Let work with the space where distributions and its inverses exist. For any distribution $F$ its inverse for any $p\in(0,1)$ is $F^{-1}(p)=\inf \{t\in \mathbb{R} : F(t) \geq p \}.$
Continuity definition: For all $\epsilon>0$, there exists a $\delta>0$ such that the following holds $$\forall x~|F(x)-G(x)|\leq\delta \implies |F^{-1}(p)-G^{-1}(p)| \leq \epsilon $$ for some $p\in(0,1)$.
I am not able to characterize the distance between inverse function assuming the distributions to be close. Any hints to solve this?
My try:
We can have $F(x)=1_{x\geq 0}$ and $G(x)=(1-\delta) 1_{x\geq 0} + \delta 1_{x\geq 1}$. This ensures $\sup_x |F(x)-G(x)| \leq \delta$. But for some $p\in(1-\delta,1)$ we have $F^{-1}(p)=0$ and $G^{-1}(p)=1$ which says $|F^{-1}(p)-G^{-1}(p)|=1.$
So, does this suffice to say that $F^{-1}$ is not satisfying this continuity? I am having trouble with the negative of this statement.