Introduction To Real Analysis Spring 2014 Lecture Notes

Introduction to Real AnalysisSpring 2014 Lecture NotesVern I. PaulsenApril 22, 2014

2

Contents1 Sequences and Series of Functions1.1 Behavior of Riemann Integrals with Limits . . . . . . . . .1.2 Uniform Convergence and Continuity . . . . . . . . . . . . .1.3 Uniform Convergence and Derivatives . . . . . . . . . . . .1.4 Series of Functions . . . . . . . . . . . . . . . . . . . . . . .1.5 An Increasing Function with a Dense Set of Discontinuities1.6 A Space Filling Curve . . . . . . . . . . . . . . . . . . . . .1.7 Power Series . . . . . . . . . . . . . . . . . . . . . . . . . . .1.8 Operations on Power Series . . . . . . . . . . . . . . . . . .1.9 Taylor Series . . . . . . . . . . . . . . . . . . . . . . . . . .1.10 Polynomial Approximation . . . . . . . . . . . . . . . . . .1.11 The Stone-Weierstrass Theorem . . . . . . . . . . . . . . . .1.12 Fourier Series . . . . . . . . . . . . . . . . . . . . . . . . . .1.13 Orthonormal Sets of Functions . . . . . . . . . . . . . . . .1.14 Fourier Series, Continued . . . . . . . . . . . . . . . . . . .1.15 Equicontinuity and the Arzela-Ascoli Theorem . . . . . . .5810131518222428343740434549532 Multivariable Differential Calculus2.1 The Total Derivative . . . . . . . . . . . . . . . . . . . .2.2 Differentiability and Continuity . . . . . . . . . . . . . .2.3 The Chain Rule . . . . . . . . . . . . . . . . . . . . . . .2.4 Multivariable Mean Value Theorem . . . . . . . . . . . .2.5 Continuously Differentiable Functions . . . . . . . . . .2.6 The Inverse Function Theorem . . . . . . . . . . . . . .2.7 The Multi-Variable Newton and Quasi-Newton Methods2.8 The Implicit Function Theorem . . . . . . . . . . . . . .2.9 Local Extrema and the Second Derivative Test . . . . .2.10 Taylor Series in Several Variables . . . . . . . . . . . . .57606465676971767884883.

4CONTENTS

Chapter 1Sequences and Series ofFunctionsIn this chapter we introduce different notions of convergence for sequenceand series of functions and then examine how integrals and derivatives behave upon taking limits of functions in these various senses. We then applythese results to power series and Fourier series.Definition 1.1. Given a set X, a metric space (Y, ρ), functions fn : X Y, n N and f : X Y, we say that the sequence of functions {fn }converges pointwise to f provided that for each x X, the sequenceof points {fn (x)} converges to the point f (x) in the metric of Y. That is,ptwprovided that limn ρ(fn (x), f (x)) 0. When this occurs we write, fn f.Note that the statement that {fn } converges pointwise to f is equivalentto the requirement that for each 0 and x X, there is a Nx so thatwhen n Nx we have that ρ(fn (x), f (x)) . That is, since pointwiseconvergence only requires convergence at each point in X, the value that wetake for N could depend on the individual point x as well as on . When thevalue for N can be picked depending only on and independent of the pointx, then we call the convergence uniform. The precise definition follows.Definition 1.2. Given a set X, a metric space (Y, ρ), functions fn : X Y,n N and f : X Y, we say that the sequence of functions {fn } convergesuniformly to f provided that for each 0, there is N, so that whenn N, then for every x X, we have ρ(fn (x), f (x)) . When this occursuwe write, fn f.Note that uniform convergence always implies pointwise convergence,since it is the stronger condition that N is independent of the point x.5

6CHAPTER 1. SEQUENCES AND SERIES OF FUNCTIONSJust as pointwise convergence is the requirement that limn ρ(fn (x), f (x)) 0, uniform convergence is equivalent to the requirement that the sequence ofnumbers sn sup{ρ(fn (x), f (x)) : x X} converges to 0. That is, providedthatlim[sup{ρ(fn (x), f (x)) : x X}] 0.nWe prove this below.Proposition 1.3. Let X be a set, let (Y, ρ) a metric space, let fn : X Y,n N and f : X Y be functions and setsn sup{ρ(fn (x), f (x)) : x X}.u f if and only if limn sn 0.Then fn u f. Let 0 be given. By the definition ofProof. First assume that fn uniform convergence, there is a N so that when n N, for every x X, wehave ρ(fn (x), f (x)) /2. But this implies that for n N, we havesn sup{ρ(fn (x), f (x)) : x X} /2 .Thus, for n N we have that sn 0 and since was arbitrary, thisproves that limn sn 0.Conversely, assume that limn sn 0. Then given 0, there is aN so that n N implies that 0 sn whihc implies that for everyn N and for every x X, we have ρ(fn (x), f (x)) . This proves thatufn f.Note that for both of these definitions, we did not need for X to be ametric space, although in many of the interesting examples, we will alsohave X a metric space.ptwuAlso note that if fn f, then fn f.Example 1.4. Let X Y ([0, 1] be endowed with the usual metric, let0 0 x 1fn (x) xn and let f (x) . Then it is easy to see that1 x 1ptwfn f. To decide if the convergence is also uniform, we computesn sup{ρ(fn (x), f (x)) : 0 x 1} sup{ xn f (x) : 0 x 1} sup{xn : 0 x 1} 1.Since, limn sup{ρ(fn (x), f (x)) : 0 x 1} 1 6 0, we conclude that {fn }does not converge uniformly to f.

7Thus although uniform convergence implies pointwise convergence, wehave that pointwise convergence does not imply uniform convergence. Forthis reason we say that uniform convergence is a stronger convergence thanpointwise convergence.In this example, each of the functions fn is continuous, but f is clearlynot continuous. Thus, this example shows that the pointwise limit of continuous functions need not be a continuous function. Later we will see thatuniform limits of continuous functions are again continuous.Example 1.5. This is a slight modification of the first example. Let X [0, B], Y [0, 1] with 0 B 1, let fn (x) xn and let f (x) 0. Nowsn sup{ fn (x) f (x) : x X} sup{xn : 0 x B} B n .u f.Since, limn B n 0, we have that fn Example 1.6. Let X Y R with the usual metric, let fn (x) ptwxnand letf (x) 0. Then fn f but {fn } does not converge to f uniformly.Example 1.7. Let X Y [ π, π] with the usual metric, let fn (x) sin(nx)and let f (x) 0. Since sup{ fn (x) f (x) : π x π} n1 wenuhave that fn f, but fn0 (0) 1, while f 0 (0) 0.Thus, even uniform convergence of functions does not guarantee convergence of their derivatives.Problem 1.8. Let X Y [0, 1] with the usual metric, let fn : X Y bedefined by fn (x) x1/n and let(0 when x 0f (x) .1 when x 6 0ptwProve that fn f and that {fn } does not converge to f uniformly.Problem 1.9. Let X R be compact, let fn : X R be defined byufn (x) nx and let f (x) 0 be the 0 function. Prove that fn f.Problem 1.10. Let f : R R be continuous at 0. Set gn (x) f ( nx ) andlet g be the constant function that is equal to f (0).ptw1. Prove that gn g as functions on R.2. Give an example of a function f such that the convergence is not uniform as functions on R.u3. Prove that if X R is any compact subset of R, then gn g asfunctions on X.

8CHAPTER 1. SEQUENCES AND SERIES OF FUNCTIONS1.1Behavior of Riemann Integrals with LimitsWe now look at how Riemann integration behaves under pointwise and uniform convergence.Example 1.11. Let X Y [0, 1] with the usual metric. Recall that therationals are countable and let {rn }n N be an enumeration of the rationalnumbers in [0, 1]. We set((1 x rk , 1 k n1 x rationalfn (x) and f (x) .0 otherwise0 x irrationalptwIt is easily checked that fn f, but for every n, sup{ fn (x) f (x) : 0 x 1} 1 and so the sequence does not converge uniformly Rto f. Now verify1that each function fn is Riemann integrable on [0, 1] with 0 fn (x)dx 0,but f is not Riemann integrable.Thus, we see that if {fn } is a sequence of Riemann integrable fuctionsthat converges pointwise to a function f, then f might not even be Riemannintegrable! In this case there is no chance that the limit of the integrals isthe integral of the limit.The next example shows that even when the limit function is Riemannintegrable, pointwise convergence is not enough to guarantee that the limitof the integrals is the integral of the limit.Example 1.12. Let X Y [0, 1], set 2 2n xfn (x) 2n 2n2 x 010 x 2n112n x n1n x 1ptwand let f (x) 0. ThenR 1 are Riemann inteR 1 fn f, all of these functionsgrable on [0, 1], but 0 fn (x)dx 1/2 for all n, while 0 f (x)dx 0.These last two examples show that the formula,Z bZ blimfn (x)dx lim fn (x)dx,naanis not valid when the limit of the functions is meant in the pointwise sense.We now prove that all of these problems “go away” when one considersuniform convergence instead.

1.1. BEHAVIOR OF RIEMANN INTEGRALS WITH LIMITS9Lemma 1.13. Let α : [a, b] R be an increasing function and let f, g :[a, b] R be two bounded functions. If δ sup{ f (x) g(x) : a x b},then for any partition P {x0 , ., xn } of [a, b] e have that U (f, P, α) U (g, P, α) δ(α(b) α(a)) and L(f, P, α) L(g, P, α) δ(α(b) α(a)).Proof. For each x, we have that g(x) δ f (x) g(x) δ. Hence, for eachsubinterval of the partition we have that mi (g) δ inf{g(x) : xi 1 x xi } δ inf{f (x) : xi 1 x xi } mi (f ) and Mi (f ) sup{f (x) : xi 1 x xi } sup{g(x) : xi 1P x xi } δ Mi (g) δ. PThus, U (f, P, α) ni 1 Mi (f )(α(xi ) α(xi 1 )) ni 1 (Mi (g) δ)(α(xi ) α(xi 1 )) U (g, P, α) δ(α(b) α(a)).The proof of the other inequality is similar.Theorem 1.14. Let α : [a, b] R be an increasing function on [a, b] andlet {fn } be a sequence of functions that are Riemann-Stieltjes integrable on[a, b] with respect to α and converge uniformly to a function f on [a, b]. Thenf is Riemann-Stieltjes integrable on [a, b] with respect to α andZlimnbZfn dα abf dα.aProof. We first prove that f satisfies the Riemann-Stieltjes integrability criterion on [a, b]. The case that α(a) α(b) is trivial, so we assume thatα(b) α(a) 0. Given 0, set δ 4(α(b) α(a))and choose an integer N so that forn N we have that sup{ f (x) fn (x) : a x b} δ.Since fN is Riemann-Stieltjes integrable, we may choose a partition P,so that U (fN , P, α) L(fN , P, α) 2 . Applying the lemma, we have thatU (f, P, α) L(f, P, α) [U (fN , P, α) δ(α(b) α(a))] [L(fN , P, α) δ(α(b) α(a))] [U (fN , P, α) L(fN , P, α)] 2δ(α(b) α(a)) .2 2Thus, f satisfies the Riemann-Stieltjes integrability criterion.Also, for any n N, applying the lemma again, we have thatZbf dα inf{U (f, P, α) : P} aZinf{U (fn , P, α) δ(α(b) α(a)) : P} ab fn dα ,4

10CHAPTER 1. SEQUENCES AND SERIES OF FUNCTIONSwhileZ bf dα sup{L(f, P, α) : P} aZsup{L(fn , P, α) δ(α(b) α(a)) : P} ab fn dα .4These two inequalities show that for n N, we have thatZ bZ b fn dα .f dα 4aaSince was arbitrary, we have thatZ bZ blimfn dα f dα.naaProblem 1.15. Prove that the functions {fn } in Example 1.7 do convergepointwise to f, prove that each fn is Riemann integrable with Riemann integral equal to 0 and prove that f is not Riemann integrable.Problem 1.16. Prove that the functions {fn } of Example 1.8 do convergepointwise to f and prove that each fn is Riemann integrable with integralequal to 1/2.1.2Uniform Convergence and ContinuityWe have already seen that pointwise limits of continuous functions need notbe continuous. In this section we will show that uniform limits preservecontinuity. Some of these results have been seen in a different form inChapter I.4.Theorem 1.17. Let (X, d) and (Y, ρ) be metric spaces, let fn : X Y, n N be a sequence of functions from X to Y that converge uniformly to afunction f : X Y and let x0 X. If fn is continuous at x0 for all n N,then f is continuous at x0 .Proof. Given 0, we may pick N so that for n N, we have thatsup{ρ(fn (x), f (x)) : x X} /3. Fix any n N, and using the fact thatfn is continuous at x0 , we may pick δ 0, so that d(x, x0 ) δ implies thatρ(fn (x), fn (x0 )) /3.