$\newcommand{\M}{\mathcal{M}}$ Consider a Riemannian manifold $\M$ and a $p$ point on it. Is the following claim true?
$$ \inf_{r \geq 0} \sup_{u,v \in \mathbb{B}(p,r)} \frac{\|\exp_p^{-1}u - \exp_p^{-1}v\|_p} {d(u,v)}=1. $$
$\newcommand{\M}{\mathcal{M}}$ Consider a Riemannian manifold $\M$ and a $p$ point on it. Is the following claim true?
$$ \inf_{r \geq 0} \sup_{u,v \in \mathbb{B}(p,r)} \frac{\|\exp_p^{-1}u - \exp_p^{-1}v\|_p} {d(u,v)}=1. $$
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Differential $(d\exp_p)_o$ at origin is isometry so that the result is followed (cf. 2.9 Proposition in Do Carmo's book : Riemannian geometry).
$f={\rm exp}_p$ is bijective on $B_r(o)$ And since $df_o$ is isometry, then $$ {\rm dil}_o\ f= \sup_v\ \lim_{t>0\rightarrow o}\ \frac{ d( f(tv),f(o)) }{|tv-o|} =1 $$ where $|v|=1$.
For any $u\in B_r(o) $, ${\rm dil}_u\ f<1+g(r),\ g(r)>0,\ \lim_r\ g(r)=0$
(1) If $u,\ v\in B_r(o)$, then $c(t)=f(tu +(1-t)v)$ Then $$ d(f(u),f(v)) \leq {\rm length}\ c \leq |u-v| {\rm sup}_u\ {\rm dil}_u\ f < |u-v|(1+g(r))$$
(2) Do the same thing for $f^{-1}$.