ ∗ (I) = ∞ and so µ ∗ (I) = ℓ(I) = ∞. Hence for unbounded interval outer measure of an interval is equal to its length. Now consider a bounded interval I. We will first prove that µ ∗ (I) ≤ ℓ(I) and then prove that µ ∗ (I) ≥ ℓ(I), thus enabling us to conclude that µ ∗ (I) = ℓ(I). Let ε > 0 and I = [a, b] be a bounded interval. We show that µ ∗ (I) = ℓ(I). Since I = [a, b] ⊆ (a − ε, b + ε) then using the definition of outer measure µ ∗ (I) ≤ b − a + 2ε , since this is true for any ε > 0 so µ ∗ (I) ≤ b − a. Now, let us consider a finite cover {In } of [a, b] by using Heine Borel property that every cover of a closed and bounded set has a finite subcover. Since a ∈ [a, b] then there exist a member of {In } say (a1 , b1 ) such that a1 < a < b1 . If b ≤ b1 then [a, b] ⊆ (a1 , b1 ) and b − a ≤ b1 − a1 . Proceeding in this way, we can show that µ ∗ (I) ≥ b − a. Which completes the proofs.

3.1 Some basic concepts of Measure Theory

53

Theorem 3.8. For any subset A of R, let A + t = {a + t : a ∈ A}. Then, the outer measure of A is translation invariant that is µ ∗ (A + t) = µ ∗ (A). Proof: If a set of intervals {In : n ≥ n} covers A then, {In +t : n ≥ n} covers the set A +t. But each interval In + t has the same length as In since the length of an interval does not change if we translate it along the real line. Therefore, the length of the cover {In : n ≥ n} of A is same as that of A+t. Since the same is true for all possible cover of A and A+t, there is a one one correspondence between the length of the covers of ZA and the length of the covers of A+t in ZA+t . Therefore, ZA = ZA+t and so in f ZA = in f ZA+t which is the same as saying that µ ∗ (A + t) = µ ∗ (A). Hence proved. Corollary 3.1. µ ∗ (E) = 0 for every singleton set E. Proof: Let A = {a} be a singleton set. Then (a − 1n , a + 1n ) covers A for each n. We know that µ ∗ (A) ≤ µ ∗ (a − 1n , a + 1n ) = n2 for all integers n. Then clearly, µ ∗ (A) = 0. Corollary 3.2. The outer measure of a finite and countable set is 0. S Proof: Let A be a countable set and let A = {a0 , a1 , a2 , ......, an , ......}.Now, A = ∞ i=1 ai and ∗ using subadditivity property, µ ∗ (A) ≤ ∑∞ i=1 ai = 0. Hence µ (A) = 0. That is, countable set has measure zero. Corollary 3.3. The set [0, 1] is uncountable. Since, µ ∗ ([0, 1]) = 1 and we know that every countable set has measure zero hence [0,1] is uncountable. Corollary 3.4. Any set with outer measure different from zero is uncountable but converse is not necessarily true. Proof: We know that the cantor set is an uncountable set but it has measure zero. So it is not necessary that every the set with measure zero is not uncountable. A set with outer measure zero is countable is not always true. Example 3.9. The cantor set C is uncountable with outer measure zero. Problem 3.5. Let A be the set of irrational numbers in the interval [0, 1]. Prove that µ ∗ (A) = 1 . T Solution: Let B = Q [0, 1], the set of rational numbers in [0, 1], and note that B is countable as a countable subset of a countable set, namely Q. We know that the outer measure of a countable set of real numbers is zero. Hence µ ∗ (B) = 0. S T Now, A B = [0, 1] and A B = φ . Thus, by the subadditivity property of the outer measure

Measure Spaces and L p Spaces

54

among other things, 1 = ℓ([0, 1]) = µ ∗ ([0, 1]) = µ ∗ (A B) ≤ µ ∗ (A) + µ ∗ (B) = µ ∗ (A) + 0 or µ ∗ (A) = 1.On the other hand, A ⊂ [0, 1], and by the monotonicity property of the outer measure,we also have µ ∗ (A) ≤ 1. Hence µ ∗ (A) = 1 S

Problem 3.6. If µ ∗ (A) = 0 then µ ∗ (A B) = µ ∗ (B). S From the subadditivity of µ ∗ (A B) ≤ µ ∗ (A) + µ ∗ (B) = µ ∗ (B). S S S Since B ⊂ (A B) again using subadditivity µ ∗ (A B) ≥ µ ∗ (B). Hence µ ∗ (A B) = µ ∗ (B) when µ ∗ (A) = 0. Important Notes : While outer measure has the advantage that it is defined for all sets in R, it is not countably additive. It becomes countably additive ; however, if we suitably reduce the family of sets on which it is defined . Perhaps the best way of doing this is to use the following definition due to Caratheodory. ´ S

3.1.15

Construction of Lebesgue Measure

A set E is said to be measurable if for each set A we have µ ∗ (A) = µ ∗ (A E c )+ µ ∗ (A E).The definition of measurability says that the measurable sets are those (bounded or unbounded) which split every set (measurable or not) into two pieces that are additive with respect to the outer measure .If E is a measurable set , we define the Lebesgue measure µ(E) to be the outer measure of E, i.e. µ(E) = µ ∗ (E). Thus, µ is the set function obtained by restricting the set function µ ∗ to the family M of measurable sets. T

T

Definition 3.2. The set function µ : M → R∗ , where M is the class of all measurable set, obtained by restricting µ ∗ to the subset M of the domain P(R) of definition of µ ∗ , called the Lebesgue measure.

3.1.16

Some Properties of Lebesgue Measure

Theorem 3.9. If E is a measurable set, then so is E c . Proof : Since, E is measurable then for any set A we have, T T µ ∗ (A) = µ ∗ (A E c ) + µ ∗ (A E) T T = µ ∗ (A E c ) + µ ∗ (A (E c )c Hence E c is measurable. Theorem 3.10. The sets 0/ and R are measurable sets. The proof of the this theorem is obvious. Theorem 3.11. If µ ∗ (E) = 0 , then E is measurable . Furthermore , every subset of E is measurable.

55

3.1 Some basic concepts of Measure Theory Theorem 3.12. Every countable set is measurable and its measure is zero.

Corollary 3.5. The cantor set C and all of its subsets are measurable and each of them has measure zero. Theorem 3.13. If E1 and E2 are measurable sets, so is E1

S

Theorem 3.14. If E1 and E2 are measurable sets, so are E1

E2 .

T

E2 , E1 − E2 and E1 △E2 .

Theorem 3.15. Let E1 , E2 , ..., En be a finite sequence of disjoint measurable sets. then for any set A, ! n n  \  \ [ [ µ∗ A Ei ) = µ ∗ A Ei i=1

i=1

Theorem 3.16. Let E1 , E2 , ..., En be a finite sequence of disjoint measurable sets. then for any set A, ! µ∗

n [

Ei

=

i=1

n [

µ ∗ (Ei )

i=1

Problem 3.7. If E1 and E2 are measurable sets such that E1 ⊃ E2 and µ ∗ (E2 ) < ∞ ,then µ ∗ (E1 − E2 ) = µ ∗ (E1 ) − µ ∗ (E2 ). Problem 3.8. Let E be a measurable set . Then any translate E + y is measurable , where y is a real number. Furthermore µ ∗ (E + y) = µ ∗ (E). Theorem 3.17. A countable union of measurable sets is a measurable set. Theorem 3.18. Let {Ei } be an infinite sequence of disjoint measurable sets. then for any set A, ! µ∗

∞ [

i=1

Ei

=

∞ [

µ ∗ (Ei )

i=1

Theorem 3.19. Every Borel set in R is measurable ,that is , B ⊂ M.

3.1.17

Existence and Construction of Nonmeasurable Set

Definition 3.3. Let {Ai }i∈I be a family of non-empty subsets of X . Then a function f : {Ai }i∈I → X is called a choice function if, for every i ∈ I, f (Ai ) ∈ Ai that is , if the image of each set is an element in the set. Essentially a choice function, for a given family of sets, chooses an element from each set in the family. Whether or not a choice function exists for an arbitrary family of sets, is a question which lies at the foundation of the theory of sets. Definition 3.4. There exists a choice function for any non-empty family of non-empty sets.

Measure Spaces and L p Spaces

56

Most of the sets we usually come across in analysis are measurable. However, there are several examples of nonmeasurable sets given by G. Vitali (1905), Van Vleck (1905), F. Bernstein (1908) and others. However all these examples have been constructed on the assumption that the axiom of choice of set theory is valid, and it was not clear until recently whether a nonmeasurable set could be constructed without assuming the validity of the axiom choice. Recently, Robert Solovay (1970) has solved this problem by proving that the existence of nonmeasurble sets cannot be established if the axiom of choice is disallowed.

Definition 3.5. Let x and y are two real number in (0,1] and them sum module 1 of x and y is denoted by

( ˙ = x+y

x+y if x + y < 1 x + y − 1 if x + y ≥ 1

If E is any measurable subset of [0,1) then translate modulo 1 of E by x is given by ˙ = {z|z = x+y ˙ for all x ∈ E}. E +x Then it is easy to verify that ˙ ∈ [0, 1) (i) x, y ∈ [0, 1) ⇒ x+y ˙ is associative. (ii) The operation + (iii) If we assign to each x ∈ [0, 1) the angle 2πx then the sum modulo one corresponds to the addition of angles, and translate modulo 1 by y corresponds to rotation through an angle of 2πy . ˙ is Theorem 3.20. [18] Let E ⊂ [0, 1) and E is measurable and y ∈ [0, 1) then the set E +y ˙ = m(E). measurable and m(E +y) T T Proof: Define E1 = E [0, 1 − y) and E2 = E [1 − y, 1) then clearly E1 , E2 are disjoin and S E1 E2 = E. Therefore m(E) = m(E1 ) + m(E2 ) ˙ = E1 + y and E2 +y ˙ = E2 + y − 1 and so E1 +y ˙ and E2 +y ˙ are disjoint and Now, E1 +y ˙ = m(E1 + y) = m(E1 ) and = m(E2 + y) = m(E2 ) since m is measurable with m(E1 +y) S ˙ = E1 +y ˙ ˙ Hence E +x ˙ is measurable with m(E +y) ˙ = translation invariant . Also E +y E2 +y. ˙ = m(E1 ) + m(E2 ) = m(E) m(E1 + y) + m(E2 +y) Theorem 3.21. There exist a nonmeasurable set in the interval [0, 1). Proof: Let I = [0, 1). If x-y is a rational number, we say that x and y are equivalent and write x ∼ y . Clearly it is an equivalence relation and the relation ′ ∼′ partitions the set I into mutually disjoint equivalence classes. That is, any two elements of the same class differ

3.2 Construction and some properties of Riemann Integration

57

by a rational number while those of the different classes differ by an irrational number. By the Axiom of Choice we can form a set P by selecting exactly only one element from each equivalence class. Clearly P ⊂ [0, 1).

3.2

Construction and some properties of Riemann Integration

Definition 3.6. Let [a,b] be an interval then by a partition of [a,b] we mean a set of points P = {x0 , x1 , x2 , ..., xn } such that a = x0 < x1 < x2 < ... < xn−1 < xn = b. Example 3.10. Consider the interval [1,3] then P = {[1, 2], [2, 3]} is a partition of [1,3]. Again P2 = {[1, 32 ], [ 23 , 2], [ 25 , 3]} is also another partition of [1,3]. Definition 3.7. Let [a,b] be an interval and P1 and P2 are two partition of [a,b] such that P1 ⊆ P2 then P2 is called a refinement of P1 . Example 3.11. Consider the interval [1,3] such that P1 = {[1, 2], [2, 3]} and P2 = {[1, 32 ], [ 32 , 2], [ 52 , 3]} be two partitions of [1,3]. Then clearly P2 is a refinement of P1 since P1 ⊆ P2 .

3.2.1

Riemann Integration

Let f be a bounded real valued function defined on a compact interval [a,b] and P = {x0 , x1 , x2 , ..., xn } of [a, b] such that a = x0 < x1 < x2 < ... < xn−1 < xn = b. Then for each partition P define, mi = in f { f (x)|xi−1 ≤ x ≤ xi } for 1 ≤ i ≤ n. Mi = sup{ f (x)|xi−1 ≤ x ≤ xi } for 1 ≤ i ≤ n. L(P, f ) = ∑ni=1 mi (xi − xi−1 ) for 1 ≤ i ≤ n. U(P, f ) = ∑ni=1 Mi (xi − xi−1 ) for 1 ≤ i ≤ n. We then define the lower and upper Riemann integrals of f over [a, b], respectively, by R

Rb

a f dx

R

Rb

or

Rb

f dx or a

a f dx

Rb a

= in f {U(P, f )| P is a partition of [a,b]} and

f dx = sup{L(P, f )| P is a partition of [a,b]},

Measure Spaces and L p Spaces

58

where supremum and infimum are taken over all partition of [a,b]. Since f is assumed to be bounded and the interval [a, b] has finite length, the lower and upper Riemann integrals are finite. The upper integral is always at least as large as the lower integral, and if the two are equal we say that f is Riemann integrable over [a, b] and call this common value the Riemann integral of f over [a, b] is denoted by, Rb a

f dx or R

Z b

Rb a

Z b

f dx = R

R

f dx, that is,

a

Z b

f dx = R a

f dx. a

Step Function A real-valued function ψ defined on [a, b] is called a step function provided there is a partition P = {x0 , x1 , x2 , ..., xn } of [a,b] and numbers c0 , c1 , c2 , ..., cn such that for for 1 ≤ i ≤ n, ψ(xi ) = ci for xi−1 ≤ x ≤ xi . n

Observe that, L(P, f ) = ∑ ci (xi − xi−1 ) = U(P, f ). i=1

From this and the definition of the upper and lower Riemann integrals, we infer that a step function ψ is Riemann integrable and Z b

R a

n

ψdx = ∑ ci (xi − xi−1 ) i=1

Now, using step function we can defined Riemann integration of a function as follows: Z b

R

f dx = sup {R

Rb a

ψdx|ψ is a step f unction and ψ ≤ f on[a, b]}

Rb

ψdx|ψ is a step f unction and ψ ≥ f on [a, b]}

a

and Z b

R a

f dx = inf {R

a

It may be observed that the definition of the Riemann integral involves the approximation of the integrand by the step functions over the partitions of the domain of the integrand. If the bounded function f defined over [a, b] is not too discontinuous, the Riemann integral of f is the limit of the integrals of step functions which approximate f. Note: Let f be a bounded function defined on a closed and bounded interval [a, b] then Z b a

n

f dx = inf {U(P, f )| P is a partition of [a,b]} = lim

n→∞

∑ f (xi)∆x

i=1

59

3.2 Construction and some properties of Riemann Integration and Z b

n

f dx = sup{L(P, f )| P is a partition of [a,b] } = lim

n→∞

a

where ∆x =

b−a n

∑ f (xi)∆x

i=1

and xi = a + i∆ for 1 ≤ i ≤ n. Hence n

Z b

f dx = lim

R

∑ f (xi)∆x. n→∞

a

i=1

Examples of Riemann Integral Example 3.12. Let f (x) = x2 be a bounded function defined on [0,1]. We are interested to Z 1

f (x)dx using Riemann integral formula. Since a = 0 , b = 1 and

find the definite integral f (x) = x2 hence ∆x =

1 n

0

and xi = ni . Hence n

Z 1

f (x)dx = lim

∑ f (xi)∆x

n→∞ i=1 n i2 1 = lim 2 )n n n→∞ i=1 1 n 2 = lim 3 i n→∞ n i=1

0

∑

∑

1 n(n + 1)(2n + 1) n→∞ n3 6 = 62 = 13

= lim

Example 3.13. Let f (x) = Zsin(x) be a bounded function defined on [0, π]. We are interested π

f (x)dx using Riemann integral formula. Since a = 0 , b = π

to find the definite integral 0

and f (x) = sin(x) hence ∆x =

π n

and xi =

iπ n.

Hence n

Z π

f (x)dx = lim

n→∞

0

∑ f (xi)∆x

i=1 n

iπ π

∑ sin( n ) n n→∞

= lim

i=1

π n iπ = lim ∑ sin( ) n→∞ n n i=1 π π = lim cot( ) n→∞ n 2n π cot( 2n ) = lim 2 π n→∞

=2

2n

Measure Spaces and L p Spaces

60 n

Since,

∑ sin(iθ ) =

θ sin( nθ 2 )sin((n + 1) 2 )

sin( θ2 )

i=0

.

Example 3.14. Let f (x) = ex be a bounded function defined on [0, 1]. We are interested to R find the definite integral 01 f (x)dx using Riemann integral formula. Since a = 0 , b = 1 and f (x) = ex hence ∆x = n

= lim

n→∞

1 n

n

Z 1

and xi = ni . Hence

f (x)dx = lim

n→∞

0

∑ f (xi)∆x

i=1

( ni ) 1

∑e

i=1

n

n

i 1 e( n ) ∑ n→∞ n i=1 1 n ( 1 ) ( (i−1) ) = lim ∑ e n e n n→∞ n i=1 1 ( 1 ) n ( (i−1) ) = lim e n ∑ e n n→∞ n i=1 1 ( 1 ) (e( 1n ))n − 1 ) = lim e n ( 1 n→∞ n en − 1 = e−1

= lim

Example 3.15. Let us consider the function f(Dirichlet’s function) defined [0,1] as follows: ( 0 i f x is an irrational in [0, 1] f (x) = 1 i f x is a rational in [0, 1] Let P be any partition of [0, 1]. By the density of rational and irrational number in [0,1] Z 1

Z 1

f dx = sup{L(P, f ) = 0} and

L(P, f ) = 0 and U(P, f ) = 1. So, 0

f dx = in f {U(P, f ) = 1.

0

Hence f is not Riemann integrable.

Theorem 3.22. A necessary and sufficient condition that a bounded function f(x) be Riemann integrable in [a,b] is that the set of discontinuities of f(x) on [a,b] have measure zero. For the proof of this theorem see [19].

3.2.2

Theorems on Riemann Integrals [19]

Theorem 3.23. Let f1 (x) and f2 (x) be Riemann integrable in [a,b], then f1 (x) + f2 (x) is Riemann integrable in [a, b] and Z b a

Z b

( f1 (x) + f2 (x))dx =

a

Z b

f1 (x)dx +

a

f2 (x)dx.

Theorem 3.24. Let f (x) be Riemann integrable in [a, b] and c is any constant, then c f is Riemann integrable in [a,b] and

3.2 Construction and some properties of Riemann Integration Z b

61

Z b

c f dx = c

f (x)dx.

a

a

Theorem 3.25. If f(x) and g(x) be Riemann integrable on [a, b], then f(x)g(x) is Riemann integrable on [a, b]. Theorem 3.26. If f (x) be Riemann integrable in [a, b], then Z b

f dx = −

Z a

Z a

f (x)dx and

a

b

f dx = 0. a

Theorem 3.27. If f (x) is bounded and Riemann integrable on [a,b] and c be any point on [a,b], then Z b

Z c

f dx =

Z b

f (x)dx +

a

a

f (x)dx . c

Theorem 3.28. If f(x) and g(x) be Riemann integrable on [a,b] and f (x) ≤ g(x) , then Z b

f dx ≤

a

Z b

g(x)dx . a

Theorem 3.29. If f(x) is Riemann integrable on [a,b] and has upper and lower bounds M and m respectively, then m(b − a) ≤

Z b

f dx ≤ M(b − a).

a

If k is any number such that m ≤ k ≤ M, then Z b

f dx = k(b − a).

a

Theorem 3.30. (Mean Value Theorem) If f(x) is continuous on [a, b], then there exists a number c ∈ [a, b] such that Z b

f dx = (b − a) f (c)

a

Theorem 3.31. If f is Riemann integrable on [a, b] and | f (x)| ≤ M for some constant M, then Z b ≤ M(b − a) f dx a Theorem 3.32. If f is Riemann integrable on [a,b] and | f (x)| is Riemann integrable on [a,b], then Z b Zb a f dx ≤ a | f (x)|dx For the proof of all these theorems see [19].

Measure Spaces and L p Spaces

62

3.2.3

Limitation of Riemann Integration

The theory of Riemann integration though very useful and adequate for solving many mathematical problems, both pure and applied, is not free from defects. It does not meet the needs of a number of important branches of mathematics and physics of comparatively recent development. First of all, the Riemann integral of a function is defined on closed interval and cannot be defined on an arbitrary set. Investigations in probability theory, partial differential equations, hydromechanics and quantum mechanics often pose problems which require integration over sets. Second and more important is the fact that the Riemann integrability depends upon the continuity of the function. Of course, there are functions which are discontinuous and yet Riemann-integrable, but these functions are continuous almost everywhere. Again, given a sequence of Reimann-integrable functions converging to some function in a domain, the limit of the sequence of integrated functions may not be the Riemann integral of the limit function. In fact, the Riemann integral of the limit function may not even exist. This is a major drawback of the Riemann theory of integration, apart from the fact that even relatively simple functions are not integrable in the sense of Riemann integration. Henri Lebesgue in his classical work, introduced the concept of an integral , known after his name the Lebesgue integral, based on the measure theory that generalizes the Riemann integral. It has the advantage that it takes care of both bounded and unbounded functions and simultaneously allows their domains to be more general sets and thereby enlarges the class of functions for which the Lebsegue integral is defined. Also, it gives more powerful and useful convergence theorems relating to the interchange of the limit and integral valid under less restrictive conditions required for the Riemann integral.

3.2.4

Construction of Lebesgue Integration and some of its properties

For a simple Z function ψZ defined on a set of finite measure E, we define the integral of ψ ψ(x)dx =

over E by, E

E

ψ = ∑ni=1 ai m(Ei ) where ψ has the canonical representation. The

elementary integral of a simple function has been defined through its canonical representation. Theorem 3.33. (Linearity and Monotonicity of Integration) Let functions Z ψ and φ be simple Z Z (aψ + bφ ) = a

defined on a set of finite measure E. Then for any a and b, Moreover if φ ≤ ψ on E, then

Z E

ψ≥

E

Z

φ. E

ψ +b E

φ. E

3.2 Construction and some properties of Riemann Integration

3.2.5

63

Lebesgue Integral

Let f be a bounded real-valued function defined on a set of finite measure E. By analogy with the Riemann integral, we define the lower and upper Lebesgue integral, respectively, of f over E to be, Z

sup {

ψ : ψ is simple and f ≥ ψ on E}

E

and inf {

Z

ψ : ψ is simple and f ≤ ψ on E}.

E

A bounded function f on a domain E of finite measure is said to be Lebesgue integrable over E provided its upper and lower Lebesgue integrals over E are equal. The common value of the upper and lower integrals Z is called the Lebesgue integral, or simply the integral, of f over E and is denoted by

f. E

Since f is assumed to be bounded, by the monotonicity property of the integral for simple functions, the lower and upper integrals are finite and the upper integral is always at least as large as the lower integral. A step function takes only a finite number of values and each interval is measurable. Thus a step function is simple. Since the measure of a singleton set is zero and the measure of an interval is its length, we infer from the linearity of Lebesgue integration for simple functions defined on sets of finite measure that the Riemann integral over a closed, bounded interval of a step function agrees with the Lebesgue integral. The key difference in the definitions of Riemann and Lebesgue integrals is that in the former the domain of f is partitioned while in the latter the range of f is partitioned.

Example 3.16. Let us consider the function f(Dirichlet’s function) defined [0,1] as follows: ( 0 i f x is an irrational in [0, 1] f (x) = 1 i f x is a rational in [0, 1] The set E of rational numbers in [0, 1] is a measurable set of measure zero. The Dirichlet function f is the restriction to [0, 1] of the characteristic function of E, χE . Thus, f is integrable over [0, 1] and hence f is Lebesgue integrable on [0, 1].

Measure Spaces and L p Spaces

64

Theorem 3.34. Let f be a bounded function defined on the closed, bounded interval [a, b]. If f is Riemann integrable over [a, b], then it is Lebesgue integrable over [a, b] and the two integrals are equal. Theorem 3.35. Let f be a bounded measurable function on a set of finite measure E. Then f is integrable over E. Theorem 3.36. Let f and g be bounded measurable functions on a set of finite measure E. Then, Z Z (a) α f = α f for all real α. ZE

EZ

( f + g) =

(b) E

Z

f+ E

E

Z

f= ZE

f≤

Z

(α f + β g) = α

g and E

Z

(c) If f = g a.e. then (d) If f ≤ g a.e. then

Z

Z

f +β E

g ZE

g

E

E

Z

Z

( f + g) =

(e) If A and B are disjoint measurable subset of E then, S

A B

If α ≤ f (x) ≤ β then αm(E) ≤

g for all real α and β . E

Z E

f ≤ β m(E).

Z

f+ A

g B

(f)

3.3 L p Space

3.2.6

65

Almost everywhere

Suppose (X, M, µ) is a measure space. Let f be a measurable function on X and E ∈ M and let P be a statement. Then the statement P holds almost everywhere on E means that there exists N ∈ M, N ⊂ E such that µ(N) = 0 and P hold every points of E − N. This concept of almost everywhere depends of course very strongly on the given measure. for example, if f and g are measurable functions on X. and if µ({x ∈ X : f (x) ̸= g(x)}) = 0, then we say that f = g almost everywhere.

3.3

L p Space

If 0 < p < ∞ and if f is a complex measurable function on X then define, ∥ f ∥p = {

Z

1

| f | p dµ} p .

X

Lp

Then, ∥ f ∥ p is called the norm of f and the collection of all complex measurable function f for which ∥ f ∥ p < ∞ is called L p space.

3.3.1

Essential Supremum

Suppose g : X → [0, ∞] is a measurable function. Let s be the set of all real α such that, µ(g−1 ((α, ∞])) = 0 .......(i) Let f : X → R be a real valued function defined on a set X. A real number a is called an upper bound for f if f (x) ≥ a for all x in X, i.e., if the set f −1 (a, ∞) = {x ∈ X : f (x) > a} is empty. Let U f = {a ∈ R : f −1 (a, ∞) = 0} / be the set of upper bounds of f. Then the supremum of f is defined by, sup f = infU f if the set of upper bounds U f is nonempty, and sup f = +∞ otherwise. Now, assume in addition that (X, ∑, µ) is a measure space and, for simplicity, assume that the function f is measurable. A number a is called an essential upper bound of f if the measurable set f − 1(a, ∞) is a set of measure zero, i.e., if f (x) = a for almost all x in X. Let U fess = {a ∈ R : µ( f −1 (a, ∞)) = 0} be the set of essential upper bounds. Then the essential supremum is defined similarly as ess sup f = infU fess if U fess ̸= 0/ , and esssup f = +∞ otherwise. Exactly in the same way one defines the essential infimum as the supremum of the essential lower bounds, that is, ess inf f = sup{b ∈ R : µ({x : f (x) < b}) = 0} if the set of essential lower bounds is nonempty, and as −∞ otherwise.

Measure Spaces and L p Spaces

66

3.3.2

Convex Function

A real function ϕ defined on (a,b), where −∞ ≤ a ≤ ∞ is said to be convex if the inequality, ϕ((1 − λ )x + λ y) ≤ (1 − λ )ϕ(x) + λ ϕ(y) holds for every 0 ≤ λ ≤ 1 and a < x < b, a < y < b. Theorem 3.37. (Jensen′ s inequality) Let µ be a positive measure on σ − algebra m in a set X, so that µ(x) = 1. If f is a real function on L1 (µ), if a < f (x) < b for all x ∈ X and if ϕ is convex on (a,b). Then Z  Z ϕ f dµ ≤ (ϕo f )dµ. X

3.3.3

X

Conjugate Exponents

If p and q are two real numbers such that p + q = pq i.e., 1 p

+ 1q = 1.

Then p and q are called conjugate exponents where 1 < p < ∞ and 1 < q < ∞. Theorem 3.38. [17] Let p, q be conjugate exponents. Let X be a measurable space with measure µ. Let f and g be measurable functions with range in [0, ∞]. Then, Z

f gdµ ≤

Z

X

 1 Z p

p

1 q

q

f dµ

...(1)

g dµ X

X

and Z

p

( f + g) dµ ≤

Z

p

1

Z

p

+

f dµ X

X

p

g dµ

1

p

...(2)

X

′ s inequality and inequality (2) is known as Minkowski′ s The inequality (1) is known as H older ¨ inequality.

Z Proof : Let A =

p

f dµ X

1

Z

p

and B =

q

g dµ

1

q

. If A = 0, then f = 0 almost

X

everywhere and hence f g = 0 almost everywhere. So (1) holds. Similarly if B = 0 then (1) holds obviously. Again, if A > 0, B = ∞, then (1) holds obviously. So we need to prove the theorem for the when 0 < A < ∞ and 0 < B < ∞. Now, put F = Af and G = Bg .. This gives,

3.3 L p Space

67 Z

F p dµ = 1......(A)

X

Similarly, Z

G p dµ = 1......(B)

X s

If x ∈ X such that 0 < F(x) < ∞ and 0 < G(x) < ∞, there exist real s and t such that F(x) = e p t and G(x) = e q . Since 1p + 1q = 1, the convexity of the exponential function implies that, s

1

1

t

e p + q = e p s+(1− p t ≤ 1p es + (1 − 1p )et = s

t

es p

Thus, e p + q ≤ s

t q

es p

t

+ eq .

t

+ eq ......(C)

es et p + q Gq (x) F p (x) p + q ......(D)

⇒ e pe ≤

⇒ FG ≤ holds for every xR ∈ X. R Z p Gq (x)dµ X F (x)dµ ⇒ FGdµ ≤ + X p q ZX 1 1 ⇒ FGdµ ≤ + p q ZX ⇒

FGdµ ≤ 1

ZX

f g dµ ≤ 1 ZX A B ⇒ f gdµ ≤ AB X Z  1 Z 1 Z p q p q ⇒ f gdµ ≤ f dµ g dµ ⇒

X

X

X

Hence (1) is proved. Since, ( f + g) p = ( f + g)( f + g) p−1 ⇒Z ( f + g) p = f ( Zf + g) p−1 + g( f +Zg) p−1 ( f + g) p =

⇒

X

Now, using

f ( f + g) p−1 +

X Holder′ s

g( f + g) p−1 ......(E)

X

inequality we have,

Z

f ( f + g)

p−1

≤{

Z

X

p

1 p

Z

1

( f + g)(p−1)q dµ} q ......(M)

f dµ} {

X

X

and similarly, Z

g( f + g) X

p−1

≤

Z

p

 1 Z p

( f + g)

g dµ X

(p−1)q

X

1

q

dµ

.......(N)

Measure Spaces and L p Spaces

68 Using (M) and (N) in (E) we have, Z

p

( f + g) dµ ≤

Z

p

1

X

1

p

p

+

f dµ

X

Z

p

.

g dµ X

Hence (2) is proved. Theorem 3.39. [17] Suppose 1 ≤ p ≤ ∞, and f ∈ L p (µ), g ∈ L p (µ) then f + g ∈ L p (µ) and ∥ f + g∥ p ≤ ∥ f ∥ p + ∥g∥ p Theorem 3.40. L p µ is a complex vector space i.e., L p (µ) is a vector space. Proof : Suppose 1 ≤ p < ∞ and f ∈ L p (µ), g ∈ L p (µ) to prove that L p − space is a vector space, show that (i) f + g ∈ L p (µ) and (ii) α f ∈ L p (µ) α is any complex number. To prove (i), for 1 < p < ∞, we have from Minkowski′ s inequality, 1 Z  1 Z Z p p p p p g dµ + ( f + g) dµ ≤ f dµ X

X

X

Since, f ∈ L p (µ) and g ∈ L p (µ) then

Z

p

1

Z

p

f dµ

< ∞ and

X

p

1

g dµ

p

< ∞.

X

Thus ∥ f + g∥ p < ∞ and therefore f + g ∈ L p (µ). Similarly, α f ∈ L p (µ). Hence L p − space is a vector space. Theorem 3.41. L p (µ) is not a metric space but it can be regarded as a metric space. Proof : Suppose f , g ∈ L p (µ), then f + g ∈ L p (µ) and ∥ f + g∥ p ≤ ∥ f ∥ p + ∥g∥ p ......(A). We define a function in L p (µ) by, d( f , g) = ∥ f − g∥ p ......(B). Then, d( f , g) = 0 if f = g. Now, d( f , g) = ∥ f − g∥ p Z 1 p p = | f − g| dµ

R

√1 n

0

(1 − x2 )n dx

(1 − nx2 )dx cn √ n

81

4.5 Bases for Polynomials Space √ where cn < n. For any δ > 0, qn (x)
0 be given. Because f is uniformly continuous, we can select δ > 0 for which | f (y) − f (x)| < ε2 whenever |y − x| < δ . Let M = sup| f (x)|. Since, for each fixed δ > 0, √ limn→∞ n(1 − δ 2 )n = 0 we can select an integer N for which 0 ≤ x ≤ 1, we have that

√ n(1 − δ 2 )n
N. Then, for n > N and

1 |pn (x) − f (x)| = | −1 [ f (x + t) − f (x)]qn (t)dt| R1 ≤ −1 | f (x + t) − f (x)|qn (t)dt Rδ Rδ R ≤ 2M −1 qn (t)dt + ε2 −δ qn (t)dt + 2M δ1 qn (t)dt R√ n(1 − δ 2 )n + ε2 ≤ ε2 + ε2 ≤ 4M =ε

R

This complete the proof of the theorem.

82

4.6

Polynomial Space

Approximation of continuous functions using piecewise cubic polynomials

In this section we are going to discuss about cubic spline and we discussed about how do we use cubic polynomials to create a little arcs to create smooth curve through a given number of points in the x-y plane. Convex Linear Combination We have already known about linear combination from our first chapter. Now, we are going to discus about convex linear combination. A convex linear combination is a linear combination of vectors which is convex. Suppose A and B be two vectors of a vector space V over a field K then, the linear combination of the form, C ≡ (1 − α)A + αB ∈ V is said to be a convex linear combination of A and B.

Convex linear combination of A and B is given in figure-A. This convex linear combination C moves through A to B as α goes from 0 to 1. When α = 0 then C = A and if α = 1 then C = B. Suppose that α = 13 then the convex linear combination C ≡ (1 − 13 )A + 13 B = 23 A + 13 B. The following figure B shows the convex linear combination with α = 13 .

4.6 Approximation of continuous functions using piecewise cubic polynomials

83

Example: Suppose that A = (1, 3) and B = (3, 7) then their convex linear combination of A and B is C = (1 − α)A + αB = (1 − α)(1, 3) + α(3, 7) = (1 + 3α, 4α). Now, take a convex combination of three vectors P0 , P1 , P2 in such a way that Q0 ≡ (1 − α)P0 + αP1 , Q1 ≡ (1 − α)P1 + αP2 and R ≡ (1 − α)Q0 + αQ1 where as α goes from 0 to 1 then R moves along a parabolic(quadratic) path from P0 to P2 . The tangents (directions) at P0 , P2 both at P1 . Algebraically, R = (1 − α)Q0 + αQ1 R = (1 − α)((1 − α)P0 + αP1 ) + α(1 − α)P1 + αP2 R = (1 − α)2 P0 + 2(1 − α)αP1 + α 2 p2 . Since (1 − α)2 + 2(1 − α)α + α 2 = 1 then R is convex linear combination of P0 , P1 and P2 provided 0 ≤ α ≤ 1 that is (1 − α) ≥ 0 , 2(1 − α)α ≥ 0 and α 2 . Convex linear combination for above three vectors is shown in the figure following figure:

84

Polynomial Space

Now, take a convex combination of four vectors P0 , P1 , P2 and P3 in such a way that Q0 ≡ (1 − α)P0 + αP1 , Q1 ≡ (1 − α)P1 + αP2 , Q2 ≡ (1 − α)P2 + αP3 R0 ≡ (1 − α)Q0 + αQ1 R1 ≡ (1 − α)Q1 + αQ2 S ≡ (1 − α)R1 + αR2 . The following figure shows the convex combination of four vectors:

4.6 Approximation of continuous functions using piecewise cubic polynomials

4.6.1

85

Cubic Spline Method

Now, we will construct cubic spline from a different view that is in view of linear algebra. For better explanation let us consider an example that we are given finite number of points (0,1), (1,3), (2,2), (3,2) and (4,1) in the x-y plane. We want to fit smooth arcs to joint these points smoothly. To do this if we have two points the we have to draw a linear polynomial passing through this two points and if we have three points then we need a quadratic polynomial. As the number of points increases the degree of the polynomial will be increased. But in cubic spline we draw a curve passing through any finite number of points in terms of a branch of smooth cubic arcs between two respective points. Since we are using a collection of cubic polynomials then we are in a space P3 of cubic polynomials. Let us suppose that p(x) = a +bx +cx2 +dx3 be the polynomials which will be fitted between first two points (0,1) and (1,3). Then p(x) will pass through (0,1) if a = 1. Again p(x) will pass through the points (1,3) if a + b + c + d = 3. Then it is clear that there are two parameter family of such cubic curve since we two equations with four unknown. Now, we can specify the tangents directions or 1st and 2nd derivative at the points (0,1) and (1,3). To relates the behavior of p(x) at x = 0 and x = 1 consider the Taylor expansion. Since p(x) = a + bx + cx2 + dx3 then, p(x + 1) = a + b(x + 1) + c(x + 1)2 + d(x + 1)3

86

Polynomial Space

= (a + b + c + d) + (b + 2c + 3d)x + (c + 3d)x2 + dx3 p(x) = (a + b + c + d) + (b + 2c + 3d)(x − 1) + (c + 3d)(x − 1)2 + d(x − 1)3 This the Taylor expansion of p(x) around x = 1 since (a + b + c + d), (b + 2c + 3d), (c + 3d) ′′

′

are the p(1), p (1) and p 2!(1) . Let us write these coefficients as vectors(Taylor coefficient vector); (p)0 = (a, b, c, d) is the Taylor coefficient vectors at x = 0. (p)1 = (a + b + c + d, b + 2c + 3d, c + 3d, d) is the Taylor coefficient vectors at x = 1.   1 0 0 0  1 1 0 0    Also, (p)1 = (p)0    1 2 1 0  1 3 3 1   1 0 0 0  1 1 0 0    ⇒ (p)1 = (a, b, c, d)    1 2 1 0  1 3 3 1  1 0 0 0  1 1 0 0    where the matrix   is the Pascal matrix. From the above it is clear that  1 2 1 0  1 3 3 1   1 0 0 0  −1 1 0 0    (p)0 = (p)1    1 −2 1 0  

   Now,   

1 x x2 x3

−1 3  1   1   =   1 1   

−3 1 0 1 2 3

0 0 1 3

0 0 0 1

    

1 x−1 (x − 1)2 (x − 1)3  0  0    0  1 

    

 1 0 0 1    −1 1 0 x      ⇒ =  2    1 −2 1 x  −1 3 −3 x3   1 1  x   x−1      ⇒ p = (p)0  2  = (p)1   Now, we have five points (0,1), (1,3), (2,2),(3,2) 2  x   (x − 1)  x3 (x − 1)3 and (4,1) in the x-y plane and have to find four polynomials in P3 which passes through the 1 x−1 (x − 1)2 (x − 1)3 

4.6 Approximation of continuous functions using piecewise cubic polynomials

87

given points and join smoothly at each end points. Since coefficients vectors tells us how the polynomials looks like at a end point and the allow two polynomials to join smoothly making tangential at the end points. So each of the polynomials Pi has two Taylor coefficients vector at two end points of its interval.So the first polynomials P1 joining first two points has two Taylor coefficients vector (P1 )0 at x = 0 and (P1 )1 at x = 1 and the second polynomials P2 joining second and third points has two Taylor coefficients vector (P2 )1 at x = 1 and (P2 )2 at x = 2 Similar for P3 and P4 . We want these Taylor coefficient vectors to agree upto order 2 so we want first common Taylor coefficient vector for each consecutive polynomials. Since the first polynomial P1 (x)a + bx + cx2 + dx3 will pass through the first two point (0,1) and (1,3) if a = 1 and a + b + c + d = 3. So we can choose (P1 )0 = (1, 2, 1, −1) Then using

Figure 4.1 Convex Linear Combination of Four Vectors the tricks we find all other taylors coefficient vectors and find a smooth curve which passes through the given data points.

Chapter 5 Approximation of Continuous Functions by Polynomials According to Weierstrass approximation theorem, every continuous function can be approximated through polynomials and Bernstein theoretically proved that every continuous functions can be approximated through Bernstein polynomials. From this motivation we are intended to verify this theorem for some well known continuous functions by Bernstein polynomials using computer programming.

5.1

Approximation of Continuous Functions

Describing and analyzing continuous functions is a very important task in Mathematics. In the Mathematical world, one common approach to describing continuous functions is by using an explicit representation in terms of one variable, y = f (x). Although, this representation has many problems with it. For example, the vertical line x = c cannot be represented using this model. In addition, sometimes we have some very complicated function that we’d like to approximate as a linear combination of polynomials so we can compute points on it efficiently and store it and transform it easily. Taylor’s Theorem will only approximate a differentiable function at specific points and can produce a Taylor series that doesn’t converge to the original function. Thus, it is necessary we to find a better technique to approximate a continuous function. So in this section we will first describe Bernstein polynomials to setup a foundation for the rest of the chapter 5. Then we’ll see how these polynomials can be used to approximate any continuous functions to verify Weierstrass Approximation Theorem.

90

Approximation of Continuous Functions by Polynomials

Bernstein basis polynomials Polynomials are incredibly useful mathematical tools as they are simply defined, can be computed very quickly, and can represent a variety of functions. They can also be integrated easily and pieced together to form spline curves that can approximate any function to any accuracy desired. The general form, definition and classification of polynomials has already been discussed in chapter 4. Now, we will discuss about Bernstain polynomials only.

A Bernstein basis polynomial of degree n is defined by ! n B j,n (x) = x j (1 − x)n− j , j = 0, 1, 2, ......, n j ! n where is the binomial coefficient and j ! n n! = j!(n− j)! j Often, for mathematical convenience, we set B j,n (x) = 0 if j < 0 or j > n (Note on Notation : Bernstein basis polynomials can be written as B j,n (x) , B j,n (x)) Example : The first few Bernstein polynomials are as follows: Bo,o (x) = 1 Bo,1 (x) = 1 − x B1,1 (x) = x 2 Bo,2 (x) = (1 − x) B1,2 (x) = 2x(1 − x) B2,2 (x) = x2 Bo,3 (x) = (1 − x)3 B1,3 (x) = 3x(1 − x)2 B2,3 (x) = 3x2 (1 − x)3 B3,3 (x) = x3 Bo,4 (x) = (1 − x)4 B1,4 (x) = 4x(1 − x)3 B2,4 (x) = 6x2 (1 − x)2 B3,4 (x) = 4x3 (1 − x) B4,4 (x) = x4 First degree, second degree and third degree Bernstein polynomials basis can be plotted on a unit square as follows:

5.2

Bernstein polynomials

Bernstein polynomial of degree n is a linear combination of Bernstein basis polynomials B j,n (t) and is defined by B j,n (t)) = ∑nj=0 α j B j,n (x)

5.2 Bernstein polynomials

Figure 5.1 Bernstein Polynomial Basis of degree 1

Figure 5.2 Bernstein Polynomial Basis of degree 2

91

92

Approximation of Continuous Functions by Polynomials

Figure 5.3 Bernstein Polynomial Basis of degree 3 where α j are Bernstein coefficient. If α j are not explicitly defined then it may assume that α j = 1. The Bernstein polynomial of a function f is defined by, Bn ( f )(x) = ∑nj=0 f ( nj )B j,n (x)

5.2.1

Recursive Definition of Bernstein Polynomials [20]

A Bernstein basis polynomials of degree n can be defined as the sum of two Bernstein basis polynomials of degree as follows: bk,n (x) = (1 − x)bk,n−1 (x) + tbk−1,n−1 (x). !

n Proof : bk,n (x) = xk (1 − x)n−k k " ! !# n−1 n−1 = + xk (1 − x)n−k k k−1 ! ! n−1 n − 1 = xk (1 − x)n−k + xk (1 − x)n−k k k−1

93

5.2 Bernstein polynomials ! n−1 = (1 − x) xk (1 − x)n−1−k + x k = (1 − x)bk,n−1 (x) + xbk−1,n−1 (x)

5.2.2

n−1 k−1

! xk−1 (1 − x)n−1−(k−1)

Some Properties of Bernstein Basis Polynomials

Bernstein basis polynomials have a number of useful properties. Here we will describe a few of them. (i) Bernstein basis polynomials are non-negative on 0 ≤ x ≤ 1 and are strictly positive on the open interval 0 < x < 1. Proof : To show this we will proceed by mathematical induction on the recursive definition of Bernstein basis polynomials as described above. Base case: b0,0 = 1 ≥ 0. Inductive hypothesis : Assume bi, j ≥ 0, ∀i, ∀ j < n for some n. By our recursive definition: bk,n (x) = (1 − x)bk,n−1 (x) + tbk−1,n−1 (x). Thus bi,n (x) ≥ 0 for 0 ≤ x ≤ 1 since the right-hand side of the equation are all non-negative components on our interval. (Remember, if i