GEOMETRIC ANALYSIS PIOTR HAJLASZ 1. The Rademacher theorem 1.1. The Rademacher theorem. Lipschitz functions defined on one dimensional inter- vals are differentiable a.e. This is a classical result that is covered in most of courses in mea- sure theory. However, Rademacher proved a much deeper result that Lipschitz functions defined on open sets in Rn are also differentiable a.e. Let us first discuss differentiability of Lipschitz functions defined on one dimensional intervals. This result is a special case of differentiability of absolutely continuous functions. Definition 1.1. We say that a function f :[a; b] ! R is absolutely continuous if for very " > 0 there is δ > 0 such that if (x1; x1 +h1);:::; (xk; xk +hk) are pairwise disjoint intervals Pk in [a; b] of total length less than δ, i=1 hi < δ, then k X jf(xi + hi) − f(xi)j < ": i=1 The definition of an absolutely continuous function reminds the definition of a uniformly continuous function. Indeed, we would obtain the definition of a uniformly continuous function if we would restrict just to a single interval (x; x+h), i.e. if we would assume that k = 1. Despite similarity, the class of absolutely continuous function is much smaller than the class of uniformly continuous function. For example Lipschitz function f :[a; b] ! R are absolutely continuous, but in general H¨oldercontinuous functions are uniformly continuous, but not absolutely continuous. Proposition 1.2. If f; g :[a; b] ! R are absolutely continuous, then also the functions f ± g and fg are absolutely continuous. If in addition g ≥ c > 0 on [a; b], then f=g is absolutely continuous. Exercise 1.3. Prove it. 1 2 PIOTR HAJLASZ It turns out that absolutely continuous functions are precisely the functions for which the fundamental theorem of calculus is satisfied. Theorem 1.4. If f 2 L1([a; b]), then the function Z x (1.1) F (x) = f(t) dt a is absolutely continuous. On the other hand, if F :[a; b] ! R is absolutely continuous, then F is differentiable a.e., F 0 2 L([a; b]) and Z x F (x) = F (a) + F 0(t) dt for all x 2 [a; b]. a Absolute continuity of the function F defined by (1.1) readily follows from the absolute continuity of the integral, but the second part of the theorem is difficult and we will not prove it here. In particular the theorem applies to Lipschitz functions. Theorem 1.5 (Integration by parts). If f; g :[a; b] ! R are absolutely continuous, then Z b Z b 0 b 0 0 f(t)g (t) dt = fgja − f (t)g (t) dt: a a Indeed, fg0 = (fg)0 − f 0g. Integrating this identity and using absolute continuity of fg we obtain Z b Z b Z b Z b 0 0 0 b 0 0 f(t)g (t) dt = (f(t)g(t)) dt − f (t)g(t) dt = fgja − f (t)g (t) dt: a a a a The aim of this section is to prove the following result of Rademacher and its general- izations - the Stepanov theorem and the Kirchheim therem. Theorem 1.6 (Rademacher). If f :Ω ! R is Lipschitz continuous, where Ω ⊂ Rn is open, the f is differentiable a.e. That is the partial derivatives exist a.e. and @f @f rf(x) = (x);:::; (x) @x1 @xn satisfies f(y) − f(x) − rf(x) · (y − x) lim = 0 for a.e. x 2 Ω. y!x jy − xj In the proof we will need the following lemma which is of independent interest. 1 n R Lemma 1.7. If f 2 Lloc(Ω), where Ω ⊂ R is open, and Ω f(x)'(x) dx = 0 for all 1 ' 2 C0 (Ω), then f = 0 a.e. GEOMETRIC ANALYSIS 3 Proof. Suppose to the contrary that f 6= 0 on a set of positive measure. Without loss of generality we may assume that f > 0 on a set of positive measure (otherwise we replace f by −f). Hence there is a compact set K ⊂ Ω and " > 0 such that f ≥ " on K. Let Gi 1 be a decreasing sequence of open sets such that K ⊂ Gi b Ω and let ' 2 C0 (Gi) be such that 0 ≤ 'i ≤ 1, 'i = 1 on K. Then Z Z 0 = f(x)'i(x) dx ≥ "jKj − jf(x)j dx ! "jKj as " ! 0 Ω GinK which is an obvious contradiction. We used here an absolute continuity of the integral: f is integrable on G1 n K and measures of the sets Gi n K ⊂ G1 n K converge to zero. Proof of Theorem 1.6. Let ν 2 Sn−1 and let d Dνf(x) = f(x + tν) dt t=0 n−1 be the directional derivative. For each ν 2 S , Dνf(x) exists a.e., because Lipschitz functions in dimension one are differentiable a.e. 1 If ' 2 C0 (Ω), then for all sufficiently small h > 0 Z f(x + hν) − f(x) Z '(x − hν) − '(x) '(x) dx = − f(x) dx: Ω h Ω −h Although this can be regarded as a sort of integration by parts, it follows easily from a linear change of variables Z Z f(x + hν)'(x) dx = f(x)'(x − hν) dx: Ω Ω The dominated convergence theorem yields1 Z Z (1.2) Dνf(x)'(x) = − f(x)Dν'(x) dx: Ω Ω This is true for any ν 2 Sn−1. In particular Z @f Z @' (x)'(x) dx = − f(x) (x) dx for i = 1; 2; : : : ; n. Ω @xi Ω @xi We want to prove that the directional derivative of f is linear in ν and that it equals rf(x) · ν. The idea is to use the fact that Dν'(x) = r'(x) · ν in (1.2) and this should 1The difference quotients of f are bounded, because f is Lipschitz. 4 PIOTR HAJLASZ somehow translate to a similar property of the derivative of f. We have Z Z Z Dνf(x)'(x) dx = − f(x)Dν'(x) dx = − f(x)(r'(x) · ν) dx Ω Ω Ω n Z n Z X @'i X @f = − f(x) ν dx = '(x) (x)ν dx @x i @x i i=1 Ω i i=1 Ω i Z = '(x)(rf(x) · ν) dx Ω 1 n−1 for all ' 2 C0 (Ω). This and Lemma 1.7 implies that for every ν 2 S Dνf(x) = rf(x) · ν a.e. n−1 Let ν1; ν2;::: be a countable and dense subset of S and let Ak = fx 2 Ω: rf(x) exists, Dνk f(x) exists, and Dνk f(x) = rf(x) · νk:g Each of the sets Ω n Ak has measure zero and hence 1 \ A = Ak satisfies jΩ n Aj = 0: k=1 Clearly Dνk f(x) = rf(x) · νk for all x 2 A and all k = 1; 2;::: We will prove that f is differentiable at every point of the set A. For each x 2 A, ν 2 Sn−1 and h > 0 we define f(x + hν) − f(x) Q(x; ν; h) = − rf(x) · ν: h It suffices to prove that if x 2 A, then for every " > 0 there is δ > 0 such that jQ(x; ν; h)j < " whenever 0 < h < δ, ν 2 Sn−1. Assume that f is L-Lipschitz. Since difference quotients for f are bounded by L we have @f p ≤ L and hence jrf(x)j ≤ nL a.e. @xi Thus for any ν; ν0 2 Sn−1 and h > 0 p jQ(x; ν; h) − Q(x; ν0; h)j ≤ ( n + 1)Ljν − ν0j: Given " > 0 let p be so large that for each ν 2 Sn−1 " jν − ν j ≤ p for some k = 1; 2; : : : ; p. k 2( n + 1)L GEOMETRIC ANALYSIS 5 Since rf(x) · νi = Dνi f(x) for x 2 A, the definition of the directional derivative yields lim Q(x; νi; h) = 0 for all x 2 A and i = 1; 2;::: h!0+ Thus given x 2 A, there is δ > 0 such that jQ(x; νi; h)j < "=2 whenever 0 < h < δ and i = 1; 2; : : : ; p. Now for 0 < h < δ and ν 2 Sn−1 we have " p jQ(x; ν; h)j ≤ jQ(x; ν ; h)j + jQ(x; ν ; h) − Q(x; ν; h)j ≤ + ( n + 1)Ljν − νj < ": k k 2 k The proof is complete. 1.2. The Stepanov theorem. There is a very elegant characterization of functions that are differentiable a.e. due to Stepanov. Theorem 1.8 (Stepanov). Let Ω ⊂ Rn be open. Then a any measurable function f :Ω ! R is differentiable at almost every point of the set jf(y) − f(x)j Af = x 2 Ω : lim sup < 1 : y!x jy − xj As an immediate consequence we obtain Corollary 1.9. A measurable function f :Ω ! R is differentiable a.e. if and only if jf(y) − f(x)j lim sup < 1 a.e. y!x jy − xj We will provide two proofs of the Stepanov theorem. The first one is natural and intuitive, while the second one is tricky and very elegant. We will need some basic properties of Lipschitz functions. Lemma 1.10. If fuαgα2I is a family of Lipschitz functions on a metric space X, then U(x) = sup uα(x) α2I is L-Lipschitz provided it is finite at one point. Also u(x) = inf uα(x) α2I if L-Lipschitz provided it is finite at one point.