Almost orthogonality

From now on we will assume that the family F defined in the previous section is not Carleson, and so we have a cellP∈Dand alternating layersBk,Qk⊂Das in the previous lemma. We will eventually show that this leads to a contradiction.

Fix K. Choose ε>0, A, α>0 and η >0 in this order and run the construction of the previous section. In this section we will be primarily interested in the flat layersQk

ignoring the non-BAUP layersPk almost entirely.

For a cellQ∈D andt>0, define

Q_t={x∈Q: dist(x,R^d+1\Q)>t`(Q)}.

Note thatµ(Q\Qt)6Ct^γµ(Q) for some fixedγ >0 (see§12). Letϕ0be anyC^∞function supported on B(0,1) and such that R

ϕ0dm=1, where m is the Lebesgue measure in R^d+1. Put

ϕQ=χQ_2ε∗ 1 (ε`(Q))^dϕ0

· ε`(Q)

.

ThenϕQ=1 onQ3εand suppϕQ⊂Qε. In particular, the diameter of suppϕQ is at most 8`(Q). In addition,

kϕQkL^∞61, k∇ϕQkL^∞6 C

ε`(Q), and k∇<sup>2</sup>ϕ<sub>Q</sub>kL<sup>∞</sup>6 C ε<sup>2</sup>`(Q)².

From now on, we will be interested only in the cellsQfrom the flat layersQk. With each such cellQwe will associate the corresponding approximating planeL(Q) containingzQ

and parallel toH and the approximating measureνQ=aQϕQm_L(Q), whereaQ is chosen so that

νQ(R^d+1) = Z

ϕQdµ.

Note that, sinceB z_Q, ¹₈−3ε

`(Q)

⊂Q_3ε andQ⊂B(z_Q,4`(Q)), both integrals Z

ϕQdm_L(Q) and Z

ϕQdµ

are comparable to `(Q)^d, provided that ε<₄₈¹, say. In particular, in this case, the nor-malizing factorsaQ are bounded by some constant.

Define

Gk= X

Q∈Qk

ϕQR^H[ϕQµ−νQ], k= 0, ..., K.

We remind the reader of our convention to understand R^H(ϕ_Qµ) as R^H_µϕ_Q on suppµ (see§8) and of Lemma1 that shows that R^Hν_Q can be viewed as a Lipschitz function

in the entire spaceR^d+1. In what follows, we will freely integrate various expressions including both R^H(ϕQµ) and R^HνQ with respect to µ, which makes sense in view of what we just said. However, we will be very careful with the integration of expressions involvingR^H(ϕQµ) with respect to νQ and always make sure that for each point xin the integration domain,xis not contained in the support of any functionϕQ for which R^H(ϕQµ) in the integrand is not multiplied by some cutoff factor vanishing in some neighborhood ofx.

Now put

F_k=G_k−G_k+1 fork= 0, ..., K−1 and F_K=G_K. Note that

K

X

m=k

Fm=Gk.

The objective of this section is to prove the following result.

Proposition 9. Assuming that ε<₄₈¹,A>5,and α<ε⁸, we have

|hFk, G_k+1i|6σ(ε, α)µ(P)

for all k=0, ..., K−1,where σ(ε, α)is some positive function such that

lim

ε!0+

lim

α!0+

σ(ε, α) = 0.

In plain English, the double limit condition on σ(ε, α) means that we can make σ(ε, α) as small as we want by first choosingε>0 small enough and then choosingα>0 small enough. The exact formula forσ(ε, α) will be of no importance for the rest of the argument, so we do not even mention it here despite it being explicitly written in the end of the proof.

The assumptionsε<₄₈¹ andA>5 are there to ensure that all the results of§9can be freely applied withϕQ in the role ofϕandνQ in the role ofν. The assumptionα<ε⁸ is just used to absorb some expressions involvingαandεinto constants instead of carrying them around all the time.

Several tricks introduced in this section will be used again and again in what follows so we suggest that the reader goes over all details of the proof because here they are presented in a relatively simple setting unobscured by any other technical considerations or logical twists. Also, there is a technical lemma in the body of the proof (Lemma10) that will be used several times later, despite the fact that it is not formally proclaimed as one of the main results of this section.

Proof. We start with showing that, under our assumptions,kGkk^p_Lp(µ)6Cµ(P) for p=2,4 and allk=0, ..., K. Since

G_k= X

Q∈Qk

ϕ_QR^H[ϕ_Qµ−ν_Q]

and the summands have pairwise disjoint supports, it will suffice to prove the inequality kϕ_QR^H(ϕ_Qµ−ν_Q)k^p_Lp(µ)6Cµ(Q)

for each individualQ∈Qk and then observe thatP

Q∈Qkµ(Q)6µ(P).

Since we shall need pretty much the same estimate in §20, we will state it as a separate lemma here.

Lemma 10. Let p=2 or p=4. For each k=0, ..., K and for each cell Q∈Qk, we have

kϕQR^HνQk^p_Lp(µ)6kχQR^HνQk^p_Lp(µ)6Cµ(Q).

As a corollary,we have

kϕQR^H(ϕQµ−νQ)k^p_Lp(µ)6kχQR^H(ϕQµ−νQ)k^p_Lp(µ)6Cµ(Q).

Proof. As we have already mentioned in§5,R^H_µ is bounded in bothL²(µ) andL⁴(µ), so we even have

kR^H_µϕ_Qk^p_Lp(µ)6CkϕQk^p_Lp(µ)6Cµ(Q)

for both values ofpwe are interested in and the cutoffsϕQ andχQ can only diminish the left-hand side. Thus, we only need to prove the first chain of inequalities in the lemma.

The left inequality is trivial becauseϕQ6χQpointwise. To prove the right inequality, fix any Lipschitz functionϕe0:R^d+1![0,1] such thatϕ0=1 onB(0,4) andϕ0=0 outside B(0,5), put

ϕeQ(x) =ϕe0

x−zQ

`(Q)

, and write

kχQR^HνQk^p_Lp(µ)= Z

Q

|R^HνQ|^pdµ6 Z

ϕeQ|R^HνQ|^pdµ.

Let

˜ aQ=

Z

ϕeQdm_L(Q) −1Z

ϕeQdµ.

Note that both integrals in the definition of ˜aQare comparable to`(Q)^d, so ˜aQ6C. Put

˜

ν_Q= ˜a_Qm_L(Q).

SinceR^H_m

L(Q) is bounded inL^p(m_L(Q)), we have Z

|R^Hν_Q|^pd˜ν_Q6C Z

|R^Hν_Q|^pdm_L(Q)6CkϕQk^p_Lp(m_L(Q))6C`(Q)^d6Cµ(Q).

On the other hand, theC²-estimates for ϕQ in the beginning of this section combined with Lemma1imply that

kR^HνQkL^∞6C

ε² and kR^HνQkLip6 C ε²`(Q).

In addition, we clearly havekϕe_QkLip6C/`(Q). Thus, whenα<ε⁸<1, Lemma3 immedi-ately yields

Z

|R^Hν_Q|^pd(ϕe_Qµ−˜ν_Q)6Cα`(Q)^d+2 1 ε^2(p−1)

1 ε²`(Q)

1

`(Q)=Cαε<sup>−2p</sup>`(Q)^d6Cµ(Q) forp=2,4, so

Z

ϕe_Q|R^Hν_Q|^pdµ= Z

|R^Hν_Q|^pd(ϕe_Qµ)

= Z

|R^HνQ|^pd˜νQ+ Z

|R^HνQ|^pd(ϕeQµ−˜νQ)6Cµ(Q), as required.

Now representFk as F_k=

X

Q∈Qk

ϕ_QR^H_µϕ_Q− X

Q∈Qk+1

ϕ_QR^H_µϕ_Q

− X

Q∈Qk

ϕ_QR^Hν_Q+ X

Q∈Qk+1

ϕ_QR^Hν_Q

=F_k⁽¹⁾−F_k⁽²⁾+F_k⁽³⁾. Note that

kR^H_µ(ϕQ−χQ)k^p_Lp(µ)6CkϕQ−χQk^p_Lp(µ)6Cµ(Q\Q3ε)6Cε^γµ(Q) forp=2,4. Also

k(ϕQ−χQ)R^H_µχQk²_L2(µ)6kϕQ−χQk²_L4(µ)kR_µ^HχQk²_L4(µ)

6CkϕQ−χQk²_L4(µ)kχQk²_L4(µ)

6Cµ(Q\Q3ε)^1/2µ(Q)^1/2 6Cε^γ/2µ(Q).

Thus,

kϕQR^H_µϕQ−χQR^H_µχQk²_L2(µ)62[kϕQR^H_µ(ϕQ−χQ)k²_L2(µ)+k(ϕQ−χQ)R_µ^HχQk²_L2(µ)] 6C[ε^γ+ε^γ/2]µ(Q)6Cε^γ/2µ(Q).

If we now set

Fe_k⁽¹⁾=

X

Q∈Q_k

χQR^H_µχQ− X

Q∈Q_k+1

χQR^H_µχQ

,

we immediately see that

kFe_k⁽¹⁾−F_k⁽¹⁾k²_L2(µ)6Cε^γ/2µ(P).

Combined with the estimatekGk+1k²_L2(µ)6Cµ(P), this yields

|hFe_k⁽¹⁾−F_k⁽¹⁾, Gk+1iµ|6kFe_k⁽¹⁾−F_k⁽¹⁾k_L2(µ)kGk+1k_L2(µ)6Cε^γ/4µ(P).

Now we can write

hFe_k⁽¹⁾, Gk+1iµ= X

Q∈Qk

Q⁰∈Q_k+1 Q⁰⊂Q

hχQR^H_µχQ−χQ⁰R^H_µχQ⁰, ϕQ⁰R^H(ϕQ⁰µ−νQ⁰)iµ

because all other scalar products correspond to pairs of functions with disjoint supports, and, thereby, evaluate to 0.

FixQ∈Qk. For each Q⁰∈Qk+1 contained inQ, we haveχQ=χQ⁰=1 on suppϕQ⁰, so, when writing the scalar product as an integral, we can leave only the factor ϕQ⁰ in front of the product of Riesz transforms, which allows us to combine two of them into one and represent the scalar product as

hR^H(χ_Q\Q⁰µ), ϕQ⁰R^H(ϕQ⁰µ−νQ⁰)iµ. The next estimate is worth stating as a separate lemma.

Lemma11. Suppose that F is any bounded function and Q∈Qk. Then X

Q⁰∈Q_k+1 Q⁰⊂Q

|hR^H(χ_Q\Q⁰F µ), ϕQ⁰R^H(ϕQ⁰µ−νQ⁰)iµ|6Cα^1/(d+2)ε⁻³kFk_L^∞_(Q)µ(Q).

Proof. Let ΨQ⁰=R^H(χ_Q\Q⁰F µ). By Lemma4, we have

|hΨQ⁰ϕQ⁰, R^H(ϕQ⁰µ−νQ⁰)iµ|

6Cα^1/(d+2)`(Q⁰)^d+2[kΨQ⁰k_L^∞_(suppϕ

Q0)+`(Q⁰)kΨQ⁰k_Lip(suppϕ

Q0)]kϕQ⁰k²_Lip. Note now that, by (2),

kΨQ⁰k_Lip(suppϕ

Q0)6 CkFk_L^∞_(Q)

dist(suppϕ_Q⁰, Q\Q⁰)6CkFk_L^∞_(Q) ε`(Q⁰) and

kϕQ⁰kLip6 C ε`(Q⁰).

Thus, in our case, the bound guaranteed by Lemma4 does not exceed Cα^1/(d+2)`(Q<sup>0</sup>)<sup>d</sup>ε<sup>−2</sup>[kΨ<sub>Q</sub><sup>0</sup>k<sub>L</sub>∞(suppϕ<sub>Q</sub>0)+ε<sup>−1</sup>kFk<sub>L</sub>∞(Q)], so, taking into account that`(Q⁰)^d6Cµ(Q⁰), we get

X

Q⁰∈Q_k+1 Q⁰⊂Q

|hR^H(χ_Q\Q⁰F µ), ϕQ⁰R^H(ϕQ⁰µ−νQ⁰)iµ|

6Cα^1/(d+2)ε⁻² X

Q⁰∈Qk+1

Q⁰⊂Q

[kΨQ⁰k_L∞(suppϕ_Q0)+ε⁻¹kFk_L∞(Q)]µ(Q⁰)

6Cα^1/(d+2)ε⁻²

"

ε⁻¹kFkL^∞(Q)µ(Q)+ X

Q⁰∈Qk+1

Q⁰⊂Q

kΨQ⁰kL^∞(suppϕ_Q0)µ(Q⁰)

# .

Since theL^∞ norm of a Lipschitz function on a set does not exceed the average of the absolute value of the function over the set plus the product of the Lipschitz norm of the function on the set and the diameter of the set, we have

kΨQ⁰k_L∞(suppϕ_Q0)6Cε⁻¹kFk_L∞(Q)+ Z

ϕQ⁰dµ −1Z

|ΨQ⁰|²ϕQ⁰dµ ^1/2

=Cε⁻¹kFk_L^∞_(Q)+J(Q⁰).

However,

Z

ϕQ⁰dµ>c`(Q⁰)^d>cµ(Q⁰) and

Z

|Ψ_Q⁰|²ϕ_Q⁰dµ62 Z

Q⁰

|R^H_µ(F χ_Q)|²dµ+

Z

Q⁰

|R_µ^H(F χ_Q⁰)|²dµ

.

SinceR^H_µ is bounded inL²(µ), we have Z

Q⁰

|R^H_µ(F χQ⁰)|²dµ6CkF χQ⁰k²_L2(µ)6CkFk²_L∞(Q)µ(Q⁰)

for eachQ⁰⊂Q, and X

Q⁰∈Q_k+1 Q⁰⊂Q

Z

Q⁰

|R^H_µ(F χQ)|²dµ6 Z

Q

|R^H_µ(F χQ)|²dµ6CkF χQk²_L2(µ)6CkFk²_L∞(Q)µ(Q).

So we get

X

Q⁰∈Qk+1

Q⁰⊂Q

J(Q⁰)²µ(Q⁰)6CkFk²_L∞(Q)µ(Q).

Now it remains to apply Cauchy–Schwarz inequality to conclude that X

Q⁰∈Qk+1

Q⁰⊂Q

J(Q⁰)µ(Q⁰)6CkFk_L^∞_(Q)µ(Q),

thus completing the proof of the lemma.

Applying this lemma withF=1, we immediately get X

Q⁰∈Qk+1

Q⁰⊂Q

|hR^H(χ_Q\Q0µ), ϕ_Q⁰R^H(ϕ_Q⁰µ−νQ⁰)iµ|6Cα^1/(d+2)ε⁻³µ(Q).

It remains to sum these bounds overQ∈Qk and to combine the result with the previously obtained estimate forhFe_k⁽¹⁾−F_k⁽¹⁾, Gk+1iµ to conclude that

|hF_k⁽¹⁾, Gk+1iµ|6C(ε^γ/4+α^1/(d+2)ε⁻³)µ(P).

To estimatehF_k⁽²⁾, Gk+1iµ, note once more that by Lemma1,R^HνQ is a Lipschitz func-tion in R^d+1 with kR^HνQkL^∞6C/ε² and kR^HνQkLip6C/ε²`(Q). Since for any two Lipschitz functionsf andg one has

kf gk_L^∞6kfk_L^∞kgk_L^∞ and kf gk_Lip6kfk_Lipkgk_L^∞+kfk_L^∞kgk_Lip, we get

kϕQR^HνQkL^∞6C

ε² and kϕQR^HνQkLip6 C ε³`(Q).

Using Lemma4again and taking into account that`(Q<sup>0</sup>)6`(Q) forQ⁰⊂Q, we get

|hϕQR^HνQ, ϕQ⁰R^H(ϕQ⁰µ−νQ⁰)iµ|6Cα^1/(d+2)`(Q⁰)^d+2

Dans le document On the uniform rectifiability of AD-regular measures with bounded Riesz transform operator: (Page 50-57)