Complex Arnol’d – Liouville Maps

Abstract

We discuss the holomorphic properties of the complex continuation of the classical Arnol’d – Liouville action-angle variables for real analytic 1 degree-of-freedom Hamiltonian systems depending on external parameters in suitable Generic Standard Form, with particular regard to the behaviour near separatrices. In particular, we show that near separatrices the actions, regarded as functions of the energy, have a special universal representation in terms of affine functions of the logarithm with coefficients analytic functions. Then, we study the analyticity radii of the action-angle variables in arbitrary neighborhoods of separatrices and describe their behaviour in terms of a (suitably rescaled) distance from separatrices. Finally, we investigate the convexity of the energy functions (defined as the inverse of the action functions) near separatrices, and prove that, in particular cases (in the outer regions outside the main separatrix, and in the case the potential is close to a cosine), the convexity is strictly defined, while in general it can be shown that inside separatrices there are inflection points.

APPENDIX A. PROOFS OF PROPOSITION 2

First, recalling (3.45) and (3.46), we define the symplectic transformation

$$\Phi_{*}:({\mathtt{p}},{\mathtt{q}})\in D_{7{\mathtt{r}}/8,7{\mathtt{s}}/8}\ \longrightarrow\ \big{(}{\mathtt{p}},{\mathtt{q}}_{1}+\theta_{2j}(\hat{\mathtt{p}}),\hat{\mathtt{q}}+{\mathtt{p}}_{1}\partial_{\hat{\mathtt{p}}}\theta_{2j}(\hat{\mathtt{p}})\big{)}\in D_{{\mathtt{r}},{\mathtt{s}}},$$

(A.1)

transforming the Hamiltonian ${\mathtt{H}}$ in (2.6) into

$${\mathtt{H}}_{*}:={\mathtt{H}}\circ\Phi_{*}({\mathtt{p}},{\mathtt{q}})=:\big{(}1+\nu_{*}({\mathtt{p}},{\mathtt{q}}_{1})\big{)}{\mathtt{p}}_{1}^{2}+{\mathtt{G}}_{*}(\hat{\mathtt{p}},{\mathtt{q}}_{1}).$$

(A.2)

By Taylor expansion at $({\mathtt{p}},{\mathtt{q}}_{1})=(0,\hat{\mathtt{p}},0)$, recalling (A.2), (3.47), (3.49) and (2.10), we get

$$\displaystyle{\mathtt{H}}_{*}=E_{2j}(\hat{\mathtt{p}})+\big{(}1+\nu_{*}(0,\hat{\mathtt{p}},0)\big{)}{\mathtt{p}}_{1}^{2}-\lambda^{2}(\hat{\mathtt{p}}){\mathtt{q}}_{1}^{2}+R_{*}({\mathtt{p}},{\mathtt{q}}_{1}),\qquad\mbox{with}$$

$$\displaystyle R_{*}({\mathtt{p}},{\mathtt{q}}_{1}):=\big{(}\nu_{*}({\mathtt{p}},{\mathtt{q}}_{1})-\nu_{*}(0,\hat{\mathtt{p}},0)\big{)}{\mathtt{p}}_{1}^{2}+{\mathtt{G}}_{*}(\hat{\mathtt{p}},{\mathtt{q}}_{1})-\frac{1}{2}\partial^{2}_{{\mathtt{q}}_{1}}{\mathtt{G}}_{*}(\hat{\mathtt{p}},0){\mathtt{q}}_{1}^{2}.$$

(A.3)

Then, the following trivial lemma holds.

Lemma 7

There exists a constant $0<{\bf c_{{{}_{0}}}}<1/8$ , depending only on $\kappa,n$ , such that, defining the symplectic transformation

$$\displaystyle\Phi_{0}:\{|Y_{1}|<{\bf c_{{{}_{0}}}}\varepsilon^{1/4}\}\times\hat{D}_{3{\mathtt{r}}/4}\times\{|X_{1}|<{\bf c_{{{}_{0}}}}\varepsilon^{1/4}\}\times\mathbb{T}^{n-1}_{3{\mathtt{s}}/4}\ \longrightarrow\ D_{7{\mathtt{r}}/8,7{\mathtt{s}}/8},$$

(A.4)

$$\displaystyle{\mathtt{p}}_{1}=\delta(\hat{Y})Y_{1},\qquad\hat{\mathtt{p}}=\hat{Y},\qquad{\mathtt{q}}_{1}=\frac{1}{\delta(\hat{Y})}X_{1},\qquad\hat{\mathtt{q}}=\hat{X}-\frac{\partial_{\hat{Y}}\delta(\hat{Y})}{\delta(\hat{Y})}Y_{1}X_{1},$$

we have that ${\mathtt{H}}_{0}:={\mathtt{H}}_{*}\circ\Phi_{0}$ has the form

$${\mathtt{H}}_{0}=E_{2j}(\hat{Y})+g(\hat{Y})(Y_{1}^{2}-X_{1}^{2})+\varepsilon R_{0}(\varepsilon^{-1/4}Y_{1},\hat{Y},\varepsilon^{-1/4}X_{1}),$$

(A.5)

where $R_{0}(\tilde{Y}_{1},\hat{Y},\tilde{X}_{1})$ is holomorphic on

$$\{|\tilde{Y}_{1}|<{\bf c_{{{}_{0}}}}\}\times\hat{D}_{3{\mathtt{r}}/4}\times\{|\tilde{X}_{1}|<{\bf c_{{{}_{0}}}}\},$$

with $|R_{0}|\lessdot 1$ and, finally, it is at least cubic in $\tilde{Y}_{1},\tilde{X}_{1}$ .

Proof

The fact that $\Phi_{0}$ is well defined on its domain follows by the explicit expression in (A.4), by (3.50), (2.10) and (2.8) (in particular, $\varepsilon\leqslant{\mathtt{r}}^{2}/2^{16}$). Eq. (A.5) follows by (A.3), setting

$$R_{0}(\tilde{Y}_{1},\hat{Y},\tilde{X}_{1}):=\varepsilon^{-1}R_{*}\left(\delta(\hat{Y})\varepsilon^{1/4}\tilde{Y}_{1},\hat{Y},\frac{\varepsilon^{1/4}}{\delta(\hat{Y})}\tilde{X}_{1}\right).$$

(A.6)

Finally, the estimate $|R_{0}|\lessdot 1$ follows from (3.47) and (3.50).

Next, we shall use the following well-known result, whose proof can be found, e.g., inFootnote

See, in particular, Lemma 0 and Appendix A.3 in [12].

[12] or in [14].

See, in particular, Lemma 0 and Appendix A.3 in [12].

Lemma 8

Given a Hamiltonian ${\mathtt{H}}_{0}$ as in (A.5) . For suitable constants $0<{\bf c}_{{{}_{1}}}<{\bf c_{{{}_{0}}}}/8n{\bf c}_{{{}_{2}}}$ , depending only on $\kappa,n$ , there exist a $($ near-identity $)$ symplectic transformation

$$\displaystyle\begin{aligned} \Phi_{1}:&\{|y_{1}|<{\bf c}_{{{}_{1}}}\varepsilon^{1/4}\}\times\hat{D}_{{\mathtt{r}}/2}\times\{|x_{1}|<{\bf c}_{{{}_{1}}}\varepsilon^{1/4}\}\times\mathbb{T}^{n-1}_{{\mathtt{s}}/2}\ \longrightarrow\\ &\{|Y_{1}|<{\bf c_{{{}_{0}}}}\varepsilon^{1/4}\}\times\hat{D}_{3{\mathtt{r}}/4}\times\{|X_{1}|<{\bf c_{{{}_{0}}}}\varepsilon^{1/4}\}\times\mathbb{T}^{n-1}_{3{\mathtt{s}}/4}, \end{aligned}$$

(A.7)

and a function $R_{\rm hp}(z,\hat{y})$ satisfying (3.52) and (3.56) , such that ${\mathtt{H}}_{\rm hp}(y,x_{1}):={\mathtt{H}}_{0}\circ\Phi_{1}(y,x)$ satisfies (3.53) . Moreover, $\Phi_{1}$ has the form

$$\displaystyle Y_{1}=y_{1}+\varepsilon^{1/4}a_{1}(\varepsilon^{-1/4}y_{1},\hat{y},\varepsilon^{-1/4}x_{1}),\quad\hat{Y}=\hat{y},$$

(A.8)

$$\displaystyle X_{1}=x_{1}+\varepsilon^{1/4}a_{2}(\varepsilon^{-1/4}y_{1},\hat{y},\varepsilon^{-1/4}x_{1}),\quad\hat{X}=\hat{x}+\sqrt{\varepsilon}{\mathtt{r}}^{-1}a_{3}(\varepsilon^{-1/4}y_{1},\hat{y},\varepsilon^{-1/4}x_{1}),$$

for suitable functions $a_{i}(\tilde{y}_{1},\hat{y},\tilde{x}_{1})$ , $i=1,2,3$ , which are holomorphic and bounded by ${\bf c}_{{{}_{2}}}$ on

$$\{|\tilde{y}_{1}|<{\bf c_{{{}_{0}}}}/2\}\times\hat{D}_{{\mathtt{r}}/2}\times\{|\tilde{x}_{1}|<{\bf c_{{{}_{0}}}}/2\},$$

moreover, $a_{1},a_{2}$ and $a_{3}$ are, respectively, at least quadratic and cubic in $\tilde{y}_{1},\tilde{x}_{1}$ .

Remark 6

${\mathtt{H}}_{\rm hp}$ is simply the hyperbolic Birkhoff normal form of ${\mathtt{H}}_{0}$. Any canonical transformation of the form $y_{1}=\alpha\tilde{y}_{1}+\beta\tilde{x}_{1},$ $x_{1}=\beta\tilde{y}_{1}+\alpha\tilde{x}_{1},$ with $\alpha^{2}-\beta^{2}=1$ and $\hat{y}=\hat{\tilde{y}}$ leaves ${\mathtt{H}}_{\rm hp}$ invariant since $y_{1}^{2}-x_{1}^{2}=\tilde{y}_{1}^{2}-\tilde{x}_{1}^{2}.$ Namely, the integrating transformation $\Phi_{1}$ is not unique. However, as is well known, the form of the integrated Hamiltonian ${\mathtt{H}}_{\rm hp}$ in (3.53) is unique, in the sense that $E_{2j}$, $g$ and $R$ are unique.

Note also that the map $\Phi_{1}$ is close to the identity, for small ${\bf c}_{{{}_{1}}}$, since its Jacobian is the identity plus a matrix whose entries are (by the Cauchy estimates) uniformly bounded on its domain in (3.51) by $2{\bf c}_{{{}_{2}}}{\bf c}_{{{}_{1}}}/{\bf c_{{{}_{0}}}}\leqslant 1/4n$.

Let us come back to the proof of Proposition 2 and let us prove (3.56).

Evaluating (3.53) for $\mu=0$, we get

$$\displaystyle{\bar{\mathtt{H}}}_{1}(y,x_{1}):={\bar{\mathtt{H}}}_{1}(y,x_{1})|_{\mu=0}={\bar{\mathtt{H}}}_{0}\circ\bar{\Phi}_{1}(y,x)$$

$$\displaystyle=\bar{E}_{2j}+\bar{g}(y_{1}^{2}-x_{1}^{2})+\varepsilon\bar{R}_{\rm hp}\left(\frac{y_{1}^{2}-x_{1}^{2}}{\sqrt{\varepsilon}}\right)=O(\varepsilon)$$

(A.9)

on the domain defined in (3.51). Let us denote ${\bar{\mathtt{H}}}_{0}:={\mathtt{H}}_{0}|_{\mu=0}$. Since by (A.6), (3.47) one has ${\mathtt{H}}_{0}-{\bar{\mathtt{H}}}_{0}=O(\varepsilon\mu)$,

$${\mathtt{H}}_{0}\circ\bar{\Phi}_{1}={\bar{\mathtt{H}}}_{0}\circ\bar{\Phi}_{1}+({\mathtt{H}}_{0}-{\bar{\mathtt{H}}}_{0})\circ\bar{\Phi}_{1}={\bar{\mathtt{H}}}_{1}+R_{1},\quad\mbox{with}\ \ \ R_{1}=O(\varepsilon\mu),$$

namely, the system is integrated up to a small term of order $\varepsilon\mu$. Note also that, since $\bar{\Phi}_{1}$ has the form in (A.8), it leaves invariant the terms of order $\leqslant 2$ in $(y_{1},x_{1})$, namely,

$${\mathtt{H}}_{0}\circ\bar{\Phi}_{1}=E_{2j}+g(y_{1}^{2}-x_{1}^{2})+\varepsilon\bar{R}+Q,\quad\mbox{with}\quad Q=O(\varepsilon\mu).$$

(A.10)

Now we want to construct a symplectic transformation $\Phi_{\mu}$ integrating ${\mathtt{H}}_{0}\circ\bar{\Phi}_{1}$. Since ${\bar{\mathtt{H}}}_{1}$ is already in normal form, we claim that the integrating transformation $\Phi_{\mu}$ is $O(\varepsilon^{1/4}\mu)$-close to the identity and

$${\mathtt{H}}_{0}\circ\bar{\Phi}_{1}\circ\Phi_{\mu}=({\bar{\mathtt{H}}}_{1}+R_{1})\circ\Phi_{\mu}=:{\mathtt{H}}_{\rm hp}^{\prime}={\bar{\mathtt{H}}}_{1}+O(\varepsilon\mu),$$

(A.11)

where ${\mathtt{H}}_{\rm hp}^{\prime}$ is in normal form, namely, as in (3.53). By the unicity of the Birkhoff normal form, we deduce that ${\mathtt{H}}_{\rm hp}={\mathtt{H}}_{\rm hp}^{\prime}={\bar{\mathtt{H}}}_{1}+O(\varepsilon\mu)$. By (3.53), (A.9), (3.50) and (3.2), we get (3.56).

It remains to prove (A.11). The crucial point here is that the generating functionFootnote

According to Lie’s series method.

$\chi$ of the integrating transformation $\Phi_{\mu}$ is $O(\sqrt{\varepsilon}\mu)$ and its gradient is, by the Cauchy estimates, $O(\varepsilon^{1/4}\mu)$ in a domain $\{|y_{1}|,|x_{1}|\lessdot\varepsilon^{1/4}\}$. The fact that $\chi=O(\sqrt{\varepsilon}\mu)$ can be easily seen by passing, as is usual in Birkhoff’s normal form, to the coordinate $\xi=(y_{1}-x_{1})/\sqrt{2}$, $\eta=(y_{1}+x_{1})/\sqrt{2}$. In these coordinates, recalling (A.10), we get

$${\mathtt{H}}_{0}\circ\bar{\Phi}_{1}=E_{2j}+2g\xi\eta+\varepsilon\bar{R}^{\prime}(\xi,\eta)+Q^{\prime}(\xi,\eta)$$

with $\bar{R}^{\prime}=\bar{R}_{\rm hp}(2\xi\eta/\sqrt{\varepsilon})=O(1)$ and $Q^{\prime}=O(\varepsilon\mu)$. Note that the Taylor expansion of $\bar{R}^{\prime}$ contains only a monomial of the form $\bar{R}^{\prime}_{hh}\xi^{h}\eta^{h}$. At the first step, we have to cancel all the monomials of $Q^{\prime}$ of the form $Q^{\prime}_{hk}\xi^{h}\eta^{k}$ with $h+k=3$. The generating function$\chi^{(3)}$ of the first step is exactly

$$\chi^{(3)}=\sum_{h+k=3}\frac{Q^{\prime}_{hk}}{2g(h-k)}\xi^{h}\eta^{k}\stackrel{{\scriptstyle{(3.50)}}}{{=}}O(\sqrt{\varepsilon}\mu).$$

After this first step the Hamiltonian becomes $E_{2j}+2g\xi\eta+\varepsilon\bar{R}^{\prime}(\xi,\eta)+Q^{\prime\prime}(\xi,\eta)$ with $Q^{\prime\prime}=O(\varepsilon\mu)$. At the second step, we have to cancel all the monomials of $Q^{\prime\prime}$ of the form $Q^{\prime\prime}_{hk}\xi^{h}\eta^{k}$ with $h+k=4$, $h\neq k$. We proceed as in the first step with analogous estimates. Analogously for the other infinite steps, obtaining

$$E_{2j}+2g\xi\eta+\varepsilon\bar{R}^{\prime}(\xi,\eta)+\bar{Q}(\xi,\eta)$$

with $\bar{Q}=O(\varepsilon\mu)$ and $\bar{Q}_{hk}=0$ for $h\neq k$, proving (A.11) (recall (3.50) and (3.2)).

According to Lie’s series method.

We can conclude the proof of Proposition 2:

The composition of the symplectic transformations defined in (A.1), (A.4), (3.51) integrates ${\mathtt{H}}$, namely, (3.53) holdsFootnote

As well as (3.52) and (3.56) by Lemma 8.

with $\Phi_{\rm hp}:=\Phi_{*}\circ\Phi_{0}\circ\Phi_{1}$ satisfying (3.51), (3.54) and (3.55). The inclusion (3.57) follows by (3.54) and (3.50).

As well as (3.52) and (3.56) by Lemma 8.

APPENDIX B. PROOFS OF TWO SIMPLE LEMMATA

B.1. Proof of Lemma 1

We know that $\partial_{\theta}{\bar{\mathtt{G}}}(\bar{\theta}_{i})=0$ and we want to solve the equation $\partial_{\theta}{\mathtt{G}}\big{(}\hat{p},\theta_{i}(\hat{p})\big{)}=0$. Equivalently, for $\mu\leqslant 2^{-8}\kappa^{-6}$, we want to find a real-analytic $y=y(\hat{p}),$ $\hat{p}\in\hat{D}_{\mathtt{r}}$, with

$$\textstyle\sup_{\hat{D}_{\mathtt{r}}}|y|\leqslant\rho:=\frac{2\varepsilon\mu}{\beta{\mathtt{s}}}\stackrel{{\scriptstyle{\rm(2.10)}}}{{\leqslant}}\frac{{\mathtt{s}}}{2},$$

(B.1)

by solving the equation

$$\partial_{\theta}{\mathtt{G}}\big{(}\hat{p},\bar{\theta}_{i}+y(\hat{p})\big{)}=0,$$

(B.2)

so that $\theta_{i}(\hat{p})=\bar{\theta}_{i}+y(\hat{p})$. We have

$$\partial_{\theta}{\mathtt{G}}(\hat{p},\bar{\theta}_{i}+y)=\partial_{\theta}{\mathtt{G}}(\hat{p},\bar{\theta}_{i})+g(\hat{p},y)y,\quad\mbox{where}\quad g(\hat{p},y):=\int_{0}^{1}\partial_{\theta}^{2}{\mathtt{G}}(\hat{p},\bar{\theta}_{i}+ty)dt.$$

Then (B.2) can be written as the fixed point equation

$$y=\Psi(y),\quad\mbox{where}\quad\Psi(y):=-\frac{\partial_{\theta}{\mathtt{G}}(\hat{p},\bar{\theta}_{i})}{g(\hat{p},y)}$$

to be solved in the closed set of the real-analytic functions $y=y(\hat{p})$ on $\hat{D}_{\mathtt{r}}$ satisfying the bound (B.1). Note that, since $\partial_{\theta}{\bar{\mathtt{G}}}(\bar{\theta}_{i})=0$, by (2.4) we have $|\partial_{\theta}^{2}{\bar{\mathtt{G}}}(\bar{\theta}_{i})|\geqslant\beta$. Moreover, by (2.8) and the Cauchy estimates, we get for $|y|\leqslant\rho$ and $\hat{p}\in\hat{D}_{\mathtt{r}}$

$$|g-\partial_{\theta}^{2}{\bar{\mathtt{G}}}(\bar{\theta}_{i})|\leqslant\frac{4\varepsilon\mu}{{\mathtt{s}}^{2}},\quad\mbox{which implies}\quad|g|\geqslant\beta-\frac{4\varepsilon\mu}{{\mathtt{s}}^{2}}\stackrel{{\scriptstyle{\rm(2.10)}}}{{\geqslant}}\frac{\beta}{2}.$$

(B.3)

Again by $\partial_{\theta}{\bar{\mathtt{G}}}(\bar{\theta}_{i})=0$, (2.8) and the Cauchy estimates, we find uniformly on $\hat{D}_{\mathtt{r}}$ that

$$|\partial_{\theta}{\mathtt{G}}(\hat{p},\bar{\theta}_{i})|\leqslant\varepsilon\mu/{\mathtt{s}}.$$

(B.4)

Then by (B.3) we obtain for $|y|\leqslant\rho$ and $\hat{p}\in\hat{D}_{\mathtt{r}}$

$$\textstyle|\Psi|\leqslant\frac{2\varepsilon\mu}{\beta{\mathtt{s}}}=\rho,$$

(B.5)

by (2.10) and (B.1). Moreover,

$$\partial_{y}\Psi(y):=\frac{\partial_{\theta}{\mathtt{G}}(\hat{p},\bar{\theta}_{i})}{\big{(}g(\hat{p},y)\big{)}^{2}}\partial_{y}g(\hat{p},y).$$

Then for $|y|\leqslant\rho$ and $\hat{p}\in\hat{D}_{\mathtt{r}}$ we get

$$\textstyle|\partial_{y}\Psi|<2^{6}\frac{\varepsilon^{2}\mu}{\beta^{2}{\mathtt{s}}^{4}}\leqslant 2^{6}\kappa^{6}\mu\leqslant 1,$$

(B.6)

by (B.4), (B.3), (2.10) and since $|\partial_{y}g(\hat{p},y)|<{16\varepsilon}/{{\mathtt{s}}^{3}}$ by (2.8) and the Cauchy estimates. In conclusion, by (B.5) and (B.6) we have that $\Psi$ is a contraction and the fixed point theorem applies proving the first estimate in (3.2).

Let us now show the second estimate in (3.2).

By (2.8), the first estimate in (3.2), (2.10) and the Cauchy estimates, we get

$$\displaystyle|E_{i}(\hat{p})-\bar{E}_{i}|\leqslant|{\mathtt{G}}\big{(}\hat{p},\theta_{j}(\hat{p})\big{)}-{\bar{\mathtt{G}}}\big{(}\theta_{j}(\hat{p})\big{)}|+|{\bar{\mathtt{G}}}\big{(}\theta_{j}(\hat{p})\big{)}-{\bar{\mathtt{G}}}(\bar{\theta}_{j})|$$

$$\displaystyle\leqslant\textstyle\varepsilon\mu+\frac{2\varepsilon^{2}\mu}{\beta{\mathtt{s}}^{2}}\leqslant 3\kappa^{3}\varepsilon\mu,$$

proving the second estimate in (3.2).

Let us prove the final claim. By (2.11) (applied to ${\bar{\mathtt{G}}}$) and by the Cauchy estimates, it follows that the minimal distance between two critical points of ${\bar{\mathtt{G}}}$ can be estimated from below by $2\beta{\mathtt{s}}^{2}/\varepsilon$. Thus, by the first estimates in (3.2), it follows that the relative order of the critical points of ${\bar{\mathtt{G}}}$ is preserved, provided $8\varepsilon^{3}\mu^{2}<\beta^{3}{\mathtt{s}}^{4}$, which, using (2.10), is implied by $2^{3}\kappa^{7}\mu^{2}<1$, which, in turn, is implied by the hypothesis $\mu\leqslant 1/(2\kappa)^{6}$.

As for critical energies, since ${\bar{\mathtt{G}}}$ is $\beta$-Morse, they are at least $\beta$ apart; hence, from the second estimate in (3.2) the claim follows provided $3\kappa^{3}\varepsilon\mu<\beta$, which, by (2.10), is implied by $\mu<1/(3\kappa^{4})$, which, again, is implied by the hypothesis.

B.2. Proof of Lemma 6

First denote $R(z):={w}(z)-\cos z,$ so that $|R|_{1}\leqslant{{\mathtt{g}}_{\rm o}}.$ We note that, on the real line, ${w}$ has exactly two critical points: a maximum $x_{M}$ (with ${w}(x_{M})=1$) and a minimum $x_{m}$ (with ${w}(x_{m})=-1$) in the interval $[-\pi/2,3\pi/2).$ Indeed, since by the Cauchy estimates $\sup_{R}|{w}^{\prime}|\leqslant{{\mathtt{g}}_{\rm o}},$ the equation ${w}^{\prime}(x)=-\sin x+R^{\prime}(x)=0$ in the interval $[-\pi/2,3\pi/2)$ has only two solutions $x_{M},x_{m}$ with $|x_{M}|,|x_{m}-\pi|\leqslant 1.0001{{\mathtt{g}}_{\rm o}}\leqslant 0.001.$ Obviously, $x_{M}+b(x_{M})=0$ and $x_{m}+b(x_{m})=\pi.$

On the real line the function $b$ is given by the $2\pi$-periodic continuousFootnote

Since $b(x_{m}-2\pi)=b(x_{m})=\pi-x_{m}$.

function defined in the interval $[x_{m}-2\pi,x_{m}]$ by the expression

$$b(x):={\rm sign}(x-x_{M})\arccos\big{(}{w}(x)\big{)}-x.$$

Let us consider first the complex domain $\Omega_{0}:=\{0.4<{\rm Re}z<\pi-0.4,\ |{\rm Im}z|<1/4\}$ where $b(z)$ is clearly extendible to a holomorphic function. Here we have $\sup_{\Omega_{0}}|\cos z|\leqslant 0.913$ and, therefore, $\sup_{\Omega_{0}}|\cos z|+|R(z)|\leqslant 0.914.$ Then for $z\in\Omega_{0}$ we get

$$|b(z)|=|\arccos\big{(}\cos z+R(z)\big{)}-z|\leqslant\int_{0}^{1}\left|\frac{R(z)}{\sqrt{1-\big{(}\cos z+tR(z)\big{)}^{2}}}\right|dt\leqslant 6.1{{\mathtt{g}}_{\rm o}}.$$

Since $b(x_{m}-2\pi)=b(x_{m})=\pi-x_{m}$.

We now prove that $b(z)$ is extendible to a holomorphic function for $|z|<1/2$. First we prove that there exists a real-analytic positive function $d$ with holomorphic extension on $|z|<1/2$ such that ${w}(z)=1-\frac{1}{2}\big{(}(z-x_{M})d(z)\big{)}^{2}$. By Taylor’s expansion at $z=x_{M}$ we have that $d^{2}(z)=-2\int_{0}^{1}(1-t){w}^{\prime\prime}\big{(}x_{M}+t(z-x_{M})\big{)}dt$ and, therefore, for $|z|<1/2$

$$|d^{2}(z)-1|\leqslant 1-\cos x_{M}+\sup_{|z|<1/2}|\sin z||z-x_{M}|+2{{\mathtt{g}}_{\rm o}}\leqslant 0.55.$$

Then we can take the principle square rootFootnote

Namely, taking a cut in the negative real line.

of $d^{2}(z)$, obtaining the function $d(z)$. Now consider the holomorphic function $a(z)$ defined for $|z|<2$ such that $a^{\prime}(z)=1/\sqrt{1-(z/2)^{2}}$ and $a(0)=0$. Then for real $x$ we get $a(x)={\rm sign}(x)\arccos(1-x^{2}/2)$ and also (with $d(x)>0$)

$$b(x):={\rm sign}(x-x_{M})\arccos\left(1-\frac{1}{2}\big{(}(x-x_{M})d(x)\big{)}^{2}\right)-x=a\big{(}(x-x_{M})d(x)\big{)}-x.$$

Then $a\big{(}(z-x_{M})d(z)\big{)}-z$ is a holomorphic extension of $b$ for $|z|<2$. An analogous argument holds for $|z-\pi|<2.$

Namely, taking a cut in the negative real line.

In the following we will estimate $b(z)$ for a strip $|z|<1/2$, analogous arguments holds for $|z-\pi|<1/2$. We will often use thatFootnote

Using that $\frac{1}{2}|z|^{2}-(\cosh|z|-1-\frac{1}{2}|z|^{2})\leqslant|1-\cos z|\leqslant\cosh|z|-1$.

$$|z|\leqslant 1\quad\Longrightarrow\quad 0.45|z|^{2}\leqslant|1-\cos z|\leqslant 0.55|z|^{2}.$$

(B.7)

Now we prove that there exists a unique function $b(z)$ defined for

$$\Omega_{1}:=\{3\sqrt{{\mathtt{g}}_{\rm o}}<|z|<1/2\}$$

satisfying $\sup_{\Omega_{1}}|b|\leqslant\frac{3}{2}\sqrt{{\mathtt{g}}_{\rm o}},$ such that ${w}(z)=\cos\big{(}z+b(z)\big{)}$, as a fixed point of the equation

$$b(z)=\Psi(b)(z):=2\arcsin\left(\frac{-R(z)}{2\sin(z+b(z)/2)}\right).$$

Indeed,

$$\textstyle\cos\big{(}z+b(z)\big{)}-\cos z=-2\sin\left(z+b(z)/2\right)\sin\left(b(z)/2\right)=R(z).$$

For $z\in$ we have $|z+b(z)/2|\geqslant\frac{3}{2}\sqrt{{\mathtt{g}}_{\rm o}}$, which impliesFootnote

Using that for $|z|<1$ we have $\frac{4}{5}|z|\leqslant|\sin z|\leqslant\frac{6}{5}|z|$.

$|\sin(z+b(z)/2)|\geqslant\frac{6}{5}\sqrt{{\mathtt{g}}_{\rm o}}$ andFootnote

Using that for $|z|\leqslant 1/2$ we have $|\arcsin z|\leqslant\frac{2}{\sqrt{3}}|z|.$

$\sup_{\Omega_{1}}|\Psi(b)(z)|<\sqrt{{\mathtt{g}}_{\rm o}}.$ Finally, $\Psi$ is a contraction sinceFootnote

Using that for $|z|\leqslant 1$ we have $|\cos z|\leqslant\sqrt{3}.$

$$\displaystyle\textstyle\sup_{\Omega_{1}}|\Psi(b)-\Psi(b^{\prime})|\leqslant\frac{2}{\sqrt{3}}{{\mathtt{g}}_{\rm o}}\sup_{\Omega_{1}}\left|\frac{1}{\sin(z+b(z)/2)}-\frac{1}{\sin(z+b^{\prime}(z)/2)}\right|$$

$$\displaystyle\textstyle\leqslant\frac{2}{\sqrt{3}}\frac{5^{2}}{6^{2}}2\left|\sin\left(\frac{b^{\prime}(z)-b(z)}{4}\right)\cos\left(z+\frac{b^{\prime}(z)+b(z)}{4}\right)\right|\leqslant\frac{5}{6}\sup_{\Omega_{1}}|b-b^{\prime}|.$$

In conclusion, we get $\sup_{\Omega_{1}}|b|\leqslant\frac{3}{2}\sqrt{{\mathtt{g}}_{\rm o}}.$

Using that $\frac{1}{2}|z|^{2}-(\cosh|z|-1-\frac{1}{2}|z|^{2})\leqslant|1-\cos z|\leqslant\cosh|z|-1$.

Using that for $|z|<1$ we have $\frac{4}{5}|z|\leqslant|\sin z|\leqslant\frac{6}{5}|z|$.

Using that for $|z|\leqslant 1/2$ we have $|\arcsin z|\leqslant\frac{2}{\sqrt{3}}|z|.$

Using that for $|z|\leqslant 1$ we have $|\cos z|\leqslant\sqrt{3}.$

Next, we claim that in the domain $\Omega_{2}:=\{|z|\leqslant 3\sqrt{{\mathtt{g}}_{\rm o}}\}$ we have that $|b(z)|<9\sqrt{{\mathtt{g}}_{\rm o}}$. Indeed, by contradiction, assume that there exists $z_{0}\in\Omega_{2}$ such that for every $|z|<|z_{0}|$ we have $|b(z)|<9\sqrt{{\mathtt{g}}_{\rm o}}$, but $|b(z_{0})|=9\sqrt{{\mathtt{g}}_{\rm o}}$. Then $|z_{0}+b(z_{0})|\leqslant 12\sqrt{{\mathtt{g}}_{\rm o}}$ and by (B.7) and since $\cos\big{(}z_{0}+b(z_{0})\big{)}-1=\cos z_{0}-1+R(z_{0})$ we get

$$\displaystyle 16{{\mathtt{g}}_{\rm o}}\leqslant 0.45(|b(z_{0})|-|z_{0}|)^{2}\leqslant 0.45|z_{0}+b(z_{0})|^{2}\leqslant|\cos\big{(}z_{0}+b(z_{0})\big{)}-1|$$

$$\displaystyle\leqslant|\cos z_{0}-1|+|R(z_{0})|\leqslant 0.55|z_{0}|^{2}+{{\mathtt{g}}_{\rm o}}\leqslant 6{{\mathtt{g}}_{\rm o}},$$

which is a contradiction. Thus, $\sup_{\Omega_{2}}|b(z)|\leqslant 9\sqrt{{\mathtt{g}}_{\rm o}}.$     $\square$