To show that 2 distances between probability measures are equivalent

Question

Here is the original V-total variation distance: $d_{V} (ν, \mu) = \sup_{φ:∥φ∥_{V}≤1}$|ν(φ)−$\mu(φ)$|, where $∥φ∥_{V}=\sup_{x}\frac{|φ(x)|}{1+V(x)}$. Now I also have a modified V norm: $∥φ∥_{V,σ}$=$\sup_{x}$ $\frac{|φ(x)|}{1+σV(x)}$ and therefore modified total variation distance: $d_{V,σ}$ (ν, $\mu$) = $sup_{φ:∥φ∥_{V,σ}≤1}$|ν(φ)−$\mu(φ)$|, for any σ>0. I can't understand why the two distances are equivalent. (I know they are equal when σ=1). Could anyone offer some intuitive explanations or mathematical proof or hints?

Answer 1

I suppose $V$ is a positive function.

If $\sigma \leq 1$ then $1+\sigma V(x) \leq 1+V(x) \leq C[1+\sigma V(x)]$ provided $C \geq 1$ and $C\sigma \geq 1$.

If $\sigma > 1$ then $1+V(x) \leq 1+\sigma V(x) \leq C'[1+ V(x)]$ provided $C' \geq 1$ and $C' \geq \sigma$.