Real Analysis II 9.1 Pointwise Convergence of Sequence of ...

1 Real Analysis IIChapter 9 Sequences and Series of Pointwise Convergence of Sequence of FunctionsDefinition ALet{fn}be a Sequence of functions defined on a set of real numbersE. We say that{fn}converges Pointwise to a functionfonEfor eachx E, the Sequence of real numbers{fn(x)}converges to the numberf(x). In other words, for eachx E, we havelimx fn(x)=f(x).Example1) Letfn(x)=xn,x [0,1]and letf(x)={0if0 x<11ifx= {fn}converges tofpointwise on [0,1].2) Letgn(x)=x1+nx,x [0, ]Then{gn}converges tog(x) = 0 Pointwise on [0, ].3) Lethn(x)=nx1+n2x2,x [0, ]Then{hn}converges toh(x) = 0 Pointwise on [0, ].4) Let n(x)={1ifx [ n,n]0 { n}converges to (x) = 1 Pointwise on [ , ]. {fn}converges Pointwise tofonE. Then given >0, and givenx E, there existsN=N(x, ) I, such that|fn(x) f(x)|< ,for alln generalNdepends on as well example, consider the Sequence of Example 1 above:f(x)=xnandf(x)=0,(0 x<1) andf(1) = =1/2, then, for eachx [0,1], there existsNsuch that|fn(x) f(x)| all (n N)( )Forx=0orx= 1 , then ( ) holds withN= 1.}}

2 Forx=3/4= , ( ) holds withN= 3 and forx= needN= claim that there is noNfor which ( ) hold for allx [0,1]. For if there is such anN, then for allx [0,1), (*) impliesxn< particular we would havexN<12for allx [0,1). Taking limit asx 1 we would have 1 1/2, which is a as given in Example 2 above :gn(x)=x1+nx,then we havegn(x) 1nfor allx [0, ) and hence for a given >0,anyNwithN>1/ will imply that|gn(x) 0|< for alln>Nand for allx [0, ).We leave to you to analyze the situations for the sequences in Examples 3 and 4 Uniform Convergence of Sequence of FunctionsDefinition {fn}be a Sequence of functions onE. We say that{fn}converges uniformly tofonEif for given >0, there existsN=N( ), depending on only, such that|fn(x) f(x)|< for alln>Nand for allx {fn}converges tofuniformly tofonE, we writefn funiformly ) Unlike the Pointwise converges, in the case of uniform Convergence , we noteNdependsonly on and not ) Iffn funiformly onE, thenfn fpointwise onE.]]]]

3 The sequencefn(x)=xnon [0,1] discussed inExample 1 of the previous section shows that the converse of the above statement is not sequencegn(x)=x1+nxconverges uniformly to 0 on [0, ). It is a good exercise to show whether the sequences of Examples 3 and 4of the previous section are uniformly convergent or following corollary is a restatement of the definition of uniform converges. It is useful to show that asequence is not uniformly Sequence {fn}doesnotconverge uniformly tofonEif and only if there exists an >0 such that there is noN>0 for which|fn(x) f(x)|, for alln>Nfor allx ) Iffn funiformly onEand >0, then there existsN>0 such that for alln>N, theentire graph ofy=fn(x) lies between the graphs ofy=f(x) andy=f(x)+.]

4 2) Iffn 0 uniformly onEand >0, then there existsN>0 such that for alln>Nand allx E,|fn(x)|< . This implies that for alln>N,supx E|fn(x)| and hencelimn supx E|fn(x)| .Since >0 is an arbitrary positive number, we conclude thatIffn 0 uniformly onE, then limn supx E|fn(x)|= converse is also true and the proof is an the sequencehn(x)=nx1+n2x2we havesupx [0, )|hn(x)| hn(1n)=12and hencelimn supx [0, )|hn(x)|6= {hn}does not converge uniformly to 0 on [0, ).Theorem funiformly onEif and only if limn supx E|fn(x) f(x)|= (Cauchy Criterion for Uniform Convergence )A Sequence {fn}converges uniformlyonEif and only if for a given >0, there existsN>0 such that for alln m>Nand for allx E,|fn(x) fm(x)|< .Theorem {fn}is a Sequence of continuous functions on a bounded and closed interval [a,b] and{fn}converges Pointwise to a continuous functionfon [a,b], thenfn funiformly on [a,b].]]]

5 Consequences of Uniform ConvergenceTheorem funiformly on [a,b], iffnare continuous atc [a,b], thenfis continuous funiformly on [a,b], iffnare continuous on [a,b], thenfis continuous on [a,b].RemarkDoesfn R[a,b] andf fpointwise on [a,b] imply thatf R[a,b]? The answer is no. Forexample, letA={r1,r2,r3, }be the set of all rational numbers in [0,1], and letAn={r1,r2, ,rn}.Let nbe the characteristic function ofAnand be the characteristic ofA. Since nis discontinuous only ata finite number of points (where ?), we see that n R[a,b]. On the other hand, is not continuouos at anypoint in [0,1] and hence 6 R[a,b]. Clearly n on [0,1] R[a,b] and iffn funiformly on [a,b], thenf R[a,b].RemarkIn Theorems and , uniform Convergence is sufficient.

6 The sequencehn(x)=nx1+n2x2forx [0, ), converges to the continuous functionh(x) = 0. Recall that the converges is not uniform. Thesequencefn(x)=xnon [0,1] can be used to show that uniform Convergence is not necessary for theorem (explain).RemarkWhen doesfn fimply baf baf? To answer this question, we consider the followingexample. Letfn(x)= 2nif1n x 2n0 otherwiseThen 10fn(x)dx= 2n1n2ndx=2n(2n 1n)= 2and hence limn 10fn(x)dx= the other hand, for fixedx [0,1], we can choose anNso thatx>2/Nand hencefn(x) = 0 for alln 0 Pointwise and hence 10limn fn(x)dx= R[a,b] and iffn funiformly on [a,b], then baf (x)=xnnon [0,1] and letf(x) = 0. Thenfn funiformly butf n(1) = 1 whilef (1) = f n(x)=f (x)does not hold atx= n(x) exists for eachnand eachx [a,b], iff nis continuous on [a,b], if{fn}convegresuniformly tofon [a,b], and if{f n}convegres uniformly togon [a,b], theng=f.]

7 Convergence and Uniform Convergence of Series of FunctionsDefiniton {un}be a Sequence of functions and letsn(x)= nk=1uk(x) be thenth partial sumof the infintie series k=1uk(x). We say k=1ukconverges Pointwise tofonEifsn fpointwise case we write k=1uk=fpointwise onEExampleLetuk=xk, 1<x<1 and letf(x)x1 x. Then k=1uk=fpointwise on ( 1,1).Definiton say that k=1ukconverges tofuniformly onEifsn funiformly onE. We write k=1uk=funiformly onETheorem k=1uk=funiformly onE, and if{uk}is continuous onE, thenfis continuous (x)=x(1 xn),(0 x 1,n=0,1,2, ),and letf(x)={1,if 0<x 10, ifx= un=fpointwise on [0,1].(Verify this) Clearlyfis not continuous atx= 0 whileunis continuousfor (Weierstrass M-Test)If{uk}is a Sequence of continuous functions such that|uk(x)| Mkfor allx Eand if Mkis convergent, then k=1uk=funiformly onENotationif{ak}and{bk}are two sequences, and ifak bk, we write k=1ak<< k=1bkThus Weierstrass Theorem states thatif k=1uk<< k=1Mk< ,then, for some functionf, k=1uk=funiformly onEExampleSince n=1sin(nx)n2<< n=11n2< ,Weierstrass Theorem implies that n=1sin(nx)n2converges uniformly the power series k=0akxkconverges forx=x0(withx06= 0), then the power seriesconverges uniformly on [ x1,x1] for anyx1 [0,|x0|].}

8 Theorem {uk}is a Sequence of continouous nonnegative functions on [a,b] and if k=0ukconvgres Pointwise to a continuous functionfon [a,b], then k=1uk=funiformly on [a,b]. Integration and Differentiation of Series of FunctionsTheorem {uk}be Sequence of fucntions inR[a,b]. and suppose k=1uk=funiformly on[a,b]. Thenf R[a,b] and baf(x)dx= ba( k=1uk(x))dx= k=1( bauk(x)dx)Theorem {uk}is differentiable on [a,b], if{u k}is continuous on [a,b], if uk=funiformly,and if u kconverges uniformly on [a,b], then k=1u k(x)=f (x)Example1)1 x+x2 x3+ =11+xuniformly on (0,1)implies that for anyy (0,1), y01dx y0xdx+ y0x2dx y0x3dx+ = y011+xdxfrom which we conclude thaty y22+y33 y44+ = log(1 +y).2)1+x+x2+x3+ =11 xuniformly on (0,1)implies that for allx ( 1,1),1+2x+3x2+4x3+ =1(1 x) the power series k=0akxkconvegres tof(x)on[ b,b] for someb>0, then for anyx [ b,b],f (x)= k=1kakxk (x)= k=0akxk,thenf(n)(x)= k=nk(k 1)(k 2) (k n+1)akxk (A Continuous Nowhere Differentiable Function)Letf0(x) = the distance fromxto the nearest integer.

9 (Thusf0( ) = andf0( ) = )Definefk(x)=fk(10kx) andF(x)= k=0fk(x) continouos everywhere and differentiable example of everywhere continuous and nowhere differentiable function is due toWeierstrass and is given byG(x)= k=0cos (3nx)2n.