A product convergence theorem for Henstock Kurzweil integrals

Similar documents
Czechoslovak Mathematical Journal, 55 (130) (2005), , Abbotsford. 1. Introduction

W. We shall do so one by one, starting with I 1, and we shall do it greedily, trying

ON THE C-INTEGRAL BENEDETTO BONGIORNO

Properties of the Riemann Integral

Fundamental Theorem of Calculus for Lebesgue Integration

Lecture 1. Functional series. Pointwise and uniform convergence.

The Henstock-Kurzweil integral

Necessary and Sufficient Conditions for Differentiating Under the Integral Sign

FUNDAMENTALS OF REAL ANALYSIS by. III.1. Measurable functions. f 1 (

Lecture 3 ( ) (translated and slightly adapted from lecture notes by Martin Klazar)

A PROOF OF THE FUNDAMENTAL THEOREM OF CALCULUS USING HAUSDORFF MEASURES

The Bochner Integral and the Weak Property (N)

Review of Riemann Integral

arxiv:math/ v2 [math.ho] 16 Dec 2003

Appendix to Notes 8 (a)

MAA 4212 Improper Integrals

A HELLY THEOREM FOR FUNCTIONS WITH VALUES IN METRIC SPACES. 1. Introduction

The Banach algebra of functions of bounded variation and the pointwise Helly selection theorem

Math 324 Course Notes: Brief description

A BRIEF INTRODUCTION TO UNIFORM CONVERGENCE. In the study of Fourier series, several questions arise naturally, such as: c n e int

Lecture 1: Introduction to integration theory and bounded variation

Advanced Calculus: MATH 410 Notes on Integrals and Integrability Professor David Levermore 17 October 2004

Homework 4. (1) If f R[a, b], show that f 3 R[a, b]. If f + (x) = max{f(x), 0}, is f + R[a, b]? Justify your answer.

Math 61CM - Solutions to homework 9

Math 554 Integration

7.2 Riemann Integrable Functions

Henstock Kurzweil delta and nabla integrals

A Convergence Theorem for the Improper Riemann Integral of Banach Space-valued Functions

f(x)dx . Show that there 1, 0 < x 1 does not exist a differentiable function g : [ 1, 1] R such that g (x) = f(x) for all

SOLUTIONS FOR ANALYSIS QUALIFYING EXAM, FALL (1 + µ(f n )) f(x) =. But we don t need the exact bound.) Set

Convergence of Fourier Series and Fejer s Theorem. Lee Ricketson

Advanced Calculus: MATH 410 Uniform Convergence of Functions Professor David Levermore 11 December 2015

The Regulated and Riemann Integrals

Riemann Sums and Riemann Integrals

Riemann Sums and Riemann Integrals

Regulated functions and the regulated integral

Riemann is the Mann! (But Lebesgue may besgue to differ.)

2 Definitions and Basic Properties of Extended Riemann Stieltjes Integrals

Principles of Real Analysis I Fall VI. Riemann Integration

Presentation Problems 5

Problem Set 4: Solutions Math 201A: Fall 2016

Positive Solutions of Operator Equations on Half-Line

Henstock/Kurzweil Fourier transforms. Erik Talvila 1

MATH 174A: PROBLEM SET 5. Suggested Solution

MATH 409 Advanced Calculus I Lecture 19: Riemann sums. Properties of integrals.

Fundamental Theorem of Calculus and Computations on Some Special Henstock-Kurzweil Integrals

NOTES AND PROBLEMS: INTEGRATION THEORY

MATH34032: Green s Functions, Integral Equations and the Calculus of Variations 1

The final exam will take place on Friday May 11th from 8am 11am in Evans room 60.

DEFINITION The inner product of two functions f 1 and f 2 on an interval [a, b] is the number. ( f 1, f 2 ) b DEFINITION 11.1.

Approximation of functions belonging to the class L p (ω) β by linear operators

Lecture 3. Limits of Functions and Continuity

S. S. Dragomir. 2, we have the inequality. b a

Week 7 Riemann Stieltjes Integration: Lectures 19-21

T b a(f) [f ] +. P b a(f) = Conclude that if f is in AC then it is the difference of two monotone absolutely continuous functions.

WHEN IS A FUNCTION NOT FLAT? 1. Introduction. {e 1 0, x = 0. f(x) =

LAPLACE TRANSFORM USING THE HENSTOCK-KURZWEIL INTEGRAL

Entrance Exam, Real Analysis September 1, 2009 Solve exactly 6 out of the 8 problems. Compute the following and justify your computation: lim

Unit #9 : Definite Integral Properties; Fundamental Theorem of Calculus

Chapter 4. Lebesgue Integration

LECTURE. INTEGRATION AND ANTIDERIVATIVE.

Functional Analysis I Solutions to Exercises. James C. Robinson

MAT612-REAL ANALYSIS RIEMANN STIELTJES INTEGRAL

HENSTOCK-KURZWEIL FOURIER TRANSFORMS

KRASNOSEL SKII TYPE FIXED POINT THEOREM FOR NONLINEAR EXPANSION

MA Handout 2: Notation and Background Concepts from Analysis

Review on Integration (Secs ) Review: Sec Origins of Calculus. Riemann Sums. New functions from old ones.

ACM 105: Applied Real and Functional Analysis. Solutions to Homework # 2.

S. S. Dragomir. 1. Introduction. In [1], Guessab and Schmeisser have proved among others, the following companion of Ostrowski s inequality:

c n φ n (x), 0 < x < L, (1) n=1

The Riemann-Lebesgue Lemma

a n = 1 58 a n+1 1 = 57a n + 1 a n = 56(a n 1) 57 so 0 a n+1 1, and the required result is true, by induction.

THE EXISTENCE-UNIQUENESS THEOREM FOR FIRST-ORDER DIFFERENTIAL EQUATIONS.

Phil Wertheimer UMD Math Qualifying Exam Solutions Analysis - January, 2015

Prof. Girardi, Math 703, Fall 2012 Homework Solutions: 1 8. Homework 1. in R, prove that. c k. sup. k n. sup. c k R = inf

UNIFORM CONVERGENCE MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE

A basic logarithmic inequality, and the logarithmic mean

11 An introduction to Riemann Integration

1 i n x i x i 1. Note that kqk kp k. In addition, if P and Q are partition of [a, b], P Q is finer than both P and Q.

Mapping the delta function and other Radon measures

Journal of Inequalities in Pure and Applied Mathematics

7 Improper Integrals, Exp, Log, Arcsin, and the Integral Test for Series

Calculus in R. Chapter Di erentiation

Generalized Riemann Integral

arxiv: v1 [math.ca] 7 Mar 2012

(4.1) D r v(t) ω(t, v(t))

On the Continuous Dependence of Solutions of Boundary Value Problems for Delay Differential Equations

New Expansion and Infinite Series

UNIFORM CONVERGENCE. Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3

arxiv: v1 [math.ca] 11 Jul 2011

Physics 116C Solution of inhomogeneous ordinary differential equations using Green s functions

Polynomial Approximations for the Natural Logarithm and Arctangent Functions. Math 230

This is a short summary of Lebesgue integration theory, which will be used in the course.

Bernoulli Numbers Jeff Morton

8 Laplace s Method and Local Limit Theorems

The Fundamental Theorem of Calculus. The Total Change Theorem and the Area Under a Curve.

Recitation 3: More Applications of the Derivative

A NOTE ON PREPARACOMPACTNESS

Review of Calculus, cont d

Theoretical foundations of Gaussian quadrature

Transcription:

A product convergence theorem for Henstock Kurzweil integrls Prsr Mohnty Erik Tlvil 1 Deprtment of Mthemticl nd Sttisticl Sciences University of Albert Edmonton AB Cnd T6G 2G1 pmohnty@mth.ulbert.c etlvil@mth.ulbert.c Abstrct. Necessry nd sufficient for b fgn b fg for ll Henstock Kurzweil integrble functions f is tht g be of bounded vrition, g n be uniformly bounded nd of uniform bounded vrition nd, on ech compct intervl in (, b), g n g in mesure or in the L 1 norm. The sme conditions re necessry nd sufficient for f(g n g) 0 for ll Henstock Kurzweil integrble functions f. If g n g.e. then convergence fg n fg for ll Henstock Kurzweil integrble functions f is equivlent to f(g n g) 0. This extends theorem due to Lee Peng-Yee. 2000 Mthemtics Subject Clssifiction: 26A39, 46E30 Key words: Henstock Kurzweil integrl, convergence theorem, Alexiewicz norm Let < b nd denote the Henstock Kurzweil integrble functions on (, b) by HK. The Alexiewicz norm of f HK is f = sup I f where I the supremum is tken over ll intervls I (, b). If g is rel-vlued function on [, b] we write V [,b] g for the vrition of g over [, b], dropping the subscript when the identity of [, b] is cler. The set of functions of normlised bounded vrition, N BV, consists of the functions on [, b] tht re of bounded vrition, re left continuous nd vnish t. It is known tht the multipliers for HK re N BV, i.e., fg HK for ll f HK if nd only if g is equivlent to function in N BV. This pper is concerned with necessry nd sufficient conditions under which b fg for ll f HK. One such set of conditions ws given by Lee Peng-Yee in [2, Theorem 12.11]. If g is of bounded vrition, chnging g on countble set will mke it n element of N BV. With this observtion, minor modifiction of Lee s theorem produces the following result. 1 Reserch prtilly supported by the Nturl Sciences nd Engineering Reserch Council of Cnd.

2 Prsr Mohnty nd Erik Tlvil Theorem 1 [2, Theorem 12.11] Let < < b <, let g n nd g be relvlued functions on [, b] with g of bounded vrition. In order for b fg n fg for ll f HK it is necessry nd sufficient tht b for ech intervl (c, d) (, b), d c g n d g s n, c for ech n 1, g n is equivlent to function h n N BV, nd there is M [0, ) such tht V h n M for ll n 1. We extend this theorem to unbounded intervls, show tht the condition d c g n d c g in (1) cn be replced by g n g on ech compct intervl in (, b) either in mesure or in the L 1 norm, nd tht this lso lets us conclude f(g n g) 0. We lso show tht if g n g in mesure or lmost everywhere then fg n fg for ll f HK if nd only if fg n fg 0 for ll f HK. One might think the conditions (1) imply g n g lmost everywhere. This is not the cse, s is illustrted by the following exmple [1, p. 61]. Exmple 2 Let g n = χ (j2 k,(j+1)2 k ] where 0 j < 2 k nd n = j + 2 k. Note tht g n = 1, g n N BV, V g n 2, nd d c g n g n = 2 k < 2/n 0, so tht (1) is stisfied with g = 0. For ech x (0, 1] we hve inf n g n (x) = 0, sup n g n (x) = 1, nd for no x (0, 1] does g n (x) hve limit. However, g n 0 in mesure since if T n = {x [0, 1] : g n (x) > ɛ} then for ech 0 < ɛ 1, we hve λ(t n ) < 2/n 0 s n (λ is Lebesgue mesure). We hve the following extension of Theorem 1. Theorem 3 Let [, b] be compct intervl in R, let g n nd g be rel-vlued functions on [, b] with g of bounded vrition. In order for b fg for ll f HK it is necessry nd sufficient tht g n g in mesure s n, for ech n 1, g n is equivlent to function h n N BV, (2) nd there is M [0, ) such tht V h n M for ll n 1. If (, b) R is unbounded, then chnge the first line of (2) by requiring g n χ I gχ I in mesure for ech compct intervl I (, b). (1) Proof: By working with g n g we cn ssume g = 0. First consider the cse when (, b) is bounded intervl. If b fg n 0 for ll f HK, then using Theorem 1 nd chnging g n on countble set, we cn ssume g n N BV, V g n M, g n M nd

Product convergence for Henstock Kurzweil integrls 3 d c g n 0 for ech intervl (c, d) (, b). Suppose g n does not converge to 0 in mesure. Then there re δ, ɛ > 0 nd n infinite index set J N such tht λ(s n ) > δ for ech n J, where S n = {x (, b) : g n (x) > ɛ}. (Or else there is corresponding set on which g n (x) < ɛ for ll n J.) Now let n J. Since g n is left continuous, if x S n there is number c n,x > 0 such tht [x c n,x, x] S n. Hence, V n := {[c, x] : x S n nd [c, x] S n } is Vitli cover of S n. So there is finite set of disjoint closed intervls, σ n V n, with λ(s n \ I σn I) < δ/2. Write (, b) \ I σn I = I τn I where τ n is set of disjoint open intervls with crd(τ n ) = crd(σ n ) + 1. Let P n = crd({i τ n : g n (x) ɛ/2 for some x I}). Ech intervl I τ n tht does not hve or b s n endpoint hs contiguous intervls on its left nd right tht re in σ n (for ech of which g n (x) > ɛ/2 for some x). The intervl I then contributes more thn (ɛ ɛ/2) + (ɛ ɛ/2) = ɛ to the vrition of g n. If I hs s n endpoint then, since g n () = 0, I contributes more thn ɛ to the vrition of g n. If I hs b s n endpoint then I contributes more thn ɛ/2 to the vrition of g n. Hence, V g n (P n 1)ɛ + ɛ/2 = (P n 1/2)ɛ. (This inequlity is still vlid if P n = 1.) But, V g n M so P n P for ll n J nd some P N. Then we hve set of intervls, U n, formed by tking unions of intervls from σ n nd those intervls in τ n on which g n > ɛ/2. Now, λ( I Un I) > δ/2, crd(u n ) P + 1 nd g n > ɛ/2 on ech intervl I U n. Therefore, there is n intervl I n U n such tht λ(i n ) > δ/[2(p + 1)]. The sequence of centres of intervls I n hs convergent subsequence. There is then n infinite index set J J with the property tht for ll n J we hve g n > ɛ/2 on n intervl I (, b) with λ(i) > δ/[3(p + 1)]. Hence, lim sup n 1 I g n > δɛ/[6(p + 1)]. This contrdicts the fct tht I g n 0, showing tht indeed g n 0 in mesure. Suppose (2) holds. As bove, we cn ssume g n N BV, V g n M, g n M nd g n 0 in mesure. Let ɛ > 0. Define T n = {x (, b) : g n (x) > ɛ}. Then b g n g n + T n g n (,b)\t n (3) Mλ(T n ) + ɛ(b ). (4) Since lim λ(t n ) = 0, it now follows tht d c g n 0 for ech (c, d) (, b). Theorem 1 now shows b fg n 0 for ll f HK. Now consider integrls on R. If fg n 0 for ll f HK then it is necessry tht b fg n 0 for ech compct intervl [, b]. By the current theorem, g n g in mesure on ech [, b]. And, it is necessry tht fg 1 n 0. The chnge of vribles x 1/x now shows it is necessry tht g n be

4 Prsr Mohnty nd Erik Tlvil equivlent to function tht is uniformly bounded nd of uniform bounded vrition on [1, ]. Similrly with 1 fg n 0. Hence, it is necessry tht g n be uniformly bounded nd of uniform bounded vrition on R. Suppose (2) holds with g n g in mesure on ech compct intervl in R. Write fg n = fg n + b fg n + fg b n. Use Lemm 24 in [4] to write fg n fχ (,) V [,] g n fχ (,) M 0 s. We cn then tke lrge enough intervl [, b] R nd pply the current theorem on [, b]. Other unbounded intervls re hndled in similr mnner. Remrk 4 If (2) holds then dominted convergence shows g n g 1 0. And, convergence in 1 implies convergence in mesure. Therefore, in the first sttement of (2) nd in the lst sttement of Theorem 3, convergence in mesure cn be replced with convergence in 1. Similr remrks pply to Theorem 6. Remrk 5 The chnge of vribles rgument in the second lst prgrph of Theorem 3 cn be replced with n ppel to the Bnch Steinhus Theorem on unbounded intervls. See [3, Lemm 7]. Similrly in the proof of Theorem 8. The sequence of Heviside step functions g n = χ (n, ] shows (2) is not necessry to hve fg n 0 for ll f HK. For then, fg n = n f 0. In this cse, g n N BV nd V g n = 1. However, λ(t n ) = for ll 0 < ɛ < 1. Note tht for ech compct intervl [, b] we hve b g n 0 nd g n 0 in mesure on [, b]. It is somewht surprising tht the conditions (2) re lso necessry nd sufficient to hve f(g n g) 0 for ll f HK. Theorem 6 Let [, b] be compct intervl in R, let g n nd g be rel-vlued functions on [, b] with g of bounded vrition. In order for f(g n g) 0 for ll f HK it is necessry nd sufficient tht g n g in mesure s n, for ech n 1, g n is equivlent to function h n N BV, nd there is M [0, ) such tht V h n M for ll n 1. If (, b) R is unbounded, then chnge the first line of (5) by requiring g n χ I gχ I in mesure for ech compct intervl I (, b). (5) Proof: Certinly (5) is necessry in order for f(g n g) 0 for ll f HK. If we hve (5), let I n be ny sequence of intervls in (, b). We cn gin ssume g = 0. Write g n = g n χ In. Then g n g n, V g n V g n +

Product convergence for Henstock Kurzweil integrls 5 2 g n nd g n 0 in mesure. The result now follows by pplying Theorem 3 to f g n. Unbounded intervls re hndled s in Theorem 3. By combining Theorem 3 nd Theorem 6 we hve the following. Theorem 7 Let (, b) R then b fg for ll f HK if nd only if fg n fg 0 for ll f HK. Note tht f(g n g) fg n fg so if f(g n g) 0 then fg n fg. Thus, (5) is sufficient to hve fg n fg for ll f HK. However, this condition is not necessry. For exmple, let [, b] = [0, 1]. Define g n (x) = ( 1) n. Then g n = 1 nd V g n = 0. Let g = g 1. For no x [ 1, 1] does the sequence g n (x) converge to g(x). For no open intervl I [0, 1] do we hve I (g n g) 0. And, g n does not converge to g in mesure. However, let f HK with f > 0. Then f(g n g) = 0 when n is odd nd when n is even, f(g n g) = 2 f. And yet, for ll n, fg n = f = fg. It is nturl to sk wht extr condition should be given so tht fg n fg will imply fg n fg 0. We hve the following. Theorem 8 Let g n g in mesure or lmost everywhere. Then fg n fg for ll f HK if nd only if fg n fg 0 for ll f HK. Proof: Let [, b] be compct intervl. If fg n fg then g is equivlent to h N BV [2, Theorem 12.9] nd for ech f HK there is constnt C f such tht fg n C f. By the Bnch Steinhus Theorem [2, Theorem 12.10], ech g n is equivlent to function h n N BV with V h n M nd h n M. Let (c, d) (, b). By dominted convergence, d c g n d g. It now follows c from Theorem 1 tht b fg for ll f HK. Hence, by Theorem 7, fg n fg 0 for ll f HK. Now suppose (, b) = R nd fg n fg for ll f HK. The chnge of vribles x 1/x shows the Bnch Steinhus Theorem still holds on R. We then hve ech g n equivlent to h n N BV with V h n M nd h n M. As with the end of the proof of Theorem 3, given ɛ > 0 we cn find c R such tht c fg n < ɛ for ll n 1. The other cses re similr. Acknowledgment. An nonymous referee provided reference [3] nd pointed out tht in plce of convergence in mesure we cn use convergence in 1 (cf. Remrk 4). References [1] G.B. Follnd, Rel nlysis, New York, Wiley, 1999.

6 Prsr Mohnty nd Erik Tlvil [2] P.-Y. Lee, Lnzhou lectures on integrtion, Singpore, World Scientific, 1989. [3] W.L.C. Srgent, On some theorems of Hhn, Bnch nd Steinhus, J. London Mth. Soc. 28 (1953) 438 451. [4] E. Tlvil, Henstock Kurzweil Fourier trnsforms, Illinois J. Mth. (to pper).