A function $p \in C^2\big([0,R),\mathbb{R}\big)$ satisfies $p(x) ≥ \alpha$ and $(1 + \beta)[p'(x)]^2\leq p''(x)p(x).$ The following is exercise IV.3.1. from Analysis I by Amann and Escher.


  Suppose that $\alpha, \beta, R &gt; 0$ and $p \in C^2\big([0,R),\mathbb{R}\big)$ satisfy $p(x) ≥ \alpha$ and $$(1 + \beta)[p'(x)]^2\leq p''(x)p(x),\quad x\geq0.$$ Show that $R &lt; \infty$ and $p(x)\to\infty$ as $x\to R−$. (Hint: The function $p^{-\beta}$ is concave. Use a tangent line to $p^{-\beta}$ to provide a lower bound for $p$ (see Application 3.9(e)).


I think I found a counterexample. Let $\alpha:=\beta:=1$ and $p:[0,\infty)\to\mathbb{R}$ with $$p(x):=\frac{x+2}{x+1},\quad x\geq0.$$ Then $p'(x)=-1/(1+x)^2$, $p''(x)=2/(1+x)^3$, so $$(1 + \beta)[p'(x)]^2=\frac{2}{(1+x)^4}=\frac{2}{(1+x)^3}\frac{1}{1+x}\leq\frac{2}{(1+x)^3}\frac{2+x}{1+x}=p''(x)p(x)$$ for all $x\geq0$. Now $p(x)\geq1$, $R=\infty$ and $p(x)\to1$ as $x\to R-$.

Maybe I'm missing something. The question has been asked here but there are no answers yet.
I wrote an email to the author of this series and the following is the screenshot of the email I received:

Since in email system, it can not show Latex, I will type it down for convenience:
$0\leq (1+\beta)p'(x)\leq p''(x)p(x)$ for $0\leq x\leq R$.
In addition, you have to add “unless p is constant” at the end of the assertion.
Now you should be able to do the exercise.
Although the question was proposed three years ago, I hope this would help you somehow.
I wrote an email to the author of this series and the following is the screenshot of the email I received:

Since in email system, it can not show Latex, I will type it down for convenience:
$0\leq (1+\beta)p'(x)\leq p''(x)p(x)$ for $0\leq x\leq R$.
In addition, you have to add “unless p is constant” at the end of the assertion.
Now you should be able to do the exercise.
Although the question was proposed three years ago, I hope this would help you somehow.