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The KAM parameter sequence: every step satisfies the smallness, truncation, parameter and contraction hypotheses

Statement

Let $n \geq 2$ , $τ > n - 1$ , $μ : = τ + 3 n + 2$ , $α > 0$ , $s_{0} > 0$ , and let $C \geq 1$ bound the constants in One KAM step: the affine generator, its canonical time-one flow, and the quadratic remainder bound and Elimination of the frequency shift by a holomorphic contraction retraction. Let $c = c (n, τ) \in (0, 1]$ be the one-step smallness constant, and fix $C_{*} \geq 3 C 2^{μ}$ . Put $q : = 4^{- (μ + 1)}, D : = \max (16 n C 4^{μ}, 1 / c) .$ Fix $0 < σ_{0} \leq \min (1, s_{0} / 32)$ , $0 < r_{0} \leq 1$ , $K_{0} \geq 1$ , and $h_{0}, E_{0} > 0$ satisfying:

$E_{0} \leq \min (c^{2}, c, σ_{0}^{2}, 1 / 64, q^{2} / C_{*}^{2})$ ;
with $A : = C_{*}^{2} E_{0}$ and $L : = ∣ \log A ∣ + n \log (1 / σ_{0}) + 2 \log C_{*} + n \log 2 + 1$ , one has $K_{0} σ_{0} \geq \max (1, 2 L)$ ;
$D α E_{0} σ_{0}^{μ} \leq h_{0} \leq \min (1, α / (2 K_{0}^{τ + 1}))$ .

Define for all integers $j \geq 0$ $σ_{j} : = σ_{0} 2^{- j}, K_{j} : = K_{0} 4^{j}, h_{j} : = h_{0} 4^{- μ j}, s_{j} : = s_{0} - 8 \sum_{i < j} σ_{i},$ $E_{j + 1} : = C_{*} E_{j}^{3 / 2}, η_{j} : = \sqrt{E_{j}}, r_{j + 1} : = η_{j} r_{j}, ε_{j} : = α E_{j} r_{j} σ_{j}^{μ}, B_{j} : = C ε_{j} / r_{j} .$ Then every $j \geq 0$ satisfies:

(a) $0 < r_{j} \leq 1$ , $0 < σ_{j} \leq \min (1, s_{j} / 8)$ , $K_{j} σ_{j} \geq 1$ , $0 < h_{j} \leq 1$ , $s_{j} \geq s_{0} / 2$ , $s_{j} - 8 σ_{j} \geq s_{0} / 2$ , and $0 < η_{j} \leq \min (1 / 8, σ_{j})$ ; (b) $E_{j} \leq c η_{j}$ , $E_{j + 1} \leq q E_{j}$ , and $E_{j} \leq q^{j} E_{0} \leq 4^{- j} E_{0}$ ; (c) $ε_{j} \leq c α η_{j} r_{j} σ_{j}^{μ}$ and $ε_{j} \leq c h_{j} r_{j}$ ; (d) $h_{j} \leq α / (2 K_{j}^{τ + 1})$ and $h_{j + 1} \leq h_{j} / 4$ ; (e) $σ_{j}^{- n} e^{- K_{j} σ_{j} / 2} \leq E_{j}$ ; (f) $B_{j} = C α E_{j} σ_{j}^{μ} \leq h_{j + 1} / (16 n)$ , so the frequency contraction applies with $B = B_{j}$ ; moreover $\frac{E_{j}}{h_{j + 1}} \leq \frac{4^{μ} E_{0}}{h_{0}} 4^{- j}, \frac{B_{j}}{h_{j + 1}} \leq \frac{C 4^{μ} α E_{0} σ_{0}^{μ}}{h_{0}} 2^{- (μ + 2) j};$ (g) all the following sums converge: $\sum_{j \geq 0} (r_{j} + σ_{j}) E_{j} \leq 4 (r_{0} + σ_{0}) E_{0}, \sum_{j \geq 0} E_{j} \leq 2 E_{0}, \sum_{j \geq 0} 2^{j} E_{j} \leq 2 E_{0}, \sum_{j \geq 0} \frac{ε_{j}}{r_{j} h_{j}} \leq \frac{4}{3} \frac{α E_{0} σ_{0}^{μ}}{h_{0}} .$ The recurrence has precisely the perturbation budget $ε_{j + 1} = C_{*} 2^{- μ} E_{j} ε_{j} \geq 3 C E_{j} ε_{j} .$ Consequently the one-step remainder estimate $∣ P_{j}^{+} ∣ \leq C [E_{j} + η_{j}^{2} + σ_{j}^{- n} e^{- K_{j} σ_{j} / 2}] ε_{j}$ on the retained action and angle domain closes with $∣ P_{j + 1} ∣ \leq ε_{j + 1}$ after frequency retraction. For fixed $n, τ, c, C, C_{*}, σ_{0}$ the initial conditions are simultaneously satisfiable for all sufficiently small $E_{0}$ and every fixed $α > 0$ ; for $0 < α \leq 1$ the allowed threshold for $E_{0}$ can be chosen independently of $α$ .

Facts & Assumptions

Given: The data, constants, initial conditions and sequences in the Statement.

[F1]

The one-step hypotheses and estimates of One KAM step: the affine generator, its canonical time-one flow, and the quadratic remainder bound with $E = ε / (α r σ^{μ})$ , margins $η, σ$ , and frequency shift bounded by $C ε / r$ . The mean frequency shift has no Fourier-loss factor: $[R] = [Q]$ and each component is the angular mean of $\partial_{I_{i}} P (0, ϑ, ν)$ , bounded by $ε / r$ by Fourier decay, Cauchy derivative bounds, lattice shell counts and the exponential small-divisor sums on action-angle neighbourhoods.

[F2]

If a holomorphic frequency shift on $O_{h}$ is bounded by $B \leq h / (16 n)$ , Elimination of the frequency shift by a holomorphic contraction retraction gives $ϕ : O_{h / 4} \to O_{h / 2}$ satisfying $ϕ (ν) + v (ϕ (ν)) = ν$ , $∣ ϕ - i d ∣ \leq B$ , and $h ∥ D ϕ - I_{n} ∥ \leq (16 n / 7) B$ .

[F3]

$2^{j} \geq 1 + j$ , $2^{j} \geq (3 / 2)^{j}$ , and the geometric-series sums $\sum_{j} 2^{- j} = 2$ , $\sum_{j} 4^{- j} = 4 / 3$ (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length). Domains and notation are from Analytic KAM data: angular tori, action-angle neighbourhoods, Fourier cutoffs and Diophantine frequency labels.

[F4]

For every integer $m \geq 0$ , $t^{m} e^{- t} \to 0$ as $t \to + \infty$ (The exponential dominates every fixed nonnegative integer power at $+ \infty$ ).

Proof

technique · direct; verify the recurrence, domain margins, cutoff and parameter estimates, and then check the initial interval

1.1givenalgebra

Since $C_{*} \sqrt{E_{0}} \leq q < 1$ , induction gives $E_{j + 1} / E_{j} = C_{*} \sqrt{E_{j}} \leq q$ and $E_{j} \leq q^{j} E_{0}$ . Thus $\sqrt{E_{j}} \leq c$ , $η_{j} \leq 1 / 8$ , and $η_{j} / σ_{j} \leq (\sqrt{E_{0}} / σ_{0}) (2 \sqrt{q})^{j} \leq 1$ . This proves (b) and the assertions about $η_{j}$ in (a). Also induction directly gives $E_{j} = A^{(3 / 2)^{j}} / C_{*}^{2}$ , so $\log E_{j} = - (3 / 2)^{j} ∣ \log A ∣ - 2 \log C_{*}$ because $0 < A \leq q^{2} < 1$ .

2.1step 1.1F3givenalgebra

The action radii decrease because $η_{j} \leq 1 / 8$ , and $σ_{j} \leq σ_{0} \leq 1$ . Further, $K_{j} σ_{j} = K_{0} σ_{0} 2^{j} \geq 1$ and $h_{j} \leq h_{0} \leq 1$ . By the geometric series, $\sum_{i < j} σ_{i} = 2 σ_{0} (1 - 2^{- j})$ , whence $s_{j} = s_{0} - 16 σ_{0} + 16 σ_{j}$ and $s_{j} - 8 σ_{j} = s_{0} - 16 σ_{0} + 8 σ_{j} \geq s_{0} / 2$ . Since $σ_{j} \leq σ_{0} \leq s_{0} / 32 \leq s_{j} / 8$ , all remaining assertions of (a) follow.

2.2step 1.1givenalgebra

The inequality $E_{j} \leq c η_{j}$ gives the first bound in (c). For the other bound, divide by $h_{j} r_{j}$ : $ε_{j} / (h_{j} r_{j}) = α E_{j} σ_{0}^{μ} 4^{μ j} 2^{- μ j} / h_{0} \leq (α E_{0} σ_{0}^{μ} / h_{0}) 2^{- (μ + 2) j} \leq 1 / D \leq c$ , using $E_{j} \leq 4^{- (μ + 1) j} E_{0}$ and condition 3. Thus (c) holds at every step.

2.3step 1.1F2givenalgebra

Since $μ > τ + 1$ , condition 3 yields $h_{j} \leq [α / (2 K_{0}^{τ + 1})] 4^{- μ j} \leq α / (2 K_{0}^{τ + 1} 4^{(τ + 1) j}) = α / (2 K_{j}^{τ + 1})$ , and $h_{j + 1} / h_{j} = 4^{- μ} \leq 1 / 4$ . Moreover $E_{j} / h_{j + 1} \leq 4^{μ} (E_{0} / h_{0}) 4^{- j}$ , and hence $B_{j} / h_{j + 1} \leq (C 4^{μ} α E_{0} σ_{0}^{μ} / h_{0}) 2^{- (μ + 2) j} \leq 1 / (16 n)$ by $D \geq 16 n C 4^{μ}$ . This proves (d) and (f). In particular [F2] applies on $O_{h_{j + 1}} \subseteq O_{h_{j} / 4}$ , and its image lies in $O_{h_{j} / 2}$ .

2.4step 1.1F3givenalgebra

Put $a : = ∣ \log A ∣$ and $b : = n \log (1 / σ_{0}) + 2 \log C_{*}$ . All these quantities are nonnegative. Condition 2 and [F3] give $K_{j} σ_{j} / 2 \geq L 2^{j} \geq a (3 / 2)^{j} + b + n j \log 2 + 1$ , using separately $2^{j} \geq (3 / 2)^{j}$ , $2^{j} \geq 1$ and $2^{j} \geq j$ . Therefore $\log (σ_{j}^{- n} e^{- K_{j} σ_{j} / 2}) \leq n \log (1 / σ_{0}) + n j \log 2 - a (3 / 2)^{j} - b - n j \log 2 - 1 = \log E_{j} - 1$ . Exponentiation proves (e).

3.1step 1.1step 2.1step 2.2F3algebra

Since $r_{j} + σ_{j} \leq r_{0} + σ_{0}$ , the first sum in (g) is at most $(r_{0} + σ_{0}) E_{0} \sum_{j} 4^{- j} \leq 4 (r_{0} + σ_{0}) E_{0}$ . Also $\sum_{j} E_{j} \leq (4 / 3) E_{0} \leq 2 E_{0}$ and $\sum_{j} 2^{j} E_{j} \leq E_{0} \sum_{j} 2^{- j} = 2 E_{0}$ . Step 2.2 gives $\sum_{j} ε_{j} / (r_{j} h_{j}) \leq (α E_{0} σ_{0}^{μ} / h_{0}) \sum_{j} 2^{- (μ + 2) j} \leq (4 / 3) α E_{0} σ_{0}^{μ} / h_{0}$ . These establish (g) without any assumption relating $r_{0}$ to $σ_{0}$ .

3.2step 2.3step 2.4F1F2givenalgebra

The definitions give $ε_{j + 1} / ε_{j} = (E_{j + 1} / E_{j}) (r_{j + 1} / r_{j}) (σ_{j + 1} / σ_{j})^{μ} = C_{*} \sqrt{E_{j}} \sqrt{E_{j}} 2^{- μ} = C_{*} 2^{- μ} E_{j} \geq 3 C E_{j}$ . By (e) and $η_{j}^{2} = E_{j}$ , the remainder bound of [F1] is at most $3 C E_{j} ε_{j} \leq ε_{j + 1}$ . Retraction only evaluates this holomorphic remainder on its parameter subdomain by step 2.3 and [F2], so it cannot increase its sup-norm. The retained widths are $r_{j + 1} = η_{j} r_{j}$ and $s_{j + 1} = s_{j} - 8 σ_{j}$ .

4.1step 1.1step 3.2F4givenchoosealgebra∎

To check feasibility, choose $K_{0} = ⌈ \max (1, 2 L) / σ_{0} ⌉$ and $h_{0} = \min (1, α / (2 K_{0}^{τ + 1}))$ . As $E_{0} ↓ 0$ , this cutoff is bounded by a constant times $1 + ∣ \log E_{0} ∣$ , so $2 D E_{0} σ_{0}^{μ} K_{0}^{τ + 1} \to 0$ . For completeness, with $t = ∣ \log E_{0} ∣$ , choose an integer $m > τ + 1$ ; for $t \geq 1$ , $(1 + t)^{τ + 1} \leq 2^{τ + 1} t^{m}$ , so [F4] implies $e^{- t} (1 + t)^{τ + 1} \to 0$ . Thus for all sufficiently small $E_{0}$ one has $D E_{0} σ_{0}^{μ} \leq 1 / (2 K_{0}^{τ + 1})$ , and, for each fixed $α$ , also $D α E_{0} σ_{0}^{μ} \leq 1$ . These are exactly the two inequalities needed for condition 3 with the chosen $h_{0}$ . If $α \leq 1$ , the second follows already from the first since $K_{0} \geq 1$ ; all thresholds are then independent of $α$ . Conditions 1 and 2 hold by further decreasing $E_{0}$ and by the cutoff choice. This proves simultaneous feasibility and completes the lemma.

Remarks

The lower parameter width is proportional to the actual frequency shift $ε_{0} / r_{0} = α E_{0} σ_{0}^{μ}$ . Replacing it by a multiple of $E_{0}$ would prevent a uniform $ε \leq c α^{2}$ application as $α ↓ 0$ . The faster decrease of $h_{j}$ preserves the finite-cutoff Diophantine bound; the still faster decrease of $E_{j}$ makes parameter derivatives summable. The angular loss of $8 σ_{j}$ agrees with the retained domain of the local one-step construction.

The KAM parameter sequence: every step satisfies the smallness, truncation, parameter and contraction hypotheses

Statement

Facts & Assumptions

Proof

Remarks

Depends on

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