Let $\mathcal{P}$ be the space of probability measure on $\mathbb{R}.$ Define $d(\varphi,\phi)=\sup_x|\varphi(x)-\phi(x)|/(1+|x|),$ where $\varphi$ and $\phi$ are the characteristic functions of two measures $\mu$ and $\sigma$ from $\mathcal{P}$. Show that this a metric and equivalent to the Lévy metric (https://en.wikipedia.org/wiki/L%C3%A9vy_metric).
it's easy to see that $d$ is a metric since, $\mu=\sigma$ implies $d(\varphi,\phi)=0$ and if $d(\varphi,\phi)=0$ then $|\varphi(x)-\phi(x)| \leq (1+|x|)d(\varphi,\phi)=0,\forall x \in \mathbb{R}.$
We also have $d(\varphi,\phi)=d(\phi,\varphi)$ and $\forall x \in \mathbb{R},| \varphi(x)-\phi(x)| \leq |\varphi(x)-h(x)|+|h(x)-\phi(x)|,$ for three characteristic functions $\varphi,\phi,h$ and so we have $d(\varphi,\phi) \leq d(\varphi,h)+d(h,\phi).$
So my question is how to prove that $d$ is equivalent to the Lévy metric, denoted $d'$
Let $d(\phi_n,\phi) \to 0$. Then $\phi_n(x) \to \phi(x)$ for every $x$ and hence the corresponding measures converge weakly. This implies convergence in the Levy metric.
Conversely if $\phi_n \to \phi$ in the Levy metric then $\phi_n \to \phi$ uniformly on compact sets. Let $\epsilon >0$ and choose $M$ such that $\frac 1 {1+|x|} <\epsilon /2$ for $|x| >M$. Choose $m$ such that $|\phi_n(x)-\phi(x)| <\epsilon $ for $|x|\leq M$ and $n \geq m$. Then $d(\phi_n,\phi) \leq \epsilon$ for $n \geq m$.