Small total variation distance between sums of random variables in finite Abelian group implies close to uniform?

Question

Let $\mathbb{G} = \mathbb{Z}/p\mathbb{Z}$ (where $p$ is a prime). Let $X,Y,Z$ be independent random variables in $\mathbb G$. For a small $\epsilon$ we have $\operatorname{dist}_{TV}(X+Y,Z+Y)<\epsilon$. Assuming $X$ and $Z$ are disjoint, I'd like to express $\operatorname{dist}_{TV}(Y,U)$ in terms of $\epsilon$, where $U$ is a variable distributed uniformly over $\mathbb G$ and TV is total variation distance, that is $$\operatorname{dist}_{TV}(A,B)=\max_D\left|\Pr(A\in D)-\Pr(B\in D) \right|$$

Something similar was done by Prof. Terrence Tao but with entropy-based distances: see Proposition 1.3 of https://arxiv.org/abs/2306.13403.

I wish to find, if possible, an upper bound for $\operatorname{dist}_{TV}(Y,U)$ that does not depend on $|\mathbb G|$ or the size of the support of $X,Y,Z.$

Thank you very much to anyone willing to help!

Answer 1

I do not think such a bound is possible. Let $p$ be large, $X$ be uniform on the set $A = \{0,2, 4, \ldots, p-3\}$, and $Z$ uniform on the set $A+1$. Moreover, let $Y = \{0, 1\}$ with equal probability.

Then $X + Y$ is uniform on the set $\{0, \ldots p-2\}$, and $Z+Y$ is uniform on $\{1, \ldots, p-1\}$. So $d_{TV}(X+Y, Z+Y) = O(1/p)$, but clearly $Y$ is very far from a uniform distribution.