Hessian comparison for distance under sectional curvature bounds

The radial data and the identification of the Hessian with $S(t_0)$. [F1, F2, F3, F4, A1, given] By [F3] there are a unique unit vector $v\in S_{p}M$ and a unique $t^{\mathrm{\prime }}\in (0,c_p(v))$ with $q={\exp }_p(t^{\mathrm{\prime }}v)$; by [F2] applied to this pair, $t^{\mathrm{\prime }}=d_g(p,q)=r(q)=t_0$, and $\gamma$ is the geodesic ${\gamma }_{p,v}$ restricted to $[0,t_0]$, with $T:=\dot{\gamma }(t_0)=\mathrm{grad}r(q),\mathrm{\mid }T\mathrm{\mid }=1,N=T^{\perp }=\{\mathrm{grad}r(q){\}}^{\perp }\mathrm{.}$ By [F3], $t_0<c_p(v)$. The published Cut time does not exceed first conjugate time gives $c_p(v)\le \tau$ for the full radial geodesic (with $\tau =+\mathrm{\infty }$ when no conjugate time exists). Thus $t_0<\tau$, so [F4] makes $A(s)$ invertible and $S(s)$ smooth for every $0<s\le t_0$. This uses the whole pre-cut interval, rather than inferring absence of earlier singularities from invertibility at $t_0$ alone. Combining [F1(c)] with $A(t_0)=J(t_0)$ and with $A=P_{t_0}\bar{A}$, $D_{t}A=P_{t_0}{\bar{A}}^{\mathrm{\prime }}$ from [F4], $H:=\mathrm{\nabla }\mathrm{grad}r=(D_{t}A(t_0))\circ A(t_0{)}^{-1}=P_{t_0}\circ S(t_0)\circ P_{t_0}^{-1}\text{on }N,H(T)=0,$ and $H$ is $g$-self-adjoint on $N$ and annihilates $T$; since $H(T)=0$ and $T\perp N$, the endomorphism $H$ is self-adjoint on all of $T_{q}M$, and by [F1(b)] the Hessian satisfies $\mathrm{Hess}r(X,Y)=g_q(HX,Y)$ for all $X,Y\in T_{q}M$. Consequently, for $X,Y\in N$ with $x:=P_{t_0}^{-1}X$ and $y:=P_{t_0}^{-1}Y$ in $N_0$, $\mathrm{Hess}r(X,Y)=g_q(HX,Y)=g_q(P_{t_0}S(t_0)x,P_{t_0}y)=⟨S(t_0)x,y⟩,$ because $P_{t_0}$ is an isometry, and $S(t_0)$ is self-adjoint by [F4]. The endpoint $t_0$ is strictly positive and finite, and the formula is asserted at $q$ only; the value of $S$ at $t=0$ and past $\tau$ is never used. Since a self-adjoint endomorphism of $N_0$ is negative (respectively positive) semidefinite if and only if its quadratic form is nonpositive (respectively nonnegative), claims (1) and (2) of the statement are equivalent to the Loewner inequalities $S(t_0)\le {\mathrm{ct}}_k(t_0){\mathrm{id}}_{N_0}\text{and}S(t_0)\ge {\mathrm{ct}}_k(t_0){\mathrm{id}}_{N_0}$ respectively, and claim (3) is exactly $H(T)=0$ from [F1(c)].

The curvature hypothesis in transported form. [F4, F5, given] For $s\in (0,t_0]$ and $w\in N_0$, the definition of $R_{\gamma }$ in [F4] and the fact that $P_s$ is an isometry give $⟨R_{\gamma }(s)w,w⟩=⟨R(P_{s}w,T(s))T(s),P_{s}w⟩=\mathrm{Rm}(P_{s}w,T(s),T(s),P_{s}w),$ which vanishes for $w=0$ and equals $\mathrm{\mid }w{\mathrm{\mid }}^2\sec (P_{s}w\wedge T(s))$ for $w\mathrm{\ne }0$ by [F5], since $P_{s}w\perp T(s)$ is nonzero. Therefore the first curvature hypothesis of the statement is equivalent to $R_{\gamma }(s)\ge k{\mathrm{id}}_{N_0}\text{for all }s\in (0,t_0],$ and the second to $R_{\gamma }(s)\le k{\mathrm{id}}_{N_0}$ for all $s\in (0,t_0]$, both in the Loewner order on the $(n-1)$-dimensional space $N_0$.

The model functions. [F6, F8, given] On its positive domain the comparison cotangent satisfies the scalar Riccati identity ${\mathrm{ct}}_k^{\mathrm{\prime }}+{\mathrm{ct}}_k^2=-k:$ by [F6], ${\mathrm{cs}}_k^{\mathrm{\prime }}={\mathrm{sn}}_k^{\mathrm{\prime }\mathrm{\prime }}=-k{\mathrm{sn}}_k$ and ${\mathrm{sn}}_k^{\mathrm{\prime }}={\mathrm{cs}}_k$, so the quotient rule gives ${\mathrm{ct}}_k^{\mathrm{\prime }}=\frac{{\mathrm{cs}}_k^{\mathrm{\prime }}{\mathrm{sn}}_k-{\mathrm{cs}}_k{\mathrm{sn}}_k^{\mathrm{\prime }}}{{\mathrm{sn}}_k^2}=\frac{-k{\mathrm{sn}}_k^2-{\mathrm{cs}}_k^2}{{\mathrm{sn}}_k^2}=-k-{\mathrm{ct}}_k^2\mathrm{.}$ Moreover ${\mathrm{sn}}_k(t)=t+O(t^3)$ and ${\mathrm{cs}}_k(t)=1+O(t^2)$ as $t\downarrow 0$: by [F6], ${\mathrm{sn}}_k(0)=0$, ${\mathrm{sn}}_k^{\mathrm{\prime }}(0)=1$ and ${\mathrm{sn}}_k^{\mathrm{\prime }\mathrm{\prime }}(0)=-k{\mathrm{sn}}_k(0)=0$, while ${\mathrm{cs}}_k(0)=1$ and ${\mathrm{cs}}_k^{\mathrm{\prime }}(0)=0$, and [F8] applied with $m=3$ to ${\mathrm{sn}}_k$ and $m=2$ to ${\mathrm{cs}}_k$ gives the two expansions. Hence there is ${\delta }_0>0$ with ${\mathrm{sn}}_k(t)=t(1+O(t^2))\mathrm{\ne }0$ and $0<{\mathrm{cs}}_k(t)\le 2,{\mathrm{ct}}_k(t)=\frac{{\mathrm{cs}}_k(t)}{{\mathrm{sn}}_k(t)}=\frac{1}{t}(1+O(t^2)),{\mathrm{ct}}_k(t)\ge \frac{1}{2t}(0<t<{\delta }_0),$ the last bound by shrinking ${\delta }_0$ so that $\mathrm{\mid }O(t^2)\mathrm{\mid }\le \frac{1}{2}$ and ${\mathrm{cs}}_k\ge \frac{1}{2}$ on $(0,{\delta }_0)$. In particular $t\mapsto {\mathrm{ct}}_k(t)$ is continuous on $(0,t_0]$, because $t_0<\pi \mathrm{/}\sqrt{k}$ when $k>0$ and ${\mathrm{ct}}_k$ is defined and smooth on all of $(0,\mathrm{\infty })$ when $k\le 0$.

The eigenvalues of $S$ and their Dini bounds. [F4, F7, given] Let ${\lambda }_{\max }(s)$ and ${\lambda }_{\min }(s)$ denote the largest and the smallest eigenvalue of the self-adjoint operator $S(s)$, $s\in (0,\tau )$. We claim that both functions are Lipschitz on compact subintervals of $(0,\tau )$ and that for every $s\in (0,t_0)$ $D^{+}{\lambda }_{\max }(s)\le -{\lambda }_{\max }(s{)}^2-⟨R_{\gamma }(s)w_{+},w_{+}⟩$ for some unit eigenvector $w_{+}$ of ${\lambda }_{\max }(s)$, and $D_{+}{\lambda }_{\min }(s)\ge -{\lambda }_{\min }(s{)}^2-⟨R_{\gamma }(s)w_{-},w_{-}⟩$ for some unit eigenvector $w_{-}$ of ${\lambda }_{\min }(s)$, where $D^{+}$ and $D_{+}$ are the upper and lower right Dini derivatives. Indeed, by [F7] and compactness of the unit sphere of the finite-dimensional space $N_0$, ${\lambda }_{\max }(s)={\max }_{\mathrm{\mid }x\mathrm{\mid }=1}⟨S(s)x,x⟩$ and ${\lambda }_{\min }(s)={\min }_{\mathrm{\mid }x\mathrm{\mid }=1}⟨S(s)x,x⟩$, so both are Lipschitz where $S$ is $C^1$ by [F4]. Fix $s\in (0,t_0)$ and let $h_n\downarrow 0$ be a null sequence along which the difference quotients of ${\lambda }_{\max }$ approach its upper right Dini derivative; for each $n$ pick a unit vector $x_n$ with ${\lambda }_{\max }(s+h_n)=⟨S(s+h_n)x_n,x_n⟩$, and pass to a subsequence with $x_n\to w$, $\mathrm{\mid }w\mathrm{\mid }=1$. Since ${\lambda }_{\max }(s)\ge ⟨S(s)x_n,x_n⟩$ and $S$ is $C^1$, ${\lambda }_{\max }(s+h_n)-{\lambda }_{\max }(s)\le {\int }_0^{h_n}⟨S^{\mathrm{\prime }}(s+u)x_n,x_n⟩du=h_n⟨S^{\mathrm{\prime }}(s)w,w⟩+o(h_n),$ so $D^{+}{\lambda }_{\max }(s)\le ⟨S^{\mathrm{\prime }}(s)w,w⟩$; moreover $⟨S(s)w,w⟩={\lim }_n⟨S(s)x_n,x_n⟩={\lambda }_{\max }(s)$, so $w$ is a unit maximizer and $S(s)w={\lambda }_{\max }(s)w$. The Riccati equation $S^{\mathrm{\prime }}=-S^2-R_{\gamma }$ of [F4] then gives $⟨S^{\mathrm{\prime }}(s)w,w⟩=-\mathrm{\mid }S(s)w{\mathrm{\mid }}^2-⟨R_{\gamma }(s)w,w⟩=-{\lambda }_{\max }(s{)}^2-⟨R_{\gamma }(s)w,w⟩\mathrm{.}$ The computation for ${\lambda }_{\min }$ is the same with maximizers replaced by minimizers: for minimizers $x_n$ at $s+h_n$ one has ${\lambda }_{\min }(s+h_n)-{\lambda }_{\min }(s)\ge ⟨S(s+h_n)x_n,x_n⟩-⟨S(s)x_n,x_n⟩$ because ${\lambda }_{\min }(s)\le ⟨S(s)x_n,x_n⟩$, a limit point $w$ of a subsequence is a unit minimizer, and the same substitution applies, so $D_{+}{\lambda }_{\min }(s)\ge ⟨S^{\mathrm{\prime }}(s)w,w⟩$.

The matched-asymptotic comparison lemma. We use the following elementary fact. Let $0<T\le t_0$, let $p:(0,T]\to \mathbb{R}$ be continuous with $p(t)\ge c\mathrm{/}t$ for $t\in (0,\delta )$ and constants $c\ge 1$, $\delta \in (0,T]$, and let $h:(0,T]\to \mathbb{R}$ be continuous with $\mathrm{\mid }h(t)\mathrm{\mid }=o(1\mathrm{/}t)$ as $t\downarrow 0$ and $D^{+}h(t)+p(t)h(t)\le 0$ for every $t\in (0,T)$. Then $h\le 0$ on $(0,T]$. Proof. We first record the Dini monotonicity principle: a continuous $\phi$ on $[a,b]$ with $D^{+}\phi \le 0$ everywhere on $[a,b)$ is nonincreasing. Indeed, $\psi :=-\phi$ has $D_{+}\psi \ge 0$ everywhere; suppose $\psi (x)>\psi (y)$ for some $x<y$, put $m:=(\psi (x)-\psi (y))\mathrm{/}(y-x)>0$ and $\mathrm{\Psi }(t):=\psi (t)-\psi (x)+\frac{m}{2}(t-x)$ on $[x,y]$. Then $\mathrm{\Psi }(x)=0$, $\mathrm{\Psi }(y)=-\frac{m}{2}(y-x)<0$, and $D_{+}\mathrm{\Psi }=D_{+}\psi +\frac{m}{2}\ge \frac{m}{2}$ everywhere. The set $A=\{t\in [x,y]:\mathrm{\Psi }(t)<0\}$ is nonempty, and with $\beta :=\inf A$ one has $\beta <y$, $\mathrm{\Psi }(\beta )=0$ (the complement of $A$ is closed, and points of $A$ approach $\beta$ from above), and along a sequence $t_n\downarrow \beta$ with $t_n\in A$ the quotients $(\mathrm{\Psi }(t_n)-\mathrm{\Psi }(\beta ))\mathrm{/}(t_n-\beta )=\mathrm{\Psi }(t_n)\mathrm{/}(t_n-\beta )$ are negative; hence $D_{+}\mathrm{\Psi }(\beta )\le 0$, contradicting $D_{+}\mathrm{\Psi }(\beta )\ge m\mathrm{/}2>0$. This proves the principle, and the product rule for Dini derivatives with the $C^1$ factor $e^M$ is exact. Now fix $t\in (0,T]$ and $0<{\epsilon }_0<t$, and put $M(s):=-{\int }_s^{t}p(u)du$ for $s\in [{\epsilon }_0,t]$. Then $e^M$ is $C^1$ with $(e^M{)}^{\mathrm{\prime }}=p{e}^M$, so $D^{+}\left(h{e}^M\right)(s)=e^{M(s)}(D^{+}h(s)+p(s)h(s))\le 0,$ and the principle gives $h(t)e^{M(t)}\le h({\epsilon }_0)e^{M({\epsilon }_0)}$, that is $h(t)\le h({\epsilon }_0)\exp (-{\int }_{{\epsilon }_0}^{t}p(u)du)\mathrm{.}$ With ${\delta }^{\mathrm{\prime }}:=\min (\delta ,t)$ and $K:=(t-{\delta }^{\mathrm{\prime }}){\max }_{[{\delta }^{\mathrm{\prime }},t]}\mathrm{\mid }p\mathrm{\mid }$ (finite, $p$ being continuous on the compact set $[{\delta }^{\mathrm{\prime }},t]\subseteq (0,T]$), ${\int }_{{\epsilon }_0}^{t}p(u)du\ge c\log \frac{{\delta }^{\mathrm{\prime }}}{{\epsilon }_0}-K\text{for }0<{\epsilon }_0<{\delta }^{\mathrm{\prime }},$ so $\exp (-{\int }_{{\epsilon }_0}^{t}p)\le e^K({\epsilon }_0\mathrm{/}{\delta }^{\mathrm{\prime }}{)}^c$ and, since $c\ge 1$ and ${\epsilon }_0\mathrm{\mid }h({\epsilon }_0)\mathrm{\mid }\to 0$, $h(t)\le e^K{\delta }^{\mathrm{\prime }-c}{\epsilon }_0^c\mathrm{\mid }h({\epsilon }_0)\mathrm{\mid }=e^K{\delta }^{\mathrm{\prime }-c}{\epsilon }_0^{c-1}\cdot {\epsilon }_0\mathrm{\mid }h({\epsilon }_0)\mathrm{\mid }⟶0({\epsilon }_0\downarrow 0)\mathrm{.}$ Hence $h(t)\le 0$, and $t$ was arbitrary.

The case $K\ge k$: the largest eigenvalue is at most ${\mathrm{ct}}_k$. [F4, step 1.1, step 1.2, step 1.3, step 1.4, step 1.5, given] Assume the first curvature hypothesis, so that $R_{\gamma }(s)\ge k{\mathrm{id}}_{N_0}$ on $(0,t_0]$ by step 1.2. Put $h(t):={\lambda }_{\max }(t)-{\mathrm{ct}}_k(t),p(t):={\lambda }_{\max }(t)+{\mathrm{ct}}_k(t)(0<t\le t_0),$ both continuous on $(0,t_0]$: ${\lambda }_{\max }$ is continuous by step 1.4 and ${\mathrm{ct}}_k$ is continuous on $(0,t_0]$ by step 1.3. By the asymptotics $S(t)=t^{-1}\mathrm{id}+O(t)$ of [F4] every eigenvalue of $S(t)$ equals $t^{-1}+O(t)$, so ${\lambda }_{\max }(t)=t^{-1}+O(t)$, while ${\mathrm{ct}}_k(t)=t^{-1}(1+O(t^2))$ by step 1.3; hence $\mathrm{\mid }h(t)\mathrm{\mid }=O(t)=o(1\mathrm{/}t),p(t)=\frac{2}{t}+O(t)\ge \frac{1}{t}(0<t<{\delta }_1)$ for some ${\delta }_1\in (0,t_0]$, after shrinking ${\delta }_0$ of step 1.3 if necessary. For $s\in (0,t_0)$, step 1.4 provides a unit eigenvector $w_{+}$ of ${\lambda }_{\max }(s)$ with $D^{+}{\lambda }_{\max }(s)\le -{\lambda }_{\max }(s{)}^2-⟨R_{\gamma }(s)w_{+},w_{+}⟩\le -{\lambda }_{\max }(s{)}^2-k,$ the last step by $R_{\gamma }(s)\ge k\mathrm{id}$; and ${\mathrm{ct}}_k^{\mathrm{\prime }}(s)=-k-{\mathrm{ct}}_k(s{)}^2$ by step 1.3, so $D^{+}h(s)+p(s)h(s)\le (-{\lambda }_{\max }^{}-k)+(k+{\mathrm{ct}}_k^2)+({\lambda }_{\max }^{}-{\mathrm{ct}}_k^2)=0.$ Step 1.5 with $T:=t_0$, the functions $p,h$ above and $c:=1$ therefore gives $h\le 0$ on $(0,t_0]$, that is ${\lambda }_{\max }(t)\le {\mathrm{ct}}_k(t)(0<t\le t_0),\text{in particular }S(t_0)\le {\mathrm{ct}}_k(t_0){\mathrm{id}}_{N_0},$ the last Loewner inequality because $S(t_0)$ is self-adjoint with largest eigenvalue at most ${\mathrm{ct}}_k(t_0)$.

The case $K\le k$: the smallest eigenvalue is at least ${\mathrm{ct}}_k$. [F4, step 1.1, step 1.2, step 1.3, step 1.4, step 1.5, given] Assume now the second curvature hypothesis, so that $R_{\gamma }(s)\le k{\mathrm{id}}_{N_0}$ on $(0,t_0]$ by step 1.2, and put $\tilde{h}(t):={\mathrm{ct}}_k(t)-{\lambda }_{\min }(t),\tilde{p}(t):={\lambda }_{\min }(t)+{\mathrm{ct}}_k(t)(0<t\le t_0)\mathrm{.}$ As above ${\lambda }_{\min }(t)=t^{-1}+O(t)$, so $\mathrm{\mid }\tilde{h}(t)\mathrm{\mid }=O(t)=o(1\mathrm{/}t)$ and $\tilde{p}(t)\ge 1\mathrm{/}t$ on $(0,{\delta }_1)$. For $s\in (0,t_0)$, step 1.4 provides a unit eigenvector $w_{-}$ of ${\lambda }_{\min }(s)$ with $D_{+}{\lambda }_{\min }(s)\ge -{\lambda }_{\min }(s{)}^2-⟨R_{\gamma }(s)w_{-},w_{-}⟩\ge -{\lambda }_{\min }(s{)}^2-k,$ the last step by $R_{\gamma }(s)\le k\mathrm{id}$; since $D^{+}\tilde{h}={\mathrm{ct}}_k^{\mathrm{\prime }}-D_{+}{\lambda }_{\min }$ and ${\mathrm{ct}}_k^{\mathrm{\prime }}=-k-{\mathrm{ct}}_k^2$, $D^{+}\tilde{h}(s)+\tilde{p}(s)\tilde{h}(s)\le (-k-{\mathrm{ct}}_k^2)+({\lambda }_{\min }^{}+k)+({\mathrm{ct}}_k^2-{\lambda }_{\min }^{})=0.$ Step 1.5 with $T:=t_0$, the functions $\tilde{p},\tilde{h}$ and $c:=1$ gives $\tilde{h}\le 0$ on $(0,t_0]$, that is ${\lambda }_{\min }(t)\ge {\mathrm{ct}}_k(t)(0<t\le t_0),\text{in particular }S(t_0)\ge {\mathrm{ct}}_k(t_0){\mathrm{id}}_{N_0}\mathrm{.}$

Conclusion and boundary cases. [step 1.1, step 2.1, step 2.2, given] Under the first curvature hypothesis, step 2.1 gives $S(t_0)\le {\mathrm{ct}}_k(t_0){\mathrm{id}}_{N_0}$; the equivalence recorded in step 1.1 then yields claim (1): $\mathrm{Hess}r(X,X)\le {\mathrm{ct}}_k(t_0)g_q(X,X)$ for all $X\in N$. Under the second hypothesis, step 2.2 gives $S(t_0)\ge {\mathrm{ct}}_k(t_0){\mathrm{id}}_{N_0}$ and hence claim (2). Claim (3) is $H(T)=0$ from step 1.1. The comparison is asserted only at the point $t_0$ of the interval $(0,\min (\tau ,\pi \mathrm{/}\sqrt{k}))$ (with $\pi \mathrm{/}\sqrt{k}$ omitted when $k\le 0$), so the singularities at $t=0$ and at the first conjugate instant are not crossed, and for $k>0$ the hypothesis $t_0<\pi \mathrm{/}\sqrt{k}$ keeps $t_0$ away from the pole of ${\mathrm{ct}}_k$, where no value of the model is defined; for $k\le 0$ no upper restriction on $t_0$ is needed and none is imposed. In dimension $n=2$ the space $N_0$ is one-dimensional, the eigenvalues in steps 2.1 and 2.2 are single numbers, and the two Loewner inequalities reduce to the scalar statements $S(t_0)\le {\mathrm{ct}}_k(t_0)$ and $S(t_0)\ge {\mathrm{ct}}_k(t_0)$ for the scalar Riccati function. If both curvature hypotheses hold (constant curvature $k$ along $\gamma$) the two claims give equality $\mathrm{Hess}r={\mathrm{ct}}_k(t_0)g$ on $N$. Under the lower curvature bound, the field-level companion of the present estimate, $\mathrm{\mid }{J}_w(t)\mathrm{\mid }\le {\mathrm{sn}}_k(t)\mathrm{\mid }w\mathrm{\mid }$ for the radial Jacobi fields, is Rauch comparison theorem first form; the estimate proved here is its derivative-level form. No compactness of $M$ is used, and no choice beyond the inherited [A1] enters: the geodesic, the parallel frame, the Riccati operator and the eigenvalues are all determined by the given data.