SlowGuess/NeurIPS_2023_OCR huggingface dataset:
Proof. Assume we are provided with  $\mathsf{U}, z, \mathsf{U}_z, \mathsf{L}$  and  $\psi$  as described in Lemma 21. Let  $h \in (0,1/\mathsf{L})$  and  $a \in \mathbb{R}^d$ . We define the map  $\psi_h: \mathsf{U}_z \to \mathbb{R}^d$  by  $\psi_h(z') = z' + h\psi(z') + a$ , for any  $z' \in \mathsf{U}_z$ . Since  $\psi \in \mathrm{C}^1(\mathsf{U}_z, \mathbb{R}^d)$ , it is clear that  $\psi_h \in \mathrm{C}^1(\mathsf{U}_z, \mathbb{R}^d)$ . We have, for any  $z' \in \mathsf{U}_z$ ,  $D\psi_h(z') = \mathrm{Id} + hD\psi(z')$ . In particular, for any  $w \in \mathbb{R}^d$ , it comes that
$$
\left\| D \psi_ {h} (z) w \right\| \geq \left\| w \right\| - h L \| w \| = (1 - h L) \| w \|.
$$
Since  $1 - h\mathsf{L} > 0$ ,  $\operatorname{Jac}[\psi_h](z)$  is invertible. We conclude the proof by using the local inverse function theorem.
We recall that the set-valued map  $\mathrm{G}_h$ , defined in (6), is the generalised Leapfrog integrator, or equivalently the Störmer-Verlet integrator, of the non-separable Hamiltonian  $H_2$  defined in Section 3.1. We state below a result of local smoothness of this integrator.
Lemma 22. Let  $(\mathsf{M},\mathfrak{g})$  be a smooth manifold of  $\mathbb{R}^d$  and  $h > 0$ . Then, for any  $(x^{(0)},x^{(1)})\in \mathsf{M}\times \mathsf{M}$ , the vector  $p^{(0)}\in \mathrm{T}_{x^{(0)}}^{\star}\mathsf{M}$  such that  $(x^{(1)},p^{(1)})\in \mathrm{G}_h(x^{(0)},p^{(0)})$  for any  $p^{(1)}\in \mathrm{T}_{x^{(1)}}^{\star}\mathsf{M}$  is uniquely determined by  $p^{(0)} = \mathrm{G}_{h,x^{(0)}}(x^{(1)})$ , where the map  $\mathrm{G}_{h,x^{(0)}}:\mathsf{M}\to \mathrm{T}_{x^{(0)}}^{\star}\mathsf{M}$  is defined by
$$
\mathrm {G} _ {h, x ^ {(0)}} (y) = p ^ {(1 / 2)} (y) - \frac {h}{4} D \mathfrak {g} \left(x ^ {(0)}\right) \left[ \mathfrak {g} \left(x ^ {(0)}\right) ^ {- 1} p ^ {(1 / 2)} (y), \mathfrak {g} \left(x ^ {(0)}\right) ^ {- 1} p ^ {(1 / 2)} (y) \right], \tag {32}
$$
with  $p^{(1 / 2)}(y) = \frac{2}{h} (\mathfrak{g}(x^{(0)})^{-1} + \mathfrak{g}(y)^{-1})^{-1}(y - x^{(0)})$ . In particular, if  $\mathfrak{g}\in \mathrm{C}^2 (\mathsf{M},\mathbb{R}^{d\times d})$ , then, for any  $x^{(0)}\in \mathsf{M}$ ,  $\mathrm{G}_{h,x^{(0)}}\in \mathrm{C}^1 (\mathsf{M},\mathrm{T}_{x^{(0)}}^\star \mathsf{M})$ .
For any  $\alpha \geq 1$ , we define  $r_{\alpha}^{\star} > 0$  by
$$
r _ {\alpha} ^ {\star} = \min  \left(1 / 5 0 0 0 0, 1 / \{1 0 0 0 \alpha (\alpha + 1) \}\right). \tag {33}
$$
We recall that we denote by  $\mathsf{W}^0 (x,r)$  the open Dikin ellipsoid (w.r.t.  $\mathfrak{g}$ ) at  $x\in \mathsf{M}$  of radius  $r > 0$  given by  $\mathsf{W}^0 (x,r) = \{y\in \mathbb{R}^d:\| y - x\|_{\mathfrak{g}(x)} < r\}$ . Assume A1, A2. Let  $x^{(0)}\in \mathsf{M}$  and  $r\in (0,r_{\alpha}^{\star}]$ , where  $\alpha$  is given by A2. Then, for any  $x\in \mathsf{W}^0 (x^{(0)},r)$ ,  $x\in \mathsf{M}$  and  $\mathrm{Jac}[\mathrm{G}_{h,x^{(0)}}](x)$  is invertible for any  $h > 0$ .
Proof. Let  $(\mathsf{M},\mathfrak{g})$  be a smooth manifold of  $\mathbb{R}^d$  and  $h > 0$ . We first prove the existence and uniqueness of the map defined in (32). Let  $(x^{(0)},x^{(1)})\in \mathsf{M}\times \mathsf{M}$ . We define for any  $y\in \mathsf{M}$
$$
p ^ {(1 / 2)} (y) = \frac {2}{h} (\mathfrak {g} (x ^ {(0)}) ^ {- 1} + \mathfrak {g} (y) ^ {- 1}) ^ {- 1} (y - x ^ {(0)}), \qquad \tilde {p} (y) = p ^ {(1 / 2)} (y) + \frac {h}{2} \partial_ {x} H _ {2} (x ^ {(0)}, p ^ {(1 / 2)} (y)).
$$
Note that these expressions are obtained by simply inverting the first two aligns of (6). Then, by definition of  $\mathrm{G}_h$ , the vector  $p \in \mathrm{T}_{x^{(0)}}^{\star} \mathsf{M}$  such that  $(x^{(1)}, p^{(1)}) \in \mathrm{G}_h(x^{(0)}, p)$  for any  $p^{(1)} \in \mathrm{T}_{x^{(1)}}^{\star} \mathsf{M}$  is uniquely determined by  $p = \tilde{p}(x^{(1)})$ . We thus obtain (32), using the expression of  $\partial_x H_2$  given in Section 3.1. Moreover, it is clear that  $\mathrm{G}_{h,x^{(0)}}$  is continuously differentiable if  $\mathfrak{g} \in \mathrm{C}^2(\mathsf{M}, \mathbb{R}^{d \times d})$ .
We are now going to prove the second result of Lemma 22. Assume A1, A2. Let  $x^{(0)} \in \mathsf{M}$  and  $r \in (0, r_{\alpha}^{\star}]$ , where  $r_{\alpha}^{\star}$  is defined in (33) and  $\alpha$  is given by A2. In particular  $r_{\alpha}^{\star} \leq 1/11$ . For the sake of clarity, we will now denote  $\mathfrak{g}(x^{(0)})$  by  $\mathfrak{g}_0$  for the rest of the proof.
We define the open Dikin ellipsoid  $\mathsf{B}_0 = \mathsf{W}^0 (x^{(0)},r)$ . Since  $r < 1$ ,  $\mathsf{B}_0\subset \mathsf{M}$  by Lemma 9-(b). Let  $h > 0$ . We now define the following maps on  $\mathsf{B}_0$
$$
\mathrm {M} _ {0} = \left\{ \begin{array}{l l} \mathsf {B} _ {0} & \to S _ {d} ^ {+ +} (\mathbb {R}) \\ y & \mapsto \mathfrak {g} _ {0} ^ {- 1} + \mathfrak {g} (y) ^ {- 1} \end{array} \right., \qquad \mathrm {K} _ {0} = \left\{ \begin{array}{l l} \mathsf {B} _ {0} & \to \mathbb {R} ^ {d} \\ y & \mapsto \mathfrak {g} _ {0} ^ {- 1} \mathrm {M} _ {0} (y) ^ {- 1} (y - x ^ {(0)}) \end{array} \right.,
$$
$$
\phi_ {0} = \left\{ \begin{array}{l l} \mathsf {B} _ {0} & \to \mathbb {R} ^ {d} \\ y & \mapsto \mathrm {M} _ {0} (y) D \mathfrak {g} _ {0} [ \mathrm {K} _ {0} (y), \mathrm {K} _ {0} (y) ] \end{array} \right., \quad \tilde {\phi} _ {0} = \left\{ \begin{array}{l l} \mathsf {B} _ {0} & \to \mathbb {R} ^ {d} \\ y & \mapsto \frac {h}{2} \mathrm {M} _ {0} (y) \mathrm {G} _ {h, x ^ {(0)}} (y) \end{array} \right..
$$
Note that we have, for any  $y \in \mathsf{B}_0$ ,  $\tilde{\phi}(y) = (y - x^{(0)}) - \frac{1}{2}\phi_0(y)$ . Since  $\mathfrak{g}$  is twice continuously differentiable on  $\mathsf{B}_0$ , the maps previously defined are continuously differentiable. We now compute the derivatives of these maps. Let  $y \in \mathsf{B}_0$  and  $w \in \mathbb{R}^d$ . We have
$$
D M _ {0} (y) [ w ] = - \mathfrak {g} (y) ^ {- 1} D \mathfrak {g} (y) [ w ] \mathfrak {g} (y) ^ {- 1},
$$
$$
\begin{array}{l} D K _ {0} (y) [ w ] = - \mathfrak {g} _ {0} ^ {- 1} \mathrm {M} _ {0} (y) ^ {- 1} D \mathrm {M} _ {0} (y) [ w ] \mathrm {M} _ {0} (y) ^ {- 1} (y - x ^ {(0)}) + \mathfrak {g} _ {0} ^ {- 1} \mathrm {M} _ {0} (y) ^ {- 1} w \\ = - \mathfrak {g} _ {0} ^ {- 1} \left\{\mathfrak {g} (y) \mathfrak {g} _ {0} ^ {- 1} + \mathrm {I} _ {d} \right\} ^ {- 1} \mathfrak {g} (y) D \mathrm {M} _ {0} (y) [ w ] \mathfrak {g} (y) \left\{\mathfrak {g} _ {0} ^ {- 1} \mathfrak {g} (y) + \mathrm {I} _ {d} \right\} ^ {- 1} (y - x ^ {(0)}) \\ + \mathfrak {g} _ {0} ^ {- 1} \mathrm {M} _ {0} (y) ^ {- 1} w \\ = \mathfrak {g} _ {0} ^ {- 1} \left\{\mathfrak {g} (y) \mathfrak {g} _ {0} ^ {- 1} + \mathrm {I} _ {d} \right\} ^ {- 1} D \mathfrak {g} (y) [ w ] \left\{\mathfrak {g} _ {0} ^ {- 1} \mathfrak {g} (y) + \mathrm {I} _ {d} \right\} ^ {- 1} \left(y - x ^ {(0)}\right) + \mathfrak {g} _ {0} ^ {- 1} \mathrm {M} _ {0} (y) ^ {- 1} w , \\ \end{array}
$$
$$
\begin{array}{l} D \phi_ {0} (y) [ w ] = D M _ {0} (y) [ w ] D \mathfrak {g} _ {0} [ K _ {0} (y), K _ {0} (y) ] + M _ {0} (y) D \mathfrak {g} _ {0} [ D K _ {0} (y) [ w ], K _ {0} (y) ] \\ + \mathrm {M} _ {0} (y) D \mathfrak {g} _ {0} \left[ \mathrm {K} _ {0} (y), D \mathrm {K} _ {0} (y) [ w ] \right], \\ \end{array}
$$
$$
D \tilde {\phi} _ {0} (y) [ w ] = I _ {d} - \frac {1}{2} D \phi_ {0} (y) [ w ].
$$
We have in particular  $\tilde{\phi}_0(x^{(0)}) = 0$  and  $D\phi_0(x^{(0)}) = 0$ . Let us first study the smoothness of  $D\phi_0$  on  $\mathsf{B}_0$ .
Smoothness of  $D\phi_0$ . We prove here that  $D\phi_0$  is  $(r_{\alpha}^{\star})^{-1}$ -Lipschitz-continuous on  $\mathsf{B}_0$  with respect to  $\| \cdot \|_{\mathfrak{g}_0}$ . Let  $(y, y') \in \mathsf{B}_0$  and  $w \in \mathbb{R}^d$ . We have
$$
\begin{array}{l} \left\| D \phi_ {0} (y) [ w ] - D \phi_ {0} \left(y ^ {\prime}\right) [ w ] \right\| _ {\mathfrak {g} _ {0}} \tag {34} \\ \leq \left\| D M _ {0} (y) [ w ] D \mathfrak {g} _ {0} \left[ K _ {0} (y), K _ {0} (y) \right] - D M _ {0} (y ^ {\prime}) [ w ] D \mathfrak {g} _ {0} \left[ K _ {0} (y ^ {\prime}), K _ {0} (y ^ {\prime}) \right] \right\| _ {\mathfrak {g} _ {0}} \\ + \left\| \mathrm {M} _ {0} (y) D \mathfrak {g} _ {0} [ D K _ {0} (y) [ w ], K _ {0} (y) ] - \mathrm {M} _ {0} (y ^ {\prime}) D \mathfrak {g} _ {0} [ D K _ {0} (y ^ {\prime}) [ w ], K _ {0} (y ^ {\prime}) ] \right\| _ {\mathfrak {g} _ {0}} \\ + \left\| \mathrm {M} _ {0} (y) D \mathfrak {g} _ {0} \left[ \mathrm {K} _ {0} (y), D \mathrm {K} _ {0} (y) [ w ] \right] - \mathrm {M} _ {0} \left(y ^ {\prime}\right) D \mathfrak {g} _ {0} \left[ \mathrm {K} _ {0} \left(y ^ {\prime}\right), D \mathrm {K} _ {0} \left(y ^ {\prime}\right) [ w ] \right] \right\| _ {\mathfrak {g} _ {0}}. \\ \end{array}
$$
Remark that the two last terms in (34) are very similar and can be bounded in the same way. The rest of the proof on the Lipschitz-continuity of  $D\phi_0$  is separated in two steps, each of them consisting of bounding from above a term of (34).
Step 1. Let us first bound the first term in (34). For any  $x \in \mathsf{B}_0$ , we denote  $\overline{\mathsf{DM}_0(x)}[w]D\mathfrak{g}_0[\mathrm{K}_0(x),\mathrm{K}_0(x)]$  by  $a_1(x)$ . We have
$$
\begin{array}{l} \mathfrak {g} _ {0} ^ {1 / 2} \left(a _ {1} (y) - a _ {1} \left(y ^ {\prime}\right)\right) \tag {35} \\ = \left\{\mathfrak {g} _ {0} ^ {1 / 2} \left(D M _ {0} (y) [ w ] - D M _ {0} (y ^ {\prime}) [ w ]\right) \mathfrak {g} _ {0} ^ {1 / 2} \right\} \left\{\mathfrak {g} _ {0} ^ {- 1 / 2} D \mathfrak {g} _ {0} \left[ K _ {0} (y), K _ {0} (y) \right] \right\} \\ + \left\{\mathfrak {g} _ {0} ^ {1 / 2} D M _ {0} (y ^ {\prime}) [ w ] \mathfrak {g} _ {0} ^ {1 / 2} \right\} \left\{\mathfrak {g} _ {0} ^ {- 1 / 2} \left(D \mathfrak {g} _ {0} [ K _ {0} (y), K _ {0} (y) ] - D \mathfrak {g} _ {0} [ K _ {0} (y ^ {\prime}), K _ {0} (y ^ {\prime}) ]\right) \right\}. \\ \end{array}
$$
We now aim to bound each one of the four terms that appear in (35).
Step 1.1. First, we have
$$
\begin{array}{l} \mathfrak {g} _ {0} ^ {1 / 2} \left(D M _ {0} (y) [ w ] - D M _ {0} \left(y ^ {\prime}\right) [ w ]\right) \mathfrak {g} _ {0} ^ {1 / 2} = \mathfrak {g} _ {0} ^ {1 / 2} \left(\mathfrak {g} \left(y ^ {\prime}\right) ^ {- 1} D \mathfrak {g} \left(y ^ {\prime}\right) [ w ] \mathfrak {g} \left(y ^ {\prime}\right) ^ {- 1} - \mathfrak {g} (y) ^ {- 1} D \mathfrak {g} (y) [ w ] \mathfrak {g} (y) ^ {- 1}\right) \mathfrak {g} _ {0} ^ {1 / 2} \\ = \mathfrak {g} _ {0} ^ {1 / 2} \{\mathfrak {g} (y ^ {\prime}) ^ {- 1} - \mathfrak {g} (y) ^ {- 1} \} D \mathfrak {g} (y ^ {\prime}) [ w ] \mathfrak {g} (y ^ {\prime}) ^ {- 1} \mathfrak {g} _ {0} ^ {1 / 2} \\ + \mathfrak {g} _ {0} ^ {1 / 2} \mathfrak {g} (y) ^ {- 1} \left\{D \mathfrak {g} (y ^ {\prime}) [ w ] - D \mathfrak {g} (y) [ w ] \right\} \mathfrak {g} (y ^ {\prime}) ^ {- 1} \mathfrak {g} _ {0} ^ {1 / 2} \\ + \mathfrak {g} _ {0} ^ {1 / 2} \mathfrak {g} (y) ^ {- 1} D \mathfrak {g} (y) [ w ] \{\mathfrak {g} (y ^ {\prime}) ^ {- 1} - \mathfrak {g} (y) ^ {- 1} \} \mathfrak {g} _ {0} ^ {1 / 2}. \\ \end{array}
$$
We recall that  $r \leq r_{\alpha}^{\star} \leq 1/11$ , and thus, we have for any  $x \in \{y, y'\}$  the following inequalities
$$
\begin{array}{l} \left\| \mathfrak {g} _ {0} ^ {1 / 2} \left\{\mathfrak {g} \left(y ^ {\prime}\right) ^ {- 1} - \mathfrak {g} (y) ^ {- 1} \right\} \mathfrak {g} _ {0} ^ {1 / 2} \right\| _ {2} \leq 3 (1 - r) ^ {- 3} \| y - y ^ {\prime} \| _ {\mathfrak {g} _ {0}}, \quad \text {(o n t h e m o d e l o f (1 8))} \\ \left\| \mathfrak {g} _ {0} ^ {1 / 2} \mathfrak {g} (x) ^ {- 1} \mathfrak {g} _ {0} ^ {1 / 2} \right\| _ {2} \leq (1 - r) ^ {- 2}, \tag {Lemma 9-(b)} \\ \| \mathfrak {g} _ {0} ^ {- 1 / 2} D \mathfrak {g} (x) [ w ] \mathfrak {g} _ {0} ^ {- 1 / 2} \| _ {2} \leq (1 - r) ^ {- 2} \| \mathfrak {g} (x) ^ {- 1 / 2} D \mathfrak {g} (x) [ w ] \mathfrak {g} (x) ^ {- 1 / 2} \| _ {2}. \\ \end{array}
$$
In particular, we have for any  $x \in \{y, y'\}$
$$
\begin{array}{l} \| \mathfrak {g} (x) ^ {- 1 / 2} D \mathfrak {g} (x) [ w ] \mathfrak {g} (x) ^ {- 1 / 2} \| _ {2} = \sup  _ {\left\{w ^ {\prime} \in \mathbb {R} ^ {d}: \| w ^ {\prime} \| _ {2} = 1 \right\}} \| D \mathfrak {g} (x) [ w, \mathfrak {g} (x) ^ {- 1 / 2} w ^ {\prime} ] \| _ {\mathfrak {g} (x) ^ {- 1}} (36) \\ \leq 2 \| w \| _ {\mathfrak {g} (x)} \sup  _ {\left\{w ^ {\prime} \in \mathbb {R} ^ {d}: \| w ^ {\prime} \| _ {2} = 1 \right\}} \| \mathfrak {g} (x) ^ {- 1 / 2} w ^ {\prime} \| _ {\mathfrak {g} (x)} \quad (\text {L e m m a 9 - (a)}) \\ \leq 2 (1 - r) ^ {- 1} \| w \| _ {\mathfrak {g} _ {0}}, (Lemma 9-(c)) \\ \end{array}
$$
and using again that  $r \leq r_{\alpha}^{\star} \leq 1/11$ , we have
$$
\begin{array}{l} \left\| \mathfrak {g} (y) ^ {- 1 / 2} \left\{D \mathfrak {g} \left(y ^ {\prime}\right) [ w ] - D \mathfrak {g} (y) [ w ] \right\} \mathfrak {g} \left(y ^ {\prime}\right) ^ {- 1 / 2} \right\| _ {2} (37) \\ = \sup  _ {\left\{w ^ {\prime} \in \mathbb {R} ^ {d}: \| w ^ {\prime} \| _ {2} = 1 \right\}} \| D \mathfrak {g} (y ^ {\prime}) [ w, \mathfrak {g} (y ^ {\prime}) ^ {- 1 / 2} w ^ {\prime} ] - D \mathfrak {g} (y) [ w, \mathfrak {g} (y ^ {\prime}) ^ {- 1 / 2} w ^ {\prime} ] \} \| _ {\mathfrak {g} (y) ^ {- 1}} \\ \leq \alpha (\alpha + 1) / 3 \| w \| _ {\mathfrak {g} (y)} \left(\left(1 - \| y - y ^ {\prime} \| _ {\mathfrak {g} (y)}\right) ^ {- 3} - 1\right) (Lemma 11-(c)) \\ \leq 5 \alpha (\alpha + 1) / 3 (1 - r) ^ {- 2} \| w \| _ {\mathfrak {g} _ {0}} \| y - y ^ {\prime} \| _ {\mathfrak {g} _ {0}}. \quad \left(\text {o n t h e m o d e l o f (2 2)}\right) \\ \end{array}
$$
Thus, by using (36) and (37), we have
$$
\| \mathfrak {g} _ {0} ^ {1 / 2} (D M _ {0} (y) [ w ] - D M _ {0} (y ^ {\prime}) [ w ]) \mathfrak {g} _ {0} ^ {1 / 2} \| _ {2} \leq (1 2 (1 - r) ^ {- 8} + 5 \alpha (\alpha + 1) / 3 (1 - r) ^ {- 6}) \| y - y ^ {\prime} \| _ {\mathfrak {g} _ {0}} \| w \| _ {\mathfrak {g} _ {0}}.
$$
Step 1.2. Secondly, we have
$$
\begin{array}{l} \left\| \mathfrak {g} _ {0} ^ {- 1 / 2} D \mathfrak {g} _ {0} \left[ \mathrm {K} _ {0} (y), \mathrm {K} _ {0} (y) \right] \right\| _ {2} \leq 2 \| \mathrm {K} _ {0} (y) \| _ {\mathfrak {g} _ {0}} ^ {2} \tag {Lemma 9-(a)} \\ \leq 2 \| \mathfrak {g} _ {0} ^ {- 1 / 2} \mathrm {M} _ {0} (y) ^ {- 1} (y - x ^ {(0)}) \| _ {2} ^ {2} \quad \text {(d e f i n i t i o n} \mathrm {K} _ {0}) \\ \leq 2 \| \mathfrak {g} _ {0} ^ {- 1 / 2} \mathrm {M} _ {0} (y) ^ {- 1} \mathfrak {g} _ {0} ^ {- 1 / 2} \| _ {2} ^ {2} \| y - x ^ {(0)} \| _ {\mathfrak {g} _ {0}} ^ {2} \\ \leq 2 r ^ {2} \left\| \mathfrak {g} _ {0} ^ {- 1 / 2} \mathrm {M} _ {0} (y) ^ {- 1} \mathfrak {g} _ {0} ^ {- 1 / 2} \right\| _ {2} ^ {2}, \\ \end{array}
$$
where we have by Lemma 9-(b)
$$
\mathfrak {g} _ {0} ^ {- 1 / 2} \mathrm {M} _ {0} (y) ^ {- 1} \mathfrak {g} _ {0} ^ {- 1 / 2} = \left(\mathrm {I} _ {d} + \mathfrak {g} _ {0} ^ {1 / 2} \mathfrak {g} (y) ^ {- 1} \mathfrak {g} _ {0} ^ {1 / 2}\right) ^ {- 1} \preceq (1 + (1 - r) ^ {2}) ^ {- 1} \mathrm {I} _ {d}. \tag {38}
$$
$$
\left\| \mathfrak {g} _ {0} ^ {- 1 / 2} D \mathfrak {g} _ {0} \left[ K _ {0} (y), K _ {0} (y) \right] \right\| _ {2} \leq 2 r ^ {2} (1 + (1 - r) ^ {2}) ^ {- 2}. \tag {39}
$$
Step 1.3. Thirdly, we have
$$
\begin{array}{l} \left\| \mathfrak {g} _ {0} ^ {1 / 2} D M _ {0} (y ^ {\prime}) [ w ] \mathfrak {g} _ {0} ^ {1 / 2} \right\| _ {2} \leq (1 - r) ^ {- 2} \left\| \mathfrak {g} (x) ^ {- 1 / 2} D \mathfrak {g} (y ^ {\prime}) [ w ] \mathfrak {g} (x) ^ {- 1 / 2} \right\| _ {2} (40) \\ \leq 2 (1 - r) ^ {- 3} \| w \| _ {\mathfrak {g} _ {0}}. (using(36)) \\ \end{array}
$$
Step 1.4. Fourthly, we have
$$
\begin{array}{l} \left\| \mathfrak {g} _ {0} ^ {- 1 / 2} \left(D \mathfrak {g} _ {0} \left[ \mathrm {K} _ {0} (y), \mathrm {K} _ {0} (y) \right] - D \mathfrak {g} _ {0} \left[ \mathrm {K} _ {0} \left(y ^ {\prime}\right), \mathrm {K} _ {0} \left(y ^ {\prime}\right) \right]\right) \| _ {2} \right. (41) \\ \leq 2 \| D \mathfrak {g} _ {0} [ K _ {0} (y) - K _ {0} (y ^ {\prime}), K _ {0} (y) ] \| _ {\mathfrak {g} _ {0} ^ {- 1}} + \| D \mathfrak {g} _ {0} [ K _ {0} (y) - K _ {0} (y ^ {\prime}), K _ {0} (y) - K _ {0} (y ^ {\prime}) ] \| _ {\mathfrak {g} _ {0} ^ {- 1}} \\ \leq 2 \| \mathrm {K} _ {0} (y) - \mathrm {K} _ {0} \left(y ^ {\prime}\right) \| _ {\mathfrak {g} _ {0}} \| \mathrm {K} _ {0} (y) \| _ {\mathfrak {g} _ {0}} + \| \mathrm {K} _ {0} (y) - \mathrm {K} _ {0} \left(y ^ {\prime}\right) \| _ {\mathfrak {g} _ {0}} \| \mathrm {K} _ {0} (y) - \mathrm {K} _ {0} \left(y ^ {\prime}\right) \| _ {\mathfrak {g} _ {0}} (Lemma9-(a)) \\ \leq 4 \| \mathrm {M} _ {0} (y) ^ {- 1} (y - x ^ {(0)}) - \mathrm {M} _ {0} (y ^ {\prime}) ^ {- 1} (y ^ {\prime} - x ^ {(0)}) \| _ {\mathfrak {g} _ {0} ^ {- 1}} \| \mathrm {M} _ {0} (y) ^ {- 1} (y - x ^ {(0)}) \| _ {\mathfrak {g} _ {0} ^ {- 1}} \\ + 2 \left\| \mathrm {M} _ {0} (y) ^ {- 1} \left(y - x ^ {(0)}\right) - \mathrm {M} _ {0} \left(y ^ {\prime}\right) ^ {- 1} \left(y ^ {\prime} - x ^ {(0)}\right) \right\| _ {\mathfrak {g} _ {0} ^ {- 1}} ^ {2}. \\ \end{array}
$$
In particular,  $\mathrm{M}_0(y)^{-1}(y - x^{(0)}) - \mathrm{M}_0(y')^{-1}(y' - x^{(0)}) = \mathrm{M}_0(y)^{-1}(y - y') + (\mathrm{M}_0(y)^{-1} - \mathrm{M}_0(y')^{-1})(y' - x^{(0)})$ , and thus we have
$$
\begin{array}{l} \left\| \mathrm {M} _ {0} (y) ^ {- 1} \left(y - x ^ {(0)}\right) - \mathrm {M} _ {0} \left(y ^ {\prime}\right) ^ {- 1} \left(y ^ {\prime} - x ^ {(0)}\right) \right\| _ {\mathfrak {g} _ {0} ^ {- 1}} (42) \\ \leq \left\| \mathfrak {g} _ {0} ^ {- 1 / 2} \mathrm {M} _ {0} (y) ^ {- 1} \mathfrak {g} _ {0} ^ {- 1 / 2} \right\| _ {2} \| y - y ^ {\prime} \| _ {\mathfrak {g} _ {0}} + r \| \mathfrak {g} _ {0} ^ {- 1 / 2} (\mathrm {M} _ {0} (y) ^ {- 1} - \mathrm {M} _ {0} (y ^ {\prime}) ^ {- 1}) \mathfrak {g} _ {0} ^ {- 1 / 2} \| _ {2} \\ \leq \left(1 + (1 - r) ^ {2}\right) ^ {- 1} \| y - y ^ {\prime} \| _ {\mathfrak {g} _ {0}} + r \| \mathfrak {g} _ {0} ^ {- 1 / 2} \left(\mathrm {M} _ {0} \left(y ^ {\prime}\right) ^ {- 1} - \mathrm {M} _ {0} \left(y ^ {\prime}\right) ^ {- 1}\right) \mathfrak {g} _ {0} ^ {- 1 / 2} \| _ {2}. (using(38)) \\ \end{array}
$$
We now aim to find an upper bound for  $\| \mathfrak{g}_0^{-1 / 2}(\mathrm{M}_0(y)^{-1} - \mathrm{M}_0(y')^{-1})\mathfrak{g}_0^{-1 / 2}\| _2$ . We have
$$
\begin{array}{l} \mathfrak {g} _ {0} ^ {- 1 / 2} \left(\mathrm {M} _ {0} (y) ^ {- 1} - \mathrm {M} _ {0} \left(y ^ {\prime}\right) ^ {- 1}\right) \mathfrak {g} _ {0} ^ {- 1 / 2} \\ = \left(\mathrm {I} _ {d} + \mathfrak {g} _ {0} ^ {1 / 2} \mathfrak {g} (y) ^ {- 1} \mathfrak {g} _ {0} ^ {1 / 2}\right) ^ {- 1} - \left(\mathrm {I} _ {d} + \mathfrak {g} _ {0} ^ {1 / 2} \mathfrak {g} \left(y ^ {\prime}\right) ^ {- 1} \mathfrak {g} _ {0} ^ {1 / 2}\right) ^ {- 1} \\ = \left(\mathrm {I} _ {d} + \mathfrak {g} _ {0} ^ {1 / 2} \mathfrak {g} (y ^ {\prime}) ^ {- 1} \mathfrak {g} _ {0} ^ {1 / 2} + \mathfrak {g} _ {0} ^ {1 / 2} \{\mathfrak {g} (y) ^ {- 1} - \mathfrak {g} (y ^ {\prime}) ^ {- 1} \} \mathfrak {g} _ {0} ^ {1 / 2}\right) ^ {- 1} - \left(\mathrm {I} _ {d} + \mathfrak {g} _ {0} ^ {1 / 2} \mathfrak {g} (y ^ {\prime}) ^ {- 1} \mathfrak {g} _ {0} ^ {1 / 2}\right) ^ {- 1} \\ = \mathrm {B} (y ^ {\prime}) ^ {- 1 / 2} \left[ \left(\mathrm {I} _ {d} + \mathrm {B} (y ^ {\prime}) ^ {- 1 / 2} \mathfrak {g} _ {0} ^ {1 / 2} \{\mathfrak {g} (y) ^ {- 1} - \mathfrak {g} (y ^ {\prime}) ^ {- 1} \} \mathfrak {g} _ {0} ^ {1 / 2} \mathrm {B} (y ^ {\prime}) ^ {- 1 / 2}\right) ^ {- 1} - \mathrm {I} _ {d} \right] \mathrm {B} (y ^ {\prime}) ^ {- 1 / 2}, \\ \end{array}
$$
where  $\mathrm{B}(y^{\prime}) = \mathrm{I}_d + \mathfrak{g}_0^{1 / 2}\mathfrak{g}(y^{\prime})^{-1}\mathfrak{g}_0^{1 / 2}$ . In particular, we have
$$
\left(1 + (1 - r) ^ {- 2}\right) ^ {- 1 / 2} \mathrm {I} _ {d} \preceq \mathrm {B} \left(y ^ {\prime}\right) ^ {- 1 / 2} \preceq \left(1 + (1 - r) ^ {2}\right) ^ {- 1 / 2} \mathrm {I} _ {d},
$$
by Lemma 9-(b). Note that (17) still holds, where  $x, x'$  and  $\bar{r}$  are respectively replaced by  $y, y'$  and  $r$ . Thus, on the model on (18), we have
$$
\{(1 - \| y ^ {\prime} - y \| _ {\mathfrak {g} (y)}) ^ {2} - 1 \} (1 - r) ^ {2} \mathrm {I} _ {d} \preceq \mathfrak {g} _ {0} ^ {1 / 2} \{\mathfrak {g} (y) ^ {- 1} - \mathfrak {g} (y ^ {\prime}) ^ {- 1} \} \mathfrak {g} _ {0} ^ {1 / 2} \preceq \{(1 - \| y ^ {\prime} - y \| _ {\mathfrak {g} (y)}) ^ {- 2} - 1 \} (1 - r) ^ {- 2} \mathrm {I} _ {d},
$$
$$
\begin{array}{l} (1 + \{(1 - \| y ^ {\prime} - y \| _ {\mathfrak {g} (y)}) ^ {2} - 1 \} (1 - r) ^ {2} (1 + (1 - r) ^ {- 2}) ^ {- 1}) I _ {d} \\ \preceq \mathrm {I} _ {d} + \mathrm {B} (y ^ {\prime}) ^ {- 1 / 2} \mathfrak {g} _ {0} ^ {1 / 2} \left\{\mathfrak {g} (y) ^ {- 1} - \mathfrak {g} (y ^ {\prime}) ^ {- 1} \right\} \mathfrak {g} _ {0} ^ {1 / 2} \mathrm {B} (y ^ {\prime}) ^ {- 1 / 2} \\ \preceq \left(1 + \left\{\left(1 - \| y ^ {\prime} - y \| _ {\mathfrak {g} (y)}\right) ^ {- 2} - 1 \right\} (1 - r) ^ {- 2} \left(1 + (1 - r) ^ {2}\right) ^ {- 1}\right) I _ {d}. \\ \end{array}
$$
Therefore, we have  $\| \mathfrak{g}_0^{-1 / 2}(\mathrm{M}_0(y')^{-1} - \mathrm{M}_0(y')^{-1})\mathfrak{g}_0^{-1 / 2}\| _2\leq (1 + (1 - r)^2)^{-1}\max \big(|f_1(r)|,f_2(r)\big)$  where
$$
f _ {1} (r) = \left(1 + \left\{\left(1 - \| y ^ {\prime} - y \| _ {\mathfrak {g} (y)}\right) ^ {- 2} - 1 \right\} (1 - r) ^ {- 2} \left(1 + (1 - r) ^ {2}\right) ^ {- 1}\right) ^ {- 1} - 1,
$$
$$
f _ {2} (r) = \left(1 + \left\{\left(1 - \| y ^ {\prime} - y \| _ {\mathfrak {g} (y)}\right) ^ {2} - 1 \right\} (1 - r) ^ {2} \left(1 + (1 - r) ^ {- 2}\right) ^ {- 1}\right) ^ {- 1} - 1.
$$
We now aim to control the upper bound  $\max (|f_1(r)|,f_2(r))$
(a) We first bound  $|f_1(r)|$ . Since  $\| y' - y \|_{\mathfrak{g}(y)} \geq 0$ , it is clear that  $f_1(r) \leq 0$ . Using Inequality (a) with  $u = \| y' - y \|_{\mathfrak{g}(y)}$ , where  $u \leq 1/5$  by (17), we obtain
$$
\begin{array}{l} \left| f _ {1} (r) \right| = - f _ {1} (r) \\ \leq 1 - (1 + 3 (1 - r) ^ {- 2} (1 + (1 - r) ^ {2}) ^ {- 1} \| y ^ {\prime} - y \| _ {\mathfrak {g} (y)}) ^ {- 1} \\ \leq 3 (1 - r) ^ {- 2} \left(1 + (1 - r) ^ {2}\right) ^ {- 1} \| y ^ {\prime} - y \| _ {\mathfrak {g} (y)} \quad \text {(I n e q u a l i t y (d))} \\ \leq 3 (1 - r) ^ {- 3} \left(1 + (1 - r) ^ {2}\right) ^ {- 1} \| y ^ {\prime} - y \| _ {\mathfrak {g} _ {0}}. \tag {Lemma9-(c)} \\ \end{array}
$$
(b) We now bound  $f_{2}(r)$ . We have
$$
\begin{array}{l} f _ {2} (r) \leq (1 - 2 (1 - r) ^ {2} (1 + (1 - r) ^ {- 2}) ^ {- 1} \| y ^ {\prime} - y \| _ {\mathfrak {g} (y)}) ^ {- 1} - 1 \\ \leq 4 (1 - r) ^ {2} \left(1 + (1 - r) ^ {- 2}\right) ^ {- 1} \| y ^ {\prime} - y \| _ {\mathfrak {g} (y)} \\ \leq 4 (1 - r) \left(1 + (1 - r) ^ {- 2}\right) ^ {- 1} \| y ^ {\prime} - y \| _ {\mathfrak {g} _ {0}}, \tag {Lemma 9-(c)} \\ \end{array}
$$
where we used (i) Inequality (d) in the first line with  $u = \| y' - y\|_{\mathfrak{g}(y)}$  and (ii) Inequality (b) in the second line with  $u = 2(1 - r)^2 (1 + (1 - r)^{-2})^{-1}\| y' - y\|_{\mathfrak{g}(y)}\leq 4r(1 - r)(1 + (1 - r)^{-2})^{-1}\leq 1 / 2$  with  $r\leq 1 / 11$ .
We recall that  $r \in (0,1)$ , and thus, we have respectively  $(1 - r)^{-3} \geq (1 - r)$  and  $(1 + (1 - r)^{2})^{-1} \geq (1 + (1 - r)^{-2})^{-1}$ . Therefore,  $\max(|f_1(r)|, f_2(r)) \leq (4/3)|f_1(r)|$  and
$$
\begin{array}{l} \left\| \mathfrak {g} _ {0} ^ {- 1 / 2} \left(\mathrm {M} _ {0} \left(y ^ {\prime}\right) ^ {- 1} - \mathrm {M} _ {0} \left(y ^ {\prime}\right) ^ {- 1}\right) \mathfrak {g} _ {0} ^ {- 1 / 2} \right\| _ {2} \leq 4 (1 - r) ^ {- 3} \left(1 + (1 - r) ^ {2}\right) ^ {- 2} \| y ^ {\prime} - y \| _ {\mathfrak {g} _ {0}}, \tag {43} \\ \| \mathrm {K} _ {0} (y) - \mathrm {K} _ {0} (y ^ {\prime}) \| _ {\mathfrak {g} _ {0}} \leq (1 + (1 - r) ^ {2}) ^ {- 1} \{1 + 4 r (1 - r) ^ {- 3} (1 + (1 - r) ^ {2}) ^ {- 1} \} \| y ^ {\prime} - y \| _ {\mathfrak {g} _ {0}}. \\ \end{array}
$$
By combining (41), (42) and (43), we finally have