About Linearization of Infinite-Dimensional Hamiltonian Systems

Abstract

This article is concerned with analytic Hamiltonian dynamical systems in infinite dimension in a neighborhood of an elliptic fixed point. Given a quadratic Hamiltonian, we consider the set of its analytic higher order perturbations. We first define the subset of elements which are formally symplectically conjugated to a (formal) Birkhoff normal form. We prove that if the quadratic Hamiltonian satisfies a Diophantine-like condition and if such a perturbation is formally symplectically conjugated to the quadratic Hamiltonian, then it is also analytically symplectically conjugated to it. Of course what is an analytic symplectic change of variables depends strongly on the choice of the phase space. Here we work on periodic functions with Gevrey regularity.

Appendix A. Technical Lemmata

In the following, we adapt material from [BMP18] to non mass conservation situation. There are no new conceptual difficulties w.r.to [BMP18] but the proofs require a bit more care and some further case analysis. For the readers convenience, we have written also the proofs of a couple of technical lemmata from [BMP18] on which we rely.

1.1 Proof of Lemmata 3.5 and 3.6

We follow here [BMP18] [Appendix B. Proof of lemma 3.1]. For any $H\in {{\mathcal {H}}}_{r,s}$ (we recall that this space depends on two extra parameters $p\ge \frac{1}{2}$ and $0<\theta \le 1$) we define a map

$$\begin{aligned} B_1(\ell ^2)\rightarrow \ell ^2 \,,\quad y={\left( y_j\right) }_{j\in \mathbb {Z}}\mapsto {\left( Y^{(j)}_{H}(y;r,s)\right) }_{j\in \mathbb {Z}} \end{aligned}$$

by setting

$$\begin{aligned} Y^{(j)}_{H}(y;r,s) := \sum _*|H_{{ {\alpha }},{ {\beta }}}| \frac{({ {\alpha }}_j+{ {\beta }}_j)}{2}c^{(j)}_{r,s}({ {\alpha }},{ {\beta }}) y^{{ {\alpha }}+{ {\beta }}-e_j} \end{aligned}$$

(32)

where $e_j$ is the j-th basis vector in ${\mathbb N}^\mathbb {Z}$, while the coefficient

$$\begin{aligned} c^{(j)}_{r,s}({\alpha },{\beta })= r^{|{\alpha }|+|{\beta }|-2} {\left( \frac{\langle j \rangle ^2}{\prod _i\langle i \rangle ^{{ {\alpha }}_i+{ {\beta }}_i}}\right) }^{p} e^{-s (\sum _i \langle i \rangle ^\theta ({ {\alpha }}_i+{ {\beta }}_i) -2\langle j \rangle ^\theta )} \end{aligned}$$

(33)

For brevity, we set

$$\begin{aligned} \sum _*:=\sum _{{ {\alpha }},{ {\beta }}: \;\pi ({\alpha },{\beta })=0}\,. \end{aligned}$$

The vector field $Y_H$ is a majorant analytic function on $\ell ^2$ which has the same norm as H. Since the majorant analytic functions on a given space have a natural ordering this gives us a natural criterion for immersions, as formalized in the following Lemma.

Lemma A.1

Let $r,{r^*}>0,\,s,s'\ge 0.$ The following properties hold.

• (1) The norm of H can be expressed as

  $$\begin{aligned} {\left\| H\right\| }_{r,s}= \sup _{|y|_{\ell ^2}\le 1}{\left| Y_H(y;r,s)\right| }_{\ell ^2} \end{aligned}$$

  (34)

• (2) Given $H^{(1)}\in {{\mathcal {H}}}_{{r^*},s'}$ and $H^{(2)}\in {{\mathcal {H}}}_{r,s}\,,$

  such that for all ${ {\alpha }},{ {\beta }}\in {\mathbb N}^\mathbb {Z}_f$ and $j\in \mathbb {Z}$ with ${ {\alpha }}_j+{ {\beta }}_j\ne 0$ one has

  $$\begin{aligned} |H^{(1)}_{{ {\alpha }},{ {\beta }}}| c^{(j)}_{{r^*},s'}({ {\alpha }},{ {\beta }}) \le c |H^{(2)}_{{ {\alpha }},{ {\beta }}}| c^{(j)}_{r,s}({ {\alpha }},{ {\beta }}), \end{aligned}$$

  for some $c>0,$ then

  $$\begin{aligned} {\left\| H^{(1)}\right\| }_{{r^*},s'} \le c {\left\| H^{(2)}\right\| }_{r,s}\,. \end{aligned}$$

Proof of Lemma 3.5

Recalling (33), we have

$$\begin{aligned} \frac{c^{(j)}_{{r^*},s}({ {\alpha }},{ {\beta }})}{c^{(j)}_{r,s}({ {\alpha }},{ {\beta }})}= {\left( \frac{{r^*}}{r}\right) }^{|{ {\alpha }}|+|{ {\beta }}|-2}\,. \end{aligned}$$

Since $|{ {\alpha }}|+|{ {\beta }}|-2\ge {\mathtt {d}}$, the inequality follows by Lemma A.1 with $H^{(1)}=H^{(2)}$ and $s=s'$. $\square$

In order to prove Lemma 3.6 we need some notations and results proven in [Bou05] and [CLSY].

Definition A.2

Given a vector $v={\left( v_i\right) }_{i\in \mathbb {Z}}\in {\mathbb N}^\mathbb {Z}_f$ with $|v|\ge 2$ we denote by $\widehat{n}=\widehat{n}(v)$ the vector ${\left( \widehat{n}_l\right) }_{l\in I}$ (where $I\subset {\mathbb N}$ is finite) which is the decreasing rearrangement of

$$\begin{aligned} \{{\mathbb N}\ni h> 1\;\; \text{ repeated }\; v_h + v_{-h}\; \text{ times } \} \cup {\left\{ 1\;\; \text{ repeated }\; v_1 + v_{-1} + v_0\; \text{ times } \right\} } \end{aligned}$$

Remark A.3

A good way of envisioning this list is as follows. Given an infinite set of variables ${\left( x_i\right) }_{i\in \mathbb {Z}}$ and a vector $v={\left( v_i\right) }_{i\in \mathbb {Z}}\in {\mathbb N}^\mathbb {Z}_f$ consider the monomial $x^v:= \prod _i x_i^{v_i}$. We can write

$$\begin{aligned} x^v= \prod _i x_i^{v_i} = x_{j_1} x_{j_2}\cdots x_{j_{|v|}}\,,\quad \text{ with }\quad j_k\in \mathbb {Z}\end{aligned}$$

then $\widehat{n}(v)$ is the decreasing rearrangement of the list ${\left( \langle j_1 \rangle ,\dots ,\langle j_{|v|} \rangle \right) }$.

Example A.4

Let us set

$$\begin{aligned} v_{-1}=2, v_{0}=3, v_1= 1, v_{3}=1, v_{4}= 2. \end{aligned}$$

Hence, 1 is repeated 6 times, 3 is repeated 1 time, and 4 is repeated 2 times :

$$\begin{aligned} {\hat{n}}_1=4,{\hat{n}}_2= 4, {\hat{n}}_3 = 3, {\hat{n}}_4=\dots ={\hat{n}}_9=1 \end{aligned}$$

Given ${ {\alpha }},{ {\beta }}\in {\mathbb N}^\mathbb {Z}_f$ with $|{ {\alpha }}|+|{ {\beta }}|\ge 2$ from now on we define

$$\begin{aligned} \widehat{n}=\widehat{n}({ {\alpha }}+{ {\beta }})\, \qquad \text{ and } \text{ set }\quad N:=|{ {\alpha }}|+|{ {\beta }}| \end{aligned}$$

which is the cardinality of $\widehat{n}.$ We observe that, $N\ge 2$ and since

$$\begin{aligned} 0= \sum _{i\in \mathbb {Z}} i{\left( { {\alpha }}_i - { {\beta }}_i\right) }= \sum _{h> 0} h {\left( { {\alpha }}_h - { {\beta }}_h - { {\alpha }}_{-h} + { {\beta }}_{-h}\right) } \,, \end{aligned}$$

(35)

there exists a choice of ${\sigma }_i = \pm 1, 0$ such that⁵

$$\begin{aligned} \sum _l \sigma _l\widehat{n}_l=0. \end{aligned}$$

(36)

with $\sigma _l \ne 0$ if $\widehat{n}_l \ne 1$. Hence,

$$\begin{aligned} \widehat{n}_1\le \sum _{l\ge 2}\widehat{n}_l. \end{aligned}$$

(37)

Indeed, if $\sigma _1 = \pm 1$, the inequality follows directly from (36); if $\sigma _1 = 0$, then $\widehat{n}_1=1$ and consequently $\widehat{n}_l = 1\, \forall l$. Since $|{\alpha }|+|{\beta }|\ge 2$, the list $\widehat{n}$ has at least two elements, so the inequality is achieved.

Lemma A.5

Given ${ {\alpha }},{ {\beta }}$ such that $\sum _i i ({ {\alpha }}_i-{ {\beta }}_i)=0$, and $|{\alpha }|+|{\beta }|\ge 2$, we have that setting $\widehat{n}=\widehat{n}({ {\alpha }}+{ {\beta }})$

$$\begin{aligned} \sum _i \langle i \rangle ^\theta ({ {\alpha }}_i+{ {\beta }}_i) =\sum _{l\ge 1} \widehat{n}_l^\theta \ge 2 \widehat{n}^\theta _1+ (2-2^\theta ) {\sum _{l\ge 3} \widehat{n}_l^\theta } . \end{aligned}$$

(38)

Proof

The lemma above was proved in [Bou05] for $\theta =\frac{1}{2}$ and for general $0<\theta <1$ in [CLSY][Lemma 2.1], in the case of zero mass and momentum. Below we give a proof, using only momentum conservation.

We start by noticing that if $|{\alpha }|+|{\beta }|= 2$ then $\widehat{n}$ has cardinality equal to two and (38) becomes $\widehat{n}_1+\widehat{n}_2 \ge 2\widehat{n}_1$. Now, by (37), momentum conservation implies that $\widehat{n}_1=\widehat{n}_2$ and hence (38).

If $|{\alpha }|+|{\beta }|\ge 3$ we write

$$\begin{aligned} \sum _i \langle i \rangle ^\theta ({ {\alpha }}_i+{ {\beta }}_i) -2\widehat{n}_1^\theta =\sum _{l\ge 2} \widehat{n}_l^\theta - \widehat{n}_1^\theta \ge \sum _{l\ge 2} \widehat{n}_l^\theta - (\sum _{l\ge 2} \widehat{n}_l)^\theta \end{aligned}$$

since the cardinality of $\widehat{n}$ is at least three we may write

$$\begin{aligned} \sum _{l\ge 2} \widehat{n}_l^\theta - (\sum _{l\ge 2} \widehat{n}_l)^\theta = \widehat{n}_2^\theta + \sum _{l\ge 3} \widehat{n}_l^\theta - (\widehat{n}_2+\sum _{l\ge 3} \widehat{n}_l)^\theta \end{aligned}$$

Now setting, for $x_i\ge 1$, $i=2,\ldots , N$,

$$\begin{aligned} f(x_2,\dots ,x_N):= x_2^\theta +(2^\theta - 1) \sum _{l\ge 3} x_l^\theta - (x_2+\sum _{l\ge 3} x_l)^\theta . \end{aligned}$$

Hence, we have $\partial _{x_2} f\ge 0$ for $x_2\ge x_3\ge 1$. Then

$$\begin{aligned} f(x_2,\dots ,x_N)\ge f(x_3,x_3,x_4,\dots ,x_N)=: f_3(x_3,\dots ,x_N)\,. \end{aligned}$$

Now we set

$$\begin{aligned}&f_n(x_n,\dots ,x_N):= f(\underbrace{x_n,\dots ,x_n}_{n-1},x_{n+1},\dots ,x_N)\\&\quad = (1+ (2^\theta -1)(n-2))x_n^\theta +\sum _{\ell \ge n+1}x_\ell - ((n-1)x_n+ \sum _{\ell \ge n+1}x_\ell )^\theta \end{aligned}$$

so that $f(x_2,\dots ,x_N)\ge f_3(x_3,\dots ,x_N)$. Assume inductively that for some $3\le n<N$, one has $f(x_2,\dots ,x_N)\ge f_3(x_3,\dots ,x_N) \ge \dots \ge f_{n}(x_n,\dots ,x_N)$. By direct computation⁶

$$\begin{aligned} \partial _{x_n} f_n&=\theta \Big [ \frac{(1+ (2^\theta -1)(n-2))}{x_n^{1-\theta }} -\frac{n-1}{((n-1)x_n+ \sum _{\ell \ge n+1}x_\ell )^{1-\theta } }\Big ]\\&\ge \theta x_n^{\theta -1 }\Big [ {(1+ (2^\theta -1)(n+2))} -{(n-1)^{\theta } }\Big ]\ge 0\,, \end{aligned}$$

so that the minimum is attained in $x_n= x_{n+1}$ and $f(x_2,\dots ,x_N) \ge f_{n+1}(x_{n+1},\dots ,x_N)$. In conclusion

$$\begin{aligned} f(x_2,\dots ,x_N)\ge f(x_N,\dots ,x_N)\ge 0 \end{aligned}$$

where the last inequality follows by recalling that $1+(2^\theta -1)k - (k+1)^\theta \ge 0$ for $k\ge 1$.

$\square$

The Lemma proved above, is fundamental in discussing the properties of ${{\mathcal {H}}}_{r}({\mathtt {h}}_{p,s,a})$ with $s>0$, indeed it implies

$$\begin{aligned} \sum _i \langle i \rangle ^\theta ({ {\alpha }}_i+{ {\beta }}_i) -2\langle j \rangle ^\theta \ge (2-2^\theta ) {\left( \sum _{l\ge 3} \widehat{n}_l^\theta \right) } \ge 0 \end{aligned}$$

(39)

for all ${ {\alpha }},{ {\beta }}$ such that ${ {\alpha }}_j+{ {\beta }}_j\ne 0$. Indeed, this follows from the fact that $\langle j \rangle \le \widehat{n}_1$.

Proof of Lemma 3.6

In all that follows we shall use systematically the fact that our Hamiltonians are momentum preserving, are zero at the origin and have no linear term so that ${\left| { {\alpha }}\right| } +{\left| { {\beta }}\right| } \ge 2$.

We need to show that

$$\begin{aligned} \frac{ c^{(j)}_{r,s+{\sigma }}({ {\alpha }},{ {\beta }})}{c^{(j)}_{r,s }({ {\alpha }},{ {\beta }})} = \exp (-{\sigma }(\sum _i \langle i \rangle ^\theta ({ {\alpha }}_i+{ {\beta }}_i) -2\langle j \rangle ^\theta ) \le 1\,. \end{aligned}$$

(40)

The first identity comes form (33), while the last inequality follows by (39) of Lemma A.5$\square$

1.2 Proof of Lemma 3.7

This is a rather classical result, the proof we give is taken from [BMP18], where it is stated under the extra hypothesis of mass conservation, which is not used in the proof.

Lemma A.6

Let $0<r_1<r.$ Let E be a Banach space endowed with the norm $|\cdot |_E$. Let $X:B_r \rightarrow E$ a vector field satisfying

$$\begin{aligned} \sup _{B_r}|X|_E\le \delta _0\,. \end{aligned}$$

Then the flow $\Phi (u,t)$ of the vector field⁷ is well defined for every

$$\begin{aligned} |t|\le T:=\frac{r-r_1}{\delta _0} \end{aligned}$$

and $u\in B_{r_1}$ with estimate

$$\begin{aligned} |\Phi (u,t)-u|_E\le \delta _0 |t|\,,\qquad \forall \, |t|\le T \,. \end{aligned}$$

Proof of Lemma 3.7

The estimate for the Poisson bracket is proven in [BBP13]. In order to prove the other estimates we use Lemma A.6, with $E\rightarrow {\mathtt {h}}_{s}$, $X\rightarrow X_S$, $\delta _0\rightarrow (r+\rho ) |S|_{r+\rho },$ $r\rightarrow r+\rho ,$ $r_1\rightarrow r,$ $T\rightarrow 8e.$ finally we do not write the dependence on s which is fixed.

Then the fact that the time 1-Hamiltonian flow $\Phi ^1_S: B_r({\mathtt {h}}_{s}) \rightarrow B_{r + \rho }({\mathtt {h}}_{s})$ is well defined, analytic, symplectic follows, since

$$\begin{aligned} \sup _{u\in B_{r+\rho }({\mathtt {h}}_{s})} |X_S|_{{\mathtt {h}}_{s}} \le (r+\rho ) |S|_{r+\rho }<\frac{\rho }{8 e}\,. \end{aligned}$$

Regarding the estimate (22), again by Lemma A.6 (choosing $t=1$), we get

$$\begin{aligned} \sup _{u\in B_{r}({\mathtt {h}}_{s})} {\left| \Phi ^1_S(u)-u\right| } _{{\mathtt {h}}_{s}} \le (r+\rho ) |S|_{r+\rho } <\frac{\rho }{8 e} \,. \end{aligned}$$

Estimates (23), (24), (25) directly follow by (26) with $h=0,1,2,$ respectively and $c_k=1/k!$, recalling that by Lie series

$$\begin{aligned} H \circ \Phi ^1_S = e^\mathrm{ad_S} H = \sum _{k=0}^\infty \frac{ \mathrm{ad}_S^k H}{k!} = \sum _{k=0}^\infty \frac{ H^{(k)}}{k!}\,, \end{aligned}$$

where $H^{(i)} := \mathrm{ad}_S^i (H)= \mathrm{ad}_S ( H^{(i-1)})$, $H^{(0)}:=H$.

Let us prove (26). Fix $k\in {\mathbb N},$ $k>0$ and set

$$\begin{aligned} r_i := r +\rho (1 - \frac{i}{k}) \, \, , \qquad i = 0,\ldots ,k \, . \end{aligned}$$

Note that, by the immersion properties of the norm in Lemma 3.5,

$$\begin{aligned} \Vert S\Vert _{r_i}\le \Vert S\Vert _{r+\rho }\,,\qquad \forall \, i = 0,\ldots ,k\,. \end{aligned}$$

(41)

Noting that

$$\begin{aligned} 1+\frac{k r_i}{\rho } \, \le k {\left( 1+\frac{r}{\rho }\right) }\,, \qquad \forall \, i=0,\ldots ,k\,, \end{aligned}$$

(42)

by using k times (20) we have

$$\begin{aligned}&\Vert {H^{(k)}}\Vert _r = \Vert \{S, {H^{(k-1)}}\} \Vert _r \le 4 (1+\frac{ k r}{\rho }) \Vert {H^{(k-1)}}\Vert _{r_{k-1}}\Vert S\Vert _{r_{k-1}}\\&\qquad \qquad {\mathop {\le }\limits ^{(42)}}\Vert H\Vert _{r+\rho } \Vert S\Vert _{r+\rho }^k 4^k \prod _{i=1}^k (1+\frac{ k r_i}{\rho }) {\mathop {\le }\limits ^{(43)}}\Vert H\Vert _{r+\rho } \left( 4k {\left( 1+\frac{r}{\rho }\right) }\Vert S\Vert _{r+\rho } \right) ^k \,. \end{aligned}$$

Then, using $k^k\le e^k k!,$ we get

$$\begin{aligned} \left\| \sum _{k\ge h} c_k {H^{(k)}}\right\| _{r}\le & {} \sum _{k\ge h} |c_k| \Vert {H^{(k)}}\Vert _{r} \le \Vert H\Vert _{r+\rho } \sum _{k\ge h} \left( 4e {\left( 1+\frac{r}{\rho }\right) }\Vert S\Vert _{r+\rho } \right) ^k \\= & {} \Vert H\Vert _{r+\rho } \sum _{k\ge h} (\Vert S\Vert _{r+\rho }/2\delta )^k {\mathop {\le }\limits ^{(22)}}2 \Vert H\Vert _{r+\rho } (\Vert S\Vert _{r+\rho }/2\delta )^h\,. \end{aligned}$$

Finally, if S and H satisfy momentum conservation so does each $\mathrm{ad}_S^k H$, $k \ge 1$, hence $H \circ \Phi ^1_S$ too. $\square$

1.3 Proof of lemma 3.8

Here we strongly use the fact that $\omega _j \sim j^2$. As we said in the introduction, it would not be hard to modify the proof in order to deal with $\omega _j\sim j^\alpha$ with $\alpha >1$, by adapting the proof of Lemmata A.9–A.10.

By Lemma A.1 (2), we have

$$\begin{aligned} {\left\| L_\omega ^{-1} R\right\| }_{r,s+{\sigma }}\le \gamma ^{-1}K {\left\| R\right\| }_{r,s} \end{aligned}$$

where

$$\begin{aligned} K=\gamma \sup _{\begin{array}{c} j: { {\alpha }}_j+{ {\beta }}_j\ne 0\\ \pi ({\alpha },{\beta })=0 \end{array}} \frac{e^{-{\sigma }{\left( \sum _i\langle i \rangle ^\theta ({ {\alpha }}_i+{ {\beta }}_i) -2\langle j \rangle ^\theta \right) }}}{{\left| \omega \cdot {{\left( { {\alpha }}- { {\beta }}\right) }}\right| }}\,. \end{aligned}$$

Therefore proving (27) amounts to showing that

$$\begin{aligned} K \le e^{{{\mathcal {C}}}_1 {\sigma }^{-\frac{3}{\theta }}}\,. \end{aligned}$$

(43)

We divide in two cases regarding whether the inequality

$$\begin{aligned} {\left| \sum _i{{\left( { {\alpha }}_i-{ {\beta }}_i\right) }i^2}\right| }\le 2 \sum _i{\left| { {\alpha }}_i-{ {\beta }}_i\right| }\,, \end{aligned}$$

(44)

holds or not. We remark that

$$\begin{aligned} {\left| \sum _i{{\left( { {\alpha }}_i-{ {\beta }}_i\right) }i^2}\right| }\ge 2 \sum _i{\left| { {\alpha }}_i-{ {\beta }}_i\right| } \qquad \Longrightarrow \qquad {\left| \omega \cdot {\left( { {\alpha }}-{ {\beta }}\right) }\right| }\ge 1\,, \end{aligned}$$

(45)

indeed denoting $\omega _j = j^2 + \xi _j$ with ${\left| \xi _j\right| }\le \frac{1}{2}$,

$$\begin{aligned} {\left| \omega \cdot {\left( { {\alpha }}-{ {\beta }}\right) }\right| } \ge 2\sum _j{\left| { {\alpha }}_j - { {\beta }}_j\right| } - \frac{1}{2}\sum _j{\left| { {\alpha }}_j - { {\beta }}_j\right| }\ge 1. \end{aligned}$$

Of course if ${\left| \omega \cdot {\left( { {\alpha }}-{ {\beta }}\right) }\right| }\ge 1$, by (39) and (40) we get

$$\begin{aligned} \gamma \frac{e^{-{\sigma }{\left( \sum _i\langle i \rangle ^\theta ({ {\alpha }}_i+{ {\beta }}_i) -2\langle j \rangle ^\theta \right) }}}{{\left| \omega \cdot {{\left( { {\alpha }}- { {\beta }}\right) }}\right| }} \le 1\, \end{aligned}$$

and the bound (43) is trivially achieved.

Otherwise, to deal with the case in which (44) holds, we need some notation. Given $u\in \mathbb {Z}^\mathbb {Z}_f$, consider the set

$$\begin{aligned} M(u):={\left\{ j\ne 0 \,,\quad \text{ repeated }\quad {\left| u_j\right| } \;\text{ times }\right\} }\,, \end{aligned}$$

where $D(u)<\infty$ is its cardinality. Define the vector $m=m(u)$ as the reordering of the elements of the set above such that $|m_1|\ge |m_2|\ge \dots \ge |m_D|\ge 1.$

Given ${ {\alpha }}\ne { {\beta }}\in {\mathbb N}^\mathbb {Z}_f$ with $|{ {\alpha }}|+|{ {\beta }}|\ge 3$ we consider $m=m({ {\alpha }}-{ {\beta }})$ and $\widehat{n}=\widehat{n}({ {\alpha }}+{ {\beta }}).$ If we denote by D the cardinality of m and N the one of $\widehat{n}$ we have

$$\begin{aligned} D+{ {\alpha }}_0+{ {\beta }}_0\le N \end{aligned}$$

(46)

and

$$\begin{aligned} (|m_1|,\dots ,|m_D|,\underbrace{1,\;\dots \;,1}_{N-D\;\mathrm {times}} )\, \preceq \, {\left( \widehat{n}_1,\dots \widehat{n}_N\right) }\,. \end{aligned}$$

(47)

Example A.7

Let set $v={\alpha }+\beta$ and $u={\alpha }-\beta$ with

$$\begin{aligned} {\alpha }_{-5}=1, {\alpha }_{-2}=2, \alpha _0= 2,\alpha _4=1&\\ \beta _{-5}=1, \beta _{-3}=2, \beta _0=3,\beta _6=1&\\ \pi ({\alpha },\beta )=(-5)(1-1)+(-3)(-2)+(-2)(2)+4(1)+6(-1)=0&\\ v_{-5}=2,v_{-3}=2, v_{-2}=2, v_{0}=5, v_{4}=1, v_{6}=1&\\ u_{-5}=0,u_{-3}=-2, u_{-2}=2, u_{0}=-1, u_{4}=1, u_{6}=-1&\\ \widehat{n}(v)=(6, 5,5,4,3,3,2,2,1,1,1,1,1), N= 13(=\text {Card}({\hat{n}}))&\\ M(u)=\{-3, -3, -2, -2,4,6\}, m(u)=\{6,4, -3, -3, -2, -2\}, D(u)=6. \end{aligned}$$

Therefore, we have $D(u)+\alpha _0+\beta _0=8\le 13=N({\hat{n}}(v))$. Hence, (46) holds.

Furthermore, $(6,4, 3, 3, 2, 2, 1,1,1,1,1,1,1)\le {\hat{n}}(v)$, that is (47).

Lemma A.8

(Lemma C.3 of [BMP18]). Assume that g defined on $\mathbb {Z}$ is non negative, even and not decreasing on ${\mathbb N}.$ Then, if ${ {\alpha }}\ne { {\beta }}$,

$$\begin{aligned} \sum _{i\in \mathbb {Z}} g(i) |{ {\alpha }}_i-{ {\beta }}_i| \le 2g(m_1)+ \sum _{l\ge 3} g(\widehat{n}_l)\,. \end{aligned}$$

(48)

Proof

By definition of $m({\alpha }-{\beta })$ and setting ${\sigma }_l= \mathrm{sign}({ {\alpha }}_{m_l}-{ {\beta }}_{m_l})\,,$ we have

$$\begin{aligned} \sum _{i\in \mathbb {Z}} g(i) ({ {\alpha }}_i-{ {\beta }}_i)= & {} g(0)({ {\alpha }}_0-{ {\beta }}_0)+ \sum _{l\ge 1} {\sigma }_l g(m_l)\,. \end{aligned}$$

(49)

Hence

$$\begin{aligned} \sum _{i\in \mathbb {Z}} g(i) |{ {\alpha }}_i-{ {\beta }}_i|= & {} g(0)|{ {\alpha }}_0-{ {\beta }}_0|+ \sum _{l\ge 1} g(m_l) \\\le & {} g(1)({ {\alpha }}_0+{ {\beta }}_0)+ 2g(m_1)+ \sum _{l\ge 3} g(m_l) \end{aligned}$$

and (48) follows by (46) and (47). $\square$

By (49)

$$\begin{aligned} 0=\sum _{i\in \mathbb {Z}}{\left( { {\alpha }}_i-{ {\beta }}_i\right) }i = \sum _l {\sigma }_l m_l \end{aligned}$$

(50)

and

$$\begin{aligned} {\sum _i{{\left( { {\alpha }}_i-{ {\beta }}_i\right) }i^2}}= \sum _l {\sigma }_l m^2_l\,. \end{aligned}$$

(51)

Analogously

$$\begin{aligned} {\sum _i{{\left| { {\alpha }}_i-{ {\beta }}_i\right| }}} = D+|{ {\alpha }}_0-{ {\beta }}_0| {\mathop {\le }\limits ^{(47)}}N\,. \end{aligned}$$

(52)

Finally note that

$$\begin{aligned} \sigma _l\sigma _{l'} =-1\qquad \Longrightarrow \qquad m_l \ne m_{l'}\,. \end{aligned}$$

(53)

Lemma A.9

Given ${ {\alpha }}\ne { {\beta }}\in {\mathbb N}^\mathbb {Z}_f,$ such that $\pi ({ {\alpha }}-{ {\beta }})=0$, $N\ge 3,D\ge 1$ and satisfying (44), we have

$$\begin{aligned} {\left| m_1\right| }\le 7\sum _{l\ge 3}\widehat{n}_l^2\,. \end{aligned}$$

(54)

Proof

The case $D=1$ is not compatible with momentum conservation. Let us now consider the case $D=2$, i.e.

$$\begin{aligned} { {\alpha }}-{ {\beta }}={\sigma }_1 \mathbf{e}_{m_1}+{\sigma }_2 \mathbf{e}_{m_2} +({ {\alpha }}_0-{ {\beta }}_0)\mathbf{e}_0\,. \end{aligned}$$

If ${\sigma }_1{\sigma }_2=-1$, momentum conservation imposes $m_1=m_2$ but this contradicts (53). In the case ${\sigma }_1{\sigma }_2=1,$ by momentum conservation we have $m_1= -m_2$. Then conditions (44) and (52) imply that

$$\begin{aligned} m_1^2 + m_2^2 \le 2 (D +|{ {\alpha }}_0-{ {\beta }}_0|) {\mathop {\le }\limits ^{(53)}} 2 N \le 6(N-2) \le 6 \sum _{l=3}^N \widehat{n}_l^2 \end{aligned}$$

since $\widehat{n}_l\ge 1$.

Let us now consider the case $D \ge 3$. By (44), (51) and (52)

$$\begin{aligned} m_1^2 +{\sigma }_1{\sigma }_2 m_2^2\le & {} 2(D+|{ {\alpha }}_0-{ {\beta }}_0|) + \sum _{l=3}^Dm_l^2 \le 2 N + \sum _{l=3}^Dm_l^2 \le 2 N + \sum _{l=3}^N\widehat{n}_l^2 {\le } 7\sum _{l=3}^N \widehat{n}_l^2\,. \end{aligned}$$

since (recall $N\ge 3$) $2 N\le 6(N-2) \le 6\sum _{l=3}^N \widehat{n}_l^2$.

If $\sigma _1\sigma _2 = 1$ then

$$\begin{aligned} {\left| m_1\right| }, {\left| m_2\right| } \le \sqrt{7\sum _{l\ge 3} \widehat{n}_l^2}. \end{aligned}$$

If ${\sigma }_1{\sigma }_2 = -1$

$$\begin{aligned} (|m_1|+|m_2|)(|m_1|-|m_2|)= m_1^2 - m_2^2 \le 7\sum _{l\ge 3} \widehat{n}_l^2. \end{aligned}$$

Now, if ${\left| m_1\right| }\ne {\left| m_2\right| }$ then

$$\begin{aligned} {\left| m_1\right| } + {\left| m_2\right| } \le 7\sum _{l\ge 3} \widehat{n}_l^2. \end{aligned}$$

Conversely, if ${\left| m_1\right| } = {\left| m_2\right| }$, by (53), $m_1\ne m_2$, hence $m_1 = - m_2$. By substituting this relation into (50), we have

$$\begin{aligned} 2{\left| m_1\right| } \le \sum _{l\ge 3}{\left| m_l\right| } \le \sum _{l\ge 3}\widehat{n}_l^2\,, \end{aligned}$$

concluding the proof. $\square$

Lemma A.10

Consider ${ {\alpha }},{ {\beta }}\in {{\mathcal {M}}}$ with ${\alpha }\ne {\beta }$ and $|{\alpha }|+|{\beta }| \ge 3$. If (44) holds then

for all j such that ${ {\alpha }}_j+{ {\beta }}_j\ne 0$ one has

$$\begin{aligned} \sum _i{\left| { {\alpha }}_i-{ {\beta }}_i\right| }\langle i \rangle ^{\theta /2} \le C_* {\left( \sum _i {\left( { {\alpha }}_i+{ {\beta }}_i\right) }\langle i \rangle ^\theta - 2\langle j \rangle ^\theta \right) }\,, \qquad C_*= \frac{7}{2-2^\theta } \end{aligned}$$

(55)

Proof

Let us first consider the case $D=0$, this means that ${\alpha }-{\beta }= ({\alpha }_0-{\beta }_0)\mathbf{e}_0$ and the left hand side of (55) reads $|{\alpha }_0-{\beta }_0|$. By (39) and $N\ge 3$ the right hand side of (55) is at least $2-2^\theta$, so if $|{\alpha }_0-{\beta }_0| \le 7$ the result is trivial. Otherwise we have two cases, if $j=0$

$$\begin{aligned} |{\alpha }_0-{\beta }_0| \le 2(|{\alpha }_0-{\beta }_0| -2\langle j \rangle ^\theta ) \le 2 {\left( \sum _i {\left( { {\alpha }}_i+{ {\beta }}_i\right) }\langle i \rangle ^\theta - 2\langle j \rangle ^\theta \right) }\,, \end{aligned}$$

Otherwise we remark that if $j\ne 0$, ${\alpha }_j+{\beta }_j\ne 0$ and ${\alpha }_j-{\beta }_j=0$, then ${\alpha }_j+{\beta }_j \ge 2$, then

$$\begin{aligned} |{\alpha }_0-{\beta }_0| \le ({\alpha }_0+{\beta }_0) + ({\alpha }_j+{\beta }_j -2) \langle j \rangle ^\theta \le \sum _i {\left( { {\alpha }}_i+{ {\beta }}_i\right) }\langle i \rangle ^\theta - 2\langle j \rangle ^\theta \,. \end{aligned}$$

Now we consider indices ${\alpha },{\beta }$ such that $N\ge 3,D\ge 1$. Here we apply Lemma A.9 Given ${ {\alpha }},{ {\beta }}\in {\mathbb N}^\mathbb {Z}_f,$ as above we consider $m=m({ {\alpha }}-{ {\beta }})$ and $\widehat{n}=\widehat{n}({ {\alpha }}+{ {\beta }}).$

We have⁸

$$\begin{aligned}&\sum _i{\left| { {\alpha }}_i-{ {\beta }}_i\right| }\langle i \rangle ^{\theta /2} {\mathop {\le }\limits ^{(49)}}2{\left| m_1\right| }^{\frac{\theta }{2}} + \sum _{l\ge 3} \widehat{n}_l^{\frac{\theta }{2}} \nonumber \\&\qquad \qquad \qquad \qquad \qquad {\mathop {\le }\limits ^{(55)}}2{\left( 7 \sum _{l\ge 3} \widehat{n}_l^2 \right) }^{\frac{\theta }{2}} + \sum _{l\ge 3} \widehat{n}_l^{\frac{\theta }{2}}\nonumber \\&\qquad \qquad \qquad \qquad \qquad \le + 2(7)^{\frac{\theta }{2}}\sum _{l\ge 3}\widehat{n}_l^\theta + \sum _{l\ge 3} \widehat{n}_l^{\frac{\theta }{2}} \nonumber \\&\qquad \qquad \qquad \qquad \qquad \le \frac{2\sqrt{7}+1}{2-2^\theta }{\left( (2-2^\theta ){\left( \sum _{l\ge 3}\widehat{n}_l^\theta \right) }\right) }\,, \end{aligned}$$

(56)

Then by Lemma A.5 and (56) we get

$$\begin{aligned} \sum _i{\left| { {\alpha }}_i-{ {\beta }}_i\right| }\langle i \rangle ^{\theta /2}\le & {} \frac{7 }{2-2^\theta }{\left( \sum _i \langle i \rangle ^\theta {\left( { {\alpha }}_i +{ {\beta }}_i\right) } - 2\widehat{n}_1^\theta \right) }\\\le & {} \frac{7 }{2-2^\theta } {\left[ \sum _i \langle i \rangle ^\theta {\left( { {\alpha }}_i +{ {\beta }}_i\right) } - 2\langle j \rangle ^\theta \right] }\,, \end{aligned}$$

proving (55). $\square$

Conclusion of the proof of Lemma 3.8 By applying Lemma A.10, since $\omega \in {\mathtt {D}_\gamma }$ we get:

$$\begin{aligned}&\gamma \frac{e^{-{\sigma }{\left( \sum _i\langle i \rangle ^\theta ({ {\alpha }}_i+{ {\beta }}_i) -2\langle j \rangle ^\theta \right) }}}{{\left| \omega \cdot {{\left( { {\alpha }}- { {\beta }}\right) }}\right| }} {\mathop {\le }\limits ^{(4)}} e^{-{\sigma }{\left( \sum _i\langle i \rangle ^\theta ({ {\alpha }}_i+{ {\beta }}_i) -2\langle j \rangle ^\theta \right) }} \prod _i{\left( 1+({ {\alpha }}_i-{ {\beta }}_i)^{2}\langle i \rangle ^{{2}}\right) }\nonumber \\&{\mathop {\le }\limits ^{(56)}} e^{-\frac{{\sigma }}{C_* }\sum _i{\left| { {\alpha }}_i-{ {\beta }}_i\right| }\langle i \rangle ^{\frac{\theta }{2}}}\prod _i{\left( 1+({ {\alpha }}_i-{ {\beta }}_i)^{2}\langle i \rangle ^{{2}}\right) } \nonumber \\&\le \exp {\sum _i{\left[ -\frac{{\sigma }}{C_*} {\left| { {\alpha }}_i - { {\beta }}_i\right| }\langle i \rangle ^{\frac{\theta }{2}} + \ln {{\left( 1 + {\left( { {\alpha }}_i - { {\beta }}_i\right) }^{2}\langle i \rangle ^{{2}}\right) }}\right] }} \nonumber \\&= \exp {\sum _i f_i({\left| { {\alpha }}_i-{ {\beta }}_i\right| })} \end{aligned}$$

(57)

where, for $0<{\sigma }\le 1$, $i\in \mathbb {Z}$ and $x\ge 0$, we defined

$$\begin{aligned} f_i(x) := -\frac{{\sigma }}{C_*} x\langle i \rangle ^{\frac{\theta }{2}} + \ln {{\left( 1 + x^{2}\langle i \rangle ^{{2}}\right) }}\,. \end{aligned}$$

$\square$

Finally, we have

Lemma A.11

(Lemma 7.2 of [BMP18]) Setting

$$\begin{aligned} i_\sharp := \left( \frac{24C_*}{{\sigma }\theta } \ln \frac{12C_*}{{\sigma }\theta } \right) ^{\frac{2}{\theta }}\,, \end{aligned}$$

we get

$$\begin{aligned} \sum _i f_i(|\ell _i|)\le 18 i_\sharp \ln i_\sharp \end{aligned}$$

(58)

for every $\ell \in \mathbb {Z}^\mathbb {Z}_f$.

Proof

First of all we note that

$$\begin{aligned} \sum _i f_i(|\ell _i|)= \sum _{i\ \text {s.t.} \ \ell _i\ne 0} f_i(|\ell _i|) \end{aligned}$$

since $f_i(0)=0.$ We have that⁹

$$\begin{aligned} f_i(x) \le -\frac{{\sigma }}{C_* }\langle i \rangle ^{\frac{\theta }{2}}x + {2} \ln (x)+ 2\ln \langle i \rangle +1\,,\qquad \forall \, x\ge 1\,. \end{aligned}$$

Now,

$$\begin{aligned} \max _{x\ge 1} \left( -\frac{{\sigma }}{C_* }\langle i \rangle ^{\frac{\theta }{2}}x + 2 \ln (x)\right) = \left\{ \begin{array}{l} \displaystyle -\frac{{\sigma }}{C_* }\langle i \rangle ^{\frac{\theta }{2}} \qquad \qquad \qquad \ \,\qquad \text {if}\quad \langle i \rangle \ge i_0\,, \\ \\ \displaystyle -{2}+{2}\ln \frac{2C_* }{{\sigma }}-\theta \ln \langle i \rangle \qquad \text {if}\quad \langle i \rangle < i_0\,, \end{array} \right. \end{aligned}$$

where

$$\begin{aligned} i_0:=\left( \frac{2C_* }{{\sigma }}\right) ^{\frac{2}{\theta }}\,, \end{aligned}$$

since the maximum is achieved for $x=1$ if $\langle i \rangle \ge i_0$ and $x=\frac{2C_* }{{\sigma }\langle i \rangle ^{\theta /2}}$ if $\langle i \rangle < i_0$. Note that $i_0\ge e.$ Then we get

$$\begin{aligned}&\sum _i f_i(|\ell _i|) = \sum _{i\ \text {s.t.} \ \ell _i\ne 0} f_i(|\ell _i|) \\&\quad \le \sum _{\langle i \rangle \ge i_0\ \text {s.t.} \ \ell _i\ne 0} \left( 2\ln \langle i \rangle +1 -\frac{{\sigma }}{C_* }\langle i \rangle ^{\frac{\theta }{2}} \right) + \sum _{\langle i \rangle < i_0} \left( 2\ln \frac{2C_* }{{\sigma }}+\Big (2 -\theta \Big ) \ln \langle i \rangle \right) \,. \end{aligned}$$

We immediately have that

$$\begin{aligned} \sum _{\langle i \rangle < i_0} \left( 2\ln \frac{2C_* }{{\sigma }}+\Big ({2} -\theta \Big ) \ln \langle i \rangle \right)\le & {} 6 i_0 \left( \ln \frac{2C_* }{{\sigma }}+ \ln i_0 \right) \\= & {} 6\left( 1+ \frac{2}{\theta }\right) \left( \frac{2C_* }{{\sigma }}\right) ^{\frac{2}{\theta }} \ln \frac{2C_* }{{\sigma }}\,. \end{aligned}$$

Moreover, in the case $\langle i \rangle \ge i_0\ge e,$

$$\begin{aligned} 2\ln \langle i \rangle +1 -\frac{{\sigma }}{C_* }\langle i \rangle ^{\frac{\theta }{2}} \le 3\ln \langle i \rangle -\frac{{\sigma }}{C_* }\langle i \rangle ^{\frac{\theta }{2}} =\frac{6}{\theta }\Big ( \ln \langle i \rangle ^{\frac{\theta }{2}}-{2\mathfrak {C} } \langle i \rangle ^{\frac{\theta }{2}} \Big ) \end{aligned}$$

where

$$\begin{aligned} {\mathfrak C }:=\frac{\theta {\sigma }(2-2^\theta )}{84 }<1 \,. \end{aligned}$$

We have that¹⁰

$$\begin{aligned} \ln \langle i \rangle ^{\frac{\theta }{2}}-2{\mathfrak C} \langle i \rangle ^{\frac{\theta }{2}} \le -{\mathfrak C} \langle i \rangle ^{\frac{\theta }{2}}\,, \qquad \text {when}\qquad \langle i \rangle \ge i_*:= \left( \frac{2}{{\mathfrak C}}\ln \frac{1}{{\mathfrak C}}\right) ^{\frac{2}{\theta }} \,. \end{aligned}$$

Note that

$$\begin{aligned} i_\sharp \ge \max \{ i_0, i_*\}\,. \end{aligned}$$

Therefore

$$\begin{aligned}&\sum _{\langle i \rangle \ge i_0\ \text {s.t.} \ \ell _i\ne 0}{\left( 2\ln \langle i \rangle +1 -\frac{{\sigma }}{C_* }\langle i \rangle ^{\frac{\theta }{2}}\right) } \le \sum _{\langle i \rangle \ge i_0\ \text {s.t.} \ \ell _i\ne 0} \frac{6}{\theta }{\left( \ln \langle i \rangle ^{\frac{\theta }{2}}-{2\mathfrak C } \langle i \rangle ^{\frac{\theta }{2}}\right) } \\&\qquad \quad \quad \le \frac{6}{\theta } \left( \sum _{\langle i \rangle <i_\sharp } \ln \langle i \rangle ^{\frac{\theta }{2}} - \sum _{\langle i \rangle \ge i_\sharp \ \text {s.t.} \ \ell _i\ne 0} \Big ( {\mathfrak C} \langle i \rangle ^{\frac{\theta }{2}} \Big ) \right) \le 9 i_\sharp \ln i_\sharp \,. \end{aligned}$$

In conclusion we get

$$\begin{aligned} \sum _i f_i(|\ell _i|)\le & {} 6\frac{2+\theta }{\theta }\left( \frac{2C_* }{{\sigma }}\right) ^{\frac{2}{\theta }} \ln \frac{2C_* }{{\sigma }} + 9 i_\sharp \ln i_\sharp \\\le & {} 9 \left( \frac{2C_* }{{\sigma }\theta }\right) ^{\frac{2}{\theta }} \ln {\left( \frac{2C_* }{{\sigma }}\right) }^{\frac{\theta }{2}} + 9 i_\sharp \ln i_\sharp \le 18 i_\sharp \ln i_\sharp \end{aligned}$$

$\square$

The inequality (27) follows from plugging (58) into (57) and evaluating the constant. $\square$