**Statement** Given a finite abelian group $G$ and two independent random variables $X,Y$ taking values in $G$ and satisfying $d\_{TV}(X,U\_G)\leqslant \delta$ and $d\_{TV}(Y,U\_G)\leqslant \delta$ (where $U\_G$ denotes a uniformly distributed over $G$ and $d\_{TV}$ is the total variation distance), we ask how close is $X+Y$ to $U$? One can show that


$$ d\_{TV}(U\_G,X+Y) \leqslant 2 \cdot d\_{TV}(X,U\_G)d\_{TV}(Y,U\_G) $$


the proof is elementary yet a little bit complex, it goes by reduction to the problem of finding the winning probability in a certain game. The constant $2$ is optimal over $G=\mathbb{Z}\_2$.


**Question** Do you have any ideas what theory could be applied here, to obtain an alternative proof and a better constant for certain group (like $G=\mathbb{Z}\_p$)? 


We assume only that the variation distance are small, cannot impose any restrictions about eigen values, second norms etc. Fourier Analysis seems to be not a good choice here, as the characters for $\mathbb{Z}\_p$ are not compatible with $\{-1,1\}$-valued functions that computes the variation distance (this is not the case of $G=\mathbb{Z}\_2^{n}$ though). I was thinking also about applying Markov Chain Theory, e.g. minorizing condition but given that conditions it is not satisfied in general for any measure.


**Edit** In applications I am thinking of $\delta >> \frac{1}{|G|}$



Convergence rate of the convolution of almost uniform measures on $\mathbb{Z}_p$
I think in general you can't expect anything better than $2$ regardless of the group. 


Saying $d\_{TV}(X, U\_G) \leq \delta$ is equivalent to saying we can write the measure corresponding to $X$ as 
$$\mu\_X=\mu\_U + (\mu\_1 - \mu\_2),$$
where $\mu\_U$ is the uniform measure, and $\mu\_1$ and $\mu\_2$ are positive measures having equal total mass at most $\delta$. Similarly, we can write 
$$\mu\_Y=\mu\_U + (\mu\_3 - \mu\_4).$$
Since $\mu\_1$ and $\mu\_2$ have the same total mass, $\mu\_U \* (\mu\_1-\mu\_2)$ is $0$. Similarly, $\mu\_U \* (\mu\_3-\mu\_4)=0$. So when we expand out the convolutions, we get 
\begin{eqnarray\*}
\mu\_X \* \mu\_Y &=& \mu\_U + (\mu\_1-\mu\_2) \* (\mu\_3 - \mu\_4) \\
&=&\mu\_U + (\mu\_1 \* \mu\_3 + \mu\_2 \* \mu\_4) - (\mu\_1 \* \mu\_4 + \mu\_2 \* \mu\_3) \end{eqnarray\*}
Both of the parenthesized terms have total mass $2 d\_{TV}(X, U\_G) d\_{TV}(Y,U\_G)$, so we have that bound on the total variation distance. 


Equality holds if the two parenthesized terms have disjoint support. This can happen in any group if $\delta$ is sufficiently small -- for example, we can take $\mu\_1$ and $\mu\_3$ to be point masses on $0$, and $\mu\_2$ and $\mu\_4$ to be point masses on some nonzero $g$ (corresponding to $$P(X=0)=P(Y=0)=1/|G|+\delta, P(X=g)=P(Y=g)=1/|G|-\delta,$$ and elsewhere uniform). 


In general maximizing the distance of $X+Y$ from uniform corresponds to minimizing the overlap between the two sums of convolutions. 


