This is a proposition in a book I'm reading whose proof I am unsure of how to fill in the details


Suppose we have two smooth functions $\rho_{1,2} : (0,b) \to \mathbb{R}$ such that $$\rho_1'+ \rho_1^2 \leq \rho_2'+ \rho_2^2 $$ then $$\rho_2 - \rho_1 \geq \limsup_{t \to 0}\,(\, \rho_2(t) - \rho_1(t)\,)$$


The author says that this follows from the easily verified fact that the function $(\rho_2 - \rho_1) e^F$ is increasing where $F$ is the antiderivative of $\rho_2 + \rho_1$ on $(0, b)$. This would be true if $\rho_2 - \rho_1$ is increasing but I don't think that this is the case. How does the lim sup inequality follow from the fact that $(\rho_2 - \rho_1) e^F$ is increasing?

  • $\begingroup$ Which book? $ $ $\endgroup$
    – Dap
    Jan 25, 2019 at 4:53
  • $\begingroup$ Riemannian Geometry by Peter Petersen. In the third edition, the Riccati Comparison Principle is on page 254. $\endgroup$
    – Cookie
    Jan 29, 2019 at 3:08

2 Answers 2


The book is wrong and it's easy to find counterexamples. For example if $ρ_1(x)=1/(x+2)$ and $ρ_2(x)=1/(x+1)$ then $ρ_i'+ρ_i^2=0$ and $ρ_2(0)-ρ_1(0)=1/2$ but $ρ_2(1)-ρ_1(1)=1/6.$

The author may have been thinking of something like the following argument which I will paraphrase from "New comparison theorems in Riemannian geometry" by Y. Han and coauthors, Theorem 5.1. Suppose $ρ_1$ and $ρ_2$ are both $1/t+O(1).$ Define $ϕ_i(t)=t\exp(\int_0^t (ρ_i(s)-1/s)ds).$ Then each $ϕ_i$ is a $C^1$ function on $[0,b)$ (i.e. the right-sided derivative exists at zero - in fact it's $1$), smooth and strictly positive on $(0,b),$ with $ϕ_i(0)=0.$ Differentiating gives $ϕ'_i(t)=ρ_i(t) ϕ_i(t)$ and hence $ϕ''_i(t)=(ρ'_i(t)+ρ_i(t)^2)ϕ_i(t).$ This leads to $$(ϕ_2'ϕ_1-ϕ_1'ϕ_2)'=ϕ_2''ϕ_1-ϕ_1''ϕ_2=((ρ'_2+ρ_2^2)-(ρ'_1+ρ_1^2))ϕ_1ϕ_2\geq 0.$$ Using the fact that $ϕ_2'(0)ϕ_1(0)-ϕ_1'(0)ϕ_2(0)=0$ we get $ϕ_2'ϕ_1\geq ϕ_1'ϕ_2.$ Using our previous equation for $ϕ_i',$ this means $ρ_2(t)\geq ρ_1(t).$


Note that $$\begin{align}\frac d{dt}\left[(\rho_2-\rho_1)e^F\right]&=(\rho'_2-\rho'_1)e^F+(\rho_2-\rho_1)(\rho_2+\rho_1)e^F\\ &=e^F\left(\rho'_2-\rho'_1+\rho_2^2-\rho_1^2\right) \end{align}$$ and since $e^F>0$, your inequality implies that the derivative of $(\rho_2-\rho_1)e^F$ is positive. So the function $(\rho_2-\rho_1)e^F$ is increasing.

On the other hand, a careless approach with some possible errors could be this: $$\begin{align} \rho_1'+ \rho_1^2 \leq \rho_2'+ \rho_2^2 &\implies\rho_1'-\rho_2' \leq \rho_2^2- \rho_1^2\\ &\implies-\dfrac{\rho_1'-\rho_2'}{\rho_1-\rho_2}\ge\rho_1+\rho_2\\ &\implies-\ln|\rho_1-\rho_2|\ge F\\ &\implies\frac 1{|\rho_1-\rho_2|}\le e^F\implies(\rho_2-\rho_1)e^F\ge 1 \end{align}$$ or something like that....

  • 2
    $\begingroup$ Right, but how does the fact that $\rho_2 - \rho_1 e^F$ is increasing imply that $\rho_2 - \rho_1 \geq limsup _{t \to 0}(\rho_2(t) - \rho_1(t))$ $\endgroup$ Jan 9, 2017 at 13:29

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