Lecture 6: Contraction mapping, inverse and implicit function theorems 1 The contraction mapping theorem De nition 11 Let X be a metric space, with metric d If f : X! X and if there is a number 2 (0; 1) such that d(f(x); f(y)) d(x; y) for all x; y 2 X, then f is said to be a contraction mapping of X into X Proposition 11 Every contraction mapping is continuous De nition 12 Suppose f : X! X and x 2 X If f(x) = x, then we say that x is a xed point of f Example 1 The function f(x) = x 2 is a contration mapping of R into R Theorem 12 If X is a complete metric space, and if f is a contraction mapping of X into X, then there exists one and only one x 2 X such that f(x) = x Proof Pick x 0 2 X arbitrarily, and de ne fx n g recursively by setting x n+1 = f(x n ); for n = 0; 1; 2; : : : : For n 1; we have d(x n+1 ; x n ) = d(f(x n ); f(x n 1 )) d(x n ; x n 1 ): Hence d(x n+1 ; x n ) n d(x 1 ; x 0 ): 1
And for n < m, d(x n ; x m ) mx i=n+1 d(x i ; x i 1 ) ( n + n+1 + + m 1 )d(x 1 ; x 0 ) n 1 m n 1 d(x 1; x 0 ) Thus fx n g is a Cauchy sequence Since X is complete, fx n g converges in X Suppose x n! x; x 2 X Since f is a contraction mapping, it is continuous Therefore f(x) = lim n f(x n ) = lim n x n+1 = x To see the uniqueness, observe that if f(x) = x and f(y) = y, then d(f(x); f(y)) = d(x; y) d(x; y) This implies d(x; y) = 0 and therefore x = y This contraction mapping theorem is also called the Banach xed point theorem It has many applications, among which one is of particular interest to economists Example 2 Consider the following equation: v(k) = max u(f(k) y) + v(y); k 2 [0; 1]: y2[0;1] This equation is called the Bellman functional equation in dynamic programming Just consider k is capital, f(k) is the production function, f(k) y is the consumption today and y is the capital investment for tomorrow Let the max function be, then the solution to this equation is just a xed point of, that is a function v such that, v = (v) We will come back to this 2
2 The inverse function theorem Lemma 21 Suppose f : X! Y has an inverse function f 1 : Y! X Suppose x 0 2 X; y 0 2 Y, and y 0 = f(x 0 ) If f is di erentiable at x 0 ; f 0 (x 0 ) 6= 0; and f 1 is continuous at y 0 ; then f 1 is di erentiable at y 0 and (f 1 ) 0 (y 0 ) = 1 f 0 (x 0 ) : We omit the proof on showing that f 1 is di erentiable From the chain rule for di erentiation, (f 1 f) 0 (x 0 ) = (f 1 ) 0 (y 0 )f 0 (x 0 ):Since (f 1 f)(x) = x; 8x; we know (f 1 f) 0 (x 0 ) = 1 This gives us the formula In fact, even if f is not invertible on the whole domain, we can argue that if f is continuously di erentiable, then it is locally invertible around any point x 0 such that f 0 (x 0 ) 6= 0 If f 0 (x 0 ) 6= 0; assume f 0 (x 0 ) > 0 Then f 0 is continuous implies that f 0 (x) > 0 for all x in some small neighborhood N " (x 0 ) of x 0 In other words, f must be strictly increasing in N " (x 0 ), hence must be invertible in N " (x 0 ) The argument above can be generalized to functions in R n function (an equation system) is 8 >< >: y 1 = f 1 (x 1 ; : : : ; x n ) y n = f n (x 1 ; : : : ; x n ) ; An example of such a where f = (f 1 ; f 2 ; : : : ; f n ) : R n! R n If y 0 = f(x 0 ), then f 1 (y 0 ) locally exists as a function around (x 0 ; y 0 ) is equivalent to saying that each x = (x 1 ; : : : ; x n ) around x 0 can be expressed as a function of (y 1 ; : : : ; y n ) = (f 1 (x); : : : ; f n (x)); with the function form f 1 That is, around x 0 ; the equation system is always solvable Theorem 22 (Inverse function theorem) Suppose f is a continuously di erentiable function of an open set E R n into R n (ie, f 2 C 1 (E)), the matrix f 0 (a) is invertible for some a 2 E, and b = f(a): Then (a) there exist open sets U and V in R n such that a 2 U; b 2 V; f is 1 1 on U; and f(u) = V ; 3
(b) if g is the inverse of f (which exists, by (a)), de ned in V by g(f(x)) = x; 8x 2 U Then g 2 C 1 (V ) Please see Rudin (1976) for proof In the proof, the contraction mapping theorem is used Part (a) shows that the inverse of f on U is indeed a function, and part (b) further proves that the inverse g is continuously di erentiable After knowing all these, we can then apply the Chain-rule of di erention to calculate g 0 (f(x)) Example 3 Consider the following linear equations system y 1 = f 1 (x) = a 11 x 1 + + a 1n x n y 2 = f 2 (x) = a 21 x 1 + + a 2n x n y n = f n (x) = a n1 x 1 + + a nn x n : Here f 0 (x) at each x is the constant matrix A = (a ij ) Thus the condition of f 0 (x) being invertible is equivalent to the existence of A 1 From the theorem above, if A 1 exists, the equation system has a solution, and from linear algebra we know that x = g(y) = A 1 y: Obviously, in this simple example, the inverse function g is continuously di erentiable and g 0 (y) = A 1 for all y 3 Implicit function theorem Any function of the form y = f(x) is called an explicit function And f(x; y) = 0 de nes y as an implicit function of x If f is a continuously di erentiable real function in the plane, then the equation f(x; y) = 0 can be solved for y in terms of x in a neighborhood of any point (a; b) at which f(a; b) = 0 and f=y 6= 0: Example 4 Consider f(x; y) = x 2 + y 2 1 The equation f(x; y) = 0 speci es the unit circle on the plane At any point (x 0 ; y 0 ) other than ( 1; 0) and (1; 0), there is an open 4
neighborhood of (x 0 ; y 0 ) on which the unit circle is part of a function y = y(x) Note that at ( 1; 0) and (1; 0); f=y = 0 c; that is, Suppose we already know that there is a C 1 solution y = g(x) to the equation f(x; y) = f(x; g(x)) = c: Then we can use the chain-rule to di erentiate f with respect to x at each x 0 such that f(x y 0; g(x 0 )) 6= 0 : x f(x 0; g(x 0 )) + y f(x 0; g(x 0 ))g 0 (x 0 ) = 0: Hence g 0 (x 0 ) = f(x x 0; g(x 0 )) f(x y 0; g(x 0 )) : Hence the calculation of g 0 (x 0 ) is easy, but when will there be a C 1 function y = y(x) that solves f(x; y) = c around (x 0 ; y 0 )? Theorem 31 (Implicit function theorem, two variable case) Let (x 0 ; y 0 ) be a point on the locus of f(x; y) = c in the plane, where f is a C 1 function of two variables If (f=y)(x 0 ; y 0 ) 6= 0; then f(x; y) = c de nes a C 1 function y = y(x) in some neighborhood of (x 0 ; y 0 ) Proof We provide only a sketch of the proof Assume c = 0 We know that (f=y)(x 0 ; y 0 ) 6= 0; assume it is positive And since f =y is continuous, there is a small neighborhood N " ((x 0 ; y 0 )) of (x 0 ; y 0 ) such that for any (x; y) 2 N " ((x 0 ; y 0 )); (f=y)(x; y) > 0 Now, rst, since f(x 0 ; y 0 ) = 0, and f is strictly increasing in y, there must be small number such that f(x 0 ; y 0 + ) > 0 and f(x 0 ; y 0 ) < 0 Second, since f is continuous around (x 0 ; y 0 + ), f(x; y 0 +) is positive for x close enough to x 0, and f(x; y 0 ) is negative for x close enough to x 0 Given each such x, there must be a unique y = y(x) between y 0 and y 0 + such that f(x; y) = 0 This is because f is continuous and strictly increasing on y The rest of the work is to show that y(x) is continuously di erentiable 5
More generally, consider the equation system 8 >< >: f 1 (x 1 ; : : : ; x n ; y 1 ; : : : ; y m ) = 0 f m (x 1 ; : : : x n ; y 1 ; : : : ; y m ) = 0 ; and for any (a; b) 2 R n+m denote 0 Df x (a; b) = B f 1 (a;b) x 1 f m(a;b) x 1 f 1 (a;b) x n f m(a;b) x n 1 C A and 0 Df y (a; b) = B f 1 (a;b) y 1 f m(a;b) y 1 f 1 (a;b) y m f m(a;b) y m 1 C A : Theorem 32 (Implicit function theorem) Let f(x; y) be a C 1 mapping of an open set E R n+m into R m ; such that f(a; b) = 0 for some point (a; b) 2 E Assume that Df y (a; b) is invertible Then there exist open sets U R n and V R m with a 2 U and b 2 V; such that there is a unique C 1 onto function g : U! V; f(x; g(x)) = 0; 8x 2 U: Given this theorem, the chain-rule gives that g 0 (a) = see Rudin (1976) for a proof Df y (a; b) 1 Df x (a; b) Again, Example 5 The indi erence curve of a utilitiy function on two consumption goods is given by the equation u(x; y) = c At any consumption bundle (x 0 ; y 0 ), the marginal rate of substitution between good x and good y; which measures the amount of good y the consumer would demand to compensate 6
for 1 unit loss of good x to keep level c of utility, is given by dy dx = u(x0;y0) x u(x 0 ;y 0 ) y 7