Radial geodesics minimize length in a normal neighborhood

Write $R=\mathrm{\mid }v{\mathrm{\mid }}_{g_p}$. By radial norm preservation in [F1], $\mathrm{\mid }{\dot{\gamma }}_v(t){\mathrm{\mid }}_g=R$, so $L({\gamma }_v)={\int }_0^{1}Rdt=R$.

Let $c:[a,b]\to U$ be piecewise smooth with $c(a)=p$, $c(b)={\exp }_p(v)$, and put $r(t)=\mathrm{\mid }{\exp }_p^{-1}(c(t)){\mathrm{\mid }}_{g_p}$. Assume first $R>0$ and fix $0<\delta <R$. By [F2], the set $K_{\delta }=\{t:r(t)=\delta \}$ is nonempty; it is closed in the compact interval and hence compact, so [F2] gives its greatest element ${\tau }_{\delta }$. Then $r(t)>\delta$ for $t>{\tau }_{\delta }$: otherwise continuity and $r(b)=R>\delta$ would produce a later point of $K_{\delta }$. Thus $c$ avoids $p$ on $[{\tau }_{\delta },b]$.

On every smooth piece of $[{\tau }_{\delta },b]$, [F1] gives $\mathrm{\mid }{c}^{\mathrm{\prime }}{\mathrm{\mid }}_g\ge \mathrm{\mid }{r}^{\mathrm{\prime }}\mathrm{\mid }$. Summing the monotone integral inequalities and using [F3] gives $L(c)\ge L(c{\mathrm{\mid }}_{[{\tau }_{\delta },b]})\ge {\int }_{{\tau }_{\delta }}^b\mathrm{\mid }{r}^{\mathrm{\prime }}\mathrm{\mid }dt\ge R-\delta \mathrm{.}$ Since this holds for every $0<\delta <R$, $L(c)\ge R$. If $R=0$, nonnegativity already gives $L(c)\ge 0=L({\gamma }_v)$. This proves minimality without differentiating $r$ at a visit to $p$.

Conversely, for a curve of the displayed form, [F1] gives $\mathrm{\mid }{c}^{\mathrm{\prime }}{\mathrm{\mid }}_g=\mathrm{\mid }{r}^{\mathrm{\prime }}\mathrm{\mid }=r^{\mathrm{\prime }}$ on each smooth piece. Newton--Leibniz and finite additivity in [F3] give $L(c)=R$, including any constant pauses. If $R=0$, equality means $L(c)=0$; [F3] forces the continuous speed to vanish on each smooth piece, so $c$ is constant.

The same argument proves the auxiliary estimate $L(d)\ge \mathrm{\mid }r(d_1)-r(d_0)\mathrm{\mid }$ for any piecewise smooth $d:[d_0,d_1]\to U$: if $d$ avoids $p$, integrate the polar inequality directly; if it meets $p$, split there and apply step 2.1 to the reversed first part and the second part.

Suppose now $R>0$ and $L(c)=R$. For any $t$, step 2.1 on the prefix and step 3.1 on the suffix give $R=L(c{\mathrm{\mid }}_{[a,t]})+L(c{\mathrm{\mid }}_{[t,b]})\ge r(t)+\mathrm{\mid }R-r(t)\mathrm{\mid }\mathrm{.}$ Hence $r(t)\le R$, and equality forces the two lower bounds to be equal. If $a<u<w<b$ and $c$ avoids $p$ on $[u,w]$, applying the prefix, middle, and suffix bounds gives $R\ge r(u)+\mathrm{\mid }r(w)-r(u)\mathrm{\mid }+R-r(w)\mathrm{.}$ Therefore $r(w)\ge r(u)$ and equality holds in the middle polar length estimate.

On a closed subinterval of one smooth piece on which $c\mathrm{\ne }p$, step 4.1 and Newton--Leibniz give zero integral for the continuous nonnegative function $\mathrm{\mid }{c}^{\mathrm{\prime }}{\mathrm{\mid }}_g-r^{\mathrm{\prime }}$. By [F3] it vanishes identically. The polar identity in [F1] then gives $c_{\perp }^{\mathrm{\prime }}=0$ and $r^{\mathrm{\prime }}\ge 0$. Writing ${\exp }_p^{-1}(c)=r\theta$ with $\mathrm{\mid }\theta \mathrm{\mid }=1$, invertibility of $d{\exp }_p$ gives $r{\theta }^{\mathrm{\prime }}=0$; hence [F3] makes $\theta$ constant on every nonzero component. A component that later returned to $p$ would have positive nondecreasing $r$ tending to zero, which is impossible. Thus $c$ is constant at $p$ until its final nonzero component, and there $\theta =v\mathrm{/}R$ and $r$ is nondecreasing. This proves the stated necessary form.

Steps 2.1, 5.1, and 2.2 prove minimality and both equality directions. In dimension zero only $v=0$ occurs; dimension one permits the two radial directions and the proof fixes the one containing $v$. Empty $M$ has no centre. The open-ball condition excludes $\mathrm{\mid }v\mathrm{\mid }=\rho$, while $v=0$, visits to the centre, subdivision endpoints, and constant pauses were treated explicitly. Assumption [A1] is used exactly through [F1] for the exponential and polar structures; last hitting times are unique maxima, and no further choice is made.