Let $\Gamma$ be a discrete abelian group. We denote the group ring of $\Gamma$ by $\mathbb{C}(\Gamma)$, which is the set of all formal sums of the form $\sum_{s\in \Gamma}a_s s$, where only finitely many of the scalar coefficients $a_s \in \mathbb{C}$ are nonzero.The reduced $C^*-$algebra of $\Gamma$, denoted by $C_{\lambda}^*(\Gamma)$, is the completion of $\mathbb{C}(\Gamma)$ with respect to the norm $$ ||x||_r=||\lambda(x)||_{B(l^2(\Gamma))}$$ where $\lambda: \Gamma \to B(l^2(\Gamma))$ is defined by $\lambda(s)(\delta_t)=\delta_{st}, \forall s,t \in \Gamma$. Show that linear span of $\{\lambda_s: s\in \Gamma\}$ is dense in $C_{\lambda}^*(\Gamma)$.
Let $T \in C_{\lambda}^*(\Gamma)$. Then there exists $\{f_n\} \in \mathbb{C}(\Gamma)$ such that $$\|\lambda(f_n)-T\|_{B(l^2(\Gamma))} \to 0, \text{ as } n\to \infty$$
Let $\epsilon \gt 0$. Then there exists $N \in \mathbb{N}$ such that $\forall n \ge N$, $$\|\lambda(f_n)-T\|_{B(l^2(\Gamma))} \lt \epsilon$$
But since $f_n \in \mathbb{C}(\Gamma),f_n=\sum_{s \in \Gamma}a_ss$, where only finitely many of the scalar coefficients $a_s \in \mathbb{C}$ are nonzero. Then $\lambda(f_n)=\sum_{s \in \Gamma}a_s\lambda_s \in \text{ linear span }\{\lambda_s: s\in \Gamma\}$. This establishes the claim.
Is this alright?
Thanks for the help!!