4.5 Appendices
4.1.1 Appendix 1. The Proof of (4.2.28)
Here, we estimate the conjugate \(\overline{C_{1}(j^{^{\prime }},\lambda _{j,\beta })}\) of \(C_{1}(j^{^{\prime }},\lambda _{j,\beta }),\) namely, we prove that
$$\begin{aligned} \sum _{(j_{1},\beta _{1})\in Q(\rho ^{\alpha },9r)}\dfrac{\overline{A(j^{^{\prime }},\beta ,j^{^{\prime }}+j_{1,}\beta +\beta _{1}})\overline{A(j^{^{\prime }} +j_{1,}\beta +\beta _{1},j,\beta })}{\lambda _{j,\beta }-\lambda _{j^{^{\prime } }+j_{1},\beta +\beta _{1}}}=O(\rho ^{-2a}r^{2}), \end{aligned}$$
(4.5.1)
[see (4.2.26)], where
$$ Q(\rho ^{\alpha },9r)=\{(j_{1},\beta _{1}):\mid j_{1}\delta \mid<9r,0<\mid \beta _{1}\mid<\rho ^{\alpha }\},j\in S_{1}(\rho ),\mid j^{^{\prime }}\delta \mid <r,r=O(\rho ^{\frac{1}{2}\alpha _{2}}). $$
The conditions on indices \(j^{^{\prime }}\), \(j_{1},\) j and (4.2.20) imply that
$$ \mu _{j^{^{\prime }}+j_{1}}=O(r^{2}),\mu _{j}=O(r^{2}). $$
These with \(\beta \notin V_{\beta _{1}}^{\delta }(\rho ^{a})))\), where \(\beta _{1}\in \Gamma _{\delta }(p\rho ^{\alpha }),\) [see (4.2.9)] give
$$\begin{aligned} \lambda _{j,\beta }-\lambda _{j^{^{\prime }}+j_{1},\beta +\beta _{1}}=-2\langle \beta ,\beta _{1}\rangle +O(r^{2}), \mid \langle \beta ,\beta _{1}\rangle \mid \,>\frac{1}{3}\rho ^{a}. \end{aligned}$$
(4.5.2)
Using this, (4.2.15), and (4.5.1), we get
$$\begin{aligned} \overline{C_{1}(j^{^{\prime }},\lambda _{j,\beta })}=\sum _{\beta _{1}} \dfrac{C^{^{\prime }}}{-2\langle \beta ,\beta _{1}\rangle }+O(\rho ^{-2a} r^{2}), \end{aligned}$$
(4.5.3)
where
$$ C^{^{\prime }}=\sum _{j_{1}}\overline{A(j^{^{\prime }},\beta ,j^{^{\prime }} +j_{1,}\beta +\beta _{1}})\overline{A(j^{^{\prime }}+j_{1,}\beta +\beta _{1},j,\beta }). $$
In Chap. 3, we proved that [see (3.3.21), (3.3.7), Lemma 3.3.3]
$$\begin{aligned} \overline{A(j^{^{\prime }},\beta ,j^{^{\prime }}+j_{1,}\beta +\beta _{1}} )=\sum \limits _{n_{1}:(n_{1},\beta _{1})\in \Gamma ^{^{\prime }}(\rho ^{\alpha } )}c(n_{1},\beta _{1})a(n_{1},\beta _{1},j^{^{\prime }},\beta ,j^{^{\prime }} +j_{1,}\beta +\beta _{1}), \end{aligned}$$
(4.5.4)
$$ \overline{A(j^{^{\prime }}+j_{1,}\beta +\beta _{1},j,\beta })=\sum \limits _{n_{2} :(n_{2},-\beta _{1})\in \Gamma ^{^{\prime }}(\rho ^{\alpha })}c(n_{2},-\beta _{1})a(n_{2},-\beta _{1},j^{^{\prime }}+j_{1,}\beta +\beta _{1},j,\beta ), $$
$$ \Gamma ^{^{\prime }}(\rho ^{\alpha })=\{(n_{1},\beta _{1}):\beta _{1}\in \Gamma _{\delta }\backslash 0,n_{1}\in \mathbb {Z},\beta _{1}+(n_{1}-(2\pi )^{-1}\langle \beta _{1},\delta ^{*}\rangle )\delta \in \Gamma (\rho ^{\alpha })\}, $$
$$\begin{aligned} c(n_{1},\beta _{1})=q_{\gamma _{1}},\gamma _{1}=\beta _{1}+(n_{1}-(2\pi )^{-1}\langle \beta _{1},\delta ^{*}\rangle )\delta \in \Gamma (\rho ^{\alpha }), \end{aligned}$$
(4.5.5)
$$ a(n_{1},\beta _{1},j^{^{\prime }},\beta ,j^{^{\prime }}+j_{1,}\beta +\beta _{1})=(\mathrm{e}^{i(n_{1}-(2\pi )^{-1}\langle \beta _{1},\delta ^{*}\rangle )\zeta }\varphi _{j^{^{\prime }},v(\beta )}(\zeta ),\varphi _{j^{^{\prime }}+j_{1} ,v(\beta +\beta _{1})}(\zeta )), $$
$$ a(n_{2},-\beta _{1},j^{^{\prime }}+j_{1,}\beta +\beta _{1},j,\beta )=(\mathrm{e}^{i(n_{2} -(2\pi )^{-1}\langle -\beta _{1},\delta ^{*}\rangle )\zeta }\varphi _{j^{^{\prime }}+j_{1},v(\beta +\beta _{1})}(\zeta ),\varphi _{j,v(\beta )}(\zeta )) $$
$$\begin{aligned} =(\varphi _{j^{^{\prime }}+j_{1},v(\beta +\beta _{1})}(\zeta ),\mathrm{e}^{-i(n_{2} -(2\pi )^{-1}\langle -\beta _{1},\delta ^{*}\rangle )\zeta }\varphi _{j,v(\beta )}(\zeta )) \end{aligned}$$
(4.5.6)
$$ =\overline{(\mathrm{e}^{-i(n_{2}-(2\pi )^{-1}\langle -\beta _{1},\delta ^{*}\rangle \zeta }\varphi _{j,v(\beta )}(\zeta ),\varphi _{j^{^{\prime }}+j_{1},v(\beta +\beta _{1})}(\zeta )),} $$
where \(\delta ^{*}\) is the element of \(\Omega \) satisfying \(\langle \delta ^{*},\delta \rangle =2\pi .\)
Now, to estimate the right-hand side of (4.5.3) we prove that
$$\begin{aligned} \sum _{j_{1}}a(n_{1},\beta _{1},j^{^{\prime }},\beta ,j^{^{\prime }}+j_{1,} \beta +\beta _{1})a(n_{2},-\beta _{1},j^{^{\prime }}+j_{1},\beta +\beta _{1} ,j,\beta ) \end{aligned}$$
(4.5.7)
$$ =a(n_{1}+n_{2},0,j^{^{\prime }},\beta ,j,\beta )+O(\rho ^{-p\alpha }). $$
By definition, we have
$$ a(n_{1}+n_{2},0,j^{^{\prime }},\beta ,j,\beta )=(\mathrm{e}^{i(n_{1}+n_{2})\zeta } \varphi _{j^{^{\prime }},v(\beta )}(\zeta ),\varphi _{j,v(\beta )}(\zeta ))= $$
$$ (\mathrm{e}^{i(n_{1}-(2\pi )^{-1}\langle \beta _{1},\delta ^{*}\rangle )\zeta } \varphi _{j^{^{\prime }},v(\beta )}(\zeta ),\mathrm{e}^{-i(n_{2}-(2\pi )^{-1}\langle -\beta _{1},\delta ^{*}\rangle )\zeta }\varphi _{j,v(\beta )}(\zeta )). $$
This, (4.5.6), and the following formulas
$$ \mathrm{e}^{i(n_{1}-(2\pi )^{-1}\langle \beta _{1},\delta ^{*}\rangle )\zeta } \varphi _{j^{^{\prime }},v(\beta )}(\zeta )= $$
$$ \sum _{\mid j_{1}\delta \mid <9r}a(n_{1},\beta _{1},j^{^{\prime }},\beta ,j^{^{\prime }}+j_{1,}\beta +\beta _{1})\varphi _{j^{^{\prime }}+j_{1} ,v(\beta +\beta _{1})}(\zeta )+O(\rho ^{-p\alpha }), $$
$$ \mathrm{e}^{-i(n_{2}-(2\pi )^{-1}\langle -\beta _{1},\delta ^{*}\rangle )\zeta } \varphi _{j,v(\beta )}(\zeta )= $$
$$ \sum _{\mid j_{1}\delta \mid <9r}\overline{a(n_{2},-\beta _{1},j^{^{\prime }} ,\beta ,j^{^{\prime }}+j_{1,}\beta +\beta _{1})}\varphi _{j^{^{\prime }} +j_{1},v(\beta +\beta _{1})}+O(\rho ^{-p\alpha }) $$
and
$$\begin{aligned} \sum \limits _{j_{_{1}}}\mid a(n_{1},\beta _{1},j^{^{\prime }},\beta ,j^{^{\prime } }+j_{1,}\beta +\beta _{1})\mid =O(1) \end{aligned}$$
(4.5.8)
[see (3.3.16), (3.3.17) of Chap. 3] give the proof of (4.5.7). Now from (4.5.7), (4.5.4), and (4.5.3), we obtain
$$ C^{^{\prime }}=\sum \limits _{n_{1}} \sum \limits _{n_{2}}(c(n_{1},\beta _{1})c(n_{2},-\beta _{1})a(n_{1}+n_{2},0,j^{^{\prime }},\beta ,j,\beta )+O(\rho ^{-p\alpha })), $$
$$ \overline{C_{1}(j^{^{\prime }},\lambda _{j,\beta })}=\sum _{\beta _{1}} \sum \limits _{n_{1}} \sum \limits _{n_{2}}C_{1}^{^{\prime }}(\beta _{1},n_{1},n_{2})+O(\rho ^{-2a}r^{2}), $$
where
$$ C_{1}^{^{\prime }}(\beta _{1},n_{1},n_{2})=\dfrac{c(n_{1},\beta _{1} )c(n_{2},-\beta _{1})a(n_{1}+n_{2},0,j^{^{\prime }},\beta ,j,\beta )}{-2\langle \beta ,\beta _{1}\rangle }. $$
It is clear that
$$\begin{aligned} C_{1}^{^{\prime }}(\beta _{1},n_{1},n_{2})+C_{1}^{^{\prime }}(-\beta _{1} ,n_{2},n_{1})=0. \end{aligned}$$
(4.5.9)
Therefore
$$ \overline{C_{1}(j^{^{\prime }},\lambda _{j,\beta })}=O(\rho ^{-2a}r^{2}). $$
4.1.2 Appendix 2. The Proof of (4.2.35)
Arguing as in the proof of (4.2.27), we see that
$$ C_{2}(\Lambda _{j,\beta })=C_{2}(\lambda _{j,\beta })+O(\rho ^{-3a}) $$
and by (4.5.4)
$$\begin{aligned} \overline{C_{2}(\lambda _{j,\beta })}&=\sum _{\beta _{1},\beta _{2}} \Bigg (\sum \limits _{n_{1},n_{2},n_{3}}\Bigg (\sum _{j_{1},j_{2}}\dfrac{c(n_{1},\beta _{1})c(n_{2},\beta _{2})c(n_{3},-\beta _{1}-\beta _{2})}{(\lambda _{j,\beta }-\lambda _{j(1),\beta (1)})(\lambda _{j,\beta }-\lambda _{j(2)_,\beta (2)} )}a(n_{1},\beta _{1},j,\beta ,j(1)_,\beta (1))\times \\&a(n_{2},\beta _{2},j(1)_,\beta (1),j(2)_,\beta (2))a(n_{3},-\beta _{1}-\beta _{2},j(2),\beta (2),j,\beta )\Bigg )\Bigg ), \end{aligned}$$
where
$$ (j_{1},\beta _{1})\in Q(\rho ^{\alpha },9r_{1}),(j_{2},\beta _{2})\in Q(\rho ^{\alpha },90r_{1}),j\in S_{1},\beta _{1}+\beta _{2}\ne 0. $$
Applying (4.5.7) two times and using (4.5.8), we get
$$\begin{aligned} \sum _{j_{1}}a(n_{1},\beta _{1},j,\beta ,j(1)_,\beta (1))&\left( \sum _{j_{2}} a(n_{2},\beta _{2},j(1)_,\beta (1),j(2)_,\beta (2))a(n_{3},-\beta _{1}\right. \\&\left. \quad -\beta _{2},j(2)_,\beta (2),j,\beta )\right) \end{aligned}$$
$$ =\sum _{j_{1}}a(n_{1},\beta _{1},j,\beta ,j(1)_,\beta (1))(a(n_{2}+n_{3} ,-\beta _{1},j(1)_,\beta (1),j,\beta )+O(\rho ^{-p\alpha })) $$
$$ \quad =a(n_{1}+n_{2}+n_{3},0,j,\beta ,j,\beta )+O(\rho ^{-p\alpha }). $$
Using this in the above expression for \(C_{2}(\lambda _{j,\beta })\) and taking into account that
$$ \lambda _{j,\beta }-\lambda _{j(1)_,\beta (1)}=-2\langle \beta ,\beta _{1} \rangle +O(\rho ^{2\alpha _{1}}),\mid \langle \beta ,\beta _{1}\rangle \mid \,>\frac{1}{3}\rho ^{a} $$
and
$$ \lambda _{j,\beta }-\lambda _{j(2)_,\beta (2)}=-2\langle \beta ,\beta _{1} +\beta _{2}\rangle +O(\rho ^{2\alpha _{1}}),\mid \langle \beta ,\beta _{1}+\beta _{2}\rangle \mid \,>\frac{1}{3}\rho ^{a} $$
which can be proved as (4.5.2), we have
$$ C_{2}(\lambda _{j,\beta })=O(\rho ^{-3a+2\alpha _{1}})+ $$
$$ \sum _{\beta _{1},\beta _{2}} \sum \limits _{n_{1},n_{2},n_{3}}\frac{c(n_{1},\beta _{1})c(n_{2},\beta _{2})c(n_{3},-\beta _{1}-\beta _{2} )a(n_{1}+n_{2}+n_{3},0,j,\beta ,j,\beta )}{4\langle \beta ,\beta _{1}\rangle \langle \beta ,\beta _{1}+\beta _{2}\rangle }. $$
Grouping the terms with the equal multiplicands
$$\begin{aligned}&c(n_{1},\beta _{1})c(n_{2},\beta _{2})c(n_{3},-\beta _{1}-\beta _{2}),\\&c(n_{2},\beta _{2})c(n_{1},\beta _{1})c(n_{3},-\beta _{1}-\beta _{2}),\\&c(n_{1},\beta _{1})c(n_{3},-\beta _{1}-\beta _{2})c(n_{2},\beta _{2}),\\&c(n_{2},\beta _{2})c(n_{3},-\beta _{1}-\beta _{2})c(n_{1},\beta _{1}),\\&c(n_{3},-\beta _{1}-\beta _{2})c(n_{1},\beta _{1})c(n_{2},\beta _{2}),\text { }\\&c(n_{3},-\beta _{1}-\beta _{2})c(n_{2},\beta _{2})c(n_{1},\beta _{1}) \end{aligned}$$
and using the obvious equality
$$ \frac{1}{\langle \beta ,\beta _{1}\rangle \langle \beta ,\beta _{1}+\beta _{2}\rangle }+\frac{1}{\langle \beta ,\beta _{2}\rangle \langle \beta ,\beta _{2}+\beta _{1}\rangle }+\frac{1}{\langle \beta ,\beta _{1}\rangle \langle \beta ,-\beta _{2}\rangle }\,\,+ $$
$$ \frac{1}{\langle \beta ,\beta _{2}\rangle \langle \beta -,\beta _{1}\rangle }+\frac{1}{\langle \beta ,-\beta _{1}-\beta _{2}\rangle \langle \beta ,-\beta _{2}\rangle }+\frac{1}{\langle \beta ,-\beta _{1}-\beta _{2}\rangle \langle \beta ,-\beta _{1}\rangle }=0, $$
we see that
$$ C_{2}(\lambda _{j,\beta })=O(\rho ^{-3a+2\alpha _{1}}). $$
4.1.3 Appendix 3. The Proof of (4.2.34)
By (4.2.27) we have
$$ C_{1}(\Lambda _{j,\beta })=C_{1}(\lambda _{j,\beta })+O(\rho ^{-3a}). $$
Therefore, we need to prove that
$$ \overline{C_{1}(\lambda _{j,\beta })}=\frac{1}{4}\int _{F}\left| f_{\delta ,\beta +\tau }(x)\right| ^{2}\left| \varphi _{j,v}^{\delta }(\langle \delta ,x\rangle )\right| ^{2}\text {dx}+O(\rho ^{-3a+2\alpha _{1}}), $$
where
$$ \overline{C_{1}(\lambda _{j,\beta })}\equiv \sum _{\beta _{1}}\sum _{j_{1} }\dfrac{\overline{A(j,\beta ,j+j_{1,}\beta +\beta _{1}})\overline{A(j+j_{1,} \beta +\beta _{1},j,\beta })}{\lambda _{j,\beta }-\lambda _{j+j_{1},\beta +\beta _{1} }}, $$
$$ (j_{1},\beta _{1})\in Q(\rho ^{\alpha },9r_{1}),j\in S_{1}, $$
and by (4.5.4)
$$ \overline{C_{1}(\lambda _{j,\beta })}=\sum _{\beta _{1}} \sum \limits _{n_{1}:(n_{1},\beta _{1})\in \Gamma ^{^{\prime }}(\rho ^{\alpha })}\text { }\sum \limits _{n_{2}:(n_{2},-\beta _{1})\in \Gamma ^{^{\prime }}(\rho ^{\alpha } )} \sum _{j_{1}}\frac{c(n_{1},\beta _{1})c(n_{2},-\beta _{1})}{\lambda _{j,\beta }-\lambda _{j+j_{1},\beta +\beta _{1}}}\times $$
$$ a(n_{1},\beta _{1},j,\beta ,j+j_{1,}\beta +\beta _{1})a(n_{2},-\beta _{1} ,j+j_{1},\beta +\beta _{1},j,\beta ). $$
Replacing \(\lambda _{j,\beta }-\lambda _{j+j_{1},\beta +\beta _{1}}\) by
$$ -(2\langle \beta +\tau ,\beta _{1}\rangle +\mid \beta _{1}\mid ^{2}+\mu _{j+j_{1} }(v(\beta +\beta _{1}))-\mu _{j}(v(\beta ))) $$
and using (4.5.7) for \(j^{^{\prime }}=j,\) we have
$$ \overline{C_{1}(j,\lambda _{j,\beta })}=\sum _{\beta _{1}} \sum \limits _{n_{1}} \sum \limits _{n_{2}}\frac{c(n_{1},\beta _{1} )c(n_{2},-\beta _{1})a(n_{1}+n_{2},0,j,\beta ,j,\beta )}{-2\langle \beta +\tau ,\beta _{1}\rangle }+ $$
$$ \sum _{\beta _{1}} \sum \limits _{n_{1}} \sum \limits _{n_{2}}\text { }\sum _{j_{1}}\frac{c(n_{1},\beta _{1})c(n_{2},-\beta _{1})a(n_{1},\beta _{1},j,\beta ,j+j_{1,}\beta +\beta _{1})}{2\langle \beta +\tau ,\beta _{1} \rangle (2\langle \beta +\tau ,\beta _{1}\rangle +\mid \beta _{1}\mid ^{2}+\mu _{j+j_{1}}-\mu _{j})}\times $$
$$ a(n_{2},-\beta _{1},j+j_{1},\beta +\beta _{1},j,\beta )(\mid \beta _{1}\mid ^{2} +\mu _{j+j_{1}}(v(\beta +\beta _{1}))-\mu _{j}(v(\beta ))). $$
The formula (4.5.9) shows that the first summation of the right-hand side of this equality is zero. Thus we need to estimate the second sum. For this, we use the following relation
$$ \mu _{j+j_{1}}(v(\beta +\beta _{1}))a(n_{1},\beta _{1},j,\beta ,j+j_{1},\beta +\beta _{1})=(\mathrm{e}^{i(n_{1}-\frac{\langle \beta _{1},\delta ^{*}\rangle }{2\pi })\zeta }\varphi _{j,v(\beta )}(\zeta ),T_{v}\varphi _{j+j_{1},v(\beta +\beta _{1} )}(\zeta )) $$
$$ =(T_{v}(\mathrm{e}^{i(n_{1}-(2\pi )^{-1}(\beta _{1},\delta ^{*}\rangle )\zeta } \varphi _{j,v(\beta )}(\zeta )),\varphi _{j+j_{1},v(\beta +\beta _{1})}(\zeta )) $$
$$ =(\mid n_{1}-(2\pi )^{-1}\langle \beta _{1},\delta ^{*}\rangle \mid ^{2} \mid \delta \mid ^{2}+\mu _{j}(v))(\mathrm{e}^{i(n_{1}-(2\pi )^{-1}\langle \beta _{1} ,\delta ^{*}\rangle )\zeta }\varphi _{j,v(\beta )}(\zeta ),\varphi _{j+j_{1} ,v(\beta +\beta _{1})}(\zeta )) $$
$$ -2i(n_{1}-(2\pi )^{-1}\langle \beta _{1},\delta ^{*}\rangle )\mid \delta \mid ^{2}(\mathrm{e}^{i(n_{1}-(2\pi )^{-1}\langle \beta _{1},\delta ^{*}\rangle )\zeta } \varphi _{j,v(\beta )}^{^{\prime }}(\zeta )),\varphi _{j+j_{1},v(\beta +\beta _{1} )}(\zeta )). $$
Using this, (4.5.7), and the formula
$$ \sum _{j_{1}}((\mathrm{e}^{i(n_{1}-(2\pi )^{-1}\langle \beta _{1},\delta ^{*}\rangle )\zeta }\varphi _{j,v(\beta )}^{^{\prime }}(\zeta )),\varphi _{j+j_{1},v(\beta +\beta _{1})}(\zeta ))a(n_{2},-\beta _{1},j+j_{1},\beta +\beta _{1},j,\beta ) $$
$$ =((\mathrm{e}^{i(n_{1}+n_{2})\zeta }\varphi _{j,v(\beta )}^{^{\prime }}(\zeta )),\varphi _{j,v(\beta )}(\zeta ))+O(\rho ^{-p\alpha }) $$
which can be proved as (4.5.7), we obtain
$$\begin{aligned} \sum _{j_{1}}\mu _{j+j_{1}}(v(\beta +\beta _{1}))a(n_{1},\beta _{1},j,\beta ,j+j_{1},\beta +\beta _{1})a(n_{2},-\beta _{1},j+j_{1},\beta +\beta _{1} ,j,\beta )=\nonumber \end{aligned}$$
$$\begin{aligned} (\mid n_{1}-(2\pi )^{-1}\langle \beta _{1},\delta ^{*}\rangle \mid ^{2} )\mid \delta \mid ^{2}+\mu _{j}(v))a(n_{1}+n_{2},0,j,\beta ,j,\beta )- \end{aligned}$$
(4.5.10)
$$ 2i(n_{1}-(2\pi )^{-1}\langle \beta _{1},\delta ^{*}\rangle )\mid \delta \mid ^{2}((\mathrm{e}^{i(n_{1}+n_{2})\zeta }\varphi _{j,v(\beta )}^{^{\prime }}(\zeta )),\varphi _{j,v(\beta )}(\zeta )). $$
Here, the last multiplicand can be estimated as follows:
$$ \ \ \mu _{j}(v)(\varphi _{j,v(\beta )}(\zeta ),\mathrm{e}^{i(n_{1}+n_{2})\zeta } \varphi _{j,v(\beta )}(\zeta ))=(\varphi _{j,v(\beta )}(\zeta ),T_{v}(\mathrm{e}^{i(n_{1} +n_{2})\zeta }\varphi _{j,v(\beta )}(\zeta ))) $$
$$ =(n_{1}+n_{2})^{2}\mid \delta \mid ^{2}(\varphi _{j,v(\beta )}(\zeta ),\mathrm{e}^{i(n_{1} +n_{2})\zeta }\varphi _{j,v(\beta )}(\zeta ))+ $$
$$ 2i(n_{1}+n_{2})\mid \delta \mid ^{2}(\varphi _{j,v(\beta )}(\zeta ),\mathrm{e}^{i(n_{1} +n_{2})\zeta }\varphi _{j,v(\beta )}^{^{\prime }}(\zeta ))+\mu _{j}(v)(\varphi _{j,v(\beta )},\mathrm{e}^{i(n_{1}+n_{2})\zeta }\varphi _{j,v(\beta )}), $$
$$ ((\mathrm{e}^{i(n_{1}+n_{2})\zeta }\varphi _{j,v(\beta )}^{^{\prime }}(\zeta )),\varphi _{j,v(\beta )}(\zeta ))=\frac{n_{1}+n_{2}}{2i}((\mathrm{e}^{i(n_{1}+n_{2})\zeta } \varphi _{j,v(\beta )}(\zeta )),\varphi _{j,v(\beta )}(\zeta )). $$
Using this, (4.5.10), and (4.5.7), we get
$$ \sum _{j_{1}}(a(n_{1},\beta _{1},j,\beta ,j+j_{1,}\beta +\beta _{1})a(n_{2} ,-\beta _{1},j+j_{1},\beta +\beta _{1},j,\beta ))\times $$
$$ (\mid \beta _{1}\mid ^{2}+\mu _{j+j_{1}}(v(\beta +\beta _{1}))-\mu _{j} (v(\beta )))=a(n_{1}+n_{2},0,j,\beta ,j,\beta )\times $$
$$ (\mid \beta _{1}\mid ^{2}+\mid n_{1}-\frac{\langle \beta _{1},\delta ^{*}\rangle }{2\pi }\mid ^{2}\mid \delta \mid ^{2}-(n_{1}-\frac{\langle \beta _{1},\delta ^{*}\rangle }{2\pi })\mid \delta \mid ^{2}(n_{1}+n_{2})) $$
$$ =\left( \mid \beta _{1}\mid ^{2}+\mid \delta \mid ^{2}\left( n_{1}-\frac{\langle \beta _{1} ,\delta ^{*}\rangle }{2\pi }\right) \left( -n_{2}-\frac{\langle \beta _{1},\delta ^{*}\rangle }{2\pi }\right) \right) a(n_{1}+n_{2},0,j,\beta ,j,\beta ). $$
Thus
$$ \overline{C_{1}(j,\lambda _{j,\beta })}=C+O(\rho ^{-3a+2\alpha _{1}}), $$
where
$$\begin{aligned} C=\sum _{\beta _{1},n_{1},n_{2}}\frac{c(n_{1},\beta _{1})c(n_{2},-\beta _{1})a(n_{1}+n_{2},0,j,\beta ,j,\beta )}{4\mid \langle \beta +\tau ,\beta _{1} \rangle \mid ^{2}}\times \end{aligned}$$
(4.5.11)
$$ \left( \mid \beta _{1}\mid ^{2}+\left( n_{1}-\frac{\langle \beta _{1},\delta ^{*}\rangle }{2\pi }\right) \left( -n_{2}-\frac{\langle \beta _{1},\delta ^{*}\rangle }{2\pi }\right) \mid \delta \mid ^{2}\right) . $$
Now we consider
$$ \int _{F}\left| f_{\delta ,\beta +\tau }(x)\right| ^{2}\left| \varphi _{n,v}(\langle \delta ,x\rangle )\right| ^{2}\text {dx}, $$
where \(f_{\delta ,\beta +\tau }\) is defined in (4.1.12). By (4.5.5)
$$ f_{\delta ,\beta +\tau }(x)=\sum _{(n_{1},\beta _{1})\in \Gamma _{\delta }^{^{\prime } }(\rho ^{\alpha })}\frac{\beta _{1}+(n_{1}-\frac{\langle \beta _{1},\delta ^{*}\rangle }{2\pi })\delta }{\langle \beta +\tau ,\beta _{1}\rangle }c(n_{1},\beta _{1})\mathrm{e}^{i\langle \beta _{1}+(n_{1}-\frac{\langle \beta _{1},\delta ^{*}\rangle }{2\pi })\delta ,x\rangle }. $$
Here \(f_{\delta ,\beta +\tau }(x)\) is a vector of \(\mathbb {R}^{d}\). Using \(\langle \beta ,\delta \rangle =0\) for \(\beta \in \Gamma _{\delta },\) we obtain
$$ \left| f_{\delta ,\beta +\tau }(x)\right| ^{2}=\sum _{(n_{1},\beta _{1}),(n_{2},\beta _{2})\in \Gamma _{\delta }^{^{\prime }}(\rho ^{\alpha })} \frac{\langle \beta _{1},\beta _{2}\rangle +(n_{1}-\frac{\langle \beta _{1} ,\delta ^{*}\rangle }{2\pi })(n_{2}-\frac{\langle \beta _{1},\delta ^{*}\rangle }{2\pi })\mid \delta \mid ^{2}}{\langle \beta +\tau ,\beta _{1}\rangle \langle \beta +\tau ,\beta _{2}\rangle }\times $$
$$ c(n_{1},\beta _{1})c(-n_{2},-\beta _{2})\mathrm{e}^{i\langle \beta _{1}-\beta _{2} +(n_{1}-n_{2}-(2\pi )^{-1}\langle \beta _{1}-\beta _{2},\delta ^{*} \rangle )\delta ,x\rangle }. $$
Since \(\varphi _{j,v}(\langle \delta ,x\rangle )\) is a function of \(\langle \delta ,x\rangle ,\) we have
$$ \int _{F}\mathrm{e}^{i\langle \beta _{1}-\beta _{2}+(n_{1}-n_{2}-(2\pi )^{-1}\langle \beta _{1}-\beta _{2},\delta ^{*}\rangle )\delta ,x\rangle }\left| \varphi _{j,v}(\langle \delta ,x\rangle )\right| ^{2}\text {dx}\,=0 $$
for \(\beta _{1}\ne \beta _{2}\). Therefore
$$ \int _{F}\left| f_{\delta ,\beta +\tau }(x)\right| ^{2}\left| \varphi _{j,v}(\langle \delta ,x\rangle )\right| ^{2}\text {dx}\,=\sum _{\beta _{1} ,n_{1},n_{2}}\frac{c(n_{1},\beta _{1})c(-n_{2},-\beta _{1})}{\mid \langle \beta +\tau ,\beta _{1}\rangle \mid ^{2}}\times $$
$$ (\mid \beta _{1}\mid ^{2}+\left( n_{1}-\frac{\langle \beta _{1},\delta ^{*}\rangle }{2\pi }\right) \left( n_{2}-\frac{\langle \beta _{1},\delta ^{*}\rangle }{2\pi }\right) \mid \delta \mid ^{2}a(n_{1}-n_{2},0,j,\beta ,j,\beta \rangle . $$
Replacing \(n_{2}\) by \(-n_{2},\) we get
$$ \int _{F}\left| f_{\delta ,\beta +\tau }(x)\right| ^{2}\left| \varphi _{n,v}(\langle \delta ,x\rangle )\right| ^{2}\text {dx}\,=4C $$
[see (4.5.11)] and(4.2.34).
4.1.4 Appendix 4. Asymptotic Formulas for \(T_{v}(Q)\)
It is well known that the large eigenvalues of \(T_{0}(Q)\) lie in \(O(\frac{1}{m^{4}})\) neighborhood of
$$ \mid m\delta \mid +\frac{1}{16\pi \mid m\delta \mid ^{3}}\int _{0}^{2\pi }\left| Q(t)\right| ^{2}\mathrm{d}t $$
for the large values of m (see [5], p. 58). This formula yields the invariant (4.1.16). Using the asymptotic formulas for solutions of the Sturm–Liouville equation (see [5], p. 63), one can easily obtain that
$$ \varphi _{n,v}(\zeta )=\mathrm{e}^{i(n+v)\zeta }\left( 1+\frac{Q_{1}(\zeta )}{2i(n+v)\mid \delta \mid ^{2}}+\frac{Q(\zeta )-Q(0)-\frac{1}{2}Q_{1}^{2}(\zeta )}{4(n+v)^{2}\mid \delta \mid ^{4}}\right) +O\left( \frac{1}{n^{3}})\right) , $$
where
$$ Q_{1}(\zeta )=\int _{0}^{\zeta }Q(t)\mathrm{d}t. $$
From this, by direct calculations, we find \(A_{0}(\zeta ),\) \(A_{1}(\zeta ),\) and \(A_{2}(\zeta )\) [see (4.1.6)] and then using these in (4.1.7), we get the invariant (4.1.15).
Now we consider the eigenfunction \(\varphi _{n,v}(\zeta )\) of \(T_{v}(p)\) in the case \(v\ne 0,\ \frac{1}{2}\) and
$$ p(\zeta )=p_{1}\mathrm{e}^{i\zeta }+p_{-1}\mathrm{e}^{-i\zeta }. $$
The eigenvalues and the eigenfunctions of \(T_{v}(0)\) are \(\ (n+v)^{2} \mid \delta \mid ^{2}\) and \(\ \mathrm{e}^{i(n+v)\zeta }\), for \(n\in \mathbb {Z}\). Since the eigenvalues of \(T_{v}(p)\) are simple for \(v\ne 0,\ \frac{1}{2},\) by the well-known perturbation formula
$$ (\varphi _{n,v}(\zeta ),\ \mathrm{e}^{i(n+v)\zeta })\varphi _{n,v}(\zeta )=\mathrm{e}^{i(n+v)\zeta }+ $$
$$\begin{aligned} \sum _{k=1,2,\ldots }\frac{(-1)^{k+1}}{2i\pi }\int \limits _{C}(T_{v}(0)-\lambda )^{-1}p(x))^{k}(T_{v}(0)-\lambda )^{-1}\mathrm{e}^{i(n+v)\zeta }\mathrm{d}\lambda , \end{aligned}$$
(4.5.12)
where C is a contour containing only the eigenvalue \((n+t)^{2}\mid \delta \mid ^{2}\). Using
$$ (T_{v}(0)-\lambda )^{-1}\mathrm{e}^{i(n+v)\zeta }=\frac{\mathrm{e}^{i(n+v)\zeta }}{(n+v)^{2} \mid \delta \mid ^{2}-\lambda }, $$
we see that the kth (\(k=1,2,3,4\)) term \(F_{k}\) of the series (4.5.12) has the form
$$ F_{1}=\frac{1}{2i\pi }\int \limits _{C}\sum _{m=1,-1}\frac{p_{m}\mathrm{e}^{i(n+m+v)\zeta } }{((n+v)^{2}\mid \delta \mid ^{2}-\lambda )((n+m+v)^{2}\mid \delta \mid ^{2} -\lambda )}\mathrm{d}\lambda , $$
$$ F_{2}=\frac{-1}{2i\pi }\int \limits _{C}\sum _{m,l=1,-1}\frac{p_{m}p_{l} \mathrm{e}^{i(n+m+l+v)\zeta }}{((n+v)^{2}\mid \delta \mid ^{2}-\lambda )}\times $$
$$ \frac{1}{((n+m+v)^{2}\mid \delta \mid ^{2}-\lambda )((n+m+l+v)^{2}\mid \delta \mid ^{2}-\lambda )}\mathrm{d}\lambda , $$
$$ F_{3}=\frac{1}{2i\pi }\int \limits _{C}\sum _{m,l,k=1,-1}\frac{p_{m}p_{l} p_{k}\mathrm{e}^{i(n+m+l+k+v)\zeta }}{((n+v)^{2}\mid \delta \mid ^{2}-\lambda )((n+m+v)^{2}\mid \delta \mid ^{2}-\lambda )}\times $$
$$ \frac{1}{((n+m+l+v)^{2}\mid \delta \mid ^{2}-\lambda )((n+m+l+k+v)^{2}\mid \delta \mid ^{2}-\lambda )}\mathrm{d}\lambda , $$
$$ F_{4}=\frac{-1}{2i\pi }\int \limits _{C}\sum _{m,l,k,r=1,-1}\frac{p_{m}p_{l} p_{k}p_{r}\mathrm{e}^{i(n+m+l+k+r+v)\zeta }}{((n+m+l+k+r+v)^{2}\mid \delta \mid ^{2}-\lambda )}\times $$
$$ \frac{1}{((n+m+v)^{2}\mid \delta \mid ^{2}-\lambda )((n+m+l+v)^{2}\mid \delta \mid ^{2}-\lambda )}\times $$
$$ \frac{1}{((n+m+l+k+v)^{2}\mid \delta \mid ^{2}-\lambda )((n+v)^{2}\mid \delta \mid ^{2}-\lambda )}\mathrm{d}\lambda . $$
Since the distance between \((n+v)^{2}\mid \delta \mid ^{2}\) and \((n^{^{\prime } }+v)^{2}\mid \delta \mid ^{2}\) for \(n^{^{\prime }}\ne n\) is greater than \(c_{17}n\), we can choose the contour C such that
$$ \frac{1}{\mid (n^{^{\prime }}+v)^{2}\mid \delta \mid ^{2}-\lambda \mid }<\frac{c_{18}}{n}, \forall \lambda \in C, \forall n^{^{\prime }}\ne n $$
and the length of C is less than \(c_{19}\). Therefore
$$ F_{5}+F_{6}+\cdots =O(n^{-5}). $$
Now we calculate the integrals in \(F_{1}\), \(F_{2}\), \(F_{3}\), and \(F_{4}\) by the Cauchy integral formula and then decompose the obtained expression in power of \(\frac{1}{n}\). Then
$$ F_{1}=\mathrm{e}^{i(n+v)\zeta }\left( (p_{1}\mathrm{e}^{i\zeta }-p_{-1}\mathrm{e}^{-i\zeta })\frac{1}{\mid \delta \mid ^{2}}\left( \frac{-1}{2n}+\frac{v}{2n^{2}}-\frac{4v^{2}+1}{8n^{3}} +O\left( \frac{1}{n^{4}}\right) \right) \right. + $$
$$ (p_{1}\mathrm{e}^{i\zeta }+p_{-1}\mathrm{e}^{-i\zeta })\frac{1}{\mid \delta \mid ^{2}}\left. \left( \frac{v}{4n^{2}}-\frac{v}{2n^{3}}+\frac{12v^{2}+1}{16n^{4}}+O\left( \frac{1}{n^{5}}\right) \right) \right) . $$
Let \(F_{2,1}\) and \(F_{2,2}\) be the sum of the terms in \(F_{2}\) for which \(m+l=\pm 2\) and \(m+l=0\), respectively, i.e.,
$$ F_{2}=F_{2,1}+F_{2,2}, $$
where
$$ F_{2,1}=\mathrm{e}^{i(n+v)\zeta }\left( ((p_{1})^{2}\mathrm{e}^{2i\zeta }+(p_{-1})^{2}\mathrm{e}^{-2i\zeta } )\frac{1}{\mid \delta \mid ^{4}}\left( \frac{-1}{8n^{2}}+\frac{-v}{4n^{3}} -\frac{12v^{2}+7}{32n^{4}}+O\left( \frac{1}{n^{5}}\right) \right) \right. + $$
$$ ((p_{1})^{2}\mathrm{e}^{2i\zeta }-(p_{-1})^{2}\mathrm{e}^{-2i\zeta })\frac{1}{\mid \delta \mid ^{4} }\left. \left( \frac{-3}{16n^{3}}+O\left( \frac{1}{n^{4}}\right) \right) \right) , $$
$$ F_{2,2}=\mathrm{e}^{i(n+v)\zeta }\left| p_{1}\right| ^{2}\left( \frac{c_{20}}{n^{2} }++\frac{c_{21}}{n^{3}}+\frac{c_{22}}{n^{4}}+O\left( \frac{1}{n^{5}}\right) \right) $$
and \(c_{20},c_{21},\) and \(c_{22}\) are the known constants. Similarly,
$$ F_{3}=F_{3,1}+F_{3,2}, $$
where \(F_{3,1}\) and \(F_{3,2}\) are the sum of the terms in \(F_{3}\) for which \(m+l+k=\pm 3\) and \(m+l+k=\pm 1\), respectively. Hence
$$ F_{3,1}=\mathrm{e}^{i(n+v)\zeta }\left( (p_{1}^{3}\mathrm{e}^{3i\zeta }-p_{-1}^{3}\mathrm{e}^{-i\zeta })\frac{1}{\mid \delta \mid ^{6}}\left( \frac{-1}{48n^{3}}+O\left( \frac{1}{n^{4}}\right) \right) \right. + $$
$$ (p_{1}^{3}\mathrm{e}^{3i\zeta }+p_{-1}^{3}\mathrm{e}^{-3i\zeta })\frac{1}{\mid \delta \mid ^{6} }\left. \left( \frac{1}{16n^{4}}+O\left( \frac{1}{n^{5}}\right) \right) \right) , $$
$$ F_{3,2}=\mathrm{e}^{i(n+v)\zeta }\left( (p_{1}\mathrm{e}^{i\zeta }-p_{-1}\mathrm{e}^{-i\zeta })\left| p_{1}\right| ^{2}\left( \frac{c_{23}}{n^{3}}+\frac{c_{24}}{n^{4}}+O\left( \frac{1}{n^{5}}\right) \right) \right. + $$
$$ (p_{1}\mathrm{e}^{i\zeta }+p_{-1}\mathrm{e}^{-i\zeta })\left| p_{1}\right| ^{2} \left. \left( \frac{c_{25}}{n^{4}}+O\left( \frac{1}{n^{5}}\right) \right) \right) . $$
Similarly
$$ F_{4}=F_{4,1}+F_{4,2}+F_{4,3}, $$
where \(F_{4,1}\), \(F_{4,2}\), \(F_{4,3}\) are the sum of the terms in \(F_{4}\) for which \(m+l+k+r=\pm 4\), \(m+l+k+r=\pm 2\), \(m+l+k+r=0\), respectively. Thus
$$ F_{4,1}=\mathrm{e}^{i(n+v)\zeta }(p_{1}^{4}\mathrm{e}^{4i\zeta }+p_{-1}^{4}\mathrm{e}^{-4i\zeta })\frac{1}{\mid \delta \mid ^{8}}\left( \frac{1}{384n^{4}}+O\left( \frac{1}{n^{5}}\right) \right) , $$
$$ F_{4,2}=\mathrm{e}^{i(n+v)\zeta }(p_{1}^{2}\mathrm{e}^{2i\zeta }+p_{-1}^{2}\mathrm{e}^{-2i\zeta })\left| p_{1}\right| ^{2}\left( \frac{c_{26}}{n^{4}}+O\left( \frac{1}{n^{5}}\right) \right) , $$
$$ F_{4,3}=\mathrm{e}^{i(n+v)\zeta }\left| p_{1}\right| ^{4}\left( \frac{c_{27}}{n^{4} }+O\left( \frac{1}{n^{5}}\right) \right) . $$
Since \({p_{-1}^{k}}\mathrm{e}^{-ik\zeta }\) is conjugate of \({p_{1}^{k}}{\mathrm{e}^{ik\zeta }},\) the real and imaginary parts of \({F_{k}\mathrm{e}^{-i(n+v)\zeta }}\) consist of terms with multiplicands
$$ p_{1}^{k}\mathrm{e}^{ik\zeta }+p_{-1}^{k}\mathrm{e}^{-ki\zeta } \quad \& \quad p_{1} ^{k}\mathrm{e}^{ik\zeta }-p_{-1}^{k}\mathrm{e}^{-ik\zeta }, $$
respectively. Taking into account this and using the above estimations, we get
$$\begin{aligned} |(\varphi _{n,v},\mathrm{e}^{i(n+v)\zeta })\varphi _{n,v}|^{2}&=2\left( \sum _{k=1,2,3,4} \mathbf {Re(}F_{k})+\mathbf {Re(}F_{1}F_{2})+\mathbf {Re(}F_{1}F_{3} )\right) \\&\quad +|F_{1}|^{2}+|F_{2}|^{2}+O(n^{-5})= \end{aligned}$$
$$ 1+\frac{1}{2n^{2}}\frac{1}{\mid \delta \mid ^{2}}(p_{1}\mathrm{e}^{i\zeta }+p_{-1} \mathrm{e}^{-i\zeta }+c_{28}|p_{1}|^{2})+\frac{1}{n^{3}}((p_{1}\mathrm{e}^{i\zeta }+p_{-1} \mathrm{e}^{-i\zeta })c_{29} $$
$$ +c_{30}|p_{1}|^{2})+\frac{1}{n^{4}}((p_{1}\mathrm{e}^{i\zeta }+p_{-1}\mathrm{e}^{-i\zeta } )c_{31}+c_{32}|p_{1}|^{2}+c_{33}|p_{1}|^{4} $$
$$ +c_{34}|p_{1}|^{2}(p_{1}\mathrm{e}^{i\zeta }+p_{-1}\mathrm{e}^{-i\zeta })+(c_{35}+c_{36} |p_{1}|^{2})(p_{1}^{2}\mathrm{e}^{2i\zeta }+p_{-1}^{2}\mathrm{e}^{-2i\zeta }))+O\left( \frac{1}{n^{5} }\right) , $$
where \(\mathbf {Re(}F)\) denotes the real part of F. On the other hand
$$ |(\varphi _{n,v}(\zeta ),\mathrm{e}^{i(n+v)\zeta })|^{2}=\left( c_{37}\frac{1}{n^{2}} +c_{38}\frac{1}{n^{3}}+c_{39}\frac{1}{n^{4}}\right) |p_{1}|^{2}+c_{40}\frac{1}{n^{4} }|p_{1}|^{4}+O\left( \frac{1}{n^{5}}\right) . $$
These equalities imply (4.1.18). The invariant (4.1.19) is a consequence of (4.1.18), (4.1.16), and (4.1.7) for \(k=2,4\). \(\blacksquare \)