The flattening lemma

The goal of this section is to present a lemma that will allow us to carry out one of the major steps in our argument: the transition from the absence of large oscillation ofR^Hµ on suppµnear some fixed pointzon scales comparable to`to the flatness ofµatz on scale`.

Proposition 6. Fix four positive parameters A,α, ˜c,and C. There exist numberse A⁰, α⁰>0 depending only on these fixed parameters and the dimension d such that the following implication holds.

Suppose that H is a linear hyperplane in R^d+1,z∈R^d+1, Lis the affine hyperplane containing z and parallel to H, `>0, and µ is a C-good finite measure ine R<sup>d+1</sup> that is AD-regular in B(z,5A<sup>0</sup>`) with the lower regularity constant c. Assume that˜ µ is geometrically (H,5A⁰, α⁰)-flat at z on the scale ` and, in addition, for every (vector-valued) Lipschitz function g with suppg⊂B(z,5A<sup>0</sup>`), kgkLip6`⁻¹, and R

g dµ=0, one has

|hR^H_µ1, giµ|6α⁰`<sup>d</sup>. Then µis (H, A, α)-flat at z on the scale `.

Before proving this proposition (which we will call the “flattening lemma” from now on), let us discuss the meaning of the assumptions. In what follows, we will apply this result to restrictions of a fixed good AD-regular measureµto open balls at various scales and locations. The restriction of a good AD-regular measure to a ball may easily fail to be AD-regular in the entire spaceR^d+1, which explains why we have introduced the local notion of Ahlfors–David regularity. Every restriction of a good measure to any set is, of course, good with the same goodness constant as the original measure.

The first step in proving the rectifiability of a measure is showing that its support is almost planar on many scales in the sense of the geometric (H,5A⁰, α⁰)-flatness in the assumptions of the flattening lemma implication. This step is not that hard and we will carry it out in§15. The second condition involving the Riesz transform means, roughly speaking, thatR_µ^H1 is almost constant on suppµ∩B(z, A⁰`) in the sense that its “wavelet coefficients” near z on the scale ` are small. There is no canonical smooth wavelet system inL²(µ) whenµis an arbitrary measure, but mean-zero Lipschitz functions serve as a reasonable substitute. The boundedness of R^H_µ in L²(µ) implies thatR^H_µ1∈L²(µ) (because for finite measuresµ, we have 1∈L²(µ)), so an appropriate version of the Bessel inequality can be used to show that large wavelet coefficients have to be rare and the balls satisfying the second assumption of the implication should also be viewed as typical.

Finally, it is worth mentioning that the full (H, A, α)-flatness condition is much stronger than just the geometric one in the sense that it allows one to get non-trivial quantitative information about the Riesz transform operatorR^H_µ. The flattening lemma thus provides the missing link between the purely geometric conditions like those in the David–Semmes monograph and analytic conditions needed to make explicit estimates.

Proof. Note that the geometric (H, A, α)-flatness of µ is ensured by the geometric (H,5A⁰, α⁰)-flatness assumption of the flattening lemma implication as soon as A⁰>A andα⁰6α. The real problem is to prove the analytic part of the flatness condition.

To this end, note first that the setup of the flattening lemma is invariant under translations and dilations, so, replacing the measureµand the test functionsf andgby

`<sup>−d</sup>µ(z+`·),f(z+`·), andg(z+`·), respectively, we may always assume without loss of generality thatz=0 and`=1.

Now fixA⁰>A. Since the setLof all Lipschitz functionsf with Lipschitz constant 1 supported on B(0, A) and having zero integral with respect to mL is pre-compact in C0(R^d+1), for everyβ >0, we can find a finite familyF inLso that every functionf∈L is uniformly β-close to some f⁰∈F. As we have the a-priori bound µ(B(0, A)) 6CAe ^d, thisβ-closeness implies that

consider only test functionsf∈Fif we do not mind showing for them a stronger inequality withαreplaced by¹₂α. SinceFis finite, we see that if the flattening lemma is false, we can find one fixed test functionf and a sequence of measuresµk satisfying the assumptions of the flattening lemma implication with our fixedA⁰ andα⁰=1/ksuch thatR

f dµk>¹2α for allk.

Split each µk as

µ_k=χ_B(0,5A0)µ_k+χ_Rd+1\B(0,5A⁰)µ_k=ν_k+η_k.

Note that νk are still C-good and AD-regular ine B(0,5A⁰) with lower AD-regularity constant ˜c. Moreover, suppνk lies within distance 1/k from L and every point in L∩B(0,5A⁰−1/k) lies within distance 1/k from suppνk. Passing to a subsequence, if necessary, we may assume that νk converge weakly to some measure ν. By the re-sults of the weak limits section (§8) this limiting measure ν is C-good and, obviously,e suppν⊂L∩B(0, 5A⁰).

Fix a pointw∈L∩B(0,5A⁰) andr>0 such thatB(w, r)⊂B(0,5A⁰). Take anyr⁰<r and consider a continuous functionh:R^d+1![0,1] that is identically 1 on B(w, r⁰) and identically 0 outsideB(w, r). Sincew∈B(0,5A⁰−1/k) for all sufficiently largek, we can find a sequence of pointswk∈suppνk so that |w−wk|61/k for all sufficiently large k.

Passing to the limit, we conclude thatν(B(w, r))>R

h dν>c(r˜ ⁰)^d. Since this inequality holds for allr⁰<r, we must haveν(B(w, r))>ecr^d. Combining this with the upper bound ν(B(w, r))6Cre ^d and the inclusion suppν⊂L, we see that, by the Radon–Nikodym the-orem applied toν andm_L, the limiting measure ν can be written asν=p m_L for some

Borel functionponL satisfyingω_d⁻¹˜c6p6ω⁻¹_d Ce almost everywhere with respect tomL

onL∩B(0,5A⁰), whereωd is thed-dimensional volume of the unit ball inR^d. Fix some non-negative Lipschitz functionh with supph⊂B(0,4A⁰) andR

h dν >0.

Take any Lipschitz vector-valued functiong supported onB(0,4A⁰) withkgkLip<1 and Rg dν=0. Since R

h dνk!R

h dν >0 as k!∞, the integralsR

h dνk stay bounded away from 0 for sufficiently largek.

Put

ak= Z

h dνk

−1Z

g dνk and gk=g−akh.

The functionsgk are well defined for all large enough kand satisfy Z

gkdµk= Z

gkdνk= 0 and suppgk⊂B(0,4A⁰).

AsR

g dνk!R

g dν=0, we conclude thatak!0 ask!∞, so kg_kk_Lip<1

for large enoughk.

Sinceµksatisfies the assumptions of the flattening lemma implication, we must have

|hR_µ^H

k1, gkiµ_k|61 k

for largek. Taking into account that suppg_k⊂B(0,4A⁰), we can rewrite this as

|hR^H_νk1, gkiν_k+hR^Hηk, gkiν_k|61 k. Note that

hR^H_ν

k1, g_ki_νk=hR^H_ν

k1, gi_νk−hR^H_ν

k1, a_khi_νk and thatR^H_ν

k1 and R^H_ν1 coincide with R^H_ν

kϕ andR^H_νϕ, respectively, for any compactly supported Lipschitz function ϕ that is identically 1 on B(0,5A⁰), say. Thus, by the results of the weak limits section (§8), we get

hR^H_νk1, giνk=hR^H_νkϕ, giνk!hR^H_νϕ, giν=hR^H_ν 1, giν. Similarly,

hR^H_νk1, heiν_k!hR^H_ν1, heiν

for every vectore∈H. Since

hR^H_νk1, akhiν_k=

d

X

j=1

hak, ejihR_ν^H_k1, hejiν_k

for every orthonormal basise1, ..., ed inH, andak!0 ask!∞, we conclude that hR^H_νk1, akhiν_k!0

and, thereby,

hR^H_ν

k1, g_ki_νk!hR^H_ν1, gi_ν ask!∞.

Note now that the measure ηk is supported outside B(0,5A⁰). Together with the cancellation propertyR

gkdνk=0, this yields

hR^Hη_k, g_ki_νk=hv_k, g_ki_νk, where

vk=R^Hηk−(R^Hηk)(0) is aC^∞function inB(0,4A⁰) satisfying v_k(0)=0 and

|(∇^jvk)(x)|6C

Z dηk(y)

|x−y|^d+j6C(j)Ce (A⁰)^j wheneverx∈B(0,4A⁰) andj >0.

Since the set of functions vanishing at the origin with three uniformly bounded derivatives is compact in C²(B(0,4A⁰)), we may (passing to a subsequence again, if necessary) assume thatv_k!v inC²(B(0,4A⁰)), which is more than enough to conclude that

hv_k, g_ki_νk!hv, gi_ν

(all we need for the latter is the uniform convergencehvk, gki!hv, gi). Thus, we have found aC²functionv inB(0,4A⁰) such that

hR^H_ν 1, gi_ν=−hv, gi_ν for all Lipschitz functionsg with suppg⊂B(0,4A⁰) andR

g dν=0. Moreover,

|∇^jv|L^∞(B(0,4A⁰))6 CCe (A⁰)^j

forj=1,2. The condition kgkLip61 can be dropped now because both sides are linear ing. This equality can be rewritten as

hR^H_mLp, pgimL=−hv, pgimL

for all Lipschitz functions g with suppg⊂B(0,4A⁰) satisfying R

Lpg dmL=0. Since pis bounded from below onL∩B(0,4A⁰), the set of such productspg is dense in the space of all mean-zero functions in L²(L∩B(0,4A⁰), mL) and we conclude thatR^H_mLpdiffers from −v only by a constant on L∩B(0,4A⁰). By the toy flattening lemma (Lemma 2) applied withA⁰ instead ofA, this means that pis Lipschitz in L∩B(0, A⁰) and

kpkLip(L∩B(0,A⁰))6CCe A⁰ .

But then for the test functionf∈F introduced at the beginning of the proof, we have ifA⁰ was chosen large enough. On the other hand, we have

This contradiction finishes the proof.