Proof of Theorem 2.3

The Covering Lemma 2.19 can be applied to Eλ,R with Q > 2R sufficiently large yielding a covering T(x_l, ξ_l, s_l)

l∈{1,...,L} such thatSL

l=13Q³T(x_l, ξ_l, s_l)⊃E_λ,Rwith the tentsT(x_l, ξ_l, s_l) that are pairwise Q⁺-disjoint. Now apply the Mass Projection Lemma 2.22 with respect to the strips D(ζ_m,3σ_m)

m∈{1,...,M} and the tents T(ξ_l, x_l, s_l)l∈{1,...,L}. The resulting ea satisfies keak_L∞(l^r0).ω while

Me⁺_2R(xl, ξl, sl)≥M⁺_R(xl, ξl, sl)≥λ.

Using the bound (2.63) and the locality property 2.21 we have that µ E_λ,R

.R L

X

l=1

s_l.λ^−r0kea1_B2sR(x)(z)k^r_L⁰r0(l^r0).s ω^r0λ^−r0. This concludes the proof.

Lemma 2.24(Variational truncations of singular integral operators [JSW08]). For any function H ∈L^p(R)andσ∈[0,∞)let us define the variational truncation operator

V_σ^rH(z) = sup

σ<t₁<···<tk<...

X

k

|H∗Υ_tk+1(z)−H∗Υ_tk(z)|^r

!^1/r

(2.71) where

Υ∈S(R), ˆ

R

Υ(z)dz= 1 andΥ_t(z) :=t⁻¹Υz t

. If r >2 and for anyp∈(1,∞), above operator satisfies the bounds

kV_σ^rHkL^p.^r,pkHkL^p (2.72)

and ifσ >0 then

V_σ^rH(z).r,p B_σ(z)

M(V_σ^rH) (z⁰)dz⁰ (2.73)

whereM is the Hardy-Littlewood maximal function. The implicit constants are allowed to depend onΥ.

We record some useful properties of so calledconvex regions of tents.

Definition 2.25(Convex regions). A convex region of a tent is a subsetΩ⊂T(x, ξ, s)of a tent of the form

Ω := [

θ∈Θ

Ωθ:=

(y, ξ+θt⁻¹, t)∈T(x, ξ, s) :t > σθ(y) . (2.74) for some function σθ(y) : Θ×Bs(x)→[0, s].

Given any tentT ∈T, any collection of stripsD, and any collection of tentsT, the set Ω =T \ [

D∈D

D ∪ [

T∈T

T

!

⊂T

is a convex region of the tent T. With the next lemma we show that the bound (2.70) on a convex regions can be extended to larger regions with scale bound σ that is Lipschitz in the space variable.

Lemma 2.26 (Lipschitz convex regions). Let T(x, ξ, s) ∈ T be a tent and Ω = S

θ∈ΘΩθ ⊂ T(x, ξ, s)be a convex region as defined in (2.74)and let us fix a constantL >2. For everyθ∈Θ such that the bound

M(y, η, t)≤λ ∀(y, η, t)∈Ωθ

holds forΩθ=

(y, ξ+θt⁻¹, t)∈T(x, ξ, s) :t > σθ(y) 6=∅

there exists a Lipschitz functioneσθ:R→R⁺ with Lipschitz constant L⁻¹<1/2 such that min 2s;L⁻¹dist(y;Bs(x))

≤eσθ(y)≤2s ∀y∈R (2.75)

eσθ(y)≤σθ(y) ∀y∈Bs(x) (2.76) and

sWs(x−y)M(y, ξ+θt⁻¹, t).^Lλ ∀y∈R, t∈(eσθ(y),3s). (2.77)

Proof. Fix θ ∈ Θ such that Ωθ is non-empty and let us drop the dependence on θ from the notation by simply writingσ(y)in place ofσθ(y). Let us set

eσ(y) := min 2s; eeσ(y)

witheeσ(y) := inf

y⁰∈Bs(x)

max

σ(y⁰); |y−y⁰| L

(2.78) Clearly, this defines a function on Rsuch that conditions (2.75) and (2.76) hold. The defined function is L⁻¹-Lipschitz. It is sufficient to show that eeσ is L⁻¹-Lipschitz: for any y ∈ R and ε >0there exitsy⁰∈B_s(x)such that

ee

σ(y)≥(1 +ε)⁻¹max

σ(y⁰); |y−y⁰| L

and thus for anyy⁰⁰∈Rone has ee

σ(y⁰⁰)≤max

σ(y⁰); |y⁰⁰−y⁰| L

≤max

σ(y⁰); |y−y⁰| L

+|y⁰⁰−y|

L

≤(1 +ε)eeσ(y) +|y⁰⁰−y|

L .

Since ε > 0 was arbitrary and one can invert the role of y⁰⁰ and y in the above reasoning we obtain that|eeσ(y⁰⁰)−σ(y)| ≤ee ^|y00_L^−y| as required.

Let us now check that (2.77) holds. Suppose thaty∈Randt∈(σ(y),e 3s]. Let us distinguish the casest∈(σ(y),e 2s)andt∈[2s,3s). In the first case there exists y⁰ ∈B_s(x)andt⁰ ∈(σ_θ(y⁰), s) such thatt⁰ ∈(t/2, t)and|y−y⁰|<2Ltand thus it follows that |x−y|<2Ls. It follows that

Wt(z−y).^LWt(z−y⁰).Wt⁰(z−y⁰) ∀z∈R thus sW_s(x−y)W_t(z−y).LW_t⁰(z−y⁰) ∀z∈R

In the case thatt∈[2s,3s]there also existsy⁰∈Bs(x)andt⁰∈(σθ(y⁰), s)such thatt⁰∈(t/2, t) sinceΩθ6=∅. It follows from (2.78) that|y⁰−y|>2Lsso |x−y| ≈L |y⁰−y|so for allz∈R

sWs(x−y)Wt(z−y).^LsWs(y⁰−y)Ws(z−y).Wt⁰(z−y⁰).

Thus, since in both cases(y⁰, ξ+θt⁰⁻¹, t⁰)∈Ωwe have by the definition (2.59) ofMthat sWs(x−y)M(y, ξ+θt⁻¹, t).^LM(y⁰, ξ+θt⁰⁻¹, t)≤λ

as required.

The next technical lemma will be used as a continuous stopping time argument. It relates the Lipschitz assumption on enlarged convex regions of the previous statement with a crucial measurability estimate.

Lemma 2.27(Continuous stopping time). Letσ:R→R⁺ be a Lipschitz function with Lipschitz constantL⁻¹<1. Then the function

ρ_σ(z) :=

ˆ

R

1

2σ(x)1_Bσ(x)(z−x)dx satisfies

1 +_L−1² −1

< ρσ(z)<1 +_L−1² and in particular for any non-negative functionh(z)the bounds

ˆ

R

h(z)dz≈L

ˆ

R B_σ(x)(x)

h(z)dzdx.

hold.

Proof. SinceσisL⁻¹-Lipschitz, for anyz∈Rwe have that

B_(1+L−1)⁻¹σ(z)(z)⊆ {x:z∈B_σ(x)(x)} ⊆B_(1−L−1)⁻¹σ(z)(z).

By the same reason on{x:z∈B_σ(x)(x)} we have that 1 +L⁻¹⁻¹

σ(z)≤σ(x)≤ 1−L⁻¹⁻¹ σ(z).

The conclusion follows.

2.5.2 Proof of Proposition 2.23

LetT =T(x, ξ, s)be a tent and suppose that K and E are as in 2.23. Since the statement of Proposition 2.23 is invariant under time and frequency translations, we may assume, without loss of generality, thatT is centered at the origin i.e. T =T(0,0, s). IfT\(K∪E) =∅ there is nothing to prove. Let us set

Θ_∗={θ∈Θ :∃(y, θt⁻¹, t)∈T(0,0, s)\(K∪E)}, Θ⁽ⁱ⁾_∗ := Θ⁽ⁱ⁾∩Θ_∗ Θ^(e)_∗ := Θ^(e)∩Θ_∗.

Forθ ∈Θ_∗, using Lemma 2.26 we may assume that there exists aL⁻¹-Lipschitz functionσ_θ : R→(0,2s], withL >4 sufficiently large to be chosen later, that satisfies condition (2.75) such that

sWs(y)M(y, θt⁻¹, t).λ ∀y∈R, θ∈Θ, t∈(σ(y),3s). (2.79) Let us setΩ =S

θ∈ΘΩθ, Ω⁽ⁱ⁾=S

θ∈Θ⁽ⁱ⁾Ωθ, andΩ^(e)=S

θ∈Θ^(e)Ωθ with Ωθ=

({(y, θt⁻¹, t)∈T(0,0, s) :t > σθ(y)} θ∈Θ_∗

∅ otherwise.

We need to show that

kA 1_X\(K∪E)kS_m(T)≤ kA 1ΩkS_m(T).λ ∀T ∈T or equivalently (see (2.42)) that

kA 1_Ω(i)k_S1(T⁽ⁱ⁾).λ kA 1ΩkS²(T).λ.

In this proof all our implicit constants depend on the choice ofL.

Let us fix a choice of left truncated wave packetsΨ^c_y,η,t^−,c+(z)in the defining expression (2.14). We will show that the statement holds in this case. The proof for right truncated wave packets is simmetric.

Comparing the definitions (2.14) and (2.59) forA andMrespectively, it follows from the bound

Ψ^c_y,η,t^k(z),ck+1(z)(z)

≤W_t(z−y)that

A(y, η, t).M(y, η, t), kA 1_Ωk_S∞(T):= sup

(y,η,t)∈Ω

A(y, η, t).λ. (2.80) This implies

1 s

˚

(y,η,t)∈Ω η<0

|A(y, η, t)|².λ 1 s

˚

(y,η,t)∈Ω η<0

|A(y, η, t)|dydηdt (2.81) and thus we may assume thatα⁻=β⁻<0< β⁺< α⁺ and we can reduce to showing

kA 1_Ω(i)k_S1(T⁽ⁱ⁾).λ kA 1_Ω(e)k_S2(T^(e)).λ. (2.82)

Proof of the first inequality of (2.82) It holds that

kA 1_Ω(i)k_S1(T⁽ⁱ⁾)≈

˚

Ω⁽ⁱ⁾

ˆ

R

X

k∈Z

ak(z)Ψ^c_y,η,t^k(z),ck+1(z)(z)dz

dydηdt

≤ 1 s

ˆ

θ∈Θ⁽ⁱ⁾∗

ˆ

y∈Bs

ˆ s t=σθ(y)

ˆ

z∈R

X

k∈Z

|ak(z)|

Ψ^c_y,θt^k(z),c−1,t^k+1(z)(z) dzdt

t dydθ.

According to (2.17) The wave-packet Ψ^c_y,θt^k(z),c−1,t^k+1(z)(z) vanishes unless θ−tc_k(z) ∈ B_ε(d) and tc_k+1(z)−θ > d⁰. Sinceθ∈Θ⁽ⁱ⁾⊂[−d⁰, d−ε], the integrand vanishes unlessc_k(z)<0<c_k+1(z).

Let k^∗_z ∈ Z be the index, if it exists, such that this inequality holds and seta^∗(z) := a_k^∗

z(z), c^∗(z) =c_k∗(z)(z). If no such index exists simply seta^∗(z) = 0.

Using that givent < s andy∈B_s one has

Ψ^c_y,θt^k(z),c−1,t^k+1(z)(z)

.sWs(z)t Wt(z−y)²≤Wt(z−y) and using the statement of Lemma 2.27 we have that

kA 1_Ω(i)k_S1(T⁽ⁱ⁾).1 s

ˆ

θ∈Θ⁽ⁱ⁾∗

ˆ

y∈Bs

ˆ s t=σθ(y)

ˆ

x∈R z∈B_σθ(x)(x)

|a^∗(z)|sWs(z)t Wt(y−z)²

×1B_ε(d)(θ−tc^∗(z))dzdx dt

t dydθ=I+II where

I:= 1 s

ˆ

θ∈Θ⁽ⁱ⁾∗

ˆ

x∈R z∈B_σθ(x)(x)

|a^∗(z)|

ˆ s

t=(1−2/L)σθ(x)

ˆ

y∈Bs

W_t(y−z)dy

×1_Bε(d)(θ−tc^∗(z))dt

t dzdxdθ II := 1

s ˆ

θ∈Θ⁽ⁱ⁾∗

ˆ

x∈R

ˆ

y∈Bs

ˆ (1−2/L)σθ(x)

t=σθ(y) z∈B_σθ(x)(x)

|a^∗(z)|s Ws(z)t Wt(y−z)²

×1B_ε(d)(θ−tc^∗(z))dz dt t dydxdθ Suppose thatL >2_α^α₊⁺_−d−ε^−α− so that for anyc∈Rone has

(θ−tc∈Bε(d)

t >(1−2/L)σθ(x) =⇒ θ−σθ(x)c∈Θ. (2.83) We begin by estimating the termI. Notice that if|x|>2Lsthen integrand vanishes. We bound I by the auxiliary embedding map (2.59) as follows:

I.1 s

ˆ

θ∈Θ⁽ⁱ⁾∗

ˆ

x∈R z∈B_σθ(x)(x)

|a^∗(z)|

ˆ s

t=(1−2/L)σθ(x)

1_Bε(d)(θ−tc^∗(z))dt

t dzdxdθ .1

s ˆ

θ∈Θ⁽ⁱ⁾∗

ˆ

x∈B2Ls B_σθ(x)(x)

|a^∗(z)|1_Θ∗(θ−σ_θ(x)c_k(z)) ln

β⁺−θ+ 3ε β⁺−θ

dzdxdθ

≤1 s

ˆ

θ∈Θ⁽ⁱ⁾∗

ˆ

x∈B2Ls

M(x, θσθ(x)⁻¹, σθ(x))dxln

β⁺−θ+ 3ε β⁺−θ

dθ.λ.

The last inequality holds since(x, θσ_θ(x)⁻¹, σ_θ(x))∈Ω_θ and follows from (2.79).

We now estimate the termII. Notice that

(1−2/L)σ_θ(x)> σ_θ(y) =⇒ |x−y| ≥L(σ_θ(x)−σ_θ(y))>2σ_θ(x) (2.84) Thus ifz∈Bσ_θ(x)(x)then|y−z|> σθ(x)> t, |x−y| ≈ |y−z|, and alsosWs(z).sWs(x)so

II . ˆ

θ∈Θ⁽ⁱ⁾∗

ˆ

x∈R

W_s(x) ˆ

y∈Bs

ˆ (1−2/L)σθ(x) t=σ_θ(y)

t W_t(y−x)

×

B_σθ(x)(x)

|a^∗(z)|Wt(y−z)1_Bε(d)(θ−tc^∗(z))dzdt t dydxdθ .

ˆ

θ∈Θ⁽ⁱ⁾∗

ˆ

x∈R

Ws(x) ˆ

y∈Bs

ˆ (1−2/L)σ_θ(x) t=σ_θ(y)

t

2σθ(x)Wt(y−x)M(y, θt⁻¹, t)dt

t dydxdθ.

Since the inmost integral vanishes unless|y−x|>2σθ(x), we have that ˆ (1−2/L)σθ(x)

t=σ_θ(y)

t

2σθ(x)Wt(y−x)dt

t .W_σθ(x)(y−x) and so using (2.79) we obtain

II .λ ˆ

θ∈Θ⁽ⁱ⁾∗

ˆ

x∈R

Ws(x) ˆ

y∈Bs

W_σθ(x)(y−x)dydxdθ .λ

ˆ

θ∈Θ⁽ⁱ⁾∗

ˆ

x∈R

Ws(x) dxdθ.λ.

This concludes the proof for the first bound of (2.82).

Proof of the second inequality of (2.82)

As noted in (2.81) we may suppose thatΘ^(e)= [β⁺, α⁺); for ease of notation setΩ^(e)= Ω∩T^(e) so the required quantity to bound becomes kA 1_Ω(e)k_S2(T^(e)). We concentrate on showing the dual bound

1 s

˚

X

h(y, η, t) ˆ

R

X

k∈Z

ak(z)Ψ^c_y,η,t^k(z),ck+1(z)(z)dzdydηdt

.λkh(y, η, t)k_L2

s^1/2 (2.85)

for anyh∈C_c^∞(Ω^(e))where thek · k_L2 is the classical LebesgueL²norm relative to the measure dydηdt. A change of variables and the Minkowski inequality give

1 s

˚

X

h(y, η, t) ˆ

R

X

k∈Z

ak(z)Ψ^c_y,η,t^k(z),ck+1(z)(z)dzdydηdt

≤ 1 s

ˆ

R

X

k∈Z

|ak(z)|

˚

Ω^(e)

h(y, η, t)Ψ^c_y,η,t^k(z),ck+1(z)(z)dydηdt

dz

≤ 1 s

ˆ

θ∈Θ^(e)∗

ˆ

R

X

k∈Z

|ak(z)|

ˆ

y∈Bs

ˆ s t=σ_θ(y)

h(y, t⁻¹θ, t)Ψ^c_y,θt^k(z),c−1,t^k+1(z)(z)dydt t

dzdθ.

On the other hand the Hölder inequality gives that ˆ

θ∈Θ^(e)∗

kh(y, t⁻¹θ, t)k_L2

(dydt/t)dθ= ˆ

θ∈Θ^(e)∗

ˆ

y∈Bs

ˆ s t=σθ(y)

|h(y, t⁻¹θ, t)|²dydt t

¹₂ dθ

.kh(y, η, t)kL²

wherek · k_L2

(dydt/t) is the classic LebesgueL²norm with respect to the measuredy^dt_t. Thus (2.85) follows by showing

1 s

ˆ

R

X

k∈Z

|ak(z)|

ˆ

y∈Bs

ˆ s t=σ_θ(y)

h(y, t⁻¹θ, t)Ψ^c_y,t^k(z),c−1θ,t^k+1(z)(z)dydt t

dz (2.86)

.λ

h(y, t⁻¹θ, t) _L2

(dydt/t)

s^1/2

with a constant uniform in θ ∈ Θ^(e)_∗ . For sake of notation from now on we will omit the dependence onθby writing

h(y, t) :=h(y, t⁻¹θ, t) Ψ^c_y,t^−,c+(z) := Ψ^c_y,t^−,c−1⁺θ,t(z) σ(x) :=σ_θ(x).

Using the above notation and Lemma 2.27 we write

1 s

ˆ

R

X

k∈Z

|ak(z)|

ˆ

y∈Bs

ˆ s t=σ(y)

h(y, t)Ψ^c_y,t^k(z),ck+1(z)(z)dydt t

dz.I+II where

I: = 1 s

ˆ

x∈B2Ls z∈Bσ(x)(x)

X

k∈Z

|ak(z)|

ˆ

y∈Bs

ˆs

t=(1−2/L)σ(x)

h(y, t)Ψ^c_y,t^k(z),ck+1(z)(z)dydt t

dzdx

II : = 1 s

ˆ

x∈R z∈Bσ(x)(x)

X

k∈Z

|ak(z)|

ˆ

y∈Bs

(1−2/L)σ(x)ˆ

t=σ(y)

h(y, t)Ψ^c_y,t^k(z),ck+1(z)(z)dydt t

dzdx

We start with boundingI. Suppose thatL >1 is chosen large enough so that (2.83) holds and recall thatΨ^c_y,t^k(z),ck+1(z)(z) = 0 unlessθ−tck(z)∈Bε(d). We thus have

I=1 s

ˆ

x∈B2Ls z∈B_σ(x)(x)

X

k∈Z

|ak(z)|

ˆ

y∈Bs

ˆs t=(1−_L²)σ(x)

h(y, t)Ψ^c_y,t^k(z),ck+1(z)(z)dydt t

dzdx

≤1 s

ˆ

x∈B_2Ls B_σ(x)(x)

X

k∈Z

|a_k(z)|^r01_Θ θ−σ(x)c_k(z)1/r⁰

H_x(z)dzdx

≤1 s

ˆ

x∈B2Ls

M(x, θσ(x)⁻¹, σ(x)) sup

z∈Bσ(x)(x)

Hx(z)dx≤ λ s ˆ

x∈B2Ls

sup

z∈Bσ(x)(x)

Hx(z)dx,

where

Hx(z) :=

X

k∈Z

ˆ

y∈Bs

ˆs t=(1−_L²)σ(x)

h(y, t)Ψ^c_y,t^k(z),ck+1(z)(z)dydt t

r^1/r .

We claim that

H_x(z).V_σ(x)^r H_s(z) +E_σ(x)(z) H_τ(z) :=

ˆ τ t=0

ˆ

y∈B_s

h(y, t)Ψ^0,+∞_y,t (z)dydt

t (2.87) E_σ(x)(z) :=

ˆ s

(1−2/L)σ(x)

|h^∗(z, t)|²dt t

!1/2

h^∗(z, t) :=

ˆ

R

|h(y, t)|W_t(z−y)dy

withV_σ(x)^r defined in Lemma 2.24, and that kE0kL² .khk_L²

(dydt/t) sup

z∈B_σ(x)(x)

Eσ(x)(z).

B_2σ(x)(z)

E0(z)dz (2.88)

kHskL² .khk_L²

(dydt/t) sup

z∈B_σ(x)(x)

Hx(z)dx.

B_2σ(x)(x)

MV^rH_s(z)dz. (2.89) This would provide us with the required bounds forI. As a matter of fact, according to Lemma 2.27 and 2.24 we have that

I.λ s

ˆ

x∈B2LsB_2σ(x)(x)

MV^rH_s(z) +E_σ(x)(z) dz.R

λ s^1/2

kMV^rH_sk_L2+kE₀(z)k_L2

. λ s^1/2

kHs(z)k_L2+kE0(z)k_L2

.λ

khk_L2 (dydt/t)

s^1/2 as required.

The first bound of (2.88) follows by the Young inequality and Fubini:

kE0k_L2 ≤

¨

R×R⁺

ˆ

R

|h(y, t)|Wt(z−y)dy

2

dzdt t

≤ ˆ

R⁺

ˆ

R

|h(y, t)|²dyˆ

W_t(z−y)dy2dt

t .khk²_L2 (dydt/t)

.

The second bound follows from the fact that for small enoughε >0and as long as|z−z⁰|< εt the bound

|h^∗(z, t)−h^∗(z⁰, t)| ≤ ˆ

R

|h(y, t)||Wt(z−y)−Wt(z⁰−y)|dy

≤2⁻¹⁰⁰ ˆ

R

|h(y, t)|Wt(z−y)dy= 2⁻¹⁰⁰h^∗(z, t) holds so similarly

E_σ(x)(z)− E_σ(x)(z⁰)

≤2⁻¹⁰⁰E_σ(x)(z) as long as|z−z⁰|< εσ(x)for some sufficiently smallε >0.

B_σ(x)(z)

E0(z⁰)dz⁰&

B_σ(x)(z)

E_σ(x)(z⁰)dz⁰&^ε

B_εσ(x)(z)

E_σ(x)(z)dz⁰=E_σ(x)(z)

and the claim follows.

The first bound of (2.89) uses standard oscillatory integral techniques: notice that fort > t⁰ one has

ˆ

R

Ψy,t(z)Ψy⁰,t⁰(z)dz . t⁰

tWt(y−y⁰) so

ˆ

|Hs(z)|²(z).2 ˆ s

t=0

ˆ t t⁰=0

ˆ

y∈Bs

ˆ

y⁰∈Bs

|h(y, t)||h(y⁰, t⁰)|Wt(y−y⁰)dydy⁰t⁰ t

dt t

dt⁰ t⁰ .khk²_L2

(dydt/t)

. The second bound follows directly from Lemma 2.24.

It remains to show inequality (2.87). Notice that Hx(z) = X

k∈Z

ˆ t⁺_k(z) t⁻_k(z)

ˆ

y∈Bs

h(y, t)Ψ^c_y,t^k(z),ck+1(z)(z)dydt t

r!1/r

where fork∈Zwe set

t⁺_k(z) := supn

t∈ (1−2/L)σ(x), s

: Ψ^c_y,t^k(z),ck+1(z)(z)6= 0o

(2.90) t⁻_k(z) := infn

t∈ (1−2/L)σ(x), s

: Ψ^c_y,t^k(z),ck+1(z)(z)6= 0o .

We have omitted writing the implicit dependence onx∈Rand we will simply ignore the indexes k∈Zfor which the above sets are empty. Notice that the intervals

t⁻_k(z), t⁺_k(z)

are disjoint.

According to the conditions (2.17) on the geometry of truncated wave packets the following bounds hold:

t⁺_k(z)ck(z)∈Bε(θ−d) t⁻_k(z)ck+1(z)≥θ+d⁰. (2.91) Using the smoothness conditions (2.15) on the wave packets and writing a Lagrange remainder term we have that

Ψ^c_y,t^k(z),ck+1(z)(z)−Ψ^0,+∞_y,t (z) ≤

|tck(z)|+ max (d⁰⁰−θ−tck+1(z); 0)

Wt(y−z) (2.92) so the bound

Hx(z)≤ Hx,1(z) +Hx,2(z)

Hx,1(z) := X

k∈Z

ˆ

y∈Bs

ˆ t⁺_k(z) t=t⁻_k(z)

h(y, t)Ψ^0,+∞_y,t (z)dydt t

r1/r

= X

k∈Z

|H_t+

k(z)(z)−H_t−

k(z)(z)|^r1/r

H_x,2(z) := X

k∈Z

ˆ t⁺_k(z) t⁻_k(z)

h^∗(z, t) |tc_k(z)|+ max(d⁰⁰−θ−tc_k+1(z),0)dt t

r1/r

holds. Notice that ˆ t⁺_k(z)

t⁻_k(z)

t²|c_k(z)|²dt

t ≤ |t⁺_k(z)ck(z)|² 2 ≤C_α+

ˆ t⁺_k(z) t⁻_k(z)

max d⁰⁰−θ−tc_k+1(z); 0dt t ≤

ˆ d⁰⁰−θ t⁻_kc_k+1(z)

dt t ≤

ˆ d⁰⁰−θ d⁰+θ

dt

t .C_d0,d,α⁺,β+

for some constantC_α+ andC_d0,d,α⁺,β+. Sincer >2, Cauchy-Schwartz gives

Hx,2(z)≤ X

k∈Z

ˆ t⁺_k(z) t⁻_k(z)

h^∗(z, t)²dt t

!¹₂

.Eσ(x)(z).

This is consistent with (2.87). To estimate Hx,1: introduce a frequency cutoff Υ ∈S(R)such that

Υb ∈C_c^∞(Bθ+b) Υb ≥0 Υ = 1b onBθ Υτ(z) :=τ⁻¹Υz τ

. According to (2.16),Ψb^0,+∞_y,t is supported onB_t−1b(t⁻¹θ)so one has the following

Ψ^0,+∞_y,t ∗Υτ(z) = Ψ^0,+∞_y,t (z) if t

τ ≥θ+b θ Ψ^0,+∞_y,t ∗Υ_τ(z) = 0 if t

τ <θ−b θ+b Ψ^0,+∞_y,t ∗Υ_τ(z)

.W_t(z−y) if θ−b θ+b ≤ t

τ < θ+b θ . Thus

Hτ−Hs∗Υτ(z) .

ˆ τ^θ+b_θ τ^θ−b_θ+b

h^∗(z, t)dt t so

H_x,1(z). X

k∈Z

|H_s∗Υ_t+

k(z)−H_s∗Υ_t− k(z)|^r1_r

+ X

k∈Z

ˆ t⁻_k(z)^θ+b_θ t⁻_k(z)^θ−b_θ+b

h^∗(z, t)dt t

r¹_r

+ X

k∈Z

ˆ t⁺_k(z)^θ+b_θ t⁺_k(z)^θ−b_θ+b

h^∗(z, t)dt t

r¹_r

.V_σ(x)^r Hs(z) +E_σ(x)(z) thus concluding the proof of (2.87) and the bound on the termI.

The estimate for the termII can be done in a manner similar to the termIIin for theS¹part of the size. Recall (2.84) so that in expression forII one has thatz∈B_σ(x)(x),|y−z|> σ(x)> t, and|x−y| ≈ |y−z|. We also have thaty∈B_sso

Ψ^c_y,t^k(z),ck+1(z)(z).sWs(x)tWt(z−y)² andΨ^c_y,t^k(z),ck+1(z)(z) = 0unlesstc_k(z)< θ < tc_k+1(z), thus

II .1 s

ˆ

x∈R

sWs(x) ˆ

y∈Bs

ˆ (1−2/L)σ(x) t=σ(y)

tWt(y−x)h(y, t)

B_σ(x)(x)

X

k∈Z

|ak(z)|

×1_Θ(θ−tc_k(z))W_t(z−y)1_{(ck(z),ck+1(z))}(t⁻¹θ)dzdt t dydx .

ˆ

x∈R

W_s(x) ˆ

y∈Bs

ˆ (1x−2/L)σ(x) t=σ(y)

t

2σ(x)W_t(y−x)h(y, t)

× ˆ

B_σ(x)(x)

X

k∈Z

|ak(z)|^r01Θ(θ−tck(z))1/r⁰

Wt(y−z) dz dt t dydx .

ˆ

x∈R

Ws(x) ˆ

y∈Bs

ˆ (1x−2/L)σ(x) t=σ(y)

t

2σ(x)Wt(y−x)h(y, t)M(y, θt⁻¹, t)dt t dydx Since the inmost integral vanishes unless|y−x|>2σθ(x), we have that

ˆ (1−2/L)σθ(x) t=σ_θ(y)

t

2σ(x)W_t(y−x)

2dt t

^1/2

.W_σ(x)(y−x) it follows that

II .λ ˆ

x∈R

Ws(x) ˆ

y∈Bs

W_σ(x)(y−x)ˆ s t=0

|h(y, t)|²dt t

1/2

dydx

.λ ˆ

x∈R

W_s(x)M ˆ s

t=0

|h(·, t)|²dt t

^1/2 (x)dx

. λkhk_L2

(dydt/t)

s^1/2 This concludes the proof.

Im Dokument Time-Frequency Analysis of the Variational Carleson Operator using outer-measure Lp spaces (Seite 69-79)