Elimination of the x -Dependence from the Highest Order Term

Ta1(ϕ,x)d x. (4.9) This will allow us to conclude the proof of Proposition3.6.

4.1 Elimination of thex-Dependence from the Highest Order Term Consider an analytic functionα(ϕ,x)(to be determined) and let

T1u(ϕ,x):=(1+αx(ϕ,x))(Au)(ϕ,x), Au(ϕ,x):=u(ϕ,x+α(ϕ,x)).

We chooseα(ϕ,x)andm3(ϕ)in such a way that (λ3+d3(ϕ,x))

1+αx(ϕ,x)3

=m3(ϕ), (4.10)

which implies

α(ϕ,x):=∂_x⁻¹ m₃(ϕ)¹³ λ3+d3(ϕ,x)¹

3

−1

, m₃(ϕ):= 1 2π

T

d x

λ3+d3(ϕ,x)¹

3

₋3

. (4.11) By (4.2), (4.5) and LemmaA.5one has

|m₃−λ3|_ρ,|α|_ρ δ (4.12)

Note that for any 0< ζ ρsuch thatδζ⁻¹ 1, by LemmaA.1,x →x+α(ϕ,x)is invertible and the inverse is given byy→y+α(ϕ,y)with

α∈H(T^∞_ρ−ζ×Tρ−ζ), |α|_ρ−ζ,|α|_ρ δ. (4.13) A direct calculations shows that

A⁻¹u(ϕ,y)=u(ϕ,y+α(ϕ,y)), T⁻¹₁ =(1+αy)A⁻¹ (4.14)

and the following conjugation rules hold:

T⁻¹₁ a(ϕ,x)T1 =A⁻¹a(ϕ,x)A=(A⁻¹a)(ϕ,y), T⁻¹₁ ∂xT1=

1+A⁻¹(αx)

∂y+(1+αy)A⁻¹(αx x),

T⁻¹₁ ω·∂_ϕT1=ω·∂_ϕ+A⁻¹(ω·∂_ϕα)∂y+(1+αy)A⁻¹(ω·∂_ϕαx).

(4.15)

Clearly one can get similar conjugation formulae for higher order derivatives, having expression similar to (3.41). In conclusion

L⁽¹⁾:=T⁻¹₁ (L+Q)T1

=ω·∂ϕ+A⁻¹

(λ3+q₃)(1+αx)³

∂³y

+b₂(ϕ,y)∂_y²+b₁(ϕ,y)∂y+b₀(ϕ,y)

=ω·∂ϕ+m₃(ϕ)∂_x³+b₁(ϕ,x)∂x+b₀(ϕ,x)

(4.16)

for some (explicitly computable) coefficientsbi, where in the last equality we used (4.10) and the fact thatT1is symplectic, so thatb2(ϕ,x)=2∂xm3(ϕ)=0.

Furthermore, the estimates (4.2), (4.3), (4.12), (4.13), CorollaryA.2and LemmataA.3, A.4imply that for 0< ζ ρ

|b_i|_ρ−2ζδ0, |b_i−a_i|_ρ−2ζ ζ^−τδ, for some τ >0. (4.17) 4.2 Elimination of the'-Dependence from the Highest Order Term

We now consider a quasi periodic reparametrization of time of the form

T2u(ϕ,x):=u(ϕ+ωβ(ϕ),x) (4.18) whereβ:T^∞_ρ−ζ →Ris an analytic function to be determined. Precisely we chooseλ⁺₃ ∈R andβ(ϕ)in such a way that

λ⁺₃

1+ω·∂_ϕβ(ϕ)

=m3(ϕ), (4.19)

obtaining thus

λ⁺₃ :=

T^∞m3(ϕ)dϕ, β(ϕ):=(ω·∂ϕ)⁻¹m3

λ⁺₃ −1

(4.20) where we recall the definition A.3. By the estimates (4.12) and by Lemma3.3, one obtains that for 0< ζ ρ

|λ⁺₃ −λ3|δ, |β|_ρ−ζ e^C0ζ−μδ. (4.21) By LemmaA.1and (4.6) we see thatϕ→ϕ+ωβ(ϕ)is invertible and the inverse is given byϑ→ϑ+ωβ(ϑ)with

β∈H(T^∞_ρ−2ζ), |β|_ρ−2ζe^C0ζ−μδ. (4.22) The inverse of the operatorT2is then given by

T⁻₂¹u(ϑ,x)=u(ϑ+ωβ(ϑ), x). (4.23)

so that

Therefore by the estimates (4.12), (4.21), (4.22) and by applying Corollary A.2, LemmaA.5, and (4.6), one gets

|r−1|_ρ−ζδ

|c_i−a_i|_ρ−ζζ^−τe^C0ζ−μδ, i=0,1. (4.26) 4.3 Time Dependent Traslation of the Space Variable

Letp:T^∞_ρ−2ζ →Rbe an analytic function to be determined and let

We want to choosep(ϕ)in such a way that thex-average ofd₁is constant. To this purpose we define (recall the definition A.3). By (4.26) and Lemma3.3one gets

|p|_ρ−2ζ ζ^−τe^2C0ζ−μδ^(4.6) ζ. (4.31)

Moreover

λ⁺₁ := 1 2π

Td₁(ϕ,x)d x=λ1+ (c₁−a₁)ϕ,x. (4.32) Finally using (4.26), (A.2) (with_α=T⁻¹₃ ), (4.31), one gets

|a_i⁺−ai|_ρ−2ζζ^−τe^2C0ζ−μδ, (4.33) for someτ>0.

4.4 Conclusion of the Proof

We start by noting thatT:=T3◦T2◦T1has the form (3.33) withp⁽ⁿ⁺¹⁾= p,β⁽ⁿ⁺¹⁾=β andξ⁽ⁿ⁺¹⁾(ϕ,x) = α(ϕ+ωβ(ϕ),x+ p(ϕ)). Hence, settingr:= r_n+1,ρ := s_n −σn, δ:=σ_n⁻⁴εn,δ0:=2ε0andζ :=σn we denote

1+An+1=λ⁺₃, ,Bn+1(ϕ,x):=a₁⁺(ϕ,x), Cn+1=a₀⁺(ϕ,x),

and thus Proposition3.6follows.

5 Proof of Proposition3.8

In order to prove Proposition3.8, we start by considering a linear Hamiltonian operator defined forω∈O⊆D_γ of the form

L=L(λ3,a₁,a₀):=ω·∂ϕ+λ3∂_x³+a₁(ϕ,x)∂x+a₀(ϕ,x). (5.1) We want to show that, for any choice of the coefficientsλ3,a1,a0satisfying some hypothe-ses (see below), it is possible to reduceLto constant coefficients. Moreover we want to show that such reduction is “Lipshitz” w.r.t. the parametersλ3,a₁,a₀, in a sense that will be clarified below.

Regarding the coefficients, we need to require that ai :=

m k=0

a_i^(k), |a^(k)_i |^O_ρk δk, ∀k =0, . . . ,m, i=0,1,

|λ3−1|^Oδ0, λ1≡λ1(a1)=

m k=0

λ^(k)₁ , λ^(k)₁ := 1 2π

Ta^(k)₁ (ϕ,x)d x=const.

(5.2)

for some 0< . . . < ρm< . . . < ρ0 and 0< . . .δm . . .δ01 so that there is a third sequenceζisuch that 0< ζi< ρiand

i≥0

ζ_i^−τe^Cζi−μδiδ0, (5.3)

for someτ,C >0.

5.1 Reduction of the First Order Term

We consider an operatorLof the form (5.1) satisfying the hypotheses above. We start by showing that it is possible to reduce it to constant coefficients up to a bounded reminder, and that such reduction is “Lipshitz” w.r.t. the parametersλ3,a1,a0.

Lemma 5.1 There exists a symplectic invertible operatorM=exp(G), withG≡G(λ3,a₁) and an operatorR0 ≡R0(λ3,a₁,a₀)satisfying

G= m i=0

G⁽ⁱ⁾, G^(i)O_ρi,−1δi,

R0= m i=0

R⁽₀ⁱ⁾, R⁽₀^i)O_ρi−ζi ζ_i^−τe^Cζi−μδi

(5.4)

for some C, τ 1, such that

L0:=M⁻¹LM=ω·∂ϕ+λ3∂_x³+λ1∂x+R0. (5.5) Proof We look forGof the form

G=π₀^⊥g(ϕ,x)∂x⁻¹

and we choose the functiong(ϕ,x)whereg=g(λ3,a1)in order to solve

3λ3∂xg(ϕ,x)+a₁(ϕ,x)=λ1. (5.6) By (5.2), one obtains that

g:= 1 3λ3∂x⁻¹

λ1−a₁

(5.7) and therefore

g= m i=0

gi, gi := 1 3λ3

∂_x⁻¹

λ⁽ⁱ⁾₁ −a₁⁽ⁱ⁾ ,

|g_i|^O_ρi δi, i =0, . . . ,m.

(5.8)

Of course we can also write the operatorG :=π₀^⊥g(ϕ,x)∂_x⁻¹ =_m

i=0Gi whereGi :=

π₀^⊥gi(ϕ,x)∂_x⁻¹and one has

Gi^O_ρi,−1δi, i=0, . . . ,m. (5.9) Again by (5.2), definingP := a1∂x +a0, one has thatP = _m

i=0Pi, wherePi :=

a₁⁽ⁱ⁾∂x+a₀⁽ⁱ⁾satisfies

Pi^O_ρi,1δi. (5.10)

Therefore

L0=M⁻¹LM=e^−Gω·∂_ϕe^G+λ3e^−G∂_x³e^G+e^−GPe^G

=ω·∂ϕ+λ3∂³_x+

3λ3gx+a1

∂x+R0 (5.6)

= ω·∂ϕ+λ3∂_x³+λ1∂x+R0

(5.11)

where R0:=

e^−Gω·∂ϕe^G−ω·∂ϕ

+λ3

e^−G∂_x³e^G−∂_x³−3gx∂x

+

e^−GPe^G−P +a0.

(5.12) Then (5.3), (5.9), (5.10) guarantee that the hypotheses of LemmataA.10-A.11are verified.

Hence, we apply LemmaA.10-(ii)to expand the operatore^−GPe^G−P, LemmaA.11-(ii)to expande^−G∂_x³e^G−∂_x³−3g_x∂xand LemmaA.11-(iii)to expande^−Gω·∂ϕe^G−ω·∂ϕ. The expansion of the multiplication operatora0is already provided by (5.2). Hence, one obtains

that there existC, τ 1 such that (5.4) is satisfied.

We now consider a “small modification” of the operatorLin the following sense. We consider an operator

L⁺=L(λ⁺₃,a⁺₁,a⁺₀):=ω·∂_ϕ+λ⁺₃∂_x³+a⁺₁(ϕ,x)∂x+a⁺₀(ϕ,x) (5.13) with

1 2π

Ta₁⁺(ϕ,x)d x =:λ⁺₁ =const, |a_i⁺−a_i|ρm+1,|λ⁺₃ −λ3|δm+1. (5.14) Of course we can apply Lemma5.1and conjugateL⁺to

L⁺₀ :=ω·∂ϕ+λ⁺₃∂x³+λ⁺₁∂x+R⁺₀ (5.15) withR⁺₀ a bounded operator. We want to show thatL⁺₀ is “close” toL0, namely the following result.

Lemma 5.2 One has

|λ⁺₁ −λ1|δm+1, R⁺₀ −R0ρm+1−ζm+1 ζ_m+1^−τ e^Cζm+1−μδm+1. (5.16) Proof The first bound follows trivially from (5.14). Regarding the second bound one can reason as follows. As in Lemma5.1, er can defineG⁺:=π₀^⊥g⁺(ϕ,x)∂⁻¹_x with

g⁺ := 1 3λ⁺₃ ∂_x⁻¹

λ⁺₁ −a₁⁺

(5.17) so that

G⁺−G_ρm+1,−1δm+1. (5.18)

DefiningP⁺:=a₁⁺∂x+a⁺₀ and recalling thatP:=a₁∂x+a₀, by (5.14), one gets

P⁺−P_ρm+1,1δm+1. (5.19)

The estimate onR⁺₀ −R0follows by applying LemmataA.13,A.14, and by the estimates

(5.14), (5.19), (5.18).

5.2 Reducibility

We now consider an operatorL0of the form

L0 ≡L0(λ1, λ3,P0):=ω·∂_ϕ+D0+P0 (5.20)

withP0a bounded operator and

D0≡D0(λ1, λ3):=i diag_j∈Z\{0}0(j), 0(j):= −λ3j³+λ1j, j∈Z\ {0}, (5.21) and we show that, under some smallness conditions specified below it is possible to reduce it to constant coefficients, and that the reduction is “Lipschitz” w.r.t. the parametersλ1, λ3,P0. In order to do so, we introduce three sequences 0< . . . < ρm < . . . < ρ0, 0< . . . δm. . .δ0and 1N0 N1 · · ·and we assume that settingi =ρi−ρi+1one

has

i≥0

^−τ_i e^C−μi δi δ0, (5.22)

e^−Nkkδk+e^C−μk δ²_k2^−kδk+1, (5.23) δk(1+N_k)^{−C N}

1+η1

k (5.24)

and

|λ3−1|^O,|λ1|^O≤δ0, P0:=

m i=0

P⁽ⁱ⁾₀ , P⁽ⁱ⁾₀ ^O_ρi ≤δi, i=0, . . . ,m, (5.25) for someτ,C >0.

We have the following result.

Lemma 5.3 Fixγ ∈ [γ0/2,2γ0]. For k = 0, . . . ,m there is a sequence of setsEk ⊆Ek−1

and a sequence of symplectic mapskdefined forω∈Ek+1such that settingL0as in(5.20) and for k≥1,

Lk:=⁻_k−1¹ Lk−1k−1, (5.26)

one has the following.

1. Lkis of the form

Lk :=ω·∂ϕ+Dk+Pk (5.27)

where

• The operatorDkis of the form

Dk:=diag_j∈Z\{0}k(j), k(j)=0(j)+rk(j) (5.28) with r0(j)=0and for k≥1, rk(j)is defined forω∈E0=Oand satisfies

sup

j∈Z\{0}|r_k(j)−r_k−1(j)|^O≤δk−1 k−1

i=1

2⁻ⁱ. (5.29)

• The operatorPkis such that for 0≤k≤m, Pk=

m i=k

P⁽ⁱ⁾_k , P⁽ⁱ⁾_k ^E_ρ^k_i ≤δi

k j=1

2^−j, ∀i=k, . . . ,m.

(5.30)

2. One hask−1:=exp(k−1), such that

k−1^E_ρ^k_k e^C−μk−1P⁽_k−1^k−1)Eρ^k−1k−1 e^C−μk−1δk−1 (5.31) 3. The setsEkare defined as

Ek:=

ω∈Ek−1 : |ω·+k−1(j)−k−1(j)| ≥ γ|j³− j³| d() ,

∀(,j,j)=(0,j,j), ||η≤N_k−1 .

(5.32)

Proof The statement is trivial fork=0 so we assume it to hold up tok<mand let us prove it fork+1. For anyk:=exp(k)one has

Lk+1=⁻¹_k Lkk=ω·∂ϕ+Dk+ω·∂ϕk+ [Dk, k] +NkP^(k)_k +Pk+1 (5.33) where the operatorPk+1is defined by

Pk+1:=^⊥_NkP^(k)_k +

p≥2

Ad^p

k(ω·∂ϕ+Dk)

p! +

m i=k+1

e^−kP⁽ⁱ⁾_k e^k

+

p≥1

Ad^p

k(P⁽_k^k)) p! .

(5.34)

Then we choosekin such a way that

ω·∂ϕk+ [Dk, k] +NkP^(k)_k =Zk,

Zk:=diag_j∈Z\{0}(P⁽_k^k))^j_j(0), (5.35)

namely forω∈E_k+1we set

(k)^j_j():=

⎧⎪

⎨

⎪⎩

(P^(k)_k )^j_j() i

ω·+k(j)−k(j), ∀(,j,j)=(0,j,j), ||_η≤Nk,

0 otherwise.

(5.36) Therefore,

|(k)^j_j()|d()|(P^(k)_k )^j_j()|, ∀ω∈Ek+1. (5.37) and by applying LemmaA.6, using the induction estimate (5.30), one obtains

k^E_ρ^k+1_k−ζ e^Cζ−μP^(k)_k ^E_ρ^k_k^(5.30) e^Cζ−μδk, (5.38) for anyζ < ρk.

We now define the diagonal partDk+1.

For any j ∈ Z\ {0}and anyω ∈ Ek one has |(P⁽_k^k))^j_j(0)| P⁽_k^k)Eρ^kk (5.30)

≤ δk. The Hamiltonian structure guarantees thatP⁽_k^k)(0)^j_j is purely imaginary and by the Kiszbraun

Theorem there exists a Lipschitz extensionω ∈ O → iz_k(j) (with z_k(j) real) of this

We now estimate the remainderPk+1in (5.34). Using (5.35) we see that Pk+1=^⊥_NkP^(k)_k + By applying (A.7) and the estimate of LemmaA.9-(iii), one obtains

and similarly

m≥1

Ad^m

k(P⁽_k^k)) m! ^Ek+1

ρk+1e^C−μk δ²_k. (5.46) In conclusion we obtained

P^(k+1)_k+1 ^Eρ^k+1k+1 ≤Ce^−Nkkδk+Ce^C−μk δ²_k+δk+1

k j=1

2^−j (5.47)

where C is an appropriate constant and the last summand is a bound for the term e^−kP⁽_k^k+1)e^k, which can be obtained reasoning as in (5.43). Thus we obtain

P⁽_k+1^k+1)E_ρ^k+1_k+1 ≤δ_k+1

k+1

j=1

2^−j (5.48)

provided

Ce^−Nkkδk+Ce^C−μk δ_k²+δk+1

k j=1

2^−j ≤δk+1

k+1 j=1

2^−j,

which is of course follows from (5.23).

Now that we reducedL0to the formLm=ω·∂ϕ+Dm+Pmwe can apply a “standard”

KAM scheme to complete the diagonalization. This is a super-exponentially convergent iterative scheme based on iterating the following KAM step.

Lemma 5.4 (The(m+1)-th step)Following the notation of Lemma5.3we define Em+1:=

ω∈Em: |ω·+m(j)−m(j)| ≥γ|j³−j³| d() ,

∀(,j,j)=(0,j,j), ||_η≤Nm

and fix anyζsuch that

e^−Nmζδm+e^Cζ−μδm² δm+1 (5.49) Then there exists a change of variablesm:=exp(m), such that

m^E_ρ^m+1_m−ζ e^Cζ−μδm (5.50)

which conjugatesLmto the operator

Lm+1=ω·∂_ϕ+Dm+1+Pm+1.

The operatorD_m+1is of the form(5.28)and satisfies(5.29), with km+1, while the operatorPm+1is such that

Pm+1^E_ρ^m+1_m−ζ ≤δm+1. (5.51)

Proof We reason similarly to Lemma5.3i.e. we fixmin such a way that ω·∂_ϕm+ [Dm, m] +N_mPm=Zm,

Zm :=diag_j∈Z\{0}(Pm)^j_j(0), (5.52)

so that we obtains

m^E_ρ^m+1_m−ζ e^Cζ−μPm^E_ρ^m_m e^Cζ−μδm, (5.53) for anyζ < ρm.

Now, for any j ∈Z\ {0}and anyω ∈Em one has|(Pm)^j_j(0)| Pm^Eρ^mm≤2δm. The Hamiltonian structure guarantees thatPm(0)^j_j is purely imaginary and by the Kiszbraun Theorem there exists a Lipschitz extensionω ∈ O → izm(j) (withzm(j)real) of this function satisfying the bound|zm(j)|^Oδm. Then, we define

Dm+1:=diag_j∈Z\{0}m+1(j),

m+1(j):=m(j)+zm(j)=0(j)+r_m+1(j), ∀j∈Z\ {0}, r_m+1(j):=r_m(j)+z_m(j)

(5.54) and (5.29), withkm+1.

In order to obtain the bound5.51we start by recalling that Pm+1:=^⊥_NmPm+

p≥2

Ad^p−1

m (Zm−NmPm)

p! +

p≥1

Ad^p

m(Pm)

p! , (5.55)

so that reasoning as in (5.47) we obtain

Pm+1^E_ρ^m+1_m−ζ ≤Ce^−Nmζδm+Ce^Cζ−μδ²_m (5.56)

and by (5.49) the assertion follows.

We now iterate the step of Lemma5.4, using at each step a smaller loss of analyticity, namely at thep-th step we takeζpwith

p≥m+1

ζp=ζ,

so that we obtain the following standard reducibility result; for a complete proof see [21].

Proposition 5.5 For any j∈Z\ {0}, the sequencek(j)=0(j)+rk(j), k≥1provided in Lemmata5.3,5.4, and defined for anyω∈Oconverges to_∞(j)=0(j)+r_∞(j)with

|r_∞(j)−rk(j)|^Oδk. Defining the Cantor set E∞:=

ω∈O: |ω·+∞(j)−∞(j)| ≥2γ|j³−j³|

d() , ∀(,j,j)=(0,j,j) (5.57) and

L∞:=ω·∂ϕ+D∞, D∞:=i diag_j∈Z\{0}∞(j), (5.58) one hasE∞⊆ ∩k≥0Ek.

Defining also

k:=0◦. . .◦k with inverse ⁻_k¹=⁻_k¹◦. . .◦⁻₀¹, (5.59)

the sequencek converges for anyω ∈E∞to a symplectic, invertible map∞w.r.t. the norm · ^E_ρ^∞_m−2ζ and^±1_∞ −Id^E_ρ^∞_m−2ζ δ0. Moreover for any ω ∈ E∞, one has that ⁻¹_∞L0∞=L∞.

5.3 Variations

We now consider an operator

L⁺₀ ≡L0(λ⁺₁, λ⁺₃,P⁺₀)=ω·∂ϕ+D⁺₀ +P⁺₀, D⁺₀ :=λ⁺₃∂_x³+λ⁺₁∂x=i diag_j∈Z\{0}⁺₀(j), ⁺₀(j):= −λ⁺₃ j³+λ⁺₁ j, j ∈Z\ {0}.

(5.60)

such that

|λ⁺₁ −λ1|^O+,|λ⁺₃ −λ3|^O+,P⁺₀ −P0^O_ρm+1⁺ ≤δm+1 (5.61) whereL,λ1,λ3,P0are given in (5.21) andO⁺⊆O. In other words,L⁺₀ is a small variation ofL0in (5.20) with alsomm+1.

Of course we can apply Proposition5.5toL⁺₀; our aim is to compare the “final frequencies”

ofL⁺_∞with those ofL∞.

To this aim, we first apply Lemma5.3withL0 L⁺₀ andγ γ+ < γ. In this way we obtain a sequence of setsE⁺_k ⊆E⁺_k−1and a sequence of symplectic maps⁺_k defined for ω∈E⁺_k+1such that settingL⁺₀ as in (5.60) and

Lk:=⁻_k−1¹ Lk−1k−1, (5.62)

one has

L⁺_k :=ω·∂ϕ+D⁺_k +P⁺_k, k≤m+1, (5.63) where

D⁺_k :=diag_j∈Z\{0}⁺_k(j), ⁺_k(j)=⁺₀(j)+r_k⁺(j) (5.64) The setsE⁺_k are defined asE⁺₀ :=O⁺and fork≥1

E⁺_k :=

ω∈E⁺_k−1 : |ω·+⁺_k−1(j)−⁺_k−1(j)| ≥γ+|j³−j³| d() ,

∀(,j,j)=(0,j,j), |η| ≤N_k−1 .

(5.65)

Moreover one has⁺_k−1:=exp(_k−1⁺ ), with

_k−1^{+ E}_ρ^+k_k e^C−μk−1δk−1. (5.66) The following lemma holds.

Lemma 5.6 For all k=1, . . . ,m+1one has

P⁺_k −Pk^E_ρ^k_k^∩E+k ≤δm+1, (5.67a)

|r_k⁺(j)−rk(j)|^O∩O+≤δm+1 (5.67b)

and

_k⁺₋₁−k−1^Eρ^kk^∩E+k δm+1, (5.68) Proof We procede differently fork=1, . . . ,mandk=m+1.

For the first case we argue by induction. Assume the statement to hold up to somek<m.

We want to prove

_k⁺−k^Eρ^k+1k+1^∩E+k+1≤δm+1. (5.69) By Lemma5.3, one has forω∈E⁺_k+1

(_k⁺)^j_j():=

⎧⎪

⎨

⎪⎩

((P⁺_k)^(k))^j_j() i

ω·+⁺_k(j)−⁺_k(j), ∀(,j,j)=(0,j,j), ||η≤N_k,

0 otherwise,

(5.70) and direct calculation shows that forω∈Ek+1∩E⁺_k+1, one has

(⁺_k(j)−⁺_k(j))−(k(j)−k(j))≤δm+1|j³−j³| (5.71) and hence

|(_k⁺)^j_j()−(k)^j_j()|^Ek+1∩E+k+1 δm+1d()³|(P⁽_k^k))^j_j()|^Ek+1∩E+k+1

+d()²|(P^(k)_k )^j_j()−((P⁺_k)^(k))^j_j()|^Ek+1∩E+k+1. (5.72) Therefore, reasoning as in (5.37)–(5.38), one uses LemmaA.6, the smallness condition (5.23) and the induction estimate (5.67a) so that (5.69) follows.

Now, from the definition ofrk+1in (5.39) it follows

|r_k+1⁺ (j)−r_k+1(j)|^Ek+1∩E+k+1 ≤δ_m+1, (5.73) and by Kiszbraun Theorem applied tor_k⁺₊₁(j)−r_k+1(j), (5.67b) holds.

The estimate ofP⁺_k+1−Pk+1follows by explicit computation the difference by using the expressions provided in (5.41), using the induction estimates (5.30), (5.67a), the estimate (5.69) and by applying LemmaA.12.

Fork =m+1 the proof can be repeated word by word, the only difference being that mis defined in (5.52) while_m⁺is defined in (5.36) withk=m.

5.4 Conclusion of the Proof

To conclude the proof of Proposition3.8we start by noting that, settingOappearing in (5.2) asO⁽ⁿ⁾appearing in (3.11), the operatorLn+1appearing in (3.18) with of coursenn+1 is of the form (5.1) with

λ3=1+A_n+1,

a₁^(k)(ϕ,x)=Bk+1(ϕ,x)−Bk(ϕ,x), a₀^(k)(ϕ,x)=C_k+1(ϕ,x)−C_k(ϕ,x).

Moreover from (3.20) we have

δk=σ_k^−τ2e^Cσk−μεk, ρk=sk−3σk

wheres_k,σkandεkare defined in (3.6), so thatL_n+1satisfies (5.2) withm=n. Thus, fixing ζk =σk, 2ζ =σk,

the smallness conditions (5.3) follows by definition. Hence we can apply Lemma5.1toLn+1

obtaining an operator of the form (5.5). In particular the conjugating operatorMsatisfies M−Id^Os_n−3σn σ₀^−τ2e^Cσ0−με0.

We are now in the setting of Sect.5.2with

ρk=sk−4σk, δk =σ_k^−τ3e^2Cσ

−1 η+

k εk

for someτ3 > 0. A direct calculation shows that the smallness conditions (5.22), (5.23), (5.24), (5.49) are satisfied provided we chooseN_kappropriately, so that we can apply Propo-sition5.5.

In conclusion we obtain an operatorMn+1 =M◦_∞(recall thatMis constructed in Lemma5.1) satisfying (3.49), (3.50), where⁽ⁿ⁺¹⁾(j):=_∞(j)andE⁽ⁿ⁺¹⁾=E_∞. Note that in particular the functions⁽ⁿ⁺¹⁾(j)turn out to be of the form (3.46).

Finally (3.47) follows from Lemmata5.2and5.6whereL+has the role ofL_n+1whileL has the role ofL_n. This means that here we are takingmn−1.

Acknowledgements Riccardo Montalto is supported by INDAM-GNFM.

Funding Open access funding provided by Università degli Studi di Milano within the CRUI-CARE Agree-ment.

Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article’s Creative Commons licence, unless indicated otherwise in a credit line to the material. If material is not included in the article’s Creative Commons licence and your intended use is not permitted by statutory regulation or exceeds the permitted use, you will need to obtain permission directly from the copyright holder.

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A Technical Lemmata

We start by recalling few results proved in [21]. Of course, as already noted in [21]-Remark 2.2, all the properties holding forH(T^∞_σ+ρ, ^∞)hold verbatim forH(T^∞_σ+ρ×Tσ+ρ, ^∞). In particular, all the estimates below hold also for the Lipschitz norms| · |_σ and · _σ. Given two Banach spacesX,Ywe denote byB(X,Y)the space of bounded linear operators from XtoY.

Proposition A.1 (Torus diffeomorphism)Letα∈H(T^∞_σ+ρ, ^∞)be real on real. Then there exists a constantδ ∈ (0,1)such that if ρ⁻¹|α|_σ+ρ ≤ δ, then the map ϕ → ϕ+α(ϕ) is an invertible diffeomorphism ofT^∞_σ (w.r.t. the^∞-topology) and its inverse is of the formϑ → ϑ+α(ϑ), whereα ∈ H(T^∞_σ+ρ

2, ^∞)is real on real and satisfies the estimate

|α|_σ+^ρ

2 |α|_σ+ρ.

Corollary A.2 Givenα∈H(T^∞_σ+ρ, ^∞)as in PropositionA.1, the operators α:H(T^∞_σ+ρ,X)→H(T^∞_σ ,X), u(ϕ)→u(ϕ+α(ϕ)), α:H(T^∞_σ+ρ

2,X)→H(T^∞_σ ,X), u(ϑ)→u(ϑ+α(ϑ)) (A.1) are bounded, satisfy

α

B

H(T^∞_σ+ρ,X),H(T^∞_σ,X)

,α

B

H(T^∞_σ+ρ,X),H(T^∞_σ,X)

≤1,

and for anyϕ∈T^∞_σ, u∈H(T^∞_σ+ρ,X), v∈H(T^∞_σ+ρ

2,X)one has α◦αu(ϕ)=u(ϕ), α◦αv(ϕ)=v(ϕ).

Moreoveris close to the identity in the sense that

_α(u)−u_σ ρ⁻¹|α|_σ|u|_σ+ρ. (A.2) Given a functionu∈H(T^∞_σ ,X), we define its average on the infinite dimensional torus

as

T^∞u(ϕ)dϕ:= lim

N→+∞

1 (2π)^N

T^Nu(ϕ)dϕ1. . .dϕN. (A.3) By Lemma 2.6 in [21], this definition is well posed and

T^∞u(ϕ)dϕ=u(0) whereu(0)is the zero-th Fourier coefficient ofu.

Lemma A.3 (Algebra)One has|uv|σ≤ |u|σ|v|σfor u, v∈H(T^∞_σ ×Tσ).

Lemma A.4 (Cauchy estimates)Let u∈H(T^∞_σ+ρ×Tσ+ρ). Then|∂^ku|σ kρ^−k|u|σ+ρ. Lemma A.5 (Moser composition lemma)Let f :B_R(0)→Cbe an holomorphic function defined in a neighbourhood of the origin BR(0)of the complex planeC. Then the composition operator F(u):= f ◦u is a well defined non linear mapH(T^∞_σ ×T_σ)→H(T^∞_σ ×T_σ) and if|u|_σ ≤r<R, one has the estimate|F(u)|_σ 1+ |u|_σ. If f has a zero of order k at 0, then for any|u|σ ≤r<R, one gets the estimate|F(u)|σ |u|^k_σ.

For any functionu∈H(T^∞_σ ,X), givenN >0, we define the projectorNuas Nu(ϕ):=

||η≤N

u()e^i·ϕ and ^⊥_Nu:=u−Nu.

Lemma A.6 (i) Letρ >0. Then sup

∈Z^∞_∗

||η<∞

i

(1+ i⁵|i|⁵)e^−ρ||η ≤e^τln τ

ρ

ρ^−1η

for some constantτ =τ(η) >0.

(ii) Letρ >0. Then

∈Z^∞_∗

e^−ρ||η e^τln τ

ρ

ρ^−1η

, for some constantτ =τ(η) >0.

(iii) Letα >0. For N 1one has sup

∈Z^∞_∗ : ||α<N

i

(1+ i⁵|i|⁵)≤(1+N)^C(α)N1+α1 (A.4) for some constant C(α) >0such that C(α)→ ∞asα→0.

Lemma A.7 Given u∈H(T^∞_σ ,X)for X some Banach space, let g be a pointwise absolutely convergent Formal Fourier series such that

|g()|X ≤

i

(1+ i⁵|i|⁵)^τ|u|X,

for someτ>0. Then for any0< ρ < σ, then g∈H(T^∞_σ−ρ,X)and satisfies

|g|σ−ρ≤e^τln τ

ρ

ρ^−η1

|u|σ

Proof Follows directly from LemmaA.6and Definition2.1.

Lemma A.8 Recalling(3.8)and the definition of||1in(1.5), one has

∈Z^∞_∗

||³₁

d() <∞. (A.5)

Proof First of all note that for all∈Z^∗_∞one has

||³₁≤

i

(1+ i|i|)³, which implies

||³₁

d() 1

i(1+ i²|i|²). (A.6)

Then we recall that (see [9])

∈Z^∗_∞

1

i(1+ i²|i|²) <∞

which implies (A.5).

Lemma A.9 Let N, σ, ρ >0, m,m∈R,R∈H(T^∞_σ ,B^σ,m),Q∈H(T^∞_σ+ρ,B^σ+ρ,m). (i) The product operatorRQ∈H(T^∞_σ ,B^σ,m+m)withRQ_σ,m+m m ρ^−|m|Rσ,m

Q_σ+ρ,m. IfR(ω),Q(ω)depend on a parameterω∈⊆D_γ, thenRQ_σ,m+mm

ρ^−(|m|+2)R_σ,mQ_σ+ρ,m. If m=m=0, one hasRQ_σ R_σQ_σ. (ii) The projected operator^⊥_NRσ,m≤e^−ρNRσ+ρ,m.

Given two linear operatorsA,B, we define for anyn≥0, the operator Adⁿ_A(B)as Ad⁰_A(B):=B, Adⁿ⁺¹_A (B):= [Adⁿ_A(B),A],

where

[B,A] :=BA−AB.

By iterating the estimate(i)of LemmaA.9, one has that for anyn≥1

Adⁿ_A(B)σ ≤CⁿAⁿ_σBσ (A.7)

for some constantC >0.

Lemma A.10 Let0 < . . . < ρn < . . . < ρ0and0< . . .δn . . . δ0. Assume that

i≥0δi <∞, choose any n≥0and letAandBbe linear operators such that A=

n i=0

Ai B= n i=0

Bi Ai_ρi,−1,Bi_ρi,1≤δi, i=0, . . . ,n.

Then for any0< ζi < ρi the following holds.

(i) For any k≥1, one has

Ad^k_A(B)= n i=0

R_i^(k) with

R_i^(k)ρi−ζi ≤C^k₀ζ_i⁻¹δi ∀i =0, . . . ,n (ii) LetR:=e^−ABe^A−B. Then

R= n i=0

Ri with Riρi−ζi ζ_i⁻¹δi ∀i =0, . . . ,n

Proof of item(i). We prove the statement by induction onk. Fork=1, one has that [B,A] =

n i=0

R⁽_i¹⁾, R⁽_i¹⁾:= [Bi,Ai] + i−1

j=0

[Bi,Aj] − [Ai,Bj] .

Since for j <ione has thatρj > ρi and so all the terms in the above sum are analytic at least in the strip of widthρi. By applying LemmaA.9-(i)one has for any 0< ζi < ρi

R_i⁽¹⁾_ρi−ζi ζ_i⁻¹ δ_i²+

i j=0

δiδj

ζ_i⁻¹δi

j≥0

δj ζ_i⁻¹δi

fori = 0, . . . ,n. Now we argue by induction. Assume that for some k ≥ 1,R^(k) :=

Ad^k_A(B)=_n

i=0R_i^(k), with

R⁽_i^k)ρi−ζi ≤C^k₀ζ_i⁻¹δi, i=0, . . . ,n for any 0< ζi < ρi. Of course this implies that for all j<ione has

R⁽_j^k)_ρi−ζi ≤C₀^kζ_i⁻¹δj, i=0, . . . ,n.

By definition

Hence by applying LemmaA.9-(i)and using the induction hypothesis, one obtains R^(k+1)_i ρi−ζi ≤C

Proof Proof of(i). One has

[∂_x³, π₀^⊥g∂_x⁻¹] =π₀^⊥(3g_x∂x+3g_{x x}+g_{x x x}∂_x⁻¹)=3g_x∂x+R, R:=

n i=0

Ri, Ri :=π₀^⊥(3(gi)x x+(gi)x x x∂_x⁻¹)−3π0(gi)x∂x. Therefore

Ri_ρi−ζi ζ_i⁻³δi.

Proof of(ii). In view of the item(i), it is enough to estimate

k≥2

Ad^k_G(∂_x³) k! . Let

B:= [∂x³,G] =3g_x∂x+R= n i=0

Bi, Bi :=3(g_i)x∂x+Ri, i=0, . . . ,n, G=

n i=0

Gi, Gi:=π₀^⊥gi(ϕ,x)∂⁻_x¹ i =0, . . . ,n.

(A.8)

One has

Biρi−ζi,1ζ_i⁻³δi, i=0, . . . ,n,

Giρi−ζi,−1≤ Giρi,−1|fi|ρi δi ≤ζ_i⁻³δi, i=0, . . . ,n (A.9) For anyk≥2 one has

Ad^k_G(∂_x³)=Ad^k−1_G ([∂_x³,G])=Ad^k−1_G (B),

hence, we can apply LemmaA.10(replacingρiwithρi−ζiandδiwithζ_i⁻³δi) obtaining Ad^k_G(∂x³)=

n i=0

R⁽_i^k)

whereR^(k)_i satisfies

R^(k)_{i ρi−}2ζi ≤C₀^kζ_i⁻⁴δi, i=0, . . . ,n (A.10) and hence by setting

R=

k≥2

Ad^k_G(∂³_x) k! =

n i=0

Ri

item(ii)follows.

Proof of item(iii). The proof can be done arguing as in the item(ii), using that e^−G(ω·∂_ϕ)e^G

=ω·∂ϕ+

k≥1

Ad^k−1_G (ω·∂ϕG)

k! , where (ω·∂ϕG):=π0^⊥ω·∂ϕg(ϕ,x)∂x⁻¹.

Lemma A.12 LetA,A+,B,B+be bounded operators w.r.t. a norm · σ, and define M_A:=max{A_+σ,A_σ}, M_B:=max{B_+σ,B_σ}. (A.11) Then the following holds.

(i) For any k≥0, one has

Ad^k_A+(B+)−Ad^k_A(B)σ ≤C_∗^kM_A^kM_B

A+−Aσ+ B+−Bσ for some constant C_∗>0.

(ii)

e^−A+B₊e^A+−e^−ABe^A_σ A₊−A_σ+ B₊−B_σ.

Proof Proof of(i). We argue by induction. Of course the result is trivial fork=0. Assume that the estimate holds for somek≥1. Then

Ad^k+1_A+(B+)−Ad^k+1_A (B)=Ad_A+

Ad^k_A+(B+)

−Ad_A

Ad^k_A(B)

=Ad_A+

Ad^k_A+(B+)−Ad^k_A(B)

−Ad_A+−A

Ad^k_A(B) . Hence, by the induction hypothesis, using (A.11), (A.7) and LemmaA.9-(i), one obtains that

Ad^k+1_A+(B₊)−Ad^k+1_A (B)_σ

A_+σAd^k_A+(B₊)−Ad^k_A(B)_σ+ A₊−A_σC^kA^k_σB_σ C_∗^kM_A^k+1M_B

A₊−A_σ+ B₊−B_σ

+C^kM_A^kM_BA₊−A_σ

≤C_∗^k+1M_A^k+1M_B

A+−Aσ+ B+−Bσ for someC_∗>0 large enough.

Proof of(ii). It follows by item(i), using that e^−A+B₊e^A+−e^−ABe^A=

k≥0

Ad^k_A+(B₊)−Ad^k_A(B)

k! .

Lemma A.13 LetA,A⁺,B,B⁺be linear operators satisfying

Aρ,−1,A⁺ρ,−1,Bρ,1,B⁺ρ,1<C₀. Then the following holds.

(i) For any k≥1,

Ad^k_A+(B+)−Ad^k_A(B)ρ−ζ ≤C^kζ⁻¹

A+−Aρ,−1+ B+−Bρ,1 for some constant C>0depending on C₀.

(ii) SettingR:=e^−ABe^A−B, andR+:=e^−A+B+e^A+−B+, one has R−R+ρ−ζ ζ⁻¹

A−A+ρ,−1+ B−B+ρ,1 .

Proof Proof of(i). We first estimate Ad_A+(B+)−Ad_A(B). One has Ad_A+(B+)−Ad_A(B)=Ad_A+(B+−B)+Ad_A+−A(B).

By LemmaA.9-(i), one has

Ad_A(B)_ρ−ζ,Ad_A+(B₊)_ρ−ζ ζ⁻¹, (A.12) and

Ad_A+(B+)−Ad_A(B)ρ−ζ ζ⁻¹

A−A+ρ,−1+ B−B+ρ,1

. (A.13) In order to estimate Ad^k_A+(B₊)−Ad^k_A(B)=Ad^k−1_A+Ad_A+(B₊)−Ad^k−1_A Ad_A(B)for any k≥2, we apply LemmaA.12-(i)where we replaceB₊with Ad_A+(B₊)andBwith Ad_A(B), together with the estimates (A.12), (A.13).

Proof of(ii). It follows by(i)using thatR+−R=

k≥1

Ad^k_A+(B+)−Ad^k_A(B)

k! .

Lemma A.14 Let g₊,g ∈ Hρ,G := π₀^⊥g(ϕ,x)∂_x⁻¹,G+ := π₀^⊥g₊(ϕ,x)∂_x⁻¹. Then the following holds.

(i) The operatorsR := e^−G∂³_xe^G−∂_x³−3gx∂x,R₊ := e^−G+∂_x³e^G+−∂_x³−3(g₊)x∂x

satisfyR₊−R_ρ−ζ ζ^−τ|g₊−g|_ρfor some constantτ >0.

(ii) The operatorsR:=e^−Gω·∂ϕe^G−ω·∂ϕandR+ :=e^−G+ω·∂ϕe^G+−ω·∂ϕsatisfy the estimateR+−Rρ−ζ ζ^−τ|g₊−g|ρ, for some constantτ >0.

Proof We only prove the item(i). The item(ii)can be proved by similar arguments. We compute

B:= [∂_x³, π₀^⊥g∂_x⁻¹] =π₀^⊥(3g_x∂x+3g_{x x}+g_{x x x}∂_x⁻¹)=3g_x∂x+RB, RB:=π₀^⊥(3gx x+gx x x∂_x⁻¹)−π0(3gx∂x),

B₊:= [∂_x³, π₀^⊥g₊∂_x⁻¹] =π₀^⊥(3(g₊)x∂x+3(g₊)x x+(g₊)x x x∂_x⁻¹)=3(g₊)x∂x+R_B+, R_B+:=π₀^⊥(3(g₊)x x+(g₊)x x x∂_x⁻¹)−π0(3(g₊)x∂x).

(A.14) Hence

R+−R=RB+−RB+

k≥2

Ad^k_G+(∂³_x)−Ad^k_G+(∂³_x) k!

(A.14)

= RB+−RB+

k≥2

Ad^k−1_G

+ (B+)−Ad^k−1_G (B)

k! .

(A.15)

By a direct calculation one can show the estimates Bρ−ζ,1ζ⁻³|g|ρ, B+ρ−ζ,1ζ⁻³|g₊|ρ, G_ρ,−1|g|_ρ, G_+ρ,−1 |g₊|_ρ,

R_B+−R_Bρ−ζζ⁻³|g₊−g|_ρ, G₊−G_ρ,−1|g₊−g|_ρ.

(A.16) The latter estimates, together with LemmaA.13-(i)allow to deduce

Ad^k−1_G

+ (B+)−Ad^k−1_G (B)ρ−ζ≤C^kζ^−τ, ∀k≥2, (A.17) for some constantτ >0. Thus (A.15)-(A.17) imply the desired bound.

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