Speed of convergence of complementary probabilities on finite group

Abstract

Let function P be a probability on a finite group G, i.e. $P(g)\geq0\ $ $(g\in G),\ \sum\limits_{g}P(g)=1$ (we write $\sum\limits_{g}$ instead of $\sum\limits_{g\in G})$. Convolution of two functions $P, \; Q$ on group $G$ is \linebreak $ (P*Q)(h)=\sum\limits_{g}P(g)Q(g^{-1}h)\ \ (h\in G)$. Let $E(g)=\frac{1}{|G|}\sum\limits_{g}g$ be the uniform (trivial) probability on the group $G$, $P^{(n)}=P*...*P$ ($n$ times) an $n$-fold convolution of $P$. Under well known mild condition probability $P^{(n)}$ converges to $E(g)$ at $n\rightarrow\infty$. A lot of papers are devoted to estimation the rate of this convergence for different norms. Any probability (and, in general, any function with values in the field $R$ of real numbers) on a group can be associated with an element of the group algebra of this group over the field $R$. It can be done as follows. Let $RG$ be a group algebra of a finite group $G$ over the field $R$. A probability $P(g)$ on the group $G$ corresponds to the element $ p = \sum\limits_{g} P(g)g $ of the algebra RG. We denote a function on the group $G$ with a capital letter and the corresponding element of $RG$ with the same (but small) letter, and call the latter a probability on $RG$. For instance, the uniform probability $E(g)$ corresponds to the element $e=\frac{1}{|G|}\sum\limits_{g}g\in RG. $ The convolution of two functions $P, Q$ on $G$ corresponds to product $pq$ of corresponding elements $p,q$ in the group algebra $RG$. For a natural number $n$, the $n$-fold convolution of the probability $P$ on $G$ corresponds to the element $p^n \in RG$. In the article we study the case when a linear combination of two probabilities in algebra $RG$ equals to the probability $e\in RG$. Such a linear combination must be convex. More exactly, we correspond to a probability $p \in RG$ another probability $p_1 \in RG$ in the following way. Two probabilities $p, p_1 \in RG$ are called complementary if their convex linear combination is $e$, i.e. $ \alpha p + (1- \alpha) p_1 = e$ for some number $\alpha$, $0 <\alpha <1$. We find conditions for existence of such $\alpha$ and compare $\parallel p ^ n-e \parallel$ and $\parallel {p_1} ^ n-e \parallel$ for an arbitrary norm ǁ·ǁ.