Introduction to Real Analysis Spring 2014 Lecture Notes

1 Introduction to real AnalysisSpring 2014 Lecture NotesVern I. PaulsenApril 22, 20142 Contents1 Sequences and Series of Behavior of Riemann Integrals with Limits .. Uniform Convergence and Continuity .. Uniform Convergence and Derivatives .. Series of Functions .. An Increasing Function with a Dense Set of Discontinuities . A Space Filling Curve .. Power Series .. Operations on Power Series .. Taylor Series .. Polynomial Approximation .. The Stone-Weierstrass Theorem .. Fourier Series .. Orthonormal Sets of Functions .. Fourier Series, Continued .. Equicontinuity and the Arzela-Ascoli Theorem .. 532 Multivariable Differential The Total Derivative.

2 Differentiability and Continuity .. The Chain Rule .. Multivariable Mean Value Theorem .. Continuously Differentiable Functions .. The Inverse Function Theorem .. The Multi-Variable Newton and Quasi-Newton Methods .. The Implicit Function Theorem .. Local Extrema and the Second Derivative Test .. Taylor Series in Several Variables .. 8834 CONTENTSC hapter 1 Sequences and Series ofFunctionsIn this chapter we introduce different notions of convergence for sequenceand series of functions and then examine how integrals and derivatives be-have upon taking limits of functions in these various senses. We then applythese results to power series and Fourier a setX,a metric space(Y, ),functionsfn:X Y, n Nandf:X Y,we say that the sequence of functions{fn}converges pointwise tof provided that for eachx X,the sequenceof points{fn(x)}converges to the pointf(x)in the metric is,provided thatlimn (fn(x),f(x)) = this occurs we write,fnptw that the statement that{fn}converges pointwise tofis equivalentto the requirement that for each >0 andx X,there is aNxso thatwhenn > Nxwe have that (fn(x),f(x))<.

3 That is, since pointwiseconvergence only requires convergence at each point inX,the value that wetake forNcould depend on the individual pointxas well as on .When thevalue forNcan be picked depending only on and independent of the pointx,then we call the precise definition a setX,a metric space(Y, ),functionsfn:X Y,n Nandf:X Y,we say that the sequence of functions{fn}convergesuniformly tofprovided that for each >0,there isN,so that whenn > N,then for everyx X,we have (fn(x),f(x))< .When this occurswe write,fnu that uniform convergence always implies pointwise convergence,since it is the stronger condition thatNis independent of the 1. SEQUENCES AND SERIES OF FUNCTIONSJust as pointwise convergence is the requirement that limn (fn(x),f(x)) =0,uniform convergence is equivalent to the requirement that the sequence ofnumberssn= sup{ (fn(x),f(x)) :x X}converges to is, providedthatlimn[sup{ (fn(x),f(x)) :x X}] = prove this a set, let(Y, )a metric space, letfn:X Y,n Nandf:X Ybe functions and setsn= sup{ (fn(x),f(x)) :x X}.

4 Thenfnu fif and only iflimn sn= assume thatfnu >0 be given. By the definition ofuniform convergence, there is aNso that whenn > N,for everyx X,wehave (fn(x),f(x))< this implies that forn > N,we havesn= sup{ (fn(x),f(x)) :x X} /2< .Thus, forn > Nwe have that|sn 0|< and since was arbitrary, thisproves that limn sn= , assume that limn sn= given >0,there is aNso thatn > Nimplies that 0 sn< whihc implies that for everyn > Nand for everyx X,we have (fn(x),f(x))< .This proves thatfnu that for both of these definitions, we did not need forXto be ametric space, although in many of the interesting examples, we will alsohaveXa metric note that iffnu f,thenfnptw [0,1]be endowed with the usual metric, letfn(x) =xnand letf(x) ={0 0 x <11x= it is easy to see thatfnptw decide if the convergence is also uniform, we computesn= sup{ (fn(x),f(x)) : 0 x 1}=sup{|xn f(x)|: 0 x 1}= sup{xn: 0 x <1}= ,limnsup{ (fn(x),f(x)) : 0 x 1}= 16= 0,we conclude that{fn}does not converge uniformly although uniform convergence implies pointwise convergence, wehave that pointwise convergence does not imply uniform convergence.}

5 Forthis reason we say that uniform convergence is astrongerconvergence thanpointwise this example, each of the functionsfnis continuous, butfis clearlynot continuous. Thus, this example shows that the pointwise limit of con-tinuous functions neednotbe a continuous function. Later we will see thatuniform limits of continuous functions are again is a slight modification of the first example. LetX=[0,B],Y= [0,1]with0< B <1,letfn(x) =xnand letf(x) = sup{|fn(x) f(x)|:x X}= sup{xn: 0 x B}= ,limn Bn= 0,we have thatfnu the usual metric, letfn(x) =xnand letf(x) = fbut{fn}does not converge [ ,+ ]with the usual metric, letfn(x) =sin(nx)nand letf(x) = {|fn(x) f(x)|: x + }=1nwehave thatfnu f,butf n(0) = 1,whilef (0) = , even uniform convergence of functions does not guarantee conver-gence of their [0,1]with the usual metric, letfn:X Ybedefined byfn(x) =x1/nand letf(x) ={0whenx= 01whenx6= thatfnptw fand that{fn}does not converge Rbe compact, letfn:X Rbe defined byfn(x) =xnand letf(x) = 0be the 0 function.}

6 Prove thatfnu :R Rbe continuous (x) =f(xn)andletgbe the constant function that is equal tof(0).1. Prove thatgnptw gas functions Give an example of a functionfsuch that the convergence is not uni-form as functions Prove that ifX Ris any compact subset ofR,thengnu gasfunctions 1. SEQUENCES AND SERIES OF Behavior of Riemann Integrals with LimitsWe now look at how Riemann integration behaves under pointwise and uni-form [0,1]with the usual metric. Recall that therationals are countable and let{rn}n Nbe an enumeration of the rationalnumbers in[0,1].We setfn(x) ={1x=rk,1 k n0otherwiseandf(x) ={ is easily checked thatfnptw f,but for everyn,sup{|fn(x) f(x)|: 0 x 1}= 1and so the sequence does not converge uniformly verifythat each functionfnis Riemann integrable on[0,1]with 10fn(x)dx= 0,butfis not Riemann , we see that if{fn}is a sequence of Riemann integrable fuctionsthat converges pointwise to a functionf,thenfmight not even be Riemannintegrable!}}

7 In this case there is no chance that the limit of the integrals isthe integral of 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 [0,1],setfn(x) = 2n2x0 x 12n2n 2n2x12n< x 1n01n< x 1and letf(x) = f,all of these functions are Riemann inte-grable on[0,1],but 10fn(x)dx= 1/2for alln,while 10f(x)dx= last two examples show that the formula,limn bafn(x)dx= balimnfn(x)dx,isnot validwhen the limit of the functions is meant in the pointwise now prove that all of these problems go away when one considersuniform convergence BEHAVIOR OF RIEMANN INTEGRALS WITH LIMITS9 Lemma : [a,b] Rbe an increasing function and letf,g:[a,b] Rbe two bounded functions.

8 If = sup{|f(x) g(x)|:a x b},then for any partitionP={x0,..,xn}of[a,b]e have thatU(f,P, ) U(g,P, ) + ( (b) (a))andL(f,P, ) L(g,P, ) ( (b) (a)). eachx,we have thatg(x) f(x) g(x) + .Hence, for eachsubinterval of the partition we have thatmi(g) = inf{g(x) :xi 1 x xi} inf{f(x) :xi 1 x xi}=mi(f) andMi(f) = sup{f(x) :xi 1 x xi} sup{g(x) :xi 1 x xi}+ =Mi(g) + .Thus,U(f,P, ) = ni=1Mi(f)( (xi) (xi 1)) ni=1(Mi(g)+ )( (xi) (xi 1)) =U(g,P, ) + ( (b) (a)).The proof of the other inequality is : [a,b] Rbe an increasing function on[a,b]andlet{fn}be a sequence of functions that are Riemann-Stieltjes integrable on[a,b]with respect to and converge uniformly to a functionfon[a,b].Thenfis Riemann-Stieltjes integrable on[a,b]with respect to andlimn bafnd = bafd.

9 First prove thatfsatisfies the Riemann-Stieltjes integrability cri-terion on [a,b].The case that (a) = (b) is trivial, so we assume that (b) (a)> >0,set = 4( (b) (a))and choose an integerNso that forn Nwe have that sup{|f(x) fn(x)|:a x b}< .SincefNis Riemann-Stieltjes integrable, we may choose a partitionP,so thatU(fN,P, ) L(fN,P, )< the lemma, we have thatU(f,P, ) L(f,P, ) [U(fN,P, ) + ( (b) (a))] [L(fN,P, ) ( (b) (a))] =[U(fN,P, ) L(fN,P, )] + 2 ( (b) (a))< 2+ ,fsatisfies the Riemann-Stieltjes integrability , for anyn N,applying the lemma again, we have that bafd = inf{U(f,P, ) :P} inf{U(fn,P, ) + ( (b) (a)) :P}= bafnd + 4,10 CHAPTER 1. SEQUENCES AND SERIES OF FUNCTIONS while bafd = sup{L(f,P, ) :P} sup{L(fn,P, ) ( (b) (a)) :P}= bafnd two inequalities show that forn N,we have that| bafd bafnd | 4<.

10 Since was arbitrary, we have thatlimn bafnd = bafd .Problem that the functions{fn}in Example do convergepointwise tof,prove that eachfnis Riemann integrable with Riemann in-tegral equal to 0 and prove thatfis not Riemann that the functions{fn}of Example do convergepointwise tofand prove that eachfnis Riemann integrable with integralequal to 1 Uniform 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 (X,d)and(Y, )be metric spaces, letfn:X Y,n Nbe a sequence of functions fromXtoYthat converge uniformly to afunctionf:X Yand letx0 continuous atx0for alln N,thenfis continuous >0,we may pickNso that forn > N,we have thatsup{ (fn(x),f(x)).}