Nowhere differentiable intrinsic Lipschitz graphs

This is a self-archived version of an original article. This version may differ from the original in pagination and typographic details. Author(s): Title: Year: Version: Copyright: Rights: Rights url: Please cite the original version: CC BY 4.0 https://creativecommons.org/licenses/by/4.0/ Nowhere differentiable intrinsic Lipschitz graphs © 2021 The Authors. Bulletin of the London Mathematical Society is copyright © London Mathematical Society. Published version Julia, Antoine; Nicolussi Golo, Sebastiano; Vittone, Davide Julia, A., Nicolussi Golo, S., & Vittone, D. (2021). Nowhere differentiable intrinsic Lipschitz graphs. Bulletin of the London Mathematical Society, 53(6), 1766-1775. https://doi.org/10.1112/blms.12540 2021 Bull. London Math. Soc. 0 (2021) 1–10 doi:10.1112/blms.12540 Nowhere differentiable intrinsic Lipschitz graphs Antoine Julia, Sebastiano Nicolussi Golo and Davide Vittone Abstract We construct intrinsic Lipschitz graphs in Carnot groups with the property that, at every point, there exist infinitely many different blow-up limits, none of which is a homogeneous subgroup. This provides counterexamples to a Rademacher theorem for intrinsic Lipschitz graphs. The notion of Lipschitz submanifolds in sub-Riemannian geometry was introduced, at least in the setting of Carnot groups, by Franchi, Serapioni and Serra Cassano in a series of seminal papers [5–7] through the theory of intrinsic Lipschitz graphs. One of the main open questions concerns the differentiability properties for such graphs: in this paper, we provide examples of intrinsic Lipschitz graphs of codimension 2 (or higher) that are nowhere differentiable, that is, that admit no homogeneous tangent subgroup at any point. Recall that a Carnot group Gis a connected, simply connected and nilpotent Lie group whose Lie algebra is stratified, that is, it can be decomposed as the direct sum ⊕s j=1Vjof subspaces such that Vj+1 =[V1,V j] for every j=1,...,s−1,[V1,V s]={0},V s={0}. We shall identify the group Gwith its Lie algebra via the exponential map exp : ⊕s j=1Vj→G, which is a diffeomorphism. In this way, for λ>0, one can introduce the homogeneous dilations δλ:G→Gas the group automorphisms defined by δλ(p)=λjpfor every p∈Vj. A subgroup of Gis said to be homogeneous if it is dilation-invariant. Assume that a splitting G=WV of Gas the product of homogeneous and complementary (that is, such that W∩V={0}) subgroups is fixed; we say that a function φ:W→Vintrinsic Lipschitz if there is an open nonempty cone Usuch that V\{0}⊂Uand pU ∩Γφ=∅for allp∈Γφ, where Γφ={wφ(w):w∈W}is the intrinsic graph of φ. We say that a set Σ ⊂Gis a blow-up of Γφat ˆp=ˆwφ(ˆw) if there exists a sequence (λn)nsuch that λn→+∞and the limit lim n→∞δλn(ˆp−1Γφ)=Σ holds with respect to the local Hausdorff convergence. It is worth recalling that, if φis intrinsic Lipschitz, then every blow-up is automatically the intrinsic Lipschitz graph of a map W→V. Eventually, we say that φis intrinsically differentiable at ˆw∈Wif the blow-up of Γφ at ˆp=ˆwφ(ˆw) is unique and it is a homogeneous subgroup of G. See [8] for details. Received 8 January 2021; revised 7 April 2021. 2020 Mathematics Subject Classification 53C17 (primary), 22E25, 58C20 (secondary). AJ has been supported by the Simons Foundation Wave Project. SNG has been supported by the Academy of Finland (grant 322898 ‘Sub-Riemannian Geometry via Metric-geometry and Lie-group Theory’). DV has been supported by FFABR 2017 of MIUR (Italy) and by GNAMPA of INdAM (Italy). All three authors have been supported by the University of Padova STARS Project ‘Sub-Riemannian Geometry and Geometric Measure Theory Issues: Old and New’. C e2021 The Authors. Bulletin of the London Mathematical Society is copyright C eLondon Mathematical Society. This is an open access article under the terms of the Creative Commons Attribution License, which permits use, distribution and reproduction in any medium, provided the original work is properly cited. 2ANTOINE JULIA, NICOLUSSI GOLO AND DAVIDE VITTONE We say that a group Galong with a splitting WV satisfies an intrinsic Rademacher Theorem if all intrinsic Lipschitz maps φ:W→Vare intrinsically differentiable almost everywhere (that is, for almost all points of Wequipped with its Haar measure). It was proved in [6] that this is the case when V≃Rand Gis of step two; other partial results for graphs with codimension 1(V≃R) are contained in [4, 9]. If Vis a normal subgroup, the Rademacher Theorem has been proved for general Gby Antonelli and Merlo in [2]. Recently, the third-named author [12] proved that Heisenberg groups (with any splitting) satisfy an intrinsic Rademacher Theorem. The question has been open for a long time if Gis the Engel group (which has step 3) and V≃R(see [1]). In this paper, we prove a result in the negative direction: namely, we provide examples of intrinsic Lipschitz graphs that are nowhere intrinsically differentiable. Let us state our main result: Theorem 1. Let Gbe a Carnot group with stratification s j=1 Vj.LetWV be a splitting of Gsuch that W∩V2⊂ [W,W]and there exists v0∈V∩V1such that v0=0 and [v0,W]=0. Then there is an intrinsic Lipschitz function φ:W→Vthat is nowhere intrinsically differentiable. Moreover, φcan be constructed in such a way that, for every p∈Γφ, the following properties hold. (a) There exist infinitely many different blow-ups of Γφat p. (b) No blow-up of Γφat pis a homogeneous subgroup. The proof of Theorem 1is postponed in order to first provide some comments. Remark 1. The simplest example of a Carnot group where Theorem 1applies is G=H×R, where His the first Heisenberg group. As customary, we consider generators X, Y,T of the Lie algebra of Hsuch that [X, Y ]=T,[X,T]=[Y,T] = 0 and fix the exponential coordinates (x, y, t)=exp(xX +yY +tT ). Using coordinates (x, y, t, r)onH×Rwith r∈R,wecan consider the splitting H×R=WV given by the vertical subgroup W={x=r=0}of H and the horizontal Abelian subgroup V={y=t=0}. Then V2∩W⊂ [W,W]={0}and v0=(0,0,0,1) commutes with W. Hence, this splitting of H×Rsatisfies the conditions of Theorem 1and it does not satisfy an intrinsic Rademacher Theorem. It is worth observing that, in this setting, the map φ:W→Vprovided in the proof of Theorem 1takes the form φ(y,t)=(0,u(t)), where uis the 1 2-H¨older continuous function constructed in the Appendix. In particular, the intrinsic graph Γφis the set {(0,y,t,u(t)) : y,t ∈R}and it is contained in the Abelian subgroup W×R. One of the properties of uis that the limit lim s→t|u(t)−u(s)| |t−s| does not exists at any t∈Rand this is the ultimate reason for the nondifferentiability of φ. Similar counterexamples can be constructed in any codimension k⩾2: in fact one can consider Hk−1×R=(Rk−1 x×Rk−1 y×Rt)×Rrwith splitting WV defined by W={x=0,r= 0},V={y=0,t=0}. It can be easily checked that the map φ(y,t)=(0,u(t)) defines an intrinsic Lipschitz graph of codimension kfor which the properties (a) and (b) in Theorem 1 hold at every point. Remark 2. The measure μ=HdΓφ, where dis the Hausdorff dimension of Wand Hdis the d-dimensional Hausdorff measure, does not have a unique tangent measure at any point. Indeed, first, any tangent measure of μis supported on a blow-up of Γφ. Second, by [7, Theorem 3.9], μand all its dilations are uniformly d-Ahlfors regular, and thus any tangent measure of μis NOWHERE DIFFERENTIABLE INTRINSIC LIPSCHITZ GRAPHS 3 d-Ahlfors regular. We then conclude that if μ1and μ2are two tangent measures of μsupported on different blow-ups of Γφ, then they are two distinct measures. Since blow-ups of Γφare not unique, so are tangent measures. Observe also that no tangent measure can be flat, that is, supported on a homogeneous subgroup. In particular, Γφis purely C1 H-unrectifiable, that is, Hd(Γφ∩Σ) = 0 for every submanifold Σ of class C1 H(see, for example, [3,§2.5 and 6.1]). Remark 3. If Wis a homogeneous subgroup of Gwith codimension 1, then the conditions of Theorem 1cannot be met because s j=2 Vj=[W,W]+[W,V]. Actually, intrinsic Lipschitz graphs of codimension 1 are boundaries of sets with finite perimeter in G(see, for example, [11, Theorem 1.2]), hence at almost every point they possess at least one blow-up which is a homogeneous subgroup of codimension 1, see [1]. Therefore, any possible counterexample to the Rademacher Theorem in codimension 1 cannot be as striking as the one provided by Theorem 1, in the sense that property (b) cannot hold on a set with positive measure. Remark 4. Following the same proof strategy, one can extend Theorem 1to the case W∩Vj⊂ [W,W] for some j>2andv0∈Vk∩V\{0}with k<jand [v0,W] = 0, by taking ak/j-H¨older analogue of the function uconstructed in the appendix. Proof of Theorem 1. Let β:W→Rbe a nonzero linear function such that W∩Vj⊂ker β whenever j=2and[W,W]⊂ker β;suchaβexists†because W∩V2⊂ [W,W]. Note that such a function βis in fact a group morphism W→R. Consider a 1/2-H¨older continuous function u:R→Rwith the following properties. First, the difference quotients Δ(s, t)= u(s)−u(t) sgn(s−t)|s−t|1/2 are bounded, namely, |Δ(s, t)|⩽1 for everys, t ∈R.(1) Second, there exist c1>0andc2>0 such that, for every t0∈Rand δ∈(0,1 ], there exist s1,s 2∈Rsuch that sgn(s1−t0)=sgn(s2−t0) c1δ⩽|s1−t0|⩽δ c1δ⩽|s2−t0|⩽δ |Δ(s1,t 0)−Δ(s2,t 0)|⩾c2. (2) Such a function exists, as we show in the Appendix. We can then define φ:W→Vas φ(w)=u(β(w))v0. Note that the condition [v0,W] = 0 implies vw =wv for allw∈Wandv∈Rv0.(3) Therefore, by the Baker–Campbell–Hausdorff formula, the intrinsic graph of φis the set of points wφ(w)=w+u(β(w))v0for w∈W. †For instance, one can consider β(x)=x, w0for some w0∈(W∩V2)\[W,W] and a scalar product on W adapted to the grading s j=1 W∩Vjof W. 4ANTOINE JULIA, NICOLUSSI GOLO AND DAVIDE VITTONE Claim 1. The map φis intrinsic Lipschitz. Fix a homogeneous norm ·on G. Note that, since β(δλx)=λ2β(x) for all x∈W, there is a constant Csuch that |β(x)|⩽Cx2, for all x∈W. We check that Γφhas the cone property for the cone (see [7, Definition 10]) U={wv :w∈W,v∈V,v>2√Cv0w}. Given ˆw,w ∈W,by(3)wehave(ˆwφ(ˆw))−1(wφ(w)) = ( ˆw−1w)(φ(ˆw)−1φ(w)) and φ(ˆw)−1φ(w)=|u(β(w)) −u(β(ˆw))|v0⩽|β(w)−β(ˆw)|1/2v0 =|β(ˆw−1w)|1/2v0⩽√Cˆw−1wv0. Thus, ( ˆwφ(ˆw))−1Γφ∩U=∅for all ˆw∈W,thatis,Γ φis an intrinsic Lipschitz graph. Claim 2. For p∈Γφ, none of the blow-ups of Γφat pis a homogeneous subgroup. We first observe that, if V0⊂V∩V1is the horizontal subgroup generated by v0and L:W→ V0parameterizes a homogeneous subgroup ΓLof G, then L|W∩V2= 0. Indeed, the homogeneity of ΓLimplies that for every w∈W∩V2one has L(2w)=√2L(w), because (2w)(√2L(w)) = δ√2(w)δ√2(L(w)) = δ√2(wL(w)) ∈ΓL, while the fact that ΓLis a subgroup (plus the fact that V0and Wcommute) gives L(2w)= 2L(w), because wwL(w)L(w)=(wL(w))(wL(w)) ∈ΓL. This proves that L=0onW∩V2. We now prove the claim. Assume by contradiction that there exist ˆp=ˆwφ(ˆw)∈Γφ, a map L:W→Vsuch that the intrinsic graph ΓLof Lis a homogeneous subgroup and a sequence (λn)nwith λn→+∞,and lim n→∞δλn(ˆp−1Γφ)=Γ L. Observe that for every w∈Wand every n δλn(( ˆwφ(ˆw))−1(wφ(w))) = δλn(ˆw−1wφ(ˆw)−1φ(w)) =δλn(ˆw−1w)u(β(w)) −u(β(ˆw)) 1/λn v0. If we set w=ˆwδ1/λnw, then β(w)=β(ˆw)+β(w)/λ2 n. Therefore, the set δλn(ˆp−1Γφ)isthe intrinsic graph of the function from Wto Vgiven by φˆp,λn(w)=u(β(ˆw)+β(w)/λ2 n)−u(β(ˆw)) 1/λn v0. Since the maps φˆp,λntake values in V0,Lis also V0-valued and, as we saw above, this implies that L|W∩V2=0. Write ˆ t=β(ˆw) and let w0∈W∩V2be such that β(w0) = 1; then for every h∈R φˆp,λn(hw0) = (sgn h)|h|1/2Δ(ˆ t+h/λ2 n,ˆ t)v0.(4) By (2), there exists a sequence (hn)nsuch that for every n |hn|∈[c1,1] and φˆp,λn(hnw0)⩾√c1c2v0/2. NOWHERE DIFFERENTIABLE INTRINSIC LIPSCHITZ GRAPHS 5 Up to passing to a subsequence we can also assume that hn→¯ hwith |¯ h|∈[c1,1 ]; since φˆp,λn(hnw0)−φˆp,λn(¯ hw0)= u(ˆ t+hn/λ2 n)−u(ˆ t+¯ h/λ2 n) 1/λnv0, ⩽|hn−¯ h|1/2v0 we obtain L(¯ hw0)= lim nφˆp,λn(¯ hw0)= lim nφˆp,λn(hnw0)⩾√c1c2v0/2. This contradicts the fact that L(¯ hw0) = 0, and the claim is proved. Claim 3. For p∈Γφ, there exist infinitely many different blow-ups of Γφat p. Let ˆp=ˆwφ(ˆw)∈Γφbe fixed and let ˆ t=β(ˆw); as before, fix also w0∈W∩V2such that β(w0)=1.By(2), we can find infinitesimal sequences (s1 n)n,(s2 n)nsuch that sgn(s1 n)=sgn(s2 n) for every n, Δ(ˆ t+s1 n,ˆ t)⩾Δ(ˆ t+s2 n,ˆ t)+c2. Up to passing to a subsequence, we can assume that there exists σ∈{1,−1}and Δ1,Δ2∈R such that sgn(s1 n)=sgn(s2 n)=σfor everyn, Δ(ˆ t+s1 n,ˆ t)→Δ1and Δ(ˆ t+s2 n,ˆ t)→Δ2as n→∞, Δ1⩾Δ2+c2. Due to the continuity of s→ Δ(ˆ t+s, ˆ t)fors= 0, given Δ ∈(Δ2,Δ1) one can find an infinitesimal sequence (sn)nsuch that, for every n,sgn(sn)=σand Δ(ˆ t+sn,ˆ t) = Δ. Now, as in (4) the set δ|sn|−1/2(ˆp−1Γφ) is the intrinsic graph of a map φˆp,|sn|−1/2:W→Vsuch that φˆp,|sn|−1/2(σw0)=σΔ(ˆ t+sn,ˆ t)v0=σΔv0. Since the family (φˆp,|sn|−1/2)nis uniformly H¨older continuous, up to extracting a subsequence it converges locally uniformly to a map ψ:W→Vsuch that ψ(σw0)=σΔv0. The arbitrariness of Δ∈(Δ2,Δ1) implies that there are infinitely many different blow-ups at ˆp, and this concludes the proof.  Appendix We are now going to construct the function uused in the proof of Theorem 1: this function, in a sense, provides a counter-example to a Rademacher property for Lipschitz functions from (R,|· |1/2)to(R,|·|). We will use a classical procedure producing a self-similar function: although these ideas are well-known (see, for example, [10] and the references therein), we prefer to include a detailed construction because we were not able to find in the literature explicit statements for the precise estimates (2) we need. We construct a function u:[0,1]→[0,1 ] whose difference quotients Δ(s, t)= u(s)−u(t) sgn(s−t)|s−t|1/2 satisfy |Δ(s, t)|⩽1 for everys, t ∈[0,1].(A.1) We will construct uin such a way that there exist c1>0andc2>0 with the property that, for every t∈[0,1] and δ∈(0,1 ], one can find s1,s 2∈[0,1 ] such that the conditions in (2) 6ANTOINE JULIA, NICOLUSSI GOLO AND DAVIDE VITTONE Figure A.1 (colour online).Four instances of the functions undefined in (A.2). hold. One can then extend uto Rby setting u(t)=u(−t)fort∈[−1,0]and u(t+2n)=u(t) for all n∈Z: this extended udoes satisfy (1) and (2). The function uis obtained as the limit of a sequence (un)n∈Nwhere u0(t)=t. The function un+1 is obtained from unon setting un+1(t)=⎧ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎩ 2 3un9 4tif t∈0,4 9, 2 3−1 3un9t−4 9 if t∈4 9,5 9, 1 3+2 3un9 4t−5 9 if t∈5 9,1. (A.2) The first few of the functions u0,u 1,u 2,... are plotted in Figure A.1. Let us note that un(0) = 0 and un(1) = 1 for every n, hence un(4/9) = 2/3andun(5/9) = 1/3 for every n⩾1. Note (see Figure A.2) that the graph of un+1 is the union of three affine copies of the graph of un, via the following maps (acting on p∈R2): A0(p)=4/90 02/3p, A4/9(p)=1/90 0−1/3p+4/9 2/3, A5/9(p)=4/90 02/3p+5/9 1/3. (A.3) Claim 1. The functions unconverge uniformly on [ 0,1 ] to a function ufor which (A.1) holds. The fact that ununiformly converge to a continuous function uis a consequence of the estimate un+1 −unC0([ 0,1]) ⩽2 3un−un−1C0([ 0,1]). NOWHERE DIFFERENTIABLE INTRINSIC LIPSCHITZ GRAPHS 7 Figure A.2 (colour online).Iterated images of the unit square under the affine maps in (A.3); dots are the images of (0,0) and (1,1), and they belong to the graph of the limit function u This estimate follows directly from the definition (A.2): for instance, for t∈[0,4/9 ], one has |un+1(t)−un(t)|=2 3|un(9t/4) −un−1(9t/4)|⩽2 3un−un−1C0([ 0,1]). Similarly, one can treat the other two cases t∈[4/9,5/9] and t∈[5/9,1]. The bound (A.1) on the difference quotients of ufollows from the fact that the same is true for all unin the sequence, as we are now going to prove by induction on n. The statement is clearly true for n= 0. Suppose that unsatisfies |un(t)−un(s)|⩽|t−s|1/2for everys, t ∈[0,1], we will prove that also |un+1(t)−un+1(s)|⩽|t−s|1/2for every s, t ∈[0,1 ]. We distinguish several cases depending on which intervals ([ 0,4/9], [4/9,5/9] or [5/9,1 ]) the points sand t belong to. We can suppose that s<t. Case 1: sand tare in the same interval. We can use (A.2) and the induction hypothesis to conclude. Case 2: s∈[0,4/9]andt∈[4/9,5/9 ]. Since 0 ⩽un⩽1, one sees from the definition of un+1 that max(un+1(s),u n+1(t)) ⩽2/3=un+1(4/9). Thus |un+1(t)−un+1(s)|⩽max(un+1(4/9) −un+1(t),u n+1(4/9) −un+1(s)) ⩽max((t−4/9)1/2,(4/9−s)1/2)⩽(t−s)1/2, where the second inequality follows from Case 1. Case 3: s∈[4/9,5/9] and t∈[5/9,1 ]. Due to the symmetry un(x)=1−un(1 −x), this is similar to Case 2. Case 4: s∈[0,4/9]andt∈[5/9,1 ]. Then either |un+1(t)−un+1(s)|⩽1/3, and we are done because |t−s|⩾1/9, or |un+1(t)−un+1(s)|>1/3, and then necessarily un+1(s)<u n+1(t) (otherwise, 0 ⩽un+1(s)−un+1(t)⩽un+1(4/9) −un+1(5/9) = 2/3−1/3=1/3) and |un+1(t)−un+1(s)|=un+1(t)−un+1(s) =un+1(t)−un+1(5/9) −1/3+un+1(4/9) −un+1(s) ⩽(t−5/9)1/2−1/3+(4/9−s)1/2, 8ANTOINE JULIA, NICOLUSSI GOLO AND DAVIDE VITTONE where in the last inequality we used Case 1. By squaring the right-hand side of the last inequality, we obtain (t−5/9)1/2−1/3+(4/9−s)1/22 =(t−s)+2(t−5/9)1/2(4/9−s)1/2−2 3(t−5/9)1/2−2 3(4/9−s)1/2 =(t−s)+(t−5/9)1/2(4/9−s)1/2−2/3+(4/9−s)1/2(t−5/9)1/2−2/3 ⩽t−s, where we used the fact that 4/9−s⩽4/9andt−5/9⩽4/9. This is enough to conclude. Claim 2. There exist d1>0andd2>0 such that, for every t0∈[0,1 ], one can find s1,s 2∈[0,1 ] such that sgn(s1−t0)=sgn(s2−t0) d1⩽|s1−t0|⩽1 d1⩽|s2−t0|⩽1 |Δ(s1,t 0)−Δ(s2,t 0)|⩾d2. (A.4) In fact, we will prove Claim 2 for d1=1/18 and d2= min 1 31−4 81 −1/2−1,7 9−3 5,1 √5. We distinguish several cases. Case 1: t0∈[0,4/9 ]. In this case, it suffices to consider s1=5/9ands2=1, as we now show. Observe that the distances of s1,s 2from t0are both greater than 1/9>d 1. If u(t0)⩾2/9, by (A.1) and the equality u(0) = 0, we have t1/2 0⩾u(t0), hence t0⩾4/81; since u(t0)⩽2/3, we obtain Δ(1,t 0)⩾ 1 3 1−4 81 and Δ(5/9,t 0)⩽ 1 3−2 9 1 9 =1 3, so that Δ(1,t 0)−Δ(5/9,t 0)⩾d2. If u(t0)⩽2/9, then (4/9−t0)1/2⩾2/3−2/9=4/9, hence 5/9−t0⩾1/9+(4/9)2= (5/9)2and Δ(1,t 0)⩾7 9and Δ(5/9,t 0)⩽ 1 3 5 9 =3 5 and again Δ(1,t 0)−Δ(5/9,t 0)⩾d2. Case 2: t0∈[4/9,1/2 ]. In this case, we take s1=5/9ands2= 1. The distances of s1,s 2 from t0are both no less than 1/18 = d1and, since 1/3⩽u(t0)⩽2/3, one gets Δ(1,t 0)⩾ 1 3 5 9 =1 √5and Δ(5/9,t 0)⩽0. Case 3: t0∈[1/2,1 ]. We proved that, if t0∈[0,1/2 ], the claim can be proved on choosing s1=5/9ands2= 1. Therefore, due to the symmetry u(x)=1−u(1 −x), when t0∈[1/2,1] it is enough to take s1=0ands2=4/9. Claim 3. There exist c1>0andc2>0 such that, for every t∈[0,1] and δ∈(0,1], one can find s1,s 2∈[0,1 ] for which the conditions in (2) hold.