THE INVERSE FUNCTION THEOREM FOR LIPSCHITZ MAPS RALPH HOWARD DEPARTMENT OF MATHEMATICS UNIVERSITY OF SOUTH CAROLINA COLUMBIA, S.C. 29208, USA [email protected] Abstract. This is an edited version of a proof, in the from of exercises with detailed hints, of the inverse function that was given to a graduate class in differential equations as homework. The goal of the assignment was not only to present the basics about inverse functions, but to re- call/introduce the basics about differential calculus in Banach spaces and to give an example of a nontrivial application of the contraction mapping principle (which is also included as one of the problems) as these were among the basic tools used in the class. Thus the presen- tation starts more or less from the definition of the derivative of maps between Banach spaces and proves not only the standard version of the inverse function, but also the version for Lipschitz maps. Contents 1. Introduction. 2 2. Derivatives of maps between Banach Spaces 2 2.1. Bounded linear maps between Banach spaces. 2 2.2. The derivative of maps between Banach spaces 5 3. The Inverse Function Theorem for Lipschitz Maps 9 4. The Inverse Function Theorem 11 5. Implicit Function Theorem: Surjective From 14 5.1. Preliminary results on surjective linear maps. 14 .2. The surjective form of the implicit function theorem for Lipschitz maps. 16 A. Appendix: Contraction Mappings and the Banach Fixed Point Theorem 16 A.1. Some Review 16 A.2. Metric spaces 17 A.3. The Banach Fixed Point Theorem 17 Date: March 1997. 1 2 RALPH HOWARD 1. Introduction. There are many motivations for the inverse function theorem. Here is one. Consider a map f : Rn Rm and a point y Rm. Then it is interesting to know when the equation!f(x) = y can be solve2 for x. Of course in complete generality there is no reasonable solution to the problem. But assume f is smooth, or at least continuously differentiable, and that f(x0) = y0. Then it is reasonable to look for conditions on f so that for y sufficiently close to y0 that f(x) = y has a solution with x close to x0. This is the problem the n m inverse function theorem answers. Let f (x0): R R be the derivative 0 ! (this is the linear map that best approximates f near x0 see 2.2 for the exact n m x definition) and assume that f 0(x0): R R is onto. Then the implicit function theorem gives us a open neighbor! hood V so that for every y V the equation f(x) = y has a solution. More than that it allows us to choose2 n a continuously differentiable function ': V R with '(y0) = x0 and such that f('(y)) = y that is not only can we solve! the equation f(x) = y, we can do it in such a way that the solution x = '(y) is a smooth function of y V . Showing that f(x) = y can be solved for x is a more or less typical application2 of the Banach Fixed Point Theorem (i.e contraction mapping principle which is included in an appendix), but is nontrivial enough to be quite interesting. Showing that the solution x = '(y) depends smoothly on y requires more work. Rather than just work with maps between the Euclidean spaces, we will work with maps between Banach spaces. The definitions and results we need on bounded linear maps and differentiable functions between are given in in the preliminary sections. The work is split into two steps. First a version of the theorem is given for Lipschitz maps (see 3) that are sufficiently close to a linear map. This version gives the existencex of an inverse ' which is continuous. This is then used (in 4) to show that if the original map is continuously differentiable then so isx the its local inverse '. One advantage to splitting the proof up in this manner is that if makes it easy to give explicit bounds on the size of the neighborhood where the local inverse is defined. 2. Derivatives of maps between Banach Spaces 2.1. Bounded linear maps between Banach spaces. Recall that a Ba- nach space is a normed vector space that is complete (i.e. Cauchy se- quences converge) with respect to the metric by the norm. Let X and Y be Banach spaces with norms X and Y. Then a linear map A: X Y is bounded iff there is a constantj · j C soj that · j ! Ax Y C x X for all x X: j j ≤ j j 2 The best constant C in this inequality is the operator norm (which we will usually just call the norm) of A and denoted by A . Thus A is k k k k THE INVERSE FUNCTION THEOREM FOR LIPSCHITZ MAPS 3 given by Ax Y A := sup j j : k k 0=x X x X 6 2 j j Exercise 2.1. Show that a linear map L: X Y is continuous if and only if it is bounded. ! Denote by (X; Y) the set of all bounded linear maps A: X Y. B ! Exercise 2.2. 1. Show that the norm makes (X; Y) into a normed k · k B linear space. That is show if A; B (X; Y) and c1 and c2 are real 2 B numbers then c1A + c2B (X; Y) and 2 B c1A + c2B c1 A + c2 B : k k ≤ j jk k j jk k 2. Show that the norm is complete on (X; Y) and so (X; Y) is k · k B B a Banach space. Hint: This can be done as follows. Let Ak 1 be a f gk=1 Cauchy sequence in X. Then M := sup Ak < . Show k k k 1 (a) For any x X the sequence Akx 1 is a Cauchy sequence and as Y is 2 f gk=1 a Banach space this implies limk Akx exists. !1 (b) Define a map A: X Y by Ax := limk Akx. Then show A is linear ! !1 and Ax Y M x X for all x X. Thus A is bounded. j j ≤ j j 2 (c) Let " > 0 and let N" be so that k; ` N" implies Ak A` " (N" ≥ k − k ≤ exists as Ak 1 is Cauchy). Then for any x X and k N" and all f gk=1 2 ≥ ` N" we have ≥ Ax Akx Y Ax A`x Y + (A` Ak)x Y j − j ≤ j − j j − j Ax A`x Y + " x X ≤ j − j j j ` !1 0 + " x X = " x X: −! j j j j This implies A Ak " for k N" and thus limk Ak = A. This shows any Cauchyk − sequencek ≤ in (≥X; Y) converges. !1 3. If Z is a third Banach space A B (X; Y) and B (Y; Z)) then BA (X; Z) and BA B 2A B . In particular2 if BA (X; X) then2 by B induction Akk k ≤A kk. kk k 2 B k k ≤ k k Remark 2.3. Some inequalities involving norms of linear maps will be used repeatedly in what follows without comment. The inequalities in question are k k Ax Y A x X; BA B A ; A A : j j ≤ k kj j k k ≤ k k k k k k ≤ k k Of course the various forms of the triangle inequality will also be used. The includes the form u v u v . j − j ≥ j j − j j As first step in understanding when nonlinear maps have inverses we look at the set of invertible linear maps The linear map A (X; Y) is invertible iff there is a B (Y; X) so 2 B 2 B that AB = IY and BA = IX (where IX is the identity map on X). The 1 map B is called the inverse of A and is denoted by B = A− . 4 RALPH HOWARD Proposition 2.4. Let X be a Banach space and A (X; X) with IX A < 1. Then the A in invertible and the inverse is given2 B by k − k 1 1 k 2 3 A− = (IX A) = IX + (IX A) + (IX A) + (IX A) + X − − − − ··· k=0 and satisfies the bound 1 1 A− : k k ≤ 1 IX A − k − k Moreover if 0 < ρ < 1 then 1 1 1 A IX ; B IX ρ implies A− B− A B k − k k − k ≤ k − k ≤ (1 ρ)2 k − k − k k k Proof. Let B := k1=0(IX A) then (IX A) IX A and as P − k − k k ≤ k − k IX A < 1 the geometric series k1=0 IX A converges. Therefore by comparisonk − k the series defining B convergesP k and− k 1 k 1 B IX A = : k k ≤ X k − k 1 IX A k=0 − k − k Now compute 1 k 1 k AB = A(IX A) = (IX (IX A))(IX A) X − X − − − k=0 k=0 1 k 1 k+1 = (IX A) (IX A) = IX: X − − X − k=0 k=0 A similar calculation shows that BA = IX (or just note A and B clearly 1 1 commute). Thus B = A− . (The formula for B = A− was motivated by 1 k the power series (1 x)− = 1 x .) − Pk=0 If A IX ; B IX ρ then by what we have just done we have k − k k − k ≤ 1 1 1 A− ; B− k k k k ≤ 1 ρ − Therefore 1 1 1 1 1 1 A− B− = A− (B A)B− A− B− B A k − k k − k ≤ k k k k k − k 1 A B : ≤ (1 ρ)2 k − k − This completes the proof.