Laplacian comparison for distance under a ricci lower bound

Setup: ${\mathrm{\Delta }}_{g}r(q)=h(t_0)$. [F1, F3, F4, F5, A1, given] By [F4] fix the unique $v\in S_{p}M$ and $t_0=r(q)\in (0,c_p(v))$ with $q={\exp }_p(t_{0}v)$; then $\gamma ={\gamma }_{p,v}$, $T=\dot{\gamma }(t_0)$, and $t_0<\tau$ by [F5]. By [F3] the operator $H=\mathrm{\nabla }\mathrm{grad}r$ satisfies $H(T)=0$ and $H=D_{t}A(t_0)\circ A(t_0{)}^{-1}$ on $N=T^{\perp }$, so $H$ maps $N$ into $N$ and annihilates $T$. Choosing a $g_q$-orthonormal basis of $T_{q}M$ whose first vector is $T$ and whose remaining $n-1$ vectors form an orthonormal basis $e_1,\dots ,e_{n-1}$ of $N$, the trace formula of [F1] gives ${\mathrm{\Delta }}_{g}r(q)=\mathrm{Hess}r(T,T)+{\sum }_{i=1}^{n-1}\mathrm{Hess}r(e_i,e_i)=0+{\sum }_{i=1}^{n-1}{g}_q(D_{t}A(t_0)A(t_0{)}^{-1}{e}_i,e_i),$ the vanishing of $\mathrm{Hess}r(T,T)=g_q(H(T),T)$ by [F3]. Since $P_{t_0}$ is an isometry and $S(t_0)=P_{t_0}^{-1}\circ (D_{t}A(t_0)A(t_0{)}^{-1})\circ P_{t_0}$ by [F5], the sum is the trace of $S(t_0)$ over the orthonormal basis $P_{t_0}^{-1}{e}_1,\dots ,P_{t_0}^{-1}{e}_{n-1}$ of $N_0$: ${\mathrm{\Delta }}_{g}r(q)=\mathrm{tr}S(t_0)=h(t_0)\mathrm{.}$

The scalar Riccati inequality and its asymptotics. [F2, F5, F6, given] By [F2] the function $h=\mathrm{tr}S$ is differentiable on $(0,\tau )$ and satisfies $h^{\mathrm{\prime }}+\frac{h^2}{n-1}+{\mathrm{Ric}}_{\gamma }(\dot{\gamma },\dot{\gamma })\le 0.$ By [F6] the Ricci lower bound gives ${\mathrm{Ric}}_{\gamma (t)}(\dot{\gamma }(t),\dot{\gamma }(t))\ge (n-1)k$ at every $t\in (0,t_0]$, so $h^{\mathrm{\prime }}(t)+\frac{h(t{)}^2}{n-1}\le -(n-1)k(0<t\le t_0)\mathrm{.}$ Put $a:=h\mathrm{/}(n-1)$, a differentiable function on $(0,\tau )$ with $a^{\mathrm{\prime }}=\frac{h^{\mathrm{\prime }}}{n-1}$ and $a^2=h^2\mathrm{/}(n-1{)}^2$; dividing the inequality by the positive number $n-1>0$ gives the scalar Riccati inequality $a^{\mathrm{\prime }}(t)+a(t{)}^2\le -k(0<t\le t_0)\mathrm{.}$ Moreover $h(t)=(n-1)t^{-1}+O(t)$ by [F5], so $a(t)=\frac{1}{t}+O(t)(t\downarrow 0)\mathrm{.}$

The scalar comparison lemma. We use the following elementary fact, the integrating-factor comparison for Riccati inequalities with matched asymptotics at the singular endpoint. Let $0<T\le t_0$ and let $a,c:(0,T]\to \mathbb{R}$ be differentiable with $a^{\mathrm{\prime }}+a^2\le -k,c^{\mathrm{\prime }}+c^2=-k\text{on }(0,T],$ both $a(t)=1\mathrm{/}t+O(t)$ and $c(t)=1\mathrm{/}t+O(t)$ as $t\downarrow 0$. Then $a\le c$ on $(0,T]$. Proof. Put $\phi :=a-c$. Subtracting the two equations, ${\phi }^{\mathrm{\prime }}=a^{\mathrm{\prime }}-c^{\mathrm{\prime }}\le (-a^2-k)-(-c^2-k)=-(a+c)\phi \mathrm{.}$ Fix $0<{\epsilon }_0<t\le T$ and define $\mathrm{\Phi }(s):=\phi (s)\exp (-{\int }_s^t(a+c))$ for $s\in [{\epsilon }_0,t]$. Since $(a+c)$ is continuous, $\mathrm{\Phi }$ is differentiable with ${\mathrm{\Phi }}^{\mathrm{\prime }}(s)=\exp (-{\int }_s^t(a+c))({\phi }^{\mathrm{\prime }}(s)+(a(s)+c(s))\phi (s))\le 0,$ so $\mathrm{\Phi }$ is nonincreasing and $\phi (t)\le \phi ({\epsilon }_0)\exp (-{\int }_{{\epsilon }_0}^t(a+c))\mathrm{.}$ By the two asymptotics, choose $\delta \in (0,T]$ so that $a+c\ge 1\mathrm{/}t$ on $(0,\delta )$. Put ${\delta }^{\mathrm{\prime }}:=\min (\delta ,t)$ and $K:=(t-{\delta }^{\mathrm{\prime }}){\max }_{[{\delta }^{\mathrm{\prime }},t]}\mathrm{\mid }a+c\mathrm{\mid }$; then ${\int }_{{\epsilon }_0}^t(a+c)\ge {\int }_{{\epsilon }_0}^{{\delta }^{\mathrm{\prime }}}\frac{du}{u}-K=\log \frac{{\delta }^{\mathrm{\prime }}}{{\epsilon }_0}-K(0<{\epsilon }_0<{\delta }^{\mathrm{\prime }}),$ and hence $\phi (t)\le e^K{\delta }^{\mathrm{\prime }-1}{\epsilon }_0\mathrm{\mid }\phi ({\epsilon }_0)\mathrm{\mid }⟶0({\epsilon }_0\downarrow 0),$ using ${\epsilon }_0\mathrm{\mid }\phi ({\epsilon }_0)\mathrm{\mid }\to 0$; therefore $\phi (t)\le 0$.

The distance Laplacian is bounded by the model trace. [step 1.1, step 1.2, step 1.3, F7, given] On the positive domain of [F7] the comparison cotangent satisfies the model Riccati equation ${\mathrm{ct}}_k^{\mathrm{\prime }}+{\mathrm{ct}}_k^2=-k$: indeed ${\mathrm{cs}}_k^{\mathrm{\prime }}={\mathrm{sn}}_k^{\mathrm{\prime }\mathrm{\prime }}=-k{\mathrm{sn}}_k$ and ${\mathrm{sn}}_k^{\mathrm{\prime }}={\mathrm{cs}}_k$, so ${\mathrm{ct}}_k^{\mathrm{\prime }}=\frac{{\mathrm{cs}}_k^{\mathrm{\prime }}{\mathrm{sn}}_k-{\mathrm{cs}}_k{\mathrm{sn}}_k^{\mathrm{\prime }}}{{\mathrm{sn}}_k^2}=\frac{-k{\mathrm{sn}}_k^2-{\mathrm{cs}}_k^2}{{\mathrm{sn}}_k^2}=-k-{\mathrm{ct}}_k^2\mathrm{.}$ Moreover, by [F8] with order $3$ for ${\mathrm{sn}}_k$ and order $2$ for ${\mathrm{cs}}_k$, using ${\mathrm{sn}}_k^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}(0)=-k$ and ${\mathrm{cs}}_k^{\mathrm{\prime }\mathrm{\prime }}(0)=-k$ together with the values ${\mathrm{sn}}_k(0)=0$, ${\mathrm{sn}}_k^{\mathrm{\prime }}(0)=1$, ${\mathrm{sn}}_k^{\mathrm{\prime }\mathrm{\prime }}(0)=-k{\mathrm{sn}}_k(0)=0$ and ${\mathrm{cs}}_k(0)=1$, ${\mathrm{cs}}_k^{\mathrm{\prime }}(0)=0$, ${\mathrm{sn}}_k(t)=t(1+O(t^2)),{\mathrm{cs}}_k(t)=1+O(t^2),{\mathrm{ct}}_k(t)=\frac{1}{t}+O(t),{\mathrm{ct}}_k(t)\ge \frac{1}{2t}$ for $0<t<{\delta }_0$ and some ${\delta }_0>0$. By step 1.2 the function $a=h\mathrm{/}(n-1)$ satisfies $a^{\mathrm{\prime }}+a^2\le -k$ on $(0,t_0]$ and $a(t)=t^{-1}+O(t)$; the model $c:={\mathrm{ct}}_k$ satisfies $c^{\mathrm{\prime }}+c^2=-k$ on $(0,t_0]$ (the interval $(0,t_0]$ lies in the positive domain of ${\mathrm{ct}}_k$ by [F7] and the hypothesis on $t_0$), $c\ge 1\mathrm{/}(2t)$ near $0$, and $\mathrm{\mid }a-c\mathrm{\mid }=O(t)=o(1\mathrm{/}t)$. Step 1.3 with $T:=t_0$ therefore gives $a(t)\le {\mathrm{ct}}_k(t)(0<t\le t_0),\text{hence }h(t_0)\le (n-1){\mathrm{ct}}_k(t_0)\mathrm{.}$

Conclusion and boundary cases. [step 1.1, step 2.1, given] Combining ${\mathrm{\Delta }}_{g}r(q)=h(t_0)$ of step 1.1 with the bound of step 2.1 gives ${\mathrm{\Delta }}_{g}r(q)\le (n-1){\mathrm{ct}}_k(r(q))$, which is the assertion. The computation takes place on the interval $(0,t_0]$, so the singular initial point $t=0$ is excluded and the first conjugate instant $\tau >t_0$ is not reached; the cut locus is excluded by the hypothesis on $q$, and the model pole at $\pi \mathrm{/}\sqrt{k}$ is excluded for $k>0$ by $t_0<\pi \mathrm{/}\sqrt{k}$. For $k\le 0$ the model ${\mathrm{ct}}_k$ is defined on all of $(0,\mathrm{\infty })$ and no restriction on $t_0$ is imposed. In dimension $n=2$ the normal space is one-dimensional and $h=S$ is the scalar Riccati function, so the trace Riccati inequality of [F2] reduces to $S^{\mathrm{\prime }}+S^2+\mathrm{Ric}(\dot{\gamma },\dot{\gamma })\le 0$ and step 2.1 bounds $h=S$ by ${\mathrm{ct}}_k$ itself, as displayed. If $\mathrm{Ric}=(n-1)kg$, the differential inequality of step 1.2 is an equality where the traced Cauchy–Schwarz step Trace riccati inequality is an equality; equality in the final comparison additionally depends on the preceding radial segment. The proof uses no information about $r$ other than the single point $q$ and requires neither completeness of $M$ beyond the results quoted, nor compactness; the only choice is the inherited [A1].