Gevrey-smoothness of invariant tori for analytic nearly integrable Hamiltonian systems under Rüssmann s non-degeneracy condition

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1 J Differential Equations 235 (27) wwwelseviercom/locate/de Gevrey-smoothness of invariant tori for analytic nearly integrable Hamiltonian systems under Rüssmann s non-degeneracy condition Junxiang Xu a,,1, Jiangong You b a Department of Mathematics, Southeast University, Naning 2196, PR China b Department of Mathematics, Naning University, Naning 2193, PR China Received 12 July 26; revised 29 November 26 Available online 1 January 27 Abstract In this paper we prove Gevrey smoothness of the persisting invariant tori for small perturbations of an analytic integrable Hamiltonian system with Rüssmann s non-degeneracy condition by an improved KAM iteration method with parameters 26 Elsevier Inc All rights reserved Keywords: Gevrey smoothness; Hamiltonian system; KAM iteration; Non-degeneracy condition 1 Introduction Consider the following Hamiltonian dynamical system: { q = Hp (q, p) = h p (p) + f p (q, p), ṗ = H q (q, p) = f q (q, p) (11) where H(q,p)= h(p) + f(q,p)is the Hamiltonian function, (q, p) T n D, with T n being the usual n-dimensional torus and D a bounded connected open domain of R n Suppose h(p) and f(q,p)are real analytic on D and D T n * Corresponding author addresses: xuun@seueducn (J Xu), you@nueducn (J You) 1 The work was supported by the National Natural Science Foundation of China (157127) /$ see front matter 26 Elsevier Inc All rights reserved doi:1116/de26121

2 61 J Xu, J You / J Differential Equations 235 (27) If f =, then the system (11) is integrable and has invariant tori T n {p } for all p D, on which there exists a linear flow, p(t) = p,q(t)= ω(p )t + q for any q T n, with the frequency ω(p ) = h p (p ) The classical KAM theorem asserts that if the frequency ω(p) is not degenerate, that is, det( ω/ p) = det(h pp ), (12) then most of the invariant tori can persist when f is sufficiently small [3 1] Later the result was extended to the case of Rüssmann s non-degeneracy [1,2,15,19], see (13) These invariant tori form a parameterized family How do the invariant tori depend on the parameter? In the analytic case, if the usual non-degeneracy condition (12) holds, Pöschel proved that the persisting invariant tori are C -smooth in the frequency parameter [11] More recently, Popov improved this result and proved that these KAM tori are Gevrey-smooth in their frequencies [12,13] For some related result, also see [14] But in the case of Rüssmann s non-degeneracy condition, no result is known about Gevrey-smoothness In this case, the frequency cannot be regarded as independent parameter and so the previous methods in [9,11,12] are not valid In this paper, by an improved KAM iteration with parameters, we prove that the Gevrey smoothness of persisting invariant tori for analytic nearly integrable Hamiltonian system is also true in the case of Rüssmann s non-degeneracy Let Π R n be a closed bounded set Denote by G μ (Π) (μ 1) the space of all Gevrey functions in a domain Π of index μ This means f G μ (Π) iff f C (Π) and there exists a constant M such that sup β ξ f(ξ) M β +1 β! μ, β = (β 1,β 2,,β n ) Z+ n, ξ Π where β =β 1 + β 2 + +β n Note that the derivatives are understood in Whitney s sense [21] Remark Obviously, analytic functions are Gevrey-functions; but Gevrey-function need not be analytic For μ = 1, the Gevrey function class G μ (Π) coincides with the class of analytic functions, but for μ>1, the Gevrey function class is larger Theorem 11 Suppose that h(p) and f(q,p)are real analytic on D and T n D, respectively, ω(p) = h p (p) = (ω 1 (p), ω 2 (p),, ω n (p)) satisfies Rüssamnn s non-degeneracy condition: (a 1,a 2,,a n ) R n \{}, a 1 ω 1 (p) + a 2 ω 2 (p) + +a n ω n (p) on D (13) Then there exist a sufficiently large positive m depending on the function h and the domain D, and a sufficiently small positive constant δ>, such that for τ>nm 1, μ>τ+ 2 and a sufficiently small α>, if f =sup T n D f(q,p) δα2, then, there is a nonempty Cantor set Π(α) D with meas (D \ Π(α)) cα m 1, and for ξ Π(α) the Hamiltonian system (11) has an invariant torus T ξ with the frequency ω (ξ) satisfying the Diophantine condition: ω (ξ), k α k τ, k = (k 1,k 2,,k n ) Z n \{},

3 J Xu, J You / J Differential Equations 235 (27) where k = k 1 + k k n Moreover, the family of invariant tori {T ξ,ξ Π(α)} is G μ -smooth in ξ That means that for each ξ Π(α), the invariant torus T ξ is an embedding torus: Φ(,ξ): T n D T n, and Φ G 1,μ (T n Π(α)), that is, Φ C (T n Π(α)) and Φ(θ,ξ) is analytic in θ on T n and G μ -smooth in ξ on Π(α) Furthermore, the frequencies ω G μ (Π(α)) Herec is a positive constant depending only on τ, μ, n and ω Remark The non-degeneracy condition (13) is the sharpest one for KAM theorem, which is first given by Rüssmann in [15] It means geometrically that the frequency vector ω does not lie on a hyperplane through the origin of R n Actually, it follows from [16,17,19] that the Rüssmann s non-degeneracy condition (13) is equivalent to that there exists a sufficiently positive integer m depending on h and D such that Rank { ω(p), β p ω(p) β m } = n for all p D (14) In Theorem 11, the m is the smallest one such that Eq (14) holds Especially, for the case m = 1, the condition (13) is equivalent to the Kolmogorov s non-degeneracy condition and our results correspond to those in [12] Also from [19] we have that the Rüssmann s non-degeneracy condition (13) is also equivalent to that there exists a point p Ω such that Rank { ω(p ), β p ω(p ) β n 1 } = n We will use KAM iteration to prove this theorem; and the outline is the same as in [9] At first we linearize the Hamiltonian system (11) at the invariant tori of the integrable system and then we will consider a parameterized Hamiltonian system instead of the Hamiltonian system (11) For any ξ D, let p = ξ + I and q = θ Under the symplectic map, H(q,p)= h(ξ) + h p (ξ), I + f h (I; ξ)+ f(θ,ξ + I) = e + ω(ξ),i + P, where e = h(ξ), ω(ξ) = h p (ξ), P = P(θ,I; ξ)= f h (I, ξ) + f(θ,ξ+ I), and ξ D is regarded as parameter Here e is an energy constant and has no influence on the Hamiltonian system, so we usually omit it; ω is called frequency vector; and P is a small perturbation term The corresponding Hamiltonian system becomes { θ = H I = ω(ξ)+ P I (θ, I; ξ), I = H θ = P θ (θ, I; ξ) (15) Thus, persistence of invariant tori for the nearly integrable system (11) is reduced to that of invariant tori for the family of Hamiltonian system (15) depending on the parameter ξ D Let D(s,r) = { (θ, I) C n C n Im θ s, I 1 r }, where Im θ = max 1 i n Im θ i, I 1 = 1 i n I i Denote Π = { ξ D } dist(ξ, D) d

4 612 J Xu, J You / J Differential Equations 235 (27) and Π d = { ξ C n dist(ξ, Π) d } with 2r d = α m 1, where m is the smallest one satisfying Eq (14) Thus, we have meas(d \ Π) = O(α m 1 ) as α We usually take r 2 = ɛ with ɛ being the small perturbation scale, thus we can put the higher order of I into the perturbation term So if ɛ is sufficiently small, 2r d always holds in the sequel This technique is usually used to put high order nonlinear terms into perturbation terms and we refer to [9] for details Now the Hamiltonian function H(θ,I; ξ) is analytic in (θ, I; ξ) on D(s,r) Π d We expand f(θ,i; ξ) as Fourier series with respect to θ and we have f(θ,i; ξ)= f k (I; ξ)e i k,θ k Z n Define f D(s,r) Πd = k,l f k r;d e s k, where f k r;d = sup I 1 r,ξ Π d f k (I; ξ) Remark The norm D(s,r) Πd was introduced in [1] In this paper, by using this norm, we simplify the estimate of the Gevrey-norm in KAM steps We write f(z; ξ) G 1,μ ( D Π) iff f C ( D Π) and f(z; ξ) is analytic in z on D and G μ -smooth in ξ on Π Let τ>nm 1, τ + 2 <μ<2τ + 3, σ = ( 2 3 ) l l+τ+1 with l = μ τ 2 Let ρ = (1 σ)s/1, r = r and I n be the n-unit matrix Denote W = diag( ρ 1 I n, r 1 I n ) Below, for simplicity we will use the same notation c to indicate different constants, which usually depend on τ,μ,n and ω With these notations and definitions we have the following result Theorem 12 Let τ, μ and W be defined as above Let H(θ,I; ξ) = ω(ξ),i +P(θ,I; ξ) Suppose ω(ξ) and P(θ,I; ξ) are analytic on Π d and D(s,r) Π d, respectively Let T = max ξ Πd ω/ ξ Suppose ω(ξ) satisfies (13) Then, there exists γ>, which is independent of ɛ, α, r, s and usually depends on τ, μ, n, d, ω, such that for any <α<1,if P D(s,r) Πd = ɛ γαrs τ+1, there is a nonempty Cantor set Π Π, and a family of symplectic mappings Φ (, ; ξ): D(s/2,r/2) D(s,r), ξ Π, satisfying Φ G 1,μ (D(s/2,r/2) Π ) and, for all β Z n +, W β ξ (Φ id) D(s/2,r/2) Π cγ 1 n+1 M β β! μ, (16)

5 J Xu, J You / J Differential Equations 235 (27) where β! =β 1!β! β n! and, M and c are constants depending on n, τ, T and μ Under the symplectic mappings, the Hamiltonian function H has the following form: H (θ, I; ξ) = H Φ (θ, I : ξ) = N (I; ξ) + P (θ, I : ξ), where N (I; ξ) = ω (ξ), I, and P (θ, I; ξ) = O(I 2 ) as I Hence, the Hamiltonian system (15) has a family of invariant tori {T ξ = Φ (T n, ; ξ) ξ Π },whichisg μ -smooth in ξ on Π, and whose frequencies ω (ξ) satisfy, for all ξ Π, β [ ξ ω (ξ) ω(ξ) ] cγ n+1 1 αs τ+1 M β β! μ, β Z+ n, (17) and ω (ξ), k α k τ, k Zn \{} (18) Moreover, we have meas(π \ Π ) cα 1/m Here the above constants c depend only on τ, μ, n and ω Remark The KAM Theorem 11 can easily follow from the KAM Theorem 12 with parameters, and this technique was first introduced by Pöschel in [9] 2 Proof of theorems In the same way as in [9], we can obtain Theorem 11 from Theorem 12 So we need only to prove Theorem 12 Our method is KAM iteration and the idea is similar to [9 12,17,18,2] 21 KAM-step The procedure of KAM-step is standard; we summarize the result for one KAM step in the following lemma Iteration Lemma 21 Let H(θ,I; ξ) = N(I; ξ) + P(θ,I; ξ) with N(I; ξ) = ω(ξ),i Let <E<1 and <ρ<s/5 Let K> satisfy e Kρ = E Suppose ω(ξ),k 2α k τ, ξ Π, k Zn, < k K, (21) where <α α is fixed Let max ξ Πd ω/ ξ T, d = P s,r;d ɛ = αrρ τ+1 E, α 2TK τ+1 Suppose that where P s,r;d = P D(s,r) Πd Then, for any ξ Π d, there exists a symplectic mapping Φ(, ; ξ): D(s +,r + ) D(s,r), such that H + (θ, I; ξ)= H Φ(θ,I; ξ)= N + (I; ξ)+ P + (θ, I; ξ), where N + (I; ξ)= ω + (ξ), I and P + satisfies P + s+,r + ;d ɛ + = α + r + ρ τ+1 + E +

6 614 J Xu, J You / J Differential Equations 235 (27) with s + = s 5ρ, η = E, ρ + = σρ, r + = ηr, E + = ce 2 3, α/2 α+ α, l where σ = ( 2 3 ) l+τ+1 with l = μ τ 2 Moreover, ω+ (ξ) ω(ξ) ɛ r, ξ Π d (22) Furthermore, let α + = α ɛ 2r Kτ+1 and denote { Π = ξ Π ω + (ξ), k < 2α } + k τ, K < k K + and Π + = Π \ Π, then ω + (ξ), k 2α + k τ, ξ Π +, k Z n with < k K +, (23) where K + > such that e ρ +K + = E + Let T + = T + 3ɛ dr and d + = α + 2T + K+ τ+1 If d + 2 3d, then max ξ Πd + ω +/ ξ T +, where Π d+ is the complex d + -neighborhood of Π + Moreover, we have P + s+,r + ;d + ɛ + Thus, the above result also holds for H + in place of H Remark The above lemma is actually one step in our KAM iteration Once this lemma holds for the Hamiltonian H, it also holds for the transformed Hamiltonian H +, and so the KAM step can iterate Proof The proof of this lemma is standard KAM step and we divide it into several parts A Truncation Let R = P(θ,; ξ) + P I (θ, ; ξ),i It follows easily that R s,r;d 2 P s,r;d 2ɛ Write R = k Z n R k(i; ξ)e i k,θ and let R K = R k (I; ξ)e i k,θ k K By definition, we have R R K s ρ,r;d 2ɛe Kρ B Construction of the symplectic map The symplectic map is generated by a Hamiltonian flow map at 1-time We will find a Hamiltonian function F and define the symplectic map by Φ = X t F t=1 It follows H Φ = N + +{N,F}+R K [R]+P +, where [R] is the average of R on T n, N + = N +[R]= I,ω +, {, } is the Poisson bracket, and P + = ( R R K) + 1 { (1 t){n,f}+r,f } X t F dt + (P R) Φ

7 J Xu, J You / J Differential Equations 235 (27) We want to find F such that {N,F}+R K [R]= (24) Let {F k } and {R k } be Fourier coefficients of F and R with respect to θ By the assumption (21), we have ω(ξ),k α k τ, ξ Π d, < k K So we have F k = and F k = with k = or k >K By Lemma A1 in Appendix A, we have C Estimates for the symplectic map Let By Lemma A1, we have 1 i ω(ξ),k R k, < k K, F(θ,I; ξ) r,s ρ;d nτ ɛ αρ τ W = diag ( ρ 1 I n,r 1 I n ) WX F r,s 2ρ;d nτ ɛ αrρ τ+1 = nτ E Thus, if <η 1 8, and nτ E 1 8, then, for all ξ Π d we have Φ = XF 1 : D(rη,s 3ρ) D(2rη,s 2ρ) So W(Φ id) s 5ρ,ηr;d n τ E, W(DΦ Id)W 1 s 5ρ,ηr;d n τ E, where D is the differentiation operator with respect to (θ, I) D Estimates of error terms Let α + = α Kτ+1 ɛ 2r If ɛ αr,wehave K τ+1 ω + (ξ), k 2α +, ξ Π, k K k τ Thus, by the definition of Π, it follows easily that (23) holds Thus, small divisor condition for the next step holds

8 616 J Xu, J You / J Differential Equations 235 (27) Let r + = ηr, ρ + = σρ By Lemmas A2, A3 and A4, it follows that [ P + s+,r + ;d <c ɛ 2 αrρ τ+1 + ( η 2 + e Kρ) ɛ ], and ω+ (ξ) ω(ξ) ɛ r, ξ Π d Suppose d + 2 3d Then, by the Cauchy estimates we have ( ω+ (ξ) ω(ξ) ) / ξ 3ɛ dr, ξ Π d + Let T + = T + 3ɛ dr Then max ξ Π d + ω +/ ξ T + Moreover, if α 2α +, it follows that P + s+,r + ;d cɛe cα + r + ρ τ+1 + E 3 2 = α+ r + ρ τ+1 + E +, where E + = ce 3 2 with c a constant depending only on n, τ Thus, it follows that P + s+,r + ;d <ɛ + Note that here the constants c only depend on n, τ, μ, and ω, and are independent of KAM steps 22 Iteration Now we choose some suitable parameters so that the above iteration can go on infinitely At the initial step, let ρ = (1 σ)s/1, r = r, ɛ = α r ρ τ+1 E Let K satisfy e K ρ = E α = α>, ω = ω, T = T = max ξ Πd ω/ ξ Denote { Π = ξ Π ω (ξ), k 2α } k τ, < k K Chose d = α m 1 Note that this choice for d is only for measure estimate for parameter and has no conflict with the assumption in Theorem 12 since we can use a smaller d Let d = α 1 d and η 2T K τ+1 = E 2 Assume the above parameters are all well defined for Then, we define ρ +1 = σρ, r +1 = η r and E +1 = ce 3 2, α +1 = α ɛ 2r K τ+1 Define ɛ +1, η +1, K +1, and d +1 in the same way as the previous step Since E = ce 3 2 1, and x = K ρ = ln E, if E is sufficiently small such that ln c/ln E (1 σ)3/2, it follows that 3/2 K +1 K 3/(2σ) Thus, d d and so the assumption d d in KAM steps hold Suppose max ξ Π d ω / ξ T Let T +1 = T + 3ɛ d r Then we have max ξ Πd+1 ω +1 / ξ T +1 τ+1 Again, by the choice of σ, it follows easily that ρ +1 x l +1 ρ τ+1 x l By induction, it is easy τ+1 to see that if E is sufficiently small such that ρ x l τ+1 1, we have ρ x l 1 for all 1

9 J Xu, J You / J Differential Equations 235 (27) ɛ Let F = d r It follows that F = 2T x τ+1 e x Suppose T T + 1 Then we have F cx 1 Thus, if E is sufficiently small such that cx 1 1 3, then T T +1 T + 1 Let { Π +1 = ξ Π ω+1 (ξ), k } 2α +1 k τ, K < k K +1 Denote Π d ={ξ C n dist(ξ, Π ) d } and D = D(s,r ) for simplicity Note that here and below the notation Π d is different from the previous one Π d By the KAM-step, for all ξ Π d we have symplectic mappings Φ (, ; ξ): D(r +1,s +1 ) D(r,s ) satisfying and W (Φ id) s+1,r +1 ;d +1 n τ E W (DΦ Id)W 1 s+1 n τ E,r +1 ;d +1 Let Φ = Φ Φ 1 Φ 1, and H = H Φ = N + P, where N = ω (ξ), I Then we have ω +1 ω ɛ r, for all ξ Π d Moreover, P s,r ;d ɛ 23 Convergence of the iteration Now we prove convergence of the KAM-iteration In the same way as in [9,1], it follows that, if c 1 2 E 1 2, then W DΦ W 1 ( D Π 1 + n τ ) E d < 2 i=1 So, we have W ( Φ Φ 1) D Π d ce, and W D ( Φ Φ 1) D Π d ce By the Cauchy s estimates we have W β ξ ( Φ Φ 1) D ce β! Π d β, W β ξ D( Φ Φ 1) D ce β! Π d β,

10 618 J Xu, J You / J Differential Equations 235 (27) and Let J β = ce β! d β β ξ (ω +1 ω ) Π cɛ β! r d β and L β = cɛ β! Now we estimate J β and Lβ for all β Zn + Again r d β α +1 = α ɛ ( K τ+1 = α 1 1 2r 2 xτ+1 It follows that if E is sufficiently small, then ( xτ+1 e x = ) e x = 1 O ( x 1 ) 1 2 Thus, 1 2 α α α Obviously, we have 1 2 α α +1 α Thus, the assumption α + /2 α + α holds By 1 2 α α +1 and the definition of α +1, it follows that ɛ α r K τ+1 and so the assumption ɛ αr holds in KAM step K τ+1 Let E = ( 1 σ 1 )τ+1 γ By the above discussion, if γ is sufficiently small, under the assumptions of Theorem 12, the assumptions of the iteration lemma hold for H at the first step Then the KAM step can go on infinitively Since ) μ 1 = τ l, d = α / ( 2T K τ+1 ), ρ x l τ+1 1 and α 2 α α, we have J β c ( 2T α ) β β!e d β ( 4(T + 1) c α cm β β! μ E 1 n+1, cβ!x (τ+1+l) β e x ) β β! [ x β / x 1 e (n+1)(μ 1) x β n / e x (n+1)(μ 1) ] μ 1 e x n+1 where M = 4(T + 1)[(n + 1)(μ 1)] μ 1 /α, and, c only depends on n, α, μ Inthesameway as the above, it follows that L β 2cαM β β! μ E 1 n+1 ρ τ+1

11 J Xu, J You / J Differential Equations 235 (27) Let D = D(, 1 2 s), Π = Π and Φ = lim Φ Thus, for any β Z+ n we have Inthesameway,wehave W β ξ (Φ id) D cm β β! μ E 1 Π n+1 W β ξ (DΦ Id) D cm β β! μ E 1 Π n+1 Since Φ is affine in I, we have convergence of β ξ Φ to Φ on D(r/2,s/2) and W β ξ (Φ id) D(s/2,r/2) Π cm β β! μ n+1 E 1, β Z n (25) Since E = ( 1 1 σ )τ+1 γ, this proves (16) Let ω = lim ω We have β ξ (ω ω) Π cαm β β! μ E 1 n+1 ρ τ+1 Moreover, ω (ξ), k 2α k τ, for all k Z n and ξ Π, where α = lim α with 1 2 α α α Thus, (17) and (18) hold Let m be the smallest integer such that Eq (14) holds and β m Since T T T + 1, from d +1 d = α +1T α T +1 ( K K +1 ) τ+1 T it follows that 2(T +1) ( 2σ 3 )τ d +1 d ( 2 3 )τ It follows that J β +1 /J β ce 1 2, where c depends on β If β m, in the same way as the above, we have W β ξ (Φ id) D(s/2,r/2) Π ce (26) Similarly, we have So, L β ce, β m 1 β ξ (ω +1 ω) Π ce, 1, β m (27) 24 Estimates of measure for the parameter sets Now we estimate the Lebesgue measure of the set Π, for which the small divisor condition holds in the KAM iteration By the KAM step, we have Π \ Π = Π, 1

12 62 J Xu, J You / J Differential Equations 235 (27) where { Π = ξ Π ω+1 (ξ), k 2α } +1 k τ, K < k K +1 and K 1 = By the equivalent Rüssmann s non-degeneracy condition (14) and the estimate (27), if E is sufficiently small, then for all the frequency ω (ξ) also satisfies (14)Soby Lemma A5 (see [19]) we have Since τ>mn 1, we deduce Appendix A meas( Π ) c [ diam(π) ] n 1 c [ diam(π) ] n 1 α 1 m K < k K +1 meas(π \ Π ) c [ diam(π) ] n 1 α 1 m c [ diam(π) ] n 1 α 1 m ( α / k τ+1) 1 m 1/ k τ+1 m K < k K +1 1/ k τ+1 m k Z n In this section we state several lemmas Some of the lemmas describe properties of the norm s,r The proofs are very similar to [1] and even simpler; so we omit them Lemma A1 Let f(θ,i) be analytic on D(s,r) Then f θ s ρ,r 1 eρ f s,r and f I s,r σ 1 σ f s,r for <ρ<sand <σ <r Lemma A2 Let f(θ,i)and g(θ,i) be analytic on D(s,r) Then fg s,r f s,r g s,r Lemma A3 Let F(θ,I) and G(θ, I) be analytic on D(s,r) For <ρ<sand <σ <r we have {F,G} s ρ,r σ 2 ρσ F s,r G s,r Lemma A4 Let F(θ,I) be analytic on D(s ρ,r) and affine linear in I Let <ρ<s/3 If F s ρ,r ρr/6e, then XF t : D(s 3ρ,r/2) D(s 2ρ,r),for t 1 Moreover, G Φ s 3ρ,r/2 2 G s,r

13 J Xu, J You / J Differential Equations 235 (27) Proof The following proof is actually given in [1] Let Φ = XF 1 By the Lie series expansion we have G Φ = l ad l F G, where ad F G = G, adl F G = { ad l 1 F G, F }, l = 1, 2, Let ρ = ρ/l, σ = r/(2l), D l = D(s 2ρ lρ,r lσ ) We conclude ad l F G s 3ρ,r/2 = ad l F G Dl F θ, ad l 1 F G Dl I + F I, ad l 1 F G Dl θ 1 ρ F 2l s ρ,r ad l 1 F r G Dl r F l s ρ,r ad l 1 F ρ G Dl 1 l ( 3 F s ρ,r /ρr ) ad l 1 F G Dl 1 [ l ( 3 F s ρ,r /ρr )] l G s,r By Stirling s formula, l l /l! e l for l 1 So we obtain G Φ s 3ρ,r/2 1 ad l l! f G s 2ρ,r/2 l l ( 3e F s ρ,r /ρr ) l G s,r 2 G s,r Lemma A5 Let k Z n and Π k = { ξ Π ω(ξ),k α/ k τ } If ω(ξ) satisfies the equivalent Rüssmann s non-degeneracy condition (14), then For the proof of this lemma see [16,19] References meas(π k ) c [ diam(π) ] n 1( α/ k τ+1 ) 1 m [1] J Bourgain, On Melnikov s persistency problem, Math Res Lett 4 (1997) [2] C-Q Cheng, Birkhoff Kolmogorov Arnold Moser tori in convex Hamiltonian systems, Comm Math Phys 177 (1996) [3] LH Eliasson, Perturbations of stable invariant tori for Hamiltonian systems, Ann Sc Norm Super Pisa 15 (1988)

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