Solutions to Assignment-3 - UCB Mathematics

1 Solutions to Assignment-31. (a) Letf: (a,b) Rbe continuous such that for somep (a,b),f(p)>0. Show that there exists a >0 such thatf(x)>0 for allx (p ,p+ ).Solution:Let >0 such thatf(p) >0 (for instance one can take =f(p)/2). Sincefiscontinuous, there exists >0 such that|x p|< = |f(x) f(p)|< .In particular, for allx (p ,p+ ),f(x)> f(p) >0.(b) LetE Rbe a subset such that there exists a sequence{xn}inEwith the property thatxn x0/ that there is an unbounded continuous functionf:E :Consider the functionf(x) =1x E, this function is continuous onE. On the other hand, by the hypothesis,limn |f(xn)|= ,and so the function is unbounded (a) Ifa,b R, show thatmax{a,b}=(a+b) +|a b| :Ifa b, then max{a,b}=b.(b) Show that iff1,f2, ,fnare continuous functions on a domainE R, theng(x) = max{f1(x), ,fn(x)}is again a continuous function :Forn= 2, use part(a) to writeg(x) =(f1(x) +f2(x)) +|f1(x) f2(x)| +f2and|f1 f2|are continuous, it follows thatgis also continuous.

2 Forn >2 andk= 2,3, ,n, letgk(x) = max(f1(x) ,fk(x)).In particulargn=g. We use induction to show thatgk(x) is continuous for allk= 2, , base casek= 2 is verified, since we have already shown thatg2(x) is continuous. For theinductive step, supposegk 1(x) is continuous. We note thatgk(x) = max(gk 1(x),fk(x)),1and again by the above argument for max of two continuous functions, we see thatgk(x) is alsocontinuous. By inductiongn(x) =g(x) is also continuous.(c) Let s explore if the infinite version of this true or not. For eachn N, definefn(x) ={1,|x| 1/nn|x|,|x|<1 computeh(x) = sup{f1(x),f2(x), ,fn(x), }. Is it continuous?Solution:For anyx6= 0, there exists anNsuch that|x|>1/nfor alln > Nand|x| 1/nforn N, and sofn(x) = 1 for alln > Nand forn N,fn(x) =n|x| 1. On the otherhand,fn(0) = 0 for alln, and henceh(x) ={1, x6= 00, x= 0,and is For each of the following, decide if the function is uniformly continuous or not. In either case, give aproof using just the definition in terms of and.}}

3 (a)f(x) = x2+ 1 on (0,1).Solution:Note that|f(x) f(y)|=| x2+ 1 y2+ 1|=|x2 y2| x2+ 1 + y2+ 1=|x y||x+y| x2+ 1 + y2+ ifx,y (0,1), then|x+y|<2, and moreoverx2+ 1,y2+ 1 1, and so|f(x) f(y)|<4|x y|.Given >0, let = /4. Then|x y|< = |f(x) f(y)|< .(b)g(x) =xsin(1/x) on (0,1).Solution: was mentioned by some students in class, this problem does not seem tohave a solution without an appeal to the mean value theorem (MVT), which we of course didnot cover last week. Below is the most canonical attempt towards a solution, and you will seethe point at which I dont think one can proceed without MVT. For an independent proof ofuniform continuity, without actually using showing the dependence of on , simply considerthe functionG: [0,1] R,G(x) ={xsin(1/x), x (0,1]0, x= have shown in class that this function is continuous on [0,1]. Since [0,1] is closed andbounded,G(x) is uniformly continuous. But thenG(x) =g(x) on (0,1), and sog(x) is alsouniformly attempt at a solution.)}

4 (x+h) sin(1x+h) xsin(1x) |x| sin(1x+h) sin(1x) +|h| sin(1x+h) |x| sin(1x+h) sin(1x) +|h|.For the first term, we use the fact thatsinA sinB= 2 sin(A B2)cos(A+B2),and so|x| sin(1x+h) sin(1x) = 2|x| sin(h2x(x+h))cos(2x+h2x(x+h)) 2|x| sin(h2x(x+h)) .At this point, we really need the fact that|sin | | |for all , and I don t know any proofof this without using the mean value theorem. This inequality also follows from the fact thatdifferentiable functions with non-negative derivatives are increasing, but this latter fact itselfis a consequence of the mean value theorem!(c)g(x) =1x2on [1, ).Solution:Ifx,y 1, then|g(x) g(y)|=|x y|(x+y)x2y2=(1xy2+1yx2)|x y|<2|x y|.So given >0, let = /2 in the definition of uniform continuity.(d)g(x) =1x2on (0,1]Solution:The function is not uniformly continuous. Consider the sequencesxn=1n, yn= |xn yn|= 1/2n <1 the other hand,|g(xn) g(yn)|=1y2n 1x2n= 3n2>3,ifn >1. This contradicts the definition of uniform continuity for = (a) Letf:E Rbe uniformly continuous.

5 If{xn}is a Cauchy sequence inE, show that{f(xn)}isalso a Cauchy :Let >0. Sincefis uniformly continuous, there exists >0 such that|x y|< = |f(x) f(y)|< .Since{xn}is Cauchy, there existsNsuch that for allm,n > N,|xn xm|< .Combining the two, ifn,m > N, then|f(xn) f(xm)|< .Since this works for all >0,{f(xn)}is Cauchy.(b) Show, by exhibiting an example, that the above statement is not true iffis merely assumed to :Letf(x) = sin(1/x). Clearlyf(x) is continuous on (0,1). But consider the sequencexn=2n .Sincexn 0, it is clearly Cauchy. Butf(xn) ={0, nis even( 1)n 12, nis odd,and hence the sequence{f(xn)}is not Cauchy.(c) Letf: (a,b) Rbe continuous. Show that there exists a continuous functionF: [a,b] RsuchthatF(x) =f(x) for allx (a,b) if and only iffis uniformly , howshould you defineF(a) andF(b)?Solution:Consider the sequencexn=a+ 1/n. For large enoughn,an (a,b). Since{an}is Cauchy, and sincefis uniformly continuous, by part(a),{f(an)}is Cauchy, and henceconverges.}

6 LetA= limn f(an).Similarly, considerbn=b 1/nand defineB= limn f(bn),and defineF(x) = A, x=af(x), x (a,b)B, x= an extension continuous on [a,b]. continuous on (a,b).To prove continuity ata, let{xn}be a sequence in(a,b) converging toa. We need to show thatF(xn) =f(xn) F(a) =A. Let >0. ThereexistsN1such that for alln > N1,|A f(an)|< proof will be complete if we can show that fornlarge enough|f(xn) f(an)|can be madesmaller than /2. This is where we use uniform continuity. By uniform continuity offin (a,b),there exists a >0 such that|x y|< = |f(x) f(y)|< , sincexn aandan=a+ 1/n, there existsN2such that for alln > N2,|xn an|< ,and hence for alln > N2,|f(xn) f(an)|< max(N1,N2),using triangle inequality, we see that ifn > N, then|f(xn) A| |f(xn) f(an)|+|f(an) A|< 2+ 2= .5. (a) Show directly from the definition of uniform continuity, that any uniformly continuous functionf: (a,b) Ris :There exists >0 such that for anyx,y (a,b)|x y|<2 = |f(x) f(y)|< (b+ 1)/2, that ispis the midpoint of (a,b).

7 The argument actually works for anyfixed point in the interval (a,b). Letmbe the first natural number such thatp+m b, andconsider the intervals,(a,p (m 1) ],[p (m 1) ,p (m 2) ], ,[p ,p],[p,p+ ], ,[p+ (m 1) ,b).Then anyxbelongs to at least one of the intervals. Moreover, for anyx,yin thesameinterval,|x y|<2 .By triangle inequality, ifx > pandx [p+ (j 1) ,j ], then|f(x) f(p)| |f(x) f(p+ (j 1) )|+|f(p+ (j 1) ) f(p+ (j 2) )|+ +|f(p+ ) f(p)| 1 + + 1 =j mWe can use a similar argument forx < p. Then by triangle inequality,|f(x)| |f(p)|+m,for allx (a,b), and hence the function is also follows directly from 4(c) above. Sincefis uniformly continuous, there is acontinuous extensionF: [a,b] R. Since [a,b] is closed and bounded, andFis continuous, byextremum value theorem,Fis bounded on [a,b]. But sinceF(x) =f(x) for allx (a,b) thisshows thatfis bounded on (a,b).(b) Iff:R Ris uniformly continuous, show that there existA,B Rsuch that|f(x)| A|x|+Bforallx apply the definition of uniform continuity with = 1.

8 For the corresponding >0, note that anyx Rcan be reached from 0 be a sequence of roughly|x|/ steps. Now applythe triangle inequality repeatedly to compare|f(x)|with|f(0)|.5 Solution:The solution is similar to the one above. By uniform continuity, there exists >0such that|y x|<2 = |g(y) g(x)|< any real numbersaandband non negativensuch that|b a|=n , we have|f(b) f(a)| loss of generality, we can assumea < band sob=a+n . for some positiveintegern. Ifn= 0, there is nothing to prove, so we can assumen >0. Then|f(b) f(a)| |f(b) f(b )|+|f(b ) f(b 2 )|+ +|f(b (n 1) ) f(a)|=n 1 k=0|f(b k ) f(b (k 1) )| see the inequality in the third line, apply the above consequence of uniform continuity tox=b k ,y=b (k 1) (so that|x y|= <2 ).Continuing with the problem, letxbe an real number. Then there is an integerm(positive ornegative) such thatm x <(m+ 1) . In particular, since|x m |=< ,|f(x) f(m )|< the other hand applying the claim toa= 0,b=m andn=|m|.

9 |f(m ) f(0)|<|m|.So by triangle inequality, we obtain|f(x) f(0)|<1 +|m|.On the other hand, sincem x <(m+ 1) , it is easy to see that|m|< 1|x|+ 1. Using thisand triangle inequality, we see that|f(x)| |f(x) f(0)|+|f(0)| 1 +|f(0)|+|m| 2 +|f(0)|+|x| A|f(x)|+B,withB= 2 +|f(0)|andA= Letf: [0,1] Rbe continuous withf(0) =f(1).(a) Show that there must existx,y [0,1] satisfying|x y|= 1/2 such thatf(x) =f(y).Solution:As in the hint, consider the functiong(x) =f(x+ 1/2) f(x) on [0,1/2]. Theng(0) =f(12) f(0)g(12)=f(1) f(12)= g(0).6sincef(0) =f(1). Eitherg(0) = 0 (and we takex0= 0), orgchanges sign between 0 and 1 the latter case, by intermediate value theorem, there is anx0 (0,1/2) such thatg(x) = either case, ify=x0+ 1/2, thenf(x) =f(y).(b) Show that for eachn N, there existxn,yn [0,1] such that|xn yn|= 1/nandf(xn) =f(yn).Solution:Now considerg(x) =f(x+ 1/n) f(x)on [0,n 1n]. Sincef(0) =f(1), it is easy to see thatg(0) +g(1n)+g(2n)+ +g(n 1n)= 0,and so all the terms cannot be of the same sign.

10 That is, either one ofg(k/n) = 0 (in which casewe letx0=k/n)) or there existsj < ksuch thatg(j/n) andg(k/n) are of opposite signs. Thenthe intermediate value theorem implies that there is anx0 (j/n,k/n) such thatg(x0) = either case, ify=x0+ 1/n, thenf(x) =f(y).(c) On the other hand, ifh [0,1/2] is not of the form 1/n, show that there does not necessarily existx,ysuch that|x y|=hwithf(x) =f(y). Give an example withh= 2 :(Due to Rahul) Consider the functionf(x) = cos(5 x) + (0) =f(1) = 1. One can check easily thatf(x+25) f(x) =45,and hence there is noxsuch thatf(x+ 2/5) =f(x).7. For each stated limit, and , find the largest possible -neighborhood that makes the definition of limitswork.(a) limx 4 x= 2, = :We need to find the set of allx, such that| x 2|<1, or equivalently, 1< x 2<1,orx (1,9). Taking = min(|4 1|,|9 14) = 3, we see that|x 4|<3 = | x 2|<1,and moreover, this is the largest possible .(b) limx bxc= 3, = :For anyx, eitherbxc, which would happen if and onlyx [3,4), or|bxc 3| we need|bxc 3|< , this is only possible ifx [3,4).]]