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Let $G$ be a (discrete) torsion free group with identity $e$. Recall that for an element $\alpha\in\mathbb C[G]$, the set of complex functions on $G$ with finite support, $\alpha^*\in\mathbb C[G]$ is defined by $$\alpha^*(g):=\overline{\alpha(g^{-1})},$$ and for $\beta\in\mathbb C[G]$, we have the convolution product defined by $$\alpha\beta(g):=\sum_{xy=g}\alpha(x)\beta(y).$$

Let $\alpha\in\mathbb C[G]$ be nontrivial, i.e., the support of $\alpha$ contains at least two elements, and put $\beta=\alpha^*\alpha$. Assume that $$ \beta(e)=1,\quad\text{and that}\quad|\beta(g)|<1\quad\text{for all $g\neq e$.}$$ Do the following limits exist for all $g\neq e$? $$\lim_{n\to\infty}\frac{|\beta^n(g)|}{|\beta^n(e)|}$$

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  • $\begingroup$ Do you have any examples where the limit is neither $0$ nor $1$? $\endgroup$
    – user44191
    Commented Jun 27, 2018 at 21:37
  • $\begingroup$ @usser44191: No, I think these limits are always equal zero. $\endgroup$
    – MSMalekan
    Commented Jun 28, 2018 at 1:34
  • $\begingroup$ I think it is sometimes $1$; consider $\alpha$ the properly scaled indicator function on $0, 1 \in \mathbb{Z}$? $\endgroup$
    – user44191
    Commented Jun 28, 2018 at 3:48
  • $\begingroup$ @user44191 What do you mean by properly scaled indicator function on 0,1∈Z? $\endgroup$
    – MSMalekan
    Commented Jun 28, 2018 at 13:47
  • $\begingroup$ $\alpha = \frac{\sqrt{2}}{2}(x_0+x_1)$; then $\beta$ is $\frac{1}{2}(x_{-1}+2x_0+x_1)$, and $\beta^n$ is a centered binomial distribution, so the limit is $1$. $\endgroup$
    – user44191
    Commented Jun 28, 2018 at 14:31

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