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If $G$ is a discrete group,the reduced $C^*$ algebra of $G$ is the completion of $\mathbb{C}[G]$ with respect to the norm $||x||_{r}=||\lambda(x)||_{\mathbb{B}(l^2(G))}$,where $\lambda$ is the left regular representation of $\mathbb{C}[G]$. There is another definition, the reduced $C^*$ algebra of $G$ is the closure of $\lambda(\mathbb{C}[G])$ with respect to $||||$$_{\mathbb{B}(l^2(G))}$.How to show these two definitions are equivalent?

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The mapping $\lambda:(\mathbb C[G],\|\cdot\|_r)\to(\mathbb B(\ell^2(G)),\|\cdot\|_{\mathbb B(\ell^2(G))})$ is an isometry into a complete metric space, hence extends to an isometry on the completion of $(\mathbb C[G],\|\cdot\|_r)$.

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  • $\begingroup$ Does this mean $\overline{\mathbb{C}[G]}^{||.||_{r}}=\overline{\lambda(\mathbb{C}[G])}^{||.||_{\mathbb{B}(l^2(G))}}$ I am a little confused! $\endgroup$ – math112358 May 17 '18 at 8:55
  • $\begingroup$ They are the same up to isometric $*$-isomorphism. $\endgroup$ – Aweygan May 17 '18 at 14:32

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