Real Analysis Math 125A, Fall 2012 Final Solutions 1. R

1 Real AnalysisMath 125A, Fall 2012 Final Solutions1.(a) Suppose thatf: [0,1] Ris continuous on the closed, boundedinterval [0,1] andf(x)>0 for every 0 x 1. Prove that the reciprocalfunction 1/f: [0,1] Ris bounded on [0,1].(b) Does this result remain true if: (i)f: [0,1] Ris not continuous on[0,1]; (ii)f: (0,1) Ris continuous on the open interval (0,1)?Solution. (a) Letm= infx [0;1]f(x).Sincef>0 on [0,1], we havem 0. Sincefis a continuous functionon a compact set, it attains its infimum at some point in [0,1], whichimplies thatm>0. Thereforef m>0 and 0<1/f 1/misbounded on [0,1]. (b) The result does not remain true in either case, sincefneed notattain its infimum. A counter-example for (i) isf(x) ={xif 0<x 1,1 ifx= counter-example for (ii) isf(x) =xfor 0<x< (a) Define uniform continuity onRfor a functionf:R R.(b) Suppose thatf,g:R Rare uniformly continuous onR. (i) Prove thatf+gis uniformly continuous onR. (ii) Give an example to show thatfgneed not be uniformly continuous (a) A functionf:R Ris uniformly continuous if for every >0there exists >0 such that|f(x) f(y)|< for allx,y Rsuch that|x y|<.}

2 ( ) Let >0. Choose 1>0 such that|f(x) f(y)|< 2for allx,y Rsuch that|x y|< 1and 2>0 such that|g(x) g(y)|< 2for allx,y Rsuch that|x y|< = min( 1, 2)>0. Then|x y|< implies that|(f+g)(x) (f+g)(y)| |f(x) f(y)|+|g(x) g(y)|< ,which proves thatf+gis uniformly continuous onR. ( ) An example isf(x) =g(x) =x. Thenf,gare uniformly contin-uous onR(take = ) but (fg)(x) =x2is not. To prove thatx2is not uniformly continuous, let >0 and choosex=1 + 2,y=1 .Then|x y|= /2< , but|x2 y2|= 1 + 24>1,so the definition of uniform continuity fails for 1. (We can showsimilarly that it fails for all >0.) that a functionf:R Ris differentiable at zero andf(1n)= 0 for alln that: (a)f(0) = 0; (b)f (0) = (a) Sincefis differentiable at 0, it is continuous at 0. The sequentialdefinition of continuity then implies thatf(0) = limn f(1n)= 0. (b) Sincefis differentiable at 0, the limitf (0) = limx 0f(x) f(0)xexists, so we can evaluate it on any sequencexn 0.

3 Using (a) andtakingxn= 1/n, we get thatf (0) = limn nf(1n)= thatf,g,h:R Rare functions such that:(a)f(x) g(x) h(x) for allx R, andf(0) =h(0);(b)f,hare differentiable at 0, andf (0) =h (0).Does it follow thatgis differentiable at 0?Solution. Yes, it does follow thatgis differentiable at 0. Condition(a) implies thatf(0) =g(0) =h(0) and therefore also thatf(x) f(0) g(x) g(0) h(x) h(0).Forx>0, we havef(x) f(0)x g(x) g(0)x h(x) h(0)x,and sincelimx 0+f(x) f(0)x=f (0) =h (0) = limx 0+h(x) h(0)x,the sandwich theorem implies thatlimx 0+g(x) g(0)x=f (0).Similarly, forx<0,f(x) f(0)x g(x) g(0)x h(x) h(0)x,and the sandwich theorem implies thatlimx 0 g(x) g(0)x=f (0).Since the left and right derivatives ofgexist and are equal, it followsthatgis differentiable at 0 andf (0) =g (0) =h (0).45.(a) Determine the Taylor polynomialPn(x) of degreencentered at 0 forthe functionex.(b) Give an expression for the remainderRn(x) in Taylor s theorem such thatex=Pn(x) +Rn(x).

4 (c) Prove thatex 1 +xfor allx R, with equality if and only ifx= 0.(d) Prove thate > a good choice ofxin (c).Solution. (a) Thekth derivative ofexisex, which is equal to 1 atx= 0, so thekth Taylor coefficient off(x) =exat zero isak=f(k)(0)k!=1k!,andPn(x) =n k=01k!xk= 1 +x+12!x2+ +1n!xn. (b) The expression for the Lagrange remainder isRn(x) =1(n+ 1)!f(n+1)( )xn+1=1(n+ 1)!e xn+1for some strictly between 0 andx. (c) Forn= 1, we getex= 1 +x+12e >0, it follows thatex 1 +x, with equality if and only ifx= 0. (d) Takex= e 1in the inequality from (c). This givese =e 1>1 + e 1 ore =ee> this inequality byeand take theeth power, to gete > that (fn) is a sequence of continuous functionsfn:R R, and(xn) is a sequence inRsuch thatxn 0 asn . Prove or disprove thefollowing statements.(a) Iffn funiformly onR, thenfn(xn) f(0) asn .(b) Iffn fpointwise onR, thenfn(xn) f(0) asn .Solution. (a) This statement is true. To prove it, we first observe thatfis con-tinuous since the uniform limit of continuous functions is continuous.

5 Let >0 be given. We write|fn(xn) f(0)| |fn(xn) f(xn)|+|f(xn) f(0)|and estimate each of the terms of the right-hand side. Sincefn funiformly, there existsN1 Nsuch that|fn(x) f(x)|< 2for allx Rifn>N1. Sincefis continuous at 0, there exists >0 such that|f(x) f(0)|< 2if|x|< ,and sincexn 0 there existsN2 Nsuch that|xn|< ifn> |f(xn) f(0)|< 2ifn>N2, LetN= max(N1,N2). Ifn>N, then it follows that|fn(xn) f(0)|< 2+ 2= ,which proves the result. (b) This statement is false. As a counter-example, letfn(x) ={1 n|x|if|x|<1/n,0if|x| 1/n,xn= continuous andfn fpointwise, wheref(x) ={1 ifx= 0,0 ifx = 0. However, we havefn(xn) = 0 for everynandf(0) = 1, sofn(xn) f(0). the power seriesf(x) = 1 + n=1anx3n= 1 +x32 3+x62 3 5 6+x92 3 5 6 8 9+..,an=12 3 5 (3n 4) (3n 3) (3n 1) 3n.(a) For whichx Rdoes the series converge?(b) Prove thatf (x) =xf(x).Solution. (a) We computer= limn an+1x3(n+1)anx3n =|x|3limn 1(3n+ 2)(3n+ 3)= ratio test implies that the power series converges for everyx R.}}

6 (Its radius of convergence isR= .) (b) The differentiation theorem for power series implies thatfis in-finitely differentiable onR, and its derivatives are the sum of the term-by-term differentiated power series. Moreover, the power series for thederivatives offalso converge onR. Thus, using the identities3 2a1= 1,3n(3n 1)an=an 1forn 2,we get thatf (x) = n=13n(3n 1)anx3n 2=x{1 + n=2an 1x3n 3}=x{1 + n=1anx3n}=xf(x)8 12 10 8 6 4 202 1 1: Graph of the Airy functionf(x) = 1 + n= of the ODEy =xyare called Airy functions. Forxlarge and positive, they behave like exponential functions, and forxlarge andnegative, they behave like algebraically-decaying trigonometric Airy functions describe a transition from oscillatory to exponential be-havior, and can t be written in terms of elementary functions. They arisein many physical applications. In optics, they describe the electromagneticfield of a light wave near a boundary between light and shadow; and in quan -tum mechanics , they describe the wavefunction of a particle near a boundarybetween classically allowed and forbidden regions.

7 (Here, the relevant Airyfunctions are the ones that decay asx , rather than grow likef.)The graph of the function in the problem is shown in Figure 1. It is worthnoting that this graph was obtained by using MATLAB s built in routine forAiry functions,notby summing the power series explicitly. Ifxis large andnegative, the power series consists of large terms with alternating signs, andthese terms almost cancel. With a machine round-off error of approximately2 16, this cancelation makes the numerical values of the sum of the powerseries completely inaccurate whenx. moral is that a power series may not be as useful as it looks away fromits central point if its rate of convergence becomes very (a) Define a metric on a setX.(b) Consider the following functions defined forx,y Rby:d1(x,y) = (x y)2,d2(x,y) =|x2 y2|,d3(x,y) =|x 2y|.For each function, determine whether or not it is a metric (a) A metricdon a setXis a functiond:X X Rsuch that forallx,y,z (x,y) 0 andd(x,y) = 0 if and only ifx=y; (x,y) =d(y,x); (x,y) d(x,z) +d(z,y).

8 (b) None of these are metrics. The functiond1is positive and symmetric, but it doesn t satisfy thetriangle inequality (1,0) +d1(0, 1) = 1 + 1 = 2<4 =d1(1, 1). The functiond2fails only becaused2(x,y) = 0 implies thatx= y, notx= (1, 1) = 0. However,d2satisfies the triangle inequality,sinced2(x,y) = x2 z2+z2 y2 x2 z2 + z2 y2 =d2(x,z)+d2(z,y),and it would define a metric on [0, ). Apart from being non-negative, the functiond3is about as poor anexcuse for a metric as you can find: it isn t symmetric;d3(x,y) = 0doesn t imply thatx= (2,1) = 0; andd3doesn t satisfy thetriangle inequality (4,2) +d3(2,1) = 0 + 0<2 =d3(4,1).10]