A The continuous case: Proof of Theorem 5
Our analysis is based on the following Lyapunov energy:
$$\begin{aligned} {\mathcal {E}}(t)=t^2(F(x(t)-F^*)+\frac{1}{2}\left\| {\lambda (x(t)-x^*)+t{\dot{x}}(t)}\right\| ^2,\quad \lambda =\frac{2\alpha }{\gamma +2} \end{aligned}$$
(78)
where the parameter \(\lambda \) is chosen accordingly to Aujol et al. [6]. Remember that the expected asymptotic convergence rate is polynomial in \(\mathcal O\left( t^{-\frac{2\alpha \gamma }{\gamma +2}}\right) \) [6] with an exponent equal to \(\lambda \gamma \). Differentiating the Lyapunov energy \({\mathcal {E}}\), we easily prove that:
$$\begin{aligned} {\mathcal {E}}'(t)+\frac{\gamma \lambda -2}{t}{\mathcal {E}}(t)= & {} \lambda \gamma t\left( F(x(t))-F^* -\frac{1}{\gamma }\langle \nabla F(x(t)),x(t)-x^*\rangle \right) \\{} & {} +\,\frac{\lambda ^2(\gamma \lambda -2)}{2t}\Vert x(t)-x^*\Vert ^2\\{} & {} + (\lambda ^2(\gamma +1) -\lambda -\alpha \lambda )\langle x(t)-x^*,{\dot{x}}(t)\rangle \\{} & {} +\, t\left( \lambda +1-\alpha + \frac{\gamma \lambda -2}{2}\right) \Vert \dot{x}(t)\Vert ^2. \end{aligned}$$
Using the flatness assumption and replacing \(\lambda =\frac{2\alpha }{\gamma +2}\), we finally get:
$$\begin{aligned} {\mathcal {E}}'(t)+\frac{\gamma \lambda -2}{t}{\mathcal {E}}(t) \leqslant K(\alpha ) \left( \frac{2\alpha }{(\gamma +2)t}\Vert x(t)-x^*\Vert ^2 + \langle x(t)-x^*,{\dot{x}}(t)\rangle \right) \nonumber \\ \end{aligned}$$
(79)
where: \(K(\alpha )=\frac{2\alpha \gamma }{(\gamma +2)^2}(\alpha -1-\frac{2}{\gamma })\). We now need to control the scalar product whose sign is unknown. Combining the following two inequalities:
$$\begin{aligned} |\langle x(t)-x^*,{\dot{x}}(t)\rangle |\leqslant \frac{\sqrt{\mu }}{2}\left\| {x(t)-x^*}\right\| ^2+\frac{1}{2\sqrt{\mu }}\left\| {\dot{x}(t)}\right\| ^2 \end{aligned}$$
(80)
where the coefficients to bound the scalar product \(\sqrt{\mu }\) are chosen to get the tightest control on the energy, and
$$\begin{aligned} t^2\left\| {\dot{x}(t)}\right\| ^2\leqslant & {} \left( 1+\theta \frac{\alpha }{t\sqrt{\mu }}\right) \left\| {\lambda (x(t)-x^*)+t\dot{x}(t)}\right\| ^2+\lambda ^2\left( 1+\frac{t\sqrt{\mu }}{\theta \alpha }\right) \left\| {x(t)-x^*}\right\| ^2\nonumber \\ \end{aligned}$$
(81)
for any \(\theta >0\), we get:
$$\begin{aligned} {\mathcal {E}}'(t)+\frac{\gamma \lambda -2}{t}{\mathcal {E}}(t)\leqslant & {} K(\alpha ) \left[ \frac{\sqrt{\mu }}{2} + \frac{2\alpha }{(\gamma +2)t} \left( 1+\frac{1}{(\gamma +2)\theta }\right) \right. \nonumber \\{} & {} \left. +\,\frac{2\alpha ^2}{(\gamma +2)^2\sqrt{\mu }t^2} \right] \Vert x(t)-x^*\Vert ^2\nonumber \\{} & {} +\, \frac{K(\alpha )}{2\sqrt{\mu }t^2}\left( 1+\theta \frac{\alpha }{t\sqrt{\mu }}\right) \left\| {\lambda (x(t)-x^*)+t\dot{x}(t)}\right\| ^2 \end{aligned}$$
(82)
$$\begin{aligned}\leqslant & {} \frac{2}{\mu }K(\alpha ) \left[ \frac{\sqrt{\mu }}{2} + \frac{2\alpha }{(\gamma +2)t} \left( 1+\frac{1}{(\gamma +2)\theta }\right) \right. \nonumber \\{} & {} \left. +\frac{2\alpha ^2}{(\gamma +2)^2\sqrt{\mu }t^2} \right] (F(x(t))-F^*)\nonumber \\{} & {} +\,\frac{K(\alpha )}{2\sqrt{\mu }t^2}\left( 1+\theta \frac{\alpha }{t\sqrt{\mu }}\right) \left\| {\lambda (x(t)-x^*)+t\dot{x}(t)}\right\| ^2\nonumber \\ \end{aligned}$$
(83)
using the growth condition \({\mathcal {G}}^2_\mu \). We then choose the parameter \(\theta \) to make equal the coefficients before \(\frac{1}{t^3}\) in \(t^2(F(x(t))-F^*)\) and \(\frac{1}{2}\left\| {\lambda (x(t)-x^*)+t\dot{x}(t)}\right\| ^2\), i.e. such that:
$$\begin{aligned} \frac{2}{\mu }\frac{2\alpha }{(\gamma +2)} \left( 1+\frac{1}{(\gamma +2)\theta }\right) = \frac{\theta \alpha }{\mu } \end{aligned}$$
(84)
or equivalently:
$$\begin{aligned} (\gamma +2)^2\theta ^2-4(\gamma +2)\theta -4=0. \end{aligned}$$
(85)
A straightforward computation shows that this last equation has exactly one positive root:
$$\begin{aligned} \theta = \frac{2}{\gamma +2} (1+\sqrt{2}). \end{aligned}$$
(86)
For these choice of parameters, we have:
$$\begin{aligned} {\mathcal {E}}'(t)+\frac{\gamma \lambda -2}{t}{\mathcal {E}}(t)\leqslant \frac{K(\alpha )}{\mu t^2}\left( \sqrt{\mu }+\frac{2\alpha }{(\gamma +2)t}(1+\sqrt{2}) +\frac{4\alpha ^2}{(\gamma +2)^2\sqrt{\mu }t^2}\right) {\mathcal {E}}(t).\nonumber \\ \end{aligned}$$
(87)
Let us now define:
$$\begin{aligned} \varphi (t):=\ \frac{K(\alpha )}{\mu t^2}\left( \sqrt{\mu }+\frac{2\alpha }{(\gamma +2)t}(1+\sqrt{2}) +\frac{4\alpha ^2}{(\gamma +2)^2\sqrt{\mu }t^2}\right) \end{aligned}$$
(88)
and: \(\Phi (t)=\int _{t}^{+\infty }\varphi (x)dx\). We so have:
$$\begin{aligned} \forall t\geqslant t_0,\quad {\mathcal {E}}'(t)+\frac{\gamma \lambda -2}{t}{\mathcal {E}}(t) \leqslant \varphi (t){\mathcal {E}}(t). \end{aligned}$$
Consequently the function \(t\mapsto {\mathcal {E}}(t)t^{\lambda \gamma -2}e^{\Phi (t)}\) is non-increasing, and for any \(t_1\in {\mathbb {R}}\), we get:
$$\begin{aligned} \forall t\geqslant t_1,\quad {\mathcal {E}}(t)\leqslant {\mathcal {E}}(t_1)\left( \frac{t_1}{t}\right) ^{\lambda \gamma -2}e^{\Phi (t_1)-\Phi (t)}. \end{aligned}$$
(89)
A good choice of \(t_1\) is one ensuring a control as tight as possible on the energy \({\mathcal {E}}\). For that, \(t_1\) is chosen such that \(t_1\) minimizes the function \(u\mapsto u^{\lambda \gamma -2}e^{\Phi (u)}\) i.e. such that \(t_1\) satisfies the equation:
$$\begin{aligned} \frac{\lambda \gamma -2}{u}-\varphi (u)=0 \end{aligned}$$
(90)
Noticing that: \(\lambda \gamma -2=\frac{\gamma +2}{\alpha } K(\alpha )\) and simplifying the equation by \(K(\alpha )\), the equation can be rewritten as:
$$\begin{aligned} \frac{\gamma +2}{\alpha u}=\frac{1}{\mu u^2} \left( \sqrt{\mu }+\frac{2\alpha }{(\gamma +2)u}(1+\sqrt{2}) +\frac{4\alpha ^2}{(\gamma +2)^2\sqrt{\mu }u^2}\right) . \end{aligned}$$
(91)
Introducing \(r=(\gamma +2)\frac{\sqrt{\mu }}{\alpha }u\), we finally have to solve:
$$\begin{aligned} r^3-r^2-2(1+\sqrt{2})r-4=0. \end{aligned}$$
(92)
A straightforward computation shows that the polynomial \(r\mapsto r^3-r^2-2(1+\sqrt{2})r-4\) has only one real root: \(r^*\simeq 3\) (for which Python gives us an analytical value).
Defining \(t_1=\frac{\alpha }{(\gamma +2)\sqrt{\mu }}r^*\), the control on the energy is given by:
$$\begin{aligned} \forall t\geqslant t_1,\quad {\mathcal {E}}(t)\leqslant {\mathcal {E}}(\frac{\alpha }{(\gamma +2)\sqrt{\mu }}r^*)\left( \frac{\alpha r^*}{t(\gamma +2)\sqrt{\mu }}\right) ^{\gamma \lambda -2}e^{\Phi (t_1)-\Phi (t)}. \end{aligned}$$
(93)
Observe now that the term \({\mathcal {E}}(\frac{\alpha }{(\gamma +2)\sqrt{\mu }}r^*)\) can be bounded by the mechanical energy of the system:
$$\begin{aligned} E_m(t)=F(x(t))-F^* + \frac{1}{2}\Vert \dot{x}(t)\Vert ^2 \end{aligned}$$
(94)
Note that this energy is non-increasing since: \(E_m'(t) = \langle \nabla F(x(t))+ \ddot{x}(t),\dot{x}(t)\rangle = -\frac{\alpha }{t}\Vert \dot{x}(t)\Vert ^2 \leqslant 0\), hence \(E_m\) is uniformly bounded on \([t_0,+\infty [\). We then have:
$$\begin{aligned} {\mathcal {E}}(t_1)&=t_1^2(F(x(t_1))-F^*)+\frac{1}{2}\left\| {\frac{2\alpha }{\gamma +2}(x(t_1)-x^*)+t_1\dot{x}(t_1))}\right\| ^2\\&= t_1^2(F(x(t_1))-F^*+\frac{1}{2}\left\| {{\dot{x}}(t_1)}\right\| ^2)\\&\quad +\frac{2\alpha ^2}{(\gamma +2)^2}\left\| {x(t_1)-x^*}\right\| ^2+\frac{2\alpha }{\gamma +2}t_1\langle x(t_1)-x^*,\dot{x}(t_1)\rangle \\&=t_1^2E_m(t_1)+\frac{2\alpha ^2}{(\gamma +2)^2}\left\| {x(t_1)-x^*}\right\| ^2+\frac{2\alpha }{\gamma +2}t_1\langle x(t_1)-x^*,\dot{x}(t_1)\rangle \end{aligned}$$
Using again (80) to control the scalar product combined with the quadratic growth condition \({\mathcal {G}}_\mu ^2\), we can prove that:
$$\begin{aligned} 2\langle x(t_1)-x^*,\dot{x}(t_1)\rangle\leqslant & {} \sqrt{\mu } \Vert x(t_1)-x^*\Vert ^2 + \frac{1}{\sqrt{\mu }}\Vert \dot{x}(t_1)\Vert ^2\\\leqslant & {} \frac{2}{\sqrt{\mu }}(F(x(t_1))-F^*)+ \frac{1}{\sqrt{\mu }}\Vert \dot{x}(t_1)\Vert ^2 = \frac{2}{\sqrt{\mu }}E_m(t_1) \end{aligned}$$
Noticing that the quadratic growth condition also implies:
$$\begin{aligned} \left\| {x(t_1)-x^*}\right\| ^2\leqslant \frac{2}{\mu }(F(x(t_1))-F^*) \leqslant \frac{2}{\mu }E_m(t_1) \end{aligned}$$
and remembering that \(t_1=\frac{\alpha }{(\gamma +2)\sqrt{\mu }}r^*\), we finally get:
$$\begin{aligned} {\mathcal {E}}(t_1)&\leqslant t_1^2E_m(t_1) + \frac{2\alpha ^2}{(\gamma +2)^2}\left\| {x(t_1)-x^*}\right\| ^2 + \frac{2\alpha }{(\gamma +2)\sqrt{\mu }}t_1E_m(t_1) \end{aligned}$$
(95)
$$\begin{aligned}&\leqslant \left[ t_1^2 + \frac{4\alpha ^2}{(\gamma +2)^2\mu } + \frac{2\alpha }{(\gamma +2)\sqrt{\mu }}t_1\right] E_m(t_1) =\left( 1+\frac{2}{r^{*}}+\frac{4}{r^{*2}}\right) t_1^2E_m(t_1) \end{aligned}$$
(96)
$$\begin{aligned}&\leqslant \left( 1+\frac{2}{r^{*}}+\frac{4}{r^{*2}}\right) t_1^2E_m(t_0) \end{aligned}$$
(97)
Observe that the primitive \(\Phi (t)=\int _{t}^{+\infty }\varphi (x)dx\) of \(\varphi \) has a simple analytic expression showing that \(\Phi \) is non-positive and:
$$\begin{aligned} \Phi (t_1)=(\gamma +2)\frac{K(\alpha )}{\alpha }\left( \frac{1}{r^*} + \frac{1+\sqrt{2}}{r^{*2}}+\frac{4}{3r^{*3}}\right) \end{aligned}$$
(98)
We finally obtain the following control on the values:
$$\begin{aligned} F(x(t))-F^*\leqslant C_1E_m(t_0)\left( \frac{\alpha r^*}{t(\gamma +2)\sqrt{\mu }}\right) ^{\frac{2\alpha \gamma }{\gamma +2}}e^{\frac{2\gamma }{\gamma +2}C_2(\alpha -1-\frac{2}{\gamma })} \end{aligned}$$
(99)
where \( C_1 = 1+\frac{2}{r^{*}}+\frac{4}{r^{*2}},~C_2 =\frac{1}{r^*} + \frac{1+\sqrt{2}}{r^{*2}}+\frac{4}{3r^{*3}}. \)
B Technical Lemmas for Theorem 6
The proof of Theorem 6 is based on the following Lyapunov energy:
$$\begin{aligned} E_n = 2sn^2(F(x_n)-F^*) + \Vert \lambda (x_{n-1}-x^*)+n(x_n-x_{n-1})\Vert ^2 \end{aligned}$$
(100)
which can be rewritten as:
$$\begin{aligned} E_n = n^2w_n + \left( \lambda ^2-\lambda n\right) h_{n-1} + \left( n^2-\lambda n\right) \delta _n +\lambda n h_n \end{aligned}$$
(101)
using the reduced notations (65).
1.1 B.1 Proof of Lemma 1.
First step: using the reduced notations (65), we prove that:
$$\begin{aligned} E_{n+1}-\left( 1-\frac{\frac{2\alpha }{3}-2}{n}\right) E_n\leqslant & {} \frac{4\alpha K(\alpha )}{3}\frac{h_n}{n} + A_1(n,\alpha ) \delta _n + B_1(n,\alpha ) (h_{n-1}-h_n)\nonumber \\{} & {} + B_3(n,\alpha )(h_{n+1}-h_n-\delta _{n+1}) \end{aligned}$$
(102)
with:
$$\begin{aligned} A_1(n,\alpha )= & {} \frac{17\alpha ^2}{9}-\frac{8\alpha }{3}+2-\alpha \frac{(10\alpha ^2-18\alpha +9)n+7\alpha ^3-12\alpha ^2+6\alpha }{3(n+\alpha )^2},\\ B_1(n,\alpha )= & {} -\frac{2}{9}\alpha ^2+\frac{4}{3}\alpha -1+\frac{1}{3}\frac{3\alpha -2\alpha ^3}{n+\alpha }+\frac{1}{27}\frac{8\alpha ^3-24\alpha ^2}{n},\quad B_3(n,\alpha )=\frac{2}{3}\alpha -1. \end{aligned}$$
Indeed:
$$\begin{aligned} E_{n+1} -\left( 1 - \frac{\frac{2\alpha }{3}-2}{n}\right) E_n{} & {} = (n+1)^2w_{n+1}-\left( 1 - \frac{\frac{2\alpha }{3}-2}{n}\right) n^2w_n \nonumber \\{} & {} \quad + \,\left( (n+1)^2-\lambda (n+1)\right) \delta _{n+1}- \left( 1 - \frac{\frac{2\alpha }{3}-2}{n}\right) \left( n^2-\lambda n\right) \delta _n\nonumber \\{} & {} \quad +\,\left( \lambda ^2-\lambda (n+1)-\lambda n \left( 1 - \frac{\frac{2\alpha }{3}-2}{n}\right) \right) h_n+\lambda (n+1)h_{n+1}\nonumber \\{} & {} \quad -\, (\lambda ^2-\lambda n)\left( 1 -\frac{\frac{2\alpha }{3}-2}{n}\right) h_{n-1} \end{aligned}$$
(103)
Observe now that:
$$\begin{aligned}{} & {} (n+1)^2w_{n+1}-\left( 1 - \frac{\frac{2\alpha }{3}-2}{n}\right) n^2w_n =\left( 1 - \frac{\frac{2\alpha }{3}-2}{n}\right) n^2(w_{n+1}-w_n)\\{} & {} \qquad +\,\left( (n+1)^2-n^2\left( 1 - \frac{\frac{2\alpha }{3}-2}{n}\right) \right) w_{n+1}\\{} & {} \quad = n\left( n -\left( \frac{2\alpha }{3}-2\right) \right) (w_{n+1}-w_n)+\left( \frac{2\alpha }{3} n +1\right) w_{n+1} \end{aligned}$$
Combining the two following inequalities
$$\begin{aligned} w_{n+1}-w_n\leqslant \alpha _n^2\delta _n-\delta _{n+1} \end{aligned}$$
(104)
from Chambolle and Dossal [14] and:
$$\begin{aligned} w_{n+1}\leqslant \Vert x_n+\alpha _n(x_n-x_{n-1})-x^*\Vert ^2 -\Vert x_{n+1}-x^*\Vert ^2 \end{aligned}$$
from Apidopoulos et al. [3], or equivalently with our notations:
$$\begin{aligned} w_{n+1}\leqslant (1+\alpha _n)h_n-\alpha _nh_{n-1}-h_{n+1}+(\alpha _n+\alpha _n^2)\delta _n \end{aligned}$$
(105)
we then deduce:
$$\begin{aligned}{} & {} (n+1)^2w_{n+1}-\left( 1 - \frac{\frac{2\alpha }{3}-2}{n}\right) n^2w_n \\{} & {} \quad \leqslant n\left( n - \frac{2\alpha }{3}+2\right) (\alpha _n^2\delta _n-\delta _{n+1})+\left( \frac{2\alpha }{3} n+1\right) \\{} & {} \qquad \left( (1+\alpha _n)h_n-\alpha _n h_{n-1}-h_{n+1}+(\alpha _n+\alpha _n^2)\delta _n\right) \end{aligned}$$
It follows:
$$\begin{aligned} E_{n+1}-\left( 1-\frac{\frac{2\alpha }{3}-2}{n}\right) E_n\leqslant & {} A_1(n,\alpha )\delta _n+A_2(n,\alpha )\delta _{n+1}+B_1(n,\alpha )h_{n-1}\nonumber \\{} & {} +B_2(n,\alpha )h_n+B_3(n,\alpha )h_{n+1} \end{aligned}$$
(106)
where:
$$\begin{aligned} A_1(n,\alpha )= & {} \frac{17\alpha ^2}{9}-\frac{8\alpha }{3}+2{-}\alpha \frac{(10\alpha ^2-18\alpha +9)n+7\alpha ^3-12\alpha ^2+6\alpha }{3(n+\alpha )^2},\\ A_2(n,\alpha )= & {} 1-\frac{2\alpha }{3}\\ B_1(n,\alpha )= & {} -\frac{2}{9}\alpha ^2+\frac{4}{3}\alpha -1+\frac{1}{3}\frac{3\alpha -2\alpha ^3}{n+\alpha }+\frac{1}{27}\frac{8\alpha ^3-24\alpha ^2}{n}, \end{aligned}$$
and
$$\begin{aligned} B_2(n,\alpha )=\frac{2}{9}\alpha ^2-2\alpha +2-\frac{1}{3}\frac{3\alpha -2\alpha ^3}{n+\alpha },\quad B_3(n,\alpha )=\frac{2}{3}\alpha -1. \end{aligned}$$
(107)
Observe now that: \(A_2(n,\alpha )=-B_3(n,\alpha )\) and:
$$\begin{aligned} B_1(n,\alpha )+B_2(n,\alpha )+B_3(n,\alpha )=\frac{8\alpha ^2}{27}\frac{\alpha -3}{n}=\frac{4\alpha K(\alpha )}{3n}. \end{aligned}$$
so that (106) becomes:
$$\begin{aligned} E_{n+1}-\left( 1-\frac{\frac{2\alpha }{3}-2}{n}\right) E_n\leqslant & {} \frac{4\alpha K(\alpha )}{3}\frac{h_n}{n} + A_1(n,\alpha ) \delta _n + B_1(n,\alpha ) (h_{n-1}-h_n)\nonumber \\{} & {} +\, B_3(n,\alpha )(h_{n+1}-h_n-\delta _{n+1}) \end{aligned}$$
(108)
Step 2: First observe that combining the growth condition \({\mathcal {G}}_\mu ^2\) with the control of the values by the energy (namely: \(E_n \geqslant n^2w_n\) for all n), we have:
$$\begin{aligned} \forall n\in {\mathbb {N}}^*, \quad \frac{h_n}{n} \leqslant \frac{w_n}{\kappa n}\leqslant \frac{E_n}{\kappa n^3}\leqslant \frac{E_n}{\kappa n(n-\frac{2\alpha }{3})^2}, \end{aligned}$$
so that applying the following Lemma whose proof is detailed in Appendix B.4:
Lemma 4
For all \(n\geqslant 1\) and any \((A,B)\in {\mathbb {R}}^2\)
$$\begin{aligned} A\delta _n+B(h_{n-1}-h_n)\leqslant \left( 2|A+B|+\frac{\sqrt{2}|B|}{\sqrt{s\mu }}\right) \left( 1+\frac{4\alpha ^2}{9s\mu n^2}\right) \frac{E_n}{\left( n-\frac{2\alpha }{3}\right) ^2}. \end{aligned}$$
we can prove that:
$$\begin{aligned} \frac{4\alpha K(\alpha )}{3}\frac{h_n}{n} + A_1(n,\alpha ) \delta _n + B_1(n,\alpha ) (h_{n-1}-h_n) \leqslant \frac{\widetilde{C}_1(n,\alpha ,\kappa )E_n}{\left( n-\frac{2\alpha }{3}\right) ^2} \end{aligned}$$
(109)
and:
$$\begin{aligned} B_3(n,\alpha )(h_{n+1}-h_n-\delta _{n+1}) \leqslant \frac{{\widetilde{C}}_2(n,\alpha ,\kappa )E_{n+1}}{(n+1-\frac{2\alpha }{3})^2} \end{aligned}$$
(110)
where:
$$\begin{aligned} \widetilde{C}_1(n,\alpha ,\kappa )= & {} 2\left| \frac{5}{3}\alpha ^2-\frac{4\alpha }{3}+1 +R(n,\alpha )\right| \nonumber \\{} & {} +\,\sqrt{2}\left( \frac{|-\frac{2\alpha ^2}{9}+\frac{4\alpha }{3}-1+Q(n,\alpha )|}{\sqrt{\kappa }}\right) \left( 1+\frac{4\alpha ^2}{9\kappa n^2}\right) +\frac{4\alpha K(\alpha )}{3\kappa n}\nonumber \\ \end{aligned}$$
(111)
with:
$$\begin{aligned} |R(\alpha ,n)|= & {} \left| A_1(n,\alpha )+B_1(n,\alpha )-\left( \frac{5}{3}\alpha ^2-\frac{4\alpha }{3}+1\right) \right| \leqslant \frac{8\alpha ^3}{n}\\ |Q(\alpha ,n)|= & {} \frac{\alpha ^3}{3n}\left| n\frac{3-2\alpha ^2}{\alpha ^2(n+\alpha )}+8\frac{\alpha -3}{9\alpha }\right| \leqslant \frac{\alpha ^3}{n}, \end{aligned}$$
and:
$$\begin{aligned} {\widetilde{C}}_2(n,\alpha ,\kappa )=\left( \frac{2\alpha }{3}-1\right) \left( 4 +\frac{\sqrt{2}}{\sqrt{\kappa }}\right) \left( 1+\frac{4\alpha ^2}{9\kappa (n+1)^2}\right) \end{aligned}$$
(112)
Finally observe that since \(\kappa \in [0,1]\), for all \(n\geqslant \frac{4\alpha }{3\sqrt{\kappa }}\), we have:
$$\begin{aligned} \frac{1}{n-\frac{2 \alpha }{3}} \leqslant \frac{1}{n} \left( 1 + \sqrt{\kappa }\right) \quad \text{ and } \quad \frac{1}{n+1-\frac{2 \alpha }{3}} \leqslant \frac{1}{n+1} \left( 1 + \sqrt{\kappa }\right) \end{aligned}$$
(113)
hence:
$$\begin{aligned} \forall n\geqslant & {} \frac{4\alpha }{3\sqrt{\kappa }},\quad E_{n+1}-\left( 1-\frac{\frac{2\alpha }{3}-2}{n}\right) E_n \nonumber \\\leqslant & {} (1+\sqrt{\kappa })^2\left( \widetilde{C}_1(n,\alpha ,\kappa )\frac{E_n}{n^2}+\widetilde{C}_2(n,\alpha ,\kappa )\frac{E_{n+1}}{(n+1)^2}\right) . \end{aligned}$$
(114)
Step 3: The last step is to uniformly bound the coefficients \({\widetilde{C}}_1(n,\alpha ,\kappa )\) and \({\widetilde{C}}_2(n,\alpha ,\kappa )\) with respect to n. For any \(n\geqslant \frac{4\alpha }{3\sqrt{\kappa }}\) and \(\alpha \geqslant 3\), we have:
$$\begin{aligned} {\widetilde{C}}_2(n,\alpha ,\kappa )= & {} \left( \frac{2\alpha }{3}-1\right) \left( 4 +\frac{\sqrt{2}}{\sqrt{\kappa }}\right) \left( 1+\frac{4\alpha ^2}{9\kappa (n+1)^2}\right) \\\leqslant & {} \frac{5}{4}\sqrt{\frac{2}{\kappa }} \left( \frac{2\alpha }{3}-1 \right) \left( 1+2 \sqrt{2\kappa }\right) \end{aligned}$$
The calculations to bound the coefficient \(\widetilde{C}_1(n,\alpha ,\kappa )\) are similar but a little more painful. For all \(n\geqslant \frac{4\alpha }{3\sqrt{\kappa }}\), we have:
$$\begin{aligned} 4\alpha \frac{K(\alpha )}{3\kappa n} \leqslant \frac{2\alpha (\alpha -3)}{9\sqrt{\kappa }}\quad \text{ and } \quad \frac{4\alpha ^2}{9\kappa n^2} \leqslant \frac{1}{4} \end{aligned}$$
so that for all \(n\geqslant \frac{4\alpha }{3\sqrt{\kappa }}\):
$$\begin{aligned} {\widetilde{C}}_1(n,\alpha ,\kappa )= & {} 2\left| \frac{5}{3}\alpha ^2-\frac{4\alpha }{3}+1+R(n,\alpha )\right| +\sqrt{\frac{2}{\kappa }}\left| -\frac{2\alpha ^2}{9} +\frac{4\alpha }{3}-1\right. \\{} & {} \left. \,+Q(n,\alpha )\right| \left( 1+\frac{4\alpha ^2}{9\kappa n^2}\right) +4\alpha \frac{K(\alpha )}{3\kappa n}\\\leqslant & {} \frac{5}{4} \sqrt{\frac{2}{\kappa }}\left[ \left| -\frac{2\alpha ^2}{9}+\frac{4\alpha }{3}-1+Q(n,\alpha )\right| +\frac{4\sqrt{2}\alpha (\alpha -3)}{45}+\frac{4}{5}\left| \frac{5}{3}\alpha ^2\right. \right. \\{} & {} \left. \left. -\,\frac{4\alpha }{3}+1+R(n,\alpha )\right| \sqrt{2\kappa }\right] \end{aligned}$$
Assuming now that \(\alpha \geqslant 3+\frac{3}{\sqrt{2}}\), we have: \(\left| -\frac{2\alpha ^2}{9}+\frac{4\alpha }{3}-1\right| = \frac{2}{9}(\alpha -3)^2-1\), and:
$$\begin{aligned} {\widetilde{C}}_1(n,\alpha ,\kappa )\leqslant & {} \frac{5}{4} \sqrt{\frac{2}{\kappa }}\left[ \frac{2}{9}(\alpha -3)^2-1 +\frac{6\alpha (\alpha -3)}{45}+|Q(n,\alpha )\right. \\{} & {} \left. +\, \frac{4}{5}\left| \frac{5}{3}\alpha ^2-\frac{4\alpha }{3}+1+R(n,\alpha )\right| \sqrt{2\kappa }\right] \end{aligned}$$
Let The coefficient \({\widetilde{C}}_1(n,\alpha ,\kappa )\) can be rewritten as:
$$\begin{aligned} \forall n\geqslant \frac{4\alpha }{3\sqrt{\kappa }}, \quad \widetilde{C}_1(n,\alpha ,\kappa )\leqslant & {} \frac{5}{4} \sqrt{\frac{2}{\kappa }}P(\alpha ) \left[ 1+ \left| \frac{Q(n,\alpha )}{P(\alpha )}\right| \right. \\{} & {} \left. +\, \left( \frac{5\alpha ^2-4\alpha +3}{3P(\alpha )}+\left| \frac{R(n,\alpha )}{P(\alpha )}\right| \right) \sqrt{2\kappa }\right] . \end{aligned}$$
Studying the variations of the functions \(\alpha \mapsto \frac{\alpha ^2}{P(\alpha )}\) and \(\alpha \mapsto \frac{5\alpha ^2-4\alpha +3}{P(\alpha )}\), we easily prove that they are uniformly bounded for any real \(\alpha \geqslant 3+\frac{3}{\sqrt{2}}\), so that there exists a real constant \(B>0\) such that:
$$\begin{aligned} \forall n\geqslant \frac{4\alpha }{3\sqrt{\kappa }}, \quad \left| \frac{Q(n,\alpha )}{P(\alpha )}\right| \leqslant \frac{\alpha ^3}{nP(\alpha )}\leqslant B\sqrt{\kappa }. \end{aligned}$$
Likewise:
$$\begin{aligned} \forall n\geqslant \frac{4\alpha }{3\sqrt{\kappa }},\quad \left| \frac{R(n,\alpha )}{P(\alpha )}\right| \leqslant 8\frac{\alpha ^3}{nP(\alpha )}\leqslant B\sqrt{\kappa }. \end{aligned}$$
It finally exists some real constants \({{\tilde{c}}}_1\) and \(\tilde{c}_2\) such that for any \(\alpha \geqslant 3+\frac{3}{\sqrt{2}}\) and any \(n\geqslant \frac{4\alpha }{3\sqrt{\kappa }}\),
$$\begin{aligned} {\widetilde{C}}_1(n,\alpha ,\kappa ) \leqslant \frac{5}{4} \sqrt{\frac{2}{\kappa }} P(\alpha ) \left( 1+ {\tilde{c}}_1 \sqrt{\kappa }{+ {\tilde{c}}_2 \kappa } \right) . \end{aligned}$$
(115)
Combining (113) and (115), the inequality (67) holds as expected for any \(\alpha \geqslant 3+\frac{3}{\sqrt{2}}\) and without any condition on \(\kappa \):
$$\begin{aligned} {\forall n\geqslant \frac{4\alpha }{3\sqrt{\kappa }}},\quad E_{n+1}-\left( 1-\frac{\frac{2\alpha }{3}-2}{n}\right) E_n \leqslant \frac{C_1(\alpha ,\kappa )E_n}{n^2}+\frac{C_2(\alpha ,\kappa )E_{n+1}}{(n+1)^2} \end{aligned}$$
with:
$$\begin{aligned} C_1(\alpha ,\kappa )= & {} \frac{5}{4} \sqrt{\frac{2}{\kappa }}\left[ \frac{2}{9}(\alpha -3)\left( \frac{8}{5}\alpha -3\right) -1\right] (1+\sqrt{\kappa })^2 \left( 1+ {\tilde{c}}_1 \sqrt{\kappa }{+ {\tilde{c}}_2 \kappa } \right) \end{aligned}$$
(116)
$$\begin{aligned} C_2(\alpha ,\kappa )= & {} \frac{5}{4}\sqrt{\frac{2}{\kappa }} \left( \frac{2\alpha }{3}-1\right) (1+\sqrt{\kappa })^2(1+2 \sqrt{2\kappa }). \end{aligned}$$
(117)
\(\square \)
1.2 B.2 Proof of Lemma 2
Assume that the energy \(E_n\) satisfies:
$$\begin{aligned} E_{n+1}-\left( 1-\frac{\frac{2\alpha }{3}-2}{n}\right) E_n \leqslant \frac{C_1(\alpha ,\kappa )E_n}{n^2}+\frac{C_2(\alpha ,\kappa )E_{n+1}}{(n+1)^2} \end{aligned}$$
i.e.:
$$\begin{aligned} \left( 1-\frac{C_2(\alpha ,\kappa )}{(n+1)^2}\right) E_{n+1}-\left( 1 -\frac{\frac{2\alpha }{3}-2}{n} +\frac{C_1(\alpha ,\kappa )}{n^2}\right) E_n\leqslant 0. \end{aligned}$$
(118)
Let \(n_0\geqslant {\frac{4\alpha }{3\sqrt{\kappa }}}\). We then deduce:
$$\begin{aligned} \forall n\geqslant n_0,\quad \log (E_{n+1})-\log (E_{n_0})\leqslant \sum _{k=n_0}^n\log \left( \frac{1-\frac{\frac{2\alpha }{3}-2}{k}+\frac{C_1(\alpha ,\kappa )}{k^2} }{1-\frac{C_2(\alpha ,\kappa )}{(k+1)^2}}\right) . \end{aligned}$$
(119)
Using now the following classical inequalities:
$$\begin{aligned} \forall x>-1,\quad \frac{x}{x+1} \leqslant \log (1+x) \leqslant x, \end{aligned}$$
(120)
we get:
$$\begin{aligned} \log \left( 1 -\frac{\frac{2\alpha }{3}-2}{k} +\frac{C_1(\alpha ,\kappa )}{k^2}\right) \leqslant -\frac{\frac{2\alpha }{3}-2}{k} +\frac{C_1(\alpha ,\kappa )}{k^2} \end{aligned}$$
(121)
and
$$\begin{aligned} -\log \left( 1-\frac{C_2(\alpha ,\kappa )}{(k+1)^2}\right) \leqslant \frac{C_2(\alpha ,\kappa )}{(k+1)^2-C_2(\alpha ,\kappa )} \end{aligned}$$
(122)
We therefore get:
$$\begin{aligned} \log \left( \frac{1 -\frac{\frac{2\alpha }{3}-2}{k} +\frac{C_1(\alpha ,\kappa )}{k^2} }{1-\frac{C_2(\alpha ,\kappa )}{(k+1)^2}}\right) \leqslant -\frac{\frac{2\alpha }{3}-2}{k} +\frac{C_1(\alpha ,\kappa )}{k^2} +\frac{C_2(\alpha ,\kappa )}{(k+1)^2-C_2(\alpha ,\kappa )}\nonumber \\ \end{aligned}$$
(123)
Hence:
$$\begin{aligned} \log (E_{n+1})-\log (E_{n_0})\leqslant \sum _{k=n_0}^n \left( -\frac{\frac{2\alpha }{3}-2}{k} +\frac{C_1(\alpha ,\kappa )}{k^2} +\frac{C_2(\alpha ,\kappa )}{(k+1)^2-C_2(\alpha ,\kappa )} \right) \nonumber \\ \end{aligned}$$
(124)
We are now going to make use of the fact that the functions \(x \mapsto \frac{1}{x}\), \(x \mapsto \frac{1}{x^2}\) and \(x \mapsto \frac{C_2(\alpha ,\kappa )}{x^2-C_2(\alpha ,\kappa )}\) are decreasing functions on \((C_2,+ \infty )\). Observe that all coefficients in the very last inequality are actually non negative since \(\alpha \geqslant \alpha _0>3\). We then have:
$$\begin{aligned} \int _{k}^{k+1} \frac{dx}{x} \leqslant \frac{1}{k},\quad \frac{1}{k^2} \leqslant \int _{k-1}^{k} \frac{dx}{x^2} \end{aligned}$$
(125)
and:
$$\begin{aligned} \frac{C_2(\alpha ,\kappa )}{(k+1)^2-C_2(\alpha ,\kappa )} \leqslant \int _{k}^{k+1} \frac{C_2(\alpha ,\kappa )}{x^2-C_2(\alpha ,\kappa )} \, dx \end{aligned}$$
(126)
so that:
$$\begin{aligned} \log (E_{n+1})-\log (E_{n_0})\leqslant & {} -\left( \frac{2 \alpha }{3} -2 \right) \int _{n_0}^{n+1} \frac{dx}{x} +C_1(\alpha ,\kappa ) \int _{n_0-1}^{n} \frac{dx}{x^2}\\{} & {} +\,C_2(\alpha ,\kappa ) \int _{n_0}^{n+1} \frac{dx}{x^2-C_2(\alpha ,\kappa )} \end{aligned}$$
Noticing that:
$$\begin{aligned} \frac{1}{x^2-C_2(\alpha ,\kappa )} = \frac{1}{2 \sqrt{C_2(\alpha ,\kappa )}} \left( \frac{1}{x - \sqrt{C_2(\alpha ,\kappa )}} - \frac{1}{x + \sqrt{C_2(\alpha ,\kappa )}} \right) , \end{aligned}$$
we eventually get:
$$\begin{aligned} \log (E_{n+1})-\log (E_{n_0})\leqslant & {} -\left( \frac{2 \alpha }{3} -2 \right) \log \left( \frac{n+1}{n_0} \right) \nonumber \\{} & {} +C_1(\alpha ,\kappa ) \left( \frac{1}{n_0-1} -\frac{1}{n} \right) + \frac{\sqrt{C_2(\alpha ,\kappa )}}{ 2}\nonumber \\{} & {} \log \left( \frac{(n+1-\sqrt{C_2(\alpha ,\kappa )}) (n_0+ \sqrt{C_2(\alpha ,\kappa )})}{(n +1+ \sqrt{C_2(\alpha ,\kappa )}) (n_0 - \sqrt{C_2(\alpha ,\kappa )})} \right) \nonumber \\ \end{aligned}$$
(127)
i.e.:
$$\begin{aligned} \log (E_{n+1})-\log (E_{n_0})\leqslant & {} -\left( \frac{2 \alpha }{3} -2 \right) \log \left( \frac{n+1}{n_0} \right) +C_1(\alpha ,\kappa ) \left( \frac{1}{n_0-1} -\frac{1}{n} \right) \nonumber \\{} & {} + \,\frac{\sqrt{C_2(\alpha ,\kappa )}}{ 2} \left( \log \left( \frac{n+1-\sqrt{C_2(\alpha ,\kappa )}}{n +1+ \sqrt{C_2(\alpha ,\kappa )} } \right) \right. \nonumber \\{} & {} \left. +\, \log \left( \frac{ n_0 + \sqrt{C_2(\alpha ,\kappa )}}{n_0 - \sqrt{C_2(\alpha ,\kappa )}}\right) \right) \end{aligned}$$
(128)
Taking the exponential, we get:
$$\begin{aligned} E_{n+1} \leqslant E_{n_0} \left( \frac{n+1}{n_0} \right) ^{-\left( \frac{2\alpha }{3}-2\right) } \exp ({\tilde{\Phi }}(n_0)- {\tilde{\Phi }}(n+1)) \end{aligned}$$
(129)
with:
$$\begin{aligned} {\tilde{\Phi }}(n) = \frac{C_1(\alpha ,\kappa ) }{n-1} +\frac{\sqrt{C_2(\alpha ,\kappa )}}{2} \log \left( \frac{ n + \sqrt{C_2(\alpha ,\kappa )}}{n- \sqrt{C_2(\alpha ,\kappa )}}\right) . \end{aligned}$$
Let us finally compute a more tractable bound on the function \({\tilde{\Phi }}(n)\): using the inequality \(\log (1+x) \leqslant x\) for \(x \leqslant 1\), we have:
$$\begin{aligned} 0\leqslant \log \left( \frac{ n + \sqrt{C_2(\alpha ,\kappa )}}{n - \sqrt{C_2(\alpha ,\kappa )}} \right) =\log \left( 1+ \frac{ 2 \sqrt{C_2(\alpha ,\kappa )}}{n - \sqrt{C_2(\alpha ,\kappa )}} \right) \leqslant \frac{2 \sqrt{C_2(\alpha ,\kappa )}}{n- \sqrt{C_2(\alpha ,\kappa )}}\nonumber \\ \end{aligned}$$
(130)
Hence we deduce that:
$$\begin{aligned} 0\leqslant \frac{\sqrt{C_2(\alpha ,\kappa )}}{2} \log \left( \frac{ n + \sqrt{C_2(\alpha ,\kappa )})}{n - \sqrt{C_2(\alpha ,\kappa )}} \right) \leqslant \frac{C_2(\alpha ,\kappa )}{n- \sqrt{C_2(\alpha ,\kappa )}} \end{aligned}$$
(131)
Now, using the definition of the coefficients \(C_1(\alpha ,\kappa )\) and \(C_2(\alpha ,\kappa )\) given in Lemma 1, we get:
$$\begin{aligned}{} & {} 0 \leqslant {\tilde{\Phi }}(n) \leqslant \frac{C_1(\alpha ,\kappa )}{n-1} +\frac{C_2(\alpha ,\kappa )}{n - \sqrt{C_2(\alpha ,\kappa )}} \leqslant {\frac{2C_1(\alpha ,\kappa )}{n}} +\frac{C_2(\alpha ,\kappa )}{n - \sqrt{C_2(\alpha ,\kappa )}} \nonumber \\{} & {} \quad \leqslant \frac{5}{4n}\sqrt{\frac{2}{\kappa }}(1 + \sqrt{\kappa })^2 \left[ 2P(\alpha ) \left( 1+ {\tilde{c}}_1 \sqrt{\kappa }+ {\tilde{c}}_2 \kappa \right) + \left( \frac{2\alpha }{3}-1\right) \frac{1+2\sqrt{2\kappa }}{1-\frac{\sqrt{C_2(\alpha ,\kappa )}}{n}}\right] \nonumber \\ \end{aligned}$$
(132)
where: \(P(\alpha )=\frac{2}{9}(\alpha -3)\left( (1+\frac{2\sqrt{2}}{5})\alpha -3\right) -1\). Observe then that for \(\kappa \) small enough and \(n\geqslant \frac{4\alpha }{3\sqrt{\kappa }}\),
$$\begin{aligned} \frac{1}{1-\frac{\sqrt{C_2(\alpha ,\kappa )}}{n}} \leqslant \frac{1}{1-\frac{3\sqrt{C_2(\alpha ,\kappa )}}{4\alpha }\sqrt{\kappa }}\leqslant 1+2\frac{\sqrt{C_2(\alpha ,\kappa )}}{\alpha }\sqrt{\kappa } \end{aligned}$$
so that there exists a real constant \({{\tilde{c}}}_3\) such that for \(\kappa \) small enough and \(\alpha \geqslant 3+\frac{3}{\sqrt{2}}\) we have:
$$\begin{aligned} \frac{1}{1-\frac{\sqrt{C_2(\alpha ,\kappa )}}{n}} \leqslant 1+{{\tilde{c}}}_3 \kappa ^{1/4}. \end{aligned}$$
Therefore we finally get for any \(n\geqslant \frac{4\alpha }{3\sqrt{\kappa }}\):
$$\begin{aligned} {\tilde{\Phi }}(n)\leqslant & {} \frac{5}{4n}\sqrt{\frac{2}{\kappa }}\left( 1 +\sqrt{\kappa }\right) ^2\nonumber \\{} & {} \left( 2P(\alpha ) \left( 1+ {\tilde{c}}_1 \sqrt{\kappa }+ {\tilde{c}}_2 \kappa \right) +\left( \frac{2\alpha }{3}-1\right) (1+2\sqrt{2\kappa }) \left( 1 + {{\tilde{c}}}_3 \kappa ^{1/4} \right) \right) .\nonumber \\ \end{aligned}$$
(133)
We then deduce that there exists \(C_3>0\) (independent to \(\alpha \)) such that
$$\begin{aligned} \forall n\geqslant \frac{4\alpha }{3\sqrt{\kappa }},\quad {\tilde{\Phi }}(n)\leqslant & {} \frac{5}{4n} \sqrt{\frac{2}{\kappa }} \left( 2P(\alpha )+\frac{2\alpha }{3}-1\right) \left( 1 + C_3 \kappa ^{1/4} \right) \\\leqslant & {} \frac{5}{4n} \sqrt{\frac{2}{\kappa }} \left( 2P(\alpha )+\frac{2\alpha }{3}\right) \left( 1 + C_3 \kappa ^{1/4} \right) \end{aligned}$$
where \(2P(\alpha )+\frac{2\alpha }{3} = \frac{2}{3}(\alpha -3)(\frac{16}{15}\alpha -1)\). Let us introduce:
$$\begin{aligned} \Phi (n)= \frac{5}{6n} \sqrt{\frac{2}{\kappa }} (\alpha -3)\left( \frac{16}{15}\alpha -1\right) \left( 1 + C_3 \kappa ^{1/4} \right) . \end{aligned}$$
Let \(\alpha \geqslant 3+\frac{3}{\sqrt{2}}\) and \(n_0\geqslant \frac{4\alpha }{3\sqrt{\kappa }}\). As expected we finally get:
$$\begin{aligned} \forall n\geqslant n_0,\quad E_{n+1} \leqslant E_{n_0} \left( \frac{n+1}{n_0} \right) ^{-\left( \frac{2\alpha }{3}-2\right) } e^{\Phi (n_0)} \end{aligned}$$
(134)
\(\square \)
1.3 B.3 Proof of Lemma 3
Let \(M_n\) the mechanical energy:
$$\begin{aligned} M_n=F(x_n) - F^* + \frac{1}{2s} \Vert x_n - x_{n-1}\Vert ^2. \end{aligned}$$
Let us prove that for any \(n\in {\mathbb {N}}\), we have:
$$\begin{aligned} \frac{E_n}{2sn^2}\leqslant & {} \left( 1 + \frac{4\alpha ^2}{9\kappa n^2} + \frac{4\alpha }{3\sqrt{\kappa }n} \right) M_n = \left( 1+ \frac{2\alpha }{3\sqrt{\kappa }n}\right) ^2M_n \end{aligned}$$
(135)
First remark that:
$$\begin{aligned} b_n= & {} \left\| \frac{2 \alpha }{3} (x_{n-1}-x^*)+n(x_n-x_{n-1}) \right\| ^2 = \left\| \frac{2 \alpha }{3} (x_{n}-x^*)+\left( n-\frac{2 \alpha }{3} \right) (x_n-x_{n-1}) \right\| ^2\\= & {} \frac{4\alpha ^2}{9}\Vert x_n-x^*\Vert ^2 + \left( n-\frac{2 \alpha }{3}\right) ^2\Vert x_n-x_{n-1}\Vert ^2 + \frac{4\alpha }{3}\left( n-\frac{2 \alpha }{3}\right) \langle x_n-x^*,x_n-x_{n-1} \rangle \\\leqslant & {} \frac{4\alpha ^2}{9}\Vert x_n-x^*\Vert ^2 + n^2\Vert x_n-x_{n-1}\Vert ^2 + \frac{4\alpha }{3}\left( n-\frac{2 \alpha }{3}\right) \langle x_n-x^*,x_n-x_{n-1} \rangle \end{aligned}$$
Using a discrete version of the inequality (80), we have:
$$\begin{aligned} |\langle x_n-x^*,x_n-x_{n-1}| \rangle \leqslant \frac{\sqrt{\kappa }}{2}\Vert x_n-x^*\Vert ^2 + \frac{1}{2\sqrt{\kappa }}\Vert x_n-x_{n-1}\Vert ^2 \end{aligned}$$
(136)
so that:
$$\begin{aligned} b_n\leqslant & {} \frac{4 \alpha ^2}{9} \left\| x_{n}-x^* \right\| ^2 +n^2 \left\| x_n-x_{n-1} \right\| ^2 \nonumber \\{} & {} + \frac{2\alpha n}{3}\left( \sqrt{\kappa }\Vert x_n-x^*\Vert ^2 + \frac{1}{\sqrt{\kappa }}\Vert x_n-x_{n-1}\Vert ^2\right) \end{aligned}$$
(137)
Hence:
$$\begin{aligned} \frac{E_n}{2sn^2}= & {} F(x_n)-F^* + \frac{1}{2sn^2}b_n\\= & {} M_n + \frac{2 \alpha ^2}{9sn^2} \left\| x_{n}-x^*\right\| ^2 + \frac{\alpha }{3sn}\left( \sqrt{\kappa }\Vert x_n-x^*\Vert ^2 + \frac{1}{\sqrt{\kappa }}\Vert x_n-x_{n-1}\Vert ^2\right) \end{aligned}$$
Using now the quadratic growth condition \({\mathcal {G}}^2_\mu \) and remembering that: \(s\mu = \kappa \), we get:
$$\begin{aligned} \frac{E_n}{2sn^2}\leqslant & {} \left( 1 + \frac{4\alpha ^2}{9\kappa n^2} + \frac{4\alpha }{3\sqrt{\kappa }n} \right) M_n = \left( 1+ \frac{2\alpha }{3\sqrt{\kappa }n}\right) ^2M_n \end{aligned}$$
1.4 B.4 Proof of Lemma 4
Let us prove that for all \(n\geqslant 1\) and any \((A,B)\in {\mathbb {R}}^2\)
$$\begin{aligned} A\delta _n+B(h_{n-1}-h_n)\leqslant \left( 2|A+B|+\frac{\sqrt{2}|B|}{\sqrt{s\mu }}\right) \left( 1+\frac{4\alpha ^2}{9s\mu n^2}\right) \frac{E_n}{(n-\frac{2\alpha }{3})^2}.\nonumber \\ \end{aligned}$$
(138)
Firstly notice that
$$\begin{aligned} A\delta _n+B(h_{n-1}-h_n)= (A+B)\delta _n+B(h_{n-1}-h_{n}-\delta _n) \end{aligned}$$
(139)
and for any \(\theta >0\)
$$\begin{aligned} |h_{n-1}-h_{n}-\delta _n|=2|\langle x_n-x_{n-1},x_{n}-x^*\rangle |\leqslant \frac{h_n}{\theta }+\theta \delta _n. \end{aligned}$$
(140)
Combining the last two inequalities, it follows that for any \(\theta >0\):
$$\begin{aligned} A\delta _n+B(h_{n-1}-h_n)\leqslant (A+B+\theta |B|)\delta _n+\frac{|B|}{\theta }h_n \end{aligned}$$
(141)
To bound the coefficient of \(\delta _n\) we use a specific expression of \(b_n\):
$$\begin{aligned} b_n=\left\| {\frac{2\alpha }{3}(x_{n}-x^*)+\left( n-\frac{2\alpha }{3}\right) (x_n-x_{n-1})}\right\| ^2 \end{aligned}$$
(142)
Applying the inequality \(\left\| {u}\right\| ^2\leqslant 2\left\| {u+v}\right\| ^2+2\left\| {v}\right\| ^2\) to \(u=\left( n-\frac{2\alpha }{3}\right) (x_n-x_{n-1})\) and \(v=\frac{2\alpha }{3}(x_{n}-x^*)\), we get:
$$\begin{aligned} \left( n-\frac{2\alpha }{3}\right) ^2\delta _n\leqslant 2b_n+\frac{8\alpha ^2}{9}h_n. \end{aligned}$$
(143)
It follows that
$$\begin{aligned} \delta _n\leqslant \frac{2}{\left( n-\frac{2\alpha }{3}\right) ^2}b_n+\frac{8\alpha ^2}{9(n-\frac{2\alpha }{9})^2}h_n. \end{aligned}$$
(144)
and thus
$$\begin{aligned} A\delta _n+B(h_{n-1}-h_n)\leqslant & {} (|A+B|+\theta |B|)\frac{2}{\left( n-\frac{2\alpha }{3}\right) ^2}b_n\nonumber \\{} & {} +\left( \frac{|B|}{\theta }+ \frac{8\alpha ^2}{9\left( n-\frac{3\alpha }{4}\right) ^2}\right) h_n \end{aligned}$$
(145)
Using now the growth condition \(h_n\leqslant \frac{1}{s\mu }w_n\) for all \(n\in {\mathbb {N}}\), we get:
$$\begin{aligned} A\delta _n+B(h_{n-1}-h_n)\leqslant & {} (|A+B|+\theta |B|)\frac{2}{\left( n-\frac{2\alpha }{3}\right) ^2}b_n\nonumber \\{} & {} +\left( \frac{|B|}{s\mu \theta }+ \frac{8\alpha ^2}{9s\mu (n-\frac{2\alpha }{3})^2}\right) w_n \end{aligned}$$
(146)
Choosing \(\theta =\frac{1}{\sqrt{2 s\mu }}\) we finally deduce:
$$\begin{aligned} A\delta _n+B(h_{n-1}-h_n)\leqslant & {} (2|A+B|+\frac{\sqrt{2}|B|}{\sqrt{s\mu }})\frac{b_n}{\left( n-\frac{2\alpha }{3}\right) ^2} +\left( \frac{\sqrt{2}|B|}{\sqrt{s\mu }}\right. \nonumber \\{} & {} \left. +(2|A+B|+\frac{\sqrt{2}|B|}{\sqrt{s\mu })}\frac{4\alpha ^2}{9s\mu \left( n-\frac{2\alpha }{3}\right) ^2}\right) w_n \end{aligned}$$
(147)
and
$$\begin{aligned} A\delta _n+B(h_{n-1}-h_n)\leqslant \left( 2|A+B|+\frac{\sqrt{2}|B|}{\sqrt{s\mu }}\right) \left( 1+\frac{4\alpha ^2}{9s\mu n^2}\right) \frac{E_n}{\left( n-\frac{2\alpha }{3}\right) ^2},\nonumber \\ \end{aligned}$$
(148)
which concludes the proof of the lemma.
C Sketch of the proof of Theorem 7
The proof of Theorem 7 follows the same line than the proof of Theorem 6, and is based on the following Lyapunov energy:
$$\begin{aligned} E_n=2s\,n^2(F(x_n)-F^*+\left\| {\frac{\alpha }{2}(x_{n-1}-x^*)+\left( n-\frac{\alpha }{4}\right) (x_n-x_{n-1})}\right\| ^2. \end{aligned}$$
(149)
As in the proof of Theorem 6, the first step of this proof consists in establishing some discrete version of the differential inequality (87):
Lemma 5
Let \(\alpha >4+2\sqrt{2}\) and \(\kappa =\frac{\mu }{L}\). There exists \(\kappa _0>0\) such that for any \(\kappa \leqslant \kappa _0\), there exists some real constants \({\tilde{c}}_1\) and \({\tilde{c}}_2\) such that:
$$\begin{aligned} \forall n\geqslant \frac{3\alpha }{2\sqrt{\kappa }},\quad E_{n+1}-\left( 1-\frac{\alpha -2}{n}\right) E_n \leqslant C_1(\alpha ,\kappa )\frac{E_n}{n^2}+C_2(\alpha ,\kappa )\frac{E_{n+1}}{(n+1)^2}\nonumber \\ \end{aligned}$$
(150)
where:
$$\begin{aligned} C_1(\alpha ,\kappa )= & {} \frac{19}{36\sqrt{2\kappa }}(\alpha -2)(2\alpha -1)\left[ 1+{\widetilde{c}}_1 \sqrt{\kappa } + {\widetilde{c}}_2 \kappa \right] (1+\sqrt{\kappa })^2 \qquad \quad \end{aligned}$$
(151)
$$\begin{aligned} C_2(\alpha ,\kappa )= & {} \frac{19(\alpha -2)^2}{72\sqrt{2\kappa }}\left( 1+11\sqrt{\kappa }\right) (1+\sqrt{\kappa })^2. \end{aligned}$$
(152)
As in the proof of Theorem 7, the next step consists in integrating the inequality (150):
Lemma 6
Let \(\alpha \geqslant 4+2\sqrt{2}\) and \(n_0\geqslant \frac{3\alpha }{2\sqrt{\kappa }}\). If \(E_n\) satisfies (150) then there exists a real constant \(C_3>0\) such that:
$$\begin{aligned} \forall n\geqslant n_0, \quad E_{n} \leqslant E_{n_0} \, \left( \frac{n_0}{n}\right) ^{\alpha -2} e^{\Phi (n_0)} \end{aligned}$$
(153)
with
$$\begin{aligned} \Phi (n)= \frac{19(\alpha -2)(3\alpha -2)}{24n \sqrt{2\kappa }}\left( 1 + C_3\kappa ^{1/4}\right) . \end{aligned}$$
(154)
The proofs of Lemmas 5 and 6 are very similar to those of Lemmas 1 and 2 and are omitted here. They can be found in [7].
A good choice for \(n_0\) is one ensuring a control as tight as possible on the values \(F(x_n)-F^*\). For that \(n_0\) is chosen such that it minimizes the function \(f:x\mapsto x^{\alpha -2}e^{\Phi (x)}\). A straightforward computation gives:
$$\begin{aligned} n_0:= \frac{19(3\alpha -2)}{24\sqrt{2 \kappa }} \left( 1+C_3 \kappa ^{1/4} \right) . \end{aligned}$$
(155)
Observe that \(f(n_0)=\left( e\,n_0\right) ^{ \alpha -2}\) and that for any \(\alpha \geqslant 4+2\sqrt{2}\), the optimized value of \(n_0\) satisfies: \(n_0>\frac{3\alpha }{2\sqrt{\kappa }}\) without any condition on \(\kappa \) and that reducing \(\kappa _0\) if needed, we get:
$$\begin{aligned} \forall \alpha \geqslant 4+2\sqrt{2}, \quad \frac{3\alpha }{2\sqrt{\kappa }} \leqslant \frac{19(3\alpha -2)}{24\sqrt{2 \kappa }}\left( 1+C_3 \kappa ^{1/4}\right) \leqslant \frac{5\alpha }{\sqrt{2\kappa }}. \end{aligned}$$
(156)
Hence:
$$\begin{aligned} \forall n\geqslant \frac{5\alpha }{2\sqrt{\kappa }},\quad F(x_n)-F(x^*) \leqslant \frac{E_{n}}{2s n^2} \leqslant \frac{E_{n_0}}{2s n_0^2} \left( \frac{n_0}{n}\right) ^{\alpha } e^{\alpha -2} \end{aligned}$$
(157)
i.e.:
$$\begin{aligned} \forall n\geqslant \frac{5\alpha }{\sqrt{2\kappa }},\quad F(x_n)-F(x^*)\leqslant & {} \frac{E_{n_0}}{2s e^2 n_0^2} \left( e \, \frac{5\alpha }{2n\sqrt{\kappa }} \right) ^{\alpha } \end{aligned}$$
(158)
Uniformly bounding the energy \(E_{n_0}\) and noticing that: \(\frac{\alpha }{2n_0\sqrt{\kappa }}\leqslant \frac{1}{3}\), we have:
$$\begin{aligned} \frac{E_{n_0}}{2s n_0^2} \leqslant \left( 1+ \frac{\alpha }{2n_0\sqrt{\kappa }}\right) ^2 M_{n_0}\leqslant \frac{16}{9}M_{n_0} \end{aligned}$$
where \(M_n\) denotes the potential energy: \(M_n=F(x_n) - F^* + \frac{1}{2} \Vert x_n - x_{n-1}\Vert ^2\). Since the mechanical energy associated to the Nesterov scheme is non-increasing (see [14, Corollary 2]) and \(x_{-1}=x_0\), we then get:
$$\begin{aligned} \forall n\geqslant \frac{5\alpha }{\sqrt{2\kappa }},\quad F(x_n)-F(x^*) \leqslant \frac{16}{9} \left( e \, \frac{5\alpha }{2n\sqrt{\kappa }} \right) ^{\alpha } e^{-2}M_0. \end{aligned}$$
(159)
