Let $G$ be a (discrete) torsion free group with identity $e$. Recall that for an element $\alpha=\sum a_gg$ in $\mathbb C[G]$ (complex functions on $G$ with compact support), $\alpha^*$ is defined to be $\sum\bar{a_g}g^{-1}$ and for $\beta=\sum b_gg\in\mathbb C[G]$, we have the (convolution) product $\alpha\beta:=\sum a_gb_hgh$. Let $\alpha=\alpha^*$ be an element in $\mathbb C[G]$, define $F(\alpha):=a_e-\sum_{g\neq e}|a_g|$. Consider the sequence $(A_n)$ defined by $$A_n:=F\left(\alpha^{2n}\right)\quad(n\in\mathbb N)$$ Does the sequence $(A_n)$ contain a nonnegative number?
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asked Jun 17, 2018 at 15:21
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$\begingroup$ According to your definition, $F(g)=-1$, yes? So $F(g^{2n}=-1$. Am I missing some thing? $\endgroup$– Ali TaghaviJun 17, 2018 at 19:11
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1$\begingroup$ @AliTaghavi I think you did not note the condition $\alpha=\alpha^*$. $\endgroup$– Meisam Soleimani MalekanJun 17, 2018 at 19:16
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$\begingroup$ Yes. I am sorry. $\endgroup$– Ali TaghaviJun 17, 2018 at 19:22
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1$\begingroup$ Let $G$ be the additive grp of the integers and $\alpha$ the fnc $\mathbf Z \to \bf C$ mapping $1$ to $i$, $-1$ to $-i$, and $\mathbf Z \setminus \{\pm 1\}$ to $0$. Then it seems to me that $\alpha = \alpha^*$ and $\alpha^2$ is the fnc $\mathbf Z \to \bf C$ with support in $\{0, \pm 2\}$ sending $0$ to $2$ and $\pm 2$ to $-1$. So $F(\alpha^2) = 2 - 2 \ge 0$. $\endgroup$– Salvo TringaliJun 17, 2018 at 19:41
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1$\begingroup$ @AliTaghavi I don't know, I was just trying to answer the question in the OP. Also, I'm not sure I can interpret the notation of your comment. What does $x_Y^{-1}$ stand for? Is $Y$ a misprint for $y$? $\endgroup$– Salvo TringaliJun 17, 2018 at 19:58
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No in general. If $G$ is an infinite cyclic group generated by $g$, and $\beta=1 +g+1/g$, the sum of coefficients of $\beta^{2 n}$ equals $9^n$, and the coefficient of 1 is $9^n$ times the probability that a symmetric random walk with steps $0,\pm 1$ ends up in the origin after $2n$ steps. This probability tends to 0. Thus certain large power $\alpha=\beta^N$ is a counterexample.
answered Jun 17, 2018 at 20:33
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$\begingroup$ It is not needed to consider that $G$ is a cyclic group! $\endgroup$ Jun 17, 2018 at 21:31