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Consider the quantile functional: $$Q_\alpha(F) = \inf \{t : F(t) \geq \alpha\}$$ A functional $\gamma$ is continuous in sup norm at $F$ if for all $\epsilon >0$ there exists a $\delta > 0$ such that: $$sup_x|F(x) - G(x)| \leq \delta \implies |\gamma(F) - \gamma(G)| \leq \epsilon$$

What I've tried

I claim that the quantile functional is not continuous in sup norm at $F$ unless $F$ is right-continuous everywhere.

Consider:

$$|Q_\alpha(F) - Q_\alpha(G)| = |\inf \{t : F(t) \geq \alpha\} - \inf \{t : G(t) \geq \alpha\}|$$

We can upper bound this difference by rewriting the second term as: $\inf_{t, \eta} \{t : F(t) + \eta \geq \alpha, |\eta| \leq \delta\}$. Unless $F$ is right-continuous everywhere the difference can be arbitrarily large.

Usually we'd restrict ourselves to distribution functions (which are right-continuous).

I could use some feedback on my reasoning as well as any other approaches.

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You didn't mention the set of functions you're considering but the problem is when you consider a function like $ F_{\epsilon}(x) = (x+1)x(x-1) - \epsilon $. Clearly that function has sup-distance $\epsilon$ to $F(x) = (x+1)x(x-1)$ but if $\alpha = F\left(-\frac{1}{\sqrt{3}}\right) = \frac{2}{3 \sqrt{3}}$ (which is a local maximum for F), $Q_{\alpha}(F) = -\frac{1}{\sqrt{3}}$ but $Q_{\alpha}(F_{\epsilon}) \ge \frac{1}{\sqrt{3}}$. This means $Q_{\alpha}$ cannot possibly be continuous at $F$.

A plot to help visualizing: https://www.wolframalpha.com/input/?i=plot%28%28x-1%29x%28x%2B1%29%2C+%28x-1%29x%28x%2B1%29+-+0.1%3B+2%2F%283+sqrt%283%29%29%29

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  • $\begingroup$ I hadn't intended to restrict the functions I was considering, but I was more focused on the case of increasing functions. The claim comes from the condition on $G$. $\endgroup$
    – dmh
    Commented Jun 7, 2020 at 22:55
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    $\begingroup$ Gotcha. For an example with an increasing function, consider $F(x) = 0$ if $x < 0$, $F(x) = x^2$ if $x \ge 0$ and $F_{\epsilon}(x) = F(x) - \epsilon$. Then $Q_{0}(F) = -\infty$ (or most likely the lower bound of your domain). However $Q_{0}(F_{\epsilon}) = \epsilon$. $\endgroup$
    – MBW
    Commented Jun 7, 2020 at 23:02
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    $\begingroup$ Looking at the way you stated, when considering $F(t) + \eta \ge \alpha$ the set $F(t) = \alpha$ can consist of discrete points or some intervals where the function is constant in such a way that adding any perturbation to $F$ will remove those points and intervals instead of shifting them. These can easily be created with local maxima or discontinuities $\endgroup$
    – MBW
    Commented Jun 7, 2020 at 23:11

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