I am trying to prove the following:
Let $h(x) = \frac{f(x)}{1-F(x)}$ the hazard rate of a distribution that has PDF $f(x)$ and CDF $F(x)$ and further, be $h$ positive and differentiable on $(x_0, \infty)$ for some $-\infty \leq x_0 < \infty$.
I need to prove the following implication $$\lim_{x\to\infty} \left(\frac{1}{h}\right)'(x) = \gamma > 0 \Longrightarrow \lim_{x\to\infty} \frac{h(x)}{(\gamma x)^{-1}} = 1,$$
i.e. that $h(x)$ is asymptotically equivalent to $(\gamma x)^{-1}$.
Alternatively can be shown that $\lim_{x\to\infty}xh(x) = 1/\gamma$ follows.
I have already tried using the proofs for Lemma 1.1 and Lemma 1.3 of "Von Mises Conditions Revisited" by Michael Falk and Frank Marohn, but I am not getting near a solution.
Any help is appreciated! Thank you!
A possible solution is the following approach:
Define the function $g(\cdot):=\frac{1}{h(\cdot)}: \left(x_0,\infty\right) \rightarrow (0,\infty)$ and be $\varepsilon > 0$.
Then, for this $\varepsilon$ exists a $x_\varepsilon > x_0 \geq -\infty$ such that for all $x \geq x_\varepsilon$ holds $\Vert g'(x) - \gamma \Vert < \varepsilon$, and therefore this holds further for any $x \geq x_\varepsilon \wedge 0$ [$:=\min(x_\varepsilon,0)$].
To prove the asymptotic equivalence, we need to show, that $\frac{g(x)}{x} = \frac{1}{xh(x)}$ converges to $\gamma$ for $x\rightarrow \infty$.
$$ \left\Vert \frac{g(x)}{x} - \gamma\right\Vert = \left\Vert \frac{g(x)-g(x_\varepsilon \wedge 0)}{x} - \gamma + \frac{g(x_\varepsilon \wedge 0)}{x}\right\Vert$$ $$ = \left\Vert \frac{1}{x} \left( \quad \int\limits_{x_\varepsilon \wedge 0}^x g'(u)\,du\,- \int\limits_{0}^x \gamma \, du\, + g(x_\varepsilon \wedge 0)\right)\right\Vert$$ $$ = \left\Vert \frac{1}{x} \left( \quad \int\limits_{x_\varepsilon \wedge 0}^x g'(u) - \gamma\,du\,- \int\limits_{0}^{x_\varepsilon \wedge 0} \gamma \, du\, + g(x_\varepsilon \wedge 0)\right)\right\Vert $$ $$\leq \frac{1}{x} \int\limits_{x_\varepsilon \wedge 0}^x \underbrace{\left\Vert g'(u) - \gamma\right\Vert}_{< \varepsilon} \, du \, + \frac{(x_\varepsilon \wedge 0)\cdot\gamma + g(x_\varepsilon \wedge 0)}{x}$$ $$< \varepsilon \cdot \frac{x-(x_\varepsilon \wedge 0)}{x} + \frac{(x_\varepsilon \wedge 0)\cdot\gamma + g(x_\varepsilon \wedge 0)}{x}$$ As we have $x_\varepsilon \wedge 0$ constant, $\frac{x-(x_\varepsilon \wedge 0)}{x} \to 0$ and $\frac{(x_\varepsilon \wedge 0)\cdot\gamma + g(x_\varepsilon \wedge 0)}{x} \to 0$ for $x \to \infty$ and therefore $$\overline{\lim\limits_{x\to\infty}}\left\Vert \frac{g(x)}{x} - \gamma\right\Vert \leq \varepsilon \text{ for any } \varepsilon > 0$$ and it the to be proven statement follows by taking $\varepsilon \to 0$.