COMPLETE METRIC SPACES AND THE CONTRACTION MAPPING THEOREM

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1 COMPLETE METRIC SPACES AND THE CONTRACTION MAPPING THEOREM A metric space (M, d) is a set M with a metric d(x, y), x, y M that has the properties d(x, y) = d(y, x), x, y M d(x, y) d(x, z) + d(z, y), x, y, z M (triangle inequality) and d(x, y) = x = y Most of you have seen these definitions and so I will not go into any details. A sequence of points x n M is a Cauchy Sequence if for any ε > there exists N(ε) such that d(x n, x m ) < ε for all n, m > N(ε). Accordingly we say that a complete metric space is complete if every Cauchy Sequence converges to some element x M, i.e., for every ε > there exists N(ε) such that d(x n, x) < ε for all n > N(ε). A function f : M M is a contraction if there exists a constant α < such that for all x, y M d(f(x), f(y)) αd(x, y). A simple consequence of these definitions is the Banach fixed point theorem: Theorem.. Let (M, d) be a complete metric space and f : M M a contraction. Then the equation x = f(x) has a unique solution x. Moreover, if x M is any initial point and x n+ = f(x n ), n =,,..., then d(x, x n ) αn α d(f(x ), x ) Proof. If x and y are two solutions then d(x, y) = d(f(x), f(y)) αd(x, y) and hence d(x, y) = and therefore x = y. Pick any n > m and use the triangle inequality to find Moreover, and so n d(x k+, x k ) d(x n, x m ) n d(x k+, x k ) d(x k+, x k ) = d(f(x k ), f(x k )) αd(x k, x k ) α k d(f(x ), x ) n α k d(f(x ), x ) = α m αn m α d(f(x ), x ) αm α d(f(x ), x ).

2 2 COMPLETE METRIC SPACES AND THE CONTRACTION MAPPING THEOREM Hence d(x n, x m ) αm α d(f(x ), x ) which proves that the sequence x n is a Cauchy sequence. The completeness of (M, d) guarantees that x n has a limit x. We have to show that x solves the equation. Pick n arbitrary and use that which is bounded above by d(f(x), x) d(f(x), x n ) + d(x n, x) = d(f(x), f(x n )) + d(x n, x) αd(x, x n ) + d(x n, x) 2αn α d(f(x ), x ). Because n is arbitrary d(f(x), x) is smaller than any positive number and hence equal to. Example : For a simple example of a metric space that is not necessarily complete consider any set S R d and consider the Euclidean distance d(x, y) = x y = d (x j y j ) 2. It is obvious that d(x, y) = d(y, x) for all x, y S. Further j= = d(x, y) = x y implies that x and y have the same components and hence are equal. The triangle inequality follows from the following facts: Schwarz inequality x y := d x j y j d x 2 j j= j= j= d yj 2 = x y Minkowski s inequality x + y x + y This follows easily from Schwarz inequality. Thus, we find that for any x, y, z S d(x, y) = x y = x z + z y x z + z y = d(x, z) + d(z, y). In this context, the following is an interesting application of the contraction mapping theorem. We start first with an easy case. A map is called Lipschitz, if there exists a constant L such that for all x, x 2 f(x ) f(x 2 ) L x x 2. Thus, a contraction is a Lipschitz map with Lipschitz constant L <. Given a map f : R d R d and consider the map h : R d R d

3 COMPLETE METRIC SPACES AND THE CONTRACTION MAPPING THEOREM 3 given by h(x) = x + f(x). Assume that f is a contraction, i.e., f(x) f(y) α x y for some constant α <. We claim that h has an inverse which is also a contraction To see this we have to show two things. a) h is injective. This follows from the fact that x + f(x ) = x 2 + f(x 2 ) entails that x x 2 = f(x 2 ) f(x ) α x x 2 which yields x = x 2. b) Next we have to show that h is onto. For any given y R d we consider the equation which we rewrite as The map φ is a contraction y = x + f(x) x = y f(x) := φ(x). φ(x ) φ(x 2 ) = f(x ) f(x 2 ) α x x 2 and hence there exists a unique fixed point a R d, i.e., a = φ(a) = y f(a). Hence h has an inverse, which we denote by g : R d R d. To show that g is Lipschitz we write y i = h(x i ), i =, 2 and note that x x 2 y y 2 + f(x ) f(x 2 ) y y 2 + α x x 2 so that x x 2 α y y 2 which shows that g is Lipschitz with Lipschitz constant. This argument can be adapted α to a more general situation. Theorem.2. Imagine an open set S R d and let f : S R d be a contraction with contraction constant α <. Then for the map h : S h(s), h(x) = x + f(x) h(s) is open and the map h has an inverse g : h(s) S which is Lipschitz with Lipschitz constant α. Proof. The fact that h is injective has the same proof as before. A priori we do not know much about the set h(s). We prove that this set is open in R d. Pick any y h(s). Then, by definition, there exists a point x S so that h(x ) = y. To arrange things in a convenient way we set U(x) = h(x + x) y = x + f(x + x) + x y = x + V (x) so that U() = i.e., U fixes the origin. Hence U : S x h(s) y,

4 4 COMPLETE METRIC SPACES AND THE CONTRACTION MAPPING THEOREM and our goal is to show that U(S x ) is an open set. Pick r > so that the closed ball B r () S x and note that V (x) = V (x) V () = f(x + x ) f(x ) α x so that V maps the ball B r () into the ball B αr () B r (). Such a radius r exists, because S x is open. Indeed pick r so that the open ball B r () S x and pick any < r < r which assures that B r () S x. Hence, V is a map of the metric space B r () into itself. Moreover, V is a contraction on B r (). Indeed for x, x 2 B r () we have that V (x ) V (x 2 ) = f(x + x ) f(x 2 + x ) α x x 2. If we can show that any point y B r () is of the form U(z) for some z B r () we are done. Thus, we have to find z B r () so that y = z + V (z) i.e., the map y V (x) has a fixed point in B r (). Note that B r () is closed and hence is a complete metric space. Thus, by the fixed point theorem there exists z B r () with the desired properties. Denoting the inverse by g : h(s) S we have for y, y 2 h(s), setting x i = g(y i ), i =, 2, x x 2 = (y f(x )) (y 2 f(x 2 )) y y 2 + f(x ) f(x 2 ) y y 2 + α x x 2 so that x x 2 α y y 2, which shows that g is a Lipschitz map with Lipschitz constant α. A consequence of this Theorem is the inverse function theorem. Theorem.3. Let S R d be an open set and F : S R d a map that is continuously differentiable. Assume that the Jacobi matrix DF (x ), x S, is invertible. Then there exists an open set U R d with x U such that F (U) is open and there exists a map g : F (U) U such that g F = id. Moreover, g is differentiable at F (x ) and we have that Dg(F (x )) = DF (x ) Proof. We have to construct U. First we normalize things conveniently. By replacing F (x) by DF (x ) F (x) we may assume that DF (x ) = I. Further, replacing F (x) by F (x+x ) F (x ) we may assume that x = and F (x ) =. Let s denote this renormalized map by h. Since h is continuously differentiable we have that which leads to where It is convenient to set h(x) h() = f(x) := d dt h(tx)dt = h(x) = x + f(x) (Dh(tx) I)dt x. M i,j (x) = (Dh(x) I) i,j. Dh(tx)dt x

5 COMPLETE METRIC SPACES AND THE CONTRACTION MAPPING THEOREM 5 Since Dh(x) is continuous at we can find r > so that Hence we have that f(x) = ( i j t max i,j sup x 3r M i,j (x) 2d. M i,j (tx)dtx j ) 2 ( 2d i j x j ) 2 = d( 2d j x j ) 2 d 2d 2 ( j x j 2 ) = 2 x and we see that f(b 2r ()) B r () and in particular f(b r ()) B r (). Further for x, x 2 B r () we have that and f(x ) f(x 2 ) = h(x ) h(x 2 ) x + x 2 = and hence ( t)x 2 + tx x 2 + t x x 2 3r f(x ) f(x 2 ) 2 x x 2. [Dh(( t)x 2 + tx ) I] dt(x x 2 ) Thus, f : B r () B r () is a contraction and therefore V = f(b r ()) is open and h : V B r () has a Lipschitz continuous inverse, which we denote again by g. To see that g is differentiable at we shall show that g(x) x = o( x ). This implies that g is differentiable at and Dg() = I as it should be. Pick any sequence x n, set y n = g(x n ). Note that y n as well and we compute Because g() =, we have that and hence g(x n ) x n x n = y n h(y n ) g(x n ) y n x n g(x n ) x n 2 g(x n ) x n lim n x n =..