Let $p$ and $q$ be probability distributions on a metric space $X=(X, d)$ with densities $dp$ and $dq$, such that there exists $0 < \alpha < \beta < \infty$ satisfying

$$ \alpha d p \le dq \le \beta dp . $$

What is an upper bound for the Wasserstein distance $W_d(p,q)$ ?

Notes: $W_d(p, q) := \sup_{\|f\|_{\text{Lip}} \le 1} |\mathbb E_{x \sim p}[f(x)] - \mathbb E_{x \sim q}[f(x)]|$

Update

It's well known (e.g see Gabe K's response below or Theorem 4 of this paper) that $W_d(p, q) \le \operatorname{diam}(X)\operatorname{TV}(p, q)$ Thus if $X$ has finite diameter, it suffices to bound $\operatorname{TV}(p, q)$.

Recall the definition of total variation,

$$ \operatorname{TV}(p, q) := \sup_{A \subseteq X} \left|\int_A dq-\int_A dp\right|. $$ Now, for any $A \subseteq X$, one has $ \int_A dq - \int_A dp \ge \int_A(\alpha-1)dp = (\alpha-1)p(A). $ Similarly, one has $\int_A dq - \int_A dp \le (\beta-1)p(A)$. Thus $$ \left|\int_A dq - \int_A dp \le (\beta-1)p(A)\right| \le \max(1-\alpha,\beta-1)p(A) \le \max(1-\alpha,\beta-1), $$ and so $\operatorname{TV}(p,q) \le \max(1-\alpha,\beta-1)$. Putting things together, we get

$$ W_d(p,q) \le \operatorname{diam}(X)\max(1-\alpha,\beta-1). $$