IMPLICI AND INVERSE FUNCION HEOREMS PAUL SCHRIMPF 1 OCOBER 25, 213 UNIVERSIY OF BRIISH COLUMBIA ECONOMICS 526 We have exensively sudied how o solve sysems of linear equaions. We know how o check wheher soluions exis and wheher hey are unique. he inverse and implici funcion heorems provide similar resuls for nonlinear equaions. 1. INVERSE FUNCIONS Suppose f : R n R m. If we know f x = y, when can we solve for x in erms of y? In oher words, when is f inverible? Well, suppose we know ha f a = b for some a R n and b R m. hen we can expand f around a, f x = f a + D f a x a + r 1 a, x a where r 1 a, x a is small. Since r 1 is small, we can hopefully ignore i hen y = f x can be rewrien as a linear equaion: f a + D f a x a = y D f a x = y f a + D f a a we know ha his equaion has a soluion if rankd f a = rank D f a y f a + D f a a. I has a soluion for any y if rankd f a = m. Moreoever, his soluion is unique if rankd f a = n. his discussion is no enirely rigorous because we have no been very careful abou wha r 1 being small means. he following heorem makes i more precise. heorem 1.1 Inverse funcion. Le f : R n R n be coninuously differeniable on an open se E. Le a E, f a = b, and D f a be inverible. hen 1 here exis open ses U and V such ha a U, b V, f is one-o-one on U and f U = V, and 1. 2 he inverse of f exiss and is coninuously differeniable on V wih derivaive D f f x 1 he open ses U and V are he areas where r 1 is small enough. he coninuiy of f and is derivaive are also needed o ensure ha r 1 is small enough. he proof of his heorem is a bi long, bu he main idea is he same as he discussion preceding he heorem. Commen 1.1. he proof uses he fac ha he space of all coninuous linear ransformaions beween wo normed vecor spaces is iself a vecor space. I do no hink we have alked abou his before. Anyway, i is a useful fac ha already came up in he proof ha coninuous Gâeaux differeniable implies Fréche differeniable las lecure. Le V 1 hanks o Dana Galizia for correcions. 1
IMPLICI AND INVERSE FUNCION HEOREMS and W be normed vecor spaces wih norms V and W. Le BLV, W denoe he se of all coninuous or equivalenly bounded linear ransformaions from V o W. hen BLV, W is a normed vecor space wih norm Av A BL sup W. v V v V his is someimes called he operaor norm on BLV, W. Las lecure, he proof ha Gâeaux differeniable implies Fréche differeniable required ha he mapping from V o BLV, W defined by D f x as a funcion of x V had o be coninuous wih respec o he above norm. We will ofen use he inequaliy, Av W A BL v V, which follows from he definiion of BL. We will also use he fac ha if V is finie dimensional and f x, v : V V W, is coninuous in x and v and linear in v for each x, hen f x, : V BLV, W is coninuous in x wih respec o BL. Proof. For any y R n, consider φ y x = x + D fa 1 for x 1, x 2 U, where a U and U is open, Noe ha φ y x 1 φ y x 2 = Dφ ȳ x x 1 x 2 y f x. By he mean value heorem Dφ ȳ x 1 =I D fa D f x =D f 1 a D f a D f x. Since D f x is coninuous as a funcion of x if we make U small enough, hen D f a D f x 1 will be near. Le λ =. Choose U small enough ha D f 2 D fa 1 a D f x < λ for all BL x U. From above, we know ha φ y x 1 φ y x 2 = D fa 1 D f a D f x x 1 x 2 Dφ y x BL D f a D f x BL x 1 x 2 1 2 x 1 x 2 1 For any y f U we can sar wih an arbirary x 1 U, hen creae a sequence by seing From 1, his sequence saisfies x i+1 = φ y x i. x i+1 x i 1 2 x i x i 1. Using his i is easy o verify ha x i form a Cauchy sequence, so i converges. he limi saisfy φ y x = x, i.e. f x = y. Moreover, his x is unique because if φ y x 1 = x 1 and 2
IMPLICI AND INVERSE FUNCION HEOREMS φ y x 2 = x 2, hen we have x 1 x 2 1 2 x 1 x 2, which is only possible if x 1 = x 2. 1 hus for each y f U, here is exacly one x such ha f x = y. ha is, f is one-o-one on U. his proves he firs par of he heorem and ha f 1 exiss. We now show ha f 1 is coninuously differeniable wih he saed derivaive. Le y, y + k V = f U. hen x, x + h U such ha y = f x and y + k = f x + h. Wih φ y as defined above, we have By 1, h D f 1 a φ y x + h φ y x =h + D fa 1 f x f x + h =h D fa 1 k k 1 2 h. I follows ha D f 1 a k 2 1 h and h 2 D f 1 k = λ 1 k. BL a Imporanly as k, we also have h. Now, f 1 y + k f 1 y D fx 1 k D fx 1 f x + h f x D f x h = k k D f x 1 λ f x + h f x D f xh h f 1 y + k f 1 y D fx 1 k lim lim D f x 1 BL k k λ f x + h f x D f xh k h Finally, since D f x is coninuous, so is D f f 1 y 1, which is he derivaive of f 1. he proof of he inverse funcion heorem migh be a bi confusing. he imporan idea is ha if he derivaive of a funcion is nonsingular a a poin, hen you can inver he funcion around ha poin because invering he sysem of linear equaions given by he mean value expansion around ha poin nearly gives he inverse of he funcion. 2. IMPLICI FUNCIONS he implici funcion heorem is a generalizaion of he inverse funcion heorem. In economics, we usually have some variables, say x, ha we wan o solve for in erms of some parameers, say β. For example, x could be a person s consumpion of a bundle of goods, and b could be he prices of each good and he parameers of he uiliy funcion. Someimes, we migh be able o separae x and β so ha we can wrie he condiions of our model as f x = bβ. hen we can use he inverse funcion heorem o compue x i β j and oher quaniies of ineres. However, i is no always easy and someimes no possible o separae x and β ono opposie sides of he equaion. In his case our model gives us equaions of he form f x, β = c. he implici funcion heorem ells us when we can solve for x in erms of β and wha x i β j will be. 1 Funcions like φ y ha have dϕx, ϕy cdx, y for c < 1 are called conracion mappings. he x wih x = ϕx is called a fixed poin of he conracion mapping. he argumen in he proof shows ha conracion mappings have a mos one fixed poin. I is no hard o show ha conracion mappings always have exacly one fixed poin. 3 =
IMPLICI AND INVERSE FUNCION HEOREMS he basic idea of he implici funcion heorem is he same as ha for he inverse funcion heorem. We will ake a firs order expansion of f and look a a linear sysem whose coefficiens are he firs derivaives of f. Le f : R n R m. Suppose f can be wrien as f x, y wih x R k and y R n k. x are endogenous variables ha we wan o solve for, and y are exogenous parameers. We have a model ha requires f x, y = c, and we know ha some paricular x and y saisfy f x, y = c. o solve for x in erms of y, we can expand f around x and y. f x, y = f x, y + D x f x,y x x + D y f x,y y y + rx, y = c In his equaion, D x f x,y is he m by k marix of firs parial derivaives of f wih respec o x evaluaed a x, y. Similary, D y f x,y is he m by n k marix of firs parial derivaives of f wih respec o y evaluaed a x, y. hen, if rx, y is small enough, we have f x, y + D x f x,y x x + D y f x,y y y c D x f x,y x x c f x, y D y f x,y y y his is jus a sysem of linear equaions wih unknowns x x. If k = m and D x f x,y is nonsingular, hen we have 1 x x + D x f x,y c f x, y D y f x,y y y which gives x approximaely as funcion of y. he implici funcion says ha you can make his approximaion exac and ge x = gy. he heorem also ells you wha he derivaive of gy is in erms of he derivaive of f. heorem 2.1 Implici funcion. Le f : R n+m R n be coninuously differeniable on some open se E and suppose f x, y = c for some x, y E, where x R n and y R m. If D x f x,y is inverible, hen here exiss open ses U R n and W R n k wih x U and y W such ha 1 For each y W here is a unique x such ha x, y U and f x, y = c. 2 Define his x as gy. hen g is coninuously differeniable on W, gy = x, f gy, y = 1 c for all y W, and Dg y = D x f x,y Dy f x,y Proof. We will show he firs par by applying he inverse funcion heorem. Define F : R n+m R n+m by Fx, y = f x, y, y. o apply he inverse funcion heorem we mus show ha F is coninuously differeniable and DF x,y is inverible. o show ha F is coninuously differeniable, noe ha Fx + h, y + k Fx, y = f x + h, y + k f x, y, k =D f x,ȳ h k, k where he second line follows from he mean value heorem. I is hen apparen ha Fx + h, y + k Fx, y Dx f x,y D y f x,y h I m k =. h, k lim h,k 4
IMPLICI AND INVERSE FUNCION HEOREMS Dx f So, DF x,y = x,y D y f x,y, which is coninuous sinve D f I x,y is coninuous. Also, m DF x,y can be shown o be inverible by using he pariioned inverse formula because D x f x,y is inveriable by assumpion. herefore, by he inverse funcion heorem, here exiss open ses U and V such ha x, y U and c, y V, and F is one-o-one on U. Le W be he se of y R m such ha c, y V. By definiion, y W. Also, W is open in R m because V is open in R n+m. We can now complee he proof of 1. If y W hen c, y = Fx, y for some x, y U. If here is anoher x, y such ha f x, y = c, hen Fx, y = c, y = Fx, y. We jus showed ha F is one-o-one on U, so x = x. We now prove 2. Define gy for y W such ha gy, y U and f gy, y = c, and Fgy, y = c, y. By he inverse funcion heorem, F has an inverse on U. Call i G. hen Gc, y = gy, y and G is coninuously differeniable, so g mus be as well. Differeniaing he above equaion wih respec o y, we have Dgy D y G c,y = On he oher hand, from he inverse funcion heorem, he derivaive of G a x, y is 1 DG x,y = DF x,y 1 Dx f = x,y D y f x,y I m Dx f 1 D = x,y x f 1 x,y D y f x,y I m I m In paricular, D y G c,y = Dx f 1 x,y D y f x,y I m Dgy = I m so Dg y = D x f 1 x,y D y f x,y. 3. CONRACION MAPPINGS One sep of he proof he of he inverse funcion heorem was o show ha φ y x 1 φ y x 2 1 2 x 1 x 2. his propery ensures ha φx = x has a unique soluion. Funcions like φ y appear quie ofen, so hey have name. 5
IMPLICI AND INVERSE FUNCION HEOREMS Definiion 3.1. Le f : R n R n. f is a conracion mapping on U R n if for all x, y U, for some c < 1. f x f y c x y If f is a conracion mapping, hen an x such ha f x = x is called a fixed poin of he conracion mapping. Any conracion mapping has a mos one fixed poin. Lemma 3.1. Le f : R n R n be a conracion mapping on U R n. If x 1 = f x 1 and x 2 = f x 2 for some x 1, x 2 U, hen x 1 = x 2. Proof. Since f is a conracion mapping, f x 1 f x 2 c x 1 x 2. f x i = x i, so x 1 x 2 c x 1 x 2. Since c < 1, he previous inequaliy can only be rue if x 1 x 2 =. hus, x 1 = x 2. Saring from any x, we can consruc a sequence, x 1 = f x, x 2 = f x 1, ec. When f is a conracion, x n x n+1 c n x 1 x, which approaches as n. hus, {x n } is a Cauchy sequence and converges o a limi. Moreover, his limi will be such ha x = f x, i.e. i will be a fixed poin. Lemma 3.2. Le f : R n R n be a conracion mapping on U R n, and suppose ha f U U. hen f has a unique fixed poin. Proof. Pick x U. As in he discussion before he lemma, consruc he sequence defined by x n = f x n 1. Each x n U because x n = f x n 1 f U and f U U by assumpion. Since f is a conracion on U, x n+1 x n c n x 1 x, so lim n x n+1 x n =, and {x n } is a Cauchy sequence. Le x = lim n x n. hen x f x x x n + f x f x n 1 x x n + c x x n 1 x n x, so for any ϵ > N, such ha if n N, hen x x n < x f x < ϵ ϵ 1+c. hen, for any ϵ >. herefore, x = f x. 4. APPLICAIONS his lecure and he previous one have been raher heoreical, so his secion goes over a couple of applicaions of wha has been covered. 6
IMPLICI AND INVERSE FUNCION HEOREMS 4.1. Roy s Ideniy. Le Vm, p be an indirec uiliy funcion. Given oal expendiure m and a vecor of prices p, he maximum uiliy ha a person can achieve is Vm, p. If U is he uiliy funcion, he indirec uiliy funcion is given by Vm, p = max Uc s.. pc m. 2 c Similarly, expendiure funcion, Eu, p, is he minimum amoun of money ha can be spen o achieve uiliy u when faced wih prices p. ha is, Eu, p = min c pc s.. Uc u. 3 We haven ye covered opimizaion, so le s jus assume ha 2 and 3 have unique soluions. In normal cases, we would expec ha VEu, p, p = u and EVm, p, p = m. Le s come up wih condiions ha ensure hese wo equaliies hold. Le s sar by working wih VEu, p, p = u. By definiion of Eu, p, here mus be some c such ha pc = Eu, p and Uc = u. Using ha same c in 2, we see ha VEu, p, p Uc = u. Suppose i were sricly greaer. hen here is some c such ha U c > u and p c pc = m. Bu if U is coninuous, hen for any ϵ, we can find δ > such ha if h < δ hen U c U c + h < ϵ and in paricular, U c + h > u. If p =, we can choose an h wih h < δ and ph <. However, hen p c + h < pc, which should no be possible given how we have defined c. hus, assuming U is coninuous and p = is enough o ensure ha VEu, p, p = u. Having esablished, VEu, p, p = u, we can differeniae wih respec o he price of i good o ge V E Eu, p, p u, p + V Eu, p, p = m p i p i E p i u, p = V p i Eu, p, p V m Eu, p, p his resul combined wih Shephard s lemma which we will prove afer sudying opimizaion gives Roy s ideniy. Shephard s lemma is and Roy s ideniy is c i u, p = E p i u, p, V ci m, p = pi m, p. V m m, p Roy s ideniy is very useful because i relaes demand, somehing ha we can observe, o he indirec uiliy funcion, somehing ha canno be direcly observed. 4.2. Comparaive saics. Consider a simple finie horizon macro model. he producion funcion is Cobb-Douglas wih only one inpu, capial k, so oupu a ime is y = A k α 7
IMPLICI AND INVERSE FUNCION HEOREMS where A is produciviy. Oupu can be eiher consumed or saved as capial for he nex period. Capial depreciaes a rae δ. he budge consrain each period is c + k +1 = 1 δk + A k α. he consumer has a sandard CRRA uiliy funcion ha is addiively separable over ime and discoun rae β. he social planner s problem is o maximize oal uiliy subjec o he budge consrains, max {c,k } = = β c1 γ 1 γ s.. c + k +1 = 1 δk + A k α. You are likely already familiar wih using Lagrangians o solve consrained maximizaion problems. If no, do no worry because we will cover i in some deail in a week or wo. Anyway, he idea is ha we can replace he consrained problem wih an unconsrained one of he form: max {c,k,λ } = = β c1 γ 1 γ + λ c + k +1 1 δk A k α where we have adding λ imes he h consrain o he objecive funcion and we are now maximizing over λ as well as c and k. We have already shown ha for c, k, λ, o be local maxima, we mus have he derivaive of he objecive funcion equal o zero. his is called he firs order condiion. Here, he firs order condiions are [c ] : β c γ =λ [k ] : λ 1 =λ 1 δ + A αk α 1 [λ ] : c + k +1 =1 δk + A k α he values of c, k, λ ha solve he maximizaion problem necessarily solve he firs order condiions as well, bu no every soluion o he firs order condiions solves he maximizaion problem. here is a second condiion ha ensures a soluion o he firs order condiion solves he maximizaion problem. We will overlook second order condiions for now, bu we will sudy hem soon. We can eliminae λ from he sysem of firs order condiions by combining [c ] and [c ] o ge β 1 c γ c c 1 1 =β c γ γ =β 1 δ + A αk α 1 1 δ + A αk α 1 Combined wih he budge consrain we now have wo nonlinear equaion for each wih wo unknowns for each. he soluion o hese equaions is he opimal sequence of c and k. c c 1 γ =β 1 δ + A αk α 1 c + k +1 =1 δk + A k α 8
IMPLICI AND INVERSE FUNCION HEOREMS Unforunaely, here is no closed form soluion o hese equaions. However, we can sill calculae cerain quaniies of ineres by using he implici funcion heorem. For example, wha is he effec of changes in produciviy, A, on consumpion, capial, and welfare? Suppose A changes unexpecedly a ime 1 for one period only, so we can rea c 2 and k 1 as consan. We wan o find he change in c 1, c, and k. Noe ha he equaions above hold for each. he relevan hree equaions here are he implici funcion heorem says ha c 1 A 1 c A 1 k A 1 = = =Fc, c 1, k, A, A 1, c 2, k 1 c 1 + k 1 δk 1 A 1 k α 1 = c 1 δk A k α c γ 1 c γ β 1 δ + A αk α 1 F 1 c 1 F 1 c F 1 k F 2 c 1 F 2 c F 2 k F 3 c 1 F 3 c F 3 k γc γ 1 1 1 F 1 A 1 F 2 A 1 F 3 A 1 1 1 1 1 δ A αk α 1 γc γ 1 β 1 δ + A αk α 1 c γ βa αα 1k α 2 1 k α 1 We can inver his marix using Gaussian eliminaion: 1 1 k α 1 1 1 δ A αk α 1 γc γ 1 β 1 δ + A αk α 1 c γ βa αα 1k α 2 γc γ 1 1 1 1 k α 1 1 1 δ A αk α 1 γc γ 1 β 1 δ + A αk α 1 c γ βa αα 1k α 2 + γc γ 1 1 γc γ 1 1 1 1 k α 1 1 1 δ A αk α 1 E γc γ 1 1 k α 1 k α 1 where E = γc γ 1 β 1 δ + A αk α 1 c γ βa αα 1k α 2 + γc γ 1 1. 9 1 δ + A αk α 1 +
IMPLICI AND INVERSE FUNCION HEOREMS his expression is quie complicaed, bu we can sill make a few observaions. Firs, we generally assume ha α 1, which ensures ha E >. hen, k = γc γ 1 1 A 1 E kα 1 So when produciviy goes up a ime 1, capial a ime increases. Also, from he second equaion, c = k 1 δ + A α k α 1 A 1 A, 1 so consumpion a ime also increases. From he firs equaion, c 1 A 1 =k α 1 k A 1 = kα 1 E γc γ 1 1 kα 1 E = kα 1 γc γ 1 c 1 A 1 < k α 1 β 1 δ + A αk α 1 > 1 δ + A αk α 1 E c γ βa αα 1k α 2 So c 1 increases when A 1 increases, bu less han he increase in oupu a ime 1. 1