H-one local length comparison for a conjugate-free geodesic

The square-integrable inverse-branch pullback. [F1, F2, given] Take the radius, partition and branches of [F2]. Their domain and gluing arguments use the continuity of $\alpha$, which follows from [F1]; no piecewise-$C^1$ hypothesis is used in that part. On each closed strip, ${\phi }_i={\beta }_i\circ \alpha$ is a smooth composition of an $H^1$ curve and hence a vector-valued primitive with $L^2$ derivative by [F1]. The finitely many ${\phi }_i$ agree at strip boundaries by [F2], and so glue to one $H^1$ vector primitive $\phi$ with $\phi (a)=0$, $\phi (b)=X$ and $\alpha ={\exp }_p\circ \phi$. The a.e. chain rule gives $\dot{\alpha }=d({\exp }_p{)}_{\phi }{\phi }^{\mathrm{\prime }}$ on each strip.

Radial comparison on every subinterval. [F1, F3, step 1.1, given] For $\sigma >0$ set $f_{\sigma }(t)=\sqrt{\mathrm{\mid }\phi (t){\mathrm{\mid }}_g^2+{\sigma }^2}$. This is a primitive with $L^2$ derivative $f_{\sigma }^{\mathrm{\prime }}=⟨\phi ,{\phi }^{\mathrm{\prime }}⟩\mathrm{/}{f}_{\sigma }$ a.e. by [F1]. At a point where $\phi \mathrm{\ne }0$, decompose ${\phi }^{\mathrm{\prime }}=c\phi +w$ with $w\perp \phi$. Gauss lemma [F3] and step 1.1 give $\mathrm{\mid }\dot{\alpha }{\mathrm{\mid }}_g^2=c^2\mathrm{\mid }\phi {\mathrm{\mid }}_g^2+\mathrm{\mid }d({\exp }_p{)}_{\phi }w{\mathrm{\mid }}_g^2\ge \mathrm{\mid }{f}_{\sigma }^{\mathrm{\prime }}{\mathrm{\mid }}^2\mathrm{.}$ At $\phi =0$, $f_{\sigma }^{\mathrm{\prime }}=0$, so the inequality remains true. Thus on every $[u,z]\subset [a,b]$, integration of the nonnegative speed and the primitive identity on the finite strips yield $L_g(\alpha {\mathrm{\mid }}_{[u,z]})\ge {\int }_u^z\mathrm{\mid }{f}_{\sigma }^{\mathrm{\prime }}\mathrm{\mid }dt\ge \mathrm{\mid }{f}_{\sigma }(z)-f_{\sigma }(u)\mathrm{\mid }\mathrm{.}$ Letting $\sigma \downarrow 0$ proves the subinterval bound $L_g(\alpha {\mathrm{\mid }}_{[u,z]})\ge \mathrm{\mid }\mathrm{\mid }\phi (z){\mathrm{\mid }}_g-\mathrm{\mid }\phi (u){\mathrm{\mid }}_g\mathrm{\mid }$ and, at the endpoints, $L_g(\alpha )\ge \mathrm{\mid }X{\mathrm{\mid }}_g=L_g(\gamma )$ by [F3].

Equality makes the radius monotone. [F1, step 1.1, step 2.1, given] Suppose $L_g(\alpha )=\mathrm{\mid }X{\mathrm{\mid }}_g$ and put $r=\mathrm{\mid }\phi {\mathrm{\mid }}_g$. Because length is an integral, it is additive across any finite subdivision. If $r(s)>r(z)$ for some $s<z$, first $r(z)\le \mathrm{\mid }X{\mathrm{\mid }}_g$: otherwise the subinterval bounds on $[a,z]$ and $[z,b]$ give $L_g(\alpha )\ge r(z)+(r(z)-\mathrm{\mid }X{\mathrm{\mid }}_g)>\mathrm{\mid }X{\mathrm{\mid }}_g$. Then step 2.1 on $[a,s]$, $[s,z]$, and $[z,b]$ gives $L_g(\alpha )\ge r(s)+(r(s)-r(z))+(\mathrm{\mid }X{\mathrm{\mid }}_g-r(z))>\mathrm{\mid }X{\mathrm{\mid }}_g$, a contradiction. Hence $r$ is nondecreasing. If $X=0$, then the same bound gives $r=0$; step 1.1 gives $\alpha ={\exp }_p(0)=\gamma$. If $X\mathrm{\ne }0$, continuity and monotonicity imply $\{r>0\}=(c,b]$ for some $c\in [a,b)$, with $r=0$ before $c$.

The radial primitive and a.e. equality. [F1, F3, F4, step 1.1, step 3.1, given] The function $r$ is itself a primitive. Indeed $f_{\sigma }\to r$ uniformly, while $f_{\sigma }^{\mathrm{\prime }}$ is dominated by $\mathrm{\mid }{\phi }^{\mathrm{\prime }}\mathrm{\mid }\in L^1$ and converges a.e. as $\sigma \downarrow 0$ to $h=⟨\phi ,{\phi }^{\mathrm{\prime }}⟩\mathrm{/}r$ on $\{r>0\}$ and $h=0$ on $\{r=0\}$. Dominated convergence [F4] in the primitive identities gives $r(t)=r(a)+{\int }_a^{t}h$, so $r^{\mathrm{\prime }}=h$ a.e. and $r^{\mathrm{\prime }}\ge 0$ a.e. On $\{r>0\}$ Gauss lemma gives, with $w={\phi }^{\mathrm{\prime }}-(r^{\mathrm{\prime }}\mathrm{/}r)\phi \perp \phi$, $\mathrm{\mid }\dot{\alpha }{\mathrm{\mid }}_g^2=(r^{\mathrm{\prime }}{)}^2+\mathrm{\mid }d({\exp }_p{)}_{\phi }w{\mathrm{\mid }}_g^2\mathrm{.}$ On $\{r=0\}$ the comparison $\mathrm{\mid }\dot{\alpha }{\mathrm{\mid }}_g\ge r^{\mathrm{\prime }}=0$ is trivial. Thus $\mathrm{\mid }\dot{\alpha }{\mathrm{\mid }}_g-r^{\mathrm{\prime }}\ge 0$ a.e.; its integral is $L_g(\alpha )-(r(b)-r(a))=\mathrm{\mid }X{\mathrm{\mid }}_g-\mathrm{\mid }X{\mathrm{\mid }}_g=0$. It follows that $\mathrm{\mid }\dot{\alpha }{\mathrm{\mid }}_g=r^{\mathrm{\prime }}$ a.e. Each inverse branch is a diffeomorphism, so $d({\exp }_p{)}_{\phi }$ is injective on its strip; the displayed identity then gives $w=0$ a.e. on $\{r>0\}$.

Constant direction and reparametrization. [F1, F2, step 1.1, step 3.1, step 4.1, given] On each compact subinterval of $(c,b]$, the map $e=\phi \mathrm{/}r$ is an $H^1$ primitive by smooth superposition [F1], and step 4.1 gives $e^{\mathrm{\prime }}=w\mathrm{/}r=0$ a.e. Its primitive identity makes $e$ constant there; overlapping subintervals make the constant common. Since $\phi (b)=X$, it is $e_0=X\mathrm{/}\mathrm{\mid }X{\mathrm{\mid }}_g=v\mathrm{/}\mathrm{\mid }v{\mathrm{\mid }}_g$. Thus $\phi (t)=r(t)e_0$ for all $t$, including $[a,c]$ where both sides vanish. Define $\tau (t)=a+r(t)\mathrm{/}\mathrm{\mid }v{\mathrm{\mid }}_g$. By step 4.1 it is an absolutely continuous nondecreasing function with $L^2$ derivative; its endpoint values are $a,b$, hence it is a surjection. The radial exponential identity in [F2] now gives $\alpha (t)={\exp }_p(r(t)e_0)=\gamma (\tau (t))$ for every $t$. Together with step 3.1 this proves both equality cases.