Disclaimer. I've moved this question from MO hoping that here is the right venue. Also, this is my first post on this channel, so please have some patience.
So, Iet $X = (X,d)$ be a Polish space, equipped with the Borel sigma-algebra and $P_0,P_1$ be probability distributions on $X$. Let $\mathcal M(X,X)$ be the measurable maps $X\rightarrow X$, and for each $T \in \mathcal M(X,X)$, let the measure $T_{\#}P_k$ be the push-forward of $P_k$ by $T$. For $\varepsilon \ge 0$, let $\mathcal T_\varepsilon$ be the set of all $T \in \mathcal M(X,X)$ such that $\sup_{x \in X}d(T(x),x) \le \varepsilon$. Define $\alpha_\varepsilon$ via the optimization problem
$$ \alpha_\varepsilon := \min_{T \in \mathcal T_\varepsilon} TV(T_{\#}P_0, T_{\#}P_1). $$
Note that by the data-processing inequality, it holds that $$ 0 \le \alpha_\varepsilon \le TV(P_0,P_1) \le 1. $$ Unfortunately, this bound is presumably very loose as it is independent of the parameter $\varepsilon$.
Question
- Is there a nontrivial upper bound on $\alpha_\varepsilon$ in terms of $P_0$, $P_1$ and $\varepsilon$ ? My wild guess (and hope!) is something of the form "$\alpha_\varepsilon \le g(\varepsilon)TV(P_0,P_1)$", where $g_\varepsilon:[0,1]\rightarrow [0, 1]$ is a nontrivial function such that $g_\varepsilon(t) < t$ for $t \in (0, 1)$.
- Same question for the special case $X=(\mathbb R^p,\|\cdot-\cdot\|_\infty)$. Note that in this case, $\alpha_\varepsilon$ can be written as $$ \alpha_\varepsilon := \inf_{T_1,\ldots,T_p \in \mathcal M(\mathbb R^p,\mathbb R) \text{ s.t } x_j - \varepsilon \le T_j(x) \le x_j + \varepsilon\;\forall j \in [\![p]\!]} TV(T_{\#}P_0,T_{\#}P_1) $$
- Same question for the Hamming cube $X = (\{0, 1\}^p, d_{\text{H}})$. In this case, we can rewrite $$ \alpha_\varepsilon = \inf_{T:\{0, 1\}^p \rightarrow \{0, 1\} \text{ s.t } \sum_{j \mid T(x)_j \ne j} 1 \le \varepsilon,\;\forall x \in \{0, 1\}^p} TV(T_{\#}P_0,T_{\#}P_1) $$