The following is from Extreme Value Theory: An Introduction by de Haan.
We assume that $F^n(a_n x + b_n) \to G(x)$, where $F$ and $G$ are distribution functions. (Here, $F^n$ refers to $F \times F \times \ldots \times F$.
We can show that the above is equivalent to $\frac{1}{n(1-F(a_n x + b_n))} \to \frac{1}{-\log G(x)}$.
Now, we define the generalized inverse for non-decreasing function $f$ by $f^{-1}(x) = \int \{y \colon f(y) \geq x \}$.
We can show that if $f_n(x) \to f(x)$ for $x$ a continuous point of $f$, and if $x$ is a continuous point of $f^{-1}$, $f_n^{-1}(x) \to f^{-1}(x)$ too.
Now, we let $U$ be the generalized inverse of $\frac{1}{1-U}$. I want to show that $\frac{1}{n(1-F(a_n x + b_n))} \to \frac{1}{-\log G(x)}$ is equivalent to $\frac{U(nx)-b_n}{a_n} \to G^{-1}(e^{\frac{-1}{x}})$ for any $x>0$.
I understand the basic idea that the inverse of $\frac{1}{-\log G(x)}$ is $G^{-1}(e^{\frac{-1}{x}})$ and the same for $\frac{1}{n(1-F(a_n x + b_n))}$, but how do we show rigorously that $(\frac{1}{-\log G(x)})^{-1} = G^{-1}(e^{\frac{-1}{x}})$ and $(\frac{1}{n(1-F(a_n x+b_n))})^{-1} = \frac{U(nx)-b_n}{a_n}$?
Thank you.
Below are all the relevant section in the textbook.
