Let $G$ be a torsion free group. Let $\alpha$ be an element in $\mathbb CG$, the group algebra of $G$, with $\|\alpha\|_1=1$ and assume that

  1. $\{1,\alpha,\alpha^2,\dotsc\}$ is linearly independent,
  2. $(\alpha^n)_{n\in\mathbb N}$ converges to 0 in strong operator topology, so in particular $\lim_n\|\alpha^n\|_2=0$.

Assume that $K$ is the closed linear span of $\{1,\alpha,\alpha^2,\dotsc\}$ in $\ell^2(G)$. Is $\{1,\alpha,\alpha^2,\dotsc\}$ a Schauder basis for $K$?

  • What is strong operator topology here? Do you consider the elements of the group algebra as operators? From where to where? – Fedor Petrov Nov 28 at 13:12
  • @FedorPetrov In the setting of my question, that's mean $\lim_n\|\alpha^n\beta\|_2=0$ for all $\beta\in\mathbb CG$. Yes, the elements of the group algebra can be considered as operator on $\ell^2(G)$. – Meisam Soleimani Malekan Nov 28 at 15:24
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    @FedorPetrov I think the reasoning here is that $L_\alpha : \ell^2(G) \to \ell^2(G)$ is a contraction and we are assuming its powers converge to zero in the SOT of ${\mathcal B}(\ell^2(G))$. However, I don't immediately see why this is enough to deduce that $\Vert \alpha^n\Vert_2 \to 0$ – Yemon Choi Nov 28 at 15:30
  • @YemonChoi neither do I. I see the opposite implication. – Fedor Petrov Nov 28 at 15:52

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