The Implicit Function Theorem for Lipschitz Functions and Applications. A Masters Thesis. presented to. the Faculty of the Graduate School

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1 The Implicit Function Theorem for Lipschitz Functions and Applications A Masters Thesis presented to the Faculty of the Graduate School University of Missouri In Partial Fulfillment of the Requirements for the Degree Master of Science by Michael Wuertz Dr. Marius Mitrea, Dissertation Supervisor MAY 2008

2 The undersigned, appointed by the Dean of the Graduate School, have examined the thesis entitled The Implicit Function Theorem for Lipschitz Functions and Applications presented by Michael Wuertz, a candidate for the degree of Master of Science in and hereby certify that in their opinion it is worthy of acceptance. Professor Marius Mitrea Professor Fritz Gesztesy Professor Steve Hofmann Professor Ioan Kosztin

3 Thanks to Professor Marius Mitrea, Courtney Bagge, Steven Wuertz, and Susan Wuertz

4 ACKNOWLEDGEMENTS It is a pleasure to thank Simona, Nick, Simon, and James for their special contribution. ii

5 Table of Contents ACKNOWLEDGEMENTS ii 1 Introduction 1 2 The Implicit and Inverse Function Theorems First Proof Using the Contraction Principle Second Proof Using the Invariance of Domain Theorem Lipschitz Functions 24 4 Rademacher s Theorem Almost Everywhere Differentiability of Lipschitz Functions Almost Everywhere Differentiability of Absolutely Continuous Functions Some Consequences of Rademacher s Theorem The Implicit Function Theorem for Lipschitz Functions 46 6 Lipschitz Domains General Considerations Some Toplogical Results Applications First Application Second Application An Application for the Classical Implicit Function Theorem Bibliography 77 iii

6 Chapter 1 Introduction A class of domains which plays an important role in the field of partial differential equations is the collection of all Lipschitz domains in R n. Recall that a function f : U R m, where U is a subset of R n, is called Lipschitz if { } f(x) f(y) Lip(f; U) := sup : (x, y) U U \ diag < (1.1) x y with Lip (f; U) called the Lipschitz constant of f in U. Next, a nonempty, proper open subset Ω of R n is called a Lipschitz domain if for every x 0 Ω there exist a new system of coordinates R n 1 R, which is isometric to the original one, along with an open upright cylinder C containing x 0, and a Lipschitz function ϕ : R n 1 R whose graph contains x 0 and is disjoint from the top and bottom lids of C, such that C Ω = C {(x, x n ) R n 1 R : x n = ϕ(x )}, (1.2) C Ω = C {(x, x n ) R n 1 R : x n > ϕ(x )}, (1.3) C (Ω) c = C {(x, x n ) R n 1 R : x n < ϕ(x )}. (1.4) Thus, informally speaking, Ω R n is a Lipschitz domain if in a neighborhood of any boundary point x 0, the surface Ω agrees (in a suitable sense) with the graph of a Lipschitz function ϕ : R n 1 R. Then the outward unit normal has an explicit formula in terms of ϕ namely, in the new system of coordinates, ν(x, ϕ(x )) = ( ϕ(x ), 1) 1 + ϕ(x ) 2, if (x, ϕ(x )) is near x 0, (1.5) where denotes the gradient with respect to x R n 1. The existence of ϕ at almost every point in R n 1 is a consequence of the celebrated Rademacher Theorem. What makes Lipschitz domains both a convenient and a resourceful environment for doing analysis is: (i) the fact that they are still retaining some of the most basic geometric properties of, say, the upper-half space (which, arguably, is the most standard type of domain studied in analysis), and (ii) the fact that this is a very large class, encompassing many irregular structures encountered in nature, such as polygonal domains, domains with corners and edges, etc. 1

7 To substantiate the claim made in (i), we wish to point to the existence of an outward unit normal a.e. on the boundary (as discussed above), the fact that the Divergence Theorem holds in Lipschitz domains, and the fact that the theory of Sobolev spaces in Lipschitz domains is virtually as rich as the one for the upper-half space. As regards (ii), it is illuminating to point out that, as opposed to the class of domains exhibiting corners and edges, the singularities in a Lipschitz domain are not necessarily isolated, and they can even accumulate. One of the main issues addressed in this thesis is determining geometrical criteria for deciding whether a given domain is Lipschitz.(1.6) As is well-known, if Ω R n is a bounded domain, then Ω is Lipschitz if and only if it satisfies a uniform cone property. This asserts that there exists an open, circular, truncated, one-component cone Γ with vertex at 0 R n such that for every x 0 Ω there exist r > 0 and a rotation R about the origin such that x + R(Γ) Ω, x B(x 0, r) Ω. (1.7) An open set Ω R n is called starlike with respect to x 0 Ω if I(x, x 0 ) Ω for all x Ω, where I(x, x 0 ) := the open segment with endpoint x and x 0. Call an open set Ω starlike with respect to a ball B Ω if Ω is starlike with respect to any point in B. One of our main results reads as follows: Theorem 1.1. Let Ω be a bounded, open, non-empty set which is starlike with respect to some ball. Then Ω is a Lipschitz domain. Q Ω P β 2

8 While this result appears to be folklore, no complete written proof has appeared as of now in the literature. Here we address this issue thoroughly and establish the aforementioned result by proceeding in two steps: Step I. If Ω R n is a bounded, open, non-empty set which is starlike with respect to a ball B Ω centered at the origin, then there exists a Lipschitz function ϕ : S n 1 (0, ), (1.8) where S n 1 is the unit sphere in R n centered at the origin, such that, in polar coordinates, Ω = { rω : ω S n 1, 0 r < ϕ(ω) }. (1.9) Step II. If Ω R n is as in (1.9), with ϕ as in (1.8) Lipschitz function, then Ω is a Lipschitz domain (in the sense of the earlier definition). Step I is of purely geometrical nature. At each vector ω S n 1 the function ϕ is defined where ϕ(ω) := the distance from the point in L ω Ω to the origin, (1.10) L ω := {rω : r > 0}. (1.11) The Lipschitz constant of ϕ then turns out to be controlled in terms of the radius of B, and the diameter of Ω. Parenthetically, let us point out that the converse of the claim in Step I also holds. That is, if Ω is as in (1.9) with ϕ as in (1.8) Lipschitz function, then Ω is a starlike domain with respect to some ball B Ω, centered as the origin. In contrast to the proof of Step I, the justification of Step II is analytical in nature. The goal is to show that, near every boundary point, the surface {ϕ(ω)ω : ω S n 1 } agrees with the graph of a real-valued Lipschitz function ψ (after a rigid relocation). Consider, as example, the point ϕ( e n )e n, corresponding to the south pole on S n 1. Careful inspection reveals that the goal just stated amounts to finding ψ : R n 1 R (1.12) such that, if x n = ψ(x ), then ( ) (x, x n ) ϕ (x, x n ) = (x, x n ). (1.13) 3

9 It is therefore natural to try to find the Lipschitz function ψ by solving (1.13) for x n as a function of x. To this end, it is convenient to rephrase (1.13) as where F (x, x n ) = 0, (x, x n ) near (0, ϕ( e n )), (1.14) ( ) (x F (x, x n ) := ϕ 2, x n ) x 2 x 2 (x n, (x, x n ) near (0, ϕ( e n )), (1.15), x n ) While the above reasoning strongly suggests using the Implicit Function Theorem in order to solve (7.10) for x n in terms of x, with x n (0 ) = ϕ( e n ), (1.16) there is one major problem in the implementation of this result. Specifically, in the classical formulation of the Implicit Function Theorem the function in question has to be of class C 1. In our case, since ϕ is Lipschitz, F given in (1.15) is only Lipschitz. This gives us the impetus to reconsider the scope of this basic result in the theory of functions of several real variables, and prove a suitable extension which is capable of handling the current situation. The extension of the Implicit Function Theorem to the class of Lipschitz functions which we establish in this thesis appears to be new, and will most surely prove useful in many other problems. To explain the genesis of this new result, let us first recall the standard form of the Implicit Function Theorem, as presented in virtually any textbook on Advanced Calculus. Specifically, we have: Theorem 1.2. The Implicit Function Theorem Let U be an open set in R m, x 0 U, V an open set in R n, y 0 V. Let F : U V R n be a continuous function such that: (α) F (x 0, y 0 ) = 0; (β) F is class C 1 in U V ; (γ) F y (x 0, y 0 ) is an invertible n n matrix. Then, there exist r > 0 and a function ϕ satisfying (I) ϕ : B(x 0, r) R n ; (II) ϕ is continuous on B(x 0, r), ϕ(x 0 ) = y 0 ; (III) F (x, ϕ(x)) = 0 for all x B(x 0, r); (IV) ϕ is unique with (I)-(III). In addition, ϕ is differentiable and ( F ) 1 (Dϕ)(x) = y (x, y) F (x, y). (1.17) x 4

10 There are several features of Theorem1.2 which stand out. Namely, it is assumed that: (i) Smoothness of F : the function F is of class C 1 (i.e., continuously differentiable in all variables collectively); (ii) Non-degeneracy of F : F y (x 0, y 0 ) is an invertible n n matrix; (iii) Regularity of ϕ: the function ϕ is of class C 1 ; (iv) Uniqueness of ϕ: the only function satisfying (I)-(III) above. Our result which extends the scope of Theorem 1.2 then reads: Theorem 1.3 (The Implicit Function Theorem for Lipschitz Functions). Let U m R m and U n R n open. Next fix a U m and b U n where U := U m U n. Consider F : U m U n R n (1.18) a Lipschitz such that F (a, b) = 0 (1.19) and with the property that there exists a constant K > 0 for which F (x, y 1 ) F (x, y 2 ) K y 1 y 2 for all (x, y j ) U where j = 1, (1.20) 2. Then there exist V m R m open, such that a V m, and a Lipschitz function ϕ : V m U n such that ϕ(a) = b, and {(x, y) V m U n : F (x, y) = 0} = {(x, ϕ(x)) : x V m }. (1.21) In particular, F (x, ϕ(x)) = 0, for all x V m. (1.22) It is of interest to draw a parallel between the features (i) (iv) of the classical formulation of the Implicit Function Theorem above and our version of this result, valid for Lipschitz functions. Concretely, properties (i) (iv) should be contrasted to the following features of Theorem 1.3: (i) Smoothness of F : the function F is Lipschitz in all variables collectively; (ii) Non-degeneracy of F : the function y F (x, y) is bi-lipschitz, uniformly in x; 5

11 (iii) Regularity of ϕ: the function ϕ is Lipschitz; (iv) Uniqueness of ϕ: the only function satisfying (1.21). As this analysis reveals, there are natural parallels between the statements and the conclusions in Theorem 1.2 and Theorem 1.3; heuristically speaking, the latter is a version of the former written by decreasing the amount of regularity assumed from C 1 to Lipschitz. Alas, this comparative analysis is misleading since there are fundamental differences in how the two theorems under discussion are proved. Indeed, while Lipschitz functions are differentiable almost everywhere (by the classical Rademacher Theorem mentioned earlier), the fact that these first-order partial derivatives are merely L functions, typically lacking continuity, renders the standard proof of the classical Implicit Function Theorem virtually useless in the more general context we are considering here. We are therefore forced to adopt a new approach, which makes more economical use of the weaker hypotheses we are adopting. Let us stress that, as already explained, the current extension of is not purely academical, but rather dictated by natural, practical considerations. One ingredient in the proof of Theorem 1.2, which has intrinsic interest, is the so-called Invariance of Domain Theorem, which is a topological result about homeomorphic subsets of Euclidean space R n. This states: Theorem 1.4 (Invariance of Domain Theorem). Let U be an open subset of R n and let f : U R n be an injective continuous function. Then V = f(u) is open and f : U V is a homeomorphism. In particular f 1 : V U is a continuous function. The theorem is due to L.E.J. Brouwer, who published it in 1912; see [1]. The proof uses tools of algebraic topology, notably Brouwer s Fixed Point Theorem. This is a very general and useful result. In fact, we show that it is possible to se it in order to provide a new proof of the Implicit Function Theorem along the line of Invariance of Domain Theorem = Inverse Function Theorem = Implicit Function Theorem. (1.23) The above is a summary of part of the arguments which go into the proof of Step II, stated near the bottom of page 3. Returning to that context, it is possible to check that the hypotheses of Theorem 1.2 are satisfied and, as a result, given x 0 Ω, there exists a Lipschitz function ψ : B R, with B a (n 1)-dimensional ball, (1.24) such that the graph of ψ agrees with Ω near x 0. While this is certainly in the spirit of condition (1.2), for the purpose of fully matching the conditions stipulated in the definition of a Lipschitz domain, we must extend the Lipschitz function ψ, originally defined in B, to the entire R n 1. To do so, we rely on the following abstract extension result (well-known to experts): 6

12 Theorem 1.5 (Extension of Lipschitz Functions). Let (X, d) be a metric space, and assume that A X is a given set and f : A R (1.25) is a Lipschitz function. That is, f satisfies { f(x) f(y) Lip(f; A) := sup d(x, y) Then there exists } : (x, y) A A \ diag <. (1.26) F : X R (1.27) such that F = f and F is Lipschitz with Lip(F ; X) = Lip(f; A). (1.28) A Let us momentarily digress for the purpose of explaining an interesting feature of the above result, which highlights the advantages of having such an extension result stated in the context of abstract metric spaces. Specifically, there holds: Corollary 1.6 (Extension of Hölder Functions). Let S R n be an arbitrary set and fix s (0, 1). Also, let f : S R (1.29) be a Hölder function of order s. That is, { } f(x) f(y) Hol s (f; S) := sup x y s : (x, y) S S \ diag <. (1.30) Then there exists F : R n R (1.31) such that F = f S and F is Hölder with Hol s (F ; R n ) = Hol s (f; S). (1.32) 7

13 Indeed, since for every s (0, 1), x z s ( s x y + y z ) x y s + y z s, for all x, y, z R(1.33) n, the space (S, d) with d(x, y) := x y s becomes a metric space. Since, in the context of (S, d), the function f satisfying (1.31) becomes Lipschitz, the conclusion in Corollary 1.6 follows from Theorem 1.5. Returning now to the mainstream discussion, having to do with the proof of Step II (on page 3), we can use Theorem 1.5 in order to ensure that (1.24) holds. In turn, this now guarantees that condition (1.2) in the definition of a Lipschitz domain has been fully checked. The next step would then, naturally, be to proceed to verify the remaining conditions in this definition, i.e., (1.3), (1.4). Instead of following this course, we choose to adopt a different point of view whose aim is to fully clarify the interrelationships amongst the conditions (1.3), (1.2), (1.4). The result we prove in this regard which is most directly relevant to the current discussion is as follows: Proposition 1.7. For a nonempty open set Ω R n which satisfies Ω = Ω, (1.34) there holds (1.2) = (1.3) and (1.4). (1.35) This is a purely topological result, which is quite useful whenever the task at hand is to show that a given domain is Lipschitz. In our situation (i.e., in the proof of Step II), given what we have proved up to this stage, and since for Ω as in (1.9) Ω = { ϕ(ω)ω : ω S n 1} = Ω, (1.36) (i.e., (1.34) holds), Proposition 1.7 can be employed to finish the proof of Step II. In turn, Step I and Step II conclude the proof of Theorem 1.1. In closing, we wish to point out that a suitable version of Theorem 1.1 can be used to actually fully characterize the class of bounded Lipschitz domains. More specifically, the following holds: Theorem 1.8. Let Ω be a bounded, open, non-empty set. Then Ω is a Lipschitz domain if and only if Ω is locally starlike with respect to balls. The latter condition means that for every x Ω there exists O R n open neighborhood of x, along with a ball B Ω O such that Ω O is starlike with respect to B. 8

14 Indeed, the implication = follows from (1.3)-(1.4), whereas the implication = can be established, without too much difficulty, using Theorem 1.1. We conclude with a brief outline of the contents of the chapters of this thesis. Chapter 2 is devoted to discussing the classical Implicit and Inverse Function Theorems. It is divided into two sections, the first of which contains a proof based on the Contraction Principle (Proposition 2.1). The actual Implicit Function Theorem is stated as Theorem 2.2. Subsequently, we addressed the Inverse Function Theorem, which appears as Theorem 2.3. The second section of Chapter 2 contains a new proof of the aforementioned theorems, based on the Invariance of Domain Theorem (Theorem 2.4). Chapter 3 contains a number of useful results about Lipschitz functions, including: (i) the fact that their distributional derivatives are in L (Proposition 3.1) and, (ii) conversely, if a distribution has a its gradient in L then this is given by a Lipschitz function (Proposition 3.2; see also Theorem 3.3); (iii) the fact that real-valued Lipschitz functions on arbitrary sets can be extended to the entire space with preservation of the Lipschitz constant (Theorem 3.4; cf. also Theorem 3.5 for a related version). Chapter 4 is devoted to presenting a proof of the famous Rademacher Theorem. It is divided into three sections, the first of which contains the actual statement and proof of this result. In the second section of this chapter, we discuss the issue of the almost everywhere differentiability of absolutely continuous functions on the real line (which is used in the proof of Rademacher s theorem). Finally, in the third section we prove the following result: Theorem 1.9. Let U R n be an open set and let f : U R m (1.37) be a bi-lipschitz function. That is, there exist 0 < C 1 C 2 < such that C 1 x y f(x) f(y) C 2 x y, (1.38) for every x, y U. Then there exists E U, of Lebesgue measure zero, such that and Df(x) exists for every x U \ E, (1.39) C 1 v Df(x)v C 2 v, for every v R n, and x U \ E. (1.40) 9

15 As a corollary, rank Df(x) = n for every x U \ E, (1.41) which showes that necessarily m n. (1.42) In addition, m = n = f(u) is an open set. (1.43) It is worth pointing out that, as the above theorem entails, there exist bi-lipschitz mappings f : R n R m m n. (1.44) Chapter 5 is where we present the main novel technical tool of this thesis, namely a version of the Implicit Function Theorem for Lipschitz functions. See Theorem 5.1 for the actual statement. The proof of this result occupies the bulk of Chapter 5. Chapter 6 consists of two sections. In the first, we collect some definitions and generalities about bounded Lipschitz domains in R n. In the second section of this chapter we then proceed to study in greater detail the degree of independence (or rather lack thereof) of the conditions appearing in the definition of a Lipschitz domain (Definition 6.5). See Proposition 6.1 and Proposition 6.2 for precise formulations. The latter result has been rephrased as Proposition 1.7 earlier in the introduction. Chapter 7 is devoted to discussing some applications of our main technical results. It is partitioned into three sections, and the main results, Theorem 7.1 and Theorem 7.2, have been reformulated as Step I and Step II, respectively, earlier in the introduction. Acknowledgments It is a pleasure to thank Simona, Nick, Simon, and James for their special contribution. 10

16 Chapter 2 The Implicit and Inverse Function Theorems 2.1 First Proof Using the Contraction Principle Proposition 2.1 (The Contraction Principle). Let x 0 R m, y 0 R n, r, ρ > 0 and assume that f : B(x 0, r) B(y 0, ρ) R n is a continuous function for which there exists α (0, 1) such that (i) f(x, y ) f(x, y ) α y y for all x B(x 0, r) and y, y B(y 0, ρ) (i.e., for every x B(x 0, r), f(x, ) is a contraction on B(y 0, ρ)), (ii) f(x, y 0 ) y 0 < ρ(1 α) for all x B(x 0, r). Then, there exists a function ϕ with the following properties: 1. ϕ : B(x 0, r) B(y 0, ρ), 2. ϕ is continuous, 3. f(x, ϕ(x)) = ϕ(x) for all x B(x 0, r), 4. ϕ is unique with these properties. Proof. We will prove this result in a sequence of 6 steps. Step I: For x B(x 0, r) arbitrary, define inductively the sequence {ϕ j (x)} j N0 ϕ 0 (x) = y 0 and ϕ j+1 (x) = f(x, ϕ j (x)) for j N 0. Observe that by ϕ j+1 (x) ϕ j (x) = f(x, ϕ j (x)) f(x, ϕ j 1 (x) α ϕ j (x) ϕ j 1 (x)... α j ϕ 1 (x) ϕ 0 (x) = α j f(x, y 0 ) y 0 α j ρ(1 α). (2.1) Step II: We claim that for each x B(x 0, r) the sequence {ϕ j (x)} j N0 is Cauchy. Indeed, if j N 0 and l N, applying the triangle inequality repeatedly and then using (2.1) we can write 11

17 ϕ j+l (x) ϕ j (x) ϕ j+l (x) ϕ j+l 1 (x) + ϕ j+l 1 (x) ϕ j+l 2 (x) + + ϕ j+1 (x) ϕ j (x) α j ρ(1 α)(α l 1 + α l α + 1) = α j ρ(1 α) 1 αl 1 α < αj ρ. (2.2) Hence, since lim j α j ρ = 0, for ε > 0 there exists j 0 N such that 0 < α j ρ < ε for j j 0, which in turn, used in (2.2), shows that if ε > 0 there exists j 0 N such that ϕ m (x) ϕ j (x) < ε for all j, m j 0. This proves the claim. Step III: As seen in Step II, for each x B(x 0, r) the sequence {ϕ j (x)} j N0 is Cauchy, thus convergent. Hence, we can define ϕ(x) := lim ϕ j (x). If we pass to the limit as j j in the identity ϕ j+1 (x) = f(x, ϕ j (x)) (recall f is continuous) it follows that ϕ(x) = f(x, ϕ(x)), proving (1) and (3) in the statement of the proposition. Step IV: (the uniqueness of ϕ) Assume that there exist two functions ϕ, ϕ satisfying (1) (3). Then, for x B(x 0, r), ϕ(x) ϕ (x) = f(x, ϕ(x)) f(x, ϕ (x)) α ϕ(x) ϕ (x), (2.3) which implies that (1 α) ϕ(x) ϕ (x) 0. The latter is only possible if ϕ(x) ϕ (x) = 0, which in turn implies ϕ(x) = ϕ (x). This completes the proof of (4). Step V: We will prove by induction that each ϕ j is continuous. Clearly the constant function ϕ(x) = y 0, x B(x 0, r), is continuous. Suppose ϕ j is continuous for some j N. Since we know that ϕ j+1 = f(x, ϕ j (x)), f is continuous, and the composition of continuous functions is continuous, it follows that ϕ j+1 must be continuous. By the Principle of Mathematical Induction, if follows that ϕ j is continuous for any j N 0. Step VI: (continuity of ϕ) Fix an arbitrary point x B(x 0, r) and ε > 0. By the triangle inequality, if x B(x 0, r), we have for each j N that ϕ(x) ϕ(x ) ϕ(x) ϕ j (x) + ϕ j (x) ϕ j (x ) + ϕ j (x ) ϕ(x ). (2.4) Since ϕ(y) = lim j ϕ j (y) for each y B(x 0, r), there exists j 1, j 2 N such that ϕ(x) ϕ j (x) ε 3 for j j 1, ϕ(x ) ϕ j (x ) ε 3 for j j 2. (2.5) Select j 0 = max{j 1, j 2 }. By the continuity of ϕ j0 at x, there exists δ > 0 such that ϕ j0 (x) ϕ j0 (x ) < ε 3 whenever x x < δ. (2.6) 12

18 Now we set j = j 0 in (2.4) and use (2.5)-(2.6) to conclude that ϕ(x) ϕ(x ) < ɛ whenever x x < δ. This proves that ϕ is continuous at x. Since x was arbitrary in B(x 0, r) we obtain that ϕ is continuous on B(x 0, r). The proof of the proposition is therefore complete at this point. Theorem 2.2 (The Implicit Function Theorem). Let U be an open set in R m, x 0 U, V an open set in R n, y 0 V. Let F : U V R n be a continuous function such that: (α) F (x 0, y 0 ) = 0, (β) F (x, y) exists and is continuous on U V, y (γ) F y (x 0, y 0 ) is an invertible n n matrix. Then, there exist r > 0 and a function ϕ satisfying (I) ϕ : B(x 0, r) R n ; (II) ϕ is continuous on B(x 0, r), ϕ(x 0 ) = y 0 ; (III) F (x, ϕ(x)) = 0 for all x B(x 0, r); (IV) ϕ is unique with (I)-(III). Furthermore, if F is differentiable at (x 0, y 0 ), then ϕ is differentiable at x 0 and ( F ) 1 (Dϕ)(x 0 ) = y (x F 0, y 0 ) x (x 0, y 0 ). (2.7) Proof. Introduce the function defined by f(x, y) = y ( F y (x 0, y 0 )) 1 (F (x, y)) for each (x, y) U V. Observe that f(x 0, y 0 ) = y 0 and, for (x, y) U V, f(x, y) = y F (x, y) = 0. Claim: There exist r, ρ > 0 such that f : B(x 0, r) B(y 0, ρ) R n satisfies the hypothesis of the Contraction Principle with α = 1 2. First we focus on condition (i) from the Contraction Principle. We have 13

19 f(x, y ) f(x, y ) ( F ) 1F ( F ) 1F = y y (x 0, y 0 ) (x, y ) y + y (x 0, y 0 ) (x, y ) ( F ) 1 [ F ] = y (x 0, y 0 ) y (x 0, y 0 ) (y y ) F (x, y ) + F (x, y ) c F y (x 0, y 0 ) (y y ) F (x, y ) + F (x, y ) = c F y (x 0, y 0 )(y y ) F y (x, y 0) (y y ) + F y (x, y 0) (y y ) F (x, y ) + F (x, y ) [ F c y (x 0, y 0 ) F ] y (x, y 0) (y y ) + c F y (x, y 0)(y y ) F (x, y ) + F (x, y ) =: A + B, where c = ( F (x y 0, y 0 )) 1 is a positive constant depending only on F, x 0, y 0, n, m. If we use the component-wise notation F = (f 1, f 2,..., f n ), x = (x 1, x 2,..., x m ) and y = (y 1, y 2,..., y n ), then the entries of the matrix F y are of the form f i for 1 i, j n. y j Hence, A cc n max 1 i,j n f i (x 0, y 0 ) f i (x, y 0 ) y y, (2.9) y j y j with c n > 0 a constant depending only on n. Since for each 1 i, j n the function f i (, y 0 ) is continuous at x 0, there exists r ij > 0 such that y j f i (x 0, y 0 ) f i (x, y 0 ) < 1 whenever x x 0 < r ij. (2.10) y j y j 4cc n If we now choose 0 < r < min r ij, then from (2.9) and (2.10) it follows that 1 i,j n A 1 4 y y whenever x x 0 < r. (2.11) Now we will proceed with the second norm in the very last expression in (2.8). By the Mean Value Theorem, F (x, y ) F (x, y ) = F (x, z) for some z on the line segment y joining y and y (we denote this by z [y, y ]). Hence, (2.8) B c y y sup z [y,y ] cc n y y sup F y (x, y 0) F y (x, y) max z [y,y ] 1 i,j n ) y y (B 1 + B 2 f i (x, y 0 ) f i (x, z) y j y j (2.12) 14

20 where and B 1 := cc n B 2 := cc n sup max z [y,y ] 1 i,j n sup max z [y,y ] 1 i,j n f i (x, y 0 ) f i (x 0, y 0 ) (2.13) y j y j f i (x, z) f i (x 0, y 0 ). (2.14) y j y j Recall that we are assuming x x 0 < r and observe that z [y, y ] B(y 0, ρ) entails z y 0 < ρ. Hence, since each f i is continuous at (x 0, y 0 ), it follows that y j we can choose r > 0 and ρ > 0 small enough such that B 1 < 1 8 and B 2 < 1 8 whenever y, y B(y 0, ρ) and x B(x 0, r). (2.15) Combining (2.8), (2.11), (2.12), and (2.15) we can conclude that f satisfies (i) with the r and ρ selected above and α = 1 2. Next we will show that (ii) in the Contraction Principle holds, that is, f(x, y 0 ) y 0 < ρ, x B(x 2 0, r). We know that f(x 0, y 0 ) = y 0. Fix ε > 0. Since f is continuous at (x 0, y 0 ), there exists δ > 0 such that f(x, y) f(x 0, y 0 ) < ɛ if (x, y) (x 0, y 0 ) < δ. In particular, f(x, y 0 ) y 0 < ɛ if x x 0 < δ. Now choosing ɛ = ρ (with ρ > 0 as 2 selected earlier in the proof) we obtain δ = δ(ρ) > 0 such that f(x, y 0 ) y 0 < ρ/2 if x x 0 < δ. At this point all we have to do is to return to the r determined so far and selected so that r δ and we see that, with this new r, f verifies the hypotheses of the Contraction Principle. Applying the Contraction Principle, we obtain the existence of a function ϕ satisfying (1) (4) in the conclusion of the Contraction Principle. So far we have that f(x 0, ϕ(x 0 )) = ϕ(x 0 ) and f(x 0, y 0 ) = y 0. To conclude that y 0 = ϕ(x 0 ), observe that the function ψ : B(x 0, r) B(y 0, ρ) defined by ψ(x) = ϕ(x) for x x 0 and ψ(x) = y 0 for x = x 0 is continuous and satisfies f(x, ψ(x)) = ψ(x) for all x B(x 0, r). By the uniqueness property 4. in the Contraction Principle, we obtain that ψ = ϕ in B(x 0, r), and in particular, y 0 = ϕ(x 0 ). Hence, conditions (I) (II) in the statement of the Implicit Function Theorem are satisfied. To conclude that F (x, ϕ(x)) = 0 for all x B(x 0, r) (condition (III) in the statement of the Implicit Function Theorem) combine the equivalence f(x, y) = y F (x, y) = 0 with (3) in the Contraction Principle. By a similar reasoning we see that the uniqueness statement in (IV ) follows. We are left with the proof of the statement regarding the differentiability of ϕ under the additional condition that F is differentiable at (x 0, y 0 ). Assume first that we have proved that if F is differentiable at (x 0, y 0 ) then ϕ is differentiable at x 0 (this 15

21 fact will be proved latter), and introduce the function G : B(x 0, r) R m R n, defined by G(x) := (x, ϕ(x)) for x B(x 0, r). Then by (III), F G = 0 on B(x 0, r), and by (II) G(x 0 ) = (x 0, y 0 ). Under the assumption that ϕ is differentiable at x 0, we obtain that G is differentiable at x 0 and, since F is differentiable at G(x 0 ), by the Chain Rule, F G is differentiable at x 0 and D(F G)(x 0 ) = (DF )(G(x 0 )) (DG)(x( 0 ). Hence, ) Im m 0 = (DF )(x 0, y 0 )(DG)(x 0 ). A direct computation reveals that DG(x 0 ) =, Dϕ(x 0 ) while DF (x 0, y 0 ) = ( F (x x 0, y 0 ) F (x y 0, y 0 )). Multiplying the two we obtain 0 = ( F x (x 0, y 0 ) ) ( ) F y (x Im m 0, y 0 ) Dϕ(x) = F x (x 0, y 0 ) + F y (x 0, y 0 ) Dϕ(x 0 ) and the formula (Dϕ)(x 0 ) = ( F y (x 0, y 0 )) 1 ( F x (x 0, y 0 )) now follows. Lastly, we will show that ϕ is differentiable at x 0 if F is differentiable at (x 0, y 0 ). Fix an arbitrary ɛ > 0. We want to find δ > 0 such that ( F ) 1( ϕ(x 0 + h) ϕ(x 0 ) + y (x F 0, y 0 ) x (x 0, y 0 )) h ɛ h (2.16) if h R m is such that h < δ. To this end, we write ( F ) 1( ϕ(x 0 + h) ϕ(x 0 ) + y (x F 0, y 0 ) x (x 0, y 0 )) h ( F ) 1 [ F ( ) = y (x 0, y 0 ) y (x 0, y 0 ) ϕ(x 0 + h) ϕ(x 0 ) + F ] x (x 0, y 0 ) h c F ( ) y (x 0, y 0 ) ϕ(x 0 + h) ϕ(x 0 ) + F x (x 0, y 0 ) h = c DF (x 0, y 0 )(h, ϕ(x 0 + h) ϕ(x 0 )), (2.17) ( ) F since DF (x 0, y 0 ) = x (x F 0, y 0 ) y (x 0, y 0 ) (an n (n + m) matrix). Note that F ((x 0, y 0 ) + (h, ϕ(x 0 + h) ϕ(x 0 ))) = F (x 0 + h, ϕ(x 0 + h)) = 0 whenever x 0 + h B(x 0, r), or equivalently, whenever h < r. Since F (x 0, y 0 ) = 0, under the assumption that h < r, we can write c (DF (x 0, y 0 )) (h, ϕ(x 0 + h) ϕ(x 0 ))) (2.18) ( ) = c F (x 0, y 0 ) + (h, ϕ(x 0 + h) ϕ(x 0 )) F (x 0, y 0 ) DF (x 0, y 0 ) (h, ϕ(x 0 + h) ϕ(x 0 )). 16

22 Because F is differentiable at (x 0, y 0 ), for each ε 1 > 0, there exists δ 1 > 0 such that if (h, ϕ(x 0 + h) ϕ(x 0 )) < δ 1 then ( ) c F (x 0, y 0 ) + (h, ϕ(x 0 + h) ϕ(x 0 )) F (x 0, y 0 ) DF (x 0, y 0 )(h, ϕ(x 0 + h) ϕ(x 0 )) c ε 1 (h, ϕ(x 0 + h) ϕ(x 0 )) c ε 1 ( h + ϕ(x 0 + h) ϕ(x 0 ) ). (2.19) If we denote the norm in the left hand-side of the estimate in (2.16) as LHS, if (h, ϕ(x 0 + h) ϕ(x 0 )) < δ 1 we can write ( F ) 1( ϕ(x 0 + h) ϕ(x 0 ) LHS + y (x F 0, y 0 ) x (x 0, y 0 ) h) c ε 1 ( h + ϕ(x 0 + h) ϕ(x 0 ) ) + c h, (2.20) ( ) 1 where c F = (x y 0, y 0 ) F (x x 0, y 0 ). For the first inequality in (2.20) we have applied the triangle inequality, while the estimate of LHS for the second inequality in (2.20) is due to a combination of (2.17), (2.18), and (2.19). Now we require that ε 1 > 0 is such that c ε 1 < 1. Then (2.20) becomes 2 ϕ(x 0 + h) ϕ(x 0 ) 1 2 h ϕ(x 0 + h) ϕ(x 0 ) + c h, which further implies that ϕ(x 0 + h) ϕ(x 0 ) (1 + 2c ) h, provided h + ϕ(x 0 + h) ϕ(x 0 ) < δ(2.21) 1. Observe that if we impose the conditions h < δ 1 2 and ϕ(x 0 + h) ϕ(x 0 ) < δ 12, then (2.21) holds. Since ϕ is continuous at x 0, there exists δ 2 > 0 such that ϕ(x 0 + h) ϕ(x 0 ) < δ 1 2 whenever h < δ 2. Returning with (2.21) in (2.19) and also recalling (2.17)-(2.18), we see that (with appropriate restrictions on the size of h ) LHS c ε 1 (2 + 2c ) h. (2.22) To arrive at the desired conclusion we impose one more condition on ε 1 to ensure that c ε 1 (2 + 2c ) < ε. The final conclusion is as follows: for each ε > 0, take 0 < ɛ 1 < min{ 1, ɛ }, which yields a corresponding value for δ 2c c(2+2c ) 1, which further provides a δ 2, and if we take δ := min{r, δ 1 2, δ 2 } then (2.16) holds for all h R m is such that h < δ. This concludes the proof of the fact that ϕ is differentiable at x 0. 17

23 Theorem 2.3 ( The Inverse Function Theorem). Let U be an open set in R n, x 0 U, f : U R n be a continuously differentiable function (equivalently f is C 1 in U), and (Df)(x 0 ) is a n n invertible matrix. Then there exists an open set V U with x 0 V, such that f(v ) is open in R n and f V : V f(v ) is a bijective function with a continuous inverse (f V ) 1. Moreover (f V ) 1 is differentiable at f(x 0 ) and D(f V ) 1 (f(x 0 )) = ((Df)(x 0 )) 1. (2.23) Proof. Let us define the function F : U R n R n by F (x, y) := y f(x) for x U, y R n. Then F is C 1 on U R n and if we set y 0 := f(x 0 ) then F (x 0, y 0 ) = 0. In addition, from the expression of F we see that F (x, y) = (Df)(x). Hence, x F (x x 0, y 0 ) = (Df)(x 0 ) is an invertible matrix. We can apply the Implicit Function Theorem to obtain r > 0 and a function ϕ satisfying (i) ϕ : B(y 0, r) R n ; (ii) ϕ is continuous on B(y 0, r), ϕ(y 0 ) = x 0 ; (iii) (ϕ(y), y) U R n for all y B(y 0, r); (iv) F (ϕ(y), y) = 0 for all y B(y 0, r); (v) ϕ is unique with (i)-(iv); (vi) ϕ is C 1 at y 0 and (Dϕ)(y 0 ) = ( F x (x 0, y 0 )) 1 ( F y (x 0, y 0 )). Now set V := ϕ(b(y 0, r)). By (iii) we have that V U and (ii) ensures that x 0 V. Making use of (iv) we see that f(ϕ(y)) = y for all y B(y 0, r). Thus, f(v ) = f(ϕ(b(y 0, r))) = B(y 0, r), which is open. So far we can conclude that the restriction mapping f V : V f(v ) is well defined and onto. To prove that this mapping is one-to-one, let x 1, x 2 V be such that f(x 1 ) = f(x 2 ). From the definition of V it follows that there exist y 1, y 2 B(y 0, r) such that x 1 = ϕ(y 1 ) and x 2 = ϕ(y 2 ). Consequently, f(ϕ(y 1 )) = f(ϕ(y 2 )). However, f(ϕ(y 1 )) = y 1 and f(ϕ(y 2 )) = y 2, thus y 1 = y 2. Applying ϕ to this last equality it follows that ϕ(y 1 ) = ϕ(y 2 ), which is the same as x 1 = x 2. This completes the proof of the fact that f V : V f(v ) is one-to-one. The mapping f V : V f(v ) being bijective is invertible. We claim that (f V ) 1 = ϕ, when we consider ϕ : f(v ) = B(y 0, r) V = ϕ(b(x 0, r)). To prove this claim, let y f(v ). Then ϕ(y) V and f(ϕ(y)) = y. We already This means that ϕ(y) is the unique z V for which f(z) = y, i.e., (f V ) 1 (y) = ϕ(y). We have already seen that F (x x 0, y 0 ) = (Df)(x 0 ). Moreover, by a direct computation, F y 0 (x 0, y 0 ) = I. Hence, ( F ) 1 D(f V ) 1 (f(x 0 )) = Dϕ(y 0 ) = x (x F 0, y 0 ) (x 0, y 0 ) = ((Df)(x 0 )) 1 (2.24). y 0 We are left with proving that V is open. To do so, fix x V. We want to find ρ > 0 such that B(x, ρ) V. From the definition of V, we see that there exists y 18

24 B(y 0, r) such that x = ϕ(y ). This implies that f(x ) = f(ϕ(y )) = y B(y 0, r). Hence, x f 1 (B(y 0, r)) (where f 1 is the preimage of f) and since f is continuous, f 1 (B(y 0, r)) is open. Thus there exists ρ > 0 such that B(x, ρ) f 1 (B(y 0, r)), which implies that f(b(x, ρ)) B(y 0, r). Now apply ϕ to the last inclusion to obtain that ϕ(f(b(x, ρ))) ϕ(b(y 0, r)) = V. If we can prove that ϕ(f(ˆx)) = ˆx for all ˆx B(x, ρ), (2.25) then the conclusion will be that B(x, ρ) V as desired. Finally, to see that (2.25) holds, define ψ : B(y 0, r) R n, by ψ(y) = ϕ(y) if y B(y 0, r), y f(ˆx), and ψ(y) = ˆx if y = f(ˆx) B(y 0, r). Then this function verifies (i)-(iv). By (v), we must have ψ = ϕ, hence ϕ(f(ˆx)) = ψ(f(ˆx)) = ˆx for all ˆx B(x, ρ) which gives (2.25). 2.2 Second Proof Using the Invariance of Domain Theorem In this subsection we present a second proof of the Implicit and Inverse Function Theorems. As a preamble to this we isolate a useful result, called the Invariance of Domain theorem, which is a theorem from topology about homeomorphic subsets of Euclidean space R n. It states: Theorem 2.4 (Invariance of Domain). Let U be an open subset of R n and let f : U R n be an injective continuous function. Then V = f(u) is open and f : U V is a homeomorphism. In particular f 1 : V U is a continuous function. The theorem and its proof are due to L.E.J. Brouwer, published in 1912; see [1]. The proof uses tools of algebraic topology, notably the Brouwer fixed point theorem. We will now prove the Inverse Function Theorem stated in Section 1. Proof of Theorem 2.3. We make the following claim. There exists r > 0 such that f is injective when restricted to B(x 0, r). Proof of Claim. Reasoning by contradiction, we may assume that for all j N, f is not injective when restricted to B(x 0, 1 ). This means that for all j N there exists j x j, y j B(x 0, 1), x j j y j and such that f(x j ) = f(y j ). Define w j := y j x j y j x j Sn 1, j N. (2.26) Also, consider the path γ j (t) := x j + t(y j x j ), 0 t 1 (2.27) and fix an arbitrary vector v R n. Then the function 19

25 [0, 1] t f(γ j (t)) v R (2.28) is of class C 1 and takes the same values at t = 0 and t = 1, i.e. [f(γ j (t)) v] = f(x j ) v t=0 = f(y j ) v = [f(γ j (t)) v]. (2.29) t=1 Rolle s Theorem then implies that there exists a number c j (0, 1) (2.30) such that d dt [f(γ j(t)) v] = 0. (2.31) t=cj Using Chain Rule, we see that d dt [f(γ j(t)) v] = (Df)(γ j (t)) γ j (t) v (2.32) so that, keeping in mind that γ j (t) = y j x j, we arrive at or (Df)(γ j (c j ))(y j x j ) v = 0, (2.33) (Df)(γ j (c j ))w j v = 0. (2.34) Now x j x 0 < 1 j, y j x 0 < 1 j implies that x j x 0, y j y 0 as j. Hence γ(x j ) = x 0 + x j (y j x j ) x 0 (2.35) Since w j S n 1, by passing the subsequence it can be assumed that w j w 0 S n 1 as j for some w 0 S n 1. Passing the limit j in (2.33) we then obtain (Df)(x 0 )w 0 v = 0 for all v R n. (2.36) 20

26 Choosing v := (Df)(x 0 )w 0, this implies (Df)(x 0 )w 0 2 = 0 (2.37) and ultimately (Df)(x 0 )w 0 = 0. (2.38) Since (Df)(x 0 ) is, by assumption, an invertible matrix, this forces w 0 = 0, (2.39) contradicting the fact that w S n 1. This contradiction finishes the proof of the claim made at the beginning of the proof of the theorem. Once this claim has been established, the Invariance of Domain Theorem implies that W := f(b(x 0, r)) (2.40) is an open set in R n and, if V := B(x 0, r) then f : V W (2.41) is a topological homeomorphism (i.e., continuous, bijective, and the inverse function is continuous as well). There remains to prove that and f 1 : W V is differentiable at f(x 0 ) (2.42) (Df) 1 (f(x 0 )) = ((Df)(x 0 )) 1. (2.43) To this end, consider f 1 (f(x 0 ) + h) f 1 (f(x 0 )) ((Df)(x 0 )) 1 h. (2.44) h Our goal is to show that this expression goes to zero as h 0. Pick x h B(x 0, r) such that 21

27 f(x h ) = f(x 0 ) + h, i.e., x h = f 1 (f(x 0 ) + h). (2.45) Then f(x h ) f(x 0 ) as h 0. Thus since f 1 is continuous, then x h = f 1 (f(x h )) f 1 (f(x 0 )) = x 0 as h. Thus, Then x h x 0 as h. (2.46) f 1 (f(x 0 ) + h) f 1 (f(x 0 ) ((Df 1 (f(x 0 ))h h (2.47) = f 1 (f(x h )) f 1 (f(x 0 ) ((Df 1 (f(x 0 ))h h = x h x 0 ((Df 1 (f(x 0 ))h h ( ) ( ) 1 (Df)(x0 )(x h x 0 ) h = (Df)(x 0 ). (2.48) h Hence, for our goals, it suffices to prove that ( ) (Df)(x0 )(x h x 0 ) h 0, as h. (2.49) h Recall that f is differentiable at x 0 means that there exists with η : [0, ) [0, ) (2.50) such that lim η(t) = 0 (2.51) t 0 + f(x 0 + k) f(x 0 ) (Df)(x 0 )k k η( k ) (2.52) for every k R n \ {0}. In our case, take 22

28 k := x h x 0 (2.53) so that (2.52) becomes (applying (2.53) as well) h (Df)(x 0 )(x h x 0 ) x h x 0 η x h x 0. (2.54) This implies x h x 0 = ((Df)(x 0 )) 1 (Df)(x 0 )(x h x 0 ) C (Df)(x 0 )(x h x 0 ) C h (Df)(x 0 )(x h x 0 ) + C h Cη( x h x 0 ) x h x 0 + C h, (2.55) by (2.54). Choose h close to 0 so that η( x h x 0 ) < 1 2C then absorb 1 x 2 h x 0 in the left-hand side to obtain (2.56) x h x 0 2C h (2.57) if h is sufficiently close to 0. Finally, using our previous results we can write the following (Df)(x 0 )(x h x 0 ) h h x h x 0 η( x h x 0 ) h (applying (2.54)) C h η( x h x 0 ) h (applying (2.57)) Cη( x h x 0 ) 0 as h 0, (2.58) since x h x 0 as h 0. (2.59) This completes the proof of Theorem 2.3. Finally, observe that Inverse Function Theorem = Implicit Function Theorem. (2.60) Indeed, the Implicit Function Theorem (as stated in Theorem 2.2) follows immediately from the Inverse Function Theorem (as stated in Theorem 2.3) applied to the function f(x, y) := (x, F (x, y)). (2.61) 23

29 Chapter 3 Lipschitz Functions We begin by recalling the class of Lipschitz functions. Definition (Definition of a Lipschitz Function). Let f : U R m where U is an open set in R n. Then call f Lipschitz if { f(x) f(y) Lip(f; U) := sup x y where Lip(f; U) is the Lipschitz constant. } : (x, y) U U \ diag < (3.1) It follows from the above definition that a function f : U R is Lipschitz if and only if there exists M 0 such that f(x) f(y) M x y for every x, y U. (3.2) The Lipschitz constant Lip(f; U) is, in fact, the smallest M for which (3.2) holds. Remark If the function f is Lipschitz then f is continuous (in fact, uniformly continuous). In turn, any continuous function f belongs L 1 loc. Also, any function f in L 1 loc induces a regular distribution (i.e., f D (R n )). Proposition 3.1. If f : R n R is Lipschitz then the components of f, computed in the sense of distributions, belong to L (R n ), and f L (R n ) M. Proof. Let θ C c (R n ) be such that θ(x)dx = 1 and θ 0. For each number ε > 0, define θ ε (x) := ε n θ ( x ε ), and let fε := f θ ε. Then the following are true: (1) f ε f as ε 0 + uniformly on compact sets; (2) f ε (x) f ε (y) M x y ; (3) f ε C (R n ) and f ε L (R n ) M. 24

30 Let us justify these claims. Claim (1) follows from the fact that f is continuous, as known results. As far as (2) is concerned, we write f ε (x) f ε (y) f(x z) f(y z) θ ε (z) dz M x y, R n (3.3) as desired. Turning our attention to (3), we first note that f ε C (R n ) since by differentiating under the integral sign we obtain α f ε (x) = R n f(y) α θ ε (x y) dy for every multiindex α N n 0. In turn, this implies that, for every j {1,..., n}, j f ε (x) = lim h 0 f ε (x + he j ) f ε (x) h (3.4) exists. Keeping this in mind and invoking (2) we also obtain j f ε L (R n ) M, finishing the proof of the claim made in (3) above. Moving on, for any test function ψ Cc (R n ), we write j f, ψ = f, j ψ = lim ε 0 + f ε, j ψ = lim ε 0 + jf ε, ψ = lim ( j f ε )(x)ψ(x) dx.(3.5) ε 0 + R n Using (3) we get ( j f ε )(x)ψ(x) dx jf ε L (R n ) ψ L 1 (R n ) M ψ L 1 (R n ). (3.6) R n Hence, from (3.6) and (3.5) we obtain that j f, ψ lim sup ( j f ε )(x)ψ(x) dx M ψ L 1 (R n ). (3.7) ε 0 + R n Consequently, j f, ψ M ψ L 1 (R n ) so that the linear assignment C c (R n ) ψ j f, ψ R (3.8) is continuous in the L 1 -norm, and has norm less or equal to M. Since C c (R n ) embeds densely into L 1 (R n ), the linear functional in (3.8) extends to a linear, bounded functional Thus, Λ Λ : L 1 (R n ) R, with norm less or equal to M. (3.9) ( L 1 (R n )) has norm less or equal to M, and since the dual space of L 1 (R n ) is L (R n ), we obtain that there exists a unique 25

31 such that g j L (R n ) with g j L (R n ) M, (3.10) Λ, η = g j (x)η(x) dx, R n η L 1 (R n ). (3.11) Granted this and keeping in mind that Λ is an extension of the linear assignment in (3.8), we arrive at the conclusion that j f, ψ = g j (x)ψ(x) dx, R n ψ Cc (R n ). (3.12) In other words, j f = g j in the sense of distributions in R n. (3.13) In concert, (3.13) and (3.10) give that j f L (R n ) for every j, and j f L (R n ) M. This completes the proof of the proposition. Proposition 3.2. If f is a distribution in R n with the property that f L (R n ), then f is a Lipschitz function. Proof. Let f D (R n ) be as in the statement of the proposition, and define f ε (x) := f, θ ε (x ), x R n, (3.14) where θ ε is as before. It can then be checked that f ε C (R n ) and f ε f in the sense of distributions. (3.15) Furthermore, f ε (x) = f, θ ε (x ) = f,. (θ ε (x )) = f, θ ε (x ) = ( f) ε (x) (3.16) and we claim that ( f) ε L (R n ) satisfies f ε L (R n ) f L (R n ). (3.17) To see this, we estimate 26

32 ( f)(x y)θ ε (y) dy f L (R n ), (3.18) R n which justifies (3.17). Going further, fix x 0 R n and consider the sequence of C functions Then g ε (x) := f ε (x) f ε (x 0 ), for x R n. (3.19) g ε (x) f L (R n ) x x 0 and g ε L (R n ) f ε L (R n ) f L (R(3.20) n ) By the Arzela-Ascoli Theorem, for any compact subset of R n there exists a subsequence {g ε } ε which converges uniformly on it. By a diagonal argument we can assume there exists a subsequence of {g ε } ε which converges uniformly on any compact subset of R n to some function g C 0 (R n ). On the other hand, g ε = f ε f ε (x 0 ) f c in the sense of distributions (where c is a constant). Then we have f = g + c. Combining the fact that f(x) f(y) = g(x) g(y) and f ε (x) f ε (y) g(x) g(y) with f ε (x) f ε (y) f ε L (R n ) x y f L (R n ) x y (3.21) we may finally conclude that f(x) f(y) f L (R n ) x y. This shows that f is a Lipschitz function, as desired. At this stage we would like to summarize the conclusions in Proposition (3.1) and Proposition (3.2) in the form of the following theorem. Theorem 3.3. For a function f : R n R the following two statements are equivalent, (i) f is a Lipschitz function; (ii) in the sense of distributions f L (R n ). Next we consider the issue of extending a given Lipschitz function originally defined on a set A to the entire space, denoted by X. The most general and natural context in which this task can be carried out is that of metric spaces. For the convenience of the reader we first summarize the definition of an abstract metric space. Specifically we have the following: Definition Call (X, d) a metric space if X is an arbitrary set and d : X X [0, ) (3.22) is a function satisfying the following properties: 27

33 (i) non-degeneracy: d(x, y) = 0 x = y; (ii) triangle inequality: d(x, y) d(x, z) + d(z, y) for every x, y, z X; (iii) symmetry: d(x, y) = d(y, x) for every x, y X. In this setting, the Extension result alluded to before reads as follows. Theorem 3.4 (Extension of Lipschitz Functions). Let (X, d) be a metric space such that f : A R is Lipschitz, for A X and A X (3.23) then there exists F : X R (3.24) such that: F = f A and F is Lipschitz with Lip(F ; X) = Lip(f; A). (3.25) Proof. Let us first define the formula where F (x) := inf {f(a) + Md(a, x) : a A} x X, (3.26) M := Lip(f; A). (3.27) Claim 1. If x A then F (x) = f(x). Proof of Claim 1. Fix three arbitrary points a 1, a A and x X. We may then write the following: f(a 1 ) = f(a) + [f(a 1 ) f(a)] (3.28) f(a) + Md(a 1, a) (3.29) f(a) + M[d(x, a 1 ) + d(a, x)] (3.30) = f(a) + Md(x, a 1 ) + Md(a, x). (3.31) Hence combining (3.28) and (3.31) we get the following f(a 1 ) Md(x, a 1 ) f(a) + Md(a, x). (3.32) 28

34 A few justifications are in order for the previous statements. In (3.28) we make use of the additive inverse property. Statement (3.30) makes used of the triangle inequality on metric spaces (X, d). The last two statements are algebraic simplifications. Since a A this implies that f(a 1 ) Md(x, a 1 ) is an arbitrary lower bound for {f(a) + Md(a, x) : a A}. Given the definition of F in (3.26), this gives: f(a 1 ) Md(x, a 1 ) F (x) f(a) + Md(a, x) for all a, a 1 X and x X. (3.33) If x A we may specialize (3.33) to the case when a 1 := x and a := x. In this scenario, (3.33), becomes f(x) 0 F (x) f(x) + 0 i.e. F (x) = f(x) if x A. (3.34) Thus this concludes the proof of Claim 1. Claim 2. F : X R is Lipschitz with Lip (F ; X) = Lip (f; A). Proof of Claim 2. Fix two points x 0, x 1 X. We want F (x 1 ) F (x 0 ) Md(x 0, x 1 ). Hence it is enough to show that for all x 0, x 1 X we have F (x 1 ) F (x 0 ) Md(x 0, x 1 ). Since we can interchange the roles of x 0 and x 1, we obtain F (x 0 ) F (x 1 ) Md(x 0, x 1 ). Thus fix an arbitrary ε > 0. Then there exists a 0 A such that and thus f(a 0 ) + Md(a 0, x 0 ) ε < F (x 0 ). (3.35) also F (x 0 ) < f(a 0 ) Md(a 0, x 0 ) + ε (3.36) F (x 1 ) f(a 0 ) + Md(a 0, x 1 ). (3.37) Combining (3.36) and (3.37) we get F (x 1 ) F (x 0 ) M[d(x 0, x)] < ε. (3.38) Since ε is arbitrary this finishes the proof of Claim 2. Thus (3.32) follows from the proof of Claim 1 and 2. 29

35 Theorem 3.5 (Extension of Lipschitz Functions on R n ). Let (X, d) be a metric space such that: f : A R m is Lipschitz, for A X and A X (3.39) then there exists F : X R m (3.40) such that F = f A and F is Lipschitz with Lip(F ; X) mlip(f; A). (3.41) Proof. Applying Theorem 2.4 we know that there exists an F j : X R such that A F j = f j and Lip(F j ; X) = Lip(f j ; A). (3.42) Now take F := (F 1,, F m ) then F : X R m and F = f. (3.43) A Also, for every x 1, x 2 X we have F (x 1 ) F (x 2 ) = ( m ) 1 2 (F j (x 1 ) F j (x 2 )) 2 j=1 = m Md(x 1, x 2 ). (3.44) This proves that F : X R m is a Lipschitz function with Lip(F ; X) m M, finishing the proof of the theorem. 30

36 Chapter 4 Rademacher s Theorem Here we present a proof of Rademacher s Theorem on the a.e. Lipschitz functions, following [2] differentiability of 4.1 Almost Everywhere Differentiability of Lipschitz Functions Theorem 4.1 (Rademacher). Let f : R n R m be locally Lipschitz. Then f is differentiable L n -a.e., ie gradf(x) = (D 1 f(x),..., D n f(x)) exists and f(y) f(x) grad f(x) (y x) lim y x y x for L n -a.e. x R n. = 0 (4.1) Proof. We assume m = 1 and f is Lipschitz. Fix v R n with v = 1, and define D v f(x) =: lim t 0 f(x + tv) f(x) t for x R n (4.2) if this limit exists. Claim 1. D v f(x) exsists for L n -a.e. x R n. Proof of Claim 1. In fact, since f is continuous, D v f(x) = lim sup t 0 f(x + tv) f(x) t (4.3) is Borel measurable. Why? Def: f is Borel measurable if f 1 (I) is a Borel set for every I R open interval. Fact 1: If f k are Borel measurable k N, then so is f(x) := lim k f k (x) if the limit exists a.e. Fact 2: If f k are Borel measurable for every k, then f(x) = sup k N f k (x) is Borel measurable. Fact 3: Every continuious function is Borel measurable. 31

37 D v f(x) = lim inf t 0 f(x + tv) f(x). (4.4) t And by using the definition of Borel measurable sets on the inverse of the function lim inf t 0 f(x + tv) f(x) f(x + tv) f(x) lim sup t t 0 t on the interval (0, ). (4.5) Thus B v =: {x R n ; D v f(x) does not exsist } = {x R n ; D v f(x) < D v f(x)} (4.6) is Borel measurable. Now for each x, v R n, with v =1, define φ : R R by φ(t) := f(x + tv), t R. (4.7) Then φ is Lipschitz, thus absolutely continous, and thus differentiable L 1 -a.e. Hence for each line L parallel to v, L 1 (B v L) = 0. (4.8) Then Fubini s Theorem implies L n (B v ) = 0. (4.9) As a consequence, we see that grad f(x) =: (D 1 f(x),..., D n f(x)) (4.10) exsists for L n -a.e. x R n. This finishes the proof of Claim 1. Claim 2. Show that D v f(x) = v grad f(x) for L n -a.e x R n. Proof of Claim 2. Assume that η C 0 (R n ) (i.e. η is compactly supported) then f(x + tv) f(x) R t n To see this, split the integrals η(x)dx = f(x) R n 32 η(x) η(x tv) dx. (4.11) t