MATH211112Qu24-25week2sol
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Jan 6, 2025
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Solutions to MATH21111 Question Sheet 22024Handed out in Week 2 for week 3 Examples Class.Problem 1Prove that the functiond(m, n) =|m−1−n−1|for anym,n∈Ndefines a metric on the set of natural numbers. Does thismetric extend toR+?Solution 1Note thatdis well defined because 0/∈N.The firstaxiom of Definition 1.1 holds because|m−1−n−1|= 0 iff 1/m= 1/n,and thereforem=n; the second is automatic because|m−1−n−1|=|n−1−m−1|. For the third axiom,|m−1−n−1|=|(m−1−p−1)−(n−1−p−1)| ≤ |m−1−p−1|+|n−1−p−1|implies thatd(m, n)≤d(m, p) +d(p, n) for allm, n, p∈N.Every step in this analysis applies just as well to positive reals aspositive integers, so the metric does extend toR+.Problem 2.Consider the set of binary sequences{0,1}∞, i.e.a0a1a2a3. . .whereai∈ {0,1}for alli≥0.a.Forx, y∈ {0,1}∞defined∗(x, y) =∞Xi=0|xi−yi|2i.Prove thatd∗is a metric on{0,1}∞.b. What is the maximum possible value ofd∗(x, y), and is it at-tained for any pair of sequences? If so, which?c.Describe the elements of the open and closed ballsB3/4(0) andB3/4(0) in{0,1}∞, with thed∗metric, where 0 denotes thezero sequence, 00. . ..Solution 2.a. The axioms ford∗follow since|x|is a metric onR.So,Axiom 1,d∗(x, y) = 0 iff|xi−yi|= 0, i.e.xi=yifor alli≥0 iffx=y.1
Axiom 2, Since|xi−yi|=|yi−xi|for alli≥0 thend∗(x, y) =d∗(y, x).Axiom 3, The ‘normal’ triangle inequality on real numbers,|xi−zi| ≤ |xi−yi|+|yi−zi|leads tod∗(x, z)≤d∗(x, y) +d∗(y, z).b. The maximum possible value ofd∗(x, y) is∑∞j=01/2j= 2. It isattained by any pair of sequences which differ in every element; e.g.d∗(000. . . ,111. . .) = 2.c. By definition,B3/4(0) contains allxsuch that∑∞j=0xj/2j<3/4.That isx0+∞Xj=1xj2j<34.This can only hold withx0= 0, leaving(1)x12+∞Xj=2xj2j<34.Two cases: ifx1= 0 the remaining sum satisfies∞Xj=2xj2j≤∞Xj=212j=12<34,for any choices ofxj. So any sequence starting 00, i.e. 00wfor anybinary sequencew, is inB3/4(0).Second case, ifx1= 1 in (1) then we need(2)x24+∞Xj=3xj2j<14.Thus we must havex2= 0. The remaining sum satisfies(3)∞Xj=3xj2j≤∞Xj=312j=34.for any choices ofxj. To get a strict inequality we need at least onexj̸= 1.So any sequence 010w w̸= 1111. . .is inB3/4(0).HenceB3/4(0) contains sequences of the form 00wfor any se-quencewand 010wfor any sequencew̸= 111111. . . .2
To find the additional elements inB3/4(0) look for equalitiesabove. We have equality in (2) ifx2= 1 andxj= 0 for allj≥3.Hence 011000. . .is an additional element.We have an equality in (3) ifxj= 1 for allj≥3.Hence0101111. . .is the second additional element.Problem 3.a.Forx, y∈ {0,1}∞recall the definition ofdmin:dmin(x, y) = 0ifx=y, or 1/2iwherei≥0 is the smallest index such thatxi̸=yi. In the lectures we showed thatdminis a metric. Provethatdminandd∗of Problem 2 are Lipschitz equivalent.b.Deduce that for allx∈ {0,1}∞andr≥0,(4)B∗r(x)⊆Bminr(x)⊆B∗2r(x).c.For eachr≤1 find an element inB∗2r(0)KBminr(0),where 0denotes the zero sequence.d.Looking at the second set inclusion in (4) is it the case thatfor allx∈ {0,1}∞and 0< r <1 we haveBminr(x)⊆B∗2r(x)?Give your reasons.e.Describe the elements of the open and closed ballsBmin3/4(0) andBmin3/4(0) in{0,1}∞, with thedminmetric, where 0 denotesthe zero sequence.Solution 3.a.We need only considerx̸=y.Letjbe the first occurrence ofxj̸=yj, so|xj−yj|= 1.Thend∗(x, y) =∞Xi=0|xi−yi|2i=|xj−yj|2j+∞Xi=j+1|xi−yi|2i=12j+∞Xi=j+1|xi−yi|2i≥12j=dmin(x, y).3
And, for an upper bound∞Xi=j+1|xi−yi|2i≤∞Xi=j+112i=12j.In which cased∗(x, y)≤12j+12j= 2dmin(x, y).Hencedmin(x, y)≤d∗(x, y)≤2dmin(x, y).b.Ify∈B∗r(x) thend∗(y, x)< rsodmin(y, x)< rand thusy∈Bminr(x).HenceB∗r(x)⊆Bminr(x).Ify∈Bminr(x) thendmin(y, x)< rsod∗(y, x)<2rand thusy∈B∗2r(x).HenceBminr(x)⊆B∗2r(x).c.For each 0< r≤1 there existsi≥0 : 1/2i+1< r≤1/2i.Leta∈ {0,1}∞be all zeros except for a 1 ini-th position. Thendmin(a,0) = 1/2i≥rsoa /∈Bminr(0).Butd∗(a,0) = 1/2i=2/2i+1<2rsoa∈B∗2r(0).d.No. Considerx= 0 andr= 1/2.Letb= 011111. . .. Thendmin(b,0) = 1/2 sob∈Bmin1/2(0).Yetd∗(b,0) = 1 and sob /∈B∗1(0).e.Bmin3/4(0) =Bmin3/4(0) and both contain elements 0wwherewis any binary sequence. Note how this result differs from theresult ford∗in the previous question.4
Problem 4.Explain why neither of the following functionsddefinesa metric on the set of continuous functions [0,1]→R:1.d(f, g) =|f(1/2)−g(1/2)|2.d(f, g) = supx∈[0,1]|f′(x)−g′(x)|, wheref′(x) :=df(x)/dx.Solution 4.1. Letf(x) =xandg(x) = 1−x. Thend(f, g) =|1/2−(1−(1/2)|= 0; so the first axiom of Definition 1.1 fails.2. If one offorgfails to be differentiable for somex∈[0,1],thend(f, g) is not defined! And even if we restrict attention tothe subset of differentiable functions, the examplesf(x) = 1andg(x) = 0 show thatd(f, g) =|0−0|= 0; so the first axiomfails again.Problem 5.InC[0,1] = (Cont[0,1], dsup) define precisely the openballB1(f) wheref(x) =x2.Letg(x) =x.Do the setsB1/2(f) andB1/2(g) have non-emptyintersection?Solution 5.From the definitionB1(f) ={h∈Cont[0,1] :dsup(h, f)<1}={h∈Cont[0,1] : sup[0,1]|h(x)−f(x)|<1}={h∈Cont[0,1] :∀x∈[0,1], x2−1< h(x)< x2+ 1}.ForB1/2(f)∩B1/2(g)̸=∅we need to findh∈Cont[0,1] suchthat bothx2−1/2< h(x)< x2+ 1/2 andx−1/2< h(x)< x+ 1/2hold. Becausex2≤xin [0,1] this reduces down tox−1/2< h(x)<x2+ 1/2.5
That is the function lies in the shaded region:00.20.40.60.81-0.500.511.5xyFor illustration I have shownx2and 3x/4 both of which lie inthis region and thus the intersection of open balls.Problem 6.i.Is 12x∈B12(x3) inC[0,3]?ii.Isx3∈B5/2(x) inL1[0,2]?iii.Isx2∈B31/30(x) inL2[0,2]?Solution 6.i.No.dsup(12x, x3) = sup[0,3]|12x−x3|.And 12x−x3has a turning point at 12 = 3x2i.e.x=±2. Only2 lies in [0,3] and the value at 2 is 16.At 0 and 3 the value is 0and 9 respectively, sodsup(12x, x3) = 16.Since 16>12 we have12x /∈B12(x3)ii.No.d1(x3, x) =Z20|t3−t|dt.Note thatt3−tis≤0 for 0≤t≤1 and>0 for 1< t≤2. HenceZ20|t3−t|dt=Z10(t−t3)dt+Z21(t3−t)dt= 5/2.6
Sinced1(x3, x) = 5/2 we have thatx3is not in theopenballB5/2(x).iii.Yes.d2(x2, x) =Z20(t2−t)2dt1/2=Z20(t4−2t3+t2)dt1/2=16151/2≈1.0327... <1.0˙3 = 31/30.Problem 7.i.Prove that the metricsdsupandd1on Cont[a, b] satisfyd1(f, g)≤(b−a)dsup(f, g)for allf, g∈Cont[a, b].Deduce thatBsup1(f)⊆Bd1b−a(f) for allf∈Cont[a, b].ii.a. For an integerm≥1 sketch the functiongm: [a, b]→R,gm(x) =0ifa≤x≤c−1/m2m(x−c) + 2ifc−1/m≤x≤c−2m(x−c) + 2ifc≤x≤c+ 1/m0ifc+ 1/m≤x≤b.b. Givenf∈Cont[a, b] can we findr >0 such thatBd1r(f)⊆Bsup1(f)?That is, ifRba|g(t)−f(t)|dtis sufficiently small issup[a,b]|g(t)−f(t)|also small? Justify your answer.Solution 7.i.d1(f, g) =Zba|f(t)−g(t)|dt≤Zbasup[a,b]|f(x)−g(x)|dt= sup[a,b]|f(x)−g(x)|Zbadt=dsup(f, g)(b−a).Assumeg∈Bsup1(f).Thendsup(g, f)<1. So, by what has just beenproved,d1(g, f)≤dsup(g, f)(b−a)< b−1.Henceg∈Bd1b−a(f).Truefor allg∈Bsup1(f) meansBsup1(f)⊆Bd1b−a(f).7
ii. a. The graph is simply a triangle2ac−1/mc+ 1/mbb.Assume that givenf∈Cont[a, b] we can findr >0 such thatBd1r(f)⊆Bsup1(f). Choosem: 1/2m < r. Considerhm=f+gm∈Cont[a, b], withgmas above. Thend1(hm, f) =Zba|hm(t)−f(t)|dt=Zba|gm(t)|dt= 2/m < r,sohm∈Bd1r(f).Butdsup(hm, f) =dsup(gm,0) = 2̸<1 sohm/∈Bsup1(f).HenceBd1r(f)̸⊆Bsup1(f). This contradiction means no suchrexists.Problem 8.i.Prove thatd∗(f, g) = sup[a,b]|f(x)−g(x)|+Zba|f(t)−g(t)|dtis a metric on Cont[a, b], the set of continuous real-valued func-tions on [a, b].HintDo not verify the axioms, but instead look back at Ques-tion Sheet 1.ii.Prove thatd∗anddsupare Lipschitz equivalent on Cont[a, b].iii.Isx3∈B40(x) in (Cont[1,3], d∗)? Justify your answer.8
Solution 8.i.d∗=dsup+d1.In Problem 2 on Sheet 1 you are askedto show that the sum of two metrics is a metric.ii. By definition of a metricd1≥0 sod∗=dsup+d1≥dsup.From Problem 7i we haved1≤(b−a)dsupsod∗≤(1 +b−a)dsup.Hencedsup≤d∗≤(1 +b−a)dsup.iii.d∗(x3, x) = sup[1,3]|x3−x|+Z31|t3−t|dt.The functionx3−xis increasing on [1,3], so sup[1,3]|x3−x|=27−3 = 24.Z31|t3−t|dt=t44−t2231=814−92−14+12= 16.Henced∗(x3, x) = 40, thusx3̸∈B40(x).Problem 9.a.Let (X, d1) be the space of sequencesa= (ai)i≥0such that∑i≥0|ai|converges, equipped with thel1metric, sod1(a, b) =∑i≥0|ai−bi|.Find an example of a pointainXsuch thata∈B1/10(0), buta /∈B1/10(0).b.Recall thed∗metric on{0,1}∞, given byd∗(x, y) =∑∞j=0|xj−yj|/2jfor allx, y∈ {0,1}∞. Find a distance preserving injection ofthe space ({0,1}∞, d∗) into (X, d1).Solution 9.a. A pointa∈Xsatisfiesa∈B1/10(0) anda /∈B1/10(0)iff∑∞j=0|aj|= 1/10. Any example such as(1/10,0,0, . . .)or(1/20,1/40, . . . ,1/(2j×10), . . .)will do!9
b. Consider the functioni:{0,1}∞→Xdefined byi(x0x1. . . xj. . .) = (x0, x1/2, . . . , xj/2j, . . .) ;the right hand side is well defined, because∞Xj=0|xj/2j| ≤∞Xj=01/2j= 2,as required. Obviouslyiis injective; it also preserves distance, be-caused∗(x, y) =∞Xj=0|xj−yj|2j=∞Xj=0|xj/2j−yj/2j|=d1(i(x), i(y)),as sought.10