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Let $g$ be a group element, let $G$ be the group generated by $g$, and let $\mathbb C G$ be the group algebra on $G$.

If we define $\tau:G\to\mathbb C$ as \begin{align*}\tag{1} \tau[g^n]=\begin{cases} 1&\text{if }n=0;\text{ and}\\ \alpha_n&\text{otherwise}, \end{cases} \end{align*} where $(\alpha_n:n\in\mathbb N)\subset\mathbb C$ are completely arbitrary, then $\tau$ can be extended to a normalized (i.e., $\tau[1]=1$) linear functional on $\mathbb C G$.

If, however, we would like $\tau$ to be positive (i.e., $\tau[pp^*]\geq0$ for all $p\in\mathbb C G$), there are certain restrictions on how we can choose the $\alpha_n$ in $(1)$ (for instance, we have to make sure that $\tau[g^n]=\overline{\tau[g^{-n}]}$ for all $n$).

In this context, the question I'm interested in is the following:

Question. How many of the $\alpha_n$ can be different from zero (and under what conditions)?

If $g$ has finite order $m\in\mathbb N$, then we can choose as many of $\alpha_1,\ldots,\alpha_{\lfloor m/2\rfloor}$ nonzero as we want, provided $|\alpha_1|,\ldots,|\alpha_{\lfloor m/2\rfloor}|$ are bounded above by a small enough quantity (this is mainly because, in this case, any element of $\mathbb C G$ can be written as $\beta_0+\beta_1g+\beta_2 g^2+\cdots+\beta_{m-1}g^{m-1}$).

However, if $g$ is of infinite order, I can't find an example of a positive $\tau$ with at least one $\alpha_n$ nonzero, but I can't show that all of the $\alpha_n$ need to be zero either. Is there a known characterization of positive linear functionals on such algebras?

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You are asking for the condition that a function $\tau$ on a cyclic group $G$ be (normalized by $\tau(e)=1$ and) positive-definite, i.e. $$ \tau(p^*p) = \sum_{i,j} \overline{p_i}p_j\,\tau(g^{-i}g^j)\geqslant 0 $$ for all $p = \sum_ip_ig^i\in\mathbf CG$ (finite sum). The answer is given by Bochner's theorem (due in this case to Herglotz, I believe) which says that such $\tau$'s are exactly the Fourier transforms of probability measures on the dual group $\smash{\hat G}$.

  • If $G$ has infinite order, we have $G \simeq \mathbf Z$ and $\hat G$ is the unit circle group $\mathbf T=\mathbf U(1)$, so there must be an (otherwise arbitrary) probability measure $\mu$ on $\mathbf T$ such that $$ \alpha_n=\tau(g^n) =\int_{\mathbf T}z^n\,d\mu(z). $$

  • If $G$ has order $m$, we have $G\simeq\mathbf Z_m$ and $\hat G$ is the subgroup $\simeq\mathbf Z_m\subset \mathbf T$ consisting of all $m$-th roots of 1, i.e. all $m$ powers of $\omega = e^{2\pi i/m}$; and there must be a probability measure $\mu$ on it, or in other words arbitrary nonnegative weights $\mu(\{\omega^k\})_{k=0}^{m-1}$ adding up to 1, such that $$ \alpha_n = \tau(g^n) =\sum_{k=0}^{m-1}\omega^{kn}\,\mu(\{\omega^k\}). $$

With that, it's easy to cook up examples where all $\alpha_n\ne0$. E.g. you can, in both cases, make all $\alpha_n=1$ by taking $\mu =$ Dirac measure at 1.

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