Let $F$ be a cumulative distribution function such that there exists a positive and measurable function $\beta$ with $$\lim_{y \uparrow x^*} \frac{\overline{F}(y + x\beta(y))}{\overline{F}(y)} = \begin{cases} (1 + \xi x)^{-1/\xi}, & \xi \neq 0 \\ \exp(-x), & \xi = 0 \end{cases}$$ for every $x \geq 0$ satisfying $1+\xi x > 0$ where $\overline{F}(x) = 1-F(x)$ is the tail distribution function.
$x^* = \sup\left\lbrace x: F(x) < 1 \right\rbrace$ shall be the right corner of our distribution and let $F$ be continous in $x^*$ if $x^*$ is finite.

Now I want to prove that $$\lim_{n \to \infty} n\left(1-F\left(a_n\right)\right) = 1$$ holds for the pseudoinverse $a_n = \inf \left\lbrace x: F(x) \geq 1 - \frac{1}{n} \right\rbrace$.

It is clear that $\limsup_{n \to \infty} n\left(1-F\left(a_n\right)\right) \leq 1$ holds by definition of $a_n$. So it would be sufficient to prove the other inequality.

I know how to prove this statement for $x < 0$. There one has $a_n + x\beta(a_n) < a_n$ which leads to the statement by letting $x$ go to zero.


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