Relative volume density comparison

The logarithmic derivative of the relative density. [F1, F2, F4, F5, given] On $I$ the functions $J_p(t,v)$ and ${\mathrm{sn}}_k(t)$ are positive, so $\log q_v=\log J_p(t,v)-(n-1)\log {\mathrm{sn}}_k(t)$ is differentiable with $\frac{d}{dt}\log q_v(t)=\mathrm{tr}{S}_v(t)-(n-1)\frac{{\mathrm{sn}}_k^{\mathrm{\prime }}(t)}{{\mathrm{sn}}_k(t)}=\mathrm{tr}{S}_v(t)-(n-1){\mathrm{ct}}_k(t),$ because ${\mathrm{sn}}_k^{\mathrm{\prime }}={\mathrm{cs}}_k$ and ${\mathrm{ct}}_k={\mathrm{cs}}_k\mathrm{/}{\mathrm{sn}}_k$ by [F4]. By [F2] applied at $q={\exp }_p(tv)$ this is also ${\mathrm{\Delta }}_g{r}_p({\exp }_p(tv))-(n-1){\mathrm{ct}}_k(t)$; the derivative is computed only for $t\in I\subseteq (0,c_p(v))$, where all factors are defined.

The trace bound $\mathrm{tr}{S}_v\le (n-1){\mathrm{ct}}_k$. [F3, F4, F6, given] Put $a:=h\mathrm{/}(n-1)$ with $h=\mathrm{tr}{S}_v$, a differentiable function on $(0,\tau )\supseteq I$. Dividing the trace Riccati inequality of [F3] by the positive number $n-1$ and inserting the Ricci bound gives $a^{\mathrm{\prime }}+a^2\le -k\text{on }I,a(t)=\frac{1}{t}+O(t)(t\downarrow 0)\mathrm{.}$ The model satisfies the companion identity and asymptotics ${\mathrm{ct}}_k^{\mathrm{\prime }}+{\mathrm{ct}}_k^2=-k,{\mathrm{ct}}_k(t)=\frac{1}{t}+O(t),{\mathrm{ct}}_k(t)\ge \frac{1}{2t}(0<t<\delta )$ for some $\delta >0$: the smooth model functions are defined on all of $\mathbb{R}$, so they satisfy the neighborhood hypothesis of [F6]. The identity follows from ${\mathrm{cs}}_k^{\mathrm{\prime }}={\mathrm{sn}}_k^{\mathrm{\prime }\mathrm{\prime }}=-k{\mathrm{sn}}_k$ and ${\mathrm{sn}}_k^{\mathrm{\prime }}={\mathrm{cs}}_k$ through the quotient rule, ${\mathrm{ct}}_k^{\mathrm{\prime }}=\frac{{\mathrm{cs}}_k^{\mathrm{\prime }}{\mathrm{sn}}_k-{\mathrm{cs}}_k{\mathrm{sn}}_k^{\mathrm{\prime }}}{{\mathrm{sn}}_k^2}=\frac{-k{\mathrm{sn}}_k^2-{\mathrm{cs}}_k^2}{{\mathrm{sn}}_k^2}=-k-{\mathrm{ct}}_k^2,$ while ${\mathrm{sn}}_k(t)=t(1+O(t^2))$ and ${\mathrm{cs}}_k(t)=1+O(t^2)$ by [F6] at order $3$ for ${\mathrm{sn}}_k$ and order $2$ for ${\mathrm{cs}}_k$, applied through ${\mathrm{sn}}_k(0)=0$, ${\mathrm{sn}}_k^{\mathrm{\prime }}(0)=1$, ${\mathrm{sn}}_k^{\mathrm{\prime }\mathrm{\prime }}(0)=0$ and ${\mathrm{cs}}_k(0)=1$, ${\mathrm{cs}}_k^{\mathrm{\prime }}(0)=0$ [F4], which gives the two asymptotic statements and the lower bound after shrinking $\delta >0$. Subtracting the two Riccati relations, the difference $\phi :=a-{\mathrm{ct}}_k$ satisfies ${\phi }^{\mathrm{\prime }}\le -(a+{\mathrm{ct}}_k)\phi \text{on }I\mathrm{.}$ Fix $0<{\epsilon }_0<t$ in $I$ and put $\mathrm{\Phi }(s):=\phi (s)\exp ({\int }_{{\epsilon }_0}^s(a+{\mathrm{ct}}_k))$ for $s\in [{\epsilon }_0,t]$; then $\mathrm{\Phi }$ is differentiable with ${\mathrm{\Phi }}^{\mathrm{\prime }}(s)=\exp ({\int }_{{\epsilon }_0}^s(a+{\mathrm{ct}}_k))({\phi }^{\mathrm{\prime }}(s)+(a(s)+{\mathrm{ct}}_k(s))\phi (s))\le 0,$ hence $\mathrm{\Phi }(t)\le \mathrm{\Phi }({\epsilon }_0)$ and $\phi (t)\le \phi ({\epsilon }_0)\exp (-{\int }_{{\epsilon }_0}^t(a+{\mathrm{ct}}_k))\mathrm{.}$ With ${\delta }^{\mathrm{\prime }}:=\min (\delta ,t)$ and $K:=(t-{\delta }^{\mathrm{\prime }}){\max }_{[{\delta }^{\mathrm{\prime }},t]}\mathrm{\mid }a+{\mathrm{ct}}_k\mathrm{\mid }$ after shrinking $\delta$ using both asymptotics one has $a(u)+{\mathrm{ct}}_k(u)\ge 1\mathrm{/}u$ on $(0,\delta )$ and therefore ${\int }_{{\epsilon }_0}^t(a+{\mathrm{ct}}_k)\ge \log \frac{{\delta }^{\mathrm{\prime }}}{{\epsilon }_0}-K,$ so $\phi (t)\le e^K{\delta }^{\mathrm{\prime }-1}\cdot {\epsilon }_0\mid \phi ({\epsilon }_0)\mid ⟶0({\epsilon }_0\downarrow 0),$ because ${\epsilon }_0\mathrm{\mid }\phi ({\epsilon }_0)\mathrm{\mid }\to 0$ by the two asymptotics. Hence $a\le {\mathrm{ct}}_k$ on $I$, i.e. $\mathrm{tr}{S}_v(t)\le (n-1){\mathrm{ct}}_k(t)(t\in I)\mathrm{.}$

The limit of the density at zero. [F1, F4, F5, F6, given] By the normalisation in [F1], $J_p(t,v)=t^{n-1}(1+o(1))$ as $t\downarrow 0$. By [F6] at order $3$ and ${\mathrm{sn}}_k(0)=0$, ${\mathrm{sn}}_k^{\mathrm{\prime }}(0)=1$, ${\mathrm{sn}}_k^{\mathrm{\prime }\mathrm{\prime }}(0)=0$ [F4] one has ${\mathrm{sn}}_k(t)=t(1+O(t^2))$, hence ${\mathrm{sn}}_k(t{)}^{n-1}=t^{n-1}(1+O(t^2))$ and $q_v(t)=\frac{J_p(t,v)}{{\mathrm{sn}}_k(t{)}^{n-1}}=\frac{1+o(1)}{1+O(t^2)}⟶1(t\downarrow 0),$ the quotient being defined and positive on $I$ by [F1] and [F4]. In particular the limit $q_v(0+)=1$ exists, independently of $k$ and of the geometry of $M$.

Conclusion: $q_v$ is nonincreasing and at most one. [step 1.1, step 1.2, step 1.3, given] By step 1.2, $\mathrm{tr}{S}_v(t)-(n-1){\mathrm{ct}}_k(t)\le 0$ on $I$; by step 1.1 this is the derivative of $\log q_v$, so $\log q_v$ is nonincreasing on $I$ (its derivative is continuous there by [F2]), and so is $q_v=\exp (\log q_v)$. Hence for $0<s<t$ in $I$, $q_v(t)\le q_v(s)\text{ and, passing to the limit }s\downarrow 0,q_v(t)\le q_v(0+)=1$ by step 1.3; equivalently $J_p(t,v)\le {\mathrm{sn}}_k(t{)}^{n-1}$. The interval is exactly the set on which the two factors are defined and positive: $J_p$ and the Riccati operator are defined for $t<c_p(v)\le \tau$, and for $k>0$ the model sine vanishes at the pole $\pi \mathrm{/}\sqrt{k}$, where the quotient is undefined; beyond it the radial model chart is no longer the stated positive-domain model; for $k\le 0$ the model factor is positive on all of $(0,\mathrm{\infty })$ and only the cut time bounds the interval. The value at $t=0$ itself is excluded, since $J_p(t,v)\sim t^{n-1}$ vanishes there; only the limit $q_v(0+)=1$ is asserted, and at the cut time and beyond no value of $q_v$ is defined. In dimension $n=2$ the exponent is $n-1=1$ and the density is $q_v=J_p\mathrm{/}{\mathrm{sn}}_k$; no step of the argument changes, the trace being one-dimensional. No completeness of $M$ beyond the quoted suppliers, no compactness, and no choice beyond the inherited [A1] is used.