immersion submanifold of dimension one or a $C^{l-1}$-embedded circle;
moreover $T_{\gamma}{\cal O}=\dot{\gamma}{\mathbb{R}}\subset
T_{\gamma}\mathcal{X}^{1}_{\tau}(M,\mathbb{I}_{g})$. Then for any reals $a<b$,
$[a,b]\cdot\bar{\gamma}$ is a $C^{4}$ embedded submanifold of dimension one.
Take $a>0$ such that $[-a,a]\cdot\bar{\gamma}\subset{\rm
Im}(\Phi_{\bar{\gamma}})$. Then

$S_{0}:=\Phi^{-1}_{{\bar{\gamma}}}\bigl{(}[-a,a]\cdot\bar{\gamma}\cap{\rm
Im}(\Phi_{\bar{\gamma}})\bigr{)}$ (6.7)

is an one-dimensional compact $C^{2}$ submanifold of
$\mathcal{X}^{1}_{\tau}(B^{n}_{2\iota}(0),{E}_{{\bar{\gamma}}})$ containing
$0$ as an interior point, and

$T_{0}S_{0}=(d\Phi_{\bar{\gamma}}(0))^{-1}(T_{\bar{\gamma}}S_{0})=(d\Phi_{\bar{\gamma}}(0))^{-1}(\mathbb{R}\dot{\bar{\gamma}})={\mathbb{R}}\zeta_{0}.$

Define the function $L^{\star}:\Lambda\times\mathbb{R}\times
B^{n}_{3\iota}(0)\times{\mathbb{R}}^{n}\to{\mathbb{R}}$ by

$\displaystyle
L^{\star}(\lambda,t,x,v)=L\bigr{(}\lambda,\phi_{{\bar{\gamma}}}(t,x),d\phi_{{\bar{\gamma}}}(t,x)[(1,v)]\bigl{)}.$

It is continuous and satisfies (2.7) with $U=B^{n}_{3\iota}(0)$ and
$E={E}_{{\bar{\gamma}}}$, i.e.,

$\displaystyle
L^{\star}(\lambda,t+\tau,x,v)=L\bigr{(}\lambda,\phi_{{\bar{\gamma}}}(t+\tau,x),d\phi_{{\bar{\gamma}}}(t+\tau,x)[(1,v)]\bigl{)}$
(6.8) $\displaystyle=$ $\displaystyle
L\left(\lambda,\mathbb{I}_{g}\left(\phi_{{\bar{\gamma}}}(t,E_{{\bar{\gamma}}}x)\right),d\mathbb{I}_{g}\left(\phi_{{\bar{\gamma}}}(t,E_{{\bar{\gamma}}}x)\right)\circ
d\phi_{{\bar{\gamma}}}(t,E_{{\bar{\gamma}}}x)[(1,E_{{\bar{\gamma}}}v)]\right)$
$\displaystyle=$ $\displaystyle
L\left(\lambda,\phi_{{\bar{\gamma}}}(t,E_{{\bar{\gamma}}}x),d\phi_{{\bar{\gamma}}}(t,E_{{\bar{\gamma}}}x)[(1,E_{{\bar{\gamma}}}v)]\right)$
$\displaystyle=$ $\displaystyle
L^{\star}(\lambda,t,E_{{\bar{\gamma}}}x,E_{{\bar{\gamma}}}v)$

for all $(\lambda,t,x,v)\in\Lambda\times\mathbb{R}\times
B^{n}_{3\iota}(0)\times{\mathbb{R}}^{n}$, and thus

$\displaystyle\partial_{x}L^{\star}(\lambda,t+\tau,x,v)$ $\displaystyle=$
$\displaystyle
E_{{\bar{\gamma}}}\partial_{x}L^{\star}\bigl{(}\lambda,t,E_{{\bar{\gamma}}}x,E_{{\bar{\gamma}}}v),$
$\displaystyle\partial_{v}L^{\star}(\lambda,t+\tau,x,v)$ $\displaystyle=$
$\displaystyle
E_{{\bar{\gamma}}}\partial_{v}L^{\star}\bigl{(}\lambda,t,E_{{\bar{\gamma}}}x,E_{{\bar{\gamma}}}v).$

(Here $\partial_{x}L^{\star}$ and $\partial_{v}L^{\star}$ denote the gradients
of $L^{\star}$ with respect to $x$ and $v$, respectively. Recall that all
vectors in $\mathbb{R}^{n}$ in this paper are understood as column vectors.)
Each $L^{\star}(\lambda,\cdot)$ is $C^{4}$ and all its partial derivatives of
order no more than two depend continuously on
$(\lambda,t,x,v)\in\Lambda\times\mathbb{R}\times
B^{n}_{3\iota}(0)\times{\mathbb{R}}^{n}$. Moreover,
$d\phi_{{\bar{\gamma}}}(t,x)[(1,v)]=\partial_{x}\phi_{{\bar{\gamma}}}(t,x)[v]+\partial_{t}\phi_{{\bar{\gamma}}}(t,x)$
implies that

$\mathbb{R}^{n}\ni v\mapsto
L^{\star}(\lambda,t,x,v)=L\bigr{(}\lambda,\phi_{{\bar{\gamma}}}(t,x),d\phi_{{\bar{\gamma}}}(t,x)[(1,v)]\bigl{)}\in\mathbb{R}$

is strictly convex.

Consider the functional
$\mathfrak{E}^{\star}_{\lambda}:\mathcal{X}^{1}_{\tau}(B^{n}_{2\iota}(0),{E}_{\bar{\gamma}})\to\mathbb{R}$,

$\mathfrak{E}^{\star}_{\lambda}(\xi)=\int^{\tau}_{0}L^{\star}(\lambda,t,\xi(t),\dot{\xi}(t))dt=\mathfrak{E}_{\lambda}\circ\Phi_{{\bar{\gamma}}}(\xi),$
(6.9)

where $\mathfrak{E}_{\lambda}$ is given by (1.19). It is $C^{2}$ by
Proposition LABEL:prop:C^2-funct. Since $\mathfrak{E}_{\lambda}$ is invariant
under the continuous $\mathbb{R}$-action on
$\mathcal{X}^{1}_{\tau}(M,\mathbb{I}_{g})$ given by

$\chi:\mathcal{X}^{1}_{\tau}(M,\mathbb{I}_{g})\times\mathbb{R}\to\mathcal{X}^{1}_{\tau}(M,\mathbb{I}_{g}),\;(s,\gamma)\mapsto
s\cdot\gamma,$ (6.10)

where $(s\cdot\gamma)(t)=\gamma(s+t)\;\forall s,t\in\mathbb{R}$, $S_{0}$ is a
critical submanifold of each $\mathfrak{E}^{\star}_{\lambda}$, and there holds

$D^{2}\mathfrak{E}^{\star}_{\lambda}(0)[\xi,\eta]=D^{2}\mathfrak{E}_{\lambda}(\bar{\gamma})\big{[}d\Phi_{{\bar{\gamma}}}(0)[\xi],d\Phi_{{\bar{\gamma}}}(0)[\eta]\big{]}\quad\forall\xi,\eta\in\mathcal{X}^{1}_{\tau}(B^{n}_{2\iota}(0),{E}_{\bar{\gamma}}),$

which imply

$m^{-}_{\tau}(\mathfrak{E}^{\star}_{\lambda},0)=m^{-}_{\tau}(\mathfrak{E}_{\lambda},\bar{\gamma})\quad\hbox{and}\quad
m^{0}_{\tau}(\mathfrak{L}^{\star}_{\lambda},0)=m^{0}_{\tau}(\mathfrak{E}_{\lambda},\bar{\gamma}).$
(6.11)

Since $S_{0}$ is compact, there exists $\rho_{0}>3\iota$ such that
$\sup_{t}|\dot{x}(t)|<\rho_{0}$ for all $x\in S_{0}$. For the Lagrangian
$L^{\star}$, as in Lemma 3.8 we can modify it to obtain:

######  Lemma 6.1.

For a given subset $\hat{\Lambda}\subset\Lambda$ which is either compact or
sequential compact, There exists a continuous function
$\check{L}:\hat{\Lambda}\times\mathbb{R}\times\bar{B}^{n}_{2\iota}(0)\times\mathbb{R}^{n}\to{\mathbb{R}}$
satisfying the following conditions for some constants $\kappa>0$ and $0<c<C$:

  1. (L0)

$\check{L}(\lambda,t+\tau,x,v)=\check{L}(\lambda,t,E_{{\bar{\gamma}}}x,E_{{\bar{\gamma}}}v)$
for all $(\lambda,t,x,v)$, and each the function
$\check{L}_{\lambda}(\cdot)=\check{L}(\lambda,\cdot)$
($\lambda\in\hat{\Lambda}$) is $C^{4}$ and partial derivatives

$\displaystyle\partial_{t}\check{L}_{\lambda}(\cdot),\quad\partial_{q}\check{L}_{\lambda}(\cdot),\quad\partial_{v}\check{L}_{\lambda}(\cdot),\quad\partial_{qv}\check{L}_{\lambda}(\cdot),\quad\partial_{qq}\check{L}_{\lambda}(\cdot),\quad\partial_{vv}\check{L}_{\lambda}(\cdot)$
$\displaystyle\hbox{and}\quad\partial_{tt}\check{L}_{\lambda}(\cdot),\quad\partial_{tq}\check{L}_{\lambda}(\cdot),\quad\partial_{tv}\check{L}_{\lambda}(\cdot)$

depend continuously on
$(\lambda,t,q,v)\in\hat{\Lambda}\times\mathbb{R}\times\bar{B}^{n}_{2\iota}(0)\times\mathbb{R}^{n}$.

  2. (L1)

$\check{L}$ and ${L}^{\star}$ are same on $\hat{\Lambda}\times\mathbb{R}\times
B^{n}_{3\iota/2}(0)\times B^{n}_{\rho_{0}}(0)$;

  3. (L2)

$\partial_{vv}\check{L}_{\lambda}(t,q,v)\geq
cI_{n},\quad\forall(\lambda,t,q,v)\in\hat{\Lambda}\times[0,\tau]\times
B^{n}_{3\iota/2}(0)\times\mathbb{R}^{n}$.

  4. (L3)

$\Bigl{|}\frac{\partial^{2}}{\partial q_{i}\partial
q_{j}}\check{L}_{\lambda}(t,q,v)\Bigr{|}\leq
C(1+|v|^{2}),\quad\Bigl{|}\frac{\partial^{2}}{\partial q_{i}\partial
v_{j}}\check{L}_{\lambda}(t,q,v)\Bigr{|}\leq C(1+|v|),\quad\hbox{and}\\\
\Bigl{|}\frac{\partial^{2}}{\partial v_{i}\partial
v_{j}}\check{L}_{\lambda}(t,q,v)\Bigr{|}\leq
C,\quad\forall(\lambda,t,q,v)\in\hat{\Lambda}\times[0,\tau]\times
B^{n}_{3\iota/2}(0)\times\mathbb{R}^{n}$.

  5. (L4)

$\check{L}(\lambda,t,q,v)\geq\kappa|v|^{2}-C,\quad\forall(\lambda,t,q,v)\in\hat{\Lambda}\times\mathbb{R}\times
B^{n}_{3\iota/2}(0)\times{\mathbb{R}}^{n}$.

  6. (L5)

$|\partial_{q}\check{L}(\lambda,t,q,v)|\leq C(1+|v|^{2})$ and
$|\partial_{q}\check{L}(\lambda,t,q,v)|\leq C(1+|v|)$ for all
$(\lambda,t,q,v)\in\hat{\Lambda}\times\mathbb{R}\times
B^{n}_{3\iota/2}(0)\times{\mathbb{R}}^{n}$.

  7. (L6)

$|\check{L}_{\lambda}(t,q,v)|\leq
C(1+|v|^{2}),\quad\forall(\lambda,t,q,v)\in\hat{\Lambda}\times\mathbb{R}\times\bar{B}^{n}_{3\iota/2}(0)\times\mathbb{R}^{n}$.

Consider the Banach space

${\bf X}:=\mathcal{X}^{1}_{\tau}(\mathbb{R}^{n},{E}_{{\bar{\gamma}}})$

with the induced norm $\|\cdot\|_{C^{1}}$ from
$C^{1}([0,\tau],\mathbb{R}^{n})$, and the Hilbert space

${\bf H}:=\\{\xi\in W^{1,2}_{\rm
loc}(\mathbb{R};\mathbb{R}^{n})\,|\,{E}^{T}_{\bar{\gamma}}(\xi(t))=\xi(t+\tau)\;\forall
t\in\mathbb{R}\\}$

equipped with $W^{1,2}$-inner product as in (2.2). Both carry a natural
$\mathbb{R}$-action given by $(\theta\cdot x)(t)=x(t+\theta)$ for
$\theta\in\mathbb{R}$. The spaces ${\bf H}$ and ${\bf X}$ have the following
$\mathbb{R}$-invariant open subsets

$\displaystyle\mathcal{U}:$ $\displaystyle=$ $\displaystyle\\{\xi\in
W^{1,2}_{\rm
loc}(\mathbb{R};B^{n}_{3\iota/2}(0))\,|\,{E}^{T}_{\bar{\gamma}}(\xi(t))=\xi(t+\tau)\;\forall
t\in\mathbb{R}\\},$ $\displaystyle\mathcal{U}^{X}:$ $\displaystyle=$
$\displaystyle\mathcal{U}\cap{\bf
X}=\mathcal{X}^{1}_{\tau}(B^{n}_{3\iota/2}(0),{E}_{{\bar{\gamma}}})$

respectively. Define a family of functionals
$\check{\mathscr{L}}_{\lambda}:\mathcal{U}\to{\mathbb{R}}$ by

$\check{\mathscr{L}}_{\lambda}(x)=\int^{\tau}_{0}\check{L}_{\lambda}(t,x(t),\dot{x}(t))dt,\quad\lambda\in\hat{\Lambda},$
(6.12)

and put ${\bf H}^{\bot}:=\\{x\in{\bf H}\,|\,(\zeta_{0},x)_{1,2}=0\\}$ and

$\displaystyle{\bf X}^{\bot}:=\\{x\in{\bf X}\,|\,(\zeta_{0},x)_{1,2}=0\\}={\bf
X}\cap{\bf H}^{\bot},$ (6.14)
$\displaystyle{\check{\mathscr{L}}}_{\lambda}^{\bot}:\mathcal{U}\cap{\bf
H}^{\bot}\to{\mathbb{R}},\;x\mapsto\check{\mathscr{L}}_{\lambda}(x).$

######  Remark 6.2.

Because of Lemma 6.1, by [6, Theorem 4.5] we deduce that every critical point
of $\check{\mathscr{L}}_{\lambda}$ is $C^{4}$.

####  6.2 Properties of functionals $\check{\mathscr{L}}$ and
$\check{\mathscr{L}}^{\bot}_{\lambda}$

In order to use the abstract theorems developed in [34, 35, 37] we need to
study some properties of the functionals $\check{\mathscr{L}}$ and
$\check{\mathscr{L}}^{\bot}_{\lambda}$ near $0$. By [32, §3] and [30], we have
(i)-(iii) of the following corresponding result of Proposition 3.9.

######  Proposition 6.3.

  1. (i)

$\check{\mathscr{L}}_{\lambda}$ is $C^{2-0}$ and the gradient map
$\nabla\check{\mathscr{L}}_{\lambda}:\mathcal{U}\to{\bf H}$ has the Gâteaux
derivative ${B}_{\lambda}(x)\in{\mathscr{L}}_{s}({\bf H})$ at
$x\in\mathcal{U}$.

  2. (ii)

$\nabla\check{\mathscr{L}}_{\lambda}$ restricts to a $C^{1}$ map
$A_{\lambda}:\mathcal{U}^{X}\to{\bf X}$.

  3. (iii)

(D1) of [34, Hypothesis 1.1] and (C) of [34, Hypothesis 1.3] hold near the
origin $0\in{\bf H}$, i.e.,

$\\{u\in{\bf H}\,|\,{B}_{\lambda}(0)u=su,\,s\leq 0\\}\subset{\bf
X}\quad\hbox{and}\quad\\{u\in{\bf H}\,|\,{B}_{\lambda}(0)u\in X\\}\subset{\bf
X}.$ (6.15)

  4. (iv)

Since $\check{L}=\tilde{L}=L^{\star}$ on
$\Lambda\times\mathbb{R}\times{B}^{n}_{3\iota/2}(0)\times
B^{n}_{\rho_{0}}(0)$, it holds that

$\check{\mathscr{L}}_{\lambda}(x)=\int^{\tau}_{0}\check{L}_{\lambda}(t,x(t),\dot{x}(t))dt=\mathscr{L}^{\star}_{\lambda}(x)$
(6.16)

for each $x$ in an open subset
$\\{x\in\mathcal{X}^{1}_{\tau}(B^{n}_{3\iota/2}(0),{E}_{{\bar{\gamma}}})\,|\,\sup_{t}|\dot{x}(t)|<\rho_{0}\\}$
of $\mathcal{U}^{X}$. Clearly, $0\in S_{0}$ has an open neighborhood $S_{00}$
in $S_{0}$ contained in the open subset. It follows that
$d\check{L}_{\lambda}(x)=0\;\forall x\in S_{00}$ and that

$m^{-}_{\tau}(\check{\mathscr{L}}_{\lambda},0)=m^{-}_{\tau}(\mathfrak{E}^{\star}_{\lambda},0)\quad\hbox{and}\quad
m^{0}_{\tau}(\check{\mathscr{L}}_{\lambda},0)=m^{0}_{\tau}(\mathfrak{E}^{\star}_{\lambda},0).$
(6.17)

By (iv) and (6.9), a point
$x\in\mathcal{X}^{1}_{\tau}(B^{n}_{3\iota/2}(0),{E}_{{\bar{\gamma}}})$ near
$0$ is a critical point of $\mathfrak{E}^{\star}_{\lambda}$ if and only if
$\gamma=\Phi_{{\bar{\gamma}}}(x)\in\mathcal{X}_{\tau}(M,\mathbb{I}_{g})$ is a
critical point of $\mathfrak{E}_{\lambda}$ near $\bar{\gamma}$. However, if
$\gamma\in{\rm Im}(\Phi_{\bar{\gamma}})$ is a critical point of
$\mathfrak{E}_{\lambda}$, so is each point in $\mathbb{R}\cdot\gamma$.
Therefore for $|s|$ small enough $(\Phi_{\bar{\gamma}})^{-1}(s\cdot\gamma)$ is
also a critical point of $\mathfrak{E}^{\star}_{\lambda}$. Such a critical
point is said to be the $\mathbb{R}$-same as $\gamma$. We need to study
behavior of $\mathbb{R}$-distinct critical points of
$\mathfrak{E}^{\star}_{\lambda}$ near $0$. Clearly,
$d\check{\mathscr{L}}_{\lambda}(0)=0$ and
$d\check{\mathscr{L}}_{\lambda}^{\bot}(0)=0\;\forall\lambda$.

Denote by $\Pi:{\bf H}\to{\bf H}^{\bot}$ the orthogonal projection. Then
$\Pi(x)=x-\frac{(x,\zeta_{0})_{1,2}}{\|\zeta_{0}\|_{1,2}}\zeta_{0}$ for
$x\in{\bf H}$, and

$\nabla{\check{\mathscr{L}}}_{\lambda}^{\bot}(x)=\nabla\check{\mathscr{L}}_{\lambda}(x)-\frac{(\nabla\check{\mathscr{L}}_{\lambda}(x),\zeta_{0})_{1,2}}{\|\zeta_{0}\|_{1,2}}\zeta_{0}\quad\forall
x\in\mathcal{U}\cap{\bf H}^{\bot},$ (6.18)

where by [32, (3.10)-(3.11)] (all vectors in $\mathbb{R}^{n}$ are understood
as column vectors now)

$\displaystyle\nabla\check{\mathscr{L}}_{\lambda}(\xi)(t)$ $\displaystyle=$
$\displaystyle\frac{1}{2}\int^{\infty}_{t}e^{t-s}\left[\partial_{q}\check{L}_{\lambda}\bigl{(}s,\xi(s),\dot{\xi}(s)\bigr{)}-\mathfrak{R}^{\xi}_{\lambda}(s)\right]\,ds$
(6.19) $\displaystyle+$
$\displaystyle\frac{1}{2}\int_{-\infty}^{t}e^{s-t}\left[\partial_{q}\check{L}_{{\lambda}}\bigl{(}s,\xi(s),\dot{\xi}(s)\bigr{)}-\mathfrak{R}^{\xi}(s)\right]\,ds+\mathfrak{R}^{\xi}_{\lambda}(t)$

where $\mathfrak{R}^{\xi}_{\lambda}$, (provided $2={\rm ord}(S_{p})>{\rm
ord}(S_{p+1})$ for some $p\in\\{0,\cdots,\sigma\\}$ in (6.3)), is given by

$\displaystyle\mathfrak{R}_{\lambda}^{\xi}(t)=\int^{t}_{0}\partial_{v}\check{L}_{\lambda}\bigl{(}s,\xi(s),\dot{\xi}(s)\bigr{)}ds+$
$\displaystyle\left[\left(\oplus_{l\leq
p}\frac{\sin\theta_{l}}{2-2\cos\theta_{l}}{\scriptscriptstyle\left(\begin{array}[]{cc}0&-1\\\
1&0\end{array}\right)}-\frac{1}{2}I_{2p}\right)\oplus{\rm
diag}(a_{p+1}(t),\cdots,a_{\sigma}(t))\right]\int^{\tau}_{0}\partial_{v}\check{L}_{\lambda}\bigl{(}s,\xi(s),\dot{\xi}(s)\bigr{)}ds$
(6.22)

with $a_{j}(t)=\frac{2t{S}_{j}+2t+1-{S}_{j}}{4}$, $j=p+1,\cdots,{\sigma}=n-p$.
(As usual $p=0$ (resp. $p={\sigma}$) means that ${\rm ord}(S_{1})=\cdots={\rm
ord}(S_{\sigma})=1$ (resp. ${\rm ord}(S_{1})=\cdots={\rm ord}(S_{\sigma})=2$)
and hence there is no the first (resp. second) term in the square brackets in
(6.2).) Note that

$\displaystyle\frac{d}{dt}\mathfrak{R}_{\lambda}^{\xi}(t)=\partial_{v}\check{L}_{\lambda}\bigl{(}t,\xi(t),\dot{\xi}(t)\bigr{)}+\mathfrak{M}\int^{\tau}_{0}\partial_{v}\check{L}_{\lambda}\bigl{(}s,\xi(s),\dot{\xi}(s)\bigr{)}ds,$
(6.24)

where $\mathfrak{M}\in\mathbb{R}^{n\times n}$ is a matrix only depending on
$E$.

Since $\zeta_{0}$ is $C^{1}$, we derive from (6.18) that
$\nabla{\check{\mathscr{L}}}_{\lambda}^{\bot}(x)\in{\bf X}^{\bot}$ for any
$x\in{\bf X}^{\bot}\cap\mathcal{U}$, and

$\mathbb{A}_{\lambda}:\mathcal{U}^{X}\cap{\bf X}^{\bot}\to{\bf
X}^{\bot},\;x\mapsto\nabla{\check{\mathscr{L}}}_{\lambda}^{\bot}(x)$ (6.25)

is $C^{1}$. By [32, §3] $\nabla{\check{\mathscr{L}}}_{\lambda}^{\bot}$ has the
Gâteaux derivative $\mathbb{B}_{\lambda}(x)\in{\mathscr{L}}_{s}({\bf
H}^{\bot})$ at $x\in\mathcal{U}\cap{\bf H}^{\bot}$ given by

$\mathbb{B}_{\lambda}(x)v=B_{\lambda}(x)v-\frac{(B_{\lambda}(x)v,\zeta_{0})_{2}}{\|\zeta_{0}\|_{2}}\zeta_{0}\quad\forall
v\in{\bf H}^{\bot},$ (6.26)

where $B_{\lambda}(\zeta)\in{\mathscr{L}}_{s}({\bf H})$ for
$\zeta\in\mathcal{U}$ is a self-adjoint Fredholm operator defined by

$\displaystyle(B_{\lambda}(\zeta)\xi,\eta)=\int_{0}^{\tau}\Bigl{(}\\!\\!\\!\\!\\!$
$\displaystyle\\!\\!\\!\\!\\!\partial_{vv}\check{L}_{{\lambda}}\bigl{(}t,\zeta(t),\dot{\zeta}(t)\bigr{)}\bigl{[}\dot{\xi}(t),\dot{\eta}(t)\bigr{]}+\partial_{qv}\check{L}_{{\lambda}}\bigl{(}t,\zeta(t),\dot{\zeta}(t)\bigr{)}\bigl{[}\xi(t),\dot{\eta}(t)\bigr{]}$
(6.27)
$\displaystyle+\partial_{vq}\check{L}_{{\lambda}}\bigl{(}t,\zeta(t),\dot{\zeta}(t)\bigr{)}\bigl{[}\dot{\xi}(t),\eta(t)\bigr{]}$
$\displaystyle+\partial_{qq}\check{L}_{{\lambda}}\bigl{(}t,\zeta(t),\dot{\zeta}(t)\bigr{)}\bigl{[}\xi(t),\eta(t)\bigr{]}\Bigr{)}\,dt$

for any $\xi,\eta\in{\bf H}$. The map ${B}_{\lambda}$ has a decomposition
${B}_{\lambda}=\textsl{{P}}_{{\lambda}}+\textsl{{Q}}_{{\lambda}}$, where
$\textsl{P}_{{\lambda}}(\zeta)\in{\mathscr{L}}_{s}(H)$ is a positive
definitive linear operator defined by

$\displaystyle(\textsl{P}_{{\lambda}}(\zeta)\xi,\eta)_{1,2}=\int_{0}^{\tau}\left(\partial_{vv}\check{L}_{{\lambda}}\bigl{(}t,\zeta(t),\dot{\zeta}(t)\bigr{)}[\dot{\xi}(t),\dot{\eta}(t)]+\bigl{(}\xi(t),\eta(t)\bigr{)}_{{\mathbb{R}}^{n}}\right)\,dt,$
(6.28)

and $\textsl{Q}_{{\lambda}}(\zeta)\in\check{\mathscr{L}}_{s}(H)$ is a compact
linear operator. Then

$\mathbb{P}_{\lambda}(x):=\Pi\circ\textsl{P}_{\lambda}(x)|_{{\bf
H}^{\bot}}\quad\hbox{and}\quad\mathbb{Q}_{\lambda}(x):=\Pi\circ\textsl{Q}_{\lambda}(x)|_{{\bf
H}^{\bot}}$ (6.29)

are positive definitive and compact, respectively, and
$\mathbb{B}_{\lambda}(x)=\mathbb{P}_{\lambda}(x)+\mathbb{Q}_{\lambda}(x)$ by
(6.26). Since ${\mathbb{R}}\zeta_{0}\subset{\rm Ker}(B_{\lambda}(0))$ we get

$\displaystyle
m^{-}_{\tau}({\check{\mathscr{L}}}_{\lambda}^{\bot},0)=m^{-}_{\tau}({\check{\mathscr{L}}}_{\lambda},0)\quad\hbox{and}\quad
m^{0}_{\tau}({\check{\mathscr{L}}}_{\lambda}^{\bot},0)=m^{0}_{\tau}({\check{\mathscr{L}}}_{\lambda},0)-1.$
(6.30)

Clearly, (L1) and (6.28) yield

$\displaystyle(\textsl{P}_{{\lambda}}(\zeta)\xi,\xi)_{1,2}\geq\min\\{c,1\\}\|\xi\|_{1,2}^{2},\quad\forall
x\in\mathcal{U},\;\forall\xi\in{\bf H},$ (6.31)

and hence

$\displaystyle(\mathbb{P}_{{\lambda}}(x)\xi,\xi)_{1,2}\geq\min\\{c,1\\}\|\xi\|_{1,2}^{2},\quad\forall\xi\in{\bf
H}^{\bot},\quad\forall x\in\mathcal{U}\cap{\bf H}^{\bot}.$ (6.32)

######  Proposition 6.4.

Let $(\lambda_{k})\subset\Lambda$ and $(\zeta_{k})\subset\mathcal{U}$ converge
to $\mu\in\Lambda$ and $0\in\mathcal{U}$, respectively. Then
$\|\textsl{P}_{{\lambda_{k}}}(\zeta_{k})\xi-\textsl{P}_{\mu}(0)\xi\|_{1,2}\to
0$ for each $\xi\in H$. In particular, if
$(\zeta_{k})\subset\mathcal{U}\cap{\bf H}^{\bot}$ converges to zero, then
$\|\mathbb{P}_{\lambda_{k}}(\zeta_{k})\xi-\mathbb{P}_{\mu}(0)\xi\|_{1,2}\to 0$
for any $\xi\in{\bf H}^{\bot}$.

###### Proof.

By (6.28) we have

$\displaystyle\|[\textsl{P}_{{\lambda_{k}}}(\zeta_{k})-\textsl{P}_{{\mu}}(0)]\xi\|^{2}_{1,2}\leq\int_{0}^{\tau}\left|[\partial_{vv}\check{L}_{{\lambda_{k}}}\bigl{(}t,\zeta_{k}(t),\dot{\zeta}_{k}(t)\bigr{)}-\partial_{vv}\check{L}_{\mu}\bigl{(}t,0,0\bigr{)}]\dot{\xi}(t)\right|^{2}_{\mathbb{R}^{n}}\,dt.$

Note that $\|\zeta_{k}\|_{1,2}\to 0$ implies $\|\zeta_{k}\|_{C^{0}}\to 0$.
Since $(\lambda,t,x,v)\mapsto\partial_{vv}\check{L}_{{\lambda}}(t,x,v)$ is
continuous, by the third inequality in (L2) in Lemma 6.1 we may apply [37,
Prop. B.9] ([35, Prop. C.1]) to

$f(t,\eta;\lambda)=\partial_{vv}\check{L}(\lambda,t,\zeta_{k}(t),\dot{\zeta}_{k}(t))\eta$

to get that

$\int_{0}^{\tau}\left|[\partial_{vv}\check{L}_{{\lambda_{k}}}\bigl{(}t,\zeta_{k}(t),\dot{\zeta}_{k}(t)\bigr{)}-\partial_{vv}\check{L}_{{\mu}}\bigl{(}t,0,0\bigr{)}]\dot{\xi}(t)\right|^{2}_{\mathbb{R}^{n}}\,dt\to
0.$

Moreover, the Lebesgue dominated convergence theorem also leads to

$\int_{0}^{\tau}\left|[\partial_{vv}\check{L}_{{\lambda_{k}}}\bigl{(}t,0,0\bigr{)}-\partial_{vv}\check{L}_{{\mu}}\bigl{(}t,0,0\bigr{)}]\dot{\xi}(t)\right|^{2}_{\mathbb{R}^{n}}\,dt\to
0.$

Hence
$\|[\textsl{P}_{{\lambda_{k}}}(\zeta_{k})-\textsl{P}_{\mu}(0)]\xi\|_{1,2}\to
0$. The final claim follows from this and (6.29). ∎

######  Proposition 6.5.

$\mathcal{U}\ni\zeta\mapsto\textsl{Q}_{{\lambda}}(\zeta)\in\mathscr{L}_{s}({\bf
H})$ is uniformly continuous at $0$ with respect to $\lambda\in\Lambda$.
Moreover, if $(\lambda_{k})\subset\Lambda$ converges to $0\in\Lambda$ then
$\|\textsl{Q}_{\lambda_{k}}(0)-\textsl{Q}_{\mu}(0)\|\to 0$.

###### Proof.

Write
$\textsl{Q}_{\lambda}(\zeta):=\textsl{Q}_{\lambda,1}(\zeta)+\textsl{Q}_{\lambda,2}(\zeta)+\textsl{Q}_{\lambda,3}(\zeta)$,
where

$\displaystyle(\textsl{Q}_{\lambda,1}(\zeta)\xi,\eta)_{1,2}=\int_{0}^{\tau}\partial_{vq}\check{L}_{\lambda}\big{(}t,\zeta(t),\dot{\zeta}(t)\big{)}[\dot{\xi}(t),\eta(t)]dt,$
$\displaystyle(\textsl{Q}_{\lambda,2}(\zeta)\xi,\eta)_{1,2}=\int_{0}^{\tau}\partial_{qv}\check{L}_{\lambda}\big{(}t,\zeta(t),\dot{\zeta}(t)\big{)}[\xi(t),\dot{\eta}(t)]dt,$
$\displaystyle(\textsl{Q}_{\lambda,3}(\zeta)\xi,\eta)_{1,2}=\int_{0}^{\tau}\Bigl{(}\partial_{qq}\check{L}_{\lambda}\big{(}t,\zeta(t),\dot{\zeta}(t)\big{)}[\xi(t),\eta(t)]-\bigl{(}\xi(t),\eta(t)\bigr{)}_{\mathbb{R}^{n}}\Bigr{)}\,dt.$

As above the first claim follows from (L2) in Lemma 6.1 and [37, Prop. B.9]
([35, Prop. C.1]) directly.

In order to prove the second claim, as in the proof of [30, page 571] we have

$\displaystyle\|\textsl{Q}_{\lambda_{k},1}(0)-\textsl{Q}_{\mu,1}(0)\|_{\mathscr{L}({\bf
H})}$ $\displaystyle\leq$ $\displaystyle
2(e^{\tau}+1)\left(\int^{\tau}_{0}\left|\partial_{vq}\check{L}_{\lambda_{k}}(s,0,0)-\partial_{vq}\check{L}_{\mu}(s,0,0)\right|^{2}ds\right)^{1/2}.$

Because of the second inequality in (L2), it follows from the Lebesgue
dominated convergence theorem that
$\|\textsl{Q}_{\lambda_{k},1}(0)-\textsl{Q}_{\mu,1}(0)\|_{\mathscr{L}({\bf
H})}\to 0$. Observe that
$(\textsl{Q}_{\lambda,2}(\zeta)\xi,\eta)_{1,2}=\bigr{(}\xi,(\textsl{Q}_{\lambda,1}(\zeta))^{\ast}\eta\bigl{)}_{1,2}$.
Hence
$\|\textsl{Q}_{\lambda_{k},2}(0)-\textsl{Q}_{\mu,2}(0)\|_{\mathscr{L}({\bf
H})}\to 0$. Finally, it is easy to deduce that

$\displaystyle\|\textsl{Q}_{\lambda_{k},3}(0)-\textsl{Q}_{\mu,3}(0)\|^{2}_{\mathscr{L}({\bf
H})}\leq\int_{0}^{\tau}\Bigl{|}\partial_{qq}\check{L}_{\lambda_{k}}\big{(}t,0,0\big{)}-\partial_{qq}\check{L}_{\mu}\big{(}t,0,0\big{)}\Bigr{|}^{2}\,dt.$

By the Lebesgue dominated convergence theorem the right side converges to
zero. Then
$\|\textsl{Q}_{\lambda_{k},3}(0)-\textsl{Q}_{\mu,3}(0)\|_{\mathscr{L}({\bf
H})}\to 0$ and therefore
$\|\textsl{Q}_{\lambda_{k}}(0)-\textsl{Q}_{\mu}(0)\|\to 0$. ∎

By (6.29) and Proposition 6.5, maps $\mathcal{U}\cap{\bf
H}^{\bot}\ni\zeta\mapsto\mathbb{Q}_{{\lambda}}(\zeta)\in\mathscr{L}_{s}({\bf
H}^{\bot})$ is uniformly continuous at $0$ with respect to
$\lambda\in\Lambda$. Moreover, if $(\lambda_{k})\subset\Lambda$ converges to
$0\in\Lambda$ then $\|\mathbb{Q}_{\lambda_{k}}(0)-\mathbb{Q}_{\mu}(0)\|\to 0$.

######  Proposition 6.6.

$({\bf H}^{\bot},{\bf
X}^{\bot},{\check{\mathscr{L}}}_{\lambda}^{\bot},\mathbb{A}_{\lambda}=\nabla{\check{\mathscr{L}}}_{\lambda}^{\bot},\mathbb{B}_{\lambda})$
satisfies (C) of [37, Hypothesis B.2] ([34, Hypothesis 1.3]) and (D1) of [37,
Hypothesis B.1] ([34, Hypothesis 1.1]) at the origin $0\in{\bf H}^{\bot}$,
namely

  1. (C)

$\\{u\in{\bf H}^{\bot}\,|\,\mathbb{B}_{\lambda}(0)u\in{\bf
X}^{\bot}\\}\subset{\bf X}^{\bot}$,

  2. (D1)

$\\{u\in{\bf H}^{\bot}\,|\,\mathbb{B}_{\lambda}(0)u=su,\;s\leq 0\\}\subset{\bf
X}^{\bot}$.

###### Proof.

In order to prove (C) let $u\in{\bf H}^{\bot}$ be such that
$v:=\mathbb{B}_{\lambda}(0)u\in{\bf X}^{\bot}$. By (6.26)

$B_{\lambda}(0)u-\frac{(B_{\lambda}(0)u,\zeta_{0})_{2}}{\|\zeta_{0}\|_{2}}\zeta_{0}=\mathbb{B}_{\lambda}(0)u=v\in{\bf
X}^{\bot}.$

Since $\zeta_{0}\in{\bf X}$, it follows from this and (6.15) that $u\in{\bf
X}$ and hence $u\in{\bf X}^{\bot}$.

Next, let $u\in{\bf H}^{\bot}$ satisfy $\mathbb{B}_{\lambda}(0)u=su$ for some
$s\leq 0$. Then (6.26) leads to

$B_{\lambda}(0)u=su+\frac{(B_{\lambda}(0)u,\zeta_{0})_{2}}{\|\zeta_{0}\|_{2}}\zeta_{0}.$

Since $T_{0}S_{0}=\mathbb{R}\zeta_{0}$ and
$\nabla{\check{\mathscr{L}}}_{\lambda}(x)=0\,\forall x\in S_{0}$, we deduce
$B_{\lambda}(0)\zeta_{0}=0$ and therefore $B_{\lambda}(0)u=su$. By (6.15) this
implies $u\in{\bf X}$ and so $u\in{\bf X}^{\bot}$.

∎

We also need the following two corresponding results of Propositions 3.10,
3.11.

######  Proposition 6.7.

Both
$\hat{\Lambda}\times\mathcal{U}\ni(\lambda,x)\mapsto\check{\mathscr{L}}_{\lambda}(x)\in{\mathbb{R}}$
and $\hat{\Lambda}\times\mathcal{U}^{X}\ni(\lambda,x)\mapsto
A_{\lambda}(x)\in{\bf X}$ are continuous.

Proofs of this proposition and the following key result are similar to those
of Propositions 3.10, 3.11. For completeness their proof are put off until
Section 6.4 because they are rather long.

######  Proposition 6.8.

Let $\bar{\varepsilon}>0$ be such that $B_{{\bf
H}}(0,\bar{\varepsilon})\subset\mathcal{U}$. For any given $\epsilon>0$ there
exists $0<\varepsilon\leq\bar{\varepsilon}$ such that if $x\in B_{{\bf
H}^{\bot}}(0,\varepsilon):=\\{x\,|\,x\in{\bf
H}^{\bot},\;\|x\|_{1,2}<\varepsilon\\}$ is a critical point of
$\check{\mathscr{L}}^{\bot}_{\lambda}$ with some $\lambda\in\hat{\Lambda}$
then $x$ is a critical point of $\check{\mathscr{L}}_{\lambda}$, belongs to
$C^{4}(\mathbb{R};\mathbb{R}^{n})$ and also satisfies
$\|x\|_{C^{2}}<\epsilon$. In particular, for $x\in B_{{\bf
H}^{\bot}}(0,\varepsilon)$, $d{\check{\mathscr{L}}}_{\lambda}^{\bot}(x)=0$ if
and only if $d{\check{\mathscr{L}}}_{\lambda}(x)=0$.

####  6.3 Completing the proofs of Theorems 1.23, 1.24, 1.25, 1.26

The ideas are the same as the proofs of [37, Theorems 1.18, 1.19, 1.21]. But
the corresponding checks and computations are much more complex and difficult.

######  Proposition 6.9 ([5, Proposition 3.5]).

Let $P$ be a finite-dimensional manifold, $N$ a (possibly infinite
dimensional) Banach manifold, $Q\subset N$ a Banach submanifold, and $A$ a
topological space. Assume that $\chi:A\times P\to N$ is a continuous function
such that there exist $a_{0}\in A$ and $m_{0}\in P$ with:

  1. (a)

$\chi(a_{0},m_{0})\in Q$;

  2. (b)

$\chi(a_{0},\cdot):P\to N$ is of class $C^{1}$;

  3. (c)

$\partial_{2}\chi(a_{0},m_{0})(T_{m_{0}}P)+T_{\chi(a_{0},m_{0})}Q=T_{\chi(a_{0},m_{0})}N$.

Then, for $a\in A$ near $a_{0}$, $\chi(a,P)\cap Q\neq\emptyset$.

For $0<\delta\leq 3\iota/2$ put

$B_{{\bf X}^{\bot}}(0,\delta):=\\{\xi\in
C^{1}_{{E}_{{\bar{\gamma}}},\tau}(\mathbb{R},B^{n}_{3\iota/2}(0))\,|\,\|\xi\|_{C^{1}}<\delta\\}\quad\hbox{and}\quad\Omega_{\delta}:=\Phi_{\bar{\gamma}}\left(B_{{\bf
X}^{\bot}}(0,\delta)\right).$

Clearly, $\Omega_{\delta}$ is a $C^{2}$ Banach submanifold of
$\mathcal{X}_{\tau}(M,\mathbb{I}_{g})$. For the action $\chi$ in (6.10), since
$\bar{\gamma}$ is nonconstant and $C^{6}$,

$\chi(\bar{\gamma},\cdot):\mathbb{R}\to\mathcal{X}_{\tau}(M,\mathbb{I}_{g}),\;s\mapsto\chi(\bar{\gamma},s)$

is a $C^{4}$ one-to-one immersion, and

$\partial_{2}\chi(\bar{\gamma},0)(T_{0}\mathbb{R})=\mathbb{R}\dot{\bar{\gamma}}=d\Phi_{\bar{\gamma}}(0)(\mathbb{R}\zeta_{0})\quad\hbox{and}\quad
T_{\chi(\bar{\gamma},0)}\Omega_{\delta}=T_{\bar{\gamma}}\Omega_{\delta}=d\Phi_{\bar{\gamma}}(0)({\bf
X}^{\bot}),$

we have
$\partial_{2}\chi(\bar{\gamma},0)(T_{0}\mathbb{R})+T_{\chi(\bar{\gamma},0)}\Omega_{\delta}=T_{\chi(\bar{\gamma},)}\mathcal{X}_{\tau}(M,\mathbb{I}_{g})$.
Applying Proposition 6.9 to $A=N=\mathcal{X}_{\tau}(M,\mathbb{I}_{g})$,
$P={\mathbb{R}}$, $Q=\Omega_{\delta}$, $a_{0}=\bar{\gamma}$, $m_{0}=0$ we get:

######  Proposition 6.10.

For any given $0<\delta\leq 2\iota$, if
$\gamma\in\mathcal{X}_{\tau}(M,\mathbb{I}_{g})$ is close to $\bar{\gamma}$,
then $({\mathbb{R}}\cdot\gamma)\cap\Omega_{\delta}\neq\emptyset$, that is,
${\mathbb{R}}\cdot\Omega_{\delta}$ is a neighborhood of the orbit
$\mathcal{O}={\mathbb{R}}\cdot\bar{\gamma}$ in
$\mathcal{X}_{\tau}(M,\mathbb{I}_{g})$.

###### Proof of Theorem 1.23.

By the assumptions there exists a sequence $(\lambda_{k})\subset\Lambda$
converging to $\mu\in\Lambda$ such that the problem (1.23) with
$\lambda=\lambda_{k}$ has solutions $\gamma_{k}$, $k=1,2,\cdots$, which are
${\mathbb{R}}$-distinct each other and satisfy
$\gamma_{k}|_{[0,\tau]}\to\bar{\gamma}|_{[0,\tau]}$ in $C^{1}([0,\tau];M)$.
Then $\hat{\Lambda}=\\{\mu,\lambda_{k}\,|\,k\in\mathbb{N}\\}$ is compact and
sequential compact. Take a decreasing sequence of positive numbers
$\delta_{m}\leq 2\iota$ such that $\delta_{m}\to 0$. For each $\delta_{m}$, by
Proposition 6.10 we have
$\gamma_{k_{m}}\in{\mathbb{R}}\cdot\Omega_{\delta_{m}}$ and thus
$\beta_{m}:=s_{m}\cdot\gamma_{k_{m}}\in\Omega_{\delta_{m}}$ for some
$s_{m}\in{\mathbb{R}}$. Note that each $\beta_{m}$ is a critical point of
${{\mathfrak{E}}}_{\lambda_{k_{m}}}$ on
$\Phi_{\bar{\gamma}}(\mathcal{U}^{X})$. Since any two of $(\gamma_{k})$ are
${\mathbb{R}}$-distinct, so are any two of $(\beta_{m})$.

Note that
$\Omega_{\delta}\subset\Phi_{\bar{\gamma}}\left(\mathcal{U}^{X}\right)\subset\Phi_{\bar{\gamma}}\left(C^{1}_{{E}_{{\bar{\gamma}}},\tau}(\mathbb{R},B^{n}_{3\iota/2}(0))\right)$,
and by (6.9) and (6.16) we have

$\check{\mathscr{L}}_{\lambda}(x)=\mathfrak{E}_{\lambda}\left(\Phi_{\bar{\gamma}}(x)\right)\quad\forall
x\in B_{{\bf X}^{\bot}}(0,\delta)$ (6.33)

because $\delta\leq 2\iota<\rho_{0}$ and $x\in B_{{\bf X}^{\bot}}(0,\delta)$
imply that for all $t\in\mathbb{R}$,

$(t,x(t),\dot{x}(t))\in\mathbb{R}\times B^{n}_{2\iota}(0)\times
B^{n}_{\rho_{0}}(0)$

and so
$\check{L}(\lambda,t,x(t),\dot{x}(t))=L^{\star}(\lambda,t,x(t),\dot{x}(t))$.

It follows that each

$x_{m}:=(\Phi_{\bar{\gamma}})^{-1}(\beta_{m})\in\\{\xi\in{\bf
X}^{\bot}\,|\,\|\xi\|_{C^{1}}<\delta_{m}\\}$

is a critical point of $\check{\mathscr{L}}_{\lambda_{k_{m}}}$ in ${\bf H}$
(and hence that of $\check{\mathscr{L}}_{\lambda_{k_{m}}}^{\bot}$ in ${\bf
H}^{\bot}$) and they are distinct each other. Moreover
$\|x_{m}\|_{1,2}\leq\sqrt{\tau}\|x_{m}\|_{C^{1}}\to 0$. Hence
$(\mu,0)\in\Lambda\times(\mathcal{U}\cap{\bf H}^{\bot})$ is a bifurcation
point of $\nabla{\check{\mathscr{L}}}_{\lambda}^{\bot}=0$ in
$\Lambda\times(\mathcal{U}\cap{\bf H}^{\bot})$.

Since $\check{\mathscr{L}}_{\lambda}$ is $C^{2-0}$ and
$\nabla\check{\mathscr{L}}_{\lambda}:\mathcal{U}\to{\bf H}$ has the Gâteaux
derivative ${B}_{\lambda}(x)\in{\mathscr{L}}_{s}({\bf H})$ at
$x\in\mathcal{U}$, ${\check{\mathscr{L}}}_{\lambda}^{\bot}$ is $C^{2-0}$ and
$\nabla{\check{\mathscr{L}}}_{\lambda}^{\bot}$ has a Gâteaux derivative
$\mathbb{B}_{\lambda}(x)\in{\mathscr{L}}_{s}({\bf H}^{\bot})$ at
$x\in\mathcal{U}\cap{\bf H}^{\bot}$ given by (6.26). From (6.29), (6.32) and
Propositions 6.4, 6.5 it easily follows that the conditions (i)-(iv) of [34,
Theorem 3.1] ([37, Theorem C.6]) are satisfied with
$\mathcal{F}_{\lambda}={\check{\mathscr{L}}}_{\lambda}^{\bot}$ and $H=X={\bf
H}^{\bot}$ and $U=\mathcal{U}\cap{\bf H}^{\bot}$. Therefore
$m^{0}_{\tau}(\check{\mathscr{L}}_{\mu}^{\bot},0)\geq 1$. This and (6.11),
(6.17) and (6.30) lead to
$m^{0}_{\tau}(\mathfrak{E}_{\mu},\bar{\gamma})=m^{0}_{\tau}(\check{\mathscr{L}}_{\mu},0)\geq
2$. ∎

Note that Propositions 6.8, 6.7 are not used in the proof of Theorem 1.23.
However, they are necessary for proofs of Theorems 1.24,1.26.

###### Proof of Theorem 1.24.

The original $\Lambda$ can be replaced by compact and sequential compact
$\hat{\Lambda}=\\{\mu,\lambda^{+}_{k},\lambda^{-}_{k}\,|\,k\in\mathbb{N}\\}$.
Follow the notations above. By Propositions 6.7, 6.4, 6.5, and (6.15) and
(6.31), the conditions of [36, Theorem 3.3] (or [34, Theorem A.3]) are
satisfied with $\mathcal{L}_{\lambda}=\check{\mathscr{L}}_{\lambda}$, $H={\bf
H}$, $X={\bf X}$, $U=\mathcal{U}$ and $\lambda^{\ast}=\mu$. From these,
(6.25)-(6.26) and (6.29), (6.32) and Proposition 6.6 it follows that
$\mathcal{L}_{\lambda}=\check{\mathscr{L}}_{\lambda}^{\bot}$, $H={\bf
H}^{\bot}$, $X={\bf X}^{\bot}$, $U=\mathcal{U}\cap{\bf H}^{\bot}$ and
$\lambda^{\ast}=\mu$ satisfy the conditions of [36, Theorem 3.3] (or [34,
Theorem A.3]).

By the assumptions (a)–(b) of Theorem 1.24 we may use (6.11), (6.17) and
(6.30) to deduce that $m^{0}_{\tau}(\check{\mathscr{L}}_{\mu}^{\bot},0)\geq 1$
and that for each $k\in\mathbb{N}$,

$[m^{-}_{\tau}(\check{\mathscr{L}}^{\bot}_{\lambda_{k}^{-}},0),m^{-}_{\tau}(\check{\mathscr{L}}^{\bot}_{\lambda_{k}^{-}},0)+m^{0}_{\tau}(\check{\mathscr{L}}^{\bot}_{\lambda_{k}^{-}},0)]\cap[m^{-}_{\tau}(\check{\mathscr{L}}^{\bot}_{\lambda_{k}^{+}},0),m^{-}_{\tau}(\check{\mathscr{L}}^{\bot}_{\lambda_{k}^{+}},0)+m^{0}_{\tau}(\check{\mathscr{L}}^{\bot}_{\lambda_{k}^{+}},0)]=\emptyset$

and either $m^{0}_{\tau}(\check{\mathscr{L}}^{\bot}_{\lambda_{k}^{-}},0)=0$ or
$m^{0}_{\tau}(\check{\mathscr{L}}^{\bot}_{\lambda_{k}^{+}},0)=0$.

Thus [37, Theorem C.4] concludes that there exists an infinite sequence
$(\lambda_{k},x_{k})\subset\hat{\Lambda}\times{\bf
H}^{\bot}\setminus\\{(\mu,0)\\}$ converging to $(\mu,0)$ such that each
$x_{k}\neq 0$ and satisfies
$\check{\mathscr{L}}^{\bot}_{\lambda_{k}}(x_{k})=0$ for all $k\in\mathbb{N}$.

Fix $0<\delta\leq 2\iota$. Let $\Omega_{\delta}$ be as in Proposition 6.10. By
Proposition 6.8, passing to a subsequence (if necessary) we may assume: each
$x_{k}$ is $C^{4}$ and a critical point of
$\check{\mathscr{L}}_{\lambda_{k}}$, $\|x_{k}\|_{C^{2}}<\delta\;\forall k$ and
$\|x_{k}\|_{C^{2}}\to 0$. Then each
$\gamma_{k}:=\Phi_{\bar{\gamma}}(x_{k})\in\Omega_{\delta}$ is a $C^{6}$
solution of the corresponding problem (1.18) with $\lambda=\lambda_{k}$,
$k=1,2,\cdots$, and $(\gamma_{k})$ converges to $\bar{\gamma}$ on any compact
interval $I\subset{\mathbb{R}}$ in $C^{2}$-topology as $k\to\infty$.

Since $d\Phi_{{\bar{\gamma}}}(0)[\zeta_{0}]=\dot{\bar{\gamma}}$ and
$T_{\bar{\gamma}}\Omega_{\delta}=d\Phi_{\bar{\gamma}}(0)({\bf X}^{\bot})$,
$\mathbb{R}\zeta_{0}+{\bf X}^{\bot}={\bf X}$ implies
$\mathbb{R}\dot{\bar{\gamma}}+T_{\bar{\gamma}}\Omega_{\delta}=T_{\bar{\gamma}}\mathcal{X}^{1}_{\tau}(M,\mathbb{I}_{g})$,
that is, the $C^{4}$ embedded circle $\mathcal{O}=\mathbb{R}\cdot\bar{\gamma}$
(because of periodicity of $\bar{\gamma}$) and $\Omega_{\delta}$ are
transversely intersecting at $\bar{\gamma}$. It follows that there exists a
neighborhood $\mathscr{V}$ of $\bar{\gamma}$ in
$\mathcal{X}^{1}_{\tau}(M,\mathbb{I}_{g})$ such that
$\mathscr{V}\cap\mathcal{O}\cap\Omega_{\delta}=\\{\bar{\gamma}\\}$. Because
$\|x_{k}\|_{C^{2}}\to 0$, there exists $k_{0}>0$ such that for each $k>k_{0}$,
$\gamma_{k}=\Phi_{\bar{\gamma}}(x_{k})\in\mathscr{V}\cap\Omega_{\delta}\setminus\\{\bar{\gamma}\\}$
and the $C^{4}$ immersed submanifold $\mathbb{R}\cdot\gamma_{k}$ transversely
intersect with $\Omega_{\delta}$ at $\gamma_{k}$. Hence

$\mathbb{R}\cdot\gamma_{k}\neq\mathcal{O}\quad\hbox{for any $k>k_{0}$.}$
(6.34)

(Otherwise, $\mathcal{O}$ and $\Omega_{\delta}$ have at least two distinct
intersecting points $\gamma_{k}$ and $\bar{\gamma}$ in $\mathscr{V}$.)

We conclude that $\\{\mathbb{R}\cdot\gamma_{k}\,|\,k\in\mathbb{N}\\}$ is an
infinite set. (Thus $(\gamma_{k})$ has a subsequence which only consists of
${\mathbb{R}}$-distinct elements. The proof is completed.) Otherwise, passing
to a subsequence we may assume that all $\gamma_{k}$ are ${\mathbb{R}}$-same,
i.e., $\gamma_{k}=s_{k}\cdot\gamma^{\ast}$ for some $s_{k}\in\mathbb{R}$,
where $\gamma^{\ast}:\mathbb{R}\to M$ satisfies (1.23) with
$\lambda=\lambda_{k}$, $k=1,2,\cdots$. Since all its partial derivatives of
$L(\lambda,\cdot)$ of order no more than two depend continuously on
$(\lambda,x,v)\in\Lambda\times TM$, it easily follows that $\gamma^{\ast}$
satisfies (1.23) with $\lambda=\mu$. Clearly, $\bar{\gamma}$ sits in the
intersection of $\mathcal{O}$ and the closure of
$\mathbb{R}\cdot\gamma^{\ast}$. By the assumption (c) of Theorem 1.24
$\mathbb{R}\cdot\gamma^{\ast}$ is closed and so equal to $\mathcal{O}$, namely
$\mathbb{R}\cdot\gamma_{k}=\mathcal{O}\;\forall k$, which contradicts (6.34).

∎

In order to prove Theorem 1.26 we also need some preparations. Consider the
$C^{4}$ Hilbert–Riemannian manifold

$\Lambda_{\tau}(M,\mathbb{I}_{g})=\\{\gamma\in
W^{1,2}_{loc}({\mathbb{R}},M)\,|\,\gamma(t+\tau)=\mathbb{I}_{g}(\gamma(t))\;\forall
t\\}$

with the natural Riemannian metric given by (1.3); see [48, Theorem 4.2] (or
[46, Theorem(8)]). Let $\|\cdot\|_{1}=\sqrt{\langle\cdot,\cdot\rangle_{1}}$ be
the induced norm.

Since ${\mathbb{R}}_{\bar{\gamma}}$ is an infinite cyclic subgroup of
${\mathbb{R}}$ with generator $p>0$, i.e., $\bar{\gamma}$ has the least period
$p$, the orbit ${\cal O}:={\mathbb{R}}\cdot\bar{\gamma}$ is an
${\mathbb{R}}$-invariant compact connected $C^{3}$ submanifold of
$\Lambda_{\tau}(M,\mathbb{I}_{g})$, precisely an $C^{3}$ embedded circle
$S^{1}(p):={\mathbb{R}}/p{\mathbb{Z}}$. Let $\pi:N{\cal O}\to{\cal O}$ be the
normal bundle of ${\cal O}$ in $\Lambda_{\tau}(M,\mathbb{I}_{g})$. It is a
$C^{2}$ Hilbert vector bundle over ${\cal O}$ (because $T{\cal O}$ is a
$C^{2}$ subbundle of $T_{\cal O}\Lambda_{\tau}(M,\mathbb{I}_{g})$), and

$XN{\cal O}:=T_{\cal O}\mathcal{X}_{\tau}(M,\mathbb{I}_{g})\cap N{\cal O}$

is a $C^{2}$ Banach vector subbundle of $T_{\cal
O}\mathcal{X}_{\tau}(M,\mathbb{I}_{g})$ by [32, Proposition 5.1]. Recall that
$3\iota$ is less than the injectivity radius of $g$ at each point on
$\bar{\gamma}(\mathbb{R})$. For $0<\nu\leq 3\iota$ we define

$N{\cal O}(\nu):=\\{(\gamma,v)\in N{\cal
O}\,|\,\|v\|_{1,2}<\nu\\}\quad\hbox{and}\quad XN{\cal
O}(\nu):=\\{(\gamma,v)\in XN{\cal O}\,|\,\|v\|_{C^{1}}<\nu\\}.$

Clearly, $XN{\cal O}(\nu)\subset N{\cal O}(\sqrt{\tau}\nu)$ and there exist
natural induced ${\mathbb{R}}$-actions on these bundles given by

$(\gamma,v)\mapsto(s\cdot\gamma,s\cdot v)\quad\forall s\in\mathbb{R}.$

Using the exponential map $\exp$ of $g$ we define the map

${\rm EXP}:T\Lambda_{\tau}(M,\mathbb{I}_{g})(\sqrt{\tau}\nu)=\\{(\gamma,v)\in
T\Lambda_{\tau}(M,\mathbb{I}_{g})\,|\,\|v\|_{1,2}<\sqrt{\tau}\nu\\}\to\Lambda_{\tau}(M,\mathbb{I}_{g})$
(6.35)

by ${\rm EXP}(\gamma,v)(t)=\exp_{\gamma(t)}v(t)\;\forall t\in{\mathbb{R}}$.
Clearly, ${\rm EXP}$ is equivariant, i.e.,

$s\cdot({\rm EXP}(\gamma,v))={\rm EXP}(s\cdot\gamma,s\cdot v)\quad\forall
s\in\mathbb{R}.$

It follows from [32, Lemma 5.2] that ${\rm EXP}$ is $C^{2}$. For sufficiently
small $\nu>0$, ${\rm EXP}$ gives rise to a $C^{2}$ diffeomorphism
$\digamma:N{\cal O}(\sqrt{\tau}\nu)\to{\cal N}({\cal O},\sqrt{\tau}\nu)$,
where ${\cal N}({\cal O},\sqrt{\tau}\nu)$ is an open neighborhood of ${\cal
O}$ in $\Lambda_{\tau}(M,\mathbb{I}_{g})$. Let ${\cal X}({\cal
O},\nu):=\digamma(XN{\cal O}(\nu))$, which is contained in ${\cal N}({\cal
O},\sqrt{\tau}\nu)$.

######  Lemma 6.11.

Suppose that $(\mathbb{I}_{g})^{l}=id_{M}$ for some $l\in\mathbb{N}$, and that
$0<\delta\leq 2\iota$ is so small that

$\Omega_{\delta}=\Phi_{\bar{\gamma}}\left(B_{{\bf
X}^{\bot}}(0,\delta)\right)\subset{\cal X}({\cal O},\nu).$

Then for any two different points $\gamma_{i}\in\Omega_{\delta}$, $i=1,2$,
either they are $\mathbb{R}$-distinct, or there exists an integer $0<m\leq l$
such that $\gamma_{2}=(m\tau)\cdot\gamma_{2}$ or
$\gamma_{1}=(m\tau)\cdot\gamma_{1}$. In particular, if $l=1$ and $\tau$ is
equal to the minimal period $p$ of $\bar{\gamma}$, then any two different
points in $\Omega_{\delta}$ are $\mathbb{R}$-distinct.

###### Proof.

Let different points $\xi_{1},\xi_{2}\in B_{{\bf X}^{\bot}}(0,\delta)$ be such
that $\gamma_{1}=\Phi_{\bar{\gamma}}(\xi_{1})$ and
$\gamma_{2}=\Phi_{\bar{\gamma}}(\xi_{2})$ are $\mathbb{R}$-same. Then we have
$s\geq 0$ such that $s\cdot\gamma_{1}=\gamma_{2}$. Since
$\mathbb{I}^{l}_{g}=id_{M}$ implies $(kl\tau)\cdot\gamma_{1}=\gamma_{1}$ and
$(s-kl\tau)\cdot\gamma_{1}=s\cdot\gamma_{1}$ for any $k\in\mathbb{Z}$ we can
assume $0\leq s<l\tau$. By (6.6),
$(\bar{\gamma},\sum^{n}_{i=1}\xi_{2}^{i}e_{i})$ and
$(s\cdot\bar{\gamma},\sum^{n}_{i=1}(s\cdot\xi_{1}^{i})(s\cdot e_{i}))$ belong
to $N{\cal O}(\sqrt{\tau}\nu)$. Note that

$\displaystyle(s\cdot\gamma_{1})(t)$ $\displaystyle=$
$\displaystyle\gamma_{1}(s+t)=\exp_{{\bar{\gamma}}(s+t)}\left(\sum^{n}_{i=1}\xi_{1}^{i}(s+t)e_{i}(s+t)\right)$
$\displaystyle=$ $\displaystyle{\rm
EXP}\left(s\cdot\bar{\gamma},\sum^{n}_{i=1}(s\cdot\xi_{1}^{i})(s\cdot
e_{i})\right)(t)$

and so

$s\cdot\gamma_{1}=\digamma\left(s\cdot\bar{\gamma},\sum^{n}_{i=1}(s\cdot\xi_{1}^{i})(s\cdot
e_{i})\right).$

Similarity, we have

$\gamma_{2}=\digamma\left(\bar{\gamma},\sum^{n}_{i=1}\xi_{2}^{i}e_{i}\right).$

Then $s\cdot\bar{\gamma}=\bar{\gamma}$ and

$\sum^{n}_{i=1}(s\cdot\xi_{1}^{i})(s\cdot
e_{i})=\sum^{n}_{i=1}\xi_{2}^{i}e_{i}.$

The former implies
$s\in\mathbb{R}_{\bar{\gamma}}\subset\\{[0],\cdots,[(l-1)\tau]\\}$, where
$[q\tau]=q\tau+l\mathbb{Z}$. Hence $s\in\\{0,\cdots,l-1\\}$. Combing with the
latter we obtain $\xi_{1}=\xi_{2}$, and therefore a contradiction.

When $l=1$ and $\tau$ is equal to the minimal period $p$ of $\bar{\gamma}$,
$\mathbb{R}_{\bar{\gamma}}=\\{0\\}$ and so $\xi_{1}=\xi_{2}$. A contradiction
is obtained.

∎

###### Proof of Theorem 1.26.

The original $\Lambda$ may be replaced by
$\hat{\Lambda}=[\mu-\varepsilon,\mu+\varepsilon]$. By the first paragraph in
the proof of Theorem 1.24 we have checked that
$\mathcal{L}_{\lambda}=\check{\mathscr{L}}_{\lambda}^{\bot}$, $H={\bf
H}^{\bot}$, $X={\bf X}^{\bot}$, $U=\mathcal{U}\cap{\bf H}^{\bot}$ and
$\lambda^{\ast}=\mu$ satisfy the conditions of [36, Theorem 3.6] except for
the condition (f).

By the assumptions of Theorem 1.26 and (6.11), (6.17) and (6.30) we get that
$m^{0}_{\tau}(\check{\mathscr{L}}_{\mu}^{\bot},0)\geq 1$ and
$m^{0}_{\tau}(\check{\mathscr{L}}^{\bot}_{\lambda},0)=0$ for each
$\lambda\in\hat{\Lambda}\setminus\\{\mu\\}$ near $\mu$, and that
$m^{-}_{\tau}(\mathscr{L}^{\bot}_{\lambda},0)$ takes, respectively, values
$m^{0}_{\tau}(\check{\mathscr{L}}^{\bot}_{\mu},0)$ and
$m^{-}_{\tau}(\check{\mathscr{L}}^{\bot}_{\mu},0)+m^{0}_{\tau}(\check{\mathscr{L}}^{\bot}_{\mu},0)-1$
as $\lambda\in\hat{\Lambda}$ varies in two deleted half neighborhoods of
$\mu$. These mean that the condition (f) of [37, Theorem C.7] ([36, Theorem
3.6]) is satisfied. Therefore one of the following alternatives occurs:

  1. (i)

There exists a sequence $(x_{k})\subset{\bf H}^{\bot}\setminus\\{0\\}$
converging to $0$ in ${\bf H}^{\bot}$ such that
$\nabla\check{\mathscr{L}}^{\bot}_{\mu}(x_{k})=0$ for all $k$.

  2. (ii)

For each $\lambda\in\Lambda\setminus\\{\mu\\}$ near $\mu$,
$\nabla\check{\mathscr{L}}^{\bot}_{\lambda}(w)=0$ has a solution
$x_{\lambda}\in{\bf X}^{\bot}$ different from $0$, which converges to $0$ in
${\bf X}^{\bot}$ as $\lambda\to\mu$.

  3. (iii)

Given a neighborhood $\mathfrak{W}$ of $0$ in ${\bf X}^{\bot}$, there is an
one-sided neighborhood $\Lambda^{0}$ of $\mu$ such that for any
$\lambda\in\Lambda^{0}\setminus\\{\mu\\}$,
$\nabla\check{\mathscr{L}}^{\bot}_{\lambda}(w)=0$ has at least two nonzero
solutions in $\mathfrak{W}$, $x_{\lambda}^{1}$ and $x_{\lambda}^{2}$, which
can also be required to satisfy
$\check{\mathscr{L}}^{\bot}_{\lambda}(x_{\lambda}^{1})\neq\check{\mathscr{L}}^{\bot}_{\lambda}(x^{2}_{\lambda})$
provided that $m^{0}_{\tau}(\check{\mathscr{L}}_{\mu}^{\bot},0)\geq 2$ and
$\nabla\check{\mathscr{L}}^{\bot}_{\lambda}(w)=0$ has only finitely many
nonzero solutions in $\mathfrak{W}$.

Let $\delta>0$ satisfy Proposition 6.10 and Lemma 6.11. By Proposition 3.11,
we obtain:

  1. $\bullet$

In case (i), passing to a subsequence (if necessary) all $x_{k}$ are $C^{4}$
and satisfy: $\nabla\check{\mathscr{L}}_{\mu}(x_{k})=0$,
$0<\|x_{k}\|_{C^{2}}<\delta$ and $\|x_{k}\|_{C^{2}}\to 0$. Therefore each
$\gamma_{k}:=\Phi_{\bar{\gamma}}(x_{k})\in\Omega_{\delta}$ is a $C^{6}$
solution of the corresponding problem (1.23) with $\lambda=\mu$, and
$(\gamma_{k})$ converges to $\bar{\gamma}$ on any compact interval
$I\subset{\mathbb{R}}$ in $C^{2}$-topology as $k\to\infty$.

  2. $\bullet$

In case (ii), when $\lambda\in\Lambda\setminus\\{\mu\\}$ is close to $\mu$,
all $x_{\lambda}$ are $C^{4}$ and satisfy:
$\nabla\check{\mathscr{L}}_{\lambda}(x_{\lambda})=0$,
$0<\|x_{\lambda}\|_{C^{2}}<\delta$ and $\|x_{\lambda}\|_{C^{2}}\to 0$ as
$\lambda\to\mu$. Hence each
$\gamma_{\lambda}:=\Phi_{\bar{\gamma}}(x_{\lambda})\in\Omega_{\delta}\setminus\\{\bar{\gamma}\\}$
is a $C^{6}$ solution of the corresponding problem (1.23), $\gamma_{\lambda}$
converges to $\bar{\gamma}$ on any compact interval $I\subset{\mathbb{R}}$ in
$C^{2}$-topology as $\lambda\to\mu$, and
$\mathbb{R}\cdot\gamma_{\lambda}\neq\mathcal{O}$ by Lemma 6.11.

  3. $\bullet$

In case (iii), we can require that the neighborhood $\mathfrak{W}$ so small
that $\Phi_{\bar{\gamma}}(\mathfrak{W})\subset\mathcal{W}$ and
$\mathfrak{W}\subset B_{{\bf X}^{\bot}}(0,\delta)$. The latter implies
$\Phi_{\bar{\gamma}}(\mathfrak{W})\subset\Omega_{\delta}$. Then all
$x_{\lambda}^{1}$ and $x_{\lambda}^{2}$ are $C^{4}$ and critical points of
$\check{\mathscr{L}}_{\lambda}$, and satisfy:
$0<\|x^{i}_{\lambda}\|_{C^{2}}<\delta$ and $\|x^{i}_{\lambda}\|_{C^{2}}\to 0$,
$i=1,2$. Consequently,
$\gamma_{\lambda}^{1}:=\Phi_{\bar{\gamma}}(x_{\lambda}^{1})$ and
$\gamma_{\lambda}^{2}:=\Phi_{\bar{\gamma}}(x_{\lambda}^{2})$ belong to
$\Omega_{\delta}\cap\mathcal{W}\setminus\mathbb{R}\cdot\gamma_{0}$, are
$C^{6}$ solutions of the corresponding problem (1.23). When
$m^{0}_{\tau}(\mathfrak{E}_{\mu},\bar{\gamma})=m^{0}_{\tau}(\check{\mathscr{L}}_{\mu}^{\bot},0)+1\geq
3$, and (1.23) with parameter value $\lambda$ has only finitely many
$\mathbb{R}$-distinct solutions in $\mathcal{W}$ which are
$\mathbb{R}$-distinct from $\bar{\gamma}$, it is clear that
$\nabla\check{\mathscr{L}}^{\bot}_{\lambda}(w)=0$ has only finitely many
nonzero solutions in $\mathfrak{W}$, and therefore we can require that
$x_{\lambda}^{1}$ and $x_{\lambda}^{2}$ satisfy
$\check{\mathscr{L}}^{\bot}_{\lambda}(x_{\lambda}^{1})\neq\check{\mathscr{L}}^{\bot}_{\lambda}(x^{2}_{\lambda})$,
which implies
$\mathfrak{E}_{\lambda}(\gamma^{1}_{\lambda})\neq\mathfrak{E}_{\lambda}(\gamma^{2}_{\lambda})$,

∎

###### Proof of Theorem 1.25.

The original $\Lambda$ may be replaced by $\hat{\Lambda}=\alpha([0,1])$. By
the first paragraph in the proof of Theorem 1.24 we have checked that
$\mathcal{L}_{\lambda}=\check{\mathscr{L}}_{\lambda}^{\bot}$, $H={\bf
H}^{\bot}$, $X={\bf X}^{\bot}$, $U=\mathcal{U}\cap{\bf H}^{\bot}$ satisfy the
assumptions a)-c) and (i)-(v) of [37, Theorem C.5] for any
$\lambda^{\ast}\in\hat{\Lambda}$. The condition (d) of Theorem 1.25 can be
translated into:

$[m^{-}_{\tau}(\check{\mathscr{L}}^{\bot}_{\lambda^{-}},0),m^{-}_{\tau}(\check{\mathscr{L}}^{\bot}_{\lambda^{-}},0)+m^{0}_{\tau}(\check{\mathscr{L}}^{\bot}_{\lambda^{-}},0)]\cap[m^{-}_{\tau}(\check{\mathscr{L}}^{\bot}_{\lambda^{+}},0),m^{-}_{\tau}(\check{\mathscr{L}}^{\bot}_{\lambda^{+}},0)+m^{0}_{\tau}(\check{\mathscr{L}}^{\bot}_{\lambda^{+}},0)]=\emptyset$

and either $m^{0}_{\tau}(\check{\mathscr{L}}^{\bot}_{\lambda^{-}},0)=0$ or
$m^{0}_{\tau}(\check{\mathscr{L}}^{\bot}_{\lambda^{+}},0)=0$.

Hence the condition (e.3) of [37, Theorem C.5] is satisfied. Thus [37, Theorem
C.4] concludes that there exists $\mu\in\alpha([0,1])$ and an infinite
sequence $(\lambda_{k},x_{k})\subset\hat{\Lambda}\times{\bf
H}^{\bot}\setminus\\{(\mu,0)\\}$ converging to $(\mu,0)$ such that each
$x_{k}\neq 0$ and satisfies
$\check{\mathscr{L}}^{\bot}_{\lambda_{k}}(x_{k})=0$ for all $k\in\mathbb{N}$.
Moreover, $\mu$ is not equal to $\lambda^{+}$ (resp. $\lambda^{-}$) if
$m^{0}_{\tau}(\check{\mathscr{L}}^{\bot}_{\lambda^{+}},0)=0$ (resp.
$m^{0}_{\tau}(\check{\mathscr{L}}^{\bot}_{\lambda^{-}},0)=0$). We can assume
$\lambda_{k}=\alpha(t_{k})$ for some $(t_{k})\subset[0,1]$ converging to
$\bar{t}\in[0,1]$.

Fix $0<\delta\leq 2\iota$. Let $\Omega_{\delta}$ satisfy Proposition 6.10 and
Lemma 6.11 By Proposition 6.8, passing to a subsequence (if necessary) we may
assume: each $x_{k}$ is $C^{4}$ and a critical point of
$\check{\mathscr{L}}_{\lambda_{k}}$, $\|x_{k}\|_{C^{2}}<\delta\;\forall k$ and
$\|x_{k}\|_{C^{2}}\to 0$. Then each
$\gamma_{k}:=\Phi_{\bar{\gamma}}(x_{k})\in\Omega_{\delta}$ is a $C^{6}$
solution of the corresponding problem (1.18) with $\lambda=\lambda_{k}$,
$k=1,2,\cdots$, and $(\gamma_{k})$ converges to $\bar{\gamma}$ on any compact
interval $I\subset{\mathbb{R}}$ in $C^{2}$-topology as $k\to\infty$. We can
assume that all $x_{k}$ are distinct each other. By Lemma 6.11 each
$\gamma_{k}$ has at most $l$ $\mathbb{R}$-same points in
$\\{\gamma_{k}\,|\,k\in\mathbb{N}\\}$. Hence $(\gamma_{k})$ has a subsequence
$(\gamma_{k_{i}})$ consisting of completely $\mathbb{R}$-distinct points. The
required assertions are proved. ∎

####  6.4 Proofs of Propositions 6.7,6.8

###### Proof of Proposition 6.7.

Step 1(Prove that
$\hat{\Lambda}\times\mathcal{U}\ni(\lambda,x)\mapsto\check{\mathscr{L}}_{\lambda}(x)\in{\mathbb{R}}$
is continuous). Indeed, for any two points $(\lambda,x)$ and
$(\lambda_{0},x_{0})$ in $\hat{\Lambda}\times\mathcal{U}$ we can write

$\displaystyle\check{\mathscr{L}}_{\lambda}(x)-\check{\mathscr{L}}_{\lambda_{0}}(x_{0})$
$\displaystyle=$
$\displaystyle\left[\int^{\tau}_{0}\check{L}_{\lambda}(t,x(t),\dot{x}(t))dt-\int^{\tau}_{0}\check{L}_{\lambda}(t,x_{0}(t),\dot{x}_{0}(t))dt\right]$
$\displaystyle+\left[\int^{\tau}_{0}\check{L}_{\lambda}(t,x_{0}(t),\dot{x}_{0}(t))dt-\int^{\tau}_{0}\check{L}_{\lambda_{0}}(t,x_{0}(t),\dot{x}_{0}(t))dt\right].$

As $(\lambda,x)\to(\lambda_{0},x_{0})$, we derive from (L6) in Lemma 6.1 and
[37, Prop.B.9] or [35, Proposition C.1] (resp. (L6) in Lemma 6.1 and the
Lebesgue dominated convergence theorem) that the first (resp. second) bracket
on the right side converges to the zero.

(Actually, we only need that
$\hat{\Lambda}\times\mathcal{U}^{X}\ni(\lambda,x)\mapsto\check{\mathscr{L}}_{\lambda}(x)\in{\mathbb{R}}$
is continuous. This can easily be proved as follows. For any fixed point
$x_{0}\in\mathcal{U}^{X}$, we can take a positive $\rho>0$ such that
$\rho>\sup_{t}|\dot{x}_{0}(t)|$. Since $E_{\bar{\gamma}}$ is an orthogonal
matrix, and
$\check{L}:\hat{\Lambda}\times\mathbb{R}\times\bar{B}^{n}_{2\iota}(0)\times\mathbb{R}^{n}\to{\mathbb{R}}$
is continuous, we deduce that $\check{L}$ is uniformly continuous in
$\hat{\Lambda}\times\mathbb{R}\times\bar{B}^{n}_{2\iota}(0)\times\bar{B}^{n}_{\rho}(0)$.
This and (6.12) lead to the desired claim.)

Step 2(Prove that $\hat{\Lambda}\times\mathcal{U}^{X}\ni(\lambda,x)\mapsto
A_{\lambda}(x)\in{\bf X}$ is continuous). For
$(\lambda_{1},x),(\lambda_{2},y)\in\hat{\Lambda}\times\mathcal{U}^{X}$, and
$\xi\in{\bf H}$, since

$\displaystyle
d\check{\mathscr{L}}_{\lambda_{1}}(x)[\xi]-d\check{\mathscr{L}}_{\lambda_{2}}(y)[\xi]$
$\displaystyle=$
$\displaystyle\int_{0}^{\tau}\left(\partial_{q}\check{L}(\lambda_{1},t,x(t),\dot{x}(t))-\partial_{q}\check{L}(\lambda_{2},t,y(t),\dot{y}(t))\right)\cdot\xi(t)dt$
$\displaystyle+$
$\displaystyle\int_{0}^{\tau}\left(\partial_{v}\check{L}\left(\lambda_{1},t,x(t),\dot{x}(t))-\partial_{v}\check{L}(\lambda_{2},t,y(t),\dot{y}(t))\right)\cdot\dot{\xi}(t)\right)\,dt,$

we have

$\displaystyle\|\nabla\check{\mathscr{L}}_{\lambda_{1}}(x)-\nabla\check{\mathscr{L}}_{\lambda_{2}}(y)\|_{1,2}$
$\displaystyle\leq$
$\displaystyle\left(\int_{0}^{\tau}\left|\partial_{q}\check{L}(\lambda_{1},t,x(t),\dot{x}(t))-\partial_{q}\check{L}(\lambda_{2},t,y(t),\dot{y}(t))\right|^{2}dt\right)^{1/2}$
$\displaystyle+$
$\displaystyle\left(\int_{0}^{\tau}\left|\partial_{v}\check{L}(\lambda_{1},t,x(t),\dot{x}(t))-\partial_{v}\check{L}(\lambda_{2},t,y(t),\dot{y}(t))\right|^{2}dt\right)^{1/2}.$

Fix a point $(\lambda_{1},x)\in\hat{\Lambda}\times\mathcal{U}^{X}$. Then
$\\{(\lambda_{1},t,x(t),\dot{x}(t))\,|\,t\in[0,\tau]\\}$ is a compact subset
of $\hat{\Lambda}\times[0,\tau]\times B^{n}_{2\iota}(0)\times\mathbb{R}^{n}$.
Since $\partial_{q}\check{L}$ and $\partial_{v}\check{L}$ are uniformly
continuous in any compact neighborhood of this compact subset we deduce that

$\|\nabla\check{\mathscr{L}}_{\lambda_{1}}(x)-\nabla\check{\mathscr{L}}_{\lambda_{2}}(y)\|_{C^{0}}\leq
C_{\tau}\|\nabla\check{\mathscr{L}}_{\lambda_{1}}(x)-\nabla\check{\mathscr{L}}_{\lambda_{2}}(y)\|_{1,2}\to
0$ (6.36)

provided $(\lambda_{2},y)\in\hat{\Lambda}\times\mathcal{U}^{X}$ converges to
$(\lambda_{1},x)$ in $\hat{\Lambda}\times\mathcal{U}^{X}$.

By (6.19) and (6.24), we have

$\displaystyle\frac{d}{dt}\nabla\check{\mathscr{L}}_{\lambda}(x)(t)$
$\displaystyle=$
$\displaystyle\frac{e^{t}}{2}\int^{\infty}_{t}e^{-s}\left(\partial_{q}\check{L}_{{\lambda}}\bigl{(}s,x(s),\dot{x}(s)\bigr{)}-\mathfrak{R}_{\lambda}^{x}(s)\right)\,ds$
(6.37)
$\displaystyle-\frac{e^{-t}}{2}\int^{t}_{-\infty}e^{s}\left(\partial_{q}\check{L}_{{\lambda}}\bigl{(}s,x(s),\dot{x}(s)\bigr{)}-\mathfrak{R}_{\lambda}^{x})(s)\right)\,ds$
$\displaystyle+\partial_{v}\check{L}_{{\lambda}}\bigl{(}t,x(t),\dot{x}(t)\bigr{)}+\mathfrak{M}\int^{\tau}_{0}\partial_{v}\check{L}_{{\lambda}}(s,x(s),\dot{x}(s))ds,$

where $\mathfrak{R}_{\lambda}^{x}$ is given by (6.2). Let
$\mathfrak{T}_{\lambda}^{x}(t)$ denote a column vector

$\displaystyle\left[\left(\oplus_{l\leq
p}\frac{\sin\theta_{l}}{2-2\cos\theta_{l}}{\scriptscriptstyle\left(\begin{array}[]{cc}0&-1\\\
1&0\end{array}\right)}-\frac{1}{2}I_{2p}\right)\oplus{\rm
diag}(a_{p+1}(t),\cdots,a_{\sigma}(t))\right]\int^{\tau}_{0}\partial_{v}\check{L}_{\lambda}\bigl{(}s,x(s),\dot{x}(s)\bigr{)}ds.$

Then we have a constant $C(E_{\bar{\gamma}})>0$ only depending on
$E_{\bar{\gamma}}$ such that for all $t$,

$\big{|}\mathfrak{T}_{\lambda_{1}}^{x}(t)-\mathfrak{T}_{\lambda_{2}}^{y}(t)\big{|}\leq
C(E_{\gamma_{0}})(1+|t|)\int^{\tau}_{0}\Big{|}\partial_{v}\check{L}_{\lambda_{1}}\bigl{(}s,x(s),\dot{x}(s)\bigr{)}-\partial_{v}\check{L}_{\lambda_{2}}\bigl{(}s,y(s),\dot{y}(s)\bigr{)}\Big{|}ds.$
(6.39)

By (6.2) and (6.24) we observe

$\displaystyle\mathfrak{R}_{\lambda}^{x}(t)=\int^{t}_{0}\partial_{v}\check{L}_{\lambda}\bigl{(}s,x(s),\dot{x}(s)\bigr{)}ds+\mathfrak{T}_{\lambda}^{x}(t),$
(6.41)
$\displaystyle\frac{d}{dt}\mathfrak{T}_{\lambda}^{x}(t)=\mathfrak{M}\int^{\tau}_{0}\partial_{v}\check{L}_{\lambda}\bigl{(}s,x(s),\dot{x}(s)\bigr{)}ds.$

For $0\leq t\leq\tau$, let

$\displaystyle\Gamma^{+}_{\lambda}(x)(t):$ $\displaystyle=$
$\displaystyle\frac{e^{t}}{2}\int^{\infty}_{t}e^{-s}\left(\partial_{q}\check{L}_{{\lambda}}\bigl{(}s,x(s),\dot{x}(s)\bigr{)}-\mathfrak{R}_{\lambda}^{x}(s)\right)ds,$
(6.42) $\displaystyle\Gamma^{-}_{\lambda}(x)(t):$ $\displaystyle=$
$\displaystyle\frac{e^{-t}}{2}\int^{t}_{-\infty}e^{s}\left(\partial_{q}\check{L}_{{\lambda}}\bigl{(}s,x(s),\dot{x}(s)\bigr{)}-\mathfrak{R}_{\lambda}^{x}(s)\right)ds.$
(6.43)

It follows from (6.37) that

$\displaystyle\left|\frac{d}{dt}\nabla\check{\mathscr{L}}_{\lambda_{1}}(x)(t)-\frac{d}{dt}\nabla\check{\mathscr{L}}_{\lambda_{2}}(y)(t)\right|$
(6.44) $\displaystyle\leq$
$\displaystyle|\Gamma^{+}_{\lambda_{1}}(x)(t)-\Gamma^{+}_{\lambda_{2}}(y)(t)|+|\Gamma^{-}_{\lambda_{1}}(x)(t)-\Gamma^{-}_{\lambda_{2}}(y)(t)|$
$\displaystyle+|\partial_{v}\check{L}_{{\lambda_{1}}}\bigl{(}t,x(t),\dot{x}(t)\bigr{)}-\partial_{v}\check{L}_{{\lambda_{2}}}\bigl{(}t,y(t),\dot{y}(t)\bigr{)}|$
$\displaystyle+|\mathfrak{M}|\int^{\tau}_{0}|\partial_{v}\check{L}_{{\lambda_{1}}}(s,x(s),\dot{x}(s))-\partial_{v}\check{L}_{{\lambda_{2}}}(s,y(s),\dot{y}(s))|ds.$

Let us estimate terms in the right side.

Suppose $k\tau\leq t\leq(k+1)\tau$ for some integer $k\geq 0$. By
$\check{L}(\lambda,t+\tau,x,v)=\check{L}(\lambda,t,E_{{\bar{\gamma}}}x,E_{{\bar{\gamma}}}v)$
and (6.39) we deduce

$\displaystyle|\mathfrak{R}_{\lambda_{1}}^{x}(t)-\mathfrak{R}_{\lambda_{2}}^{y}(t)|$
(6.45) $\displaystyle\leq$
$\displaystyle\int^{(k+1)\tau}_{0}\left|\partial_{v}\check{L}_{\lambda_{1}}\bigl{(}s,x(s),\dot{x}(s)\bigr{)}-\partial_{v}\check{L}_{\lambda_{2}}\bigl{(}s,y(s),\dot{y}(s)\bigr{)}\right|ds+\big{|}\mathfrak{T}_{\lambda_{1}}^{x}(t)-\mathfrak{T}_{\lambda_{2}}^{y}(t)\big{|}$
$\displaystyle\leq$
$\displaystyle(k+1)\int^{\tau}_{0}\left|\partial_{v}\check{L}_{\lambda_{1}}\bigl{(}s,x(s),\dot{x}(s)\bigr{)}-\partial_{v}\check{L}_{\lambda_{2}}\bigl{(}s,y(s),\dot{y}(s)\bigr{)}\right|ds$
$\displaystyle+$ $\displaystyle
C(E_{\bar{\gamma}})(1+(k+1)\tau)\int^{\tau}_{0}\Big{|}\partial_{v}\check{L}_{\lambda_{1}}\bigl{(}s,x(s),\dot{x}(s)\bigr{)}-\partial_{v}\check{L}_{\lambda_{2}}\bigl{(}s,y(s),\dot{y}(s)\bigr{)}\Big{|}ds.$

Similarly, for $(-k-1)\tau\leq t\leq-k\tau$ for some integer $k\geq 0$. We
have also

$\displaystyle|\mathfrak{R}_{\lambda_{1}}^{x}(t)-\mathfrak{R}_{\lambda_{2}}^{y}(t)|$
(6.46) $\displaystyle\leq$
$\displaystyle\int_{(-k-1)\tau}^{0}\left|\partial_{v}\check{L}_{\lambda_{1}}\bigl{(}s,x(s),\dot{x}(s)\bigr{)}-\partial_{v}\check{L}_{\lambda_{2}}\bigl{(}s,y(s),\dot{y}(s)\bigr{)}\right|ds+\big{|}\mathfrak{T}_{\lambda_{1}}^{x}(t)-\mathfrak{T}_{\lambda_{2}}^{y}(t)\big{|}$
$\displaystyle\leq$
$\displaystyle(k+1)\int^{\tau}_{0}\left|\partial_{v}\check{L}_{\lambda_{1}}\bigl{(}s,x(s),\dot{x}(s)\bigr{)}-\partial_{v}\check{L}_{\lambda_{2}}\bigl{(}s,y(s),\dot{y}(s)\bigr{)}\right|ds$
$\displaystyle+$ $\displaystyle
C(E_{\bar{\gamma}})(1+(k+1)\tau)\int^{\tau}_{0}\Big{|}\partial_{v}\check{L}_{\lambda_{1}}\bigl{(}s,x(s),\dot{x}(s)\bigr{)}-\partial_{v}\check{L}_{\lambda_{2}}\bigl{(}s,y(s),\dot{y}(s)\bigr{)}\Big{|}ds.$

For $0\leq t\leq\tau$, it follows from (6.42) and (6.45) that

$\displaystyle|\Gamma^{+}_{\lambda_{1}}(x)(t)-\Gamma^{+}_{\lambda_{2}}(y)(t)|$
$\displaystyle\leq$
$\displaystyle\frac{e^{\tau}}{2}\sum^{\infty}_{k=0}\int^{(k+1)\tau}_{k\tau}e^{-s}\left|\partial_{q}\check{L}_{{\lambda_{1}}}\bigl{(}s,x(s),\dot{x}(s)\bigr{)}-\partial_{q}\check{L}_{{\lambda_{2}}}\bigl{(}s,y(s),\dot{y}(s)\bigr{)}\right|ds$
$\displaystyle+\frac{e^{\tau}}{2}\sum^{\infty}_{k=0}\int^{(k+1)\tau}_{k\tau}e^{-s}\left|\mathfrak{R}_{\lambda_{1}}^{x}(s)-\mathfrak{R}_{\lambda_{2}}^{y}(s)\right|ds$
$\displaystyle\leq$
$\displaystyle\frac{e^{\tau}}{2}\sum^{\infty}_{k=0}e^{-k\tau}\int^{\tau}_{0}\left|\partial_{q}\check{L}_{{\lambda_{1}}}\bigl{(}s,x(s),\dot{x}(s)\bigr{)}-\partial_{q}\check{L}_{{\lambda_{2}}}\bigl{(}s,y(s),\dot{y}(s)\bigr{)}\right|ds$
$\displaystyle+\frac{\tau
e^{\tau}}{2}\sum^{\infty}_{k=0}e^{-k\tau}(k+1)\int^{\tau}_{0}\left|\partial_{v}\check{L}_{\lambda_{1}}\bigl{(}s,x(s),\dot{x}(s)\bigr{)}-\partial_{v}\check{L}_{\lambda_{2}}\bigl{(}s,y(s),\dot{y}(s)\bigr{)}\right|ds$
$\displaystyle+C(E_{\bar{\gamma}})\frac{\tau
e^{\tau}}{2}\sum^{\infty}_{k=0}e^{-k\tau}(1+(k+1)\tau)\int^{\tau}_{0}\Big{|}\partial_{v}\check{L}_{\lambda_{1}}\bigl{(}s,x(s),\dot{x}(s)\bigr{)}-\partial_{v}\check{L}_{\lambda_{2}}\bigl{(}s,y(s),\dot{y}(s)\bigr{)}\Big{|}ds.$

Similarly, for $0\leq t\leq\tau$, (6.43) and (6.46) lead to

$\displaystyle|\Gamma^{-}_{\lambda_{1}}(x)(t)-\Gamma^{-}_{\lambda_{2}}(y)(t)|$
$\displaystyle\leq$
$\displaystyle\frac{1}{2}\sum^{\infty}_{k=0}\int^{(-k+1)\tau}_{-k\tau}e^{s}\left|\partial_{q}\check{L}_{{\lambda_{1}}}\bigl{(}s,x(s),\dot{x}(s)\bigr{)}-\partial_{q}\check{L}_{{\lambda_{2}}}\bigl{(}s,y(s),\dot{y}(s)\bigr{)}\right|ds$
$\displaystyle+\frac{1}{2}\sum^{\infty}_{k=0}\int^{(-k+1)\tau}_{-k\tau}e^{s}\left|\mathfrak{R}_{\lambda_{1}}^{x}(s)-\mathfrak{R}_{\lambda_{2}}^{y}(s)\right|ds$
$\displaystyle\leq$
$\displaystyle\frac{e^{\tau}}{2}\sum^{\infty}_{k=0}e^{-k\tau}\int^{\tau}_{0}\left|\partial_{q}\check{L}_{{\lambda_{1}}}\bigl{(}s,x(s),\dot{x}(s)\bigr{)}-\partial_{q}\check{L}_{{\lambda_{2}}}\bigl{(}s,y(s),\dot{y}(s)\bigr{)}\right|ds$
$\displaystyle+\frac{\tau
e^{\tau}}{2}\sum^{\infty}_{k=0}e^{-k\tau}(k+1)\int^{\tau}_{0}\left|\partial_{v}\check{L}_{\lambda_{1}}\bigl{(}s,x(s),\dot{x}(s)\bigr{)}-\partial_{v}\check{L}_{\lambda_{2}}\bigl{(}s,y(s),\dot{y}(s)\bigr{)}\right|ds$
$\displaystyle+C(E_{\bar{\gamma}})\frac{\tau
e^{\tau}}{2}\sum^{\infty}_{k=0}e^{-k\tau}(1+(k+1)\tau)\int^{\tau}_{0}\Big{|}\partial_{v}\check{L}_{\lambda_{1}}\bigl{(}s,x(s),\dot{x}(s)\bigr{)}-\partial_{v}\check{L}_{\lambda_{2}}\bigl{(}s,y(s),\dot{y}(s)\bigr{)}\Big{|}ds.$

From these and (6.44) we get

$\displaystyle\left|\frac{d}{dt}\nabla\check{\mathscr{L}}_{\lambda_{1}}(x)(t)-\frac{d}{dt}\nabla\check{\mathscr{L}}_{\lambda_{2}}(y)(t)\right|$
(6.49) $\displaystyle\leq$
$\displaystyle|\partial_{v}\check{L}_{{\lambda_{1}}}\bigl{(}t,x(t),\dot{x}(t)\bigr{)}-\partial_{v}\check{L}_{{\lambda_{2}}}\bigl{(}t,y(t),\dot{y}(t)\bigr{)}|$
$\displaystyle+(C(E_{\bar{\gamma}})C_{\tau}+|\mathfrak{M}|)\int^{\tau}_{0}|\partial_{v}\check{L}_{{\lambda_{1}}}(s,x(s),\dot{x}(s))-\partial_{v}\check{L}_{{\lambda_{2}}}(s,y(s),\dot{y}(s))|ds$
$\displaystyle+C^{\ast}_{\tau}\int^{\tau}_{0}|\partial_{q}\check{L}_{{\lambda_{1}}}(s,x(s),\dot{x}(s))-\partial_{q}\check{L}_{{\lambda_{2}}}(s,y(s),\dot{y}(s))|ds$

for some constant $C^{\ast}_{\tau}>0$ and for all $0\leq t\leq\tau$.

As above, for a fixed $(\lambda_{1},x)\in\hat{\Lambda}\times\mathcal{U}^{X}$,
by uniform continuity of $\partial_{q}\check{L}$ and $\partial_{v}\check{L}$
on a compact neighborhood of a compact subset
$\\{(\lambda_{1},t,x(t),\dot{x}(t))\,|\,t\in[0,\tau]\\}$ of
$\hat{\Lambda}\times[0,\tau]\times B^{n}_{2\iota}(0)\times\mathbb{R}^{n}$, we
can derive from these and (6.49) that

$\displaystyle\left\|\frac{d}{dt}\nabla\check{\mathscr{L}}_{\lambda_{1}}(x)-\frac{d}{dt}\nabla\check{\mathscr{L}}_{\lambda_{2}}(y)\right\|_{C^{0}}\to
0$

provided $(\lambda_{2},y)\in\hat{\Lambda}\times\mathcal{U}^{X}$ converges to
$(\lambda_{1},x)$ in $\hat{\Lambda}\times\mathcal{U}^{X}$. This and (6.36)
lead to the second claim. ∎

###### Proof of Proposition 3.11.

It is enough to prove sufficiency. Since $x\in B_{{\bf
H}^{\bot}}(0,\varepsilon)$ is a critical point of the restriction of
$\check{\mathscr{L}}_{\lambda}$ to $B_{{\bf H}^{\bot}}(0,\varepsilon)$,

$d\check{\mathscr{L}}_{\lambda}(x)[\xi]=0\;\;\forall\xi\in T_{x}B_{{\bf
H}^{\bot}}(0,\varepsilon)={\bf H}^{\bot}.$ (6.50)

We shall prove $d\check{\mathscr{L}}_{\lambda}(x)=0$ in four steps.

Step 1. Prove that $x$ is $C^{4}$. By (6.50) we have
$\mu(\lambda,x)\in{\mathbb{R}}$ such that

$\nabla\check{\mathscr{L}}_{\lambda}(x)=\mu(\lambda,x)\zeta_{0}.$ (6.51)

That is,
$d\check{\mathscr{L}}_{\lambda}(x)[\xi]-\mu(\lambda,x)(\zeta_{0},\xi)_{1,2}=0$
for all $\xi\in{\bf H}$. It follows that

$\displaystyle\int^{\tau}_{0}[(\partial_{q}\check{L}_{\lambda}(x(t),\dot{x}(t)),\xi(t))_{\mathbb{R}^{n}}+(\partial_{v}\check{L}_{\lambda}(x(t),\dot{x}(t),\dot{\xi}(t))_{\mathbb{R}^{n}}]dt$
$\displaystyle-\int^{\tau}_{0}[\mu(\lambda,x)(\zeta_{0}(t),\xi(t))_{\mathbb{R}^{n}}+\mu(\lambda,x)(\dot{\zeta}_{0}(t),\dot{\xi}(t))_{\mathbb{R}^{n}}]dt=0\quad\forall\xi\in{\bf
H}.$

Define ${\bf
L}_{\lambda}(t,q,v)=\check{L}_{\lambda}(t,q,v)-\mu(\lambda,x)(\zeta_{0}(t),q)_{\mathbb{R}^{n}}+\mu(\lambda,x)(\dot{\zeta}_{0}(t),v)_{\mathbb{R}^{n}}$.
Then $x(t)$ satisfies

$\displaystyle\int^{\tau}_{0}[(\partial_{q}{\bf
L}_{\lambda}(t,x(t),\dot{x}(t)),\xi(t))_{\mathbb{R}^{n}}+(\partial_{v}{\bf
L}_{\lambda}(t,x(t),\dot{x}(t)),\dot{\xi}(t))_{\mathbb{R}^{n}}]dt=0\quad\forall\xi\in{\bf
H}.$ (6.52)

Since $\zeta_{0}$ is $C^{5}$, ${\bf L}_{\lambda}$ is $C^{4}$ and satisfies the
conditions in Lemma 6.1, by Remark 6.2 we obtain that $x$ is $C^{4}$.

Step 2. For any $\epsilon>0$ there exists $\delta>0$ such that
$\|x\|_{1,2}<\delta$ implies $|\mu(\lambda,x)|<\epsilon$. Since
$\hat{\Lambda}\subset{\mathbb{R}}$ is compact and sequential compact,
$\check{L}(\lambda,t+\tau,q,v)=\check{L}(\lambda,t,E_{{\bar{\gamma}}}q,E_{{\bar{\gamma}}}v)$,
and partial derivatives

$\partial_{q}\check{L}_{\lambda}(\cdot),\quad\partial_{v}\check{L}_{\lambda}(\cdot),\quad\partial_{qv}\check{L}_{\lambda}(\cdot),\quad\partial_{qq}\check{L}_{\lambda}(\cdot),\quad\partial_{vv}\check{L}_{\lambda}(\cdot),\quad\partial_{vt}\check{L}_{\lambda}(\cdot)$

depend continuously on
$(\lambda,t,q,v)\in\hat{\Lambda}\times\mathbb{R}\times\bar{B}^{n}_{2\iota}(0)\times\mathbb{R}^{n}$,
by shrinking $\iota>0$ (if necessary) it follows from (L3) in Lemma 6.1 that

$\displaystyle|\partial_{q}\check{L}_{\lambda}(t,q,v)-\partial_{q}\check{L}_{\lambda}(t,0,0)|$
$\displaystyle\leq$
$\displaystyle|\partial_{q}\check{L}_{\lambda}(t,q,v)-\partial_{q}\check{L}_{\lambda}(t,q,0)|+|\partial_{q}\check{L}_{\lambda}(t,q,0)-\partial_{q}\check{L}_{\lambda}(t,0,0)|$
$\displaystyle\leq$ $\displaystyle\sup_{0\leq s\leq
1}|\partial_{qv}\check{L}_{\lambda}(t,q,sv)|\cdot|v|+\sup_{0\leq s\leq
1}|\partial_{qq}\check{L}_{\lambda}(t,sq,0)|\cdot|q|$ $\displaystyle\leq$
$\displaystyle C(|v|+|v|^{2})+C|q|.$

Hence we have a constant $C^{\prime}>0$ such that

$|\partial_{q}\check{L}_{\lambda}(t,q,v)|\leq
C^{\prime}(1+|v|^{2}),\quad\forall(\lambda,t,q,v)\in\hat{\Lambda}\times\mathbb{R}\times\bar{B}^{n}_{2\iota}(0)\times\mathbb{R}^{n}.$
(6.53)

Similarly, we can increase the constant $C^{\prime}>0$ so that

$|\partial_{v}\check{L}_{\lambda}(t,q,v)|\leq
C^{\prime}(1+|v|),\quad\forall(\lambda,t,q,v)\in\hat{\Lambda}\times\mathbb{R}\times\bar{B}^{n}_{2\iota}(0)\times\mathbb{R}^{n}.$
(6.54)

Since $\nabla\check{\mathscr{L}}_{\lambda}(0)=0$, by (6.51) we have

$\displaystyle\mu(\lambda,x)(\zeta_{0},\xi)_{1,2}$ $\displaystyle=$
$\displaystyle
d\check{\mathscr{L}}_{\lambda}(x)[\xi]-d\check{\mathscr{L}}_{\lambda}(0)[\xi]$
$\displaystyle=$
$\displaystyle\int^{\tau}_{0}[(\partial_{q}\check{L}_{\lambda}(t,x(t),\dot{x}(t)),\xi(t))_{\mathbb{R}^{n}}-(\partial_{q}\check{L}_{\lambda}(t,0,0),\xi(t))_{\mathbb{R}^{n}}]dt$
$\displaystyle+\int^{\tau}_{0}[(\partial_{v}\check{L}_{\lambda}(t,x(t),\dot{x}(t),\dot{\xi}(t))_{\mathbb{R}^{n}}-(\partial_{v}\check{L}_{\lambda}(t,0,0),\dot{\xi}(t))_{\mathbb{R}^{n}}]dt.$

For the first integral, by the mean value theorem, (L3) in Lemma 6.1 and (2.3)
we derive

$\displaystyle\Big{|}\int^{\tau}_{0}[(\partial_{q}\check{L}_{\lambda}(t,x(t),\dot{x}(t))-\partial_{q}\check{L}_{\lambda}(t,0,0),\xi(t))_{\mathbb{R}^{n}}]dt\Big{|}$
(6.55) $\displaystyle\leq$
$\displaystyle\Big{|}\int^{\tau}_{0}\int^{1}_{0}[(\partial_{qq}\check{L}_{\lambda}(t,sx(t),s\dot{x}(t))x(t),\xi(t))_{\mathbb{R}^{n}}]dtds\Big{|}$
$\displaystyle\qquad+\Big{|}\int^{\tau}_{0}\int^{1}_{0}[(\partial_{qv}\check{L}_{\lambda}(t,sx(t),s\dot{x}(t))\dot{x}(t),\xi(t))_{\mathbb{R}^{n}}]dtds\Big{|}$
$\displaystyle\leq$ $\displaystyle
C\int^{\tau}_{0}\int^{1}_{0}(1+|s\dot{x}(t))|^{2})|x(t)||\xi(t)|dtds+C\int^{\tau}_{0}\int^{1}_{0}(1+|s\dot{x}(t))|)|\dot{x}(t)||\xi(t)|dtds$
$\displaystyle\leq$ $\displaystyle
C(C_{\tau})^{2}\|x\|_{1,2}\|\xi\|_{1,2}\big{(}\tau+2\|x\|_{1,2}^{2}\big{)}+CC_{\tau}\|\xi\|_{1,2}(\sqrt{\tau}\|x\|_{1,2}+\|x\|_{1,2}^{2}).$

Similarly, we may estimate the second integral as follows:

$\displaystyle\Big{|}\int^{\tau}_{0}[(\partial_{v}\check{L}_{\lambda}(t,x(t),\dot{x}(t),\dot{\xi}(t))_{\mathbb{R}^{n}}-(\partial_{v}\check{L}_{\lambda}(t,0,0),\dot{\xi}(t))_{\mathbb{R}^{n}}]dt\Big{|}$
(6.56) $\displaystyle\leq$
$\displaystyle\Big{|}\int^{\tau}_{0}\int^{1}_{0}[(\partial_{vv}\check{L}_{\lambda}(t,sx(t),s\dot{x}(t))\dot{x}(t),\dot{\xi}(t))_{\mathbb{R}^{n}}]dtds\Big{|}$
$\displaystyle\qquad+\Big{|}\int^{\tau}_{0}\int^{1}_{0}[(\partial_{qv}\check{L}_{\lambda}(t,sx(t),s\dot{x}(t)){x}(t),\dot{\xi}(t))_{\mathbb{R}^{n}}]dtds\Big{|}$
$\displaystyle\leq$ $\displaystyle
C\|x\|_{1,2}\|\xi\|_{1,2}+CC_{\tau}\|x\|_{1,2}(\sqrt{\tau}\|\xi\|_{1,2}+\|x\|_{1,2}\|\xi\|_{1,2}).$

Hence we get

$\displaystyle|\mu(\lambda,x)|\|\zeta_{0}\|_{1,2}$ $\displaystyle\leq$
$\displaystyle
C(C_{\tau})^{2}\|x\|_{1,2}\big{(}\tau+2\|x\|_{1,2}^{2}\big{)}+CC_{\tau}(\sqrt{\tau}\|x\|_{1,2}+\|x\|_{1,2}^{2})$
$\displaystyle+C\|x\|_{1,2}+CC_{\tau}\|x\|_{1,2}(\sqrt{\tau}+\|x\|_{1,2}).$

The desired claim immediately follows because $\zeta_{0}\neq 0$.

Step 3. For any $\epsilon>0$ there exists $\varepsilon>0$ such that
$\|x\|_{1,2}\leq\varepsilon$ implies $\|x\|_{C^{2}}<\epsilon$. By (L2) in
Lemma 6.1 and the mean value theorem of integrals we deduce

$\displaystyle c|v|^{2}$ $\displaystyle\leq$
$\displaystyle\int^{1}_{0}\left(\partial_{vv}\check{L}_{\lambda}(t,q,sv)[v],v\right)_{\mathbb{R}^{n}}ds$
$\displaystyle=$
$\displaystyle\left(\partial_{v}\check{L}_{\lambda}(t,q,v)-\partial_{v}\check{L}_{\lambda}(t,q,0),v\right)_{\mathbb{R}^{n}}$

and so

$c|v|\leq\left|\partial_{v}\check{L}_{\lambda}(t,q,v)-\partial_{v}\check{L}_{\lambda}(t,q,0)\right|$
(6.57)

for any
$(\lambda,t,q,v)\in\hat{\Lambda}\times\mathbb{R}\times\bar{B}^{n}_{2\iota}(0)\times\mathbb{R}^{n}$.
Since we have proved that $x$ is $C^{4}$ in Step 1, (6.51) also holds in the
sense of pointwise, i.e.,

$\nabla\check{\mathscr{L}}_{\lambda}(x)(t)=\mu(\lambda,x)\zeta_{0}(t)\;\forall
t.$ (6.58)

From this and (6.37) it follows that

$\displaystyle\mu(\lambda,x)\dot{\zeta}_{0}(t)$ $\displaystyle=$
$\displaystyle\frac{e^{t}}{2}\int^{\infty}_{t}e^{-s}\left(\partial_{q}\check{L}_{{\lambda}}\bigl{(}s,x(s),\dot{x}(s)\bigr{)}-\mathfrak{R}_{\lambda}^{x}(s)\right)\,ds$
(6.59)
$\displaystyle-\frac{e^{-t}}{2}\int^{t}_{-\infty}e^{s}\left(\partial_{q}\check{L}_{{\lambda}}\bigl{(}s,x(s),\dot{x}(s)\bigr{)}-\mathfrak{R}_{\lambda}^{x})(s)\right)\,ds$
$\displaystyle+\partial_{v}\check{L}_{{\lambda}}\bigl{(}t,x(t),\dot{x}(t)\bigr{)}+\mathfrak{M}\int^{\tau}_{0}\partial_{v}\check{L}_{{\lambda}}(s,x(s),\dot{x}(s))ds,$

where $\mathfrak{R}_{\lambda}^{x}$ is given by (6.2). In particular, taking
$x=0$ we get

$\displaystyle\mu(\lambda,0)\dot{\zeta}_{0}(t)$ $\displaystyle=$
$\displaystyle\frac{e^{t}}{2}\int^{\infty}_{t}e^{-s}\left(\partial_{q}\check{L}_{{\lambda}}\bigl{(}s,0,0\bigr{)}-\mathfrak{R}_{\lambda}^{0}(s)\right)\,ds$
(6.60)
$\displaystyle-\frac{e^{-t}}{2}\int^{t}_{-\infty}e^{s}\left(\partial_{q}\check{L}_{{\lambda}}\bigl{(}s,0,0\bigr{)}-\mathfrak{R}_{\lambda}^{0})(s)\right)\,ds$
$\displaystyle+\partial_{v}\check{L}_{{\lambda}}\bigl{(}t,0,0\bigr{)}+\mathfrak{M}\int^{\tau}_{0}\partial_{v}\check{L}_{{\lambda}}(s,0,0)ds.$

(6.59) minus (6.60) gives rise to

$\displaystyle\partial_{v}\check{L}_{{\lambda}}\bigl{(}t,0,0\bigr{)}-\partial_{v}\check{L}_{{\lambda}}\bigl{(}t,x(t),\dot{x}(t)\bigr{)}$
(6.61) $\displaystyle=$
$\displaystyle\frac{e^{t}}{2}\int^{\infty}_{t}e^{-s}\left(\partial_{q}\check{L}_{{\lambda}}\bigl{(}s,x(s),\dot{x}(s)\bigr{)}-\partial_{q}\check{L}_{{\lambda}}\bigl{(}s,0,0\bigr{)}\right)\,ds+\frac{e^{t}}{2}\int^{\infty}_{t}e^{-s}\left(\mathfrak{R}_{\lambda}^{0}(s)-\mathfrak{R}_{\lambda}^{x}(s)\right)\,ds$
$\displaystyle-\frac{e^{-t}}{2}\int^{t}_{-\infty}e^{s}\left(\partial_{q}\check{L}_{{\lambda}}\bigl{(}s,x(s),\dot{x}(s)\bigr{)}-\partial_{q}\check{L}_{{\lambda}}\bigl{(}s,0,0\bigr{)}\right)\,ds+\frac{e^{-t}}{2}\int^{t}_{-\infty}e^{s}\left(\mathfrak{R}_{\lambda}^{x}(s)-\mathfrak{R}_{\lambda}^{0}(s)\right)\,ds$
$\displaystyle+\mathfrak{M}\int^{\tau}_{0}[\partial_{v}\check{L}_{{\lambda}}(s,x(s),\dot{x}(s))-\partial_{v}\check{L}_{{\lambda}}(s,0,0)]ds$
$\displaystyle+\mu(\lambda,0)\dot{\zeta}_{0}(t)-\mu(\lambda,x)\dot{\zeta}_{0}(t)$

For $0\leq t\leq\tau$, it follows from (6.4) and (6.4) that

$\displaystyle\bigg{|}\frac{e^{t}}{2}\int^{\infty}_{t}e^{-s}\left(\partial_{q}\check{L}_{{\lambda}}\bigl{(}s,x(s),\dot{x}(s)\bigr{)}-\partial_{q}\check{L}_{{\lambda}}\bigl{(}s,0,0\bigr{)}\right)\,ds+\frac{e^{t}}{2}\int^{\infty}_{t}e^{-s}\left(\mathfrak{R}_{\lambda}^{0}(s)-\mathfrak{R}_{\lambda}^{x}(s)\right)\,ds\bigg{|}$
(6.62) $\displaystyle\leq$
$\displaystyle\biggl{[}\frac{e^{\tau}}{2}\sum^{\infty}_{k=0}e^{-k\tau}+\frac{\tau
e^{\tau}}{2}\sum^{\infty}_{k=0}e^{-k\tau}(k+1)+C(E_{\bar{\gamma}})\frac{\tau
e^{\tau}}{2}\sum^{\infty}_{k=0}e^{-k\tau}(1+(k+1)\tau)\biggr{]}\times$
$\displaystyle\times\Big{|}\int^{\tau}_{0}\big{[}\partial_{v}\check{L}_{\lambda}\bigl{(}s,x(s),\dot{x}(s)\bigr{)}-\partial_{v}\check{L}_{\lambda}\bigl{(}s,0,0\bigr{)}\big{]}ds\Big{|}$

and

$\displaystyle\bigg{|}-\frac{e^{-t}}{2}\int^{t}_{-\infty}e^{s}\left(\partial_{q}\check{L}_{{\lambda}}\bigl{(}s,x(s),\dot{x}(s)\bigr{)}-\partial_{q}\check{L}_{{\lambda}}\bigl{(}s,0,0\bigr{)}\right)\,ds+\frac{e^{-t}}{2}\int^{t}_{-\infty}e^{s}\left(\mathfrak{R}_{\lambda}^{x}(s)-\mathfrak{R}_{\lambda}^{0}(s)\right)\,ds\bigg{|}$
(6.63) $\displaystyle\leq$
$\displaystyle\biggl{[}\frac{e^{\tau}}{2}\sum^{\infty}_{k=0}e^{-k\tau}+\frac{\tau
e^{\tau}}{2}\sum^{\infty}_{k=0}e^{-k\tau}(k+1)+C(E_{\bar{\gamma}})\frac{\tau
e^{\tau}}{2}\sum^{\infty}_{k=0}e^{-k\tau}(1+(k+1)\tau)\biggr{]}\times$
$\displaystyle\times\Big{|}\int^{\tau}_{0}\big{[}\partial_{v}\check{L}_{\lambda}\bigl{(}s,x(s),\dot{x}(s)\bigr{)}-\partial_{v}\check{L}_{\lambda}\bigl{(}s,0,0\bigr{)}\big{]}ds\Big{|}.$

An elementary calculation yields

$\displaystyle\biggl{[}\frac{e^{\tau}}{2}\sum^{\infty}_{k=0}e^{-k\tau}+\frac{\tau
e^{\tau}}{2}\sum^{\infty}_{k=0}e^{-k\tau}(k+1)+C(E_{\bar{\gamma}})\frac{\tau
e^{\tau}}{2}\sum^{\infty}_{k=0}e^{-k\tau}(1+(k+1)\tau)\biggr{]}$
$\displaystyle<$ $\displaystyle
C(E_{\bar{\gamma}},\tau):=(2+C(E_{\bar{\gamma}})\tau+C(E_{\bar{\gamma}})\tau^{2})\frac{e^{2\tau}}{e^{\tau}-1}+(1+C(E_{\bar{\gamma}})\tau^{2})\frac{e^{4\tau}}{(e^{\tau}-1)^{2}}.$

Hence (6.61), (6.62) and (6.63) lead to

$\displaystyle\left|\partial_{v}\check{L}_{{\lambda}}\bigl{(}t,x(t),\dot{x}(t)\bigr{)}-\partial_{v}\check{L}_{{\lambda}}\bigl{(}t,0,0\bigr{)}\right|$
$\displaystyle\leq$ $\displaystyle
2C(E_{\bar{\gamma}},\tau)\Big{|}\int^{\tau}_{0}\big{[}\partial_{v}\check{L}_{\lambda}\bigl{(}s,x(s),\dot{x}(s)\bigr{)}-\partial_{v}\check{L}_{\lambda}\bigl{(}s,0,0\bigr{)}\big{]}ds\Big{|}$
$\displaystyle+|\mathfrak{M}|\Big{|}\int^{\tau}_{0}\big{[}\partial_{v}\check{L}_{\lambda}\bigl{(}s,x(s),\dot{x}(s)\bigr{)}-\partial_{v}\check{L}_{\lambda}\bigl{(}s,0,0\bigr{)}\big{]}ds\Big{|}+|\mu(\lambda,0)-\mu(\lambda,x)|\cdot|\dot{\zeta}_{0}(t)|.$

From this, (6.57) and (6.61), (6.62) and (6.63) we deduce

$\displaystyle c|\dot{x}(t)|$ $\displaystyle\leq$
$\displaystyle\left|\partial_{v}\check{L}_{{\lambda}}\bigl{(}t,x(t),\dot{x}(t)\bigr{)}-D_{v}\check{L}_{\lambda}(t,x(t),0)\right|$
(6.64) $\displaystyle\leq$
$\displaystyle\left|\partial_{v}\check{L}_{{\lambda}}\bigl{(}t,x(t),\dot{x}(t)\bigr{)}-\partial_{v}\check{L}_{{\lambda}}\bigl{(}t,0,0\bigr{)}\right|+\left|D_{v}\check{L}_{\lambda}(t,x(t),0)-\partial_{v}\check{L}_{{\lambda}}\bigl{(}t,0,0\bigr{)}\right|$
$\displaystyle\leq$
$\displaystyle(2C(E_{\bar{\gamma}},\tau)+|\mathfrak{M}|)\Big{|}\int^{\tau}_{0}\big{[}\partial_{v}\check{L}_{\lambda}\bigl{(}s,x(s),\dot{x}(s)\bigr{)}-\partial_{v}\check{L}_{\lambda}\bigl{(}s,0,0\bigr{)}\big{]}ds\Big{|}$
$\displaystyle+|\mu(\lambda,0)-\mu(\lambda,x)|\cdot|\dot{\zeta}_{0}(t)|+\left|D_{v}\check{L}_{\lambda}(t,x(t),0)-\partial_{v}\check{L}_{{\lambda}}\bigl{(}t,0,0\bigr{)}\right|$

for all $0\leq t\leq\tau$.

Since $|\dot{\zeta}_{0}(t)|$ are bounded on $[0,\tau]$, by Step 2 we have
$|\mu(\lambda,0)-\mu(\lambda,x)|\cdot\sup_{t}|\dot{\zeta}_{0}(t)|\to 0$ as
$\|x\|_{1,2}\to 0$. It follows from (6.54) and [37, Prop.B.9] ([35,
Proposition C.1]) that

$\Big{|}\int^{\tau}_{0}\big{[}\partial_{v}\check{L}_{\lambda}\bigl{(}s,x(s),\dot{x}(s)\bigr{)}-\partial_{v}\check{L}_{\lambda}\bigl{(}s,0,0\bigr{)}\big{]}ds\Big{|}\to
0$

uniformly in $\lambda$ as $\|x\|_{1,2}\to 0$. These and (6.64) show:

For any $\epsilon>0$ there exists $\varepsilon>0$ such that
$\|x\|_{1,2}\leq\varepsilon$ implies $\|x\|_{C^{1}}<\epsilon$. (6.65)

By (6.52) we have

$\displaystyle 0$ $\displaystyle=$
$\displaystyle\frac{d}{dt}\Big{(}\partial_{v}{\bf
L}_{\lambda}(t,x(t),\dot{x}(t))\Big{)}-\partial_{q}{\bf
L}_{\lambda}(t,x(t),\dot{x}(t))$ (6.66) $\displaystyle=$
$\displaystyle\frac{d}{dt}\Big{(}\partial_{v}\check{L}_{\lambda}(t,x(t),\dot{x}(t))+\mu(\lambda,x)\dot{\zeta}_{0}(t)\Big{)}-\partial_{q}\check{L}_{\lambda}(t,x(t),\dot{x}(t))+\mu(\lambda,x)\zeta_{0}(t)$
$\displaystyle=$
$\displaystyle\partial_{vv}\check{L}_{\lambda}(t,x(t),\dot{x}(t))\ddot{x}(t)+\partial_{vq}\check{L}_{\lambda}(t,x(t),\dot{x}(t))\dot{x}(t)+\partial_{vt}\check{L}_{\lambda}(t,x(t),\dot{x}(t))$
$\displaystyle-\partial_{q}\check{L}_{\lambda}(t,x(t),\dot{x}(t))+\mu(\lambda,x)\ddot{\zeta}_{0}(t)+\mu(\lambda,x)\zeta_{0}(t).$

In particular, taking $x=0$ we get

$\displaystyle
0=\partial_{vt}\check{L}_{\lambda}(t,0,0)-\partial_{q}\check{L}_{\lambda}(t,0,0)+\mu(\lambda,0)\ddot{\zeta}_{0}(t)+\mu(\lambda,0)\zeta_{0}(t).$
(6.67)

(6.66) minus (6.67) gives rise to

$\displaystyle 0$ $\displaystyle=$
$\displaystyle\partial_{vv}\check{L}_{\lambda}(t,x(t),\dot{x}(t))\ddot{x}(t)+\partial_{vq}\check{L}_{\lambda}(t,x(t),\dot{x}(t))\dot{x}(t)$
(6.68)
$\displaystyle+\partial_{vt}\check{L}_{\lambda}(t,x(t),\dot{x}(t))-\partial_{vt}\check{L}_{\lambda}(t,0,0)$
$\displaystyle-\partial_{q}\check{L}_{\lambda}(t,x(t),\dot{x}(t))+\partial_{q}\check{L}_{\lambda}(t,0,0)$
$\displaystyle+\mu(\lambda,x)\ddot{\zeta}_{0}(t)-\mu(\lambda,0)\ddot{\zeta}_{0}(t)+\mu(\lambda,x)\zeta_{0}(t)-\mu(\lambda,0)\zeta_{0}(t).$

Note that (L2) of Lemma 6.1 implies
$|[\partial_{vv}\check{L}_{\lambda}(t,x(t),\dot{x}(t))]^{-1}\xi|\leq\frac{1}{c}|\xi|\;\forall\xi\in{\mathbb{R}^{n}}$.
(6.68) leads to

$\displaystyle|\ddot{x}(t)|$ $\displaystyle\leq$
$\displaystyle\frac{1}{c}|\partial_{vq}\check{L}_{\lambda}(t,x(t),\dot{x}(t))|\cdot|\dot{x}(t)|+\frac{1}{c}|\partial_{q}\check{L}_{\lambda}(t,x(t),\dot{x}(t))-\partial_{q}\check{L}_{\lambda}(t,0,0)|$
(6.69)
$\displaystyle+\frac{1}{c}|\partial_{vt}\check{L}_{\lambda}(t,x(t),\dot{x}(t))-\partial_{vt}\check{L}_{\lambda}(t,0,0)|+\frac{1}{c}|\partial_{q}\check{L}_{\lambda}(t,x(t),\dot{x}(t))-\partial_{q}\check{L}_{\lambda}(t,0,0)|$
$\displaystyle+|\mu(\lambda,x)-\mu(\lambda,0)||\ddot{\zeta}_{0}(t)|+|\mu(\lambda,x)-\mu(\lambda,0)||\zeta_{0}(t)|.$

Since $|{\zeta}_{0}(t)|$ and $|\ddot{\zeta}_{0}(t)|$ are bounded on
$[0,\tau]$, the desired claim may follow from this, Step 2 and (6.65).

Step 4. Prove $d\check{\mathscr{L}}_{\lambda}(x)=0$. Since
$\|\zeta_{0}\|_{C^{0}}>0$, by Step 3 we get $\varepsilon>0$ such that
$\|x\|_{1,2}\leq\varepsilon$ implies $\|x\|_{C^{2}}<\rho_{0}$. In particular,
$x\in\mathcal{U}^{X}$. Since $\check{L}=L^{\star}$ on
$\hat{\Lambda}\times\mathbb{R}\times B^{n}_{3\iota/2}(0)\times
B^{n}_{\rho_{0}}(0)$, it holds that

$\check{\mathscr{L}}_{\lambda}(x)=\int^{\tau}_{0}\check{L}_{\lambda}(t,x(t),\dot{x}(t))dt=\mathscr{L}^{\star}_{\lambda}(x)=\mathfrak{E}_{\lambda}(\Phi_{\bar{\gamma}}(x)).$

Note that $\Phi_{\bar{\gamma}}(x)\in\mathcal{X}^{6}_{\tau}(M,\mathbb{I}_{g})$.
$[-a,a]\cdot\Phi_{\bar{\gamma}}(x)$ is a $C^{4}$ submanifold of dimension one.
Therefore by these and the arguments below (6.4) we get that

$S_{x}:=\Phi^{-1}_{{\bar{\gamma}}}\bigl{(}[-a,a]\cdot\Phi_{\bar{\gamma}}(x)\cap{\rm
Im}(\Phi_{\bar{\gamma}})\cap\mathcal{X}^{1}_{\tau}(M,\mathbb{I}_{g})\big{)}$

is a $C^{2}$ submanifold of
$\mathcal{X}^{1}_{\tau}(B^{n}_{2\iota}(0),{E}_{\bar{\gamma}})$ containing $0$
as an interior point. Observe that

$T_{\Phi_{\bar{\gamma}}(x)}\big{(}[-a,a]\cdot\Phi_{\bar{\gamma}}(x)\cap{\rm
Im}(\Phi_{\bar{\gamma}})\cap\mathcal{X}^{1}_{\tau}(M,\mathbb{I}_{g})\bigr{)}=\mathbb{R}\frac{d}{ds}(s\cdot\Phi_{\bar{\gamma}}(x))\Big{|}_{s=0}=\mathbb{R}(\Phi_{\bar{\gamma}}(x))^{\cdot}$

and that $\mathfrak{E}_{\lambda}$ is constant on
$[-a,a]\cdot\Phi_{\bar{\gamma}}(x)$. Hence

$d\mathfrak{E}_{\lambda}(\Phi_{\bar{\gamma}}(x))[(\Phi_{\bar{\gamma}}(x))^{\cdot}]=0.$

Let
$\zeta_{x}:=(d\Phi_{\bar{\gamma}}(x)))^{-1}((\Phi_{\bar{\gamma}}(x))^{\cdot})\in
T_{x}S_{x}$. Then

$d\check{\mathscr{L}}_{\lambda}(x)[\zeta_{x}]=d\mathfrak{E}_{\lambda}(\Phi_{\bar{\gamma}}(x))[(\Phi_{\bar{\gamma}}(x))^{\cdot}]=0.$
(6.70)

Since $\Phi_{{\bar{\gamma}}}(\xi)(t)=\phi_{{\bar{\gamma}}}(t,\xi(t))$ and
$d\phi_{{\bar{\gamma}}}(t,p)[(1,v)]=\partial_{2}\phi_{{\bar{\gamma}}}(t,p)[v]+\partial_{1}\phi_{{\bar{\gamma}}}(t,p)$
we get

$\displaystyle\partial_{2}\phi_{{\bar{\gamma}}}(t,x(t))[\dot{x}(t)]+\partial_{1}\phi_{{\bar{\gamma}}}(t,x(t))=(\Phi_{\bar{\gamma}}(x))^{\cdot}(t)=(d\Phi_{\bar{\gamma}}(x)[\zeta_{x}])(t)$
$\displaystyle=$
$\displaystyle\frac{d}{ds}\Big{|}_{s=0}\Phi_{\bar{\gamma}}(x+s\zeta_{x})(t)=\frac{d}{ds}\Big{|}_{s=0}\phi_{\bar{\gamma}}(t,x(t)+s\zeta_{x}(t))$
$\displaystyle=$
$\displaystyle\partial_{2}\phi_{\bar{\gamma}}(t,x(t))[\zeta_{x}(t)]$

and thus

$\partial_{1}\phi_{{\bar{\gamma}}}(t,x(t))=\partial_{2}\phi_{\bar{\gamma}}(t,x(t))[\zeta_{x}(t)-\dot{x}(t)].$
(6.71)

As $\|x\|_{C^{1}}\to 0$ we deduce that $\|\zeta_{x}-\zeta_{0}\|_{C^{0}}\to 0$,
that is,

$\zeta_{x}(t)=\dot{x}(t)+(\partial_{2}\phi_{\bar{\gamma}}(t,x(t)))^{-1}\bigl{(}\partial_{1}\phi_{{\bar{\gamma}}}(t,x(t))\bigr{)}\to(\partial_{2}\phi_{\bar{\gamma}}(t,0))^{-1}\bigl{(}\dot{\bar{\gamma}}(t)\bigr{)}=\zeta_{0}(t)$
(6.72)

uniformly on $[0,\tau]$.

We hope to prove that $\|\zeta_{x}-\zeta_{0}\|_{C^{1}}\to 0$ as
$\|x\|_{1,2}\to 0$.

Fix $\bar{t}\in[0,\tau]$ and a $C^{7}$ coordinate chart $(\Theta,W)$ around
$\bar{\gamma}(\bar{t})$ on $M$, where $W$ is an open neighborhood of
$\bar{\gamma}(\bar{t})$ and $\Theta$ is a $C^{7}$ diffeomorphism from $W$ to
an open subset in $\mathbb{R}^{n}$. Then there exists a closed neighborhood
$J$ of $\bar{t}$ in $[0,\tau]$ such that $\bar{\gamma}(t)\in W$ for any $t\in
J$.

Shrinking $J$ we have a positive number $\nu<2\iota$ such that
$x([0,\tau])\subset B^{n}_{\nu}(0)$ and that

$\phi_{\bar{\gamma},t}:=\phi_{\bar{\gamma}}(t,\cdot)\;\hbox{maps
$B^{n}_{\nu}(0)$ into $W$ for each $t\in J$}.$ (6.73)

Define $\Upsilon:J\times
B^{n}_{\nu}(0)\to\mathbb{R}^{n},\;(t,p)\mapsto\Theta(\phi_{\bar{\gamma}}(t,p))$.
It is $C^{5}$, and for each $t\in J$, $\Upsilon(t,\cdot)$ is a $C^{5}$
diffeomorphism from $B^{n}_{\nu}(0)$ onto an open subset in $\mathbb{R}^{n}$.
Let $\partial_{1}\Upsilon$ and $\partial_{2}\Upsilon$ denote differentials of
$\Upsilon$ with respect to $t$ and $p$, respectively. For $t\in J$, since

$\displaystyle
d\Theta(\phi_{\bar{\gamma}}(t,x(t)))\circ(\partial_{2}\phi_{\bar{\gamma}}(t,x(t)))=d(\Theta\circ\phi_{\bar{\gamma},t})(x(t))=\partial_{2}\Upsilon(t,x(t)),$
$\displaystyle
d\Theta(\phi_{\bar{\gamma}}(t,x(t)))[\partial_{1}\phi_{{\bar{\gamma}}}(t,x(t))]=\frac{d}{ds}\Theta(\phi_{\bar{\gamma}}(s,p))\Big{|}_{(s,p)=(t,x(t))}=\partial_{1}\Upsilon(t,x(t)),$

composing $d\Theta(\phi_{\bar{\gamma}}(t,x(t)))$ in two sides of (6.71) we
obtain

$\partial_{1}\Upsilon(t,x(t))=\partial_{2}\Upsilon(t,x(t))[\zeta_{x}(t)-\dot{x}(t)],\quad\forall
t\in J.$ (6.74)

Differentiating this equality with respect to $t$ gives rise to

$\displaystyle\partial_{1}\partial_{1}\Upsilon(t,x(t))+\partial_{2}\partial_{1}\Upsilon(t,x(t))[\dot{x}(t)]$
$\displaystyle=$
$\displaystyle\partial_{1}\partial_{2}\Upsilon(t,x(t))[\zeta_{x}(t)-\dot{x}(t)]+\partial_{2}\partial_{2}\Upsilon(t,x(t))[\dot{x}(t),\zeta_{x}(t)-\dot{x}(t)]$
$\displaystyle+\partial_{2}\Upsilon(t,x(t))[\dot{\zeta}_{x}(t)-\ddot{x}(t)]$

and so

$\displaystyle\dot{\zeta}_{x}(t)$ $\displaystyle=$
$\displaystyle\ddot{x}(t)-\big{(}\partial_{2}\Upsilon(t,x(t))\big{)}^{-1}\partial_{1}\partial_{2}\Upsilon(t,x(t))[\zeta_{x}(t)-\dot{x}(t)]$
$\displaystyle-$
$\displaystyle\big{(}\partial_{2}\Upsilon(t,x(t))\big{)}^{-1}\partial_{2}\partial_{2}\Upsilon(t,x(t))[\dot{x}(t),\zeta_{x}(t)-\dot{x}(t)]$
$\displaystyle+$
$\displaystyle\big{(}\partial_{2}\Upsilon(t,x(t))\big{)}^{-1}\partial_{1}\partial_{1}\Upsilon(t,x(t))+\big{(}\partial_{2}\Upsilon(t,x(t))\big{)}^{-1}\partial_{2}\partial_{1}\Upsilon(t,x(t))[\dot{x}(t)],$

where $\partial_{2}\Upsilon(t,x(t))$ may be understand as matrixes. As
$\|x\|_{C^{2}}\to 0$ it follows that

$\displaystyle\dot{\zeta}_{x}(t)$ $\displaystyle\to$
$\displaystyle\big{(}\partial_{2}\Upsilon(t,0))\big{)}^{-1}\partial_{1}\partial_{1}\Upsilon(t,0)=\dot{\zeta}_{0}(t)$

uniformly on $J$, and hence that $\|\zeta_{x}-\zeta_{0}\|_{C^{1}(J)}\to 0$ by
(6.72).

Note that $[0,\tau]$ can be covered by finitely many neighborhoods of form
$J$. We arrive at $\|\zeta_{x}-\zeta_{0}\|_{C^{1}}\to 0$, in particular,
$\|\zeta_{x}-\zeta_{0}\|_{1,2}\to 0$. Then for $\varepsilon>0$ small enough,
the orthogonal decomposition ${\bf H}=({\mathbb{R}}\zeta_{0})\oplus{\bf
H}^{\bot}$ of Hilbert spaces implies a direct sum decomposition of Banach
spaces ${\bf H}=(\mathbb{R}\zeta_{x})\dot{+}{\bf H}^{\bot}$. From this, (6.70)
and (6.50) we may deduce that $d\check{\mathscr{L}}_{\lambda}(x)=0$ because
$T_{x}B_{{\bf H}^{\bot}}(0,\varepsilon)=T_{0}B_{{\bf
H}^{\bot}}(0,\varepsilon)={\bf H}^{\bot}$.

∎

####  6.5 Strengthening of Theorems 1.23, 1.24, 1.25, 1.26 for Lagrangian
systems on $\mathbb{R}^{n}$

When $M$ is an open subset $U$ of $\mathbb{R}^{n}$ and $\mathbb{I}_{g}$ is an
orthogonal matrix $E$ of order $n$ which maintains $U$ invariant, Assumption
1.21 in Theorems 1.23, 1.24, 1.25, 1.26 can be replaced by the following
weaker:

######  Assumption 6.12.

For an orthogonal matrix $E$ of order $n$, and an $E$-invariant open subset
$U\subset\mathbb{R}^{n}$ let $L:\Lambda\times
U\times\mathbb{R}^{n}\to{\mathbb{R}}$ be a continuous function such that for
each $\lambda\in\Lambda$ the function $L_{\lambda}(\cdot)=L(\lambda,\cdot)$ is
$C^{4}$ and partial derivatives

$\partial_{q}L_{\lambda}(\cdot),\quad\partial_{v}L_{\lambda}(\cdot),\quad\partial_{qv}L_{\lambda}(\cdot),\quad\partial_{qq}L_{\lambda}(\cdot),\quad\partial_{vv}L_{\lambda}(\cdot)$

depend continuously on $(\lambda,q,v)\in\Lambda\times U\times\mathbb{R}^{n}$.
Moreover, for each $(\lambda,q)\in\Lambda\times U$, ${L}(\lambda,q,v)$ is
convex in $v$, and satisfies

$\displaystyle
L(\lambda,Eq,Ev)=L(\lambda,q,v)\quad\forall(\lambda,q,v)\in\Lambda\times
U\times\mathbb{R}^{n}.$ (6.75)

A nonconstant $C^{2}$ map $\bar{\gamma}:{\mathbb{R}}\to U$ satisfies (5.8) for
each $\lambda\in\Lambda$ (so $\bar{\gamma}$ is $C^{4}$), and the closure of
$\bar{\gamma}(\mathbb{R})$ has an $E$-invariant compact neighborhood $U_{0}$
contained in $U$ (thus there exists $\nu_{0}>0$ such that
$Cl(\bar{\gamma}(\mathbb{R}))+\bar{B}^{n}_{\nu_{0}}(0)\subset U_{0})$. For
some real $\rho>\sup_{t}|\dot{\bar{\gamma}}(t)|$ and each
$(\lambda,q)\in\Lambda\times U$, ${L}(\lambda,q,v)$ is strictly convex in $v$
in $B^{n}_{\rho}(0)$.

In fact, taking an orthogonal matrix $\Xi$ such that $\Xi^{-1}E\Xi$ is equal
to the right side of (6.2) and replacing $L_{\lambda}$ and $\bar{\gamma}$ by

$(\Xi^{-1}U)\times\mathbb{R}^{n}\to\mathbb{R},\;(x,v)\mapsto L_{\lambda}(\Xi
x,\Xi v)$

and $\Xi^{-1}\bar{\gamma}$, respectively, we may assume $E={\rm
diag}(S_{1},\cdots,S_{\sigma})\in{\mathbb{R}}^{n\times n}$ as in (6.3).

Take $\iota=\nu_{0}/3$ and define $\phi_{{\bar{\gamma}}}:\mathbb{R}\times
B^{n}_{3\iota}(0)\to U$ by $\phi_{{\bar{\gamma}}}(t,x)=\bar{\gamma}(t)+x$,
which is clearly $C^{4}$. Then $\phi_{{\bar{\gamma}}}$ gives rise to a
$C^{\infty}$ coordinate chart around ${\bar{\gamma}}$ on the $C^{\infty}$
Banach manifold $\mathcal{X}_{\tau}(U,E)$,

$\displaystyle\Phi_{{\bar{\gamma}}}:\mathcal{X}^{1}(B^{n}_{2\iota}(0),{E}^{T})\to\mathcal{X}^{1}_{\tau}(U,E)$

given by $\Phi_{{\bar{\gamma}}}(\xi)(t)=\bar{\gamma}(t)+\xi(t)$. For any
$\xi\in\mathcal{X}^{1}(B^{n}_{2\iota}(0),{E}^{T})$ and
$\gamma\in\mathcal{X}^{1}_{\tau}(U,E)$ there holds

$T_{\xi}\mathcal{X}^{1}(B^{n}_{2\iota}(0),{E}^{T})=T_{\gamma}\mathcal{X}^{1}_{\tau}(U,E)={\bf
X}:=\mathcal{X}^{1}(\mathbb{R}^{n},E^{T}).$

Clearly, $d\Phi_{{\bar{\gamma}}}(0)={id}_{\bf X}$. Since
$d\phi_{{\bar{\gamma}}}(t,x)[(1,v)]=\dot{\bar{\gamma}}(t)+v$, we define

$\displaystyle L^{\star}:\Lambda\times\mathbb{R}\times
B^{n}_{3\iota}(0)\times{\mathbb{R}}^{n}\to{\mathbb{R}},\;(\lambda,t,x,v)\mapsto
L\bigr{(}\lambda,\bar{\gamma}(t)+x,\dot{\bar{\gamma}}(t)+v\bigl{)}.$

It is continuous and satisfies (6.8) with $E_{\bar{\gamma}}=E$. Moreover, each
$L^{\star}_{\lambda}$ is $C^{4}$. Take $\rho_{0}>3\iota$ such that
$\rho>\rho_{0}>\sup_{t}|\dot{\bar{\gamma}}(t)|$. Then for a given subset
$\hat{\Lambda}\subset\Lambda$ which is either compact or sequential compact,
Lemma 6.1 yields a continuous function
$\check{L}:\hat{\Lambda}\times\mathbb{R}\times\bar{B}^{n}_{2\iota}(0)\times\mathbb{R}^{n}\to{\mathbb{R}}$.
Put

$\displaystyle{\bf H}:=\\{\xi\in W^{1,2}_{\rm
loc}(\mathbb{R};\mathbb{R}^{n})\,|\,{E}(\xi(t))=\xi(t+\tau)\;\forall
t\in\mathbb{R}\\}\quad\hbox{and}$ $\displaystyle\mathcal{U}:=\\{\xi\in
W^{1,2}_{\rm
loc}(\mathbb{R};B^{n}_{2\iota}(0))\,|\,{E}(\xi(t))=\xi(t+\tau)\;\forall
t\in\mathbb{R}\\},\quad\mathcal{U}^{X}:=\mathcal{U}\cap{\bf X}.$

Define $\check{\mathscr{L}}_{\lambda}:\mathcal{U}\to{\mathbb{R}}$ as in
(6.12), and ${\check{\mathscr{L}}}_{\lambda}^{\bot}:\mathcal{U}\cap{\bf
H}^{\bot}\to{\mathbb{R}}$ as in (6.14), where

${\bf H}^{\bot}:=\\{x\in{\bf
H}\,|\,(\dot{\bar{\gamma}},x)_{1,2}=0\\}\quad\hbox{and}\quad{\bf
X}^{\bot}:={\bf X}\cap{\bf H}^{\bot}.$

The other arguments are same, except that “$C^{6}$” is replaced by “$C^{4}$”,
$\mathbb{I}_{g}$ by $E$, etc. Actually, the proof of Proposition 3.11 are much
simpler in this situation.

###  7 Proofs of Theorems 1.29, 1.30, 1.31

Following the paragraph above (3.2) we take a path $\overline{\gamma}\in
EC^{7}(S_{\tau};M)$ such that

$\displaystyle{\rm
dist}_{g}(\gamma_{\mu}(t),\overline{\gamma}(t))<\iota\;\forall
t\in\mathbb{R}.$ (7.1)

We first assume:

$\displaystyle
d_{g}(\gamma_{\lambda}(t),\overline{\gamma}(t))<\iota,\quad\forall(\lambda,t)\in\Lambda\times\mathbb{R}.$
(7.2)

(For cases of Theorems 1.29, 1.31, by contradiction we may use nets to prove
that (7.2) is satisfied after shrinking $\Lambda$ toward $\mu$.) Then (7.1)
and (7.2) imply

$\displaystyle d_{g}(\gamma_{\lambda}(t),\gamma_{\mu}([0,\tau]))\leq
d_{g}(\gamma_{\lambda}(t),\gamma_{\mu}(t))<2\iota,\quad\forall(\lambda,t)\in\Lambda\times\mathbb{R}.$
(7.3)

By Claim A.2 we have a unit orthogonal parallel $C^{5}$ frame field
${\mathbb{R}}\to\overline{\gamma}^{\ast}TM,\;t\mapsto(e_{1}(t),\cdots,e_{n}(t))$
to satisfy

$\displaystyle(e_{1}(\tau\pm t),\cdots,e_{n}(\tau\pm
t))=(e_{1}(t),\cdots,e_{n}(t))\quad\forall t\in{\mathbb{R}}.$ (7.4)

Let $B^{n}_{2\iota}(0):=\\{x\in{\mathbb{R}}^{n}\,|\,|x|<2\iota\\}$ and $\exp$
denote the exponential map of $g$. Then

$\displaystyle\phi_{\overline{\gamma}}:{\mathbb{R}}\times B^{n}_{2\iota}(0)\to
M,\;(t,x)\mapsto\exp_{\overline{\gamma}(t)}\left(\sum^{n}_{i=1}x_{i}e_{i}(t)\right)$

is a $C^{5}$ map and satisfies

$\phi_{\overline{\gamma}}(\tau\pm
t,x)=\phi_{\overline{\gamma}}(t,x)\quad\hbox{and}\quad\frac{d}{dt}\phi_{\overline{\gamma}}(\tau\pm
t,x)=\pm(\phi_{\overline{\gamma}})^{\prime}_{1}(\tau\pm t,x)$ (7.5)

for any $(t,x)\in{\mathbb{R}}\times B^{n}_{2\iota}(0)$. By [48, Theorem 4.3],
we have a $C^{2}$ coordinate chart around $\overline{\gamma}$ on the $C^{4}$
Banach manifold $EC^{1}(S_{\tau};M)$,

$\displaystyle\Phi_{\overline{\gamma}}:EC^{1}(S_{\tau};B^{n}_{2\iota}(0))=\\{\xi\in
C^{1}(S_{\tau};\mathbb{R}^{n})\,|\,\|\xi\|_{C^{0}}<2\iota\\}\to
EC^{1}(S_{\tau};M)$ (7.6)

given by
$\Phi_{\overline{\gamma}}(\xi)(t)=\phi_{\overline{\gamma}}(t,\xi(t))$, and

$d\Phi_{\overline{\gamma}}(0):EC^{1}(S_{\tau};\mathbb{R}^{n})\to
T_{\overline{\gamma}}EC^{1}(S_{\tau};M),\;\xi\mapsto\sum^{n}_{j=1}\xi_{j}e_{j}.$

It is easy to proved that

$\Phi_{\overline{\gamma}}(EC^{1}(S_{\tau};B^{n}_{2\iota}(0)))=\mathcal{U}(\overline{\gamma},2\iota):=\\{\gamma\in
EC^{1}(S_{\tau},M)\,|\,d_{\infty}(\gamma,\overline{\gamma})<2\iota\\},$

where $d_{\infty}(\gamma_{1},\gamma_{2})=\max_{t\in S_{\tau}}{\rm
dist}_{g}(\gamma_{1}(t),\gamma_{2}(t))$ for $\gamma_{i}\in
EC^{1}(S_{\tau},M)$, $i=1,2$.

Since the injectivity radius of $g$ at each point on
$\overline{\gamma}(S_{\tau})$ is at least $2\iota$, by (7.2) there exists a
unique map ${\bf u}_{\lambda}:{\mathbb{R}}\to B^{n}_{\iota}(0)$ such that

$\gamma_{\lambda}(t)=\phi_{\overline{\gamma}}(t,{\bf
u}_{\lambda}(t))=\exp_{\overline{\gamma}(t)}\left(\sum^{n}_{i=1}{\bf
u}_{\lambda}^{i}(t)e_{i}(t)\right)\quad\forall t\in{\mathbb{R}},$

Note that $\overline{\gamma}$ is even and $\tau$-periodic. From (7.4) we
derive

$\gamma_{\lambda}(\tau\pm t)=\exp_{\overline{\gamma}(\tau\pm
t)}\left(\sum^{n}_{i=1}{\bf u}_{\lambda}^{i}(\tau\pm t)e_{i}(\tau\pm
t)\right)=\exp_{\overline{\gamma}(t)}\left(\sum^{n}_{i=1}{\bf
u}_{\lambda}^{i}(\tau\pm t)e_{i}(t)\right).$

Hence ${\bf u}_{\lambda}\in EC^{1}(S_{\tau};B^{n}_{\iota}(0))\cap
C^{2}(S_{\tau};B^{n}_{\iota}(0))$. By Lemma 3.1, 3.2 we also see that

$(\lambda,t)\mapsto{\bf u}_{\lambda}(t),\quad(\lambda,t)\mapsto\dot{\bf
u}_{\lambda}(t)\quad\hbox{and}\quad(\lambda,t)\mapsto\ddot{\bf
u}_{\lambda}(t)$

are continuous maps from $\Lambda\times{\mathbb{R}}$ to ${\mathbb{R}}^{n}$.

Define $L^{\ast}_{\lambda}:{\mathbb{R}}\times
B^{n}_{2\iota}(0)\times{\mathbb{R}}^{n}\to{\mathbb{R}}$ by

$L^{\ast}_{\lambda}(t,x,v)=L_{\lambda}\left(t,\phi_{\overline{\gamma}}(t,x),(\phi_{\overline{\gamma}})^{\prime}_{1}(t,x)+(\phi_{\overline{\gamma}})^{\prime}_{2}(t,x)[v]\right).$
(7.7)

Then it follows from (7.5) that
$L^{\ast}_{\lambda}(\tau+t,x,v)=L^{\ast}_{\lambda}(t,x,v)$ and

$\displaystyle L^{\ast}_{\lambda}(\tau-t,x,-v)$ $\displaystyle=$
$\displaystyle
L_{\lambda}\left(\tau-t,\phi_{\overline{\gamma}}(\tau-t,x),(\phi_{\overline{\gamma}})^{\prime}_{1}(\tau-t,x)+(\phi_{\overline{\gamma}})^{\prime}_{2}(\tau-t,x)[-v]\right)$
(7.8) $\displaystyle=$ $\displaystyle
L_{\lambda}\left(\tau-t,\phi_{\overline{\gamma}}(t,x),-(\phi_{\overline{\gamma}})^{\prime}_{1}(t,x)-(\phi_{\overline{\gamma}})^{\prime}_{2}(t,x)[v]\right)$
$\displaystyle=$ $\displaystyle
L_{\lambda}\left(t,\phi_{\overline{\gamma}}(t,x),(\phi_{\overline{\gamma}})^{\prime}_{1}(t,x)+(\phi_{\overline{\gamma}})^{\prime}_{2}(t,x)[v]\right)$
$\displaystyle=$ $\displaystyle
L^{\ast}_{\lambda}(t,x,v)\quad\forall(t,x,v)\in{\mathbb{R}}\times
B^{n}_{2\iota}(0)\times{\mathbb{R}}^{n}.$

As before we also define $\tilde{L}^{\ast}:\Lambda\times\mathbb{R}\times
B^{n}_{\iota}(0)\times{\mathbb{R}}^{n}\to{\mathbb{R}}$ by

$\tilde{L}^{\ast}(\lambda,t,q,v)=\tilde{L}^{\ast}_{\lambda}(t,q,v)={L}^{\ast}(\lambda,t,q+{\bf
u}_{\lambda}(t),v+\dot{\bf u}_{\lambda}(t)).$ (7.9)

It satisfies Proposition 3.3 after the interval $[0,\tau]$ is replaced by
$\mathbb{R}$. Moreover, (7.8) leads to

$\displaystyle\tilde{L}^{\ast}(\lambda,\tau-t,q,-v)$ $\displaystyle=$
$\displaystyle{L}^{\ast}(\lambda,\tau-t,q+{\bf
u}_{\lambda}(\tau-t),-v+\dot{\bf u}_{\lambda}(\tau-t))$ $\displaystyle=$
$\displaystyle{L}^{\ast}(\lambda,\tau-t,q+{\bf u}_{\lambda}(t),-v-\dot{\bf
u}_{\lambda}(t))$ $\displaystyle=$ $\displaystyle{L}^{\ast}(\lambda,t,q+{\bf
u}_{\lambda}(t),v+\dot{\bf u}_{\lambda}(t))$ $\displaystyle=$
$\displaystyle\tilde{L}^{\ast}(\lambda,t,q,v)$

because ${\bf u}_{\lambda}(\tau-t)={\bf u}_{\lambda}(t)$ implies $-\dot{\bf
u}_{\lambda}(\tau-t)=\dot{\bf u}_{\lambda}(t)$.

Now we have a family of $C^{2}$ functionals

$\tilde{\mathcal{L}}^{E}_{\lambda}:EC^{1}(S_{\tau};B^{n}_{\iota}(0))\to{\mathbb{R}},\;x\mapsto\int^{\tau}_{0}\tilde{L}^{\ast}_{\lambda}(t,x(t),\dot{x}(t))dt,\quad\lambda\in\Lambda$

(see the proof of the first claim in [35, Proposition 4.2]). For $\xi\in
EC^{1}(S_{\tau};B^{n}_{\iota}(0))$, since

$\displaystyle\frac{d}{dt}\Phi_{\overline{\gamma}}(\xi+{\bf
u}_{\lambda})(t)=(\phi_{\overline{\gamma}})^{\prime}_{1}(t,\xi(t)+{\bf
u}_{\lambda}(t))+(\phi_{\overline{\gamma}})^{\prime}_{2}(t,\xi(t)+{\bf
u}_{\lambda}(t))[\dot{\xi}(t)+\dot{\bf u}_{\lambda}(t)]\quad\forall
t\in{\mathbb{R}},$ (7.10)

we deduce

$\tilde{\cal L}^{E}_{\lambda}(\xi)={\cal
L}^{E}_{\lambda}\left(\Phi_{\overline{\gamma}}(\xi+{\bf
u}_{\lambda})\right)\quad\forall\xi\in EC^{1}(S_{\tau};B^{n}_{\iota}(0))$
(7.11)

and therefore that

$m^{-}_{\tau}(\tilde{\mathcal{L}}^{E}_{\lambda},0)=m^{-}_{\tau}(\mathcal{L}^{E}_{\lambda},\gamma_{\lambda})\quad\hbox{and}\quad
m^{0}_{\tau}(\tilde{\mathcal{L}}^{E}_{\lambda},0)=m^{0}_{\tau}(\mathcal{L}^{E}_{\lambda},\gamma_{\lambda}).$

For the function $\tilde{L}^{\ast}$ in (7.9), a positive number $\rho_{0}>0$
and a subset $\hat{\Lambda}\subset\Lambda$ which is either compact or
sequential compact, as in Lemma 3.8 using Lemma 2.4, 2.7 we can construct a
continuous function $\check{L}:\hat{\Lambda}\times\mathbb{R}\times
B^{n}_{3\iota/4}(0)\times\mathbb{R}^{n}\to{\mathbb{R}}$ satisfying the
properties (L0)-(L6) in Lemma 3.8 on $\hat{\Lambda}\times\mathbb{R}\times
B^{n}_{3\iota/4}(0)\times\mathbb{R}^{n}$ and the following equality

$\displaystyle\check{L}(\lambda,-t,q,-v)=\check{L}(\lambda,t,q,v)=\check{L}(\lambda,\tau+t,q,v)$

for all $(\lambda,t,q,v)\in\hat{\Lambda}\times\mathbb{R}\times
B^{n}_{3\iota/4}(0)\times\mathbb{R}^{n}$. Let us write

$\displaystyle{\bf H}:=W^{1,2}(S_{\tau};{\mathbb{R}}^{n}),\quad{\bf
X}:=C^{1}(S_{\tau};{\mathbb{R}}^{n}),$
$\displaystyle\mathcal{U}:=W^{1,2}(S_{\tau};B^{n}_{\iota/2}(0),\quad\mathcal{U}^{X}:=\mathcal{U}\cap{\bf
X}=EC^{1}\left(S_{\tau};B^{n}_{\iota/2}(0)\right),$ $\displaystyle{\bf
H}_{e}:=EW^{1,2}(S_{\tau};{\mathbb{R}}^{n})=\\{x\in
W^{1,2}(S_{\tau};{\mathbb{R}}^{n})\,|\,x(-t)=x(t)\;\forall t\\},\quad{\bf
X}_{e}:=EC^{1}(S_{\tau};{\mathbb{R}}^{n}),$
$\displaystyle\mathcal{U}_{e}:=\left\\{u\in
W^{1,2}\left(S_{\tau};B^{n}_{\iota/2}(0)\right)\,\big{|}\,u(-t)=u(t)\;\forall
t\right\\},$ $\displaystyle\mathcal{U}^{X}_{e}:=\mathcal{U}\cap{\bf
X}=\left\\{u\in
C^{1}\left(S_{\tau};B^{n}_{\iota/2}(0)\right)\,\big{|}\,u(-t)=u(t)\;\forall
t\right\\}.$

For each $\lambda\in\hat{\Lambda}$ define functionals

$\displaystyle\check{\mathcal{L}}_{\lambda}:\mathcal{U}\to{\mathbb{R}},\;x\mapsto\int^{\tau}_{0}\check{L}_{\lambda}(t,x(t),\dot{x}(t))dt\quad\hbox{and}\quad\check{\mathcal{L}}_{\lambda}^{E}:\mathcal{U}_{e}\to{\mathbb{R}},\;x\mapsto\int^{\tau}_{0}\check{L}_{\lambda}(t,x(t),\dot{x}(t))dt.$

Clearly,
$\check{\mathcal{L}}_{\lambda}|_{\mathcal{U}_{e}}=\check{\mathcal{L}}_{\lambda}^{E}$.
As a special case,
$\\{(\check{\mathcal{L}}_{\lambda},\mathcal{U},\mathcal{U}^{X})\,|\,\lambda\in\hat{\Lambda}\\}$
has the same properties as
$\\{(\check{\mathcal{E}}_{\lambda},\mathcal{U},\mathcal{U}^{X})\,|\,\lambda\in\hat{\Lambda}\\}$
in Section 4.2. By the arguments in [29],
$\\{(\check{\mathcal{L}}_{\lambda}^{E},\mathcal{U}_{e},\mathcal{U}^{X}_{e})\,|\,\lambda\in\hat{\Lambda}\\}$
has also the same properties. (In fact, from the expression of
$\nabla\check{\mathcal{L}}_{\lambda}$ it is not hard to prove that
$\nabla\check{\mathcal{L}}_{\lambda}(x)$ is even for each
$x\in\mathcal{U}_{e}$. Therefore
$\nabla\check{\mathcal{L}}_{\lambda}^{E}(x)=\nabla\check{\mathcal{L}}_{\lambda}(x)$
for each $x\in\mathcal{U}_{e}$.) This implies that for each
$x\in\mathcal{U}_{e}$ the operator $B_{\lambda}(x)\in\mathscr{L}_{s}({\bf H})$
defined by (6.27) maps ${\bf H}_{e}$ into ${\bf H}_{e}$. Almost repeating the
proofs in Section 3 we can obtain the required results. Of course, the next
section also provides a way.

###  8 An alternate method for bifurcations of Lagrangian systems on open
subsets in $\mathbb{R}^{n}$

In this section, we will show that for the main results in Section 1, except
for Theorems 1.23, 1.24, 1.25, 1.26, all others, when restricting to
Lagrangian systems on open subsets in $\mathbb{R}^{n}$, may be almost derived
from bifurcation results for Hamiltonian systems in [37].

######  Assumption 8.1.

Let Assumption 2.2, 2.5 be satisfied and let $E$ be a real orthogonal matrix
of order $n$ such that $(EU)\cap U\neq\emptyset$. Consider the Lagrangian
boundary value problem (2.5)-(2.6) on $U$. Suppose that each $x_{\lambda}$ in
Assumption 2.5 also satisfies (2.6).

Then $C^{1}_{E}([0,\tau];U)$ is a nonempty open subset in
$C^{1}_{{E}}([0,\tau];\mathbb{R}^{n})$ [cf. (4.3)],

$\mathscr{L}_{\lambda}:C^{1}_{E}([0,\tau];U)\to{\mathbb{R}},\;x\mapsto\int^{\tau}_{0}L_{\lambda}(t,x(t),\dot{x}(t))dt$
(8.1)

is a $C^{2}$ functional, each $x_{\lambda}\in C^{2}_{E}([0,\tau];U)$ satisfies
$d\mathfrak{L}_{\lambda}(x_{\lambda})=0$, and the second variation at
$x_{\lambda}$ is given by

$\mathscr{L}^{\prime\prime}_{\lambda}(x_{\lambda})[y,z]=\int^{\tau}_{0}\left[(\textsf{P}_{\lambda}\dot{y}+\textsf{Q}_{\lambda}y)\cdot\dot{z}+\textsf{
Q}_{\lambda}^{T}\dot{y}\cdot z+\textsf{R}_{\lambda}y\cdot z\right]dt$ (8.2)

for all $y,z\in C^{1}_{E}([0,\tau];{\mathbb{R}}^{n})$, where
$\textsf{P}_{\lambda}(t)=\partial_{vv}L_{\lambda}\left(t,x_{\lambda}(t),\dot{x}_{\lambda}(t)\right)$
and

$\textsf{Q}_{\lambda}(t)=\partial_{xv}L_{\lambda}\left(t,x_{\lambda}(t),\dot{x}_{\lambda}(t)\right),\qquad\textsf{R}_{\lambda}(t)=\partial_{xx}L_{\lambda}\left(t,x_{\lambda}(t),\dot{x}_{\lambda}(t)\right).$
(8.3)

The form $\mathscr{L}^{\prime\prime}_{\lambda}(x_{\lambda})$ can be extended
into a continuous symmetric bilinear form on
$W^{1,2}_{E}([0,\tau];\mathbb{R}^{n})$, whose Morse index and nullity are
called the  Morse index and nullity of $\mathscr{L}_{\lambda}$ at
$x_{\lambda}$, denoted by

$m^{-}_{\tau}(\mathscr{L}_{\lambda},x_{\lambda})\quad\hbox{and}\quad
m^{0}_{\tau}(\mathscr{L}_{\lambda},x_{\lambda})$ (8.4)

respectively. Both are finite because $\textsf{P}_{\lambda}(t)$ is positive
definite.

For a given compact or sequential compact subset $\hat{\Lambda}\subset\Lambda$
let $\hat{L}:\hat{\Lambda}\times[0,\tau]\times
B^{n}_{\delta}(0)\times{\mathbb{R}}^{n}\to{\mathbb{R}}$ be given by (2.14).
Define

$\hat{\mathscr{L}}_{\lambda}:C^{1}_{E}([0,\tau];B^{n}_{\delta}(0))\to{\mathbb{R}},\;x\mapsto\int^{\tau}_{0}\hat{L}_{\lambda}(t,x(t),\dot{x}(t))dt.$
(8.5)

It is $C^{2}$ because
$\hat{\mathscr{L}}_{\lambda}(x)={\mathscr{L}}_{\lambda}(x+x_{\lambda})$ for
all $x\in C^{1}_{E}([0,\tau];B^{n}_{\delta}(0))$. It is clear that
$x^{0}:[0,\tau]\to{\mathbb{R}}^{n},\;t\mapsto 0$, satisfies the following
boundary value problem:

$\displaystyle\frac{d}{dt}\left(\partial_{v}\hat{L}_{\lambda}(t,x(t),\dot{x}(t))\right)-\partial_{q}\hat{L}_{\lambda}(t,x(t),\dot{x}(t))=0,$
(8.6) $\displaystyle
E(x(0))=x(\tau)\quad\hbox{and}\quad(E^{T})^{-1}\left[\partial_{v}\hat{L}_{\lambda}(0,x(0),\dot{x}(0))\right]=\partial_{v}\hat{L}_{\lambda}(\tau,x(\tau),\dot{x}(\tau)).$
(8.7)

Therefore $d\hat{\mathscr{L}}_{\lambda}(x^{0})=0$,
$\hat{\mathscr{L}}^{\prime\prime}_{\lambda}(x^{0})={\mathscr{L}}^{\prime\prime}_{\lambda}(x_{\lambda})$
and

$m^{-}_{\tau}(\hat{\mathscr{L}}_{\lambda},x^{0})=m^{-}_{\tau}(\mathscr{L}_{\lambda},x_{\lambda})\quad\hbox{and}\quad
m^{0}_{\tau}(\hat{\mathscr{L}}_{\lambda},x^{0})=m^{0}_{\tau}(\mathscr{L}_{\lambda},x_{\lambda}).$
(8.8)

The following is easily proved.

######  Claim 8.2.

For $\lambda\in\hat{\Lambda}$, a curve $x\in C^{2}([0,\tau],U)$ satisfies
(2.5)-(2.6) and $\|x-x_{\lambda}\|_{C^{1}}<\rho_{0}-\rho_{00}$ if and only if
$x-x_{\lambda}\in C^{2}([0,\tau],{\mathbb{R}}^{n})$ is a solution of
(8.6)–(8.7) satisfying $\|x-x_{\lambda}-x^{0}\|_{C^{1}}<\rho_{0}-\rho_{00}$.
Specially, the bifurcation problem (2.5)–(2.6) around $(\mu,x_{\mu})$ with
respect to the trivial branch
$\\{(\lambda,x_{\lambda})\,|\,\lambda\in\Lambda\\}$ is equivalent to that of
the corresponding problem (8.6)–(8.7) around $(\mu,x^{0})$ with respect to the
trivial branch $\\{(\lambda,x^{0})\,|\,\lambda\in\hat{\Lambda}\\}$.

For each $(\lambda,t,q)\in\hat{\Lambda}\times[0,\tau]\times
B^{n}_{\delta}(0)$, by [12, Exercise 1.3.4]), (2.15) implies that
$v\mapsto\hat{L}_{\lambda}(t,q,v)$ is superlinear, and therefore the
associated Legendre transform

${\mathfrak{L}}_{\lambda,t,q}:\mathbb{R}^{n}\to(\mathbb{R}^{n})^{\ast}\equiv\mathbb{R}^{n},\;v\mapsto\partial_{v}\hat{L}_{\lambda}(t,q,v)$

is a $C^{1}$ diffeomorphism ([12, Corollary 1.4.7]). (Here the dual space
$(\mathbb{R}^{n})^{\ast}$ of $\mathbb{R}^{n}$ is naturally identified with
$\mathbb{R}^{n}$.) Since

$\hat{\Lambda}\times[0,\tau]\times
B^{n}_{\delta}(0)\times\mathbb{R}^{n}\to\mathbb{R}^{n},\;(\lambda,t,q,v)\mapsto\partial_{v}\hat{L}_{\lambda}(t,q,v)$

is continuous, we derive from [12, Lemma 2.7.2] that

${\mathfrak{L}}:\hat{\Lambda}\times[0,\tau]\times
B^{n}_{\delta}(0)\times\mathbb{R}^{n}\to\hat{\Lambda}\times[0,\tau]\times
B^{n}_{\delta}(0)\times\mathbb{R}^{n},\;(\lambda,t,q,v)\mapsto(\lambda,t,q,\partial_{v}\hat{L}_{\lambda}(t,q,v))$

is a homeomorphism. Similarly, for each $\lambda\in\hat{\Lambda}$, the
Legendre transform

${\mathfrak{L}}_{\lambda}:[0,\tau]\times
B^{n}_{\delta}(0)\times\mathbb{R}^{n}\to[0,\tau]\times
B^{n}_{\delta}(0)\times\mathbb{R}^{n},\;(t,q,v)\mapsto(t,q,\partial_{v}\hat{L}_{\lambda}(t,q,v)),$

is a $C^{1}$ diffeomorphism, and
$(\lambda,t,q,v)\mapsto(t,q,\partial_{v}\hat{L}_{\lambda}(t,q,v))$ is
continuous by the implicit function theorem. Hence for
$(\lambda,t,q,p)\in\hat{\Lambda}\times[0,\tau]\times
B^{n}_{\delta}(0)\times\mathbb{R}^{n}$ we have a unique $v\in\mathbb{R}^{n}$
such that $\partial_{v}\hat{L}_{\lambda}(t,q,v)=p$. Define
$\hat{H}:\hat{\Lambda}\times[0,\tau]\times\mathbb{R}^{n}\times
B^{n}_{\delta}(0)\to\mathbb{R}$ by

$\hat{H}(\lambda,t,p,q)=(p,v)_{{\mathbb{R}}^{n}}-\hat{L}_{\lambda}(t,q,v).$
(8.9)

######  Claim 8.3.

$\hat{H}$ is continuous, each $\hat{H}({\lambda},t,\cdot)$ is $C^{2}$ and all
possible partial derivatives of it depend continuously on
$(\lambda,t,p,q)\in\hat{\Lambda}\times[0,\tau]\times\mathbb{R}^{n}\times
B^{n}_{\delta}(0)$. For $\lambda\in\hat{\Lambda}$, if a curve $x\in
C^{2}([0,\tau],U)$ satisfies (2.5)-(2.6) and
$\|x-x_{\lambda}\|_{C^{1}}<\rho_{0}-\rho_{00}$, then

$[0,\tau]\ni t\mapsto
z_{\lambda}(t):=\left(\begin{array}[]{c}y_{\lambda}(t)\\\
x(t)-x_{\lambda}(t)\end{array}\right)\in\mathbb{R}^{2n}$

where
$y_{\lambda}(t):=\partial_{v}\hat{L}_{\lambda}(t,x(t)-x_{\lambda}(t),\dot{x}(t)-\dot{x}_{\lambda}(t))=\partial_{v}{L}_{\lambda}(t,x(t),\dot{x}(t))$,
satisfies

$\dot{z}(t)=J\nabla\hat{H}_{\lambda,t}(z(t))\quad\hbox{and}\quad\mathbb{E}z(0)=z(\tau)$
(8.10)

write $\mathbb{E}={\scriptscriptstyle\left(\begin{array}[]{cc}E&0\\\
0&E\end{array}\right)}$ (a symplectic orthogonal matrix of order $2n$), and

$\displaystyle\int^{\tau}_{0}\left[\frac{1}{2}(J\dot{z}_{\lambda}(t),z_{\lambda}(t))_{\mathbb{R}^{2n}}+\hat{H}(\lambda,t,{z}_{\lambda}(t))\right]dt=-\mathscr{L}_{\lambda}(x)=-\int^{\tau}_{0}L_{\lambda}(t,x(t),\dot{x}(t))dt;$
(8.11)

in particular, for
$y_{0,\lambda}(t):=\partial_{v}\hat{L}_{\lambda}(t,0,0)=\partial_{v}\hat{L}_{\lambda}(t,x^{0},\dot{x}^{0})=\partial_{v}{L}_{\lambda}(t,x_{\lambda}(t),\dot{x}_{\lambda}(t))$,
the curve

$\displaystyle[0,\tau]\ni t\mapsto
u_{0,\lambda}(t):=(y_{0,\lambda}(t)^{T},0)^{T}\in\mathbb{R}^{2n}$ (8.12)

satisfies (8.10). Conversely, if $z(t)=(p(t)^{T},x(t)^{T})^{T}$ is a solution
of (8.10) near $u_{0,\lambda}$ (which can be required to be close in
$C^{1}$-topology by [37, Proposition 1.3]), then $x+x_{\lambda}$ satisfies
(2.5)-(2.6) and

$\displaystyle\mathscr{L}_{\lambda}(x+x_{\lambda})=\hat{\mathscr{L}}_{\lambda}(x)=-\int^{\tau}_{0}\left[\frac{1}{2}(J\dot{z}_{\lambda}(t),z_{\lambda}(t))_{\mathbb{R}^{2n}}+\hat{H}(\lambda,t,{z}_{\lambda}(t))\right]dt.$
(8.13)

###### Proof.

Step 1[Prove the assertions for $\hat{H}$]. Because of dependence on $\lambda$
we cannot directly use [12, Proposition 2.6.3]. But the proof is almost same
as that of [12, Proposition 2.6.3]. By (8.9), it holds that for any
$(\lambda,t,q,v)\in\hat{\Lambda}\times[0,\tau]\times
B^{n}_{\delta}(0)\times\mathbb{R}^{n}$,

$\hat{H}(\lambda,t,\partial_{v}\hat{L}_{\lambda}(t,q,v),q)=\sum^{n}_{j=1}\partial_{v_{j}}\hat{L}_{\lambda}(t,q,v)v_{j}-\hat{L}_{\lambda}(t,q,v).$
(8.14)

Differentiating both sides with respect to the variable $v_{i}$, we get

$\sum^{n}_{j=1}\partial_{p_{j}}\hat{H}(\lambda,t,\partial_{v}\hat{L}_{\lambda}(t,q,v),q)\partial_{v_{i}v_{j}}\hat{L}_{\lambda}(t,q,v)=\sum^{n}_{j=1}v_{j}\partial_{v_{i}v_{j}}\hat{L}_{\lambda}(t,q,v),\quad
i=1,\cdots,n.$

(Note: hereafter we understand
$\partial_{v_{i}v_{j}}L=\partial_{v_{i}}(\partial_{v_{j}}L)$. Otherwise, the
desired result cannot be derived.) But $\partial_{vv}\hat{L}_{\lambda}(t,q,v)$
is invertible. It follows that

$\partial_{p_{j}}\hat{H}(\lambda,t,\partial_{v}\hat{L}_{\lambda}(t,q,v),q)=v_{j},\quad
j=1,\cdots,n.$ (8.15)

Differentiating both sides of (8.14) with respect to the variable $q_{i}$, and
using (8.15) we obtain

$\displaystyle\partial_{q_{i}}\hat{H}(\lambda,t,\partial_{v}\hat{L}_{\lambda}(t,q,v),q)+\sum^{n}_{j=1}v_{j}\partial_{q_{i}v_{j}}\hat{L}_{\lambda}(t,q,v)=\sum^{n}_{j=1}v_{j}\partial_{q_{i}v_{j}}\hat{L}_{\lambda}(t,q,v)-\partial_{q_{i}}\hat{L}_{\lambda}(t,q,v)$

and hence

$\displaystyle\partial_{q_{i}}\hat{H}(\lambda,t,\partial_{v}\hat{L}_{\lambda}(t,q,v),q)=-\partial_{q_{i}}\hat{L}_{\lambda}(t,q,v),\quad
i=1,\cdots,n.$ (8.16)

Differentiating both sides of (8.14) with respect to the variable $t$ and
using (8.15) lead to

$\displaystyle\partial_{t}\hat{H}(\lambda,t,\partial_{v}\hat{L}_{\lambda}(t,q,v),q)+\sum^{n}_{j=1}v_{j}\partial_{tv_{j}}\hat{L}_{\lambda}(t,q,v)=\sum^{n}_{j=1}v_{j}\partial_{tv_{j}}\hat{L}_{\lambda}(t,q,v)-\partial_{t}\hat{L}_{\lambda}(t,q,v)$

and so

$\displaystyle\partial_{t}\hat{H}(\lambda,t,p,q)=-\partial_{t}\hat{L}_{\lambda}(t,q,v).$
(8.17)

By the implicit function theorem we have a continuous map

$v:\hat{\Lambda}\times[0,\tau]\times
B^{n}_{\delta}(0)\times\mathbb{R}^{n}\to{\mathbb{R}}^{n},\;(\lambda,t,q,p)\mapsto
v(\lambda,t,q,p),$

such that the following holds:

  1. (i)

$\partial_{v}\hat{L}_{\lambda}(t,q,v(\lambda,t,q,p))=p$ for all
$(\lambda,t,q,p)\in\hat{\Lambda}\times[0,\tau]\times
B^{n}_{\delta}(0)\times\mathbb{R}^{n}$;

  2. (ii)

$v(\lambda,t,q,p)$ is $C^{1}$ in $(t,q,p)$, and
$\partial_{t}v(\lambda,t,q,p)$, $\partial_{q_{i}}v(\lambda,t,q,p)$ and
$\partial_{p_{i}}v(\lambda,t,q,p)$ are continuous $(\lambda,t,q,p)$.

  3. (iii)

$\partial_{q}v(\lambda,t,q,p)=-[\partial_{vv}\hat{L}_{\lambda}(t,q,v(\lambda,t,q,p))]^{-1}\partial_{qv}\hat{L}_{\lambda}(t,q,v(\lambda,t,q,p))$
and
$\partial_{p}v(\lambda,t,q,p)=[\partial_{vv}\hat{L}_{\lambda}(t,q,v(\lambda,t,q,p))]^{-1}$.

It follows from (8.15) and (8.16) and (8.17) that

$\displaystyle\partial_{p_{j}}\hat{H}(\lambda,t,p,q)=v_{j}(\lambda,t,q,p),$
(8.18)
$\displaystyle\partial_{q_{j}}\hat{H}(\lambda,t,p,q)=-\partial_{q_{j}}\hat{L}_{\lambda}(t,q,v(\lambda,t,q,p)),$
(8.19)
$\displaystyle\partial_{t}\hat{H}(\lambda,t,p,q)=-\partial_{t}\hat{L}_{\lambda}(t,q,v(\lambda,t,q,p))$
(8.20)

for $(\lambda,t,p,q)\in\hat{\Lambda}\times[0,\tau]\times\mathbb{R}^{n}\times
B^{n}_{\delta}(0)$ and $j=1,\cdots,n$. These imply that

$\partial_{t}\hat{H}(\lambda,t,p,q),\quad\partial_{p_{j}}\hat{H}(\lambda,t,p,q)\quad\hbox{and}\quad\partial_{q_{j}}\hat{H}(\lambda,t,p,q)$

are continuous in
$(\lambda,t,p,q)\in\hat{\Lambda}\times[0,\tau]\times\mathbb{R}^{n}\times
B^{n}_{\delta}(0)$, and $C^{1}$ in $(p,q)$, and hence that

$\hat{H}$ is continuous, and $C^{2}$ in $(p,q)$ and and all its partial
derivatives
depend continuously on
$(\lambda,t,p,q)\in\hat{\Lambda}\times[0,\tau]\times\mathbb{R}^{n}\times
B^{n}_{\delta}(0)$.

Step 2[Prove that $z_{\lambda}$ satisfies (8.10)-(8.11)]. Suppose a curve
$x\in C^{2}([0,\tau],U)$ satisfies (2.5)-(2.6) and
$\|x-x_{\lambda}\|_{C^{1}}<\rho_{0}-\rho_{00}$ By Claim 8.2, $x-x_{\lambda}\in
C^{2}_{E}([0,\tau],{\mathbb{R}}^{n})$ is a solution of (8.6)–(8.7) satisfying
$\|x-x_{\lambda}\|_{C^{1}}<\rho_{0}-\rho_{00}$. Put

$\displaystyle
y_{\lambda,i}(t)=\partial_{v_{i}}\hat{L}_{\lambda}(t,x(t)-x_{\lambda}(t),\dot{x}(t)-\dot{x}_{\lambda}(t)),\quad
i=1,\cdots,n.$ (8.21)

Then
$v_{i}(\lambda,t,x(t)-x_{\lambda}(t),y_{\lambda}(t))=\dot{x}_{\lambda,i}(t)-\dot{x}_{\lambda,i}(t)$,
$i=1,\cdots,n$. From (8.18), (8.19) and (8.6) we derive

$\displaystyle\dot{x}_{i}(t)-\dot{x}_{\lambda,i}(t)=\partial_{p_{i}}\hat{H}(\lambda,t,y_{\lambda}(t),x(t)-x_{\lambda}(t)),$
(8.22)
$\displaystyle\dot{y}_{\lambda,i}(t)=-\partial_{q_{i}}\hat{H}(\lambda,t,y_{\lambda}(t),x(t)-x_{\lambda}(t),)$
(8.23)

for $i=1,\cdots,n$, that is,
$z_{\lambda}(t):=(y_{\lambda}(t)^{T},x(t)^{T}-(x_{\lambda}(t))^{T})^{T}$
satisfies the first equation in (8.10). Moreover, that $x-x_{\lambda}$
satisfies (8.6)–(8.7) implies
$Ey_{\lambda}(0)=(E^{T})^{-1}y_{\lambda}(0)=y_{\lambda}(\tau)$ by (8.21). This
and $E(x(0)-x_{\lambda}(0))=x(\tau)-x_{\lambda}(0)$ show that $z_{\lambda}(t)$
satisfies the second equation in (8.10).

In order to prove (8.11), note that (8.14) and (8.21) yield

$\displaystyle\hat{L}_{\lambda}(t,x(t)-x_{\lambda}(t),\dot{x}(t)-\dot{x}_{\lambda}(t))=(x(t)-x_{\lambda}(t),y_{\lambda}(t))_{\mathbb{R}^{n}}-\hat{H}_{\lambda}(t,\dot{x}(t)-\dot{x}_{\lambda}(t),x(t)-x_{\lambda}(t)).$
(8.24)

Since $E$ is an orthogonal matrix,
$E(x(0)-x_{\lambda}(0))=x(\tau)-x_{\lambda}(0)$ and
$Ey_{\lambda}(0)=y_{\lambda}(\tau)$, a direct computation leads to

$\displaystyle\int^{\tau}_{0}(\dot{x}(t)-\dot{x}_{\lambda}(t),y_{\lambda}(t))_{\mathbb{R}^{n}}dt=-\frac{1}{2}\int^{\tau}_{0}(J\dot{z}_{\lambda}(t),z_{\lambda}(t))_{\mathbb{R}^{2n}}dt$

as in [43, pages 36-37]. Then (8.11) follows from this and (8.24).

Step 3[Prove the converse part]. Suppose that $z(t)=(p(t)^{T},x(t)^{T})^{T}$
near $u_{0,\lambda}$ is a solution of (8.10), i.e., $\mathbb{E}z(0)=z(\tau)$,
and for $i=1,\cdots,n$ it holds that

$\displaystyle\dot{x}_{i}(t)=\partial_{p_{i}}\hat{H}(\lambda,t,p(t),x(t))\quad\hbox{and}\quad\dot{p}_{i}(t)=-\partial_{q_{i}}\hat{H}(\lambda,t,p(t),x(t)).$
(8.25)

Then the former and (8.18) lead to $\dot{x}(t)=v(\lambda,t,x(t),p(t))$, and so

$\displaystyle\partial_{v}\hat{L}_{\lambda}(t,x(t),\dot{x}(t))=\partial_{v}\hat{L}_{\lambda}(t,x(t),v(\lambda,t,x(t),p(t)))=p(t)$
(8.26)

by (i) below (8.17). From this, (8.25) and (8.19) we derive

$\frac{d}{dt}\left(\partial_{v}\hat{L}_{\lambda}(t,x(t),\dot{x}(t))\right)=\dot{p}(t)=\partial_{q}\hat{L}_{\lambda}(t,x(t),v(\lambda,t,x(t),p(t)))=\partial_{q}\hat{L}_{\lambda}(t,x(t),\dot{x}(t)).$

Moreover, $E$ is an orthogonal matrix, and $\mathbb{E}z(0)=z(\tau)$ if and
only if

$Ex(0)=x(\tau)\quad\hbox{and}\quad
E\partial_{v}\hat{L}_{\lambda}(0,x(0),\dot{x}(0))=\partial_{v}\hat{L}_{\lambda}(\tau,x(\tau),\dot{x}(\tau))$

by (8.26). Therefore $x$ satisfies (8.7). By Claim 8.2 $x+x_{\lambda}$
satisfies (2.5)-(2.6). With the same proof as (8.11) we may obtain (8.13). ∎

Let
$\textsf{P}(\lambda,t,p,q)=\partial_{vv}\hat{L}_{\lambda}\left(t,q,v\right)$,
$\textsf{Q}(\lambda,t,p,q)=\partial_{qv}\hat{L}_{\lambda}\left(t,q,v\right)$
and
$\textsf{R}(\lambda,t,p,q)=\partial_{qq}\hat{L}_{\lambda}\left(t,q,v\right)$.
with $v=v(\lambda,t,q,p)$. It follows from (iii) in Step 1 and (8.18)-(8.19)
that

$\displaystyle\partial_{qp}\hat{H}(\lambda,t,p,q)=\partial_{q}(\partial_{p}\hat{H}(\lambda,t,p,q))=\partial_{q}v(\lambda,t,q,p)=-[\textsf{P}(\lambda,t,p,q)]^{-1}\textsf{Q}(\lambda,t,p,q),$
$\displaystyle\partial_{pp}\hat{H}(\lambda,t,p,q)=\partial_{p}(\partial_{p}\hat{H}(\lambda,t,p,q))=\partial_{p}v(\lambda,t,q,p)=[\textsf{P}(\lambda,t,p,q)]^{-1},$
$\displaystyle\partial_{pq}\hat{H}(\lambda,t,p,q)=\partial_{p}(\partial_{q}\hat{H}(\lambda,t,p,q))=-\textsf{Q}(\lambda,t,p,q)^{T}[\textsf{P}(\lambda,t,p,q)]^{-1},$
$\displaystyle\partial_{qq}\hat{H}(\lambda,t,p,q)=[\textsf{Q}(\lambda,t,p,q)]^{T}[\textsf{P}(\lambda,t,p,q)]^{-1}\textsf{Q}(\lambda,t,p,q)-\textsf{R}(\lambda,t,p,q).$

Therefore

$\displaystyle\hat{H}^{\prime\prime}_{(p,q)}(\lambda,t,p,q)={\scriptscriptstyle\left(\begin{array}[]{cc}\partial_{pp}\hat{H}(\lambda,t,p,q)&\partial_{qp}\hat{H}(\lambda,t,p,q)\\\
\partial_{pq}\hat{H}(\lambda,t,p,q)&\partial_{qq}\hat{H}(\lambda,t,p,q)\end{array}\right)}$
(8.29)
$\displaystyle=\left(\begin{array}[]{cc}[\textsf{P}(\lambda,t,p,q)]^{-1}&-[\textsf{P}(\lambda,t,p,q)]^{-1}\textsf{Q}(\lambda,t,p,q)\\\
-[\textsf{Q}(\lambda,t,p,q)]^{T}[\textsf{P}(\lambda,t,p,q)]^{-1}&[\textsf{Q}(\lambda,t,p,q)]^{T}[\textsf{P}(\lambda,t,p,q)]^{-1}\textsf{Q}(\lambda,t,p,q)-\textsf{R}(\lambda,t,p,q)\end{array}\right).$
(8.32)

By Claim 8.3, $u_{0,\lambda}(t)=(y_{0,\lambda}(t)^{T},0)^{T}$ satisfies
(8.10). It follows that $u_{0,\lambda}$ is $C^{2}$ and that
$u_{0,\lambda}(t)$, $\dot{u}_{0,\lambda}(t)$ and $\ddot{u}_{0,\lambda}(t)$ are
continuous in $(\lambda,t)$. These imply that the Hamiltonian
${H}:\hat{\Lambda}\times[0,\tau]\times\mathbb{R}^{n}\times
B^{n}_{\delta}(0)\to\mathbb{R}$ defined by

${H}(\lambda,t,z)={H}_{\lambda,t}(z):=(J\dot{u}_{0,\lambda}(t),z)_{\mathbb{R}^{2n}}+\hat{H}_{\lambda}(t,z+u_{0,\lambda}(t))$
(8.34)

is continuous and each ${H}_{\lambda,t}(\cdot)$ is $C^{2}$ and all its partial
derivatives depend continuously on $(\lambda,t,z)$. Clearly, (8.34) implies

$\displaystyle\nabla
H_{\lambda,t}(z)=J\dot{u}_{0,\lambda}(t)+\nabla\hat{H}_{\lambda,t}(z+u_{0,\lambda}(t)),$
(8.35) $\displaystyle
H_{\lambda,t}^{\prime\prime}(z)=\hat{H}_{\lambda,t}^{\prime\prime}(z+u_{0,\lambda}(t)).$
(8.36)

Recall that
$y_{0,\lambda}(t):=\partial_{v}\hat{L}_{\lambda}(t,0,0)=\partial_{v}\hat{L}_{\lambda}(t,x^{0},\dot{x}^{0})=\partial_{v}{L}_{\lambda}(t,x_{\lambda}(t),\dot{x}_{\lambda}(t))$.
We have $q=0$, $p=y_{0,\lambda}(t)$ and $v(\lambda,t,0,y_{0,\lambda}(t))=0$,
and hence

$\displaystyle\textsf{P}(\lambda,t,y_{0,\lambda}(t),0)=\partial_{vv}\hat{L}_{\lambda}\left(t,0,0\right)=\partial_{vv}{L}_{\lambda}(t,x_{\lambda}(t),\dot{x}_{\lambda}(t))=\textsf{P}_{\lambda}(t),$
$\displaystyle\textsf{Q}(\lambda,t,y_{0,\lambda}(t),0)=\partial_{qv}\hat{L}_{\lambda}\left(t,0,0\right)=\partial_{qv}{L}_{\lambda}(t,x_{\lambda}(t),\dot{x}_{\lambda}(t))=\textsf{Q}_{\lambda}(t),$
$\displaystyle\textsf{R}(\lambda,t,y_{0,\lambda}(t),0)=\partial_{qq}\hat{L}_{\lambda}\left(t,0,0\right)=\partial_{qq}{L}_{\lambda}(t,x_{\lambda}(t),\dot{x}_{\lambda}(t))=\textsf{R}_{\lambda}(t).$

by (8.3). These, (8.29) and (8.36) lead to

$\displaystyle
H_{\lambda,t}^{\prime\prime}(0)=\hat{H}_{\lambda,t}^{\prime\prime}(u_{0,\lambda}(t))=\hat{H}^{\prime\prime}_{(p,q)}(\lambda,t,y_{0,\lambda}(t),0)$
$\displaystyle=\left(\begin{array}[]{cc}\textsf{P}_{\lambda}(t)^{-1}&-\textsf{P}_{\lambda}(t)^{-1}\textsf{Q}_{\lambda}(t)\\\
-\textsf{Q}_{\lambda}(t)^{T}\textsf{P}_{\lambda}(t)^{-1}&\textsf{Q}_{\lambda}(t)^{T}\textsf{P}_{\lambda}(t)^{-1}\textsf{Q}_{\lambda}(t)-\textsf{R}_{\lambda}(t)\end{array}\right).$
(8.39)

It follows from Claim 8.3 and (8.35) that

$\displaystyle\dot{z}_{\lambda}(t)-\dot{u}_{0,\lambda}(t)$ $\displaystyle=$
$\displaystyle
J\nabla\hat{H}_{\lambda,t}(z_{\lambda}(t))-\dot{u}_{0,\lambda}(t)$
$\displaystyle=$ $\displaystyle
J\nabla\hat{H}_{\lambda,t}\big{(}z_{\lambda}(t)-u_{0,\lambda}(t)+u_{0,\lambda}(t)\big{)}-\dot{u}_{0,\lambda}(t)$
$\displaystyle=$ $\displaystyle
J\big{(}\nabla{H}_{\lambda,t}(z_{\lambda}(t)-u_{0,\lambda}(t))-J\dot{u}_{0,\lambda}(t)\big{)}-\dot{u}_{0,\lambda}(t)$
$\displaystyle=$ $\displaystyle
J\nabla{H}_{\lambda,t}(z_{\lambda}(t)-u_{0,\lambda}(t)).$

Hence we obtain:

######  Claim 8.4.

For any $\lambda\in\hat{\Lambda}$, by (8.35) the constant path
$z^{0}:[0,\tau]\to\mathbb{R}^{2n},\;t\mapsto 0$, satisfies

$\dot{z}(t)=J\nabla{H}_{\lambda,t}(z(t))\quad\hbox{and}\quad\mathbb{E}z(0)=z(\tau);$
(8.40)

and $(\mu,x_{\mu})$ is a bifurcation point of the problem (2.5)–(2.6) in
$\hat{\Lambda}\times C^{1}_{E}([0,\tau];\mathbb{R}^{n})$ with respect to the
trivial branch $\\{(\lambda,x_{\lambda})\,|\,\lambda\in\hat{\Lambda}\\}$ if
and only if $(\mu,z^{0})$ is that of (8.40) in $\hat{\Lambda}\times
C^{1}_{\mathbb{E}}([0,\tau];\mathbb{R}^{2n})$ (or equivalently
$\hat{\Lambda}\times W^{1,2}_{\mathbb{E}}([0,\tau];\mathbb{R}^{2n})$) with
respect to the trivial branch
$\\{(\lambda,z^{0})\,|\,\lambda\in\hat{\Lambda}\\}$. Precisely, if a sequence
$(\lambda_{k},x^{k})$ in $\hat{\Lambda}\times C^{2}([0,\tau],U)$ converges to
$(\mu,x_{\mu})$ in $\hat{\Lambda}\times C^{1}([0,\tau],U)$ and each $x^{k}$
satisfies (2.5)-(2.6) with $\lambda=\lambda_{k}$, $k=1,2,\cdots$, then

$z^{k}(t):=\left(\begin{array}[]{c}x^{k}(t)-x_{\lambda^{k}}(t)\\\
\partial_{v}{L}_{\lambda}(t,x^{k}(t),\dot{x}^{k}(t))\end{array}\right)-u_{0,\lambda_{k}}(t)=\left(\begin{array}[]{c}x^{k}(t)-x_{\lambda_{k}}(t)\\\
\partial_{v}{L}_{\lambda}(t,x^{k}(t),\dot{x}^{k}(t))-y_{0,\lambda_{k}}(t)\end{array}\right)$

satisfies (8.40) with $\lambda=\lambda_{k}$ and

$\displaystyle-\int^{\tau}_{0}\left[\frac{1}{2}(J\dot{z}^{k}(t),z^{k}(t))_{\mathbb{R}^{2n}}+H(\lambda_{k},t,{z}^{k}(t))\right]dt=\mathscr{L}_{\lambda}(x)=\int^{\tau}_{0}L_{\lambda_{k}}(t,x^{k}(t),\dot{x}^{k}(t))dt$

for each $k\in\mathbb{N}$, and $z^{k}\neq z^{0}$ (if $x^{k}\neq
x_{\lambda_{k}}$), $z^{k}\to z^{0}$ in
$C^{1}_{\mathbb{E}}([0,\tau];\mathbb{R}^{2n})$ (by [37, Proposition 1.3]);
conversely, if a sequence $z^{k}(t)=(p^{k}(t)^{T},x^{k}(t)^{T})^{T}$ in
$C^{1}_{\mathbb{E}}([0,\tau];\mathbb{R}^{2n})$ converges to $z^{0}$ in
$C^{1}_{\mathbb{E}}([0,\tau];\mathbb{R}^{2n})$ and each $z^{k}$ satisfies
(8.40) with $\lambda=\lambda_{k}$, then $x^{k}+x_{\lambda_{k}}$ satisfies
(2.5)-(2.6) with $\lambda=\lambda_{k}$ and

$\displaystyle\mathscr{L}_{\lambda_{k}}(x^{k}+x_{\lambda_{k}})=-\int^{\tau}_{0}\left[\frac{1}{2}(J\dot{z}^{k}(t),z^{k}(t))_{\mathbb{R}^{2n}}+H(\lambda_{k},t,{z}^{k}(t))\right]dt,$
(8.41) $\displaystyle
p^{k}(t)+y_{0,\lambda_{k}}(t)=\partial_{v}\hat{L}_{\lambda_{k}}(t,x^{k}(t),\dot{x}^{k}(t))=\partial_{v}{L}_{\lambda_{k}}(t,x^{k}(t)+x_{\lambda_{k}}(t),\dot{x}^{k}(t)+\dot{x}_{\lambda_{k}}(t)).$
(8.42)

(The latter implies that $p^{k}=0$ if and only if $x^{k}=0$, and hence that
$z^{k}=0$ if and only if $x^{k}=0$.)

Let $\delta^{\prime}\in(0,\delta)$ be close to $\delta$. We can choose a
$C^{\infty}$ function $\phi:{\mathbb{R}}^{n}\to[0,1]$ such that it is equal to
$1$ in $B^{n}_{\delta^{\prime}}(0)$, has support ${\rm supp}(\phi)\subset
B^{n}_{\delta}(0)$ and satisfies $\phi(q)=\phi(|q|)$ for all
$q\in{\mathbb{R}}^{n}$. Clearly, Claim 8.4 is still true if $H$ is replaced by

$\Lambda\times[0,\tau]\times\mathbb{R}^{n}\times{\mathbb{R}}^{n}\to{\mathbb{R}},\;(\lambda,t,p,q)\mapsto\left\\{\begin{array}[]{ll}\phi(q)H(\lambda,t,p,q),&\quad
q\in B^{n}_{\delta}(0),\\\ 0,&\quad q\in{\mathbb{R}}^{n}\setminus
B^{n}_{\delta}(0).\end{array}\right.$ (8.43)

Following the notations below (8.1) let $S_{\lambda}(t)$ denote the matrix in
(8). We have obtained
$H_{\lambda,t}^{\prime\prime}(z^{0}(t))=\hat{H}_{\lambda,t}^{\prime\prime}(u_{0,\lambda}(t))=S_{\lambda}(t)$.
Let $\Upsilon_{\lambda}:[0,\tau]\to{\rm Sp}(2n,{\mathbb{R}})$ be the
fundamental solution of the problem $\dot{\bf u}(t)=JS_{\lambda}(t){\bf u}$.
Then

$m^{0}_{\tau}(\mathscr{L}_{\lambda},x_{\lambda})=m^{0}_{\tau}(\hat{\mathscr{L}}_{\lambda},x^{0})=\dim{\rm
Ker}(\Upsilon_{\lambda}(\tau)-\mathbb{E})$ (8.44)

by [18, Lemma 3.1], and $\Upsilon_{\lambda}$ gives rise to a path of
Lagrangian subspaces

$[0,\tau]\ni t\mapsto{\rm
Gr}(\Upsilon_{\lambda})(t):=\\{(v^{T},(\Upsilon_{\lambda}(t)v)^{T})^{T}\,|\,v\in\mathbb{R}^{2n}\\}$

in the symplectic vector space
$(F,\Omega):=(\mathbb{R}^{2n}\oplus\mathbb{R}^{2n},(-\omega_{0})\oplus\omega_{0})$.
(Recall that all vectors in ${\mathbb{R}}^{m}$ are understand as column
vectors.) Let $\mathscr{L}(F,\Omega)$ denote the manifold of Lagrangian
subspaces of $(F,\Omega)$. Then ${\rm
Gr}(\mathbb{E})\in\mathscr{L}(F,\Omega)$. Recall that the Cappell-Lee-Miller
index $\mu^{\rm CLM}$ characterized by properties I-VI of [7, pp. 127-128]
assigns an integer $\mu^{\rm CLM}_{F}(\Lambda,\Lambda^{\prime})$ to every pair
of Lagrangian paths $\Lambda,\Lambda^{\prime}:[a,b]\to\mathscr{L}(F,\Omega)$.

Let $\mathcal{P}_{\tau}(2n)=\\{\Upsilon\in C([0,\tau],{\rm
Sp}(2n,{\mathbb{R}}))\,|\,\Upsilon(0)=I_{2n}\\}$. As extensions of the Maslov-
type index $(i_{\tau}(\Upsilon),\nu_{\tau}(\Upsilon))$ of
$\Upsilon\in\mathcal{P}_{\tau}(2n)$, Dong [10] and Liu [24] defined the Maslov
type index of $\Upsilon\in\mathcal{P}_{\tau}(2n)$ relative to $P\in{\rm
Sp}(2n,{\mathbb{R}})$ via different methods, respectively denoted by

$(i_{\tau,P}(\Upsilon),\nu_{\tau,P}(\Upsilon))\quad\hbox{and}\quad(i_{\tau}^{P}(\Upsilon),\nu_{\tau}^{P}(\Upsilon))$
(8.45)

for the sake of clearness (though both were written as
$(i_{P}(\Upsilon),\nu_{P}(\Upsilon))$ in [10] and [24]), where

$\nu_{\tau,P}(\Upsilon)=\dim{\rm
Ker}(\Upsilon(\tau)-P)=\nu_{\tau}^{P}(\Upsilon).$ (8.46)

######  Lemma 8.5.

  1. (i)

$m^{-}_{\tau}(\mathscr{L}_{\lambda},x_{\lambda})+\dim{\rm
Ker}(E-I_{n})=\mu^{\rm CLM}_{F}({\rm Gr}(\mathbb{E}),{\rm
Gr}(\Upsilon_{\lambda}))$.

  2. (ii)

$i^{P}_{\tau}(\Upsilon)=\mu^{\rm CLM}_{F}({\rm Gr}(P),{\rm
Gr}(\Upsilon))\;\;\forall(P,\Upsilon)\in{\rm
Sp}(2n,{\mathbb{R}})\times\mathcal{P}_{\tau}(2n)$.

  3. (iii)

$i_{\tau,P}(\Upsilon)=i^{P}_{\tau}(\Upsilon)\;\;\forall(P,\Upsilon)\in{\rm
Sp}(2n,{\mathbb{R}})\times\mathcal{P}_{\tau}(2n)$.

These three equalities come from [18, Theorem 1.2], [26, Theorem 5.18(2)] and
[27], respectively. Therefore from this lemma, (8.44) and (8.46) we derive

$\left.\begin{array}[]{ll}&m^{0}_{\tau}(\mathscr{L}_{\lambda},x_{\lambda})=m^{0}_{\tau}(\hat{\mathscr{L}}_{\lambda},x^{0})=\nu_{\tau,\mathbb{E}}(\Upsilon_{\lambda}),\\\
&m^{-}_{\tau}(\mathscr{L}_{\lambda},x_{\lambda})+\dim{\rm
Ker}(E-I_{n})=i_{\tau,\mathbb{E}}(\Upsilon_{\lambda}).\end{array}\right\\}$
(8.47)

######  Remark 8.6.

  1. (i)

Under Assumptions 2.2, 2.5, suppose also that $U$ is a symmetric open
neighborhood of the origin in $\mathbb{R}^{n}$, and that for each
$(\lambda,t)$ the function $L(\lambda,t,q,v)$ is even in $(q,v)$. If
$x_{\lambda}=0\;\forall\lambda$, then $H$ defined by (8.34) can be required to
be even in $z$. In fact, by Lemma 2.4(vi) $\tilde{L}(\lambda,t,q,v)$ can be
chosen to be even in $(q,v)$. Since $x_{\lambda}=0\;\forall\lambda$, the
function $\hat{L}$ defined by (2.14) is even in $(q,v)$, and so
$\partial_{v}\hat{L}_{\lambda}(t,-q,-v)=-\partial_{v}\hat{L}_{\lambda}(t,q,v)$.
(This implies $y_{0,\lambda}(t)=\partial_{v}\hat{L}_{\lambda}(t,0,0)=0$ and
hence $u_{0,\lambda}(t)=(y_{0,\lambda}(t)^{T},0)^{T}=0$.) Suppose that
$\partial_{v}\hat{L}_{\lambda}(t,q,v)=p$. Then
$-p=\partial_{v}\hat{L}_{\lambda}(t,-q,-v)$. By (8.9) we deduce that

$\displaystyle\hat{H}(\lambda,t,-p,-q)$ $\displaystyle=$
$\displaystyle(-p,-v)_{{\mathbb{R}}^{n}}-\hat{L}_{\lambda}(t,-q,-v)$
$\displaystyle=$
$\displaystyle(p,v)_{{\mathbb{R}}^{n}}-\hat{L}_{\lambda}(t,q,v)=\hat{H}(\lambda,t,p,q)$

and therefore that ${H}(\lambda,t,z)$ is even in $z$ by (8.34).

  2. (ii)

We make the following assumption:

######  Assumption 8.7.

Under Assumption 2.3 with $E=I_{n}$, suppose that $L$ also satisfies

$L(\lambda,-t,q,-v)=L(\lambda,t,q,v)\quad\forall(t,q,v)\in\Lambda\times{\mathbb{R}}\times
U\times\mathbb{R}^{n}.$ (8.48)

For each $\lambda\in\Lambda$, let $x_{\lambda}:{\mathbb{R}}\to U$ be a $C^{2}$
map satisfying

$\left.\begin{array}[]{ll}&\frac{d}{dt}\Big{(}\frac{\partial
L_{\lambda}}{\partial v}(t,x(t),\dot{x}(t))\Big{)}-\frac{\partial
L_{\lambda}}{\partial q}(t,x(t),\dot{x}(t))=0\;\forall t\in\mathbb{R}\\\
&x(-t)=x(t)=x(t+\tau)\quad\forall t\in\mathbb{R}\end{array}\right\\}$ (8.49)

and such that maps $\Lambda\times{\mathbb{R}}\ni(\lambda,t)\to
x_{\lambda}(t)\in U$ and
$\Lambda\times{\mathbb{R}}\ni(\lambda,t)\mapsto\dot{x}_{\lambda}(t)\in{\mathbb{R}}^{n}$
are also continuous. Moreover, for any compact or sequential compact subset
$\hat{\Lambda}\subset\Lambda$ there exists $\rho>0$ such that
$\sup\\{|\dot{x}_{\lambda}(t)|\,|\,(\lambda,t)\in\hat{\Lambda}\times[0,\tau]\\}<\rho$
and that $\hat{\Lambda}\times[0,\tau]\times U\times
B^{n}_{\rho}(0)\ni(\lambda,t,q,v)\mapsto{L}_{\lambda}(t,q,v)$ is strictly
convex with respect to $v$.

Then we have a corresponding result to Lemma 2.4, in which $[0,\tau]$ is
replaced by ${\mathbb{R}}$ and $\tilde{L}:\Lambda\times{\mathbb{R}}\times
U\times\mathbb{R}^{n}\to{\mathbb{R}}$ is defined by

$\tilde{L}(\lambda,t,q,v)={L}(\lambda,t,q,v)+\psi_{\rho_{0},\rho_{1}}(|v|^{2})$
(8.50)

as in (2.8). Clearly, it is $\tau$-periodic in $t$ and (8.48) implies

$\tilde{L}(\lambda,-t,q,-v)=\tilde{L}(\lambda,t,x,v)\quad\forall(\lambda,t,q,v)\in\Lambda\times{\mathbb{R}}\times
U\times\mathbb{R}^{n}.$ (8.51)

As in (2.14) we define $\hat{L}:\Lambda\times{\mathbb{R}}\times
B^{n}_{\delta}(0)\times{\mathbb{R}}^{n}\to{\mathbb{R}}$ by

$\hat{L}(\lambda,t,q,v)\mapsto\tilde{L}(\lambda,t,q+x_{\lambda}(t),v+\dot{x}_{\lambda}(t)).$
(8.52)

By Assumption 8.7 and (8.51) it is $\tau$-periodic and

$\displaystyle\hat{L}(\lambda,-t,q,-v)$ $\displaystyle=$
$\displaystyle{L}(\lambda,-t,q+x_{\lambda}(-t),-v-\dot{x}_{\lambda}(t))$
(8.53) $\displaystyle=$
$\displaystyle{L}(\lambda,t,q+x_{\lambda}(t),v+\dot{x}_{\lambda}(t))=\hat{L}(\lambda,t,q,v)$

for any $(\lambda,t,q,v)\in\Lambda\times{\mathbb{R}}\times
B^{n}_{\delta}(0)\times{\mathbb{R}}^{n}$. Thus
$\partial_{v}\hat{L}(\lambda,-t,q,-v)=-\partial_{v}\hat{L}(\lambda,t,q,v)$.
Let

$v:\Lambda\times{\mathbb{R}}\times
B^{n}_{\delta}(0)\times\mathbb{R}^{n}\to{\mathbb{R}}^{n},\;(\lambda,t,q,p)\mapsto
v(\lambda,t,q,p)$

be the unique solution of

$\displaystyle\partial_{v}\hat{L}_{\lambda}(t,q,v(\lambda,t,q,p))=p\quad\forall(\lambda,t,q,p)\in\Lambda\times{\mathbb{R}}\times
B^{n}_{\delta}(0)\times\mathbb{R}^{n}$ (8.54)

obtained by the implicit function theorem as in (8.54). Then we have

$\displaystyle
v(\lambda,-t,q,-p)=-v(\lambda,t,q,p)\quad\forall(\lambda,t,q,p)\in\Lambda\times{\mathbb{R}}\times
B^{n}_{\delta}(0)\times\mathbb{R}^{n}.$ (8.55)

As in (8.9) we define
$\hat{H}:\Lambda\times{\mathbb{R}}\times\mathbb{R}^{n}\times
B^{n}_{\delta}(0)\to{\mathbb{R}}$ by

$\hat{H}(\lambda,t,p,q)=(p,v)_{{\mathbb{R}}^{n}}-\hat{L}_{\lambda}(t,q,v),$

where $(\lambda,t,q,v)\in\Lambda\times{\mathbb{R}}\times
B^{n}_{\delta}(0)\times\mathbb{R}^{n}$ satisfies
$\partial_{v}\hat{L}_{\lambda}(t,q,v)=p$, that is,

$\hat{H}(\lambda,t,p,q)=(p,v(\lambda,t,q,p))_{{\mathbb{R}}^{n}}-\hat{L}_{\lambda}(t,q,v(\lambda,t,q,p)).$
(8.56)

Then from (8.53) and (8.55) we derive

$\displaystyle\hat{H}(\lambda,-t,-p,q,)$ $\displaystyle=$
$\displaystyle(-p,v(\lambda,-t,q,-p))_{{\mathbb{R}}^{n}}-\hat{L}_{\lambda}(-t,q,v(\lambda,-t,q,-p))$
(8.57) $\displaystyle=$
$\displaystyle(p,v(\lambda,t,q,p))_{{\mathbb{R}}^{n}}-\hat{L}_{\lambda}(-t,q,-v(\lambda,t,q,p))$
$\displaystyle=$
$\displaystyle(p,v(\lambda,t,q,p))_{{\mathbb{R}}^{n}}-\hat{L}_{\lambda}(t,q,v(\lambda,t,q,p))$
$\displaystyle=$ $\displaystyle\hat{H}(\lambda,t,p,q)$

for any $(\lambda,t,p,q)\in\Lambda\times{\mathbb{R}}\times\mathbb{R}^{n}\times
B^{n}_{\delta}(0)$. Let us define
${H}:\Lambda\times{\mathbb{R}}\times\mathbb{R}^{n}\times
B^{n}_{\delta}(0)\to\mathbb{R}$ by

${H}(\lambda,t,z):=(J\dot{u}_{0,\lambda}(t),z)_{\mathbb{R}^{2n}}+\hat{H}_{\lambda}(t,z+u_{0,\lambda}(t))$
(8.58)

as in (8.34), where $z=(q^{T},p^{T})^{T}$ and