Existence of Hyperbolic Motions to a Class of Hamiltonians and Generalized N-Body System via a Geometric Approach

Abstract

For the classical N-body problem in ${{\mathbb {R}}}^d$ with $d\geqq 2$, Maderna–Venturelli in their remarkable paper (Ann Math 192:499–550, 2020) proved the existence of hyperbolic motions with any positive energy constant, starting from any configuration and along any non-collision configuration. Their original proof relies on the long time behavior of solutions by Chazy 1922 and Marchal-Saari 1976, on the Hölder estimate for Mañé’s potential by Maderna 2012, and on the weak KAM theory. We give a new and completely different proof for the above existence of hyperbolic motions. The central idea is that, via some geometric observation, we build up uniform estimates for Euclidean length and angle of geodesics of Mañé’s potential starting from a given configuration and ending at the ray along a given non-collision configuration. Moreover, our geometric approach works for Hamiltonians $\frac{1}{2}\Vert p\Vert ^2-F(x)$, where $F(x)\geqq 0$ is lower semicontinuous and decreases very slowly to 0 faraway from collisions. We therefore obtain the existence of hyperbolic motions to such Hamiltonians with any positive energy constant, starting from any admissible configuration and along any non-collision configuration. Consequently, for several important potentials $F\in C^{2}(\Omega )$, we get similar existence of hyperbolic motions to the generalized N-body system $\ddot{x} = \nabla _x F(x)$, which is an extension of Maderna–Venturelli [Ann Math 2020].

Appendix A Properties of Hamiltonians and Mañé’s potentials

In the appendix we always assume that $m_i=1$ for all $1\leqq i\leqq N$. For general masses, all of the following conclusions still hold, up to some obvious modifications. We omit the details.

Let F be as in (1.4). Apriori, F is only defined in the set $\Omega$ since $F_{ij}$ in only defined in ${{\mathbb {R}}}^{d}\times {{\mathbb {R}}}^d{\setminus }\Delta$ for all $1\leqq i<j\leqq \infty$. Now we extend the definition of F to the whole ${{\mathbb {R}}}^{dN}$ by defining the value of $F_{ij}$ for all $1\leqq i<j\leqq \infty$ as follows:

$$\begin{aligned} F_{ij}(z,z)=\liminf _{(x_i,x_j)\rightarrow (z,z)}F_{ij}(x_i,x_j), \quad \forall (z,z)\in \Delta . \end{aligned}$$

It may happen that $F_{ij}(z,z)=\infty$ for some $z\in {{\mathbb {R}}}^d$, and hence $F(x)=\infty$ for some $x\in \Sigma$. But one may directly check that that $F_{ij}$ is lower semicontinuous in ${{\mathbb {R}}}^d\times {{\mathbb {R}}}^d$, that is, the set $\{(x_i,x_j)\in {{\mathbb {R}}}^d\times {{\mathbb {R}}}^d: F_{ij}(x_i,x_j)>\delta \}$ is open for all $\delta >0$. Thus F is also lower semicontinuous in whole ${{\mathbb {R}}}^{dN}$.

For any $\lambda >0$, Mañé’s potential $m_\lambda$ is defined in (1.5). Observe that

$$\begin{aligned} m_\lambda (x,y)\geqq \sqrt{2\lambda }|x-y| \text{ for } \text{ all } x,y\in {{\mathbb {R}}}^{dN}. \end{aligned}$$

(A.1)

Indeed, for any ${\gamma }\in {\mathcal{A}\mathcal{C}}(x,y;[0,\sigma ],{{\mathbb {R}}}^{dN})$, by the Cauchy-Schwartz inequality, its Euclidean length

$$\begin{aligned} l({\gamma }) \leqq (2\sigma )^{1/2}\left( \int _0^\sigma \frac{1}{2}|{\dot{{\gamma }}}(s)|^2 \,ds\right) ^{1/2}\leqq (2\sigma )^{1/2}\left( A_\lambda -\lambda \sigma \right) ^{1/2}, \end{aligned}$$

and hence by $|x-y|\leqq l({\gamma })$, one has

$$\begin{aligned} A_\lambda ({\gamma })\geqq \frac{1}{2\sigma } l({\gamma }) ^2 +\lambda \sigma \geqq \frac{1}{2\sigma } |x-y| ^2 +\lambda \sigma . \end{aligned}$$

Since $\frac{1}{2\sigma } |x-y| ^2 +\lambda \sigma$ reaches its minimal value at $\sigma = |x-y|/\sqrt{2\lambda }$, we have $A_\lambda ({\gamma })\geqq \sqrt{2\lambda }|x-y|$ as desired.

On the other hand, we have

$$\begin{aligned} m_\lambda (x,y) <\infty , \, \, \quad \forall x,y\in \Omega . \end{aligned}$$

Indeed, since $\Omega$ is path connected, one can always find a smooth curve ${\gamma }_{x,y}\in \mathcal{A}\mathcal{C}(x,y;[0,\sigma ];\Omega )$ for some $\sigma >0$. By the semicontinuity of F and $F<\infty$ in $\Omega$, we know that F is bounded in the compact set ${\gamma }([0,\sigma ])$ and hence we know that $A_\lambda ({\gamma })<\infty$. But for $x\in \Omega$ and $y\notin \Omega$, it is not clear whether $m_\lambda (x,y)$ is finite or not. This is indeed determined by the behaviour of F around $\Sigma$.

Set

$$\begin{aligned} \Omega _\lambda :=\{y\in {{\mathbb {R}}}^{dN}: m_\lambda (x,y)<\infty , \ \forall x\in \Omega \}. \end{aligned}$$

Obviously, $\Omega \subset \Omega _\lambda \subset {{\mathbb {R}}}^{dN}$. We observe that $\Omega _\lambda =\Omega _\mu$ for all $0<\lambda<\mu <\infty$. To see this, it suffices to show that

$$\begin{aligned} m_\lambda (x,y)<\infty \text{ if } \text{ and } \text{ only } \text{ if } m_\mu (x,y)<\infty \text{ for } \text{ any } x\in \Omega ,y\in {{\mathbb {R}}}^{dN}. \end{aligned}$$

Since $A_\lambda ({\gamma })\leqq A_\mu ({\gamma })$ for all possible ${\gamma }$, by definition, one always has $m_\lambda (x,y)\leqq m_{\mu }(x,y)$. On the other hand, if $m_\lambda (x,y)<\infty$, we have $A_\lambda ({\gamma })<\infty$ for some ${\gamma }\in {\mathcal{A}\mathcal{C}}(x,y;[0,t];{{\mathbb {R}}}^{dN})$. Thus

$$\begin{aligned} A_\mu ({\gamma })\leqq A_\lambda ({\gamma })+ (\mu -\lambda ) t<\infty , \end{aligned}$$

which implies $m_\mu (x,y)<\infty$ as desired. Recalling (1.6), we write $\widetilde{\Omega }=\Omega _\lambda$ for any $\lambda >0$.

One may directly check that $m_\lambda$ is a distance in $\widetilde{\Omega }$, and hence $(\widetilde{\Omega },m_\lambda )$ is a metric space. Observe that $(\widetilde{\Omega },m_\lambda )$ is always complete. Indeed, let $\{y_n\}_{n\in {{\mathbb {N}}}}\subset \widetilde{\Omega }$ be a Cauchy sequence with respect to $m_\lambda$. Recall that (A.1) gives $|z-w|\leqq \frac{1}{\sqrt{2\lambda }}m_\lambda (z,w)$ for all $z,w\in {{\mathbb {R}}}^{dN}$. It follows that $\{y_n\}_{n\in {{\mathbb {N}}}}$ is also a Cauchy sequence with respect to the Euclidean distance, and hence $|y_n-y|\rightarrow 0$ as $n\rightarrow \infty$ for some $y\in {{\mathbb {R}}}^{dN}$. It then suffices to show that $y\in \widetilde{\Omega }$ and $m_\lambda (y_n,y)\rightarrow 0$ as $n\rightarrow \infty$. If there exists $m\in {{\mathbb {N}}}$ such that $y_n=y$ for all $n\geqq m$, then $y_m\in \widetilde{\Omega }$ implies $y\in \widetilde{\Omega }$ and $m_\lambda (y_n,y)=0$ for all $n\geqq m$ as desired. Now we assume that there are infinitely many k such that $y_k\ne y$. Thanks to this, it is standard to find a subsequence $\{y_{n_k}\}_{k\in {{\mathbb {N}}}}$ of $\{y_n\}_{n\in {{\mathbb {N}}}}$ such that $0<m_\lambda (y_{n_k},y_{n_{k+1}})\leqq 2^{-k}$. For each $k\in {{\mathbb {N}}}$, there is ${\gamma }_k \subset {\mathcal{A}\mathcal{C}}(y_{n_k},y_{n_{k+1}};[0,\sigma _k];{{\mathbb {R}}}^{dN})$ such that $A_\lambda ({\gamma }_k)\leqq 2m_\lambda (y_{n_k},y_{n_{k+1}}) \leqq 2^{-k+1}$. Fix an arbitrary $x \in \Omega$. There is ${\gamma }_0 \subset {\mathcal{A}\mathcal{C}}(x,y_{n_1};[0,\sigma _0];{{\mathbb {R}}}^{dN})$ such that $A_\lambda ({\gamma }_0)\leqq 2m_\lambda (x,y_1) <\infty$. Noting that $\sigma _k\leqq \frac{1}{\lambda }A_\lambda ({\gamma }_k)$ for all $k\in {{\mathbb {N}}}\cup \{0\}$, it implies that

$$\begin{aligned} \sigma :=\sum _{k=0}^\infty \sigma _k \leqq \frac{1}{\lambda }\sum _{k=0}^\infty A_\lambda ({\gamma }_k) \leqq \frac{1}{\lambda }\sum _{k=0}^\infty 2^{-k+1}<\infty . \end{aligned}$$

Then the concatenation ${\gamma }$ of these curves ${\gamma }_k$ (that is, ${\gamma }=*_{k}{\gamma }_k$) belongs to ${\mathcal{A}\mathcal{C}}(x,y;[0,\sigma ];{{\mathbb {R}}}^{dN})$ and satisfies $A_\lambda ({\gamma })\leqq \sum _{k=0}^\infty A_\lambda ({\gamma }_k)<\infty$. Thus $m_\lambda (x,y)<\infty$, that is, $y\in \widetilde{\Omega }$, and moreover

$$\begin{aligned} m_{\lambda }(y_{n_k},y)\leqq \sum _{j=k}^\infty A_\lambda ({\gamma }_j) \leqq \sum _{j=k}^\infty 2^{-j+2}= 2^{-k+3}, \end{aligned}$$

which implies $m_{\lambda }(y_{n_k},y)\rightarrow 0$ as $k\rightarrow \infty$. Since $\{y_n\}_{n\in {{\mathbb {N}}}}$ is a Cauchy sequence with respect to $m_\lambda$, by a standard argument, one further gets $m_\lambda (y_n,y)\rightarrow 0$ as $n\rightarrow \infty$ as desired.

Recall that $\Omega \subset \widetilde{\Omega }$, and $\Omega$ is an open subset of $\widetilde{\Omega }$ with respect to $m_\lambda$. Thus $(\Omega , m_\lambda )$ is also a metric space as a subspace of $(\widetilde{\Omega },m_\lambda )$. Since there is no any other assumption made on the behaviour of F around $\Sigma$ (except the lower semicontinuity), $\Omega$ is not necessarily closed or complete in general. It is interesting to determine the relation between $(\Omega ,m_\lambda )$, or its completion/closure, and $(\widetilde{\Omega }, m_\lambda )$. In the following remark, we have some discussion about this issue.

Remark A.1

(i) It is easy to see that if

$$\begin{aligned} F_{ij}(x_i,x_j) \geqq C |x_i-x_j|^{-2} \text{ when } |x_i-x_j|\leqq \frac{1}{2} \text{ for } \text{ all } 1\leqq i<j\leqq N, \end{aligned}$$

then $\widetilde{\Omega }=\Omega$, and hence, $(\Omega ,m_\lambda )$ is complete. One may also check that if

$$\begin{aligned}&| F_{ij}(x_i,x_j)|\leqq C|x_i-x_j|^{-\alpha } \text{ when } |x_i-x_j|\\ {}&\leqq \frac{1}{2} \text{ for } \text{ all } 1\leqq i<j\leqq N \text{ and } \alpha \in (0, 2), \end{aligned}$$

then $\Omega \subsetneqq \widetilde{\Omega }={{{{\mathbb {R}}}}^n}$. But in general, it is difficult to determine the set $\widetilde{\Omega }$. For example, see [4, 7, 10, 11, 15, 21, 29, 33] and references therein.

(ii) If $F=+\infty$ in $\Sigma = {{\mathbb {R}}}^{dN}{\setminus } \Omega$, then $\Omega$ is dense in $(\widetilde{\Omega },m_\lambda )$, equivalently, the completion of $( \Omega ,m_\lambda )$ is $(\widetilde{\Omega }, m_\lambda )$. Indeed, if $\Omega \subsetneqq \widetilde{\Omega }$, given any $y \in \widetilde{\Omega }{\setminus } \Omega$ and $x \in \Omega$, there is ${\gamma }\in {\mathcal{A}\mathcal{C}}(x,y;[0,\sigma ];{{\mathbb {R}}}^{dN})$ with $A_\lambda ({\gamma })<\infty$. Since $F=\infty$ in $\Sigma$, we know that ${\gamma }(t)\subset \Omega$ for almost all $t\in [0,\sigma ]$. Therefore there exists a sequence $\{t_n\}\subset [0,\sigma ]$ so that $t_n\rightarrow \sigma$ with ${\gamma }(t_n)\in \Omega$, and $A_\lambda ({\gamma }|_{[t_n,\sigma ]})\rightarrow 0$, which implies $m_\lambda ({\gamma }(t_n), y) \rightarrow 0$ as $n \rightarrow \infty$, that is, y is the limit of the Cauchy sequence $\{{\gamma }(t_n)\} \subset \Omega$ with respect to $m_\lambda$. This shows the closure ${\overline{\Omega }}^{m_\lambda }$ of $\Omega$ with respect to $m_\lambda$ contains $\widetilde{\Omega }$. On the other hand, since $\Omega \subset \widetilde{\Omega }$ and $\widetilde{\Omega }$ is complete with respect to $m_\lambda$, it follows that ${\overline{\Omega }}^{m_\lambda } \subset \widetilde{\Omega }$. We conclude that ${\overline{\Omega }}^{m_\lambda }=\widetilde{\Omega }$.

(iii) Without the assumption $F=+\infty$ in $\Sigma$, in general one can not expect that $\Omega$ is dense in $(\widetilde{\Omega },m_\lambda )$, equivalently, the completion of $( \Omega ,m_\lambda )$ is $(\widetilde{\Omega }, m_\lambda )$. Indeed, we construct a potential $F:{{\mathbb {R}}}^{dN}\rightarrow (0,\infty )$ so that $\Omega$ is not dense in $(\widetilde{\Omega },m_\lambda )$; see the following 3 steps.

Step 1. Constructions of $h_\pm :[0,\infty )\rightarrow (0,\infty )$ by modifying $r^{-2}$.

Let $h_\pm :[0,\infty )\rightarrow (0,\infty )$ be two functions defined by

$$\begin{aligned} h_+(r)=\left\{ \begin{aligned} 1\ \quad \quad&\text{ when }\ 0\leqq r\leqq 1;\\ \frac{1}{r^2}\quad \quad&\text{ when }\ 1<r<\infty \end{aligned}\right. \end{aligned}$$

and

$$\begin{aligned} h_-(r)=\left\{ \begin{aligned}&1 \quad&\text{ when } \ r\in \Lambda :=\{0\}\cup (\cup _{k\geqq 3} [2^{-k^2-1}, 2^{-k^2+1}]); \\&\frac{1}{r^2}\quad&\text{ when } \ r\in (0,\infty ){\setminus }\Lambda . \end{aligned}\right. \end{aligned}$$

Obviously, $h_+\in C^0([0,\infty ))$, and $h_-$ is lower semicontinuous in $[0,\infty )$ with

$$\begin{aligned} \displaystyle h_-(r)=\liminf _{t\rightarrow r}h_-(t)\quad \forall \ 0\leqq r<\infty . \end{aligned}$$

moreover, $h_+(r)\leqq h_-(r)\leqq r^{-2}$ for all $0<r<\infty$, and $h_+=h_-$ in $\Lambda \cup [1,+\infty )$.

Step 2. Construction of a potential $F:{{\mathbb {R}}}^{dN}\rightarrow (0,\infty )$ via $h_\pm$.

For any nonempty subset $E \subset {{\mathbb {R}}}^{2d}$, recall the standard characteristic function $\chi _{E}: {{\mathbb {R}}}^{2d} \rightarrow \{0, 1\}$, which maps the elements of E to 1, and all other elements to 0. Define

$$\begin{aligned} F(x):= F_{12}(x_1,x_2) \text{ for } \text{ all } x=(x_1, \ldots , x_N) \in {{\mathbb {R}}}^{dN} \end{aligned}$$

with

$$\begin{aligned} F_{12}(x_1,x_2):= h_-(|x_1-x_2|)\chi _{ \Gamma \times \Gamma }+ h_+(|x_1-x_2|)\chi _{{{\mathbb {R}}}^{d}\times {{\mathbb {R}}}^d {\setminus } [\Gamma \times \Gamma ]}, \end{aligned}$$

where

$$\begin{aligned} x_i=(x_i^1, \ldots , x_i^d) \in {{\mathbb {R}}}^{d } \text{ and } \Gamma := \{x_i \in {{\mathbb {R}}}^{d }: x_i^j < 0, \ \forall 1 \leqq j\leqq d \}. \end{aligned}$$

Obviously, $F_{12}$ is locally bounded in ${{\mathbb {R}}}^{d}\times {{\mathbb {R}}}^d{\setminus }\{(x_1,x_2)\in {{\mathbb {R}}}^{d}\times {{\mathbb {R}}}^d: x_1=x_2, \, x_1\in \overline{\Gamma }\},$ and $F_{12}(x_1,x_2)= 1$ whenever $|x_1-x_2|\in \Lambda$. Moreover, $F_{12}$ is lower semicontinuous in ${{\mathbb {R}}}^d\times {{\mathbb {R}}}^d$ with

$$\begin{aligned} F_{12}(x_1,x_2)=\liminf _{(z_1,z_2)\rightarrow (x_1,x_2)}F_{12}(z_1,z_2) \text{ for } \text{ all } (x_1,x_2)\in {{\mathbb {R}}}^{2d}. \end{aligned}$$

In particular, F satisfies the condition (1.4).

Note that

$$\begin{aligned} \limsup _{(z_1,z_2)\rightarrow (x_1,x_2)} F_{12}(z_1,z_2)=+\infty \quad \text{ whenever } x_1=x_2 \text{ and } x_1\in \Gamma . \end{aligned}$$

Below we write

$$\begin{aligned} \Sigma ^-_{12}:=\{x=(x_1,\ldots ,x_N)\in {{\mathbb {R}}}^{dN}: x_1=x_2\in \Gamma \} \subset \Gamma \times \Gamma \times {{\mathbb {R}}}^{d(N-2)} \end{aligned}$$

and

$$\begin{aligned} \Sigma _{12}:=\{x=(x_1,\ldots ,x_N)\in {{\mathbb {R}}}^{dN}: x_1=x_2\}. \end{aligned}$$

Notice that the local boundedness of $F_{12}$ in ${{\mathbb {R}}}^{d}\times {{\mathbb {R}}}^d{\setminus }\{(x_1,x_2)\in {{\mathbb {R}}}^{d}\times {{\mathbb {R}}}^d: x_1=x_2, \, x_1\in \overline{\Gamma }\}$ implies the local boundedness of F in ${{\mathbb {R}}}^{dN}{\setminus }\overline{\Sigma _{12}^-}$. In particular, F is locally bounded in $\Sigma {\setminus } \Sigma _{12}$.

Step 3. We show that $\widetilde{\Omega }={{\mathbb {R}}}^{dN}$ and $\Sigma _{12}^{-}\subset \widetilde{\Omega }{\setminus } {\overline{\Omega }}^{m_\lambda }$.

We first show that $\widetilde{\Omega }={{\mathbb {R}}}^{dN}$. It suffices to show that $\Sigma \subset \widetilde{\Omega }$. We consider three cases: $x \in \Sigma {\setminus } \Sigma _{12}$, $x \in \Sigma _{12} {\setminus } \overline{\Sigma _{12}^-}$ and $x \in \overline{\Sigma _{12}^-}$.

Case $x \in \Sigma {\setminus } \Sigma _{12}$. By Step 2, we know that F is locally bounded at x. Since ${{\mathbb {R}}}^{dN}{\setminus } \Sigma _{12}$ is a connected open subset, there exists an Euclidean ball $B(x,r_x) \subset {{\mathbb {R}}}^{dN}{\setminus } \Sigma _{12}$ for some $r_x>0$ such that $F|_{B(x,r_x)} \leqq M_x$ for some $M_x>0$. Noting that $\Sigma ={{\mathbb {R}}}^{dN}{\setminus }\Omega$ is closed and does not have interior points, we can find a curve ${\gamma }_x \in C^1([0,t], B(x,r_x))$ with end point x such that ${\gamma }_x {\setminus } \{x\} \subset \Omega$ and $F|_{{\gamma }_x} \leqq M_x$. Therefore, we see that $m_\lambda ({\gamma }_x(0),x)\leqq A_\lambda ({\gamma }_x)<\infty$, that is, $x \in \widetilde{\Omega }$.

Case $x \in \Sigma _{12} {\setminus } \overline{\Sigma _{12}^-}$. By the definition of F, we know $F(x)=1$. Note that ${{\mathbb {R}}}^{dN}{\setminus } \overline{\Sigma _{12}^-}$ is a connected open subset of ${{\mathbb {R}}}^{dN}$. Similar to the first case, there is an Euclidean ball $B(x,r_x) \subset {{\mathbb {R}}}^{dN}{\setminus } \overline{\Sigma _{12}^-}$ for some $r_x>0$ such that $F|_{B(x,r_x)} \leqq 2$, and moreover, we can find a curve ${\gamma }_x \in C^1([0,t], B(x,r_x))$ with end point x such that ${\gamma }_x {\setminus } \{x\} \subset \Omega$ and $F|_{{\gamma }_x} \leqq 2$. Therefore, we see that $m_\lambda ({\gamma }_x(0),x)\leqq A_\lambda ({\gamma }_x)<\infty$, that is, $x \in \widetilde{\Omega }$.

Case $x \in \overline{\Sigma _{12}^-}$. Letting $y = (y_i)_{1}^N \in \Sigma _{12}{\setminus } \overline{\Sigma _{12}^-}$ such that $y_1=y_2=(1,\ldots ,1) \in {{\mathbb {R}}}^d$ and $y_i=x_i \in {{\mathbb {R}}}^d$ for $i=3,\ldots ,N$, taking ${\gamma }$ as the line-segment joining y and x, then ${\gamma }\subset \Sigma _{12}$ and hence $F|_{{\gamma }}\equiv 1$. We have $m_\lambda (x,y)<A_\lambda ({\gamma })<\infty$. Since $y \in \Sigma _{12}{\setminus } \overline{\Sigma _{12}^-} \subset \widetilde{\Omega }$, there exists some $x_0 \in \Omega$, such that $m_\lambda (x,x_0) \leqq m_\lambda (x,y) + m_\lambda (y, x_0)<\infty$. Hence, $x \in \widetilde{\Omega }.$

Finally, we show that $\Sigma _{12}^{-}\subset \widetilde{\Omega }{\setminus } {\overline{\Omega }}^{m_\lambda }$. Let $x\in \Sigma _{12}^{-}$ be an arbitrary point. It then suffices to show that

$$\begin{aligned} \inf _{z\in \Omega } m_\lambda (z,x) >0. \end{aligned}$$

(A.2)

To prove this, we need some properties of $m_\lambda$-geodesics joining z, x which will be introduced/proved below. So we postpone the proof of (A.2) to the end of this appendix.

Moreover, we know that $(\widetilde{\Omega }, m_\lambda )$ is a geodesic space, that is, for any $x, y \in \widetilde{\Omega }$ with $x\ne y$, there exists a $\eta \in {\mathcal{A}\mathcal{C}}(x, y; [0,m_\lambda (x,y)];{{\mathbb {R}}}^{dN})$ such that

$$\begin{aligned} \eta ([0,m_\lambda (x,y)])\subset \widetilde{\Omega }, \text{ and } m_\lambda (\eta (s),\eta (t))=t-s \text{ for } \text{ all } 0\leqq s<t\leqq m_\lambda (x,y). \end{aligned}$$

Indeed, thanks to Lemma A.2 below, the desired geodesic ${\gamma }$ comes from the arc length (with respect to $m_\lambda$) parametrisation of the following minimizer of $A_\lambda$ in the class of $\cup _{\sigma >0}{\mathcal{A}\mathcal{C}}(x, y; [0,\sigma ];\widetilde{\Omega })$.

Lemma A.2

Given $\lambda >0$ and $x, y \in \widetilde{\Omega }$ with $x\ne y$, there is a curve ${\gamma }\in {\mathcal{A}\mathcal{C}}(x, y; [0,\sigma ];\widetilde{\Omega })$ such that $A_{\lambda }({\gamma }) = m_\lambda (x, y)$. Moreover, for any ${\gamma } \in {\mathcal{A}\mathcal{C}}(x, y; [0,\sigma ];\widetilde{\Omega })$ satisfying $A_{\lambda }({\gamma }) = m_\lambda (x, y)$, we have

$$\begin{aligned} m_\lambda ({\gamma }(s ),{\gamma }(t))=A_\lambda ({\gamma }|_{[s,t ]}), \quad \forall 0\leqq s <t\leqq \sigma . \end{aligned}$$

The proof of Lemma A.2 is standard, for readers’ convenience we give it later.

Considering the geodesic nature, as in the introduction we call a minimizer ${\gamma }:[0,\sigma ]\rightarrow \widetilde{\Omega }$ of $A_\lambda$ with given endpoints x, y as an $m_\lambda$-geodesic with the canonical parameter joining x, y. We also call a ray ${\gamma }:[0,\infty )\rightarrow \widetilde{\Omega }$ as an $m_\lambda$-geodesic ray with the canonical parameter if ${\gamma }|_{[0,\sigma ]}$ is an $m_\lambda$-geodesic with the canonical parameter for any $\sigma >0$.

Next we prove the following result:

Lemma A.3

Given any $x,y\in \widetilde{\Omega }$ and $\lambda >0$, let ${\gamma }\in {\mathcal{A}\mathcal{C}}(x,y;[0,\sigma ], {{\mathbb {R}}}^{dN})$ be any $m_\lambda$-geodesic with canonical parameter. Then ${\gamma }$ has energy constant $\lambda$, that is,

$$\begin{aligned} |{\dot{{\gamma }}}(s)|=\sqrt{2(F({\gamma }(s))+\lambda )} \text{ almost } \text{ all } s\in [0,\sigma ]. \end{aligned}$$

Moreover,

$$\begin{aligned} m_\lambda (x,y)= \widetilde{m}_\lambda (x,y)=\widetilde{A}_\lambda ({\gamma }), \end{aligned}$$

where

$$\begin{aligned} \widetilde{m}_\lambda (x,y):=\inf \{ \widetilde{A}_\lambda (\eta ):\eta \in \cup _{\tau >0}{\mathcal{A}\mathcal{C}}(x,y;[0,\tau ], {{\mathbb {R}}}^{dN})\}\quad \forall x,y\in {{\mathbb {R}}}^{dN}, \end{aligned}$$

and

$$\begin{aligned} \widetilde{A}_\lambda (\eta ):=\int _0^{\tau } |{\dot{\eta }}(s)|\sqrt{2 F\circ \eta +2 \lambda }\,ds. \end{aligned}$$

Here in the above inequality, we use the convention that $|{\dot{\eta }}(s)|\sqrt{2 F\circ \eta +2 \lambda } =0$ when $|{\dot{\eta }}(s)|=0$ and $F\circ \eta (s) =\infty$.

Consequently, we have

Lemma A.4

Under additional assumption $F\in C^{2}(\Omega )$, for any $\lambda >0$, if ${\gamma }\in {\mathcal{A}\mathcal{C}}(x,y;[0,\sigma ], {{\mathbb {R}}}^{dN})$ is an $m_\lambda$-geodesic with canonical parameter joining x, y and also is interiorly collision-free (that is, ${\gamma }|_{(0,\sigma )}\subset \Omega$), then ${\gamma }$ is a solution to $\ddot{x}=\nabla F$ in $(0,\sigma )$ starting from x and ending at y.

Consequently, if ${\gamma }:[0,\infty )\rightarrow {{\mathbb {R}}}^{dN}$ is an $m_\lambda$-geodesic ray with canonical parameter starting from x and also is interiorly collision-free (that is, ${\gamma }|_{(0,\infty )}\subset \Omega$), then ${\gamma }$ is a solution to $\ddot{x}=\nabla F$ in $(0,\infty )$ starting from x.

Below we prove Lemma A.2-Lemma A.4.

Proof of Lemma A.2

Let $x, y \in \widetilde{\Omega }$ be two given configurations, with $x \ne y$. Since $0<m_\lambda (x,y)<\infty$, there exist a sequence of curves $\{ \delta _n \in {\mathcal{A}\mathcal{C}}(x,y;[0,\sigma _n];{{\mathbb {R}}}^{dN})\}_{n\in {{\mathbb {N}}}}$ such that

$$\begin{aligned} \lim _{n \rightarrow \infty } A_\lambda (\delta _n) = m_\lambda (x,y). \end{aligned}$$

Thus for n large enough we have $A_\lambda (\delta _n) \leqq m_\lambda (x,y)+1$. Without loss of generality, we may assume that for all n, this holds. This also implies that $\delta _n \in {\mathcal{A}\mathcal{C}}(x,y;[0,\sigma _n];\widetilde{\Omega })$; otherwise $\delta _n(t)\notin \widetilde{\Omega }$ for some $0<t<\sigma _n$ and hence

$$\begin{aligned} m_\lambda (x,\delta _n(t))<A_\lambda (\delta _n|_{[0,t]})<A_\lambda (\delta _n )<\infty , \end{aligned}$$

which is a contradiction with the definition of $\widetilde{\Omega }$.

Next, for every n, since

$$\begin{aligned} \frac{1}{2\sigma _n} |x-y| ^2 +\lambda \sigma _n\leqq A_\lambda (\delta _n) \leqq 1+m_\lambda (x,y), \end{aligned}$$

we know

$$\begin{aligned} 0< \liminf _{n\rightarrow \infty } \sigma _n \leqq \limsup _{n \rightarrow \infty }\sigma _n < +\infty . \end{aligned}$$

and therefore up to considering subsequence, we may assume that $\sigma _n \rightarrow \sigma _0$ as $n \rightarrow \infty$.

Now, for each $n > 0$, we parameterize $\delta _n$ linearly as below. Define ${\gamma }^{(n)} (t) = \delta _n(\sigma _n\sigma _0^{-1} t)$ for $t\in [0,\sigma _0]$. A direct calculation leads to

$$\begin{aligned} \lim _{n \rightarrow \infty } A_\lambda ({\gamma }^{(n)} ) = \lim _{n \rightarrow \infty } A_\lambda (\delta _n) = m_\lambda (x,y). \end{aligned}$$

We may also assume that

$$\begin{aligned} A_\lambda ({\gamma }^{(n)} )\leqq m_\lambda (x,y)+1, \quad \forall n\geqq 1. \end{aligned}$$

It is easy to see that $\{{\gamma }^{(n)} \}$ is uniform bounded and equicontinuous. Indeed, for any $s,t\in [0,\sigma _0]$ with $s<t$, one has

$$\begin{aligned} |{\gamma }^{(n)} (s)-{\gamma }^{(n)} (t)|\leqq & {} \int _s^t|{\dot{{\gamma }}}^{(n)} (\tau )|\,d\tau \leqq \left( \int _0^{\sigma _0}|{\dot{{\gamma }}}^{(n)} (\tau )|^2\,d\tau \right) ^{1/2}|s-t|^{1/2}\\\leqq & {} [2(m_\lambda (x,y)+1)]^{1/2}|s-t|^{1/2}. \end{aligned}$$

In particular,

$$\begin{aligned} |{\gamma }^{(n)} (0)-{\gamma }^{(n)} (t)|\leqq [2(m_\lambda (x,y)+1)]^{1/2}| t|^{1/2}\leqq [2(m_\lambda (x,y)+1)]^{1/2} |\sigma _0|^{1/2}. \end{aligned}$$

By Arzela-Ascoli Theorem, up to some subsequence, we may assume that the sequence $\{{\gamma }^{(n)} \}$ converges uniformly to a curve ${\gamma }\in C^0( [0,\sigma _0];{{\mathbb {R}}}^{dN})$ with ${\gamma }(0)=x,{\gamma }(\sigma _0)=y$. Note that ${\gamma }$ is absolutely continuous. Indeed, for any $\varepsilon >0$ and for any family $\{[s_i,t_i]\}_{1\leqq i\leqq k}$ of mutually disjoint intervals such that $\sum _{i=1}^k|s_i-t_i| <\varepsilon$, applying the Cauchy-Schwarz inequality, we have

$$\begin{aligned} \sum _{i=1}^k|{\gamma }^{(n)} (s_i)-{\gamma }^{(n)} (t_i)|&\leqq \int _{\cup _{1\leqq i\leqq k} [s_i,t_i]}|{\dot{{\gamma }}}^{(n)} (\tau )|\,d\tau \\&\leqq \left[ \int _{\cup _{1\leqq i\leqq k} [s_i,t_i]} |{\dot{{\gamma }}}^{(n)} (\tau )|^2\,d\tau \right] ^{1/2} \left[ \sum _{i=1}^k|s_i-t_i|\right] ^{1/2} \\&\leqq 2^{1/2} [m_\lambda (x,y)+1]^{1/2}\left[ \sum _{i=1}^k|s_i-t_i|\right] ^{1/2} \\&<2^{1/2} [m_\lambda (x,y)+1]^{1/2}\varepsilon ^{1/2}. \end{aligned}$$

Thus ${\gamma }\in \mathcal{A}\mathcal{C} (x, y;[0,\sigma _0];{{\mathbb {R}}}^{dN})$. Apply Tonelli’s Theorem for convex Lagrangians to get

$$\begin{aligned} \frac{1}{2} \int _0^{\sigma _0} |{\dot{{\gamma }}}(s)|^2 \, ds \leqq \frac{1}{2} \liminf _{n \rightarrow \infty } \int _0^{\sigma _0} |{\dot{{\gamma }}}^{(n)} (s)|^2 \, ds \end{aligned}$$

and Fatou’s Lemma to obtain that

$$\begin{aligned} \int _0^{\sigma _0} F({\gamma }) \, ds \leqq \liminf _{n \rightarrow \infty } \int _0^{\sigma _0} F({\gamma }^{(n)} ) \, ds. \end{aligned}$$

Therefore $A_\lambda ({\gamma }) \leqq m_\lambda (x, y)$, which is only possible if the equality holds. Note that $A_\lambda ({\gamma })<\infty$ also implies that ${\gamma }\in \mathcal{A}\mathcal{C} (x, y;[0,\sigma _0];\widetilde{\Omega })$.

Next for any ${\gamma }\in {\mathcal{A}\mathcal{C}}(x, y; [0,\sigma ];\widetilde{\Omega })$ such that $A_{\lambda }({\gamma }) = m_\lambda (x, y)$ and for any $0\leqq s <t\leqq \sigma$, we claim that

$$\begin{aligned} m_\lambda ({\gamma }(s),{\gamma }(t))\leqq A_\lambda ({\gamma }|_{[s,t]}). \end{aligned}$$

We show this by contradiction. If

$$\begin{aligned} m_\lambda ({\gamma }(s),{\gamma }(t))< A_\lambda ({\gamma }|_ {[s,t]}), \end{aligned}$$

we find ${\gamma }_0 \in {\mathcal{A}\mathcal{C}}({\gamma }(s), {\gamma }(t); [0,\sigma _0];\widetilde{\Omega })$ such that

$$\begin{aligned} A_{\lambda }({\gamma }_0 ) = m_\lambda ({\gamma }(s), {\gamma }(t))=m_\lambda ({\gamma }(s),{\gamma }(t))< A_\lambda ({\gamma }|_ {[s,t]}). \end{aligned}$$

The concatenation of ${\gamma }|_{[0,s]}$,${\gamma }_0$ and ${\gamma }|_{[t,\sigma ]}$ gives a curve $\eta \in {\mathcal{A}\mathcal{C}}(x, y; [0,\sigma _0+\sigma - (t-s) ];\widetilde{\Omega })$ such that

$$\begin{aligned}m_\lambda (x,y)&\leqq A_\lambda (\eta )= A_\lambda ({\gamma }|_{[0,s]})+A_\lambda ({\gamma }_0)+A_\lambda ({\gamma }|_{[t,\sigma ]})\\&< A_\lambda ({\gamma }|_{[0,s]})+A_\lambda ({\gamma }|_{[s,t]})+A_\lambda ({\gamma }|_{[t,\sigma ]})\\&=A_\lambda ({\gamma })=m_\lambda (x,y), \end{aligned}$$

which is a contradiction. $\square$

To prove lemma A.3, we need the following auxiliary Lemma A.5 - Lemma A.8.

Lemma A.5

Given any $\lambda >0$ and $x,y\in \Omega$, let ${\gamma }\in {\mathcal{A}\mathcal{C}}(x,y;[0,\sigma ];\widetilde{\Omega })$ satisfying $m_\lambda (x,y)=A_\lambda ({\gamma })$. Then $F\circ {\gamma }<\infty$ and $|{\dot{{\gamma }}}|>0$ almost everywhere.

To prove Lemma A.5 and for later use, we recall the following two change of variable formulas. We refer to for example [18, Section 3.3.3, Theorem 2] for the first one, which comes from the area formula and works for Lipschitz maps; and refer to [9, Proposition 2.2.18] for the second one, which works for absolute continuous maps.

Lemma A.6

(i) Let $f:{{\mathbb {R}}}\rightarrow {{\mathbb {R}}}$ be Lipschitz. For any $g\in L^1({{\mathbb {R}}})$, one has

$$\begin{aligned} \left| \sum _{s\in f^{-1}(t)} g (s)\right| \leqq \sum _{s\in f^{-1}(t)}|g|(s)\in L^1({{\mathbb {R}}}), \end{aligned}$$

where $f^{-1}(t)$ is at most countable for almost all $t\in {{\mathbb {R}}}$, and

$$\begin{aligned} \int _{{\mathbb {R}}}g(s)|f'(s)|\,ds=\int _{{\mathbb {R}}}\left[ \sum _{s\in f^{-1}(t)}g(s)\right] \,dt. \end{aligned}$$

If f is injective in addition, then $g\circ f^{-1}\in L^1({{\mathbb {R}}})$ and

$$\begin{aligned} \int _{{\mathbb {R}}}g(s)|f'(s)|\,ds=\int _{{\mathbb {R}}}g\circ f^{-1}(t) \,dt. \end{aligned}$$

(ii) Let $f:[\alpha ,\beta ]\rightarrow [a,b]$ be absolutely continuous and increasing, and satisfy $a=f(\alpha )$ and $b=f(\beta )$. Then for any $g \in L^1([a,b])$, one has $(g\circ f)f'\in L^1([\alpha ,\beta ])$ and

$$\begin{aligned} \int _a^b g (t)\,dt=\int _{\alpha }^{\beta }g (f(s))f'(s) \,ds. \end{aligned}$$

Proof of Lemma A.5

Note that $A_\lambda ({\gamma })<\infty$ implies the integrability of $F\circ {\gamma }$ in $[0,\sigma ]$, and hence $F\circ {\gamma }(s)<\infty$ for almost all $s\in [0,\sigma ]$. Moreover, denote by E the set of $t\in [0,\sigma ]$ such that $|{\dot{{\gamma }}}(t)|=0$. We prove $|E|=0$ by contradiction. Assuming that $|E|>0$ below.

First, we show that there is no interval $[s_0, s_1] \subset [0, \sigma ]$ such that almost all points in $[s_0, s_1]$ are in E. Otherwise, assume that $|{\dot{{\gamma }}} |=0$ in the interval $[s_0,s_1]\subset [0,\sigma ]$ with $s_0<s_1$. Then ${\gamma }(s)= {\gamma }(s_0)$ for $s\in [s_0,s_1]$. Let

$$\begin{aligned} \eta (s)={\gamma }(s) \text{ if } 0\leqq s\leqq s_0, \text{ and } \eta (s)={\gamma }(s+(s_1-s_0)) \text{ if } \sigma -(s_1-s_0) \geqq s\geqq s_0. \end{aligned}$$

Obviously, $\eta \in {\mathcal{A}\mathcal{C}}(x,y;[0,{\sigma }-(s_1-s_0)];\widetilde{\Omega })$ and $A(\eta )<A({\gamma })$, which is a contradiction.

Next define

$$\begin{aligned} \phi (s):=\int ^s_0\chi _{[0,\sigma ]{\setminus } E}(\delta )\,d\delta . \end{aligned}$$

It is obvious that $\phi$ is Lipschitz, and hence absolutely continuous, $\phi '(s)=\chi _{[0,\sigma ]{\setminus } E}(s)$ for almost all $s\in [0,{\sigma }]$. Moreover, $|\phi (E)|=0$ and $\phi ([0,\sigma ])=[0,\sigma -|E|]$. Since E does not contain any interval, we know that $\phi$ is strictly increasing in $[0,\sigma ]$, and hence injective. Thus the inverse $\phi ^{-1}:[0,\sigma -|E|]\rightarrow [0,\sigma ]$ is well-defined.

For any $g\in L^1([0,\sigma ])$, we claim that

$$\begin{aligned} g \circ \phi ^{-1} \in L^1 ([0,\sigma -|E|]) \text{ and } \int _0^\sigma g\chi _{[0,\sigma ]{\setminus } E} \,ds = \int _0^{\sigma -|E|}g \circ \phi ^{-1}\,ds \end{aligned}$$

(A.3)

Indeed, let $\widetilde{\phi }(t)=\int _0^t\chi _{E^\complement }$ for $t\in {{\mathbb {R}}}$. Then $\widetilde{\phi }$ is Lipschitz and strictly increasing in whole ${{\mathbb {R}}}$, $\widetilde{\phi }|_{[0,\sigma ]}=\phi$ and $(\widetilde{\phi })^{-1}|_{[0,\sigma -|E|]}=\phi ^{-1}$. Applying the change of variable formula in Lemma A.6(i) to $g\chi _{[0,\sigma ]}$ and $\widetilde{\phi }$, noting that $\phi [0,\sigma ]=[0,\sigma -|E|]$ and $\phi '=\chi _{[0,\sigma ]{\setminus } E}$ almost everywhere, one has

$$\begin{aligned}&\int _0^\sigma g\chi _{[0,\sigma ]{\setminus } E} \,ds= \int _{{\mathbb {R}}}(g\chi _{[0,\sigma ] }) |(\widetilde{\phi })'|\,ds\\ {}&= \int _{{\mathbb {R}}}(g\chi _{[0,\sigma ] })\circ (\widetilde{\phi })^{-1}\,ds= \int _0^{\sigma -|E|}g \circ \phi ^{-1}\,ds \end{aligned}$$

as desired.

Write $\eta (t)=\gamma (\phi ^{-1}(t))$ for $t\in [0,\sigma -|E|]$. Since ${\gamma }$ is absolutely continuous (hence ${\dot{{\gamma }}}\in L^1([0,\sigma ])$) and ${\dot{{\gamma }}}= 0$ in E, for any $t\in [0,\sigma -|E|]$ we have

$$\begin{aligned} \eta (t )-\eta (0)&= \gamma (\phi ^{-1}(t))-\gamma (0)= \int _{0}^{\phi ^{-1}(t)}{\dot{{\gamma }}}(s)\,ds\\ {}&= \int _{0}^{\sigma } ({\dot{{\gamma }}} \chi _{[0,\phi ^{-1}(t)]}\chi _{[0,\sigma ]\setminus E})(s) \,ds. \end{aligned}$$

Applying (A.3) to ${\dot{{\gamma }}} \chi _{[0,\phi ^{-1}(t)]}$ for all $t\in [0,\sigma -|E|]$, one has ${\dot{{\gamma }}} \circ \phi ^{-1} \in L^1 ([0,\sigma -|E|])$ and

$$\begin{aligned} \eta (t )-\eta (0) = \int _{0}^{\sigma -|E|} ({\dot{{\gamma }}} \chi _{[0,\phi ^{-1}(t)]})\circ \phi ^{-1}(s) \,ds. \end{aligned}$$

Thanks to $\chi _{[0,\phi ^{-1}(t)]}\circ \phi ^{-1}=\chi _{[0, t]}$, we obtain

$$\begin{aligned} \eta (t )-\eta (0) =\int _{0}^{\sigma -|E|}{\dot{{\gamma }}} \circ \phi ^{-1}(\delta ) \chi _{[0,t]} (\delta ) \,d\delta =\int _0^t {\dot{{\gamma }}} \circ \phi ^{-1}(\delta ) \,d\delta . \end{aligned}$$

Thus $\eta \in {\mathcal{A}\mathcal{C}}(x,y;[0,\sigma -|E|],\widetilde{\Omega })$ and hence differentiable almost everywhere in $[0,\sigma -|E|]$ with

$$\begin{aligned} {\dot{\eta }} = {\dot{{\gamma }}} \circ \phi ^{-1} \in L^1([0,\sigma -|E|]). \end{aligned}$$

(A.4)

By (A.4) and $|{\dot{{\gamma }}}|^2\in L^1([0,\sigma ])$, applying (A.3) to $|{\dot{{\gamma }}}|^2$ and $F\circ {\gamma }\in L^1([0,\sigma ])$ one has

$$\begin{aligned} A_\lambda (\eta )&= \int _0^{\sigma -|E|}\left[ \frac{1}{2}| {\dot{\eta }}(t)|^2+F\circ \eta (t) +\lambda \right] \,dt \\&= \int _0^{\sigma -|E|}\left[ \frac{1}{2}|{\dot{{\gamma }}}\circ \phi ^{-1} (t)|^2+(F\circ \gamma )\circ \phi ^{-1} (t) +\lambda \right] \,dt \\&= \int _0^{\sigma }\left[ \frac{1}{2}| {\dot{{\gamma }}} (t)|^2 \chi _{[0,\sigma ]\setminus E} (t)+(F\circ \gamma ) (t) \chi _{[0,\sigma ]\setminus E} (t) \right] \,dt + \lambda (\sigma -|E|) \end{aligned}$$

Since $|[0,\sigma ]\setminus E|=\sigma -|E|$, one has

$$\begin{aligned} A_\lambda (\eta )&= \int _0^{\sigma }[ \frac{1}{2}|{\dot{{\gamma }}} (s)|^2+F\circ {\gamma }(s) +\lambda ]\chi _{[0,\sigma ]\setminus E}(s)\,ds <A_\lambda ({\gamma }). \end{aligned}$$

Hence $A_\lambda (\eta ) < m_\lambda (x,y)$, which contradicts to the definition of $m_\lambda (x,y)$. $\square$

Lemma A.7

Let ${\gamma }\in {\mathcal{A}\mathcal{C}}(x,y;[0,\sigma ];{{\mathbb {R}}}^{dN})$ with $\widetilde{A}_\lambda ({\gamma })<\infty$. We can reparameterize ${\gamma }$ to get a new curve $\xi \in {\mathcal{A}\mathcal{C}}(x,y;[0,\tau ];{{\mathbb {R}}}^{dN})$ so that $|{\dot{\xi }}|>0$ almost everywhere and $\widetilde{A}_\lambda (\xi )=\widetilde{A}_\lambda ({\gamma })$.

Proof

The proof is much similar to that of Lemma A.5. $\square$

Lemma A.8

Let ${\gamma }\in {\mathcal{A}\mathcal{C}}(x,y;[0,\sigma ]; {{\mathbb {R}}}^{dN})$ with $\widetilde{A}_\lambda ({\gamma })<\infty$, and $|{\dot{{\gamma }}}|>0$ almost everywhere. Then ${\gamma }\in {\mathcal{A}\mathcal{C}}(x,y;[0,\sigma ]; \widetilde{\Omega })$, and we can reparameterize ${\gamma }$ to get a new curve $\eta \in {\mathcal{A}\mathcal{C}}(x,y;[0,\tau ];\widetilde{\Omega })$ such that

$$\begin{aligned} |{\dot{\eta }}(t)|=\sqrt{2(F\circ \eta (t) +\lambda )}\quad \text{ almost } \text{ everywhere. } \end{aligned}$$

Moreover,

$$\begin{aligned} A_\lambda (\eta )=\widetilde{A}_\lambda (\eta )=\widetilde{A}_\lambda ({\gamma })\leqq A_\lambda ({\gamma }). \end{aligned}$$

Proof

Write

$$\begin{aligned} \psi (s)=\int _0^s\frac{|{\dot{{\gamma }}}(\delta )|}{\sqrt{2 F\circ {\gamma }(\delta )+2\lambda }}\,d\delta \quad \forall s\in [0,\sigma ]. \end{aligned}$$

Note that $\psi (0)=0$ and $\psi (\sigma )<\infty$. Obviously, $\psi$ is absolutely continuous.

$$\begin{aligned} \psi '(s)= \frac{|{\dot{{\gamma }}}(s)|}{\sqrt{2 F\circ {\gamma }(s)+2\lambda }}\quad \text{ for } \text{ almost } \text{ all } s\in [0,\sigma ]. \end{aligned}$$

(A.5)

Note that $|{\dot{{\gamma }}}(s)|\sqrt{2 F\circ {\gamma }(s)+2\lambda } \in L^1([0,\sigma ])$ and $|{\dot{{\gamma }}}|>0$ almost everywhere, hence $F\circ {\gamma }<\infty$ almost everywhere. Since $|{\dot{{\gamma }}}|>0$ almost everywhere, we have $\psi ' >0$ almost everywhere. Thus $\psi$ is continuous and strictly increasing. Therefore $\psi ([0,\sigma ])=[0,\psi (\sigma )]$, and $\psi ^{-1}:[0,\psi (\sigma )]\rightarrow [0,\sigma ]$ is also continuous, strictly increasing.

Next we show that $\psi ^{-1}$ is absolutely continuous, that is, $|\psi ^{-1}(E)|=0$ whenever $E \subset [0,\psi (\sigma )]$ has measure $|E|=0$. We only need to prove that for any set $E\subset [0,\sigma ]$, if $\psi (E)$ has measure 0, then E has measure 0. Indeed, $\psi (E)$ must be contained in a $G_\delta$-set H which has measure 0, and hence $E\subset \psi ^{-1}(H)$. So if $\psi ^{-1}(H)$ has measure 0, then E has measure 0. Since H is a $G_\delta$-set, we know that $\psi ^{-1}(H)$ is also a $G_\delta$-set, one can see that

$$\begin{aligned} 0=|H|=\int _{\psi ^{-1}(H)}\psi '(s)\,ds. \end{aligned}$$

Since $\psi '>0$ almost everywhere, we have $|\psi ^{-1}(H)|=0$ as desired.

Let $\eta (t)={\gamma }(\psi ^{-1}(t))$ for $t\in [0,\psi (\sigma )]$. Thus, for any $t\in [0,\psi (\sigma )]$, one has

$$\begin{aligned} \eta (t)-\eta (0)=\int _0^{\psi ^{-1}(t)}{\dot{{\gamma }}}(\delta )\,d\delta =\int _0^\sigma ({\dot{{\gamma }}}\chi _{[0,\psi ^{-1}(t)]}) (\delta )\,d\delta . \end{aligned}$$

Applying the change of variable formula given in Lemma A.6(ii) to ${\dot{{\gamma }}}\chi _{[0,\psi ^{-1}]}\in L^1([0,\sigma ])$ and $\psi ^{-1}$, noting $\chi _{[0,\psi ^{-1}(t)]}\circ \psi ^{-1} =\chi _{[0,t]}$, one has ${\dot{{\gamma }}} \circ \psi ^{-1} (\psi ^{-1})'\in L^1([0,t])$ and

$$\begin{aligned}{} & {} \eta (t)-\eta (0)=\int _0^{\psi (\sigma )}({\dot{{\gamma }}}\chi _{[0,\psi ^{-1}(t)]})(\psi ^{-1}(s))(\psi ^{-1})'(s) \,ds\\ {}{} & {} =\int _0^t {\dot{{\gamma }}} (\psi ^{-1}(s))(\psi ^{-1})'(s) \,ds. \end{aligned}$$

Thus $\eta \in \mathcal{A}\mathcal{C}(x,y;[0,\psi (\sigma )];\widetilde{\Omega })$ with ${\dot{\eta }} ={\dot{{\gamma }}}\circ \psi ^{-1} (\psi ^{-1})'$ almost everywhere.

Since $t=\psi \circ \psi ^{-1}(t)$ for all $t\in [0,\psi (\sigma )]$, by the chain rule we have $1=\psi '\circ \psi ^{-1} (\psi ^{-1})'$ almost everywhere in $[0,\psi (\sigma )]$. Denote by E the set where $\psi$ is differentiable and $\psi '>0$. Since $|[0,\sigma ]{\setminus } E|=0$, by the absolute continuity of $\psi$, we have $|[0,\psi (\sigma )]\setminus \psi (E)|=|\psi ([0,\sigma ]\setminus E)|=0.$ Thus $\psi '\circ \psi ^{-1}>0$ in $\psi ( E)$, and hence, almost everywhere in $[0,\psi (\sigma )]$. Recalling (A.5), we obtain

$$\begin{aligned} (\psi ^{-1})'(t) =\frac{1}{\psi '\circ \psi ^{-1}(t)}= \frac{\sqrt{2 F\circ {\gamma }\circ \psi ^{-1}(t)+2\lambda }}{|{\dot{{\gamma }}}\circ \psi ^{-1}(t)|}\quad \text{ for } \text{ almost } \text{ all } t\in [0,\psi (\sigma )], \end{aligned}$$

and hence

$$\begin{aligned} |{\dot{\eta }}(t)| ={\sqrt{2 F\circ {\gamma }(\psi ^{-1}(t))+2\lambda }}={\sqrt{2 F\circ \eta ( t )+2\lambda }} \quad \text{ for } \text{ almost } \text{ all } t\in [0,\psi (\sigma )]. \end{aligned}$$

Thus

$$\begin{aligned}{}[\frac{1}{2}|{\dot{\eta }}(t)|^2+F\circ \eta (t)+\lambda ]= |{\dot{\eta }}(t)|\sqrt{2 F\circ \eta (t)+2\lambda } \quad \text{ for } \text{ almost } \text{ all } t\in [0,\psi (\sigma )]. \end{aligned}$$

We therefore obtain that

$$\begin{aligned} A_\lambda (\eta )=\widetilde{A}_\lambda (\eta )=\widetilde{A}_\lambda ({\gamma }) \end{aligned}$$

as desired. $\square$

We are now in a position to show

Proof of Lemma A.3

Note that for any ${\gamma }\in {\mathcal{A}\mathcal{C}}(x,y;[0,\sigma ];{{\mathbb {R}}}^{dN})$, by Cauchy-Schwartz inequality, one has

$$\begin{aligned} \frac{1}{2} |{\dot{{\gamma }}} (s)|^2+ F({\gamma }(s))+\lambda \geqq 2\sqrt{F({\gamma }(s))+\lambda } \sqrt{\frac{1}{2} |{\dot{{\gamma }}} (s)|^2} = \sqrt{2(F({\gamma }(s))+\lambda ) } |{\dot{{\gamma }}} (s) |, \end{aligned}$$

which gives $A_\lambda ({\gamma })\geqq \widetilde{A}_\lambda ({\gamma }).$ Thus $m_\lambda (x,y)\geqq \widetilde{m}_\lambda (x,y).$

We prove by contradiction that $|{\dot{{\gamma }}}(t)|=\sqrt{2(F\circ {\gamma }(t) +\lambda )} \,\, \text{ almost } \text{ everywhere. }$ Suppose that this is not correct. Write

$$\begin{aligned} E=\{ s\in [0,\tau ]: |{\dot{{\gamma }}} (s)|\ne \sqrt{2(F({\gamma }(s))+\lambda )}, F({\gamma }(s))<\infty \}. \end{aligned}$$

Then $|E|>0$. Moreover for $s\in E$, one has

$$\begin{aligned} \frac{1}{2} |{\dot{{\gamma }}} (s)|^2+ F({\gamma }(s))+\lambda > 2\sqrt{F({\gamma }(s))+\lambda } \sqrt{\frac{1}{2} |{\dot{{\gamma }}} (s)|^2} = \sqrt{2(F({\gamma }(s))+\lambda ) } |{\dot{{\gamma }}} (s) |. \end{aligned}$$

Thus

$$\begin{aligned} m_\lambda (x,y)=A_\lambda ({\gamma })>\widetilde{A}_\lambda ({\gamma }). \end{aligned}$$

Reparameterize ${\gamma }$ to get $\eta$ as in Lemma A.5 we know that

$$\begin{aligned} A_\lambda (\eta )=\widetilde{A}_\lambda (\eta ) =\widetilde{A}_\lambda ({\gamma }). \end{aligned}$$

Note that $m_\lambda (x,y)\leqq A_\lambda (\eta )$. This is a contradiction.

It is easy to see that $m_\lambda (x,y)\geqq \widetilde{m}_\lambda (x,y).$ We prove $\widetilde{m}_\lambda (x,y)= m_\lambda (x,y)$ by contradiction. Suppose this is not correct. Then $\widetilde{m}_\lambda (x,y)< m_\lambda (x,y)$. There exists a curve ${\gamma }$ such that

$$\begin{aligned}\widetilde{m}_\lambda (x,y)\leqq \widetilde{A}_\lambda ({\gamma })< m_\lambda (x,y). \end{aligned}$$

Reparameterizing ${\gamma }$ like in Lemma A.5, we find a curve $\eta$ such that

$$\begin{aligned} A_\lambda (\eta )=\widetilde{A}_\lambda (\eta )=\widetilde{A}_\lambda ({\gamma }). \end{aligned}$$

Since $m_\lambda (x,y)\leqq A_\lambda (\eta )$, this is a contradiction. $\square$

Proof of Lemma A.4

We only need to prove Lemma A.4 for $m_\lambda$-geodesics with canonical parameter. Assume that ${\gamma }\in \mathcal (x,y;[0,\sigma ];\widetilde{\Omega })$ is an $m_\lambda$-geodesic with canonical parameter joining x, y, and moreover ${\gamma }|_{(0,\sigma )}\subset \Omega$. Up to considering ${\gamma }_{[\delta ,\sigma -\delta ]} \subset \Omega$ for any sufficiently small $\delta >0$, we may assume that ${\gamma }\subset \Omega$. Recall that

$$\begin{aligned} m_\lambda (x,y)=A_\lambda ({\gamma })\leqq A_\lambda ({\gamma }+\varepsilon \xi ),\quad \forall \xi \in C^1(0,0;[0,\sigma ];{{\mathbb {R}}}^{dN}). \end{aligned}$$

Since ${\gamma }\subset \Omega$, there exist $\varepsilon _0$ depending on ${\gamma }$ and $\xi$ such that $A_\lambda ({\gamma }+\varepsilon \xi )<\infty$ for any $\varepsilon <\varepsilon _0$. Thus

$$\begin{aligned} 0=\frac{d}{d\varepsilon }|_{\varepsilon =0} A_\lambda ({\gamma }+\varepsilon \xi ), \end{aligned}$$

that is,

$$\begin{aligned} 0=\frac{d}{d\varepsilon }|_{\varepsilon =0} \int _0^\sigma ( \frac{1}{2} |({\dot{{\gamma }}} +\varepsilon {\dot{\xi }})(s)|^2+ F\circ ({\gamma }+\varepsilon \xi )(s) +\lambda ) \,ds. \end{aligned}$$

A direct calculation gives

$$\begin{aligned}{} & {} \frac{d}{d\varepsilon }|_{\varepsilon =0} [ \frac{1}{2} |({\dot{{\gamma }}} +\varepsilon {\dot{\xi }})(s)|^2+ F\circ ({\gamma }+\varepsilon \xi )(s) +\lambda ]\\ {}{} & {} = \langle \langle {\dot{{\gamma }}} (s), {\dot{\xi }}(s) \rangle \rangle + \langle \langle \nabla F({\gamma }(s)), \xi (s) \rangle \rangle . \end{aligned}$$

Therefore we obtain

$$\begin{aligned} \int _0^\sigma {[ \langle \langle {\dot{{\gamma }}} (s), {\dot{\xi }}(s) \rangle \rangle + \langle \langle \nabla F({\gamma }(s)), \xi (s)\rangle \rangle ] }\,ds=0, \quad \forall \xi \in C^1(0,0,[0,\sigma ];{{\mathbb {R}}}^{dN}). \end{aligned}$$

Note that ${\gamma }\subset \Omega$ is free of collision. Since $F \in C^2(\Omega )$, it follows from [19, Chapter 3] that ${\gamma }\in C^2$ , and hence,

$$\begin{aligned} \int _0^\tau {\langle \langle [-\ddot{\gamma }(s)+ \nabla F({\gamma }(s)) ], \xi (s)\rangle \rangle }\,ds=0. \end{aligned}$$

By the arbitrariness of $\xi$, we have $\ddot{\gamma }(s)=\nabla F({\gamma }(s))$ for all $s \in (0,\sigma )$ as desired. $\square$

Finally, we prove (A.2) in the Step 3 in Remark A.1(iii).

Proof of (A.2)

Let all notations and notions be as in Remark A.1(iii). In particular, recall

$$\begin{aligned}{} & {} \Sigma ^-_{12} =\{x=(x_1,\ldots ,x_N)\in {{\mathbb {R}}}^{dN}: x_1=x_2\in \Gamma \} \text{ and } \\{} & {} \Gamma = \{x_i \in {{\mathbb {R}}}^{d }: x_i^j < 0, \ \forall 1 \leqq j\leqq d \}. \end{aligned}$$

Let $x\in \Sigma _{12}^-$ be an arbitrary point. We have $x_2=x_1\in \Gamma$, that is, $x_2^j=x_1^j< 0$ for all $1\leqq j\leqq d$. Set

$$\begin{aligned} c_*(x):= \min \{1,|x_1^1|, \ldots ,|x_1^d| \}=\min \{1,-x_1^1, \ldots ,-x_1^d \}. \end{aligned}$$

Then $c_*(x)>0$. To get (A.2), that is, $\inf _{z\in \Omega } m_\lambda (z,x)>0$, obviously it suffices to prove

$$\begin{aligned} \inf _{z\in \Omega } m_\lambda (z,x) \geqq \sqrt{2\lambda } \frac{1}{2}c_*(x). \end{aligned}$$

(A.6)

To see (A.6), given any $z\in \Omega$, let ${\gamma }\in {\mathcal{A}\mathcal{C}}(z,x;[0,\sigma ];\widetilde{\Omega })$ for some $\sigma >0$ be an $m_\lambda$-geodesic with canonical parameter so that $A_\lambda ({\gamma })=m_\lambda (z,x)<\infty$; see Lemma A.2 for its existence. From Lemma A.2 and (A.1) it follows that

$$\begin{aligned} m_\lambda (z,x)= & {} A_\lambda ({\gamma }) \geqq A_\lambda ({\gamma }|_{[s,\sigma ]})= m_\lambda ({\gamma }(s),x)\nonumber \\\geqq & {} \sqrt{ 2\lambda } |{\gamma }(s)-x|\quad \forall s\in [0,\sigma ). \end{aligned}$$

(A.7)

We then claim that

$$\begin{aligned} \text{ there } \text{ exists } \text{ an } s\in [0,\sigma ) \text{ such } \text{ that }\ |{\gamma }(s)-x|\geqq \frac{1}{2}c_*(x). \end{aligned}$$

(A.8)

Obviously, from (A.7) and the claim (A.8) it follows that $m_\lambda (z,x) \geqq \frac{1}{2}\sqrt{ 2\lambda }c_*(x).$ Thus (A.6) holds as desired.

Below we prove the above claim (A.8) by contradiction. Assume that the claim (A.8) is not correct, in other words, assume that

$$\begin{aligned} |{\gamma }(s)-x|< \frac{1}{2} c_*(x)\quad \text{ for } \text{ all } s\in [0,\sigma ]. \end{aligned}$$

(A.9)

Write ${\gamma }=({\gamma }_1,\ldots ,{\gamma }_N)$ and ${\gamma }_i=({\gamma }^1_i,\ldots ,{\gamma }_i^d)$. Given any $i=1,2$ and $1\leqq j\leqq d$, by the hypothesis (A.9) one has

$$\begin{aligned}|{\gamma }_i^j(s)- x_i^j|\leqq |{\gamma }(s)-x|< \frac{1}{2} c_*(x) \quad \text{ and } \text{ hence } \quad {\gamma }_i^j(s)\leqq x_i^j+ \frac{1}{2} c_*(x). \end{aligned}$$

Noting $0<c_*(x)\leqq -x_i^j$, we get

$$\begin{aligned} {\gamma }_i^j(s)\leqq \frac{1}{2} x_i^j<0. \end{aligned}$$

Thus, $({\gamma }_1(s),{\gamma }_2(s))\in \Gamma \times \Gamma$ and hence, by the definition of F, we have

$$\begin{aligned} F\circ {\gamma }= h_-\circ (|{\gamma }_1-{\gamma }_2|). \end{aligned}$$

Define an auxiliary function

$$\begin{aligned} v(t)=|{\gamma }_1(t)-{\gamma }_2(t)| \text{ for } \text{ all } t\in [0,\sigma ]. \end{aligned}$$

Then $F\circ {\gamma }= h_-\circ v.$ Moreover, note that $v\in C^0([0,\sigma ])$. By $x_1=x_2$,

$$\begin{aligned} v(\sigma )=|{\gamma }_1(\sigma )-{\gamma }_2(\sigma )|=|x_1-x_2|=0; \end{aligned}$$

while by $z\in \Omega$, we have $z_1\ne z_2$ and hence

$$\begin{aligned} v(0)=|{\gamma }_1(0)-{\gamma }_2(0)|=|z_1-z_2|>0. \end{aligned}$$

Set

$$\begin{aligned} \kappa :=\min \{s\in [0,\sigma ] \ | \ v(s)=0\}. \end{aligned}$$

Then $0<\kappa \leqq \sigma$. Below we will prove that

$$\begin{aligned} \int _0^{\kappa } |{\dot{{\gamma }}} |\sqrt{ h_-\circ v }\,ds=\infty . \end{aligned}$$

(A.10)

Before we give the proof of (A.10), we show that (A.10) leads to a contradiction. Indeed, Lemma A.2 and Lemma A.3 yield

$$\begin{aligned} m_\lambda (z,x)\geqq m_\lambda (z,{\gamma }(\kappa ) )\geqq \widetilde{A}_\lambda ({\gamma }|_{[0,\kappa ]}). \end{aligned}$$

Write

$$\begin{aligned} \widetilde{A}_\lambda ({\gamma }|_{[0,\kappa ]})= & {} \int _0^{\kappa } |{\dot{{\gamma }}} |\sqrt{2 F\circ {\gamma }+2 \lambda }\,ds =\int _0^{\kappa } |{\dot{{\gamma }}} |\sqrt{2 h_-\circ v+2 \lambda }\,ds\\\geqq & {} \sqrt{2}\int _0^{\kappa } |{\dot{{\gamma }}} |\sqrt{ h_-\circ v }\,ds. \end{aligned}$$

We see that (A.10) implies $\widetilde{A}_\lambda ({\gamma }|_{[0,\kappa ]})=\infty$ and hence $m_\lambda (z,x)=\infty$. However, since $z\in \Omega$, we have $m_\lambda (z,x)<\infty$. This is a contradiction. Therefore, we conclude that the hypothesis (A.9) is not correct, and hence, the claim (A.8) must hold as desired.

Finally, we prove (A.10) via the following 4 substeps.

Substep 1. From the definition of $\kappa$ it follows that

$$\begin{aligned} v(\kappa )=0, v(t)>0 \text{ for } t\in [0,\kappa ), {\gamma }(\kappa )\in \Sigma ^{-}_{12} \text{ and } {\gamma }|_{[0,\kappa )}\subset {{\mathbb {R}}}^{dN}\setminus \Sigma ^{-}_{12}. \end{aligned}$$

Set

$$\begin{aligned} \tau =\sup _{s\in [0,\kappa ]} v(s). \end{aligned}$$

The choice of $\kappa$ and the continuity of v give that $v([0,\kappa ])=[0,\tau ]$. Moreover, we observe that $\tau \leqq c_*(x)\leqq 1$. Indeed, by $x_1=x_2$ and the triangle inequality, for any $s\in [0,\kappa ]$ we have

$$\begin{aligned} v(s)=|{\gamma }_1 (s)-{\gamma }_2 (s)|\leqq |{\gamma }_1 (s)-x_1 (s)| +|x_2 (s)-{\gamma }_2 (s)| \leqq 2|{\gamma }-x| \end{aligned}$$

and hence, by the hypothesis (A.9), $v(s)\leqq c_*(x)$.

For any $s\in [0,\kappa ]$, recall that

$$\begin{aligned} h_-\circ v(s) =\left\{ \begin{aligned}&1 \quad&\text{ if } v(s)\in \Lambda =\{0\}\cup (\cup _{k\geqq 3} [2^{-k^2-1}, 2^{-k^2+1}]); \\&\frac{1}{v(s)^2}\ {}&\text{ if } \quad v(s)\in [0,\tau ]\setminus \Lambda . \end{aligned}\right. \end{aligned}$$

Wrtie $E:=\{s\in [0,\kappa ]|v(s)\in [0,\tau ]\setminus \Lambda \}$.Then

$$\begin{aligned} h_-(v(s))=\frac{1}{v(s)^2}\quad \text{ for } s\in E \end{aligned}$$

and

$$\begin{aligned} \chi _E(s) = \chi _{[0,\tau ]\setminus \Lambda }(v(s))= \chi _{[0,\tau ]\setminus \Lambda }\circ v(s)\quad \forall s\in [0,\kappa ]. \end{aligned}$$

Thus

$$\begin{aligned} \int _0^{\kappa } |{\dot{{\gamma }}} |\sqrt{ h_-\circ v }\,ds\geqq & {} \int _0^\kappa \chi _E(s) |{\dot{{\gamma }}}(s) |\frac{1}{v(s)}\,ds\nonumber \\= & {} \int _0^\kappa \chi _{[0,\tau ]\setminus \Lambda }\circ v(s) |{\dot{{\gamma }}} (s)| \frac{1}{v(s)}\,ds. \end{aligned}$$

(A.11)

Substep 2. Since both of ${\gamma }_1$ and ${\gamma }_2$ are absolutely continuous in $[0,\kappa ]$, we know that v is also absolutely continuous in $[0,\kappa ]$, and hence $\dot{v}\in L^1([0,\kappa ])$. One further has

$$\begin{aligned} |\dot{v}(t)|= & {} \frac{|\langle ({\dot{{\gamma }}}_1-{\dot{{\gamma }}}_2)(t), {\gamma }_1(t)- {\gamma }_2 (t)\rangle |}{v(t)}\nonumber \\\leqq & {} |({\dot{{\gamma }}}_1-{\dot{{\gamma }}}_2)(t)|\leqq 2 |{\dot{{\gamma }}}(t)| \quad \text{ for } \text{ almost } \text{ all } t\in [0,\kappa ]. \end{aligned}$$

(A.12)

Applying this in (A.11), we obtain

$$\begin{aligned} \int _0^{\kappa } |{\dot{{\gamma }}} |\sqrt{ h_-\circ v }\,ds\geqq \frac{1}{2}\int _0^\kappa \chi _{[0,\tau ]\setminus \Lambda }\circ v(s) |\dot{v} (s)| \frac{1}{v(s)}\,ds. \end{aligned}$$

(A.13)

Substep 3. We claim that

$$\begin{aligned} \int _{0}^{\kappa } \frac{1}{v }|\dot{v} | \chi _{[0,\tau ]\setminus \Lambda }\circ v \,dt&\geqq \int _{[0,\tau ]\setminus {\Lambda }} \frac{1}{t}\,dt, \end{aligned}$$

(A.14)

whose proof will be given in Substep 4. Assume that the claim (A.14) holds for the moment. By (A.13) and the claim (A.14), we obtain

$$\begin{aligned}\int _0^{\kappa } |{\dot{{\gamma }}} |\sqrt{ h_-\circ v }\,ds> \int _{[0,\tau ]\setminus {\Lambda }} \frac{1}{t}\,dt. \end{aligned}$$

Let $k_\tau \geqq 3$ such that $2^{-k_\tau ^2+1}\leqq \tau$. Then

$$\begin{aligned} \int _{[0,\tau ]\setminus {\Lambda }} \frac{1}{t}\,dt\geqq & {} \sum _{k\geqq k_\tau } \int _{2^{-k^2+1}}^{2^{-(k-1)^2-1}}\frac{1}{t}\,dt= \sum _{k\geqq k_\tau } \ln \frac{2^{-(k-1)^2-1}}{2^{-k^2+1}}\\\geqq & {} \sum _{k\geqq k_\tau } \ln 2^{2k-3} \geqq (\ln 2 )\sum _{k\geqq k_\tau } k=\infty . \end{aligned}$$

Thus

$$\begin{aligned} \int _0^{\kappa } |{\dot{{\gamma }}} |\sqrt{ h_-\circ v }\,ds=\infty , \end{aligned}$$

which gives (A.10).

Substep 4. We prove the claim (A.14) via Lemma A.6(i). Since v is only defined in $[0,\sigma ]$ and also not necessarily Lipschitz in $[0,\kappa ]$, we can not use Lemma A.6(i) directly.

To overcome this difficulty, we show that the restriction of v in subintervals $[0,{\kappa }-\varepsilon ]$ is Lipschitz for all sufficiently small $\varepsilon >0$, and extend them to ${{\mathbb {R}}}$ via the McShane’s extension. To be precise, by Lemma A.3 and $F\circ {\gamma }=h_-\circ v$, one has

$$\begin{aligned} |{\dot{{\gamma }}} |= \sqrt{2(F\circ {\gamma }+\lambda )} = \sqrt{2(h_-\circ v+\lambda )} \quad \text { almost everywhere in } [0,\kappa ]. \end{aligned}$$

Recall that $0<v \leqq 1$ in $[0,\kappa )$, $\lambda >0$, and $h_-(t)\leqq \frac{1}{t^2}$ by definition, we have

$$\begin{aligned} h_-\circ v+\lambda \leqq (1+\lambda ) \frac{1}{v^2}\quad \text{ in } [0,\kappa ]. \end{aligned}$$

Thus, by (A.12),

$$\begin{aligned} |\dot{v} | \leqq 2|{\dot{{\gamma }}}|\leqq 2\sqrt{2(1+\lambda ) }\frac{1}{v } \quad \text{ almost } \text{ everywhere } \text{ in } [0, \kappa ]. \end{aligned}$$

For $0<\varepsilon <{\kappa }$, set

$$\begin{aligned} \delta _\varepsilon :=\min \{v(t),t\in [0,{\kappa } -\varepsilon ]\}. \end{aligned}$$

Since v is continuous in $[0,{\kappa } ]$ and $v>0$ in $[0,\kappa )$, we know that $\delta _\varepsilon >0$. Thus

$$\begin{aligned}|\dot{v}|\leqq 2\sqrt{2 (1+\lambda ) } \frac{1}{\delta _\varepsilon } \text{ almost } \text{ everywhere } \text{ in } [0,\kappa -\varepsilon ]. \end{aligned}$$

From this and the absolute continuity of v in $[0,\kappa ]$, we know that v is Lipschitz in $[0,\kappa -\varepsilon ]$ with

$$\begin{aligned} |v(s)-v(t)|=|\int _s^t\dot{v}(\xi )\,d\xi |\leqq 2\sqrt{2 (1+\lambda ) } \frac{1}{\delta _\varepsilon }|s-t|,\quad \forall s,t\in [0,\kappa -\varepsilon ]. \end{aligned}$$

Denote by $\widetilde{v^\varepsilon }$ the McShane’s extension of $v|_{[0,\kappa -\varepsilon ]}$ into ${{\mathbb {R}}}$, that is,

$$\begin{aligned} \widetilde{v^\varepsilon }(t)=\inf \left\{ v (s)+2\sqrt{2 (1+\lambda ) } \frac{1}{\delta _\varepsilon }|t-s| \ \bigg | \ s\in [0,\kappa -\varepsilon ] \right\} , \quad \forall t\in {{\mathbb {R}}}.\qquad \end{aligned}$$

(A.15)

Then

$$\begin{aligned} |\widetilde{v^\varepsilon }(s)-\widetilde{v^\varepsilon }(t)|\leqq 2\sqrt{2 (1+\lambda ) } \frac{1}{\delta _\varepsilon }|s-t|,\quad \forall s,t\in {{\mathbb {R}}}, \end{aligned}$$

that is, $\widetilde{v^\varepsilon }$ is Lipschitz in ${{\mathbb {R}}}$, and moreover, $\widetilde{v^\varepsilon }|_{[0,\kappa -\varepsilon ]}=v|_{[0,\kappa -\varepsilon ]}$; see for example [23,Chapter 6].

Moreover, write

$$\begin{aligned}\int _{0}^{\kappa } \frac{1}{v }|\dot{v} | \chi _{[0,\tau ]\setminus \Lambda }\circ v \,dt= \lim _{\varepsilon \rightarrow 0} \int _{0}^{{\kappa } -\varepsilon } \frac{1}{v }|\dot{v} |\chi _{[0,\tau ]\setminus {\Lambda }}\circ v \,ds. \end{aligned}$$

Since v is continuous in $[0,{\kappa } ]$, $v>0$ in $[0,\sigma )$ and $\varphi ({\kappa } )=0$, we know that $\delta _\varepsilon \rightarrow 0$ as $\varepsilon \rightarrow 0$. When $\varepsilon >0$ is sufficiently small, one also has $v([0,\kappa -\varepsilon ])=[\delta _\varepsilon ,\tau ]$, that is,

$$\begin{aligned} \chi _{[0,\tau ]\setminus {\Lambda }}\circ v(s)= \chi _{[\delta _\varepsilon ,\tau ]\setminus {\Lambda }}\circ v (s)\quad \quad \text{ for } \text{ any } s\in [0,{\kappa } -\varepsilon ]. \end{aligned}$$

Therefore, for all sufficiently small $\varepsilon >0$, one has

$$\begin{aligned} \int _{0}^{{\kappa } -\varepsilon } \frac{1}{v }|\dot{v} |\chi _{[0,\tau ]\setminus {\Lambda }}\circ v \,ds=\int _{0}^{{\kappa } -\varepsilon } \frac{1}{v }|\dot{v} |\chi _{[\delta _\varepsilon ,\tau ]\setminus {\Lambda }}\circ v \,ds. \end{aligned}$$

Denote by $\widetilde{v}^\varepsilon$ the McShane’s extension of $v|_{[0,\kappa -\varepsilon ]}$ into ${{\mathbb {R}}}$ as in (A.15). Since $\widetilde{v^\varepsilon }|_{[0,\kappa -\varepsilon ]}=v|_{[0,\kappa -\varepsilon ]}$, it follows that

$$\begin{aligned} \int _{0}^{{\kappa } -\varepsilon } \frac{1}{v }|\dot{v} |\chi _{[0,\tau ]\setminus {\Lambda }}\circ v \,ds= \int _{{\mathbb {R}}}\chi _{[0,\kappa -\varepsilon ]}(s)\chi _{[\delta _\varepsilon ,\tau ]\setminus {\Lambda }}\circ \widetilde{v^\varepsilon } (s) \frac{1}{ \widetilde{v^\varepsilon } (s)}|(\widetilde{v^\varepsilon })'(s)|\,ds. \end{aligned}$$

Since $\widetilde{v^\varepsilon }$ is Lipschitz in ${{\mathbb {R}}}$ and $\chi _{[0,\kappa -\varepsilon ]}\frac{1}{ \widetilde{v^\varepsilon } }\chi _{[\delta _\varepsilon ,\tau ]{\setminus } {\Lambda }}\circ \widetilde{v^\varepsilon } \in L^1({{\mathbb {R}}})$, we are able to apply the change of variable formula in Lemma A.6(i) with $f=\widetilde{v^\varepsilon }$ in ${{\mathbb {R}}}$ and $g=\chi _{[0,\kappa -\varepsilon ]}\frac{1}{ \widetilde{v^\varepsilon } }\chi _{[\delta _\varepsilon ,\tau ]{\setminus } {\Lambda }}\circ \widetilde{v^\varepsilon }$ therein to get

$$\begin{aligned} \int _{0}^{{\kappa } -\varepsilon } \frac{1}{v }|\dot{v} |\chi _{[0,\tau ]\setminus {\Lambda }}\circ v \,ds=\int _{{{\mathbb {R}}}}\sum _{s\in (\widetilde{v^\varepsilon })^{-1}(\{ t\})} \left[ \chi _{[0,\kappa -\varepsilon ]}(s)\chi _{[\delta _\varepsilon ,\tau ]\setminus {\Lambda }}\circ \widetilde{v^\varepsilon } (s) \frac{1}{\widetilde{v^\varepsilon }(s)}\right] \,dt. \end{aligned}$$

(A.16)

For any sufficiently small $\varepsilon >0$, since $v([0,\kappa -\varepsilon ])=[\delta _\varepsilon ,\tau ]$, given any $t\in [\delta _\varepsilon ,\tau ]{\setminus } \Lambda$, one can find at least one $s\in [0,\kappa -\varepsilon ]$ such that $\widetilde{v}^\varepsilon (s)= v(s)=t$, and hence,

$$\begin{aligned} \sum _{s\in (\widetilde{v^\varepsilon })^{-1}(\{ t\})} \left[ \chi _{[0,\kappa -\varepsilon ]}(s)\chi _{[\delta _\varepsilon ,\tau ]\setminus {\Lambda }}\circ \widetilde{v^\varepsilon }(s) \frac{1}{ \widetilde{v^\varepsilon }(s)}\right] \geqq \frac{1}{t}. \end{aligned}$$

From this and (A.16) one deduces that

$$\begin{aligned} \int _{0}^{{\kappa } -\varepsilon } \frac{1}{v }|\dot{v} |\chi _{[0,\tau ]\setminus {\Lambda }}\circ v \,ds \geqq \int _{[\delta _\varepsilon ,\tau ]\setminus \Lambda }\frac{1}{t}\,dt. \end{aligned}$$

Sending $\varepsilon \rightarrow 0$, by $\delta _\varepsilon \rightarrow 0$ one gets

$$\begin{aligned}\int _{0}^{\kappa } \frac{1}{v }|\dot{v} | \chi _{[0,\tau ]\setminus \Lambda }\circ v \,dt&\geqq \int _{[0,\tau ]\setminus {\Lambda }} \frac{1}{t}\,dt, \end{aligned}$$

which gives the claim (A.14) as desired. The proof is complete. $\square$