# Is this basis a Schauder basis?

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
• @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