3.1 Notations
We introduce the following notations and spaces.
For given \(\sigma>0\), \(r>0\), \(\nu=1,2,\ldots\) , we define sequences \(\{ \sigma_{\nu}\}\) and \(\{r_{\nu}\}\):
-
(1)
\(\sigma_{0}=\sigma\), \(\sigma_{\nu}=\sigma_{0}(1-\tau_{\nu})\) with \(\tau_{0}=0\) and \(\tau_{\nu}=\frac{\sum_{j=1}^{\nu}j^{-2}}{2\sum_{j=1}^{\infty}j^{-2}}\). It is easy to see \(\sigma_{0}>\sigma_{\nu}>\sigma_{\nu+1}>\sigma/2\).
-
(2)
\(r_{0}=r\), \(r_{\nu}=r_{0}(1-\tau_{\nu})\), \(\varepsilon_{0}=\varepsilon\), \(\varepsilon_{\nu}=\varepsilon^{(1+\delta)^{\nu}}\). It is easy to see \(r_{0}>r_{\nu}>r_{\nu+1}>r/2\).
Denote
$$\begin{aligned} \Theta(\sigma)= \bigl\{ \theta=(\theta_{1},\ldots,\theta_{m}) \in \Bbb{C}^{m}/2\pi\Bbb{Z}^{m}: \vert \operatorname{Im} \theta \vert < \sigma \bigr\} \end{aligned}$$
and
$$\begin{aligned} D^{a,\rho} =&D^{a,\rho}(\sigma,r)\\ =& \bigl\{ (\theta ,J,z,\bar{z})\in { \Bbb{C}^{m}/2\pi\Bbb{Z}^{m}}\times \Bbb{C}^{m} \times{l^{a,\rho}\times l^{a,\rho}}: \\ & {}\vert \operatorname{Im}\theta \vert < \sigma, \vert J \vert < r^{2}, \Vert z \Vert _{a,\rho}< r, \Vert \bar{z} \Vert _{a,\rho}< r \bigr\} . \end{aligned}$$
Thus, we get a family of domains:
$$\begin{aligned} \Theta(\sigma_{0})\supset\Theta(\sigma_{1})\supset\cdots \supset\Theta(\sigma_{\nu})\supset\Theta(\sigma_{\nu+1}) \supset \cdots\supset\Theta\biggl(\frac{\sigma_{0}}{2}\biggr) \end{aligned}$$
and
$$D^{a,\rho}(\sigma_{0},r_{0})\supset\cdots\supset D^{a,\rho}(\sigma_{\nu},r_{\nu})\supset D^{a,\rho}(\sigma_{\nu+1},r_{\nu+1}) \supset\cdots\supset D^{a,\rho}\biggl(\frac{\sigma_{0}}{2},\frac{r_{0}}{2}\biggr). $$
We rewrite \(\Theta_{l}:=\Theta(\sigma_{l})\), \(D^{a,\rho}_{l}=D^{a,\rho}(\sigma_{l},r_{l})\), \(l=0,1,\ldots\) .
For a one order Whitney smooth function \(F(\omega)\) on closed bounded set Ĵ, we define
$$\Vert F \Vert ^{*}_{\hat{J}}=\max \Bigl\{ \sup_{\omega\in\hat{J}} \vert F \vert ,\sup_{\omega\in \hat{J}} \vert \partial_{\omega} F \vert \Bigr\} . $$
If \(F(\omega)\) is a vector function from Ĵ to \(l^{a,\rho}\) (or \(\Bbb{R}^{b_{1}\times b_{2}}\)) which is one order Whitney smooth on Ĵ, we define
$$\Vert F \Vert ^{*}_{a,\rho,\hat{J}}= \bigl\Vert \bigl( \bigl\Vert F_{i}(\omega) \bigr\Vert ^{*}_{\hat{J}}\bigr)_{i} \bigr\Vert _{a,\rho}\quad \biggl(\mbox{or } \Vert F \Vert ^{*}_{J^{*}}=\max_{1\leq i_{1}\leq b_{1}}\sum _{1\leq i_{2}\leq b_{2}}\bigl( \bigl\Vert F_{i_{1}i_{2}}(\omega) \bigr\Vert ^{*}_{J^{*}}\bigr) \biggr). $$
Let \(\tilde{w}= (\theta,J,z,\bar{z})\in D^{a,\rho}\), we denote the weighted norm for w̃
$$\vert \tilde{w} \vert _{a,\rho}= \vert \theta \vert + \frac{1}{r^{2}} \vert J \vert +\frac{1}{r} \Vert z \Vert _{a,\rho} +\frac{1}{r} \Vert \bar{z} \Vert _{a,\rho}. $$
If \(F(\tilde{w};\omega)\) is a vector function from \(D^{a,\rho}\times\hat{J}\) to \(l^{a,\rho}\) which is one order Whitney smooth on ω, we define
$$\Vert F \Vert ^{*}_{a,\rho,D^{a,\rho}\times \hat{J}}=\sup_{\tilde{w}\in D^{a,\rho}} \Vert F \Vert ^{*}_{a,\rho, \hat{J}} \quad\mbox{and}\quad \Vert F \Vert ^{*}_{D^{a,\rho}\times \hat{J}}=\sup_{\tilde{w}\in D^{a,\rho}} \Vert F \Vert ^{*}_{ \hat{J}}. $$
To the function \(F(\theta,J,z,\bar{z})\), associate a hamiltonian vector field defined as \(X_{F}=\{F_{J},-F_{\theta}, \mathrm{i} F_{\bar{z}},-\mathrm{i} F_{z}\}\), we denote the weighted norm for \(X_{F}\) by letting
$$ \vert X_{F} \vert ^{*}_{a,\rho,D^{a,\rho }\times\hat{J}} = \Vert F_{J} \Vert ^{*}_{D^{a,\rho}\times\hat{J}} +\frac{1}{r^{2}} \Vert F_{\theta} \Vert ^{*}_{D^{a,\rho}\times\hat{J}} +\frac{1}{r} \bigl( \Vert F_{\bar{z}} \Vert ^{*}_{a,\rho,D^{a,\rho}\times \hat{J}}+ \Vert F_{z} \Vert ^{*}_{a,\rho,D^{a,\rho}\times\hat{J}} \bigr). $$
(16)
Let \(B(\tilde{w};\omega)\) be an operator from \(D^{a,\rho}\) to \(D^{a,\bar{\rho}}\) for \((\tilde{w};\omega)\in D^{a,\rho}\times\hat{J}\), we define the operator norm
$$\begin{aligned} & \bigl\vert B(\tilde{w};\omega) \bigr\vert ^{op}_{a,\bar{\rho},\rho,D^{a,\rho }\times\hat{J}}= \sup_{(\tilde{w};\omega)\in D^{a,\rho}\times\hat{J}}\sup_{\tilde{w}\neq0}\frac{ \vert B(\tilde{w};\omega)\tilde{w} \vert _{a,\bar{\rho}}}{ \vert \tilde{w} \vert _{a,\rho}}, \\ & \bigl\vert B(\tilde{w};\omega) \bigr\vert ^{\ast op}_{a,\bar{\rho},\rho,D^{a,\rho }\times\hat{J}}= \max\bigl\{ \vert B \vert ^{op}_{a,\bar{\rho},\rho,D^{a,\rho }\times \hat{J}}, \vert \partial_{\omega} B \vert ^{op}_{a,\bar{\rho},\rho ,D^{a,\rho}\times \hat{J}}\bigr\} . \end{aligned}$$
3.2 Reducibility
Now, we are ready to introduce our reducibility and prove it via the KAM iteration.
Lemma 3.1
For the Hamiltonian
H
in (10), there are a
\(0<\varepsilon^{*}\ll1\)
and a set
\(\overline{J}\subset\hat{J}\)
with
\(\operatorname {meas}\overline{J}\geq\operatorname{meas}\hat{J} (1-\mathcal {O}(\varrho^{m}) ) \), such that, for any
\(0<\varepsilon<\varepsilon^{*},\omega\in\overline{J}\), there is a linear symplectic transformation
$$\Sigma^{\infty}: D^{a,\rho}(\sigma/2,r/2)\times \overline{J} \rightarrow D^{a,\rho}(\sigma,r) $$
such that the following statements hold:
-
(i)
There are two absolute constants
\(C>0\)
and
\(0<\delta<1\)
such that
$$\bigl\vert \Sigma^{\infty}-id \bigr\vert ^{*}_{a,\rho+1, D^{a,\rho}(\sigma /2,r/2)\times \overline{J}}\leq C \varepsilon^{\frac{\delta}{2}}, $$
where
id
is the identity mapping.
-
(ii)
The transformation
\(\Sigma^{\infty}\)
changes Hamiltonian (10) into
$$\begin{aligned} H^{\infty}:= H\circ \Sigma^{\infty}= \langle\omega,J \rangle+\sum_{n\in\mathbb{Z}^{d}}\mu_{n} z_{n}\bar{z}_{n}, \end{aligned}$$
(17)
where
$$ \mu_{n}=\lambda_{n}+\sum _{s=0}^{\infty}\varepsilon_{s} \tilde{ \lambda}_{n,s},\quad \tilde{\lambda}_{n,0}= \frac{\psi_{0}}{2\lambda_{n}} \textit{ and }\vert \tilde{\lambda}_{ns} \vert \leq C, s=1,2,3,\ldots. $$
(18)
3.3 Proof the Lemma 3.1
Proof
We first introduce the measure estimates to treat small divisors in reducing, which will be proved in Lemma 4.1. For \(k\in\Bbb{Z}^{m}\), \(n_{1},n_{2}\in\mathbb{Z}^{d}\), there exists a family of closed subsets \(\overline{J}_{l}(l=0,\ldots,\nu)\)
$$\overline{J}_{\nu}\subset\cdots\subset\overline{J}_{l+1} \subset\overline{J}_{l} \subset\cdots\subset\overline{J}_{0} \subset\hat{J}\subset J^{0} $$
such that, for \(\omega\in\overline{J}_{l}\),
$$\begin{aligned} \bigl\vert \langle k,\omega\rangle\pm(\lambda _{n_{1},\nu} \pm\lambda_{n_{2},\nu}) \bigr\vert \geq& \frac{\varrho\operatorname{meas}\hat{J}}{(1+l^{2})( \vert k \vert +1)^{m+3}} \end{aligned}$$
(19)
and
$$\begin{aligned} \operatorname{meas}\overline{J}_{l}\geq \operatorname{meas}\hat{J} \Biggl(1-C\varrho^{m}\sum _{i=0}^{l}\frac{1}{1+i^{2}} \Biggr), \end{aligned}$$
(20)
where C is a constant depending on d. Moreover, let \(\overline{J}=\bigcap_{l=0}^{\infty}\overline{J}_{l}\), then
$$\begin{aligned} \operatorname{meas}\overline{J}\geq \operatorname{meas}\hat{J} \bigl(1-\mathcal{O}\bigl(\varrho^{m}\bigr) \bigr) \end{aligned}$$
(21)
provided that ϱ is small enough.
Let \((m_{1},m_{2})=\{(2,0),(1,1),(0,2)\}\). We construct an iterative series \(\{H_{l}\}\) of Hamiltonian functions of the form
$$H_{l}=\langle\omega,J\rangle+\sum_{n\in\mathbb{Z}^{d}} \lambda_{n,l}z_{n}\bar{z}_{n}+ \varepsilon_{l}P_{l}(\theta,J,z,\bar{z};\omega),\quad l=0,1,\ldots,\nu,\hspace{93pt} (E)_{l} $$
where
$$\begin{aligned} P_{l}(\theta,J,z,\bar{z})= \sum_{k\in{\Bbb {Z}}^{m}, n_{1},n_{2}\in\mathbb{Z}^{d} } \bigl(\zeta _{n_{1}n_{2},l}^{k20}z_{n_{1}}z_{n_{2}} + \zeta_{n_{1}n_{2},l}^{k11}z_{n_{1}}\bar{z}_{n_{2}} + \zeta_{n_{1}n_{2},l}^{k02}\bar{z}_{n_{1}}\bar{z}_{n_{2}} \bigr)e^{\mathrm {i}\langle k,\theta\rangle} \end{aligned}$$
with
$$\zeta_{n_{1}n_{2},l}^{m_{1}m_{2}}=\sum_{k\in{\Bbb {Z}}^{m},n_{1},n_{2}\in\mathbb {Z}^{d}} \zeta_{n_{1}n_{2},l}^{km_{1}m_{2}}e^{\mathrm{i}\langle k,\theta\rangle} $$
and
$$\textstyle\begin{array}{@{}l@{}} \displaystyle\zeta_{n_{1}n_{2},l}^{k20}=0,\qquad \zeta_{n_{1}n_{2},l}^{k02}=0,\quad \mbox{if } k=0, n_{1}+n_{2}=0; \\ \displaystyle\zeta_{n_{1}n_{2},l}^{k11}=0,\quad \mbox{if } k=0, n_{1}-n_{2}=0. \end{array}\displaystyle \hspace{131pt} (3.0)_{l} $$
Furthermore, the functions \(\zeta_{n_{1}n_{2},l}^{m_{1}m_{2}}\) are analytic on the domain \(\Theta_{l}\times\overline{J}_{l}\),
$$\zeta_{n_{1}n_{2},l}^{m_{1}m_{2}}=\zeta_{n_{1}n_{2},l}^{m_{1}m_{2}*},\qquad \bigl\Vert \zeta_{n_{1},n_{2},l}^{m_{1}m_{2}*} \bigr\Vert ^{*}_{\Theta_{l}\times\overline {J}_{l}} \leq C,\quad n_{1},n_{2}\in\mathbb{Z}^{d}, l=0,1, \ldots,\nu,\hspace{49pt} (3.1)_{l} $$
and
$$\lambda_{n,0}=\lambda_{n},\qquad \lambda_{n,l}= \lambda_{n}+\sum_{s=0}^{l-1} \varepsilon_{s}\tilde{\lambda}_{n,s},\quad l=1,2,3,\ldots, \nu,\hspace{112pt} (3.2)_{l} $$
with
$$ \tilde{\lambda}_{n,s}=\frac{1}{(2\pi)^{m}} \int_{{\Bbb {T}}^{m}}\zeta_{nn,s}^{1,1}(\theta,\omega) \,d\theta, \quad s=0,1,2,\ldots,l-1. $$
(22)
Clearly, we have \(H_{l}|_{l=0}=H\). For \(l=0\), we have \(P_{0}(\theta ,J,z,\bar{z})=P(\theta,J,z,\bar{z})\) defined in (11). From (12) and Assumption (A2), the functions \(\zeta _{n_{1}n_{2},0}^{m_{1}m_{2}}(\theta;\omega)\) are analytic on the domain \(\Theta_{0}\times\overline{J}_{0}\) and satisfy \((3.1)_{0}\). Thus, we get
$$\tilde{\lambda}_{n,0}=\frac{1}{(2\pi)^{m}} \int_{{\Bbb {T}}^{m}}\zeta _{nn,0}^{11}(\theta;\omega) \,d\theta =\frac{\psi_{0}}{2\lambda_{n}}. $$
This implies that \((3.2)_{0}\) is satisfied.
We look for a change of variables \(S_{\nu}\) defined in a domain \(D_{\nu+1}^{a,\rho}\) by the time-one map \(X_{\mathcal{F}_{\nu}}^{1}\) of the Hamiltonian vector field \(X_{\mathcal{F}_{\nu}}\), such that the system \((E)_{\nu}\) is transformed into the form \((E)_{\nu+1}\) and satisfies \((3.1)_{\nu+1}\), \((3.2)_{\nu+1}\). In fact, the new Hamiltonian \(H_{\nu+1}\) can be written as
$$\begin{aligned} H_{\nu+1} :=&H_{\nu}\circ X^{1}_{\mathcal{F}_{\nu}} \\ =&\langle\omega,J\rangle+\sum_{n\in\mathbb{Z}^{d}} \lambda_{n,\nu}z_{n}\bar{z}_{n}+ \biggl\{ \langle \omega,J\rangle +\sum_{n\in\mathbb{Z}^{d}}\lambda_{n,\nu}z_{n} \bar{z}_{n},{\mathcal{F}_{\nu}} \biggr\} \\ &{}+\varepsilon_{\nu}P_{\nu}(\theta,J,z,\bar{z};\omega) \bigl\{ \varepsilon_{\nu}P_{\nu}(\theta,J,z,\bar{z};\omega),{\mathcal {F}_{\nu}} \bigr\} \\ &{}+ \int_{0}^{1}(1-t) \bigl\{ \{H_{\nu}, \mathcal{F}_{\nu}\},\mathcal{F}_{\nu}\bigr\} \circ X^{t}_{\mathcal{F}_{\nu}}\,dt. \end{aligned}$$
(23)
Let \(\mathcal{\mathcal{F}}_{\nu}=\varepsilon_{\nu}F_{\nu}\), and \(F_{n_{1}n_{2},\nu}^{m_{1}m_{2}}=\sum_{k\in{\Bbb {Z}}^{m}}F_{n_{1}n_{2},\nu }^{km_{1}m_{2}}e^{\mathrm{i}\langle k,\theta\rangle}\) with
$$\begin{aligned} &F_{n_{1}n_{2},\nu}^{k20}=0,\qquad F_{n_{1}n_{2},\nu}^{k02}=0, \quad \mbox{if } k=0, n_{1}+n_{2}=0; \\ &F_{n_{1}n_{2},\nu}^{k11}=0,\quad \mbox{if } k=0, n_{1}-n_{2}=0. \end{aligned}$$
We shall find a function F of the form
$$\begin{aligned} F_{\nu}(\theta,J,z,\bar{z};\omega) =&\sum _{k\in{\Bbb {Z}}^{m},n_{1},n_{2}\in \mathbb{Z}^{d},\atop \vert k \vert + \vert \vert n_{1} \vert - \vert n_{2} \vert \vert \neq0}F_{n_{1}n_{2},\nu}^{k11}z_{n_{1}} \bar{z}_{n_{2}}e^{\mathrm {i}\langle k,\theta\rangle} \\ &{}+\sum_{k\in{\Bbb {Z}}^{m}, n_{1},n_{2}\in\mathbb{Z}^{d} } \bigl(F_{n_{1}n_{2},\nu }^{k20}z_{n_{1}}z_{n_{2}} +F_{n_{1}n_{2},\nu}^{k02}\bar{z}_{n_{1}}\bar{z}_{n_{2}} \bigr)e^{\mathrm {i}\langle k,\theta\rangle} \end{aligned}$$
(24)
satisfying the homological equation
$$\begin{aligned} \biggl\{ \langle\omega,J\rangle+\sum_{n\in\mathbb{Z}^{d}} \lambda_{n,\nu }z_{n}\bar{z}_{n},F_{\nu}\biggr\} +P_{\nu}(\theta,J,z,\bar{z};\omega) =\sum _{n\in\mathbb{Z}^{d}}\bigl[\zeta_{nn,\nu}^{11} \bigr]z_{n}\bar{z}_{n}. \end{aligned}$$
(25)
It follows that
$$\begin{aligned} & \biggl\{ \langle\omega,J\rangle+\sum _{n\in\mathbb{Z}^{d}}\lambda_{n,\nu }z_{n} \bar{z}_{n},F_{\nu}\biggr\} \\ &\quad =\mathrm{i}\sum_{ k\in{\Bbb {Z}}^{m},n_{1},n_{2}\in\mathbb{Z}^{d}} \bigl(\langle k, \omega\rangle-\lambda_{n_{1},\nu}-\lambda_{n_{2},\nu} \bigr)F_{n_{1}n_{2},\nu}^{k20}z_{n_{1}} z_{n_{2}}e^{\mathrm{i}\langle k,\theta \rangle} \\ &\qquad {}+\mathrm{i}\sum_{ k\in{\Bbb {Z}}^{m},n_{1},n_{2}\in\mathbb{Z}^{d}} \bigl(\langle k, \omega\rangle+\lambda_{n_{1},\nu}+\lambda_{n_{2},\nu} \bigr)F_{n_{1}n_{2},\nu}^{k02}z_{n_{1}} z_{n_{2}}e^{\mathrm{i}\langle k,\theta \rangle} \\ &\qquad {}+\mathrm{i}\sum_{ k\in{\Bbb {Z}}^{m},n_{1},n_{2}\in\mathbb{Z}^{d},\atop \vert k \vert + \vert \vert n_{1} \vert - \vert n_{2} \vert \vert \neq0} \bigl(\langle k, \omega\rangle-\lambda_{n_{1},\nu}+\lambda_{n_{2},\nu} \bigr)F_{n_{1}n_{2},\nu}^{k11}z_{n_{1}} \bar{z}_{n_{2}}e^{\mathrm{i}\langle k,\theta \rangle}. \end{aligned}$$
(26)
By (25), it follows that, for \(k\in\mathbb{Z}^{m},n_{1},n_{2}\in \mathbb{Z}^{d}\), the \(F_{n_{1}n_{2},\nu}^{km_{1}m_{2}}\) are determined by the following linear algebraic system:
$$\begin{aligned} \begin{aligned} & \bigl(\langle k,\omega\rangle-\lambda_{n_{1},\nu}- \lambda_{n_{2},\nu} \bigr)F_{n_{1}n_{2},\nu}^{k20}=\mathrm{i} \varsigma_{n_{1}n_{2},\nu }^{k20}, \\ & \bigl(\langle k,\omega\rangle-\lambda_{n_{1},\nu}+\lambda_{n_{2},\nu} \bigr)F_{n_{1}n_{2},\nu}^{k11}=\mathrm{i}\varsigma_{n_{1}n_{2},\nu}^{k11}, \quad \vert k \vert + \bigl\vert \vert n_{1} \vert - \vert n_{2} \vert \bigr\vert \neq 0, \\ & \bigl(\langle k,\omega\rangle+\lambda_{n_{1},\nu}+\lambda_{n_{2},\nu} \bigr)F_{n_{1}n_{2},\nu}^{k02}=\mathrm{i}\varsigma_{n_{1}n_{2},\nu}^{k02}. \end{aligned} \end{aligned}$$
(27)
For \(n_{1},n_{2}\in\mathbb{Z}^{d}\), we get
$$ \begin{aligned} &F_{n_{1}n_{2},\nu}^{11}=\sum _{ k\in{\Bbb {Z}}^{m},\atop \vert k \vert + \vert \vert n_{1} \vert - \vert n_{2} \vert \vert \neq0} \frac{\mathrm{i}\zeta_{n_{1}n_{2},\nu}^{k11}}{\langle k,\omega\rangle -\lambda_{n_{1},\nu}+\lambda_{n_{2},\nu} }e^{\mathrm{i}\langle k,\theta \rangle}, \\ &F_{n_{1}n_{2},\nu}^{20}=\sum_{ k\in{\Bbb {Z}}^{m}} \frac{\mathrm{i}\zeta_{n_{1}n_{2},\nu}^{k20}}{\langle k,\omega\rangle -\lambda_{n_{1},\nu}-\lambda_{n_{2},\nu}}e^{\mathrm{i}\langle k,\theta \rangle}, \\ &F_{n_{1}n_{2},\nu}^{02}=\sum_{k\in{\Bbb {Z}}^{m}} \frac{\mathrm{i}\zeta_{n_{1}n_{2},\nu}^{k02}}{\langle k,\omega\rangle +\lambda_{n_{1},\nu}+\lambda_{n_{2},\nu} }e^{\mathrm{i}\langle k,\theta \rangle}. \end{aligned} $$
(28)
By Cauchy’s estimate and \((3.1)_{\nu}\), we get
$$\begin{aligned} \bigl\vert \zeta_{n_{1}n_{2},\nu}^{km_{1}m_{2}} \bigr\vert \leq \bigl\Vert \zeta_{n_{1}n_{2},\nu }^{m_{1}m_{2}} \bigr\Vert ^{*}_{\Theta_{\nu}\times\overline{J}_{\nu}} e^{- \vert k \vert \sigma_{\nu}} \end{aligned}$$
(29)
and
$$\begin{aligned} \bigl\vert \partial_{\omega}\zeta_{n_{1}n_{2},\nu}^{km_{1}m_{2}} \bigr\vert \leq \bigl\Vert \zeta_{n_{1}n_{2},\nu}^{m_{1}m_{2}} \bigr\Vert ^{*}_{\Theta_{\nu}\times\overline{J}_{\nu}} e^{- \vert k \vert \sigma_{\nu}}. \end{aligned}$$
(30)
Note that (19) and (28), for \(n_{1},n_{2}\in\mathbb {Z}^{d}\), we get
$$\begin{aligned} &\sup_{(\theta;\omega)\in{ \Theta_{\nu+1}\times\overline{J}_{\nu}}} \bigl\vert F_{n_{1}n_{2},\nu}^{11} \bigr\vert \leq C \bigl\Vert \zeta_{n_{1}n_{2},\nu}^{11} \bigr\Vert ^{*}_{\Theta_{\nu}\times \overline{J}_{\nu}} \bigl(1+\nu^{2}\bigr) \biggl(\sum _{ \vert k \vert + \vert \vert n_{1} \vert - \vert n_{2} \vert \vert \neq0,\atop k\in{\Bbb {Z}}^{m}}\bigl( \vert k \vert +1\bigr)^{m+3}e^{- \vert k \vert (\sigma_{\nu}-\sigma _{\nu+1})} \biggr), \end{aligned}$$
and for \((m_{1},m_{2})=\{(2,0),(0,2)\}\),
$$\begin{aligned} \displaystyle\sup_{(\theta;\omega)\in{ \Theta_{\nu+1}\times\overline{J}_{\nu}}} \bigl\vert F_{n_{1}n_{2},\nu }^{m_{1}m_{2}} \bigr\vert \leq C \bigl\Vert \zeta_{n_{1}n_{2},\nu}^{m_{1}m_{2}} \bigr\Vert ^{*}_{\Theta_{\nu}\times \overline{J}_{\nu}} \bigl(1+\nu^{2}\bigr)\sum _{ k\in{\Bbb {Z}}^{m}}\bigl( \vert k \vert +1\bigr)^{m+3}e^{- \vert k \vert (\sigma_{\nu}-\sigma _{\nu+1})}. \end{aligned}$$
Furthermore, using Lemma 3.3 in [14], for \((\theta;\omega)\in \Theta_{\nu+1}\times\overline{J}_{\nu}\), we get
$$\begin{aligned} \bigl\vert F_{n_{1}n_{2},\nu}^{m_{1}m_{2}} \bigr\vert \leq C \bigl\Vert \zeta_{n_{1}n_{2},\nu }^{m_{1}m_{2}} \bigr\Vert ^{*}_{\Theta_{\nu}\times\overline{J}_{\nu}}( \nu +1)^{6m+24},\quad (m_{1},m_{2})=\bigl\{ (2,0),(1,1),(0,2)\bigr\} . \end{aligned}$$
(31)
From \((3.2)_{l}\), we have
$$\bigl\vert \partial_{\omega}(\lambda_{n_{1},\nu}\pm \lambda_{n_{2},\nu}) \bigr\vert \leq C\varepsilon,\qquad \vert \partial_{\omega}\lambda_{n_{1},\nu} \vert \leq C\varepsilon . $$
Thus, in view of (28)-(31), and using Lemma 3.3 in [14], we have, for \((\theta;\omega)\in \Theta_{\nu+1}\times\overline{J}_{\nu}\), \(n_{1},n_{2}\in\mathbb{Z}^{d}\),
$$\begin{aligned} \bigl\vert \partial_{\omega}F_{n_{1}n_{2},\nu}^{11} \bigr\vert \leq&\sum_{k\in{\Bbb {Z}}^{m},\atop \vert k \vert + \vert \vert n_{1} \vert - \vert n_{2} \vert \vert \neq0} \biggl\vert \frac{\partial_{\omega}\zeta_{n_{1}n_{2},\nu}^{k11}}{ \langle k,\omega\rangle-\lambda_{n_{1},\nu}+\lambda_{n_{2},\nu}} \biggr\vert \bigl\vert e^{\mathrm{i}\langle k,\theta\rangle} \bigr\vert \\ &{}+\sum_{k\in{\Bbb {Z}}^{m},\atop \vert k \vert + \vert \vert n_{1} \vert - \vert n_{2} \vert \vert \neq0} \biggl\vert \frac{\zeta_{n_{1}n_{2},\nu}^{k11} \partial_{\omega}(\langle k,\omega\rangle-\lambda_{n_{1},\nu}+\lambda _{n_{2},\nu})}{ (\langle k,\omega\rangle-\lambda_{n_{1},\nu}+\lambda_{n_{2},\nu})^{2}} \biggr\vert \bigl\vert e^{\mathrm{i}\langle k,\theta\rangle} \bigr\vert \\ \leq& C \bigl\Vert \zeta_{n_{1}n_{2},\nu}^{11} \bigr\Vert ^{*}_{\Theta_{\nu}\times\overline {J}_{\nu}}\bigl(1+l^{2}\bigr)^{2}\sum _{ k\in{\Bbb {Z}}^{m}}2\bigl( \vert k \vert +C\varepsilon\bigr) \bigl( \vert k \vert +1\bigr)^{2m+6}e^{- \vert k \vert (\sigma_{\nu}-\sigma_{\nu+1})} \\ \leq& C \bigl\Vert \zeta_{n_{1}n_{2},\nu}^{11} \bigr\Vert ^{*}_{\Theta_{\nu}\times \overline{J}_{\nu}}(\nu+1)^{6m+24}, \end{aligned}$$
(32)
and for \((m_{1},m_{2})=\{(2,0),(0,2)\}\),
$$\begin{aligned} \bigl\vert \partial_{\omega}F_{n_{1}n_{2},\nu}^{m_{1}m_{2}} \bigr\vert \leq&\sum_{k\in {\Bbb {Z}}^{m}} \biggl\vert \frac{\partial_{\omega}\zeta_{n_{1}n_{2},\nu}^{km_{1}m_{2}}}{ \langle k,\omega\rangle\pm(\lambda_{n_{1},\nu}+\lambda_{n_{2},\nu})} \biggr\vert \bigl\vert e^{\mathrm{i}\langle k,\theta\rangle} \bigr\vert \\ &{}+\sum_{k\in{\Bbb {Z}}^{m}} \biggl\vert \frac{\zeta_{n_{1}n_{2},\nu}^{km_{1}m_{2}} \partial_{\omega}(\langle k,\omega\rangle\pm(\lambda_{n_{1},\nu}+\lambda _{n_{2},\nu}))}{ (\langle k,\omega\rangle\pm(\lambda_{n_{1},\nu}+\lambda_{n_{2},\nu }))^{2}} \biggr\vert \bigl\vert e^{\mathrm{i}\langle k,\theta\rangle} \bigr\vert \\ \leq& C \bigl\Vert \zeta_{n_{1}n_{2},\nu}^{m_{1}m_{2}} \bigr\Vert ^{*}_{\Theta_{\nu}\times\overline {J}_{\nu}}\bigl(1+l^{2}\bigr)^{2}\sum _{ k\in{\Bbb {Z}}^{m}}2\bigl( \vert k \vert +C\varepsilon\bigr) \bigl( \vert k \vert +1\bigr)^{2m+6}e^{- \vert k \vert (\sigma_{\nu}-\sigma_{\nu+1})} \\ \leq& C \bigl\Vert \zeta_{n_{1}n_{2},\nu}^{m_{1}m_{2}} \bigr\Vert ^{*}_{\Theta_{\nu}\times \overline{J}_{\nu}}(\nu+1)^{6m+24}. \end{aligned}$$
(33)
It follows immediately that, for \((m_{1},m_{2})=\{(2,0),(1,1),(0,2)\}\),
$$\begin{aligned} \bigl\vert \partial_{\omega}F_{n_{1}n_{2},\nu}^{m_{1}m_{2}} \bigr\vert \leq C \bigl\Vert \zeta _{n_{1}n_{2},\nu}^{m_{1}m_{2}} \bigr\Vert ^{*}_{\Theta_{\nu}\times\overline{J}_{\nu}}(\nu +1)^{6m+24}. \end{aligned}$$
(34)
From (31), (33), (32) and (34), we have, for \((m_{1},m_{2})=\{(2,0),(1,1),(0,2)\} \),
$$\begin{aligned} \bigl\Vert F_{n_{1}n_{2},\nu}^{m_{1}m_{2}} \bigr\Vert ^{*}_{\Theta_{\nu+1}\times\overline {J}_{\nu}} \leq C \bigl\Vert \zeta_{n_{1}n_{2},\nu}^{m_{1}m_{2}} \bigr\Vert ^{*}_{\Theta_{\nu}\times \overline{J}_{\nu}}(\nu+1)^{6m+24}. \end{aligned}$$
(35)
In view of (28), we have
$$\begin{aligned} &\partial_{\theta}F_{n_{1}n_{2},\nu}^{11}=\sum _{ \vert k \vert + \vert \vert n_{1} \vert - \vert n_{2} \vert \vert \neq0} \frac{-\zeta_{n_{1}n_{2},\nu}^{k11}}{\langle k,\omega\rangle-\lambda _{n_{1},\nu}+\lambda_{n_{2},\nu} }e^{\mathrm{i}\langle k,\theta\rangle }\cdot k , \\ &\partial_{\theta}F_{n_{1}n_{2},\nu}^{20}=\sum _{ k\in{\Bbb {Z}}^{m}} \frac{-\zeta_{n_{1}n_{2},\nu}^{k20}}{\langle k,\omega\rangle-\lambda _{n_{1},\nu}-\lambda_{n_{2},\nu}}e^{\mathrm{i}\langle k,\theta\rangle }\cdot k, \\ &\partial_{\theta}F_{n_{1}n_{2},\nu}^{02}=\sum _{k\in{\Bbb {Z}}^{m}} \frac{-\zeta_{n_{1}n_{2},\nu}^{k02}}{\langle k,\omega\rangle+\lambda _{n_{1},\nu}+\lambda_{n_{2},\nu} }e^{\mathrm{i}\langle k,\theta\rangle }\cdot k, \\ &\partial_{\theta\theta}F_{n_{1}n_{2},\nu}^{11}=\sum _{ \vert k \vert + \vert \vert n_{1} \vert - \vert n_{2} \vert \vert \neq0} \frac{-\zeta_{n_{1}n_{2},\nu}^{k11}}{\langle k,\omega\rangle-\lambda _{n_{1},\nu}+\lambda_{n_{2},\nu} }e^{\mathrm{i}\langle k,\theta\rangle }\cdot \mathrm{i}kk^{T}, \\ &\partial_{\theta\theta} F_{n_{1}n_{2},\nu}^{20}=\sum _{ k\in{\Bbb {Z}}^{m}} \frac{-\zeta_{n_{1}n_{2},\nu}^{k20}}{\langle k,\omega\rangle-\lambda _{n_{1},\nu}-\lambda_{n_{2},\nu}}e^{\mathrm{i}\langle k,\theta\rangle }\cdot \mathrm{i}kk^{T}, \\ &\partial_{\theta\theta}F_{n_{1}n_{2},\nu}^{02}=\sum _{k\in{\Bbb {Z}}^{m}} \frac{-\zeta_{n_{1}n_{2},\nu}^{k02}}{\langle k,\omega\rangle+\lambda _{n_{1},\nu}+\lambda_{n_{2},\nu} }e^{\mathrm{i}\langle k,\theta\rangle }\cdot \mathrm{i}kk^{T}, \end{aligned}$$
where k is a m column vector and \(kk^{T}\) is a \(m\times m\) matrix. Similar to the above discussion, we get the following estimates:
$$ \begin{aligned} & \bigl\Vert \partial_{\theta}F_{n_{1}n_{2},\nu}^{m_{1}m_{2}} \bigr\Vert ^{*}_{\Theta_{\nu +1}\times\overline{J}_{\nu}} \leq C \bigl\Vert \zeta_{n_{1}n_{2},\nu}^{m_{1}m_{2}} \bigr\Vert ^{*}_{\Theta_{\nu}\times \overline{J}_{\nu}}{(\nu+1)}^{6m+24}, \\ & \bigl\Vert \partial_{\theta\theta}F_{n_{1}n_{2},\nu}^{m_{1}m_{2}} \bigr\Vert ^{*}_{\Theta_{\nu+1}\times\overline{J}_{\nu}}\leq C \bigl\Vert \zeta_{n_{1}n_{2},\nu }^{m_{1}m_{2}} \bigr\Vert ^{*}_{\Theta_{\nu}\times\overline{J}_{\nu}}{(\nu+1)}^{6m+24}. \end{aligned} $$
(36)
We will give estimates of the flow \(X^{t}_{\mathcal{F}_{\nu}}\).
For \(\nu=0,1,\ldots\) , there exists a constant \(0<\delta<1\) such that
$$\begin{aligned} \bigl\vert \varepsilon_{\nu}^{1-\delta}{( \nu+1)}^{6m+24} \bigr\vert \leq C, \end{aligned}$$
(37)
as \(\varepsilon<1\), where C is an absolute constant independent on \(\nu,\varepsilon\). From (35), for \((\theta,\omega)\in\Theta_{\nu+1}\times\overline{J}_{\nu}\), we obtain
$$\begin{aligned} \bigl\Vert \varepsilon_{\nu}F_{\nu}(\theta ,J,z,\bar{z};\omega) \bigr\Vert ^{*}_{\Theta_{\nu+1}\times\overline{J}_{\nu}} \leq& C\varepsilon_{\nu}^{\delta}\bigl\Vert P_{\nu}(\theta,J,z,\bar{z};\omega) \bigr\Vert ^{*}_{\Theta_{\nu}\times\overline{J}_{\nu}}. \end{aligned}$$
(38)
By (24), (36) and (37) we obtain
$$\begin{aligned} \bigl\Vert \varepsilon_{\nu}\partial_{\theta}F_{\nu}(\theta,J,z,\bar{z};\omega) \bigr\Vert ^{*}_{\Theta_{\nu+1}\times\overline {J}_{\nu}}\leq C\varepsilon_{\nu}^{\delta}\bigl\Vert P_{\nu}( \theta,J,z,\bar{z};\omega) \bigr\Vert ^{*}_{\Theta_{\nu}\times\overline{J}_{\nu}} \end{aligned}$$
(39)
and
$$\begin{aligned} \bigl\Vert \varepsilon_{\nu}\partial_{\theta \theta}F_{\nu}( \theta,J,z,\bar{z};\omega) \bigr\Vert ^{*}_{\Theta_{\nu+1}\times\overline{J}_{\nu}} \leq C \varepsilon_{\nu}^{\delta}\bigl\Vert P_{\nu}( \theta,J,z,\bar{z};\omega) \bigr\Vert ^{*}_{\Theta_{\nu}\times\overline{J}_{\nu}}. \end{aligned}$$
(40)
It follows from (24) and (38) that
$$\begin{aligned} & \bigl\Vert (\mathcal{F}_{\nu})_{{z}_{n_{1}}} \bigr\Vert ^{\ast}_{{\Theta_{\nu +1}\times\overline{J}_{\nu}}} \\ &\quad =\varepsilon_{\nu}\biggl\Vert \sum_{k\in\mathbb{Z}^{m},n_{1},n_{2}\in\mathbb {Z}^{d},\atop \vert k \vert + \vert \vert n_{1} \vert - \vert n_{2} \vert \vert \neq0} F_{n_{1}n_{2},\nu}^{k11}\bar{z}_{n_{2}}e^{\mathrm{i}\langle k,\theta\rangle} + \varepsilon_{\nu}\sum_{k\in{\Bbb {Z}}^{m},n_{1},n_{2}\in\mathbb{Z}^{d} } \bigl(F_{n_{1}n_{2},\nu}^{k20}+F_{n_{2} n_{1},\nu}^{k20} \bigr)z_{n_{2}}e^{\mathrm {i}\langle k,\theta\rangle} \biggr\Vert ^{\ast}_{{\Theta_{\nu+1}\times\overline{J}_{\nu}}} \\ &\quad \leq \bigl\Vert \varepsilon_{\nu}F_{\nu}(\theta,J,z, \bar{z};\omega) \bigr\Vert ^{*}_{\Theta_{\nu+1}\times\overline{J}_{\nu}} \\ &\quad \leq C\varepsilon_{\nu}^{\delta}\bigl\Vert P_{\nu}(\theta,J,z,\bar{z};\omega) \bigr\Vert ^{*}_{\Theta_{\nu}\times\overline{J}_{\nu}} \end{aligned}$$
(41)
and similarly
$$\begin{aligned} \bigl\Vert (\mathcal{F}_{\nu})_{\bar{z}_{n_{1}}} \bigr\Vert ^{\ast}_{{\Theta_{\nu +1}\times\overline{J}_{\nu}}}\leq C\varepsilon_{\nu}^{\delta}\bigl\Vert P_{\nu}(\theta,J,z,\bar{z};\omega) \bigr\Vert ^{*}_{\Theta_{\nu}\times\overline{J}_{\nu}}. \end{aligned}$$
(42)
Therefore, by using of (39)-(42), we obtain
$$\begin{aligned} \vert X_{\mathcal{F}_{\nu}} \vert ^{\ast}_{a,\rho+1,D^{a,\rho}_{\nu+1}\times \overline{J}_{\nu+1}} =& \bigl\Vert (\mathcal{F}_{\nu})_{J} \bigr\Vert ^{\ast}_{D^{a,\rho}_{\nu+1}\times \overline{J}_{\nu+1}} +\frac{1}{r_{\nu+1}^{2}} \bigl\Vert (\mathcal {F}_{\nu})_{\theta}\bigr\Vert ^{\ast}_{D^{a,\rho}_{\nu+1}\times\overline{J}_{\nu +1}} \\ &{} +\frac{1}{r_{\nu+1}}\bigl( \Vert F_{\bar{z}} \Vert ^{*}_{a,\rho,D^{a,\rho }_{\nu+1}\times\overline{J}_{\nu+1}}+ \Vert F_{z} \Vert ^{*}_{a,\rho ,D^{a,\rho}_{\nu+1}\times\overline{J}_{\nu+1}}\bigr) \\ \leq&C\varepsilon_{\nu}^{\delta}\bigl\Vert X_{P_{\nu}}(\theta,J,z,\bar{z};\omega ) \bigr\Vert ^{*}_{\Theta_{\nu}\times\overline{J}_{\nu}} \\ \leq& C\varepsilon_{\nu}^{\frac{\delta}{2}}, \end{aligned}$$
(43)
by \(\varepsilon_{\nu}^{{\frac{\delta}{2}}} \Vert X_{R_{\nu}}(\theta,J,z,\bar {z};\omega) \Vert ^{*}_{\Theta_{\nu}\times\overline{J}_{\nu}}\leq C\), as \(\varepsilon<1\), where C is an absolute constant independent on \(\nu,\varepsilon\).
To get the estimates for \(X_{\mathcal{F}_{\nu}}^{t}\), we consider the integral equation
$$X_{\mathcal{F}_{\nu}}^{t}=id+ \int_{0}^{t}X_{\mathcal{F}_{\nu}}\circ X_{\mathcal {F}_{\nu}}^{s}\,ds,\quad0\leq t\leq1. $$
Hence, we obtain from (43)
$$\begin{aligned} \bigl\vert X_{\mathcal{F}_{\nu}}^{1} -id \bigr\vert ^{*}_{a,\rho+1,D^{a,\rho}_{\nu+1}\times\overline{J}_{\nu+1}}\leq \vert X_{\mathcal{F}_{\nu}} \vert ^{\ast}_{a,\rho+1,D^{a,\rho}_{\nu+1}\times \overline{J}_{\nu+1}}\leq C\varepsilon_{\nu}^{\frac{\delta}{2}}. \end{aligned}$$
(44)
Let
$$\begin{aligned} &\bigl\vert D^{s}F \bigr\vert _{a,\rho+1,\rho,D^{a,\rho}_{\nu +1}\times\overline{J}_{\nu+1}}^{op} \\ &\quad =\max \biggl\{ \biggl\vert \frac{\partial^{ \vert j \vert + \vert i \vert + \vert \alpha \vert + \vert \beta \vert }}{\partial J^{j}\theta^{i}\partial z_{n}^{\alpha}\partial\bar{z}_{n}^{\beta}} F \biggr\vert _{a,\rho+1,\rho,D^{a,\rho}_{\nu+1}\times\overline{J}_{\nu +1}}^{op}, \vert j \vert + \vert i \vert + \vert \alpha \vert + \vert \beta \vert =s\geq2 \biggr\} . \end{aligned}$$
Notice that F is a polynomial of degree 2 in \(z,\bar{z}\). By (16), (43) and the Cauchy inequality, it follows that, for any \(s\geq2\),
$$\begin{aligned} \bigl\vert D^{s}F \bigr\vert _{a,\rho+1,\rho,D^{a,\rho }_{\nu+1}\times\overline{J}_{\nu+1}}^{op} \leq C\varepsilon_{\nu}^{\frac{\delta}{2}}. \end{aligned}$$
(45)
From \(\phi^{t}_{F}=id +\int_{0}^{t}X_{F}\circ\phi^{s}_{F}\,ds\), we have \(\phi^{t}_{F}:D_{\nu +1}^{a,\rho}\rightarrow D_{\nu}^{a,\rho}\), \(-1\leq t\leq1\), which follows directly from (45). Since
$$D\phi^{t}_{F}=\mathrm{Id}+ \int_{0}^{t}(DX_{F})D \phi^{s}_{F}\,ds=\mathrm{Id}+ \int _{0}^{t}\mathcal{J}\bigl(D^{2}F \bigr)D\phi^{s}_{F}\,ds, $$
where \(\mathcal{J}=\bigl({\scriptsize\begin{matrix}{} 0&-I\cr I&0 \end{matrix}} \bigr)\), it follows that
$$\begin{aligned} \bigl\vert D\phi^{t}_{F}-\mathrm{Id} \bigr\vert _{a,\rho +1,\rho,D^{a,\rho}_{\nu+1}\times\overline{J}_{\nu+1}}^{op}\leq2 \bigl\vert D^{2}F \bigr\vert _{a,\rho+1,\rho,D^{a,\rho}_{\nu+1}\times\overline{J}_{\nu +1}}^{op}\leq C\varepsilon_{\nu}^{\frac{\delta}{2}}. \end{aligned}$$
(46)
Similarly,
$$\begin{aligned} \bigl\vert DX_{\mathcal{F}_{\nu}}^{1} - \mathrm{Id} \bigr\vert ^{*op}_{a,\rho+1,\rho ,D^{a,\rho}_{\nu+1}\times\overline{J}_{\nu+1}}\leq C\varepsilon_{\nu}^{\frac{\delta}{2}}. \end{aligned}$$
(47)
We now estimate the smaller term \(P_{\nu+1}\) and we will finish one cycle of the iteration. Let
$${\tilde{\lambda}}_{n,\nu}=\frac{1}{(2\pi)^{m}} \int_{{\Bbb {T}}^{m}}\zeta_{nn,\nu}^{11}(\theta;\omega) \,d\theta $$
and
$${\lambda}_{n,\nu+1}= {\lambda }_{n,\nu}+\varepsilon_{\nu}{ \tilde{\lambda} }_{n,\nu}, $$
then it is easy to see that \({\lambda}_{n,\nu+1}\) satisfies the conditions \((3.2)_{\nu+1}\). Moreover, from (23) and (25), we know that
$$\begin{aligned} H_{\nu+1} =\langle\omega,J\rangle+\sum_{n\in\mathbb{Z}^{d}} \lambda_{n,\nu +1}z_{n}\bar{z}_{n} + \varepsilon_{\nu+1} P_{\nu+1}(\theta,J,z,\bar{z};\omega), \end{aligned}$$
where
$$\begin{aligned} &\varepsilon_{\nu+1} P_{\nu+1}(\theta,J,z, \bar{z};\omega) \\ &\quad = \bigl\{ \varepsilon_{\nu}P_{\nu}(\theta,J,z,\bar{z}; \omega),{\mathcal{F}_{\nu}} \bigr\} + \int _{0}^{1}(1-t) \bigl\{ \{H_{\nu}, \mathcal{F}_{\nu}\},\mathcal{F}_{\nu}\bigr\} \circ X^{t}_{\mathcal{F}_{\nu}}\,dt. \end{aligned}$$
(48)
By a direct calculation we get
$$\varepsilon_{\nu+1} P_{\nu+1}(\theta,J,z,\bar{z};\omega) = \varepsilon_{\nu}^{2}\sum_{n_{1},n_{2}\in\mathbb{Z}^{d}} \sum_{m_{1},m_{2}}\tilde {\zeta}_{n_{1}n_{2},\nu+1}^{m_{1}m_{2}} (\theta;\omega)z_{n_{1}}^{m_{1}}\bar{z}_{n_{2}}^{m_{2}}, $$
where \({\tilde{\zeta}_{n_{1}n_{2},\nu+1}^{m_{1}m_{2}}(\theta;\omega})^{,}s\) are a linear combination of the product of \(F_{n_{1}n_{2},\nu}^{m_{1}m_{2}}(\theta;\omega)\) and \({\zeta_{n_{1}n_{2},\nu}^{\tilde{n}_{1}\tilde{n}_{2}}(\theta;\omega)}^{,}s\), with \((m_{1},m_{2})\) or \((\tilde{n}_{1},\tilde{n}_{2})=\{ (2,0),(1,1),(0,2)\}\). Thus, by using of \((3.1)_{\nu}\), (31) and (35),
$$\begin{aligned} \bigl\Vert \tilde{\zeta}_{n_{1}n_{2},\nu+1}^{m_{1}m_{2}} \bigr\Vert ^{\ast}_{\Theta_{\nu +1}\times\overline{J}_{\nu}} \leq C \bigl\Vert \zeta_{n_{1}n_{2},\nu}^{m_{1}m_{2}} \bigr\Vert ^{\ast}_{\Theta_{\nu +1}\times\overline{J}_{\nu}}{(\nu+1)}^{6m+24} \end{aligned}$$
(49)
is true. In view of \(\varepsilon_{\nu}^{2-(1-\delta)}=\varepsilon_{\nu +1}\), and \(C\varepsilon_{\nu}^{1-\delta} \Vert \zeta_{n_{1}n_{2},\nu +1}^{m_{1}m_{2}} \Vert ^{\ast}_{\Theta_{\nu+1}\times\overline{J}_{\nu}}{(\nu +1)}^{6m+24}\leq1\), as \(\varepsilon<1\), we can suppose that
$$\begin{aligned} \zeta_{n_{1}n_{2},\nu+1}^{m_{1}m_{2}}:=\varepsilon_{\nu}^{1-\delta} \tilde{\zeta}_{n_{1}n_{2},\nu+1}^{m_{1}m_{2}}. \end{aligned}$$
(50)
It follows from (49) that
$$\bigl\Vert \zeta_{n_{1}n_{2},\nu+1}^{m_{1}m_{2}} \bigr\Vert ^{\ast}_{\Theta_{\nu+1}\times \overline{J}_{\nu}}\leq C. $$
This implies \((E)_{\nu+1}\) as defined in \(D^{a,\rho}_{\nu+1}\) and the \({\zeta_{n_{1}n_{2},\nu+1}^{m_{1}m_{2}}}\) satisfy \((3.1)_{\nu +1}\).
The perturbation \(P_{l}\) satisfying \((3.0)_{l}\) is used to guarantee that the normal form at each KAM step has the same form as in the first step. In order to complete one KAM step, we need to prove that the new perturbation \(P_{\nu+1}\) still has the special form \((3.0)_{\nu+1}\).
From (25), we get
$$\begin{aligned} \biggl\{ \langle\omega,J\rangle+\sum_{n\in\mathbb{Z}^{d}} \lambda_{n,\nu }z_{n}\bar{z}_{n},F_{\nu}\biggr\} =-P_{\nu}(\theta,J,z,\bar{z};\omega) +\sum _{n\in\mathbb{Z}^{d}}\bigl[\zeta_{nn,\nu}^{11} \bigr]z_{n}\bar{z}_{n}. \end{aligned}$$
It is easy to see that \(\{\langle\omega,J\rangle+\sum_{n\in\mathbb {Z}^{d}}\lambda_{n,\nu}z_{n}\bar{z}_{n},F_{\nu}\} \) satisfies \((3.0)_{\nu+1}\). Thus, from (48), we only to consider \(\{P_{\nu},F_{\nu}\} \) satisfies \((3.0)_{\nu+1}\). Let \(B_{\nu}=\{P_{\nu},F_{\nu}\}\). Taking \((\alpha_{1},\beta_{1}),(\alpha_{2},\beta _{2})\in\{(e_{n_{1}}+e_{n_{2}},0), (e_{n_{1}},e_{n_{2}}), (0,e_{n_{1}}+e_{n_{2}})\} \), we can assume that
$$\begin{aligned} P_{\nu}=\sum_{k,\alpha_{1},\beta_{1}} \zeta_{k\alpha_{1}\beta_{1},\nu}e^{\mathrm{i}\langle k,\theta\rangle}z^{\alpha_{1}}\bar{z}^{\beta_{1}},\qquad F_{\nu}= \sum_{k,\alpha_{2},\beta_{2}}F_{k\alpha_{2}\beta_{2},\nu}e^{\mathrm{i}\langle k,\theta\rangle}z^{\alpha_{2}} \bar{z}^{\beta_{2}}, \end{aligned}$$
(51)
with
$$\begin{aligned} &\zeta_{k\alpha_{1}\beta_{1},\nu}=0,\quad \mbox{if } k=0, \sum _{n\in\mathbb {Z}^{d}}(\alpha_{1n}-\beta_{1n})n=0; \\ &F_{k\alpha_{1}\beta_{1},\nu}=0,\quad \mbox{if } k=0, \sum_{n\in\mathbb {Z}^{d}}( \alpha_{1n}-\beta_{1n})n=0. \end{aligned}$$
Since
$$\begin{aligned} \{P_{\nu},F_{\nu}\} =&\mathrm{i} \sum _{m}\sum_{a_{1}} \zeta_{k\alpha_{1}\beta _{1},\nu} F_{k\alpha_{1}\beta_{1},\nu}e^{2\mathrm{i}\langle k,\theta\rangle }z^{\alpha_{1}-e_{m}} \bar{z}^{\beta_{1}}z^{\alpha_{2}}\bar{z}^{\beta_{2}-e_{m}} \\ &{}-\mathrm{i} \sum_{m}\sum _{a_{2}} \zeta_{k\alpha_{1}\beta_{1},\nu} F_{k\alpha _{1}\beta_{1},\nu}e^{2\mathrm{i}\langle k,\theta\rangle}z^{\alpha_{1}} \bar {z}^{\beta_{1}-e_{m}}z^{\alpha_{2}-e_{m}}\bar{z}^{\beta_{2}} \\ &{}+\mathrm{i} \sum_{m}\sum _{k\neq0} \zeta_{k\alpha_{1}\beta_{1},\nu} F_{k\alpha_{1}\beta_{1},\nu}e^{2\mathrm{i}\langle k,\theta\rangle}z^{\alpha _{1}-e_{m}} \bar{z}^{\beta_{1}}z^{\alpha_{2}}\bar{z}^{\beta_{2}-e_{m}} \\ &{}-\mathrm{i} \sum_{m}\sum _{k\neq0} \zeta_{k\alpha_{1}\beta_{1},\nu} F_{k\alpha_{1}\beta_{1},\nu}e^{2\mathrm{i}\langle k,\theta\rangle}z^{\alpha _{1}} \bar{z}^{\beta_{1}-e_{m}}z^{\alpha_{2}-e_{m}}\bar{z}^{\beta_{2}} \\ =&\mathrm{i} \sum_{m}\sum _{a_{3}} B_{k(\alpha_{1}+\alpha_{2}-e_{m})(\beta _{1}+\beta_{2}-e_{m}),\nu} e^{2\mathrm{i}\langle k,\theta\rangle}z^{\alpha _{1}+\alpha_{2}-e_{m}} \bar{z}^{\beta_{1}+\beta_{2}-e_{m}} \\ &{}+\mathrm{i} \sum_{m}\sum _{k\neq0} B_{k(\alpha_{1}+\alpha_{2}-e_{m})(\beta _{1}+\beta_{2}-e_{m}),\nu} e^{2\mathrm{i}\langle k,\theta\rangle}z^{\alpha _{1}+\alpha_{2}-e_{m}} \bar{z}^{\beta_{1}+\beta_{2}-e_{m}}, \end{aligned}$$
where \(a_{1}\) denotes
$$\begin{aligned} &k=0,\quad (\alpha_{1m}-1-\beta_{1m})m+\sum _{n\in\mathbb{Z}^{d}\backslash\{m\} }(\alpha_{1n}-\beta_{1n})n=-m, \\ &\bigl(\alpha_{2m}-(\beta_{2m}-1)\bigr)m+\sum _{n\in\mathbb{Z}^{d}\backslash\{m\} }(\alpha_{2n}-\beta_{2n})n=m, \end{aligned}$$
\(a_{2}\) denotes
$$\begin{aligned} &k=0,\quad \bigl(\alpha_{1m}-(\beta_{1m}-1)\bigr)m+\sum _{n\in\mathbb{Z}^{d}\backslash\{m\} }(\alpha_{1n}-\beta_{1n})n=m, \\ &(\alpha_{2m}-1-\beta_{2m})m+\sum _{n\in\mathbb{Z}^{d}\backslash\{m\} }(\alpha_{2n}-\beta_{2n})n=-m, \end{aligned}$$
\(a_{3}\) denotes
$$k=0,\quad \bigl((\alpha_{1m}+\alpha_{2m}-1)-( \beta_{1m}+\beta_{2m}-1)\bigr)m+\sum _{n\in \mathbb{Z}^{d}\backslash\{m\}}(\alpha_{1n}+\alpha_{2n}- \beta_{1n}-\beta_{2n})n=0. $$
Thus, one finds that \(\{P_{\nu},F_{\nu}\} \) satisfies \((3.0)_{\nu+1}\). Moreover, \(P_{\nu+1} \) satisfies \((3.0)_{\nu+1}\).
Finally, we consider the convergence of transformations \(\Sigma^{N}\).
In view of (43) and (47), by letting
$$\begin{aligned} S_{\nu}=X_{\mathcal{F}_{\nu}}^{1}: D^{a,\rho }_{\nu+1}\times\overline{J}_{\nu+1}\longmapsto D^{a,\rho}_{\nu}\times\overline{J}_{\nu} \end{aligned}$$
(52)
we have
$$\begin{aligned} \vert S_{\nu} -id \vert ^{*}_{a,\rho+1,D^{a,\rho}_{\nu+1}\times\overline{J}_{\nu +1}} \leq C\varepsilon_{\nu}^{\frac{\delta}{2}}, \qquad \vert DS_{\nu} - \mathrm{Id} \vert ^{*op}_{a,\rho+1,\rho,D^{a,\rho}_{\nu+1}\times \overline{J}_{\nu+1}}\leq C\varepsilon_{\nu}^{\frac{\delta}{2}} . \end{aligned}$$
(53)
Now we are ready to prove the limiting transformation \(S_{0}\circ S_{1}\circ\cdots\) convergent to a linear symplectic transformation \(\Sigma^{\infty}\), which integrates equation (10). For any \(\omega\in\overline{J}, N\geq1\), we denote by \(\Sigma^{N}\) the map
$$\Sigma^{N}(\cdot;\omega)=S_{0}(\cdot;\omega)\circ\cdots \circ S_{N-1}(\cdot;\omega):D_{N}^{a,\rho}\longmapsto D^{a,\rho}(\sigma,r) $$
as usual, \(\Sigma^{0}\) is the identity mapping. From the second inequality of (53), we have
$$\begin{aligned} \bigl\vert D\Sigma^{N} \bigr\vert _{a,\rho+1,\rho,D_{N}^{a,\rho}\times\overline{J}}^{*op} \leq\prod_{\mu=0}^{N-1} \vert D S_{\mu} \vert _{a,\rho+1,\rho,D_{\mu +1}^{a,\rho}\times\overline{J}}^{*op} \leq\prod _{\mu\geq0}\bigl(1+C\varepsilon_{N}^{\frac{\delta}{2}}\bigr) \leq2 \end{aligned}$$
provided that ε is small enough. Thus, by using the first inequality of (53), we have
$$\begin{aligned} \displaystyle{ \bigl\vert \Sigma^{N+1}-\Sigma^{N} \bigr\vert }_{a,\rho +1,D_{N+1}^{a,\rho}\times\overline{J}}^{*} \leq& \bigl\vert D\Sigma^{N} \bigr\vert _{a,\rho+1,\rho,D_{N}^{a,\rho}\times\overline {J}}^{*op}\cdot \vert S_{N}-id \vert _{a,\rho+1,D_{N+1}^{a,\rho} \times\overline{J}}^{*} \\ \leq& C\varepsilon_{N}^{\frac{\delta}{2}}. \end{aligned}$$
So the sequence \(\{\Sigma^{N}\}\) converges uniformly in \(D^{a,s}_{N}\) to an analytic map
$$\Sigma^{\infty}:D^{a,\rho}(\sigma/2,r/2)\longmapsto D^{a,\rho}(\sigma,r). $$
We remark that the Hamiltonian (10) satisfies \((E)_{\nu}\), \((3.1)_{\nu}\) and \((3.2)_{\nu}\) with \(\nu=0\), the above iterative procedure can run repeatedly. So
$$\begin{aligned} \mu_{n}=\lambda_{n}+\frac{\varepsilon\psi_{0}}{2\lambda_{n}}+ \sum _{k=1}^{\infty}\varepsilon_{k}\tilde{ \lambda}_{n,k}, \end{aligned}$$
where \(\vert \tilde{\lambda}_{n,k} \vert \leq C,k=1, 2,3,\ldots\) . So (i) and (ii) are obtained. This completes the proof. □