Let's have $X,Y \in \mathbb{R}$ with probability measures $\mu, \nu$, then the Kolmogorov-Smirnov distance is defined as follows $$ d_K(X,Y)=\underset{x \in \mathbb{R}}{sup}\{|F_X(x) - F_Y(x)|\} $$
where $F_X(x)$ is the comulative distribution function of X.
If $X,Y$ have finite p-momentum then the p-Wasserstein distance is defined as follows $$d_{W_p}(X,Y) = \Biggr(\underset{\pi \in \mathcal{J}(\mu, \nu)}{inf}\int|x-y|^p d\pi(x,y)\Biggr)^{\frac{1}{p}}$$
where $\mathcal{J}(\mu, \nu)$ denote all joint distribution $\pi$ for $(X,Y)$ that have marginals $\mu$ e $\nu$.
In this paper (arXiv link) it tells
if Y is a real-valued random variable with Lebesgue density bounded above by C > 0, then for any real-valued random variable X
$$ d_K(X,Y) \leq \sqrt{2Cd_{W_1}(X,Y)} $$ and in this paper (arXiv link) I've found that, when $Y \sim \mathcal{N}(0,1)$, $C = 2$, so $$ d_K(X,Y) \leq 2\cdot\sqrt{d_{W_1}(X,Y)} $$
My questions is:
- What is the value of $C$ when $Y$ is a centered non standard gaussian so $Y \sim \mathcal{N}(0,\sigma^2)$? (it should be the upper bound to the Lebesgue density of Y, but I don't know what the Lebesgue density is and how to compute it for the centered non standard gaussian distribution)