LecturesonOptimizationA.BanerjiJuly26,2015Chapter1Introduction1.1 SomeExamplesWebrieyintroduceour frameworkforoptimization,andthendiscusssomepreliminary concepts and results that well need to analyze specic problems.Ouroptimizationexamplescanall becouchedinthefollowinggeneralframework:Suppose Vis a vector space and S V . Suppose F: V 1. We wish tondx Ss.t. F(x) F(x), x S,orx Ss.t. F(x) F(x), x S.x, xarerespectivelycalledamaximumandaminimumofFonS.Indierent applications, V canbenite- or innite-dimensional. Thelatter need more sophisticated optimization tools such as optimal control; wewillkeepthatsortofstuinabeyancefornow. Inourapplications, Fwillbecontinuous,andprettymuchalsodierentiable;oftentwicecontinuouslydierentiable. Swillbespeciedmostoftenusingconstraints.Example1Let U: 1k+ 1beautilityfunction, p1, ..., pk, I bepositiveprices andwealth. MaximizeUs.t. xi 0, i =1, ..., k, and

ki=1pixi p.x I.Here,theobjectivefunctionisU,and1S= x 1k: xi 0i = 1, ..., k, and0 p.x I.Example2Expenditureminimization. Samesettingas above. Minimizep.xs.t. xi 0i=1, ..., kandU(x) U, whereUisanon-negativerealnumber.HeretheobjectivefunctionF: 1k1isF(x) = p.xandS= x 1k: xi 0i = 1, ..., k, andU(x) UExample3ProtMaximization. Givenpositiveoutputpricesp1, ..., psandinput prices w1, ..., wk, and a production function f: 1k+ 1s(transformingkinputsintosproducts),Maximize

sj=1pjfj(x)

ki=1wixi, s.t. xi 0, i =1, ..., k. fj(x)istheoutputofproductjasafunctionofavectorxofthekinputs.Here, theobjectivefunctionis prots : 1k+ 1denedby(x) =

sj=1pjfj(x)

ki=1wixi,andS= x 1k: xi 0, i = 1, ..., kExample4Intertemporal utility maximization. (Continuous time, nitehorizon):Aworker withaknownlife spanT, earning aconstant wage w, andreceiving interest at rate r on accumulated savings, or paying the same rate on2accumulated debts,wishes to decide optimal consumption path c(t), t [0, T].Letaccumulatedassets/debtsattimetbedenotedbyk(t). Hisinstantaneousutility from consumption is u(c(t)), u

> 0, u

< 0, and future consumption isdiscounted at rate . So the problem is to choose functions c(t), k(t), t [0, T]toMaximizeF(c) =_T0etu(c(t))dts.t.(i)c(t) 0, t [0, T](ii)k

(t) = w + rk(t) c(t)(iii)k(0)=k(T)=0(iii)assumesthat theindividual hasnoinheritanceandwishestoleavenobequest.Here, theobjectivefunctionF is afunctionof aninnitedimensionalvector, theconsumptionfunctionc. TheconstraintsetSadmitsonlythosefunctionsc(t)thatsatisfyconditions(i)to(iii).Example5Intertemporal utility or welfare maximization, discrete time, in-nitehorizon:Arepresentative,innitelylivedagentmustchoose(ct, kt+1)t=0toMaximize

t=0tu(ct)subjecttoct + kt+1= f(kt) + (1 )kt,ct 0, kt+1 0, k0= k0.Here,is the rate of depreciation of capital,and f(kt) is output in periodt.Example6Game inStrategic Form. G = N, (Si), (ui), where N=1, ..., n is the set of Players, and for each player i, Siis her set of strategies,andui: ni=1Si 1isherpayofunction.3Inagame, i

spayocandependonthechoices/strategies(s1, ..., sn)ofeveryone.ANashEquilibrium(s1, ..., sn) is astrategy prole suchthat for eachplayeri,sisolvesthefollowingmaximizationproblem:Maximizeui(s1, .., si1, si, si+1, .., sn)s.t.si Si.1.1.1 OptimizationProblemsinParametricFormParametersareheldconstantintheoptimizationexercise. Forinstance, inExample 1 (Utility Maximization), (p, I) 1k+1+is the parameter vector heldconstant. Thebudgetsetinfactdependsonthisparameter, andwemaywriteS(p, I) for thebudget set toshowthis dependence. Themaximumvaluethattheutilityfunctiontakesonthisset(ifthemaximumexists),i.e.V (p, I) = MaxU(x)[x S(p, I)thereforetypicallydependsontheparameter(p, I), andwedenotethisdependenceof themaximumbythevalue functionV (p, I). Inconsumertheory,wecallthistheindirectutilityfunction. Thisisafunctionbecauseto each point (p, I) in the admissible parameter space, V (p, I) assigns a singlevalue,equaltothemaximumofU(x)overallx S(p, I).Notethatforagiven(p, I),thesetofbundlesthatmaximizeutlitymaynot be unique: we candenote this relationshipbyx(p, I), the set of allbundles that maximize utility given (p, I). If the optimal bundle were uniquefor every(p, I), thenx(p, I) is afunction(the Walrasianor Marshalliandemandfunction),andthereforeV (p, I) = U(x(p, I)).In other problems, not just the feasible set Sbut also the objective func-tiondependsontheparameter. Forinstance, inExample2(ExpenditureMinimization), theparameter is (p, U), andtheobjectivefunctionp xde-pends on the parameter, via the price vector p, while the constraint U(x) UdependsonitviatheutilitylevelU.4Ingeneral, wemaywritedowntheoptimizationprobleminparametricform as follows: A parameter is part of some admissible set of parameters,whereisasubsetof somevectorspaceW(niteorinnite-dimensional;e.g. inexample1,(p, I) 1n++1+,thislattersetisthus. ThefeasiblesetisS(),whichdependsonandisasubsetofsome(other)vectorspaceV (e.g. inexample1, S(p, I) 1n+). TheobjectivefunctionFmapsfromsomesubsetofV Wtotherealline: wewriteF(x, ). TheproblemistomaximizeorminimizeF(x, )s.t. x S().Please read the other examples in Sundaram. We give one nal one here.Example7IdentifyingParetoOptimaThereare2individuals,withutilityfunctionsu1(x1), u2(x2)respectively,that map from 1n+, the n-good space, to 1. There is an endowment 1n+tobeallocatedbetweenthem. Anallocation(x1, x2)(vectorsofthegoodsgiventothetwoagents)isfeasibleifx1, x2 0andx1 + x2 . Wewill callF() = (x1, x2)[x1, x2 0, x1 + x2 thefeasiblesetorsetoffeasibleallocations.Anallocation(y1, y2)Paretodominates(x1, x2)ifui(yi) ui(xi)i = 1, 2, with >

forsomeiAnallocation(x1, x2)isParetooptimal if thereisnofeasibleallocationthatParetodominatesit.Let a (0, 1) andconsider thesocial welfarefunction U(x1, x2, a) au1(x1) + (1 a)u2(x2). Thenif(z1, z2)isanyallocationthatsolvesMaximizeU(x1, x2, a)s.t.(x1, x2) F()5it is aParetooptimal allocation. For, if (z1, z2) is inthis set of solu-tionsbutisnotParetooptimal,thenthereisafeasibleallocation(y1, y2)s.t.u1(y1) u1(z1), u2(y2) u2(z2), with >

forat least oneof these. Mul-tiplyingtherst inequalitybya, thesecondby(1 a) andadding, wegetU(y1, y2, a) > U(z1, z2, a),contradictingthat(z1, z2)isamaximizer.If we assume that the utility functions ui(xi) are concave,then the converseholds: everyParetooptimal allocationisasolutiontoMaximizeU(x1, x2, a)s.t.(x1, x2) F()forsomechoiceofa [0, 1]).1.2 Some Concepts and Results fromRealAnalysisWe will now discuss some concepts that we will need, such as the compactnessof thesetSabove, andthecontinuityanddierentiabilityof theobjectivefunctionF. Wewill workinnormedlinearspaces. Intheabsenceof anyotherspecication, thespacewewill beinis 1nwiththeEuclideannorm[[x[[ =(

ni=1x2i)1/2. (Theres abunchof other norms that wouldworkequallywell. Recallthatanormin 1nisdenedtobeafunctionassigningtoeachvectorxanon-negativerealnumber [[x[[,s.t. (i)forallx, [[x[[ 0with=

ix=0(0

beingthezerovector); (ii)Ifc 1, [[cx[[= [c[[[x[[.(iii) [[x + y[[ [[x[[ + [[y[[. Thelastrequirement, thetriangleinequality,followsfortheEuclideannormfromtheCauchy-Schwarzinequality).Oneexampleintheprevioussectionusedanothernormedlinearspace,namelythespaceof boundedcontinuous functions denedonanintervalof real numbers, withthesupnorm. But infurther workinthis part of6thecourse, wewill sticktousingnitedimensional spaces. Someof theconceptsbelowapplytobothniteandinnitedimensional spaces, sowewill sometimes call the underlying space V . But mostly, it will help to thinkofV assimply 1n,andtovisualizestuin 12.We will measure distance between vectors using [[xy[[ =_ni=1 (xiyi)2_1/2.ThisisourintuitivenotionofdistanceusingPythagorastheorem. Further-more,itsatisesthethreepropertiesofametric,viz.,(i) [[x y[[ 0,with=ix = y;(ii) [[x y[[ = [[y x[[;(iii) [[x z[[ [[x y[[ +[[y z[[.Notethatproperty(iii)forthemetricfollowsfromthatforthetriangleinequality for the norm, since [[xz[[ = [[(xy)+(yz)[[ [[xy[[+[[yz[[.OpenandClosedSetsLet> 0 and x V . The openball centered at x with radiusis denedasB(x, ) = y: [[x y[[ < We see that if V = 1, B(x, ) is the openinterval (x , x+). IfV= 12, it is an open disk centered at x. The boundary of the disk is tracedbyPythagorastheorem.Exercise1Showthat [[x y[[denedbymax[x1y1[, . . . , [xnyn[,forall x, y 1nis a metric (i.e. satises the three requirements of a metric). Inthespace 12,sketchB(0, 1),theopenballcenteredat0,theorigin,ofradius1,inthismetric.LetS V . xisaninteriorpointofSifB(x, ) S,forsome > 0. SisanopensetifallpointsofSareinteriorpoints. Ontheotherhand,SisaclosedsetiScisanopenset.Example. Openin 1vs. openin 12.Thereisanalternative,equivalent,convenientwaytodeneclosedsets.x is an adherentpoint of S, or adheres to S, if every B(x, ) contains a point7belongingtoS. Notethat this does not necessarilymeanthat xis inS.(However,ifx SthenxadherestoSofcourse).Example. Singletonandnitesets;countablesetsneednotbeopen.Lemma1AsetSisclosediitcontainsall itsadherentpoints.ProofSupposeSisclosed, soScisopen. LetxadheretoS. Wanttoshowthat x S. Suppose not. Then x Sc, and since Scis open, x is an interiorpointofSc. Sothereissome > 0s.t. B(x, ) Sc;thisdoesnothaveanypointsfromS. SoxcannotbeanadherentpointofS. Contradiction.Conversely, suppose Scontains all its adherent points. ToshowSisclosed, weshowScisopen. Weshowthatall thepointsinScareinteriorpoints. Letx Sc. SincexdoesnotadheretoS, itmustbethecasethatforsome > 0,B(x, ) Sc. .Moreexamplesofclosed(andopen)sets.Nowwewill relateclosednesstoconvergenceof sequences. Recall thatformally, asequenceinV isafunctionx:N V . Butinsteadofwritingx(1), x(2), ...astheimagesormembersof thesequence, wewriteeitherx1, x2, ...or x1, x2, ....Denition1Convergence:Asequence(xk)k=1ofpointsinV convergestoxifforevery > 0thereexistsapositiveintegerNs.t. k Nimplies [[xkx[[ < .NotethatthisisthesameassayingthatforeveryopenballB(x, ),wecan nd Ns.t. for all points xkfollowing xN, xklies in B(x, ). This impliesthatwhenxkconvergestox(notation: xkx), allbutanitenumberofpointsin(xk)liearbitrarilyclosetox.Examples. xk= 1/k, k = 1, 2, ... is a sequence of real numbers convergingto zero. xk= (1/k, 1/k), k = 1, 2, ... is a sequence of vectors in 12convergingtotheorigin. Moregenerally, asequenceconvergesin 1nifandonlyifallthecoordinatesequencesconverge,ascanbevisualizedintheexamplehereusinghypotenusesandlegsoftriangles.8Theorem2(xk) xin 1nifor everyi 1, . . . , n, the coordinatesequence(xki) xi.Proof. Since(xki xi)2n

j=1(xkj xj)2,takingsquarerootsimplies [xki xi[ [[xkx[[,soforeveryk Ns.t.[[xkx[[ < , [xki xi[ < .Conversely, if all thecoordinatesequences convergetothecoordinatesof thepoint x, thenthereexistsapositiveinteger Ns.t. k Nimplies[xki xi[0, thereisnonumbera Ss.t. a>sup S . So, everya Smust thensatisfya sup S . But then,sup S isanupperboundofSthatislessthansup S. Thisimpliesthatsup SisnotinfactthesupremumofS. Contradiction. .Lemma4If aset Sof real numbersisboundedaboveandclosed, thenithasamaximum.Proof. Since it is bounded above, it has a supremum, sup S. sup Sis anadherentpointofS(bytheabovelemma). Sisclosedsoitcontainsallitsadherentpoints,includingsup S. Hencesup SisthemaxofS. .CorrespondingtothenotionofsupremumorleastupperboundofasetSofrealnumbers,isthenotionofinmumorgreatestlowerboundofS.AnumberlisalowerboundofSifl a, a S. TheinmumofSisanumber s s.t. s is a lower boundof S,and s l,for alllower bounds lof S.WecalltheinmumofS,inf S.Let Sbethesetofnumbersoftheform a,foralla S.Fact. sup S= inf(S).So,supandinfareintimatelyrelated.By the completeness property of real numbers, if S 1 is bounded below,(i.e., thereexistsms.t. m 0. By the above discussion,thereexistssomexN (a , a]. Andsince(xn)isanincreasingsequence, wehavethatforallk N,xk (a , a]. So(xn) a.Asimilarconclusionholdsfordecreasingboundedsequences. And:Theorem6Everysequenceofreal numbershasamonotonesubsequence.Proof. For the bounded sequence (xk), let An= xk[k n, n = 1, 2, ....If anyoneof thesesetsAndoesnothaveamaximum, wecanpull outanincreasingsequence. Forinstance, supposeA1doesnothaveamax. Thenlet xk1= x1. Let xk2be the rst member of the sequence (xk) that is greaterthanx1,andsoon.On the other hand, if all An have maxes, then we can pull out a decreasingsubsequence. Letxk1= maxA1,xk2= maxAk1 + 1, xk3= maxAk2 + 1andsoon.Itfollowsfromtheabovetwotheorems,thatwehaveTheorem7Bolzano-WeierstrassTheorem.Everyboundedsequenceofreal numbershasaconvergentsubsequence.Finally,asanapplicationtotheideasofmonotonesequences,wehaveTheorem8CantorsNestedIntervalstheorem.If [a1, b1] [a2, b2] . . . is anestedsequenceof closedintervals, thenm=1[am, bm]isnonempty. Moreover,ifbmam 0,thenthisintersectionisasinglepoint.12Proof. Becauseof thenesting, a1 a2 . . . b2 b1. So, (ak)isboundedandincreasingandsohasasupremum,saya;(bk)isboundedanddecreasingandhasaninmum, sayb; anda b. So, [a, b] [am, bm], m=1, 2, . . ., and therefore lies in the intersection m=1[am, bm]; which is thereforenonempty. Moreover, ifbm am 0, thenbysandwiching, a=bandtheintersectionisasinglepoint.CompactSets.Suppose(xn) is asequenceinV . (Notethechangeinnotation, fromsuperscript tosubscript. This is just bythe way; most places have thissubscriptnotation, butRangarajanSundaramattimeshasthesuperscriptnotationinordertoleavesubscriptstodenoteco-ordinatesofavector).Let m(k) be anincreasingfunctionfromthe natural numbers tothenatural numbers. So, l>nimpliesm(l)>m(n). Asubsequence(xm(k))of(xn)isaninnitesequencewhosekthmemberisthem(k)thmemberoftheoriginalsequence.Giveanexample. Theideaisthattogetasubsequencefrom(xn), youstrike out some members, keeping the remaining members positions thesame.Fact. If a sequence (xn) converges to x, then all its subsequences convergetox.Proof. Takeanarbitrary>0. So, thereexistsNs.t. n Nimplies[[xnx[[ < . This implies, for any subsequence (xm(k)), that k Nimplies[[xm(k)x[[ < . .However, if a sequence does not converge anywhere, it can still have (lotsof) subsequences that converge. For example, let (xn) ((1)n), n = 1, 2, ....Then, (xn)doesnotconverge; butthesubsequences(ym)= 1, 1, 1, ....and(zm) =1, 1, 1, ... bothconverge, todierent limits. (Suchpoints arecalledlimitpointsofthemothersequence(xn)).Compactsetshaveapropertyrelatedtothisfact.Denition2Aset S V iscompactif everysequence(xn)inShasasubsequencethatconvergestoapointinS.13Theorem9SupposeS 1n. ThenSiscompactifandonlyifitisclosedandbounded.Proof(Sketch).SupposeSis closedandbounded. Wecanshowits compact usingapigeonhole-likeargument; letssketchithere. SinceSisbounded, wecancoveritinaclosedrectangleR0=I1. . .In, whereIi, i =1, ..., nareclosedintervals. Takeasequence(xn)inS. Dividetherectangleintwo:I11 . . . InandI21 ... In,whereI11 I21= I1istheunionof2intervals.Then,theresaninnityofmembersof(xn)inatleastoneofthesesmallerrectangles,callthisR1. DivideR1into2smallerrectangles,saybydividingI2into 2 smaller intervals; well nd an innity of members of (xn) in at leastone of these rectangles, call it R2. This process goes on ad innitum, and wend an innity of members of (xn) in the rectangles R0 R1 R2 .... BytheCantorIntersectionTheorem, i=0Riisasinglepoint;callthispointx.Now we can choose points yi Ri, i = 1, 2, ... s.t. each yiis some memberof (xn); becausetheRis collapsetox, it is easytoshowthat (ym) is asubsequence that converges to x. Moreover, the yis lie in S, and Sis closed;sox S.Conversely,supposeSiscompact.(i)Thenitisbounded. Forsupposenot. Thenwecanconstructase-quence(xn)inSs.t. foreveryn=1, 2, ..., [[xn[[>n. Butthen, nosubse-quenceof(xn)canconvergetoapointinS. Indeed, takeanypointx Sandanysubsequence(xm(n))of(xn). Then[[xm(n)[[ = [[xm(n)x + x[[ [[xm(n)x[[ +[[x[[(Theinequalityaboveisduetothetriangleinequality).So,[[xm(n)x[[ [[xm(n)[[ [[x[[ n [[x[[andtheRHSbecomeslargerwithn. So(xm(n))doesnotconvergetox.14(ii)Sisalsoclosed. Takeanysequence(xn)inSthatconvergestox.Then, all subsequencesof(xn)convergetox, andsinceSiscompact, (xn)hasasubsequenceconvergingtoapointinS. So, thispointof limitisx,andx S. So,Sisclosed. .ContinuityofFunctionsDenition3AfunctionF : 1n 1mis continuous at x 1n, if foreverysequence(xk)that convergestoxin 1n, theimagesequence(f(xk))convergestof(x)in 1m.Exampleofpointdiscontinuity.Exampleofcontinuousfunctionondiscretespace.FiscontinuousonS 1n,ifitiscontinuousateverypointx S.Examples. Thereal-valuedfunctionF(x)=xiscontinuoususingthisdenition, almost trivially, since (xk) and x are identical to (F(xk)) and F(x)respectively.F(x)=x2iscontinuous. Wewanttoshowthatif (xk)convergestox,then(F(xk))=x2kconvergestoF(x)=x2. Thisfollowsfromtheexerciseaboveonlimits: xk x,xk ximpliesxkxk x.x = x2.Byextension,polynomialsarecontinuousfunctions.MaytalkalittleaboutthecoordinatefunctionsofF: 1n1m:(F1(x1, ..., xn), ..., Fm(x1, ..., xn)).Example: F(x1, x2)=(x1 + x2, x21 + x22). Thisiscontinuousbecause(i)F1andF2arecontinuous; e.g. letxkx. Thenthecoordinatesxk1 x1andxk2 x2. SoF1(xk) = xk1 + xk2 x1 + x2= F1(x).(ii)SincethecoordinatesequencesF1(xk) F1(x)andF2(xk) F2(x),F(xk) (F1(xk), F2(xk)) F(x) = (F1(x), F2(x)).Thereisanequivalent,(, )denitionofcontinuity.Denition4A function F: 1n1mis continuous at x 1n,if for every>0, thereexists>0s.t. if foranyy 1nwehave [[x y[[ 0 s.t. for every > 0, there exists a ywith [[x y[[ < and [[F(x) F(y)[[ . Then for this particular , we canchooseasequenceofk=1/kandxkwith [[x xk[[0s.t. for everypositive integer N, there exists k Nfor which [[F(xk) F(x)[[ . Then,for this specic , there does not exist any > 0 s.t. for all y with [[xy[[ < wehave [[F(x) F(y)[[; forwecanndforanysuch, oneofthexkss.t.[[xkx[[,so [[F(xk) F(x)[[ .Here is an immediate upshot of the latter denition. Suppose F: 1 1iscontinuousatx. IfF(x) > 0,thenthereisanopeninterval(x , x + )s.t. ifyisinthisinterval, thenF(y)>0. Theideaisthatwecantakean = F(x)/2,say,andusethe(, )denition. AsimilarstatementwillholdifF(x) < 0.Weusethisfactnowinthefollowingresult.Theorem11IntermediateValueTheoremSupposeF: 1 1 iscontinuouson aninterval [a, b]andF(a)andF(b)areofoppositesigns. Thenthereexistsc (a, b)s.t. F(c) = 0.Proof. SupposeWLOGthatF(a) > 0, F(b) < 0(i.e. fortheothercasejustconsiderthefunction F). ThenthesetS= x [a, b][F(x) 016is bounded above. Indeed, b is an upper bound of Ssince F(b) is not 0.By the completeness property of real numbers, Shas a supremum, sup S= c,say.It cant be that F(c) > 0, for then by continuity, there is an h S, h > c,s.t. F(h) > 0 so c is not an upper bound of S. It cant be that F(c) < 0. For,ifcisanupperboundofSwithF(c) < 0,thenwehaveforeveryx [a, b]withF(x) 0,x c. However,bycontinuity,thereisaninterval(c , c]s.t. everyyinthisintervalsatisesF(y) 0.B(z, )isintersectedwithSsinceitmaynot,byitself,lieentirelyinS.However, if zisintheinteriorof S, wecandiscardthat. zissaidtobeaninteriorlocalmaximum,ofminimum,offonSiff(z) ()f(x), x B(z, ),forsome > 0.We now give a necessary condition for a point to be an interior local maxormin; namely, thatitsderivativeshouldbezero. Forif not, thenwecanincreaseor decreasethefunctionvaluebymovingawayslightlyfromthepoint.FirstOrderNecessaryCondition.Theorem14Let f: 1n 1, S 1n, andlet xbealocal maxorlocalminof f onS, lyingintheinteriorof S. If f isdierentiableat x, thenDf(x) = .23Here, = (0, ..., 0) is the origin, and Df(x) = (f(x)/x1, . . . , f(x)/xn).Proof. Letxbeaninteriorlocalmax(minproofisdonealongsimilarlines).Step1: Supposen=1. Takeanysequence(yk), ykx, zk x. Sincexisalocal max, wehave, fork KandKlargeenough,f(zk) f(x)zkx 0 f(yk) f(x)ykxTakinglimits preserves theseinequalities since(, 0] and[0, ) areclosedsetsandtheratiosequenceslieintheseclosedsets. So,f

(x) 0 f

(x)sof

(x) = 0.Step2. Supposen>1. Takeanyjthaxisdirection, andletg: 1 1bedenedbyg(t) = f(x +tej). Notethatg(0) = f(x). Now,sincexisalocal max of f, f(x) f(x+tej), for t smaller than some cuto value: i.e.,g(0) g(t)fortsmallerthanthiscutovalue, i.e., g(0)isalocal interiormaximum. (Sincet 0arebothallowed). gis dierentiableat 0sinceg(0) =f((0)) =f(x), andf is dierentiableat xandisdierentiableatt=0. (Here, (t)=x + tej, soD(t)=ej, t). So, gisdierentiableat0,g

(0) = 0,andbytheChainRule,g

(0) = Df((0))D(0) = Df(x)ej=f(x)xj.Notethatthisisnecessarybutnotsucientforalocalmaxormin,e.g.f(x) =x3has avanishingrst derivativeat x=0, whichis not alocaloptimum.SecondOrderConditions24Denition. xisastrictlocalmaximumoffonSiff(x) > f(y),forally B(x, ) S, y ,= x,forsome > 0.We will represent the Hessian or second derivative (matrix) of fby D2f.Theorem15Supposef: 1n 1isC2onS 1n, andxisaninteriorpointofS.1. (necessary)If f hasalocal max(resp. local min)atx, thenD2f(x)isn.s.d. (resp. p.s.d.).2. (sucient)IfDf(x) = andD2f(x)isn.d. (resp. p.d.) atx,thenxisastrictlocal max(resp. min)offonS.TheresultsintheabovetheoremfollowfromtakingaTaylorseriesap-proximationoforder2aroundthelocalmaxorlocalmin. Forexample,f(x) = f(x) + Df(x)(x x) +12(x x)TD2f(x)(x x) + R2(x x)whereR2()isaremainderofordersmallerthantwo. Ifxisaninteriorlocal max or min, then Df(x) = 0 (a vector of zeros), so the quadratic forminthesecond-ordertermwillsharethesignof(f(x) f(x)).Examples to illustrate: (i) SONC are not sucient. f(x) = x3. (ii) Semi-denitenesscannotbereplacedbydeniteness. f(x)=x4. (iii). Theseareconditionsforlocal,notglobaloptima. f(x)=2x3 3x2. (iv)Strategyforusingtheconditionstoidentifyglobal optima. f(x)=4x3 5x2+ 2xonS= [0, 1].25Chapter4OptimizationwithEqualityConstraints4.1 IntroductionWearegivenanobjectivefunctionf : RnRtomaximizeorminimize,subject tok constraints. That is, there are k functions, g1: RnR,g2: RnR,... ,gk: RnR,andwewishtoMaximizef(x)overallx Rnsuchthatg1(x) = 0, . . . , gk(x) = 0.More compactly, collect the constraint functions (looking at them as com-ponent functions) into one function g: RnRk, where g(x) = (g1(x), . . . , gk(x)).ThenwhatwewantistoMaximizef(x)overallx Rnsuchthatg(x)1k= 1k.The Theorem of Lagrange provides necessary conditions for a local opti-mumx. Bylocalwemeanthatthevaluef(x)isamaxormincomparedtoothervaluesf(x)forall xcontainedinsomeopensetUcontainingxsuchthatxsatisesthekconstraints. ThustheproblemitconsidersistoprovidenecessaryconditionsforaMaxoraMinoff(x)overallx S,whereS= Ux Rn[g(x) = ,forsomeopensetU.26Thefollowingexampleillustratestheprincipleofnoarbitrageunderly-ingamaximum. Amoregeneral illustration, withmorethan1constraint,requiresalittlebitof themachineryof linearinequalities, whichwell notcover. TheideahereisthattheLagrangemultipliercaptureshowthecon-straintisdistributedacrossthevariables.Example1. SupposexsolvesMaxU(x1, x2)s.t. I p1x1 p2x2=0andsupposex>> .Thenreallocatingasmallamountofincomefromonegoodtotheotherdoes not increase utility. Say income dI> 0 is shifted from good 1 to good 2.Sodx1=(dI/p1)>0anddx2= (dI/p1) 0shiftedfromgood1togood2doesnotincreaseutility,sothat[(U1/p1) + (U2/p2)]dI 0,or(U1/p1) + (U2/p2) 0 (2)Eq. (1)and(2)imply(U1(x)/p1) = (U2(x)/p2) = (3)That is, the marginal utility of the last bit of income equals (U1(x)/p1) =(U2(x)/p2attheoptimum. Also,(3)impliesU1(x) = p1,U2(x) = p2.Along with p1x1+p2x2= I, these are the FONC of the Lagrangean functionMaxL(x, ) = U(x1, x2) + [I p1x1p2x2]Moregenerally, supposeF:RnRandG:RnR, andsupposexsolvesMaxF(x)s.t. c G(x) = 0. Thispartisskippable.ContemplateachangedxinxthatrespectstheconstraintG(x)=c.Thatis,dG = G1dx1 + G2dx2= 0. Therefore,G1dx1= G2dx2=dc, say. Sodx1=(dc/G1), dx2= (dc/G2). Ifdc > 0, then our change dx implies dx1> 0, dx2< 0. Fdoes not increase atthemaximum,x. So27dF= F1dx1 + f2dx2 0,or[(F1/G1) (F2/G2)]dc 0. Similarly, 0canbeshownsimilarly.Therefore,(F1(x)/G1(x)) = (F2(x)/G2(x)) = (4)Caveat: We have assumed that G1(x) and G2(x) are not both zero atx. Thisiscalledtheconstraintqualication.Again,notethat(4)canbegotastheFONCoftheproblemMaxL(x, ) = F(x) + [c G(x)].On.Letsgobacktotheutilityexample. Attheoptimum(x, ), supposeyouincrease income byI. Buyingmore x1implies utilityincreases by(U1(x)/p1)I,approximately.Buyingmorex2impliesutilityincreasesby(U2(x)/p2)IAttheoptimum,(U1(x)/p1) = (U2(x)/p2) = .Soineithercase, utilityincreasesbyI. Sogivesthechangeinthe objective (here, the objective is utility), at an optimum, that results fromrelaxingtheconstraintalittlebit.The interpretation is the same in the more general case: If G(x) = c, andcisincreasedbyc,supposex1aloneisthenincreased.SoG=g1dx1=c, or dx1=(c/G1). Soat x, F increases bydF= F1dx1= (F1(x)/G1(x))c = c.If instead x2 is changed, Fincreases by dF= F2dx2, = (F2(x)/G2(x))c =c.4.2 TheTheoremofLagrangeThesetupisthefollowing. f: RnRistheobjectivefunction,gi: RnR, i = 1, . . . , k, k < naretheconstraintfunctions.Letg: RnRkbethefunctiongivenbyg(x) = (g1(x), . . . , gk(x)).Df(x) = (f(x)/x1, . . . , f(x)/xn).Dg(x) =___(g1(x)/x1) . . . (g1(x)/xn).........(gk(x)/x1) . . . (gk(x)/xn)___=___Dg1(x)...Dgk(x)___28SoDg(x)isak nmatrix.The theorembelowprovides anecessaryconditionfor alocal maxorlocal min. Note that xis a local max (resp. min) of fon the constraint setx Rn[gi(x) = 0, i = 1, . . . , k if f(x) f(x) (resp. f(x)) for all x UforsomeopensetUcontainingx,s.t. gi(x) = 0, i = 1, . . . , k. ThusxisaMaxonthesetS= U x Rn[gi(x) = 0, i = 1, . . . , k.Theorem16(Theoremof Lagrange). Let f : RnRandgi: RnR, i=1, . . . , k, k0, andfor (x, y) =(, ), 0,r = 0, 1, . . . , k 1.(2b)xTD2L(x, )x > 0,for all nonzero x N(Dg(x)), i (1)kdet(BH(L; n+k r)) > 0,r = 0, 1, . . . , k 1.34Note. (1)Forthenegativedeniteorsemidenitenesssubjecttocon-straints cases, thedeterminant of borderedHessianwithlast r rows andcolumnsdeletedmustbeofthesamesignas(1)nr. Thesignof(1)nrswitcheswitheachsuccessiveincreaseinrfromr=0tor=k 1. Sothecorresponding borderedHessiansswitchsigns. Inthe usualtextbook case of2variablesandoneconstraint, k=1, k 1=0, sowejustneedtocheckthesignforr=0, thatis, thesignofthedeterminantofthebigborderedHessian. Youshouldbeclearaboutwhatthissignshouldbeifitistobeasucient condition for a strict local max or min. For the necessary condition,weneedtochecksigns or 0, foronepermutedmatrixaswell, inthiscase. Whatisthispermutedmatrix?(2) As in the unconstrained case, the suciency conditions do not requirecheckingweakinequalitiesforpermutedmatrices.(3)Inthep.s.d. andp.d. cases, thesignsof theprincipal minorsmustbeallpositive, ifthenumberkofconstraintsiseven, andallnegative, ifkisodd.(4) If we know that a global max or min exists, where the CQ is satised,andwegetauniquesolutionx RnthatsolvestheFONC, thenwemayuse a second order condition to check whether it is a max or a min. However,weakinequalitiesdemonstratingn.s.d. orp.s.d. (subjecttoconstraints)ofD2(L)donotimplyamaxormin; thesearenecessaryconditions. Strictinequalitiesareuseful; theyimply(strict)maxormin. Ifhowever,aglobalmaxorminexists, theCQissatisedeverywhere, andthereismorethanonesolutionoftheFONC, thentheonegivingthehighestvalueoff(x)isthemax. Inthiscase, wedontneedsecondorderconditionstoconcludethatitistheglobalmax.VII.4. TwoExamplesExample1.AconsumerwithincomeI>0facespricesp1>0, p2>0, andwishestomaximizeU(x1, x2)=x1x2. Sotheproblemis: Maxx1x2s.t. x1 0,x2 0,andp1x1 + p2x2 I.To be able to use the Theorem of Lagrange, we need equality constraints.35Now, it is easytosee that if (x1, x2) solves the above problem, then(i)(x1, x2) > (0, 0). If xi= 0 for some i, then utility equals zero; clearly, we candobetterbyallocatingsomeincometothepurchaseofeachgood; and(ii)thebudgetconstraintbindsat(x1, x2). Forifp1x1 + p2x2< I,thenwecanallocatesomeof theremainingincometobothgoods, andincreaseutilityfurther.Weconcludefromthisthatasolution(x1, x2)willalsobeasolutiontotheproblemMaxx1x2s.t. x1> 0,x2> 0,andp1x1 + p2x2= I.That is, Maximize U(x1, x2) = x1x2 over the set S= R2++ (x1, x2)[I p1x1 p2x2=0. Sincethebudgetsetinthisproblemiscompactandtheutilityfunctioniscontinuous, Uattainsamaximumonthebudgetset(byWeierstrass Theorem). Moreover, we argued above that at such a maximumx,xi> 0, i = 1, 2andthebudgetconstraintbinds. So,x S.Furthermore,Dg(x) = (p1, p2),soRank(Dg(x)) = 1,atallpointsinthebudgetset. SotheCQismet. Therefore,theglobalmaxwillbeamongthecriticalpointsofL(x1, x2, ) = x1x2 + (I p1x1p2x2).FONC:(L/x1) = x2p1= 0 (1)(L/x2) = x1p2= 0 (2)(L/) = I p1x1p2x2= 0 (3) ,= 0, (otherwise (1) and (2) imply that x1= x2= 0, which violates (3)).Therefore, from(1)and(2), =(x2/p1)=(x1/p2), orp1x1=p2x2. So(3)impliesI 2p1x1= 0,orp1x1= (I/2),whichisthestandardCobb-Douglasutility result that the budget share of a good is proportional to the exponentw.r.t. itintheutilityfunction. Sowegetxi= (I/2pi), i = 1, 2,and= (I/2p1p2).Wearguedthat theglobal maxwouldbeoneof thecritical points ofL(x, )inthisexample; (note, however, thattheglobal min(whichoccursat(x1, x2) = (0, 0)isnotacriticalpoint). Sincewehaveonlyitonecriticalpoint, it follows that this must be the global max! (We know that x1= x2= 0istheglobalmin,andnotthepointthatwehavelocated).Takeatime-out toconsider thealternativetocheckingtheconstraint36qualication. WecanstartbysettingupaLagrangeanL(x, ) = U(x) + (I px)andcrankingouttheFOCs. ThefactthattheCQholdsshowsupas ,=0intheFOC. (Convinceyourselfthatinanyproblem, theCQfailingatapointmanifestsas = 0intheFOC).TheFOCsarenow:(L/x1) = x2p1= 0 (1)(L/x2) = x1p2= 0 (2)(L/) = I p1x1p2x2= 0 (3)If =0, then(1)and(2)imply=0. All multipliersequal tozeroisnotapermittedsolutionintheTheoremofLagrange. So ,=0; wecannormalize (divide through by ) and set it equal to 1, and proceed as before.IftherewerelotsofsolutionstotheFOCs,thesucientSOCsmayhelpin narrowing down the set of points one needs to evaluate for the global max.LetscheckSOCsfortheaboveexample,althoughthisisnotnecessary.Dg(x) = (p1, p2)D2L(x, ) = D2U(x) + D2g(x)=_U11(x) U12(x)U21(x) U22(x)_+ _g11(x) g12(x)g21(x) g22(x)_=_0 11 0_+ _0 00 0_=_0 11 0_NowevaluatethequadraticformzTD2L(x, )z=2z1z2atany(z1, z2)that is orthogonal to Dg(x) =(p1, p2). So, p1z1 p2z2=0 orz1=(p2/p1)z2. Forsuch(z1, z2), zTD2L(x, )z=(2p2/p1)z22 0. Thisisthesignof(1)n= (1)2. Therefore,thereisastrictlocalmaxatx.Example2.Findglobal maximaandminimaoff(x, y)=x2 y2ontheunitcirclein 12,i.e.,ontheset (x, y) 12[g(x, y) 1 x2y2= 0.Constrainedmaximaandminimaexist, byWeirerstrass theorem, as fiscontinuousandtheunitcircleisclosedandbounded. Bounded, asitisentirelycontainedin, say, B(0, 2). Closedaswell, visually, wecanseethattheconstraintsetcontainsall itsadherentpoints. Moreformally, suppose(xk, yk)k=1isasequenceofpointsontheunitcircleconvergingtothelimit(x, y). Sinceg is continuous, and(xk, yk) (x, y), wehaveg(xk, yk) g(x, y). Since g(xk, yk) = 0, k, their limit is 0, i.e. g(x, y) = 0 or (x, y) is ontheunitcircle,andsotheunitcircleisclosed.ConstraintQualication: Dg(x, y)=(2x 2y). Therankofthisrowmatrixiszeroonlyat (x, y) =(0, 0). But theorigindoesnot satisfytheconstraint. Everywhereontheconstraint,atleastoneofxoryisnotzero,andtherankofDg(x, y)is1.So, the maxandminwill be solutions tothe FOCs of the usual La-grangean.L(x, y, ) = x2y2+ (1 x2y2)FOC.2x 2x = 0 (1)2y 2y= 0 (2)x2+ y2= 1 (3)(1) and(2) imply 2x(1 ) = 0 and2y(1 +) = 0respectively. Suppose ,= 1or1. Then(x, y) = (0, 0),violating(3). If = 1,y= 0,sox2= 1,andsoon. Sothefoursolutions(x, y, )totheFOCsformthesolutionset(1, 0, 1), (1, 0, 1), (0, 1, 1), (0, 1, 1). Evaluating the function values at38these points, we have that fhas a constrained max at (1, 0) and (1, 0) andconstrainedminat(0, 1)and(0, 1).Althoughunnecessary, letspracticesecond-orderconditionsforthisex-ample. Df(x, y) = (2x2y),Dg(x, y) = (2x2y).D2f(x, y) =_2 00 2_D2g(x, y) =_2 00 2_With= 1,forinstance,D2L(x, y, )evaluatestoD2f(x, y) + D2g(x, y) =_0 00 4_.(x, y) orthogonal to Dg(x, y) = (2 0) (at (x, y) = (1, 0) implies that(x, y)mustsatisfy(2, 0)(x, y) = 0or 2x + 0y= 0. So,x = 0,andyisfreetotakeanyvalue. Soconsiderthequadraticformwith(x, y) = (0, y).( 0 y )_0 00 4__0y_=4y2< 0forall(x, y) ,= (0, 0). Sonegativedenitenessholds,andweareatastrictlocalmax.SomeDerivatives(1). Let J: 1n 1nbedenedby J(x) =x, x 1n. Incomponentfunctionnotation, wehave J(x) =(J1(x), . . . , Jn(x)) =(x1, . . . , xn). So,DJi(x) = ei,i.e. thevectorwith1intheithplaceandzeroselsewhere. So,DJ(x) = Inn,theidentitymatrix.By similar work, we can show that if f(x) = Ax, where A is an mn ma-trix,thenDf(x) = A. Indeed,thejthcomponentfunctionfj(x) = aj1x1 +. . . + ajnxn, soitsmatrixofpartialderivativeswithrespecttox1, . . . , xnisDfj(x) = (aj1. . . ajn).(2). Letf: 1n 1mandg: 1n 1m. Bywayof convention, considerf(x)andg(x)tobecolumnvectors, andconsiderthefunctionh: 1n 139denedbyh(x) = f(x)Tg(x). Then,Dh(x) = g(x)TDf(x) + f(x)TDg(x)Indeed,D_f(x)Tg(x)_= D_m

i=1fi(x)gi(x)_=

iD(fi(x)gi(x)) =

i[gi(x)Dfi(x) + fi(x)Dgi(x)]ThethirdequalityaboveisbecausethedierentialoperatorDislinear.Notethat

igi(x)Dfi(x) = (g1(x), . . . , gn(x))_____Df1(x)Df2(x)

Dfn(x)_____= g(x)TDf(x),andsoon.Wetakeastepbackandderivethisinamoreexpandedfashion. Sinceh(x) =

mi=1fi(x)gi(x),itspartialderivativewithrespecttoxjis:h(x)/xj=m

i=1[gi(x) (fi(x)/xj) + fi(x) (gi(x)/xj)]= g(x)TDf(x)[, j] + f(x)TDg(x)[, j]whereforanymatrixA, wewriteA[, j] torepresentitsjthcolumn. SoDh(x) = (h(x)/x1. . . h(x)/xn)equals_g(x)TDf(x)[, 1] + f(x)TDg(x)[, 1], . . . , g(x)TDf(x)[, n] + f(x)TDg(x)[, n]_= g(x)TDf(x) + f(x)TDg(x)40.As an application, let h(x) = xTx. Then Dh(x) equals xTD(x)+xTDx =xTI + xTI= 2xT.OntheChainRuleWesawanexample(intheproof of the1storderconditioninuncon-strainedoptimization)of theChainRuleatwork; youveseenthisbefore.Namely, if h: 1 1nandf : 1n 1aredierentiableattherelevantpoints,thenthecompositiong(t) = f(h(t))isdierentiableattandg

(t) = Df(h(t))Dh(t) =n

j=1f(h(t))xjh

j(t)Youmayhave encounteredthis before innotationf(h1(t), . . . , hn(t)),withsomeuseof total dierentiationorsomething. Similarly, supposeh:1p1nand f: 1n1mare dierentiable at the relevant points, then thecompositiong(x) = f(h(x)), g: 1p1misdierentiableatx,andDg(x) = Df(h(x))Dh(x).Here,ontheRHSanmnmatrixmultipliesann pmatrix,toresultinthempmatrixontheLHS.Theintuitionfor theChainRuleis perhaps this. Let z =h(x). If xchangesbydx,therst-orderchangeinzisdz= Dh(x)dx. Therst-orderchangeinf(z)isthenDf(z)dz. Substitutingfordz, therst-orderchangeinf(h(x))equals[Df(h(x))Dh(x)] dx.Inthe formula, things are actuallyquite similar tothe familiar case.The(i, j)thelementof thematrixDg(x)isgi(x)/xj, wheregiistheithcomponent function of gand xjis the jthvariable. Since this is equal to thedot product of the ithrow of Df(h(x)) and the jthcolumn of Dh(x), we have41gi(x)/xj=n

k=1fi(h(x))hkhk(x)xjOntheImplicitFunctionTheoremTheorem19SupposeF: 1n+m 1misC1, andsupposeF(x, y)=0forsomey 1mandsomex 1n. Supposealsothat DFy(x, y) hasrankm. ThenthereareopensetsUcontainingxandV containingyandaC1functionf: U V s.t.F(x, f(x)) = 0 x U.Moreover,Df(x) = [DFy(x, y)]1DFx(x, y)NotethatwecouldalternativelylookattheequationF(x, y) =c, forsomegivenc 1m, withoutchanginganything. Theproofofthistheoremstarts goingdeep, sowill not bepart of this course. Theproof for then = m = 1 case, however, is provided at the end of this Chapter. But notice,thatapplyingtheChainRuletodierentiateF(x, f(x)) = 0yieldsDFx(x, y) + DFy(x, y)Df(x) = 0(*)42whencetheexpressionforDf(x).More tediously in terms of compositions, if h(x) = (x, f(x)), then Dh(x) =_IDf(x)_,whereasDF(.)=(DFx(.)[DFy(.)),somatrixmultiplicationusingparti-tionsyieldsEq.(*).Interpretations(1)IndierenceCurve.By way of motivation,think of a utility function Fdened over 2 goods,evaluatedatsomeutilityvaluec. SoF(x, y) = c. Let(x, y)beasolutionof thisequation, i.e. F(x, y)=c. Undertheassumptionsof theImplicitFunctionTheoremonF, thereexistsafunctionfs.t. if xisapointclosetox, thenF(x, f(x))=c. Thatis, aswevaryxclosetox, thereexistsauniqueys.t. F(x, y)=c. Becauseof theuniqueness, wehavey=f(x)i.e. afunctional relationship. DrawanindierencecurvecorrespondingtoF(x, y) =c tosee this visually. Moreover, the theoremasserts that thederivativeoftheimplicitfunctionf

(x) = Fx/FywhereFx, Fyarethepartial derivativesofF, evaluatedat(x, y). Themarginal rate of substitution between the two goods (LHS) equals the ratio ofthe marginal utilities (RHS). In fact, when we say under some assumptionsonF,oneoftheassumptionsisthatFyevaluatedat(x, y)isnotzero.Themnemonicforgettingthederivative: FromF(x, y)=c,wetotallydierentiate to get Fxdx+Fydy= 0, and rearrange to get dy/dx = Fx/Fy.(2). ComparativeStatics.Wethenmovetothevectorcasebyanalogy. SupposeF(x, y) = cwherexisann-vector, yanm-vector, cagivenm-vector. Let(x, y)solve F(x, y) =c. Thinkof xbeingexogenous, sothis is aset of m43equationsinthemendogenous variablesy=(y1, ..., ym). Youcanstacktheseequationsvertically;forlaziness,IwritethemnowasF1(x1, ..., xn, y1, ..., ym) = c1, ..., Fm(x1, ..., xn, y1, ..., ym) = cmSothevectorfunctionFhasmcomponentfunctionsF1, ..., Fm.NowItotallydierentiate:DFxdx + DFydy= 0asbefore; onlynow, dx=(dx1, ..., dxn), dy=(dy1, ..., dym), DFxisthemnmatrixofpartialderivativeswhoseithrowis(Fi/x1, ..., Fi/xn),andDFyisthemmmatrixwhoseithrowis(Fi/y1, ..., Fi/ym).So DFydy= DFxdx. From here, we can work out the eect of changinganyxjontheendogenous variablesy1, ..., ym. Supposewesetall dxi=0fori ,=j. Then DFxdxbecomes dxjtimesthejthcolumnofDFx. Wedividebothsidesbydxj,gettingDFy(y1/xj, ...., ym/xj)T= (F1/xj, ...., Fm/xj)T(thesuperscriptTisfortransposetogetcolumnvectors).So,(y1/xj, ...., ym/xj)T= DF1y(F1/xj, ...., Fm/xj)TWhereinthescalarcasewedividedbythenumberFy,herewemultiplybytheinverseofthematrixDFy. Now, ifwestackhorizontallythepartialderivativesofy1, ..., ymw.r.t. x1, ..., xn,ontheLHS,wehaveDf(x);andonthe RHS, the appropriate columns give Fx, so we have (DFy(x, y))1DFx,whichisthejustimplicitfunctionderivativeformula.(3)Application: CournotDuopoly.Firms 1and2haveconstant unit costs c1andc2, andfacethetwicecontinuously dierentiable inverse demand function P(Q), where Q = q1+q2isindustryoutput. Soprotsaregivenby441= P(q1 + q2)q1c1q1and2= P(q1 + q2)q2c2q2If prots are concave in own output, then the rst-order conditions belowcharacterizeCournot-Nashequilibrium(q1, q2).1/q1= P

(q1 + q2)q1 + P(q1 + q2) c1= 02/q2= P

(q1 + q2)q2 + P(q1 + q2) c2= 0The concavityof prot w.r.t. ownoutput conditions followfromtheconditionbelow: Forallq1, q221/q21= P

(q1 + q2)Q + 2P

(q1 + q2) 0, i = 1, 2Thetworst-orderconditionscanbewrittenasthevectorequationF(q1, q2, c1, c2) = 0.Wewanttoknow: HowdotheCournotoutputschangeasaresultofachangeinunitcosts? If c1decreases, forinstance, doesq1increaseandq2decrease? TheimplicitfunctiontheoremsaysthatifDFq1,q2(q1, q2, c1, c2)isof full rank(rank=2), then, locallyaroundthissolution, q=(q1, q2)isanimplicitfunctionofc = (c1, c2),withF(f(c), c) = 0. And45Df(c) = [DFq(q, c)]1DFc(q, c)NotethatDFq(.) =_F1/q1F1/q2F2/q1F2/q2_For brevity, let P

and P

be the derivative and second derivative of P(.)evaluatedattheequilibrium. ThenDFq(.) =_P

q1 + 2P

P

q1 + P

P

q2 + P

P

q2 + 2P

_Thedeterminantofthismatrixworksouttobe(P

)2+P

(P

(q1 +q2) + 2P

) > 0sinceP

< 0andtheconcavityinownoutput condition is assumed to be met. So the implicit function theorem canbeappliedNoticealsothatDFc(.) =_1 00 1_.ThuswecanworkoutDf(c), thechangesinequilibriumoutputsasaresultofchangesinunitcosts. Itwouldbeagoodexerciseforyoutoworktheseout,andsignthese.ProofoftheTheoremofLagrangeBeforetheformalproof,notethatwellusethetangencyofthecontoursetsoftheobjectiveandtheconstraintapproach,whichinotherwordsusestheimplicitfunctiontheorem. Forexample, considermaximizingF(x1, x2)s.t. G(x1, x2) =0. If G1 ,=0(this is theconstraint qualicationinthiscase), we have at a tangency point of contour sets, G1f

(x2) +G2= 0 (wherex1=f(x2) is the implicit functionthat keeps the points (x1, x2) ontheconstraint);sof

(x2) = G2/G1.On the other hand, if we vary x2and adjust x1to stay on the constraint,thefunctionvalueF(x1, x2)=F(f(x2), x2)doesnotincrease; thereforelo-callyaroundtheoptimum,F1f

(x2) + F2= 0. Substituting,F1(G2/G1) +F2= 0. Ifwenowput46F1/G1= ,wehavebothF1+ G1=0bydenition, andG2+ F2=0, thetwoFONC.TheProof:Without loss of generality, let the leading principal k kminor matrix ofDg(x)belinearlyindependent. Wewritex = (w, z)withwbeingtherstkcoordinatesofxandzbeingthelast(n k)coordinates. Soshowingtheexistenceof(a1 nvector)thatsolvesDf(x) + Dg(x) = isthesameasshowingthatthe2equationsbelowholdforthis; theequationsareofdimension1 kand1 (n k)respectively:Dfw(w, z) + Dgw(w, z) = (*)Dfz(w, z) + Dgz(w, z) = (*)SinceDgw(w, z)issquareandoffullrank,Eq.(*)yields= Dfw(w, z) [Dgw(w, z)]1(**)Weshowsolves(*)aswell. Thisneedstwosteps.First,g(h(z), z) = forsomeimplicitfunctionh,so47Dh(z) = [Dgw(w, z)]1Dgz(w, z)Second, deneF(z)=f(h(z), z). Sincetheresaconstrainedoptimumat(h(z), z), varyingzwhilekeepingw=h(z)will notincreasethevalueofF(z). SoDF(z) = Dfw(w, z)Dh(z) + Dfz(w, z) = SubstitutingforDh(z),Dfw(.)[Dgw(.)]1Dgz(.) + Dfz(.) = Thatis,Dgz(.) + Dfz(.) = SimpleImplicitFunctionTheoremwithProofTheorem20SupposeF: 121,F(a, b) = 0,FisC1,andDF2(a, b) ,= 0.Thenthereexistsanopeninterval Ucontaininga, andanopeninterval Vcontaining b, and a C1function f: U Vs.t. F(x, f(x)) = 0 for all x V .Proof. Well avoidthe proof of continuous dierentiability. SupposeWLOGDF2(a, b)0s.t.DF2(a, y) < 0forally (b h, b + h). So,F(a, b h) > 0 > F(a, b + h).Now, sinceFiscontinuous, thereexistsaninterval I containinga, s.t.F(x, b h)>0x I; andaninterval I

containingas.t. F(x, b + h) 0 > F(x, b + h)Therefore, by the Intermediate Value Theorem, there exists a y s.t. F(x, y) =0. ThisyisuniquebecauseDF2(.) < 0inthisinterval. So,wecanpulloutauniquefunctionf(x)s.t. F(x, f(x)) = 0x U.Digressionabout(discussionof ?) EnvelopeTheoremsSupposewehaveanobjectivefunctionf: 1n+1 1,thatisafunctionof thevector variable x 1n, andalsoaparameter a 1that is heldconstant whenmaximizingf onsomefeasibleset S 1n. Supposethatforeveryadmissiblevalueofa, thereisauniqueinteriormaximizer, sowecansaythatx(a)isthefunctionthatrepresentsthisrelationshipbetweenparameterandmaximizer. Supposefissmoothandx(a)isdierentiable.LetV (a)bethevaluefunctionforthisproblem, thatgivesthemaximumvaluethat f(x, a) obtains whentheparameter is at thelevel a. That is,V (a) f(x(a), a).WewishtoknowhowV (a)changeswithachangeina. Asachanges,x(a)changesaswell, butthischangehasnorst-ordereectonV (a): sotherst-orderchangeinV (a)issolelythroughthedirect eect of aonf.thisistheimplicationoftheenvelopetheorem.Indeed, usingtheChainRuleonV (a) f(x(a), a), wehaveV

(a)=Dfx(x, a)Dx(a)+f(x, a)/a. But because x is an interior Max, Dfx(x, a) =1n. So,V

(a) = f/a.Now suppose we want to maximize an objective function f(x), which doesnotdependona, butsubjecttoaconstraintg(x, a)=a G(x)=0thatdoesdependona. Underniceconditions,attheMax,Df(x) + Dg(x, a) = (i)Alsonotethatifachanges,thevalueofg(x(a), a)mustcontinuetobezero,so49Dgx(x(a), a)Dx(a) + g/a = 0 (ii)Now, V (a) f(x(a)), soV

(a)=Df(x)Dx(a). Using(i)tosubsti-tutefor Df(x), this equals Dg(x, a)Dx(a), whichequals, using(ii),g/a = . Sohere,V

(a) = ,thevalueofthemultiplierattheoptimumistherateofchangeoftheobjectivewithrespecttotheparameterabeingrelaxed.Suppose nowthat we have anobjective functionf(x, a) tomaximizesubject to g(x, a) = 0. Along similar lines, we can show that V

(a) = f/a+g/a, i.e. the direct eect of a on the Lagrangian function. As an exercise,pleasederiveRoysIdentityusingtheindirectutilityfunctionV (p, I).50Chapter5OptimizationwithInequalityConstraints5.1 IntroductionThe problem is to nd the Maximum or the Minimum of f: RnRon theset x Rn[gi(x) 0, i = 1, . . . , k, where gi: RnR are the kconstraintfunctions. Attheoptimum, theconstraintsarenowallowedtobebinding(or tight or eective),i.e. gi(x) = 0,as before,or slack (or non-binding),i.e.gi(x) > 0.Example: MaxU(x1, x2)s.t. x1 0,x2 0,I p1x1 p2x2 0. Ifwedo not know whether xi= 0, for some i, at the utility maximum, or whetherxi>0, thenclearlywecannot usetheTheoremof Lagrange. Similarly,if thereisablisspoint, thenwedonotknowinadvancewhetherthereatthebudgetconstrainedoptimumthebudgetconstraintisbindingorslack.Again,wecannotthenusetheTheoremofLagrange,tousewhichweneedtobeassuredthattheconstraintisbinding.Note the general nature of a constraint of the form gi(x) 0. If we have aconstrainth(x) 0,thisisequivalenttog(x) h(x) 0. Andsomethinglikeh(x) cisequivalenttog(x) c h(x) 0.51WeuseKuhn-TuckerTheorytoaddressoptimizationproblemswithin-equalityconstraints. Themainresult is arst order necessaryconditionthatissomewhatdierentforthatof theTheoremof Lagrange; onemaindierence is that the rst order conditions gi(x) = 0, i = 1, . . . , k in the The-oremofLagrangearereplacedbytheconditionsigi(x) = 0, i = 1, . . . , kinKuhn-Tuckertheory.Inordertomotivatethisdierence,letusdiscussasimplesetting. Con-sider anobjective functionf :12 1. We want tomaximize f(x) orf(x1, x2)overallx 12thatsatisfyG(x) a,whereG : 12 1. Wewillalternativelywriteg(x)=a G(x) 0. Forthisexample, letusassumethat G(x) is strictly increasing. We can view a as the total resource available;suchasthetotalincomeavailableforspendingongoods. Drawapicture.A maximum xcan occur either in the interior (i.e. G(x) < a or g(x) >0), or at theboundary( G(x) =aor g(x) =0). If it happens intheinterior, it implies Df(x) =. If it happens onthe boundary, it mustbe that reducing the parameter value a does not increase f(x);for whatevervector x you choose as maximizer after the reduction of a was available before,at the higher value of a, and was not chosen as the maximizer. Consider thensettinguptheLagrangianL(x, ) = f(x) + g(x)andconsidertherstorderconditionDf(x) + Dg(x) = .Whenwouldthisrst-orderconditionmakesense?(i)First,forthistocoincidewithDf(x) = whenxisintheinterior,wemusthavethatg(x) > 0(orG(x) < a)implies = 0.(ii)NowletV (a)bethevaluefunctionforthisproblem: themaximumvalueoff(x)whentheparameterintheconstraintequalsa. Considerthe52interpretation that = V

(a),the change in the value of the objective as wechangea. Ifthemaximizerxisontheboundary(G(x) = a)andweweretoreducea, themaximumvalueof f wouldgetreduced(ornotincrease);so, 0.(iii)Finally,supposethatthemaximizerxalongwithsolvetherst-orderconditionDf(x) + Dg(x)=, andsupposeDg(x) ,=. Suppose > 0. Then from the rst-order condition, we conclude Df() ,= . But thatmeansxisnotintheinteriorof thefeasibleset; itsontheboundary, sog(x) = 0. Alternatively, interpreting > 0 as the decrease in the maximizedvalueoff(x)ifaisdecreasedalittle, itmustmeang(x)=0(orG(x)=a). Forif xwerenotontheboundary, decreasingawouldnotaectthismaximumvalue.So in addition to Df(x) +Dg(x) = , its implied that 0, g(x) 0,andg(x) = 0. Notethatthisimpliesthatifg(x) > 0,then = 0,andif > 0,theng(x) = 0.5.2 Kuhn-TuckerTheoryRecall that theproblemis tondtheMaximumor theMinimumof f :RnRontheset x Rn[gi(x) 0, i = 1, . . . , k,wheregi: RnRarethekconstraintfunctions. Themaintheoremdealswithlocal maximaandminima,though.Supposethatlofthekconstraintsbindattheoptimumx. Denotethecorrespondingconstraintfunctionsas(gi)i, whereisthesetofindexesof the bindingconstraints. Let g: RnRlbe the functionwhose lcomponentsaretheconstraintfunctionsofthebindingconstraints. Thatisg(x) = (gi(x))i.Dg(x) =___Dgi(x)...Dgm(x)___, wherei, . . . , maretheindexes of thebindingconstraints. SoDg(x)isanl nmatrix.53WenowstateFONCfortheproblem. TheTheorembelowisaconsoli-dationoftheFritz-JohnandtheKuhn-TuckerTheorems.Theorem21(TheKuhn-Tucker (KT) Theorem). Let f : RnR, andgi:RnR, i=1, . . . , kbeC1functions. SupposexisaMaximumof fonthesetS=U x Rn[gi(x) 0, i=1, . . . , k, forsomeopensetU Rn. Thenthereexistreal numbers, 1, . . . , k,notall zerosuchthatDf(x) +

ki=1iDgi(x) = 1n.Moreover,ifgi(x) > 0forsomei,theni= 0.If,inaddition,RankDg(x)=l,thenwemaytaketobeequalto1.Furthermore,i 0, i = 1, . . . , k,andi> 0forsomeiimpliesgi(x) = 0.Supposetheconstraint qualication, RankDg(x) =l, is met at theoptimum. Then the KT equations are the following (n +k) equations in then + kvariablesx1, . . . , xn, 1, . . . , k:igi(x) =0, i =1, . . . , k, i0, gi(x) 0withcomplementaryslackness. (1)Df(x) +

ki=1iDgi(x) = (2)If xis a localminimum of fon S, then f(x) attains a local maximumvalue. Thusforminimization, whileEq.(1)staysthesame, Eq.(2)changestoDf(x) +

ki=1iDgi(x) = (2)Equation(1)and(2)areknownastheKuhn-Tuckerconditions.Notenallythatsincetheconditionsof theKuhn-TuckerTheoremarenotsucientconditionsforlocal optima; theremaybepointsthatsatisfyEquations(1)and(2)or(2)withoutbeinglocaloptima. Forexample,youmaycheckthatfortheproblemMaxf(x)=x3s.t. g(x)=x 0, thevaluesx==0satisfytheKTFONC(1)and(2)foralocalmaximumbutdonotyieldamaximum.545.3 UsingtheKuhn-TuckerTheoremWe want to maximize f(x) over the set x Rn[g(x) 1k, where g(x) =(g1(x), . . . , gk(x)).SetupL(x, ) = f(x) +

ki=1igi(x)(Ifwewanttominimizef(x),setupL(x, ) = f(x) +

ki=1igi(x))ToensurethattheKTFONCwill holdattheglobal max, verifythat(1)aglobal maxexistsand(2)Theconstraintqualicationismetatthemaximum.Thissecondisnotpossibletodoifwedontknowwherethemaximumis. WhatwedoinsteadistocheckwhethertheCQholdseverywhereinthedomain, andif not, wenotepointswhereitfails. TheCQinthetheoremdependsonwhatconstraintsarebindingatthemaximum. Againsincewedontknowthemaximum,wedontknowwhatconstraintsbindatit.Withkconstraint functions, thereare2kproles of bindingandnon-bindingconstraintspossible, eachoftheseprolesimplyingadierentCQ.Weeithercheckallofthem, orweruleoutsomeprolesusingcleverargu-ments.If bothchecks arene, thenwendall solutions (x

,

) totheset ofequations:i(L(x, )/i) = 0,i 0,(L(x, )/i) 0,i = 1, . . . , k,withCS.(L(x, )/xj) = 0, j= 1, . . . , n.Fromthesetofallsolutions(x

,

),pickthatsolution(x, )forwhichf(x) is maximum. Note that this method does not require checking for con-cavity of objective functions and constraints, and doesnot require checkinganysecondordercondition.ThemethodmayfailifaglobalmaxdoesnotexistoriftheCQfailsatthemaximum. TheexampleMaxf(x) = x3s.t. g(x) = x 0isonewherenoglobalmaxexists,andwesawearlierthatthemethodfails.Anexample inwhichCQfails: Maxf(x) =2x33x2, s.t. g(x) =(3 x)3 0.55Suppose the constraint does not bind at the maximum; then we dont havetocheckaCQ. Butsupposeitdoes. Thatis, supposetheoptimumoccursatx = 3. Dg(x) = 3(3 x)2= 0atx = 3. TheCQfailshere. Youcouldcheck that the KT FONC will not isolate the maximum.In fact, in this babyexample, it is easy to see that x = 3 is the max, as (3x)3 0 iff(3x) 0,so we may work with the latter constraint function, with which CQ does notfail. It is a good exercise to visualize f(x) and see that x = 3 is the maximum,ratherthanmerelycrankingoutthealgebranow.Alternatively, we may use the more general FONCs stated in the theorem.Df(x) + Dg(x) = 0,with, notbothzero.(6x26x) + (3(3 x)2) = 0,and (1)(3 x)3 0,withstrictinequalityimplying = 0. (2)If (3x)3> 0, then = 0, which from Eq.(1) implies either = 0, whichviolatestheFONC,orx = 1. Atx = 1,f(x) = 1.Ontheotherhand, if (3 x)3=0, thatisx=3, thenEq(1)implies=0, soitmustbethat>0. Atx=3, f(x)=27. sox=3isthemaximum.TwoSimpleUtilityMaximizationProblemsExample1. Thisisareal babyexamplemeantpurelyforillustration.No one expects you to use the heavy Kuhn-Tucker machinery for such simpleproblems. Inthisexample, oneexpectsinsteadthatyouwouldusereason-ingaboutthemarginalutilityperrupeeratios(U1/p1), (U2/p2)tosolvetheproblem.MaxU(x1, x2)=x1 + x2, overtheset x=(x1, x2) R2[x1 0, x2 0, I p1x1p2x2 0,whereI> 0,p1> 0andp2> 0aregiven.Sothereare3inequalityconstraints:g1(x1, x2) = x1 0,g2(x1, x2) = x2 0,andg3(x1, x2) = I p1x1p2x2 0At the maximum x, any combination of these three could bind;so thereare8possibilities. However, sinceUisstrictlyincreasing, thebudgetcon-straint binds at the maximum (g3(x) = 0). Moreover, g1(x) = g2(x) = 0 is56notpossiblesinceconsuming0ofbothgoodsgivesutilityequalto0,whichisclearlynotamaximum.Sowehavetocheckjustthreepossibilitiesoutoftheeight.Case(1)g1(x) > 0, g2(x) > 0, g3(x) = 0Case(2)g1(x) = 0, g2(x) > 0, g3(x) = 0Case(3)g1(x) > 0, g2(x) = 0, g3(x) = 0BeforeusingtheKTconditions, weverifythat(i)aglobal maxexists(here, becausetheutilityfunctioniscontinuousandthebudgetsetiscom-pact), andthat(ii)theCQholdsatall 3relevantcominationsof bindingconstraintsdescribedabove.Indeed,forCase(1),Dg(x) = Dg3(x) = (p1, p2),soRank[Dg3(x)] =1,soCQholds.ForCase(2), Dg(x)=_Dg1(x)Dg3(x)_=_1 0p1p2_, soRank[Dg(x)]=2.ForCase(3), Dg(x)=_Dg2(x)Dg3(x)_=_0 1p1p2_, soRank[Dg(x)]=2.Thusforthemaximum,x,thereexistsasuchthat(x, )willbeasolution to the KT FONCs. Of course,there could be other (x, )

s that aresolutions as well, but a simple comparison of U(x) for all candidate solutionswillisolateforustheMaximum.L(x, ) = x1 + x2 + 1x1 + 2x2 + 3(I p1x1p2x2)TheKTconditionsare1(L/1) = 1x1= 0,1 0, x1 0,withCS (1)2(L/2) = 2x2= 0,2 0, x2 0,withCS (2)3(L/3) = 3(I p1x1p2x2) = 0, 3 0, I p1x1p2x2 0, withCS (3)(L/x1) = 1 + 13p1= 0 (4)(L/x2) = 1 + 23p2= 0 (5)Sincewedontknowwhichofthethreecasesselecttheconstraintsthatbindatthemaximum,wemusttryallthree.Case(1). Sincex1>0, x2>0,(1)and(2)imply1=2=0.Pluggingthese in Eq(4) and (5), we have 1 = 3p1= 3p2. This implies 3> 0. (Also57note that this is consistent with the fact that since utility is strictly increasing,relaxingthebudgetconstraintwill increaseutility. Sothemarginal utilityofincome,3> 0. Thus3p1= 3p2impliesp1= p2.Soifatalocalmaxbothx1andx2arestrictlypositive,thenitmustbethattheirpricesareequal. All (x1, x2)thatsolveEq(3)aresolutions. Theutilityinanysuchcaseequalsx1+ (I p1x1/p2)=I/p, wherep=p1=p2. Notethatinthiscase,(U1/p1) = (U2/p2) = 1/p.Case2. x1=0implies, fromEq(3), that x2=(I/p2). Sincethis isgreaterthan0,Eq(2)implies2= 0. HencefromEq(5),3p2= 1.Since1 0,Eq(4)and(5)imply3p1= 1 +1 1 = 3p2. Moreover,since3> 0,thisimpliesp1 p2.That is, if it is the case that at the maximum, x1= 0, x2> 0, then it mustbe that p1 p2. Note that in this case, (U2/p2) = (1/p2) (U1/p1) = (1/p1).For completeness sake, Eq(5) implies 3=(1/p2). SofromEq(4),1= (p1/p2) 1. SotheuniquecriticalpointofL(x, )is(x, ) = (x1, x2, 1, 2, 3) = (0, (I/p2), (p1/p2) 1, 0, (1/p2)).Case(3). This case is similar, and we get that x2= 0, x1> 0 occurs onlyifp1 p2. Wehave(x, ) = ((I/p1), 0, 0, (p2/p1) 1, 1/p1).We see that which of the cases applies depends upon the price ratio p1/p2.If p1= p2, then all three cases are relevant, and all (x1, x2) R2+ such that thebudget constraint binds are utility maxima. But if p1> p2, then only Case(2)isapplies,becauseifCase(1)hadapplied,wewouldhavehadp1= p2,andif Case(3)hadapplied, thatwouldhaveimpliedp1 p2. Thesolutiontothe KT conditions in that case is the utility maximum. Similarly,if p1< p2,onlyCase(3)applies.Example2. MaxU(x1, x2) =(x1/1 + x1) + x2/1 + x2), s.t. x1 0,x2 0,p1x1 + p2x2 I.Check that the indierence curves are downward sloping, convex and thatthey cut the axes (show all this). This last is due to the additive form of the58utilityfunction,andmayresultin0consumptionofoneofthegoodsattheutilitymaximum.ExactlyasinExample1, weareassuredthataglobal maxexists, thattheCQismetattheoptimum, andthatthereareonly3relevantcasesofbindingconstraintstocheck.theKuhn-Tuckerconditionsare:1(L/1) = 1x1= 0,1 0, x1 0,withCS (1)2(L/2) = 2x2= 0,2 0, x2 0,withCS (2)3(L/3) = 3(I p1x1p2x2) = 0, 3 0, I p1x1p2x2 0, withCS (3)(L/x1) = (1/(1 + x1)2) + 13p1= 0 (4)(L/x2) = (1/(1 + x2)2) + 23p2= 0 (5)Case(1). x1> 0,x2> 0implies1= 2= 0. Eq(4)implies3> 0,sothatEq(4)and(5)give((1 +x2)/(1 + x1)) = (p1/p2)1/2.UsingEq(3),whichgivesx2= ((I p1x1)/p2),above,weget((p2 + I p1x1)/(p2(1 + x1)) = (p1/p2)1/2,sosimplecomputationsyieldx1= ((I + p2(p1p2)1/2)/(p1 + (p1p2)1/2)),x2= ((I + p1(p1p2)1/2)/(p2 + (p1p2)1/2)),3= (1/p1(1 + x1)2).x1> 0, x2> 0impliesI> (p1p2)1/2 p1,I> (p1p2)1/2 p2. Ifeitherofthesefails,thenwearenotintheregimeofCase1.Case(2)x1=0withEq(3)impliesx2=I/p2. Sincethisispositive,2= 0,soEq(5)implies3= 1/(1 + (I/p2))2p2.1= 3p11(fromx1= 0andEq(4)).1= p1p2/(p2 + I)21. Forthistobe 0,itisrequiredthatp1p2/(p2 + I)2 1,thatisI (p1p2)1/2p2.Utilityequalsx2/(1 + x2) = I/(p2 + I).(x1, x2, 1, 2, 3) = (0, I/p2, 1 + ((p1p2)/(p2 + I)2), 0, p2/(p2 + I)2).Case(3)Bysymmetry,thesolutionis(x1, x2, 1, 2, 3) = (I/p1, 0, 0, 1 + ((p1p2)/(p1 + I)2), p1/(p1 + I)2)Andfor this Casetoholdit is necessarythat p1p2/(p1+ I)21, orI (p1p2)1/2p1.59Tosummarize,supposep1= p2= p,then(p1p2)1/2p1= (p1p2)1/2p2equals0. SosinceI> 0,weareintheregimeofCaseI,andx1= x2= I/2patthemaximum.Suppose on the other hand that p1< p2(the contrary can be worked outsimilarly),thenp2> (p1p2)1/2> p1,sothat(p1p2)1/2p1> 0 > (p1p2)1/2p2. ThuseitherI> (p1p2)1/2p1,inwhichcaseweuseCase(1),orI (p1p2)1/2p1in which case we use Case(3). Case(2), that in which apositive amount of good2 and zero of good1 is consumed at the maximum,doesnotapply.5.4 Miscellaneous(1) For problems wheresomeconstraints areof the formgi(x) =0, andothersof theformgj(x) 0, onlythelattergiverisetoKuhn-Tuckerlikecomplementaryslacknessconditions(j 0, gj(x) 0, jgj(x) = 0).(2)Iftheobjectivetobemaximized, f, andtheconstraintsgi, i=1, . . . , k(whereconstraintsareoftheformgi(x) 0)areallconcavefunctions,andif Slatersconstraintqualicationholds(i.e., thereexistssomex 1ns.t.gi(x) >0, i =1, . . . , k), thenthe Kuhn-Tucker conditions become bothnecessaryandsucientforaglobalmax.(3) Suppose f andall the gis are quasiconcave. Thenthe Kuhn-Tuckerconditionsarealmostsucientforaglobal max: AnxandthatsatisfytheKuhn-Tuckerconditionsindicatethatxisaglobal maxprovidedthatinadditiontotheabove,eitherDf(x) ,= ,orfisconcave.60AppendixCompletenessPropertyofRealNumbers61