Toponogov distance support inequality

The relevant distances lie in the domain of $f$. For $0\le t\le L$ the triangle inequality gives $d_g(o,\gamma (t))\le \min \{a+t,b+(L-t)\},$ and adding the two estimates yields $2d_g(o,\gamma (t))\le a+b+L$. The same argument in the model gives $2d_k(\bar{o},\bar{\gamma }(t))\le a+b+L$. Hence every occurring distance is at most $(a+b+L)\mathrm{/}2$, which is $<\pi \mathrm{/}\sqrt{k}$ when $k>0$ by [F1] and finite in general; and every occurring distance is positive, because $d_g(o,\gamma (t))=0$ would put $o=\gamma (t)$ on the side and force $a+b=t+(L-t)=L$, contradicting the strict triangle inequality, and likewise in the model. By [F2], $f$ is strictly increasing on $[0,\mathrm{\infty })$ for $k\le 0$ and on $[0,\pi \mathrm{/}\sqrt{k})$ for $k>0$; all occurring distance values lie in this interval.

Endpoint values of $\delta$. Define $\rho :=d_g(o,\cdot ),\sigma :=f\circ \rho ,\bar{\rho }:=d_k(\bar{o},\cdot ),\bar{\sigma }:=f\circ \bar{\rho },$ and $\delta :=\sigma \circ \gamma -\bar{\sigma }\circ \bar{\gamma }\text{on }[0,L]\mathrm{.}$ Since $d_g(o,\gamma (0))=a=d_k(\bar{o},\bar{x})=\bar{\rho }(\bar{\gamma }(0))$ and $d_g(o,\gamma (L))=b=d_k(\bar{o},\bar{y})=\bar{\rho }(\bar{\gamma }(L))$ by [F1], $f(\rho (\gamma (0)))=f(a)=f(\bar{\rho }(\bar{\gamma }(0))),f(\rho (\gamma (L)))=f(b)=f(\bar{\rho }(\bar{\gamma }(L))),$ that is $\delta (0)=\delta (L)=0$.

The differential inequality for $\sigma$. At a point $x\mathrm{\notin }\{o\}\cup \mathrm{Cut}(o)$, $\rho$ is smooth and $\mathrm{grad}\rho$ is a unit vector. The chain rule of [F9] applied to $\sigma =f\circ \rho$ gives $\mathrm{Hess}\sigma =f^{\mathrm{\prime }\mathrm{\prime }}(\rho )d\rho \otimes d\rho +f^{\mathrm{\prime }}(\rho )\mathrm{Hess}\rho \mathrm{.}$ Write $X=\alpha \mathrm{grad}\rho +X^{\perp }$ with $\alpha =d\rho (X)$ and $X^{\perp }\perp \mathrm{grad}\rho$. Since $\mathrm{Hess}\rho (\mathrm{grad}\rho ,\cdot )=0$ by [F3] and $K\ge k$ holds on radial planes, [F3] yields $\mathrm{Hess}\rho (X^{\perp },X^{\perp })\le {\mathrm{ct}}_k(\rho )\mathrm{\mid }{X}^{\perp }{\mathrm{\mid }}^2$. Because $d\rho \otimes d\rho$ vanishes on $X^{\perp }$ and $f^{\mathrm{\prime }}={\mathrm{sn}}_k$, $f^{\mathrm{\prime }\mathrm{\prime }}={\mathrm{cs}}_k$: $\mathrm{Hess}\sigma (X,X)=f^{\mathrm{\prime }\mathrm{\prime }}(\rho ){\alpha }^2+{\mathrm{sn}}_k(\rho )\mathrm{Hess}\rho (X^{\perp },X^{\perp })\le {\mathrm{cs}}_k(\rho ){\alpha }^2+{\mathrm{sn}}_k(\rho ){\mathrm{ct}}_k(\rho )\mathrm{\mid }{X}^{\perp }{\mathrm{\mid }}^2\mathrm{.}$ Since ${\mathrm{sn}}_k{\mathrm{ct}}_k={\mathrm{cs}}_k$ by [F2] and ${\mathrm{cs}}_k=-kf+1=f^{\mathrm{\prime }\mathrm{\prime }}$, the right-hand side equals ${\mathrm{cs}}_k(\rho )({\alpha }^2+\mathrm{\mid }{X}^{\perp }{\mathrm{\mid }}^2)=(-k\sigma (x)+1)g_x(X,X),$ where $\mathrm{\mid }X{\mathrm{\mid }}^2={\alpha }^2+\mathrm{\mid }{X}^{\perp }{\mathrm{\mid }}^2$ and $\sigma (x)=f(\rho (x))$. Hence $\mathrm{Hess}\sigma \le (-k\sigma +1)g\text{on }M\setminus (\{o\}\cup \mathrm{Cut}(o))\mathrm{.}$

The model identity. In the model, $\bar{\rho }$ is smooth along $\bar{\gamma }$: by step 1.1 $0<\bar{\rho }(\bar{\gamma }(t))<\pi \mathrm{/}\sqrt{k}$ when $k>0$ and $\bar{\rho }(\bar{\gamma }(t))>0$ always; for $k>0$ the model is the round sphere and points at distance $<\pi \mathrm{/}\sqrt{k}$ from $\bar{o}$ are off $\mathrm{Cut}(\bar{o})$ by [F5], while for $k\le 0$ the model has empty cut loci by [F5]. Since the model has constant curvature $k$, [F3] applies there in both directions and $\mathrm{Hess}\bar{\rho }={\mathrm{ct}}_k(\bar{\rho })(g-d\bar{\rho }\otimes d\bar{\rho })\text{on the normal space},\mathrm{Hess}\bar{\rho }(\mathrm{grad}\bar{\rho },\cdot )=0.$ Repeating the computation of step 1.3 with equalities in place of both inequalities gives, at every point of the model side, $\mathrm{Hess}\bar{\sigma }=(-k\bar{\sigma }+1)g\mathrm{.}$ Since $\bar{\gamma }$ is a unit-speed geodesic, the second-derivative identity of [F9] gives $(\bar{\sigma }\circ \bar{\gamma }{)}^{\mathrm{\prime }\mathrm{\prime }}(t)=\mathrm{Hess}\bar{\sigma }(\dot{\bar{\gamma }},\dot{\bar{\gamma }})=-k\bar{\sigma }(\bar{\gamma }(t))+1(0\le t\le L)\mathrm{.}$

The strict barrier. [F1, F2] Suppose $m<0$ is the minimum from step 2.2. Choose $k^{\mathrm{\prime }}>k$ with $k^{\mathrm{\prime }}>0$ and $L<\pi \mathrm{/}\sqrt{k^{\mathrm{\prime }}}$: if $k>0$, take $k<k^{\mathrm{\prime }}<(\pi \mathrm{/}L{)}^2$; if $k\le 0$, choose a sufficiently small $k^{\mathrm{\prime }}>0$. Then choose $\tau >0$ with $L+\tau <\pi \mathrm{/}\sqrt{k^{\mathrm{\prime }}}$, and on $[-\tau ,L]$ define $a_0(t):=m\frac{\sin (\sqrt{k^{\mathrm{\prime }}}(t+\tau ))}{{\min }_{0\le s\le L}\sin (\sqrt{k^{\mathrm{\prime }}}(s+\tau ))}\mathrm{.}$ The denominator is positive, $a_0(t)\le m<0$ on $[0,L]$, $a_0^{\mathrm{\prime }\mathrm{\prime }}+k^{\mathrm{\prime }}{a}_0=0$, and $a_0(-\tau )=0$.

The differential inequality for $\delta$. At every $t\in (0,L)$ with $\gamma (t)\mathrm{\notin }\mathrm{Cut}(o)$ the function $\sigma \circ \gamma$ is $C^2$ near $t$ by [F4], and the second-derivative identity of [F9] with the unit-speed geodesic $\gamma$ gives $(\sigma \circ \gamma {)}^{\mathrm{\prime }\mathrm{\prime }}(t)=\mathrm{Hess}\sigma (\dot{\gamma },\dot{\gamma })\le -k\sigma (\gamma (t))+1$ by step 1.3. Subtracting the identity of step 1.4, ${\delta }^{\mathrm{\prime }\mathrm{\prime }}(t)\le -k\delta (t)\text{whenever }\gamma (t)\mathrm{\notin }\mathrm{Cut}(o)\mathrm{.}$ Note that $\gamma (t)\mathrm{\ne }o$ for all $t$ by step 1.1, so the excluded set is exactly ${\gamma }^{-1}(\mathrm{Cut}(o))$.

Assumption of contradiction and an interior minimizer. Assume that $\delta (t_1)<0$ for some $t_1\in [0,L]$. Since $\delta$ is continuous on the compact interval $[0,L]$ and $\delta (0)=\delta (L)=0$ by step 1.2, the minimum $m:={\min }_{[0,L]}\delta <0$ is attained at some point $t_0\in (0,L)$; fix such a point, so $\delta (t_0)=m<0$.

The barrier below $\delta$. [step 1.5, step 2.2] Take $a_0$ as in step 1.5 and define the continuous ratio $\delta \mathrm{/}{a}_0$ on $[0,L]$. It is zero at the endpoints and positive at the minimum point $t_0$ from step 2.2, so its positive maximum $\lambda :={\max }_{t\in [0,L]}\frac{\delta (t)}{a_0(t)}>0$ is attained at some $t_{\ast }\in (0,L)$. Put $\eta :=\lambda a_0$ and $m_{\ast }:=\delta (t_{\ast })=\eta (t_{\ast })<0$. Since $a_0<0$, the definition of $\lambda$ gives $\eta (t)\le \delta (t)$ for all $t$, with equality at $t_{\ast }$; moreover ${\eta }^{\mathrm{\prime }\mathrm{\prime }}=-k^{\mathrm{\prime }}\eta$ on $[0,L]$.

Case 1: $\gamma (t_{\ast })$ is not a cut point of $o$. [F4, F6, F7, step 2.1, step 3.1] If $\gamma (t_{\ast })\mathrm{\notin }\mathrm{Cut}(o)$, then $\delta$ is $C^2$ near $t_{\ast }$. The local minimum of $\delta -\eta$ at $t_{\ast }$ gives $(\delta -\eta {)}^{\mathrm{\prime }\mathrm{\prime }}(t_{\ast })\ge 0$, whereas step 2.1 and ${\eta }^{\mathrm{\prime }\mathrm{\prime }}=-k^{\mathrm{\prime }}\eta$ give $(\delta -\eta {)}^{\mathrm{\prime }\mathrm{\prime }}(t_{\ast })\le -k\delta (t_{\ast })+k^{\mathrm{\prime }}\eta (t_{\ast })=(k^{\mathrm{\prime }}-k)m_{\ast }<0,$ a contradiction.

Case 2: $\gamma (t_{\ast })$ is a cut point of $o$. Let $\beta :[0,L_0]\to M$, $L_0:=d_g(o,\gamma (t_{\ast }))$, be a minimizing unit-speed geodesic from $o$ to $\gamma (t_{\ast })$. For $k>0$ put $M_0:={\max }_{[0,L]}\rho \circ \gamma <\pi \mathrm{/}\sqrt{k}$ as in step 1.1 and choose $0<\epsilon <\min \{L_0,\pi \mathrm{/}\sqrt{k}-M_0\}$; for $k\le 0$ choose $0<\epsilon <L_0$. Set $o_{\epsilon }:=\beta (\epsilon ),{\rho }_{\epsilon }:=d_g(o_{\epsilon },\cdot )+\epsilon \mathrm{.}$ We claim $\gamma (t_{\ast })\mathrm{\notin }\mathrm{Cut}(o_{\epsilon })$. First, no conjugate pair along a minimizing segment $\sigma :[0,\mathrm{\ell }]\to M$ can occur at times $0<s_1<s_2\le \mathrm{\ell }$: if $s_2<\mathrm{\ell }$, the minimizing geodesic from $\sigma (s_1)$ continues past its conjugate point $\sigma (s_2)$; if $s_2=\mathrm{\ell }$, reverse $\sigma$ and use the conjugate pair at times $0$ and $\mathrm{\ell }-s_1$, past which the reversed segment still minimizes. Both contradict A geodesic does not minimize past its first conjugate point [F7]. Second, $\beta {\mathrm{\mid }}_{[\epsilon ,L_0]}$ is the unique minimizing segment from $o_{\epsilon }$ to $\gamma (t_{\ast })$: concatenating any such segment with $\beta {\mathrm{\mid }}_{[0,\epsilon ]}$ gives a minimizing curve from $o$, which is smooth at $o_{\epsilon }$ and has the initial data of $\beta$, hence equals $\beta$ by geodesic uniqueness [F8]. If $\gamma (t_{\ast })$ were a cut point of $o_{\epsilon }$, the cut-point characterization [F4] would give either a conjugate pair on this minimizing segment or two distinct minimizing segments, contradicting one of these facts.

Case 2, concluded: the support estimate. [F3, F9, step 2.1, step 3.1, step 4.2] The function ${\rho }_{\epsilon }$ is smooth near $\gamma (t_{\ast })$ by step 4.2 and is an upper support of $\rho$: the triangle inequality gives ${\rho }_{\epsilon }\ge \rho$ everywhere, with equality at $\gamma (t_{\ast })$ since $\beta$ is minimizing. As $f$ is increasing on the relevant model interval, ${\sigma }_{\epsilon }:=f\circ {\rho }_{\epsilon }$ satisfies ${\sigma }_{\epsilon }\ge \sigma$ with equality at $\gamma (t_{\ast })$. Thus ${\delta }_{\epsilon }:={\sigma }_{\epsilon }\circ \gamma -\bar{\sigma }\circ \bar{\gamma }\ge \delta \ge \eta ,$ with equality ${\delta }_{\epsilon }(t_{\ast })=\eta (t_{\ast })=m_{\ast }$. Hence ${\delta }_{\epsilon }-\eta$ has a local minimum at $t_{\ast }$. Put $u:=d_g(o_{\epsilon },\cdot )$ and $\bar{t}:=u(\gamma (t_{\ast }))=L_0-\epsilon$. By the chain rule [F9], $\mathrm{Hess}{\sigma }_{\epsilon }=f^{\mathrm{\prime }\mathrm{\prime }}(\bar{t}+\epsilon )du\otimes du+f^{\mathrm{\prime }}(\bar{t}+\epsilon )\mathrm{Hess}u\mathrm{.}$ At the contact point, Hessian comparison [F3] applied from $o_{\epsilon }$ gives, for $X=\alpha \mathrm{grad}u+X^{\perp }$, $\mathrm{Hess}{\sigma }_{\epsilon }(X,X)\le {\mathrm{cs}}_k(\bar{t}+\epsilon ){\alpha }^2+{\mathrm{sn}}_k(\bar{t}+\epsilon ){\mathrm{ct}}_k(\bar{t})\mathrm{\mid }{X}^{\perp }{\mathrm{\mid }}^2\mathrm{.}$ The model addition formula gives ${\mathrm{sn}}_k(\bar{t}+\epsilon ){\mathrm{ct}}_k(\bar{t})={\mathrm{cs}}_k(\bar{t}+\epsilon )+{\beta }_{\epsilon },{\beta }_{\epsilon }:=\frac{{\mathrm{sn}}_k(\epsilon )}{{\mathrm{sn}}_k(\bar{t})}>0,$ and ${\beta }_{\epsilon }\to 0$ as $\epsilon \downarrow 0$. Since $f^{\mathrm{\prime }\mathrm{\prime }}=-kf+1$, this yields $\mathrm{Hess}{\sigma }_{\epsilon }\le (-k{\sigma }_{\epsilon }+1+{\beta }_{\epsilon })g$ at the contact point. Combining with the model identity of step 1.4 and ${\eta }^{\mathrm{\prime }\mathrm{\prime }}=-k^{\mathrm{\prime }}\eta$ gives $({\delta }_{\epsilon }-\eta {)}^{\mathrm{\prime }\mathrm{\prime }}(t_{\ast })\le (k^{\mathrm{\prime }}-k)m_{\ast }+{\beta }_{\epsilon }<0$ for small enough $\epsilon >0$, because $m_{\ast }<0$ and $k^{\mathrm{\prime }}>k$. This contradicts the local minimum.

Conclusion. Both contact cases being impossible, the assumption of step 2.2 is false: $\delta (t)\ge 0(0\le t\le L)\mathrm{.}$ Since $f$ is strictly increasing on the interval that contains both $\rho (\gamma (t))=d_g(o,\gamma (t))$ and $\bar{\rho }(\bar{\gamma }(t))=d_k(\bar{o},\bar{\gamma }(t))$ by step 1.1, the inequality $f(\rho (\gamma (t)))\ge f(\bar{\rho }(\bar{\gamma }(t)))$ forces $d_g(o,\gamma (t))\ge d_k(\bar{o},\bar{\gamma }(t))(0\le t\le L),$ which is the assertion. The barrier was used at an interior point of $(0,L)$ in both cases, so the endpoints need no separate treatment beyond $\delta (0)=\delta (L)=0$ of step 1.2; the strict triangle inequalities of [F1] were used exactly to make $a,b>0$, to exclude $o\in \gamma ([0,L])$, and to give the comparison triangle; and no choice beyond the inherited ${\mathrm{A}\mathrm{C}}_{\omega }$ of [A1] was used.