Pointwise convergence of generalized inverse function

Question

I am reading Resnick's Extreme Values, Regular Variation, and Point Processes. In chapter 0.2 he writes about the generalized inverse of a non-decrasing function F: $$F^{\leftarrow}(y):=\inf\{x:F(x)\ge y\}.$$ In Proposition 0.1 he proofs, that for $F_n,G$ non-decrasing with $\lim\limits_{n\to\infty}F_n(x)=G(x)$ for all $x$, where $G$ is continous we have $\lim\limits_{n\to\infty}F_n^{\leftarrow}(y)=G^{\leftarrow}(y)$ for all $y$, where $G^{\leftarrow}$ is continous. This is all clear. I want to show the converse, but I struggle with the details. Here is what I have: Given $F_n,G$ non-decrasing and $\lim\limits_{n\to\infty}F_n^{\leftarrow}(y)=G^{\leftarrow}(y)$ it follows from what we already know, that $$\lim\limits_{n\to\infty}(F_n^{\leftarrow})^{\leftarrow}(x)=(G^{\leftarrow})^{\leftarrow}(x),$$ i.e. this problem reduces to showing that $(F^{\leftarrow})^{\leftarrow}(x)=\inf\{y:F^{\leftarrow}(y)\ge x\}=F(x)$. For this we have to show two things: 1. $F(x)$ is a lower bound of the set $A(y):=\{y:F^{\leftarrow}(y)\ge x\}$, i.e. $F(x)\le y$ for all $y$ with $F^{\leftarrow}(y)\ge x$ and 2. $F(x)$ is the greatest upper bound of $A(y)$. How do I show 1. and 2.?

Answer 1

This answer has three parts: first I show that $(F^{\leftarrow})^{\leftarrow}(x) = F(x)$ need not always hold. Next I develop the fact that $(F^{\leftarrow})^{\leftarrow}(x) = F(x^-)$. Finally I use the approach in the question and this fact to argue that the statement you had set out to show does indeed hold true.

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Consider the function $$ F(x) = \begin{cases} 1 & x > 0 \\ 1/2 & x = 0 \\ 0 & x < 0\end{cases}.$$ Then we have $$ F^{\leftarrow}(y) = \begin{cases} \infty & y > 1\\ 0 & 0 < y \le 1 \\ -\infty & y \le 0\end{cases}.$$

Thus, $$ (F^{\leftarrow})^{\leftarrow}(x) = \begin{cases} 1 & x > 0 \\ 0 & x \le 0 \end{cases} .$$ In particular, note that $(F^{\leftarrow})^\leftarrow (0) \neq F(0)$. In effect, we've lost information about the atypical value $F(0)$ in doing the generalised inverse. Clearly nothing is speacial about $1/2$ - the same story holds as long as $F(0) > 0$.

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What is, however, true is that in the way you've defined $F^{\leftarrow},$ $$ (F^{\leftarrow})^{\leftarrow}(x) = F(x^-) := \lim_{z \nearrow x} F(z).$$

I first claim that for every $\varepsilon > 0, F(x^-) + \varepsilon \in \{ y : F^{\leftarrow}(y) \ge x\}$. This will hold if $\inf\{z : F(z) \ge F(x^-) + \varepsilon\} \ge x$. But we know that for any $u < x, F(u) \le F(x^-) < F(z),$ so $z$ cannot lie in $(-\infty, x)$.

But then this means that $( F(x^-), \infty) \subset \{y : F^{\leftarrow}(y) \ge x\},$ which means that $(F^{\leftarrow})^\leftarrow(x) \le F(x^-)$.

Next, we show that $(F^{\leftarrow})^{\leftarrow}(x) \ge F(x^-)$. For this, observe that if $F^{\leftarrow}(y) \ge x,$ then $y \ge F(x^-)$. Indeed, if $y < F(x^-),$ then there exists some $\varepsilon > 0$ such that $y \le F(x - \varepsilon).$ But then $x - \varepsilon \in \{z : F(z) \ge y\} \implies F^{\leftarrow}(y) \le x -\varepsilon < x.$ This of course means that $\inf \{y : F^{\leftarrow}(y) \ge x\} \ge F(x^-)$, and the claim follows.

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To come back to your original question, your argument tells us that at continuity point of $G$, it holds that $F_n(x^-) \to G$. Given this, it suffices to show the following claim:

Suppose for a sequence of nondecreasing functions $F_n$, and a nondecreasing function $G$, at any continuity point $x$ of $G$, it holds that $F_n(x^-) \to G(x)$. Then at every such $x, F_n(x) \to G(x)$ as well.

This statement does hold, as I argue below.

Fix any $\varepsilon > 0.$ We want to show that for $n$ large enough, $F_n(x) \in G(x) \pm \varepsilon$. Since $x$ is a continuity point of $G,$ we can find a $\delta$ so that $G(x \pm \delta) \in G(x) \pm \varepsilon/2$. Further, $(x, x+\delta)$ must contain at least one continuity point of $G$ (in fact many more: the set of discontinuities of nondecreasing functions is countable). Pick such an $x_U \in (x, x+\delta).$

Since $x_U$ is a continuity point of $G$, we know that $F_n(x_U^-) \to G(x_U).$ Also, since $x < x_U, F_n(x) \le F_n(x_U^-)$. So for $n$ large enough, we have $$ F_n(x) \le F_n(x_U^-) \le G(x_U) + \varepsilon/2 \le G(x) + \varepsilon.$$

Further, since $F_n(x^-) \to G(x),$ we know that for large enough $n,$ $F_n(x) \ge F_n(x^-) \ge G(x) - \varepsilon$. QED.