I need some confirmation for my reasoning in this exercise. This is the first exercise of Analysis I of Amann and Escher, on page 347.
Suppose that $\alpha,\beta,R>0$ and $p\in C^2([0,R),\Bbb R)$ satisfy $$p(x)\ge\alpha,\quad\text{ and }\quad(1+\beta)[p'(x)]^2\le p''(x)p(x),\quad\forall x\ge0$$ Show that $R<\infty$ and $p(x)\to\infty$ when $x\to R-$. (Hint: the function $p^{-\beta}$ is concave. Use a tangent line to $p^{-\beta}$ to provide a lower bound of $p$.)
From the inequalities given we knows that $p$ is convex because $p''\ge 0$. After a bit of algebra and using the hint I get the inequality
$$0< \frac{\alpha}{p(y)}\le\frac{p(x)}{p(y)}\le\left(1-\beta\frac{p'(x)}{p(x)}(y-x)\right)^{1/\beta},\quad\forall x,y\in[0,R)\tag{1}$$
First part: observe that if we suppose that $R=\infty$ then $p(x)=C\ge\alpha,\forall x\in[0,\infty)$ is a solution, contradicting what is supposed that $R$ must be. Please correct me if Im wrong here.
Second part: now suppose that $R<\infty$, then using (1) and taking limits both sides we get
$$\frac{p(x)}{\lim_{y\to R^-}p(y)}\le\left(1-\beta\frac{p'(x)}{p(x)}(R-x)\right)^{1/\beta},\quad\forall x\in[0,R)\tag{2}$$
Now suppose that $\lim_{y\to R}p(y)=M>\alpha$ finite, then we get from (2)
$$\alpha\le p(x)\le M\left(1-\beta\frac{p'(x)}{p(x)}(R-x)\right)^{1/\beta},\quad\forall x\in[0,R)$$
and from here I cant see a reason to say that this cannot be possible for suitable choices of $\alpha,\beta,p$. Can you help me? Thank you.
UPDATE: I deduced the inequality
$$p(x)+p'(y)(y-x)\ge p(y)\ge\left(\frac{p(x)^{\beta+1}}{p(x)-\beta p'(x)(y-x)}\right)^{1/\beta},\quad\forall x,y\in[0,R)\tag{3}$$
from the concavity of $p^{-\beta}$ and the convexity of $p$. From here is easy to check that for $R<\infty$ if $\lim_{y\to R^-}p(y)=\infty$ then this imply, in the LHS, that it must be the case that $\lim_{y\to R^-}p'(y)=\infty$, i.e.
$$p(x)+\lim_{y\to R^-}p'(y)(R-x)\ge\infty\ge\left(\frac{p(x)^{\beta+1}}{p(x)-\beta p'(x)(R-x)}\right)^{1/\beta},\quad\forall x\in[0,\infty)$$
then taking limits again for $x\to R$ we have
$$0\le\lim_{x\to R^-}\frac{p(x)^{\beta+1}}{p(x)-\beta p'(x)(R-x)}\le\infty$$
If Im not wrong, this last inequality imply that
$$\lim_{x\to R^-}[p(x)-\beta p'(x)(R-x)]\ge 0$$
but I get stuck here. The other case (considering that $\lim p(y)<\infty$) seem more complicate to analyze/solve.
UPDATE 2: it seems that the exercise is wrong, observe that if we have $p(x)=be^{ax}$ then the inequality of the exercise become
$$(1+\beta)[abe^{ax}]^2\ge be^{ax}\cdot a^2be^{ax}\implies 1+\beta\ge 1$$
what is true for the conditions of the exercise if $b>0$, but then $R=\infty$ and $\lim_{x\to\infty} p(x)=\infty$ for $a>0$, or $\lim_{x\to\infty} p(x)=b$ for $a=0$, or $\lim_{x\to\infty} p(x)=0$ for $a<0$.
If we take functions as $\tan(x+b)$ with $b\in(0,\pi/2)$ or $1/(1-x)$ they fit for some $R<\infty$ but we need that $\beta\ge 1$.