We say the functional $\gamma$ is continuous if for all $\varepsilon>0$ and two cdfs $F$ and $G$, there exists a $\delta>0$ such that $\|G-F\|_\infty=\sup_{t\in\mathbb{R}}|G(t)-F(t)|\leq\delta$ implies $|\gamma(G)-\gamma(F)|\leq\varepsilon$. Now try to prove or disprove the continuity of mean and quatile functional, which are
- mean: $F\to\int xdF(x)$;
- quantile: $Q_{\alpha}(F)=\inf \{t\in\mathbb{R}|F(t)\geq \alpha\}$.
Intuitively, I think mean functional is continuous however I don't know how to bound the integral as it contains the derivative and then control it via the sup-norm $\|\cdot\|_\infty$ between $F$ and $G$. While quantile seems to be more complicated and I wonder if there exist some tricks to transfer the definition to a more direct form with respect the the sup-norm. Thanks for the reading.:)
For the mean, you can consider the distribution functions
$$ F=1_{[0,\infty)} \quad \text{and} \quad F_n= \frac{1}{n}1_{[-n,\infty)}+\frac{n-1}{n}1_{[0,\infty)} \text{ for all $n\in\Bbb N$},$$
where $1_A$ denotes the indicator function of the set $A$, i.e. $1_A(x)=1$ for $x\in A$ and $1_A(x)=0$ for $x\notin A$. Then $\|F-F_n\|_\infty\to0$ for $n\to\infty$, but
$$ \lim_{n\to\infty} \int_\Bbb R xdF_n(x)= \lim_{n\to\infty} -1 \neq 0 = \int_\Bbb R xdF(x).$$
Hence, the mean functional is not continuous.
The quantile functional on the other hand should be continuous.