On the weak convergence of probability measures on $\mathbb R$

Question

Let $\mathcal P(\mathbb R)$ be the set of probability measures. Set for $\mu,\nu\in\mathcal P(\mathbb R)$

$$d(\mu,\nu) := \inf\left\{\varepsilon>0:~ F_{\mu}(x-\varepsilon)-\varepsilon \le F_{\nu}(x)\le F_{\mu}(x+\varepsilon)+\varepsilon,~ \forall x\in\mathbb R\right\}$$

and

$$\rho(\mu,\nu) := \sup\left\{\int fd\mu- \int fd\nu:~ f \mbox{ is } 1-\mbox{Lipschitz and uniformly bounded by } 1 \right\},$$

where $F_{\mu}$ (resp. $F_{\nu}$) denotes the cumulative distribution function of $\mu$ (resp. $\nu$). It is known that the weak convergence on $\mathcal P(\mathbb R)$ is equivalent to the convergence under $d$ and the convergence under $\rho$. Does there exist $C>0$ s.t.

$$\frac{d(\mu,\nu)}{C}~\le~ \rho(\mu,\nu)~\le~ Cd(\mu,\nu),\quad \forall \mu,\nu\in\mathcal P(\mathbb R)?$$

Answer 1

For every $\varepsilon\in(0,1)$ there are $\mu$, $\nu$ such that $d(\mu,\nu)=\varepsilon$ and $\rho(\mu,\nu)=\varepsilon^2$.

For example, $\mu=\varepsilon\delta_0 + (1-\varepsilon)\delta_2$ and $\nu=\varepsilon\delta_\varepsilon + (1-\varepsilon)\delta_2$.

Hence there is no $C>0$ such that $d(\mu,\nu)\leq C \rho(\mu,\nu)$ for all $\mu,\nu$.

(Addition to address GJC20's comment:) However, $\frac{2d^2}{2+d}\leq\rho\leq 3d$; see Theorem 8.10.43 in Bogachev, Measure Theory (Springer 2007).