and BV loc R N ; R d)

Transcription

1 Necessary and sufficient conditions for the chain rule in W 1,1 loc R N ; R d) and BV loc R N ; R d) Giovanni Leoni Massimiliano Morini July 25, 2005 Abstract In this paper we prove necessary and sufficient conditions for the validity of the classical chain rule in the Sobolev space W 1,1 loc R N ; R d and in the space of functions of bounded variation BV loc R N ; R d. 1 Introduction The purpose of this paper is to settle a classical problem in the theory of Sobolev spaces, namely the validity of the chain rule in W 1,1 ) loc Ω; R d in the vectorial case d > 1. Since the problem is local, in the remaining of the paper we assume, without loss of generality, that Ω = R N. In 1979 Marcus and Mizel [17] proved that given a Borel function f : R d R, the superposition operator u f u maps W 1,1 loc R N ; R d) into W 1,1 ) loc R N if and only if f is Lipschitz continuous resp. locally Lipschitz if N = 1). Since f u W 1,1 ) loc R N the next step is to find a formula for the partial derivatives of f u. In the scalar case, that is when d = 1, the problem has been completely solved in W 1,1 ) loc R N by Serrin [21] in an unpublished paper see also [23], [7] and [15]), where he showed that if f : R R is a Lipschitz continuous function, then for every function u W 1,1 ) loc R N f u) x) = f u x)) u x) for L N a.e. x Ω, 1.1) where the right side of 1.1) is always well defined provided f u x)) u x) is interpreted to be zero whenever u x) = 0, irrespective of whether f u x)) is Department of Mathematical Sciences, Carnegie-Mellon University, Pittsburgh, PA, U.S.A., giovanni@andrew.cmu.edu Department of Mathematical Sciences, Carnegie-Mellon University, Pittsburgh, PA, U.S.A.; Current address: SISSA, Trieste, Italy, morini@sissa.it 1

2 defined. The validity of 1.1) relies on the fact that by Rademacher s Theorem the set Σ f := {u R : f u) does not exist} is L 1 -null and hence, by a result of Serrin and Varberg [22], u x) = 0 for L N a.e. x u 1 Σ f ), 1.2) for every u W 1,1 ) loc R N. The situation is significantly more complicated in the vectorial case, namely when f : R d R is a Lipschitz continuous function with d > 1. In this case, if we fix a basis {e 1,..., e d } in R d not necessarily orthonormal) then the analog of 1.1) becomes 1 x j f u) x) = d u x)) u i x j x), 1.3) where u x)) ui x j x) is interpreted to be zero whenever ui x j x) = 0. By Rademacher s Theorem the set Σ f := { u R d : f is not differentiable at u } 1.4) is L d -null, but the analog of 1.2) is false in general. Hence the right hand side of 1.3) may be nowhere defined. Indeed, let d = 2, N = 1, and consider the functions cf. [15]) f u) := max {u 1, u 2 }, and u x) := x, x) for x R. Then v x) := f u) x) = x so that v x) = 1 while the right-hand side of 1.3) is nowhere defined since u x) = 1, 1). Nevertheless, as shown by Ambrosio and Dal Maso [2], the following weaker form of the chain rule holds for any Lipschitz continuous function f : R d R: 2 Theorem 1.1 Let f : R d R be a Lipschitz continuous function. Then for every function u W 1,1 loc R N ; R d) the composite function v = f u belongs to W 1,1 ) loc R N and for L N a.e. x R N the restriction of the function f to the affine space T u x := { w R d : w = u x) + u x) z for some z R N} is differentiable at u x) and 1 Here for every u R d we write f u) x) = u f T u x ) u x)) u x). 1.5) u = u 1 e u d e d. 2 Theorem 1.1 follows from a more general version for functions of bounded variation. We refer to [2] for the precise statement. 2

3 An alternative proof of the previous result when f is a piecewise C 1 function has been given in [19], where using the special structure of f it is possible to give an explicit formula for the right hand side of 1.5). Theorem 1.1 leaves us with an important open problem: to establish under which additional conditions on the function f the right side of 1.5) coincides with the right side of 1.3), in other words, to find necessary and sufficient conditions on f for the classical chain rule 1.3) to hold. To be precise, when we say that 1.3) holds for a function u W 1,1 loc the partial derivative R N ; R d), we also mean that exists at u x) for L N a.e. x in the set where u i x j does not vanish. The main purpose of this paper is to thoroughly investigate the relation between the validity of the chain rule 1.3) and the structure of the singular set Σ f defined in 1.4)) of the Lipschitz continuous function f. It is clear from the definition that 1.3) depends on the choice of basis in R d. To illustrate this, we begin by considering the special case d = 2. Fix a basis {e 1, e 2 } in R 2 not necessarily orthonormal and let f : R 2 R be a Lipschitz continuous function. It turns out that the classical chain rule 1.3) holds if and only if the singular set Σ f has a one dimensional rectifiable part only in the directions {e 1, e 2 }. Precisely we prove the following result: Theorem 1.2 The classical chain rule 1.3) holds in W 1,1 loc R N ; R 2) with respect to the coordinate system {e 1, e 2 } if and only if for every H 1 -rectifiable set E Σ f and for H 1 a.e. u E either Tan 1 E, u) = span {e 1 } or Tan 1 E, u) = span {e 2 }. 1.6) Here Tan 1 E, u) is the approximate tangent space to the set E at the point u. The deeper part of the result is the necessary condition, whose proof relies on some new differentiability results for Lipschitz functions see Theorems 3.1 and 3.3 below), inspired by recent work of Bessis and Clarke [6]. We recall that an H k measurable set E R d is called countably) H k -rectifiable, 0 k d, if there exists a sequence of Lipschitz functions w n : R k R d such that ) H E k \ w n R k ) = 0. n=1 The analog of condition 1.6) is still sufficient when d 3. Indeed we show that: Theorem 1.3 Let f : R d R be a Lipschitz continuous function, and let {e 1,..., e d } be a basis in R d. Assume that for every countable H 1 -rectifiable set E Σ f and for H 1 a.e. u E, there exists i = 1,..., d depending on u such that Tan 1 E, u) = span {e i }. 1.7) Then the classical chain rule 1.3) holds in W 1,1 loc R N ; R d) with respect to the coordinate system {e 1,..., e d }. 3

4 Condition 1.7) is no longer necessary when d 3 since in this case the chain rule may hold for Lipschitz functions whose singular set is H k -rectifiable with 1 k d 2. Indeed we can prove the following result Theorem 1.4 Let {e 1,..., e d } be an orthonormal basis and let E R d, d 3, be a H d 2 -rectifiable Borel set. Then there exists a Lipschitz continuous function f : R d R such that the singular set Σ f E and H d 2 E \ Σ ) f = 0, and for which the chain rule holds in W 1,1 loc R N ; R d) for any N N with respect to the coordinate system {e 1,..., e d }. It is actually possible to construct f in such a way that the chain rule holds in W 1,1 loc R N ; R d) with respect to any finite family of bases in R d. Note that the case k = d 2 represents the worst possible situation. Indeed we can prove that a necessary but not sufficient) condition for the validity of the chain rule in W 1,1 loc R N ; R d) is that for every H d 1 -rectifiable set E Σ f and for H d 1 a.e. u E, there exists i = 1,..., d 1 depending on u such that Tan d 1 E, u) = span {{e 1,..., e d } \ {e i }}. Nevertheless, we show that this condition becomes necessary and sufficient for the validity of the chain rule in the smaller class A d 1 R N ; R d) of all functions u W 1,1 loc R N ; R d) such that rank u x)) is either zero or greater than or equal to d 1 for L N a.e. x R N see Theorem 4.1 below). Note that A 1 R N ; R 2) = W 1,1 loc R N ; R 2) so that in particular we recover Theorem 1.2. It is important to remark again that all the results presented so far depend on the particular choice of the coordinate system {e 1,..., e d }. We address next the case where the chain rule holds with respect to every coordinate system. In this case Theorem 1.2 clearly indicates that a necessary condition is that the singular set has no H 1 -rectifiable part, that is, it is purely H 1 -unrectifiable. Indeed the second main result of the paper is given by the following theorem Theorem 1.5 Let f : R d R be a Lipschitz continuous function. Then a necessary and sufficient condition for the chain rule 1.3) to hold in W 1,1 loc R N ; R d) with respect to every coordinate system in R d is that Σ f is purely H 1 -unrectifiable. We remark that the sufficiency part of the theorem was already known see [15], [16], [17]), while the necessity part, which in our opinion is the most interesting, is completely new. Note that in the two dimensional case the conclusion of the theorem follows directly from Theorem 1.2, whereas in the higher dimensional case the proof is significantly more involved. A similar result holds in the class A k R N ; R d) of all functions u in the space W 1,1 loc R N ; R d) such that rank u x)) is either zero or greater than or equal to k for L N a.e. x R N. In this case the appropriate necessary and sufficient condition is the pure H k -unrectifiability of the singular set Σ f. Indeed we can show the following: 4

5 Theorem 1.6 Let f : R d R be a Lipschitz continuous function, let 1 k min {N, d}. Then a necessary and sufficient condition for the chain rule 1.3) to hold in A k R N ; R d) with respect to every coordinate system is that Σ f is purely H k -unrectifiable. The final part of the paper is devoted to the extension of some of the results presented above to the space of functions of bounded variation. More precisely, we prove necessary and sufficient conditions for the validity of the classical chain rule in the space of functions of bounded variation BV loc R N ; R d). Besides the intrinsic interest of these results, we hope that the techniques introduced in this paper will be useful in the study of transport equations and hyperbolic systems of conservation laws in several space dimensions, where one is often led to the problem of justifying some kind of chain rule for functions with low regularity, and for which there has been a remarkable and renewed interest in the last few years see e.g. [3] and [8]). 2 Preliminaries In this section we collect some preliminary results which will be used in the sequel. We start with some notation. Here L k and H k are, respectively, the k dimensional Lebesgue measure and the k dimensional Hausdorff measure in Euclidean spaces. We denote by S d 1 the unit sphere in R d. Given f : R d R, for every u, v R d the directional derivative v u) is defined by u) := lim v t 0 f u + tv) f u). t Given a basis {e 1,, e d } in R d we denote by u 1,..., u d ) the components of a given u R d, that is, u = u 1 e u d e d. The directional derivatives in the direction e i are also denoted u i u). If all the derivatives u i u) exist at u R d, we define the vector f u) R d by f u) := u),, ) u). u 1 u d Of course, the existence of f u) does not imply the differentiability of f at u when d > 1. For 1 k < d we shall use the standard notation for ordered multi-indeces I k, d) := { α = α 1,, α k ) N k : 1 α 1 < < α k d }. If α I k, d) we denote by ᾱ I d k, d) the multi-index which complements α in {1,, d} in the natural increasing order. With an abuse of notation we write u = u α, uᾱ) R k R d k, 5

6 where u α := u α1,, u αk ) and uᾱ := ) uᾱ1,, uᾱd k. Given f : R d R, we define ) α f :=,,, ᾱf :=,, u α1 uᾱ1 u αk uᾱd k Next we introduce some basic ingredients in geometric measure theory that will be useful in the remaining of the paper. We refer to [4], [14] and [18] for more details. An H k measurable set E R d is called countably) H k -rectifiable, 0 k d, if there exists a sequence of Lipschitz functions w n : R k R d such that ) H E k \ w n R k ) = 0. n=1 It can be shown that E is H k -rectifiable if and only there exists a sequence {M n } of k-dimensional C 1 manifolds such that H k E \ ). ) M n = ) n=1 Moreover, if E is H k -rectifiable then it admits an approximate tangent space see Def in [4]), which we denote Tan k E, u), for H k a.e. u E, and it can be shown that for H k a.e. u E M n the approximate tangent space to E at u coincides with the tangent space to the manifold M n at u, that is Tan k E, u) = Tan k M n, u). 2.2) We refer to [4] for more details. Let f : R d R be Lipschitz and let M R d be a k-manifold of class C 1. We say that f is tangentially differentiable at u M if the restriction to the affine space u + Tan k M, u) is differentiable at u. The tangential differential, denoted by d M f u), is a linear map between the space Tan k M, u) and R. Remark 2.1 Let f : R d R be Lipschitz and let M R d be a k-manifold of class C 1. If f is tangentially differentiable at u M then for every curve 3 γ : δ, δ) R d with γ 0) = u and γ 0) Tan k M, u) we have f γ) 0) = d M f u) [γ 0)]. Indeed by taking δ smaller if necessary we can write γ t) = γ 1 t) + ot) where γ 1 : δ, δ) M. Since f is Lipschitz we have f γ) t) = f γ 1 ) t) + ot) from which the conclusion follows. 3 Note that the support of γ is not contained in M. 6

7 Theorem 2.2 Let f : R d R be Lipschitz and let M R d be a k-manifold of class C 1. Then f is tangentially differentiable at H k a.e. u M. Proof. For every u M consider a local parametrization ψ : D M of M, with ψ0) = u, where D R k is an open neighborhood of the origin. By Rademacher s Theorem the function f ψ is differentiable for L k almost everywhere in D. Using the Lipschitz continuity, it is easy to see that if f ψ is differentiable at v D, then f is tangentially differentiable at ψv) M. Hence f is tangentially differentiable H k almost everywhere in ψ D). In a similar way we may define tangential differentiability of a Lipschitz function f at points u E where E R d is an H k -rectifiable set, 1 k < d. In this case, the tangential differential, denoted by d E f u), is a linear map between the space Tan k E, u) and R. It can be shown that d E f u) exists for H k a.e. u E. Moreover if f : R d R m with m k then the following Generalized Area Formula holds: Jk E f u) dh k u) = H 0 f 1 v) E ) dh k v), 2.3) E fe) where J E k f u) := det d E f u)) d E f u) ), with d E f u) ) the adjoint of d E f u). An H k measurable set E R d is purely H k -unrectifiable if H k E wr k )) = 0 for any Lipschitz function w : R k R d. Theorem 2.3 Consider a set E R d of finite H k measure. Then E can be decomposed into the disjoint union of a Borel H k -rectifiable set E k rect and of a purely H k -unrectifiable set E k unrect. The decomposition is unique, up to sets of H k measure zero. Purely H k -unrectifiable sets with finite or σ-finite) H k measure may be characterized in a simple way by virtue of the Structure Theorem of Besicovitch- Federer see [18]). In what follows for 0 < k < d we denote by γ d,k the Haar measure defined on the Grasmannian manifold G d, k) of all k-dimensional planes in R d see [18]). We identify each element L G d, k) with the orthogonal projection π L : R d L. Theorem 2.4 Structure Theorem) Let E R d be an H k measurable set with H k E) <. Then i) E is H k -rectifiable if and only if H k π L E 1 ) > 0 for γ d,k a.e. L G d, k), for all H k measurable subsets E 1 of E with H k E 1 ) > 0. 7

8 ii) E is purely H k -unrectifiable if and only if H k π L E) = 0 for γ d,k a.e. L G d, k). Remark 2.5 If Γ is a rectifiable curve on the plane R 2 then it may be proved that any H 1 measurable subset of Γ with positive H 1 measure can project into a set of length zero in at most one direction. Hence if E R 2 is H 1 measurable with H 1 E) < and if there exist two lines L, L 1 G 2, 1) such that H 1 π L E) = H 1 π L1 E) = 0 then E is purely H 1 -unrectifiable. From the previous theorem we then deduce that H 1 π L E) = 0 for γ 2,1 a.e. L G 2, 1). Next we present some simple properties on the differentiability of Lipschitz functions. It is well known that if a function f is differentiable at some point, say the origin, then necessarily f is continuous at 0; the directional derivatives v 0) exist for every v Sd 1 ; for every v S d 1 v 0) = d 0) v i. 2.4) These properties are in general not sufficient to guarantee differentiability at 0. Indeed the function f : R 2 R defined by { v1 if v f v 1, v 1 ) := 2 = v 1 ) 2 ; 0 otherwise is clearly continuous at 0, v 0) = 0 for every v Sd 1, but f is not differentiable at the origin. The situation is quite different if the function f is Lipschitz continuous, as in this case it is easy to verify that if 2.4) holds for every v in a dense subset of S d 1 then f is differentiable at the origin. More precisely we have: Proposition 2.6 Let f : R d R be a Lipschitz function. Then the following conditions are equivalent: 1. f is differentiable at 0; 2. there exists a linear operator L : R d R such that the limit fhν) f0) lim = Lν) h 0 + h exists for all ν in a countable dense subset of R d ; 8

9 3. there exists a countable dense family B of orthonormal bases in S d 1 such that d d 0) e i = 0) ɛ i ɛ i for any two bases {e 1,..., e d } and {ɛ 1,..., ɛ d } in B. Finally the following Lusin-type theorem holds see [14]) Theorem 2.7 Let u W 1,1 R N ). Given λ > 0 there exist a closed set C λ and a function v λ C 1 R N ) such that u = v λ, u = v λ on C λ, R N \C λ CN, d) u λ W 1,1 R N ), v λ W 1, λ. Moreover v λ W 1, R N ) R N \C λ 0, u g λ W 1,1 0 as λ. 3 Differentiability criteria for Lipschitz functions In this section we prove some differentiability criteria for Lipschitz functions. Theorem 3.1 Let f : R d R, d > 1, be a Lipschitz function, let {e 1,..., e d } be a basis in R d and let 1 k < d. Then the following two conditions are equivalent: 1. for every α I k, d) the set Σ f α := { u = u α, uᾱ) R k R d k } : f u α, ) is not differentiable at uᾱ 3.1) is purely H k -unrectifiable; 2. the singular set is purely H k -unrectifiable. Σ f := { u R d : f is not differentiable at u } 3.2) Proof. The implication 2) 1) is trivial. To prove the converse implication, let M R d be a k-manifold of class C 1. We claim that H k M Σ f ) = 0. Fix u 0 M Σ f. By the Implicit Function Theorem, we can find an open neighborhood U of u 0 such that M U { u = u α, uᾱ) R k R d k : uᾱ = ϕu α ) }, 9

10 for some ϕ : R k R d k of class C 1 and for some α I k, d). To prove the claim it is enough to show that H k M Σ f U ) = 0. By hypothesis 1), for H k a.e. u α, ϕu α )) R k R d k, the function f u α, ) is differentiable at ϕu α ). Since the projection is Lipschitz continuous, it follows that for L k a.e. u α R k, the function f u α, ) is differentiable at ϕu α ). Consider the change of variables h : R k R d k R k R d k given by u α, uᾱ) w α, wᾱ) := u α, uᾱ ϕu α )), define the function g : R k R d k R by g w α, wᾱ) := f w α, ϕw α ) + wᾱ), and let L > 0 denote its Lipschitz constant. It is clear that L k { w α R k : the function g w α, ) is not differentiable at 0 }) = ) We claim that L k { w α R k : the function g is not differentiable at w α, 0) }) = ) Fix ν = ν α, νᾱ) S d 1, r R, and n N. Following [6] we define the set { C ν, r, n) := w α R k : > r where g w α, tνᾱ) g w α, 0) t > ν g w α, 0) d α g w α, 0) ν α ) for all t g w α + sν α, sνᾱ) g w α, 0) ν g w α, 0) := lim inf, s 0 + s 0, 1 )}, n and d α g w α, 0) denotes the differential of the function g, 0) at the point w α, which exists for L k a.e. w α R k, by Rademacher s Theorem and the fact that g is Lipschitz. We claim that for every w α C ν, r, n) there exist a constant λ 0, 1) and a sequence t j 0 such that B k w α + t j ν α, λt j ) C ν, r, n) =, 3.5) where B k w α, ρ) denotes the open ball in R k with center w α and radius ρ. The proof of the claim follows closely the argument of Bessis and Clarke see [6]). We present it here for the convenience of the reader. Fix w α C ν, r, n) and let t j 0 be such that g w α + t j ν α, t j νᾱ) g w α, 0) ν g w α, 0) = lim. j t j 10

11 By definition of C ν, r, n) we can find 0 < δ < 2L such that for all j sufficiently large r > 2δ + ν g w α, 0) d α g w α, 0) ν α ) > δ + g w α + t j ν α, t j νᾱ) g w α, 0) t j g w α + t j ν α, 0) g w α, 0) t j 3.6) = δ + g w α + t j ν α, t j νᾱ) g w α + t j ν α, 0) t j, while for z α C ν, r, n) we have g z α, t j νᾱ) g z α, 0) t j > r for all t j < 1 n. Combining this inequality with 3.6) we obtain δ < g z α, t j νᾱ) g w α + t j ν α, t j νᾱ) t j + g w α + t j ν α, 0) g z α, 0) t j 2L t j z α w α + t j ν α ), which gives 3.5), with λ := δ/2l). Let now w α Cν,r,n) be a Lebesgue point for the characteristic function χ Cν,r,n). Since B k w α + t j ν α, λt j ) B k w α, 2t j ), by 3.5) we have 0 = 1 χ Cν,r,n) w α ) = lim j 1 Lk B k w α, 2t j ) C ν, r, n)) L k B k w α, 2t j )) L k B k w α, 2t j ) \ C ν, r, n)) = lim j L k B k w α, 2t j )) lim sup j = lim sup j L k B k w α + t j ν α, λt j ) \ C ν, r, n)) L k B k w α, 2t j )) L k ) k B k w α, λt j )) λ L k B k w α, 2t j )) =, 2 which is clearly a contradiction. Hence L k C ν, r, n)) = ) Let E S d 1 be a countable dense set. In view of 3.7) the set { wα R k : g w α, 0) ν > ν g w α, 0) for all ν E } C ν, r, n) ν E r Q n N 11

12 has zero L k measure. By applying the same argument to the function g and taking into account 3.3), for L k a.e. w α R k we have that g w α, 0) ν = ν g w α, 0) for all ν E. Using now the fact that g is Lipschitz and Proposition 2.6 yields 3.4). Since and ϕ is of class C 1 we have where f u α, uᾱ) = g h 1) u α, uᾱ), H k M Σ f U ) H k h 1 { w α, 0) : w α R k} Σ f ) This concludes the proof. Lip h 1) H k { w α, 0) : w α R k} Σ g) = Lip h 1) L k { w α R k : w α, 0) Σ g}) = 0, Σ g := { w R d : g is not differentiable at w }. Remark 3.2 From the proof of the previous theorem it is clear that if M R d is a k-manifold of class C 1 and Π is a d k)-plane such that f restricted to u + Π is differentiable at u for H k a.e. u M and Tan k M, u) + Π = R d for every u M, then f is differentiable at H k a.e. u M. For the applications to the chain rule in Sobolev spaces we will need the following variant of the previous theorem. Theorem 3.3 Let f : R d R, d > 1, be a Lipschitz function, let {e 1,..., e d } be a basis in R d and let 1 k < d. Assume that for every α I k, d) and for every H k -rectifiable set E Σ f α, where Σ f α is the set defined in 3.1), we have Tan k E, u) = span {e α1,..., e αk } 3.8) for H k a.e. u E. Then for every H k -rectifiable set E Σ f and for H k a.e. u E, there exists α I k, d) depending on u such that Tan k E, u) = span {e α1,..., e αk }. 3.9) Proof. Fix an H k -rectifiable set E Σ f. By 2.1) and 2.2), we may assume, without loss of generality, that E M, where M is a k-manifold of class C 1, and that Tan k E, u) = Tan k M, u) for all u E. Thus to prove 3.9) it suffices to show that the set { } E 1 := u E : Tan k M, u) span {e α1,..., e αk } for every α I k, d) 12

13 has H k measure zero. Fix u E 1. Since M is of class C 1, we may find an open neighborhood U of u such that Tan k M, z) span {e α1,..., e αk } 3.10) for all z M U and for every α I k, d). It is enough to show that H k Σ f M U ) = 0. By hypothesis, it follows from 3.10) that H k Σ f α M U ) = 0 for every α I k, d). We can now continue as in the proof of Theorem 3.1. Remark 3.4 We remark that if k = 1 then condition 3.8) is equivalent to the following one: for every α I 1, d) and for every H 1 -rectifiable set E Σ f α we have H 1 Πᾱ E)) = 0, 3.11) where Πᾱ : R d R d 1 is the projection defined by u = u α, uᾱ) uᾱ. This follows from the Generalized Area Formula 2.3) J 1 Πᾱ u) dh 1 u) = H 0 Π 1 ᾱ v) E ) dh 1 v). 3.12) E A simple calculation shows that Π αe) J 1 Πᾱ u) = τ u))ᾱ, 3.13) where τ u) is the tangent unit vector to E at u. Assume now that 3.11) holds, then from 3.12) and 3.13) we have that τ u))ᾱ dh 1 u) = 0 E which implies that τ u))ᾱ = 0 for H 1 a.e. u E, that is 3.8). Conversely if 3.8) is satisfied then τ u))ᾱ = 0 for H 1 a.e. u E therefore H 0 Π 1 ᾱ v) E ) dh 1 v) = 0. ΠᾱE) Since H 0 Π 1 ᾱ v) E ) 1 for every v Πᾱ E), we deduce that 3.11) holds. 13

14 4 Chain rule in W 1,1 loc R N ; R d) In this section we prove the main results of the paper, namely we give sufficient and necessary conditions for the validity of the chain rule in W 1,1 loc R N ; R d). We begin by studying the validity of the chain rule with respect to a fixed basis {e 1,..., e d } in R d. We recall that A k R N ; R d) is the class of all functions u in the space W 1,1 loc R N ; R d) such that rank u x)) is either zero or greater than or equal to k for L N a.e. x R N. Theorem 4.1 Let f : R d R be a Lipschitz continuous function, {e 1,..., e d } be a basis in R d, and let 1 k min {N, d}. Assume that for every H k - rectifiable set E Σ f and for H k a.e. u E there exists α I k, d) depending on u such that Tan k E, u) = span {e α1,..., e αk }. 4.1) Then the classical chain rule 1.3) holds in the class A k R N ; R d) with respect to the coordinate system {e 1,..., e d }. Moreover, if k = d 1, then condition 4.1) is also necessary for the validity of 1.3) in the class A d 1 R N ; R d). Remark 4.2 i) Since H d = L d it follows by Rademacher s Theorem that H d Σ f ) = 0 and so condition 4.1) is automatically satisfied for every Lipschitz function f. Hence Theorem 4.1 implies in particular that for any Lipschitz function f the classical chain rule always hold in A d R N ; R d). ii) Note that when Σ f has σ-finite H k measure then in view of Theorem 2.3 it suffices to verify condition 4.1) for the H k -rectifiable part of Σ f, that is for Σ f ) k rect. We begin with some preliminary lemmas. Lemma 4.3 Under the hypotheses of Theorem 4.1 the chain rule holds in the class A k R N ; R d) if and only if for every function u C 1 R N ; R d) it holds L N a.e. in the set { x R N : either rank u x)) k or u x) = 0 }. 4.2) Proof. Assume that the chain rule holds in the class A k R N ; R d) and let u C 1 R N ; R d). Since f is Lipschitz, for every x R N such that u x) = 0 { it is clear that f u) x) = 0, so that the chain rule always holds on the set x R N : u x) = 0 }. To prove it in the set A := { x R N : rank u x)) k } fix x 0 A and let m := rank u x 0 )). We claim that there exists B x 0, r) A and a function v A k R N ; R d) such that u v on B x 0, r) and rank v) m in R N. Indeed it is enough to take v x) := ϕ x) u x) + 1 ϕ x)) u x 0 ) + u x 0 ) x x 0 )) 14

15 where ϕ C 1 c B x 0, 2r)) is such that ϕ 1 on B x 0, r) and ϕ C r. Then v x) = u x 0 ) + ϕ x) u x) u x 0 )) 4.3) + ϕ x) u x) u x 0 ) u x 0 ) x x 0 )). Clearly rank v x)) = m for all x R N \ B x 0, 2r). Since u is of class C 1 for every ε > 0 we may find r > 0 so small that u x) u x 0 ) ε, u x) u x 0 ) u x 0 ) x x 0 ) ε x x 0 for all x B x 0, 2r). Hence from 4.3) we obtain that v x) u x 0 ) Cε 4.4) for all x B x 0, 2r). Find α Im, N) and β Im, d) such that, with the usual notation x = x α, xᾱ) R m R N m and u = ) u β, u β R m R d m det α u β x 0 ) 0. Using the inequality det A det B C m) A B A m 1 + B m 1) which holds for A, B R m m, from 4.3) and 4.4) we obtain det α v β x) det α u β x 0 ) Cε for all x B x 0, 2r). By taking ε and in turn r) sufficiently small, we conclude that rank v x)) m for all x R N. To prove the opposite implication assume by contradiction that there exists a function u A k R N ; R d) for which the chain rule fails in a set F R N of positive measure. By Theorem 2.7 there exists a function v C 1 R N ; R d) which coincides with u in a subset of F of positive measure. This is clearly a contradiction. Lemma 4.4 Under the hypotheses of Theorem 4.1 if Σ f contains an H m - rectifiable subset E with H m E) > 0 then, necessarily, m k. Proof. Indeed let E be as above and assume by contradiction that m > k. By 2.1) and 2.2), we may assume, without loss of generality, that E M, where M is an m-dimensional manifold of class C 1, and that Tan m E, u) = Tan m M, u) for all u E. Clearly, H m M Σ f ) > ) 15

16 After a translation and a rotation, and by taking M smaller if necessary we may assume that 0 is a point of H m density 1 in M Σ f and that where M Graph g, 4.6) Graph g := { v = v β, v β) R m R d m : v β = g v β ) }, and g : R m R d m is a function of class C 1 with g 0) = 0, g 0) = 0, and β = 1,..., m). Without loss of generality it is also clear that we may assume that the k-plane L 0 := { v = v 1,, v k, 0,, 0) R d : v 1,, v k R } is not a coordinate plane with respect to the old coordinate system {e 1,..., e d }. For w = w k+1,, w m ) R m k let L w and M w denote respectively the k-plane L w := { v = v 1,, v k, w k+1,, w m, 0,, 0) R d : v 1,, v k R } and the k-manifold M w := M Graph g Lw ). Since Tan k M 0, 0) = L 0, by continuity we can assume that Tan k M w, u) is not a coordinate plane with respect to the old coordinate system {e 1,..., e d } for all w R m k and all u M w with u, w < ε 0 for some ε 0 > 0. To conclude the proof it suffices to show that there is w R m k with w < ε 0 such that H k M w Σ f B 0, ε 0 ) ) > 0. Indeed this would contradict 4.1). By condition 4.5) and by the fact that 0 is a point of H m density 1 in M Σ f it follows that the projection P of M Σ f on the m-plane v β = 0 has positive L m measure and that 0 R m is a point of L m density 1 for P. By Fubini s Theorem there exists w = w k+1,, w m ) R m k with w < ε 0 such that H k L w P B 0, ε 0 )) > 0. Hence H k M w Σ f B 0, ε 0 ) ) > 0 and the proof is concluded. Lemma 4.5 Let u W 1,1 loc R N ; R d) and assume that there exists a measurable set F of positive measure such that rank u x)) k for all x F, for some 1 k min {N, d}. Then there exists a k-dimensional manifold M R d of class C 1 such that H k M u F )) > 0. Proof. As in Lemma 4.3 we may assume without loss of generality that u C 1 R N ; R d). Let x F be a Lebesgue point for the characteristic function χ F. Since rank u x)) k we may find α Ik, N) and ε > 0 such that, with the usual notation x = x α, xᾱ) R k R N k, rank α u x α, xᾱ)) = k 4.7) 16

17 for every x α B k x α, ε) and xᾱ B N k xᾱ, ε). Set A := B k x α, ε) B N k xᾱ, ε). Note that since x is a Lebesgue point we have that L N A F ) > ) By Fubini s Theorem and 4.8) it is easy to see that there exists ˆxᾱ B N k xᾱ, ε) such that L k {x α B k x α, ε) : x α, ˆxᾱ) F }) > ) It is clear that is a k-dimensional manifold such that M := {u x α, ˆxᾱ) : x α B k x α, ε)} H k M u F )) > 0. As a corollary of the previous lemma we obtain the following characterization of purely H k -unrectifiable Borel sets, which can be considered as an extension of a classical result of Serrin and Varberg [22]. Corollary 4.6 A Borel set E R d is purely H k -unrectifiable, 1 k d, if and only if for every N k for every u A k R N ; R d). u = 0 L N a.e. in u 1 E) Proof. Assume that E R d is purely H k -unrectifiable. We claim that L N u 1 E) { u 0} ) = ) for every u A k R N ; R d). Indeed if L N u 1 E) { u 0} ) > 0 for some u A k R N ; R d) then, since rank u) k L N a.e. in u 1 E) { u 0}, by the previous lemma, we may find a k-dimensional manifold M R d such that H k M E) > 0, which contradicts the fact that E is purely H k -unrectifiable. Conversely assume that 4.10) holds and let M R d be a k-dimensional manifold. We claim that H k M E) = 0. If not then we can find a local parametrization ψ : D R k M of class C 1 such that H k ψ D) E) > 0 and rank u) = k. 17

18 This implies that L k D ψ 1 E) ) > 0. Reasoning as in the first part of the proof of Lemma 4.3, without loss of generality we may assume that ψ A k R k ; R d) and thus we have a contradiction to 4.10). We are now ready to prove Theorem 4.1. Proof of Theorem 4.1. By Lemma 4.3 it suffices to prove that for every u C 1 R N ; R d) the chain rule holds L N a.e. in the set { x R N : either rank u x)) k or u x) = 0 }. Since f is Lipschitz it is clear that x j f u) x) = 0 whenever u x) = 0. Moreover if u x) / Σ f Hence it remains to show the chain rule in set then there is nothing to prove. R k := u 1 Σ f ) { x R N : rank u x)) k }. 4.11) We claim that for L N a.e. x R k the rank of u x) is k. Indeed if this is not the case then there exist a set F R k with L N F ) > 0 and m > k such that rank u x)) = m in F. Then by Lemma 4.5 there exists an m-dimensional manifold M R d of class C 1 such that H m M Σ f ) > 0, which contradicts Lemma 4.4. Next we prove that for L N a.e. x R k the affine space T u x := { w R d : w = ux) + u x) v for some v R N} is parallel to a coordinate k-plane. For every x R k there exists β Ik, N) such that rank β u x)) = k. Since u is of class C 1 it follows that rank β u x)) = k for all x A := B k x β, ε) B N k x β, ε ) for some ε > 0. By the previous claim it follows that for L N a.e. x R k A T u x = { w R d : w = ux) + β u x) v for some v R k}. Therefore by Fubini s Theorem for L N k a.e. z B N k x β, ε ) u y, z) + Tan k M z, u y, z)) = T u y,z) 4.12) for L k a.e. y B k x β, ε ) such that y, z) R k, where M z is the k-dimensional manifold M z := {u y, z) : y B k x β, ε)}. Fix z B N k x β, ε ) for which 4.12) holds. By the assumption 4.1) it follows that for L k a.e. y B k x β, ε ) with y, z) R k there exists α I k, d) such that T u y,z) = uy, z) + span {e α 1,..., e αk }. 4.13) 18

19 Moreover by Theorem 2.2 we may assume that for the same set of y s there exists the tangential differential d M z f u y, z)). Hence, by Remark 2.1 applied to the curve γ t) := u x + te j ), by 4.12), and 4.13), for all such points x = y, z) we have [ ] u f u) x) = d Mz f u x)) x) x j x j which, since by 4.13) = k u l x j x) = 0 ux)) u α i x) u αi x j for all l / {α 1,..., α k }, implies that the chain rule holds for L k a.e. y B k x β, ε ) with x = y, z) R k. As this is true for L N k a.e. z B N k x β, ε ), the proof of the first part of the theorem follows from Fubini s Theorem. Finally we show that if k = d 1, then condition 4.1) is also necessary for the classical chain rule to hold in the class A d 1 R N ; R d). By Theorem 3.3 it is enough to show that for every α I d 1, d) and for every H d 1 -rectifiable set F Σ f α, where we have Σ f α := { u R d : uᾱ } does not exists at u, Tan d 1 F, u) = span { e α1,..., e αd 1 } 4.14) for H d 1 a.e. u F. Assume by contradiction that there exist α I d 1, d) and a d 1)- dimensional manifold M R d such that H d 1 M Σ f α) > 0 and 4.14) fails on a subset E M Σ f α with H d 1 E) > 0. Consider a local parametrization ψ : D R d 1 M of class C 1 such that and in D. Since 4.14) fails in E we have that H d 1 ψ D) E) > ) rank ψ) = d ) ψᾱ 0 in ψ 1 E). 4.17) Reasoning as in the first part of the proof of Lemma 4.3, without loss of generality we may assume that ψ A d 1 R d 1 ; R d). Let u x) := ψ x 1,..., x d 1 ) x R N. 4.18) 19

20 Then u A d 1 R N ; R d) and by 4.17) uᾱ 0 in ψ 1 E) R N d+1, while by 4.15), 4.16) we have that L N ψ 1 E) R N d+1) > 0. Hence in this set the chain rule fails since by definition of Σ f α the partial derivative uᾱ does not exists at u x). From the previous proof it follows that Corollary 4.7 Let F be a Borel subset of R d and let α I k, d). Then the following two conditions are equivalent: i) for every H k -rectifiable subset E F and for H k a.e. u E we have Tan k E, u) = span {e α1,..., e αk } ; 4.19) ii) for every u A k R N ; R d) and for L N a.e. x u 1 F ) we have uᾱx) = 0. Corollary 4.8 Let f : R 2 R be a Lipschitz continuous function and let {e 1, e 2 } be a basis in R 2. Then the classical chain rule holds in W 1,1 loc R N ; R 2) with respect to the coordinate system {e 1, e 2 } if and only if for every H 1 -rectifiable set E Σ f and for H 1 a.e. u E either Tan 1 E, u) = span {e 1 } or Tan 1 E, u) = span {e 2 }. 4.20) Remark 4.9 By Remark 3.4 condition 4.20) is equivalent to requiring that for all H 1 -rectifiable sets E i Σ f i, i = 1, 2, where Π i : R 2 R is the projection H 1 Π 1 E 2 )) = 0 and H 1 Π 2 E 1 )) = 0 u = u 1, u 2 ) u i, and, we recall Σ f i := {u R 2 : } does not exist at u. If k < d 1, then condition 4.1) is not necessary for the validity of the chain rule, as the following theorem shows Theorem 4.10 Let {e 1,..., e d } be a basis and let E R d, d 3, be an H d 2 - rectifiable Borel set. Then there exists a bounded Lipschitz continuous function f : R d R such that the singular set Σ f E and H d 2 E \ Σ ) f = 0, and for which the chain rule holds in W 1,1 loc R N ; R d) for any N N with respect to the coordinate system {e 1,..., e d }. 20

21 Proof. For simplicity we assume that {e 1,..., e d } is an orthonormal basis. Step 1: Assume first that E is a compact set contained in { u R d : ϕu) = 0 }, where ϕ : R d R 2 is a function of class C 1 with rank ϕ) = 2 in E. We show that for every u 0 E it is possible to construct a bounded Lipschitz continuous function f such that Σ f = E B u 0, r) for some small r and for which the chain rule holds. Consider a bounded Lipschitz continuous function g : R 3 R, with g C 1 R 3 \ {0} ), such that g,, 0) is not differentiable at the origin but admits all directional derivatives, and such that d ) for all v Λ i {0} where g 0) = g 0) v 4.21) v Λ i := {w = cos θ, sin θ) : θ θ i < ε, θ + θ i < ε}, θ i are the angles corresponding to the vectors ϕ u i u 0 ), and ε is chosen so small that the sets Λ i are pairwise disjoint. 4 Since ϕ is of class C 1 for r sufficiently small we have that ϕ u u) i ϕ Λ i 4.22) u i u) for all u B u 0, r). For u B u 0, 2r) define f u) := g ϕ u), δ 2 u) ), where δ ) is the regularized distance from the set K := E B u 0, r) see [24]). It is well known that δ is a Lipschitz continuous function with δ C R d \K ), for all u R d \ K. 1 dist u, K) δ u) C dist u, K) C 4 An example of a function satisfying all the desired properties is given by g x, y, z) := 8 >< >: Q x 2 y 3 d y cosθ i +ε) x sinθ i +ε)) 2 y cosθ i ε) x sinθ i ε)) 2 if x, y) / D, x 4+4d +y 4+4d +z 2 0 otherwise, where D is the set of all points of R 2 whose angle belongs to Λ i for some i. 21

22 Extend f as a bounded C 1 function on R d \B u 0, r). We claim that Σ f = K. It is clear that Σ f K. To prove the opposite inequality fix ū K and let ξ S d 1. Then ) g ϕ ū) + t ϕ ξ ū) + ot), ot) g ϕ ū), 0) f ū + tξ) f ū) t = ) t g ϕ ū) + t ϕ ξ ū), 0 g ϕ ū), 0) = + o1) t 4.23) where we have used the facts that g is Lipschitz, ϕ is of class C 1, and δ 2 C 1 R d) with δ 2 = 0 on K. Hence g ū) = ϕ ū), 0) ϕ ξ ν ϕ ξ ū), 4.24) where ϕ ν ϕ := ξ ū) ϕ, 0. ξ ū) Since g,, 0) is not differentiable at the origin by Proposition 2.6 we may find a direction ν = w, 0) S 2 \ d Λ i such that g 0) g 0) ν. ν Using the fact that rank ϕ) ū) = 2 we may find a direction ξ 0 S d 1 such that ϕ ξ 0 ū) = w ϕ ξ 0 ū) 0. Moreover by 4.21), 4.22), and 4.24) ) ϕ f ū) ξ 0 = g 0) ū), 0 = g 0) ν ϕ ξ 0 ū) ξ 0 g ν 0) ϕ ū) ξ 0 = ū) ξ 0 and so f is not differentiable at ū. This shows that Σ f = K. Next we prove that the chain rule holds for f. As in the proof of Theorem 4.1, to show the validity of the chain rule in W 1,1 loc R N ; R d) it is enough to prove it for any u C 1 R N ; R d). Moreover it is clear that we can assume, without loss of generality, that N = 1. If u x) / Σ f then there is nothing to prove, hence assume that u x) Σ f. As in 4.23) we have that f u x + t)) f u x)) t = g ϕ u) x) + t ϕ u) x), 0 ) g ϕ u) x), 0) + o1) t g ϕ u) x + t), 0) g ϕ u) x), 0) = + o1). t 22

23 Hence f u) x) = g,, 0) ϕ u)) x) for every x u 1 Σ f ). Since the chain rule is valid for g,, 0) by Theorem 4.1, it follows that for L 1 a.e. x u 1 Σ f ) f u) x) = α g ϕ u) x), 0) ϕ u) x) = α g ϕ u) x), 0) ϕ u x))) u x) = f u x)) u x) where α = 1, 2) and in the last equality we have used the fact that u x)) = α g ϕ u) x), 0) ϕ u x), 0) u i u i which follows from 4.21), 4.22), and 4.24). Therefore the chain rule holds. Step 2: Assume now that E R d is a H d 2 -rectifiable set, that is E = K j N j=1 where H d 2 N ) = 0 and the sets K j are disjoint compact subsets of d 2)- manifolds of class C 1. Fix j N. By Step 1 it is clear that for each u K j we can find r u > 0 such that for all 0 < r < r u there exists a bounded Lipschitz continuous function f u,r with Σ f u,r = K j B u, r) and for which the chain rule holds. The union of all such balls for u K j is a fine cover for K j and hence, by Vitali-Besicovitch covering Theorem see e.g. Thm in [4]) we can find a countable sequence of disjoint closed balls B u n, r n ) such that ) H K d 2 j \ B u n, r n ) = 0. n=1 By repeating the same procedure for each K j it is clear that we can find a sequence of bounded Lipschitz continuous functions f n for which the chain rule holds, and such that the sets Σ f n are pairwise disjoint compact subsets of E with ) H E d 2 \ = 0. n=1 Moreover we can assume that the partial derivatives of every f n exist everywhere in R d see Step 1). We may now define f u) := n=1 Σ f n 1 2 n L n f n u), L n := f n W 1, R d ). It is clear that f is bounded and Lipschitz continuous. We claim that Σ f = Σ f n. 4.25) n=1 23

24 Indeed, if u / Σ f n n=1 then for any ξ S d 1 ξ u) = n=1 1 2 n L n n ξ u) = n=1 1 2 n L n f n u) ξ = f u) ξ, 4.26) where we repeatedly used the fact that we can differentiate term-by-term. By Proposition 2.6 and since f is Lipschitz continuous it follows that u / Σ f. Conversely if u Σ fn then there exist a unique n 0 N such that u Σ fn 0. Write n=1 f = f n0 + n n n L n f n. Arguing as in 4.26) we deduce that the function 1 2 n L n f n is differentiable n n 0 at u. Since u Σ fn 0 it follows that u Σ f. Hence 4.25) holds. Finally to show that the chain rule holds for f let u C 1 R; R d). Since for L 1 a.e. x R the functions f n u are differentiable at x and f n u) x) = f n u x)) u x) for all n, using once more term-by-term differentiation we conclude that f u) x) = f u x)) u x) for L 1 a.e. x R. This concludes the proof. Remark 4.11 i) It is clear from the previous proof that given any finite family E of bases in R d and any H d 2 -rectifiable set E R d, d 3, one can construct a Lipschitz continuous function f : R d R such that the singular set Σ f E and H d 2 E \ Σ ) f = 0, and for which the chain rule holds in W 1,1 loc R N ; R d) with respect to every coordinate system {e 1,..., e d } in E. ii) It is clear that the above construction can be carried on for every 1 k d 2. We considered the case k = d 2 because it represents somehow the worst possible situation. All the results presented so far depend on the particular choice of the coordinate system {e 1,..., e d }. We address next the case where the chain rule holds with respect to every coordinate system. Theorem 4.12 Let f : R d R be a Lipschitz continuous function, let 1 k min {N, d}. Assume that the set Σ f is not purely H k -unrectifiable. Then there exists a coordinate system in R d for which the classical chain rule 1.3) fails in the class A k R N ; R d). Hence a necessary and sufficient condition for the chain rule 1.3) to hold in A k R N ; R d) with respect to every coordinate system is that Σ f is purely H k -unrectifiable. 24

25 Remark 4.13 Note that if k = d 1 then the result follows immediately from Theorem 4.1. Indeed, since 4.1) must hold for every coordinate system, it is clear that Σ f must be purely H d 1 -unrectifiable. Proof of Theorem The sufficiency part of the statement follows immediately from Theorem 4.1. The rest of the proof is devoted to showing the necessity part. Step 1: We consider the case k = 1. Let γ : [0, 1] R d be a C 1 regular curve. For every t [0, 1] denote the unit tangent vector by τ t) := Fix t 0 [0, 1] and a unit vector e different from ±τ t 0 ) and not orthogonal to τ t 0 ). Let e be the hyperplane orthogonal to e and let σ t) be the unit vector obtained by projecting τ t) on e and normalizing. We claim that for L 1 a.e. t in a neighborhood I of t 0 the function f restricted to the plane γ t) + span {e, τ t)} is differentiable at γ t). Indeed considering an orthonormal basis {e 1,..., e d 1, e} by the validity of the chain rule and the fact that e τ t 0 ) 0 it follows that the partial derivative e γ t)) exists for L1 a.e. t near t 0. If τ t) τ t 0 ) for all t near t 0 then since γ t) γ t). by Rademacher s Theorem τt γ t)) exists for 0) L1 a.e. t near t 0 the conclusion follows from Remark 3.2 applied to f restricted to the plane γ t) + span {e, τ t)} = γ t 0 ) + span {e, τ t 0 )}. If τ t) is not constant near t 0 then as in the proof of Theorem 3.1 we can reduce to the previous case by using a local diffeomorphism. We omit the details. Hence the claim holds and therefore for L 1 a.e. t I we have γ t)) = γ t)) τ t) e) + γ t)) τ t) σ t)). 4.27) τ t) e σ t) Moreover by the chain rule for every orthonormal basis {e 1,..., e d 1, e} and for L 1 a.e. t I f γ) t) = e γ t)) d 1 γ t) e) + γ t)) γ t) e i ) or, equivalently γ t)) = τ t) = e γ t)) d 1 γ t) e) + γ t)) σ t) e i ) γ t) σ t)), e i d 1 e γ t)) τ t) e) + γ t)) σ t) e i ) τ t) σ t)). Hence also by 4.27) for every orthonormal basis {e 1,..., e d 1, e} and for L 1 a.e. t I we have that d 1 σ t) γ t)) = γ t)) σ t) e i ). 4.28) 25

26 Define the invertible linear transformation L : R d R d as L u) := u + u e) w, 4.29) where w e is a unit vector different from ±σ t 0 ) and let f L := f L, γ L := L 1 γ. Since the chain rule 1.3) holds in W 1,1 loc R N ; R d) with respect to every coordinate system it is clear that it also holds for f L with respect to every coordinate system. Therefore we can also assume that for every orthonormal basis {e 1,..., e d 1, e} and for L 1 a.e. t [0, 1] L σ L t) γ d 1 L L t)) = γ L t)) σ L t) e i ), e i where as before σ L t) is the unit vector obtained by projecting the tangent unit vector τ L t) on e and normalizing. Since by 4.29) for any ν e S d 1 we have that L ν γ L t)) = γ t)), ν the above identity can be rewritten as d 1 σ L t) γ t)) = γ t)) σ L t) e i ). 4.30) Let B be a countable dense family of orthonormal bases in e, and, for every t I, let B t) be the dense subfamily composed by all bases {e 1,..., e d 1 } B such that σ t) e i 0 for i = 1,..., d 1. From 4.28) and from our conventions on the validity of the chain rule it follows that γ t)) exists for L 1 a.e. t I, for every {e 1,..., e d 1 } B t), and for every i = 1,..., d 1. By 4.28) and 4.30) there exists a set N I with L 1 N ) = 0 such that for all t I \ N and for any two bases {e 1,..., e d 1 } and {ɛ 1,..., ɛ d 1 } in B t) we have that d 1 d 1 d 1 γ t)) σ t) e i ) = γ t)) σ t) ɛ i ) ɛ i d 1 γ t)) σ L t) e i ) = γ t)) σ L t) ɛ i ). ɛ i Since for every t near t 0 the space e is generated by vectors of the form σ L t) σ t) for a suitable choice of linear maps L satisfying 4.29), it follows that d 1 d 1 γ t)) e i = γ t)) ɛ i ɛ i 26

27 for all t I \ N. As this holds for any pair of orthonormal bases in B t) by Proposition 2.6 it follows that f restricted to the hyperplane γ t) + e is differentiable for all t I \N. By Remark 3.2 this implies that f is differentiable at γ t) for L 1 a.e. t near t 0. Since this is true for any t 0 [0, 1] a compactness argument allows to conclude that f is differentiable at H 1 a.e. point of γ. Given the arbitrariness of the curve γ we may conclude that Σ f is purely H 1 - unrectifiable. Step 2: If 1 < k < d 1, let M R d be a k-dimensional manifold such that M Σ f and let ū M Σ f. We claim that there exists ε > 0 such that f is differentiable for H k a.e. u B ū; ε) M. Fix any ū M Σ f and consider a local regular parametrization ψ : D R k M of class C 1 such that ū ψ D), where D is an open set and ψ 0) = ū. Let e S d 1 be a vector transversal to Tan k M, ū), that is e is not orthogonal and it does not belong to Tan k M, ū). Taking a smaller D, if necessary, we may assume that ψ x 1 x) is not parallel to e for every x D. Let e be the hyperplane orthogonal to e and let B be a countable dense family of orthonormal bases in e, and, for every x D, let B x) be the dense subfamily composed of all bases {e 1,..., e d 1 } B such that ψ x 1 x) e i 0 for i = 1,..., d 1. Let σ x) be the vector obtained by projecting ψ x 1 x) on e and normalizing. Define the invertible linear transformation L : R d R d as L u) := u + u e) w, where w e is a unit vector different from ±σ 0) and let f L := f L and ψ L := L 1 ψ. Without loss of generality assume that D = B 1 0; r) B k 1 0; r), where r > 0 is so small that w ±σ x) for every x D and write x = x 1, x ) R R k 1. Since f and f L satisfy the chain rule in the class A k R k ; R d) with respect to every coordinate system of the form {e 1,..., e d 1, e d } where e d := e and {e 1,..., e d 1 } is in B, we have x 1 f ψ) x) = x 1 f L ψ) x) = d d ψ x)) ψ i x 1 x), L ψ x)) ψ i x 1 x) and for L k a.e. x D, where ψ x)) exist near x := ψ 1 ū), whenever {e 1,..., e d 1 } B x), as ψi x 1 x) 0 by the definition of B x). By Fubini s Theorem and using the fact that B is countable, there exists a set M B k 1 0; r) with L k 1 M) = 0 such that for all x B k 1 0; r) \ M and 27

28 for every coordinate system of the form {e 1,..., e d 1, e d }, where e d := e and {e 1,..., e d 1 } is in B, we have x 1 f ψ) x 1, x ) = x 1 f L ψ) x 1, x ) = d d ψ x 1, x )) ψ i x 1 x 1, x ) L ψ x 1, x )) ψ i x 1 x 1, x ) for L 1 a.e. x 1 B 1 0; r). Fix x B k 1 0; r) \ M and consider the curve ψ, x ). Reasoning as in the previous step it follows from the above identities that for L 1 a.e. x 1 B 1 0; r) and for every coordinate system {e 1,..., e d 1 } in B we have σ x 1, x ) ψ x d 1 1, x )) = ψ x 1, x )) σ x 1, x ) e i ), e i L σ x 1, x ) ψ d 1 L x 1, x L )) = ψ L x 1, x )) σ L x 1, x ) e i ). e i We may continue as in the previous step to conclude that for L 1 a.e. x 1 B 1 0; r) the function f is differentiable at ψ x 1, x ). Since this is true for all x B k 1 0; r) \ M by Fubini s Theorem we have that f is differentiable at ψ x) for L k a.e. x D. Hence f is differentiable H k almost everywhere in ψ D). This concludes the proof. Remark 4.14 It is clear from the previous proof that for Σ f to be purely H k - unrectifiable it is enough to assume that the chain rule 1.3) holds in A k R N ; R d) with respect to a dense set of coordinate systems. 5 Chain rule in BV Ω; R d) In this section we extend the results of the previous section to the space of functions of bounded variation. We refer to [4] for the definition and main properties. As already mentioned in the introduction a weak form of the chain rule in BV loc R N ; R d ) was established by Ambrosio and Dal Maso in [2] for any Lipschitz continuous function f : R d R see also [10] for a different proof in the scalar case d = 1). We study here the classical chain rule. Since by a result of Alberti [1] the Cantor part of the distributional derivative of a function of bounded variation has rank one, to extend the results of the previous section we can only consider the case k = 1. Theorem 5.1 Let f : R d R be a Lipschitz continuous function, let {e 1,..., e d } be a basis in R d, and assume that for every H 1 -rectifiable set E Σ f and for H 1 a.e. u E, there exists i {1,..., d} depending on u such that Tan 1 E, u) = span {e i }. 5.1) 28