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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 $\lambda(s)(\delta_t)=\delta_{st}, \forall s,t \in \Gamma.$

I want to identify $C_{\lambda}^*(\Gamma)$ with $C(\hat{\Gamma})$ using Pontryagin Duality. I know that there is a canonical isomorphism between $\Gamma$ and $\hat{\hat{\Gamma}}$ by the Pontryagin Duality Theorem.

How do I do it??

Thanks for the help!!

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    Not sure why you need Pontrjagin duality here. Since $L^1(\Gamma)$ is a commutative Banach algebra, all its irreducible representations are one dimensional, and so they correspond to the elements of $\widehat{\Gamma}$. The Gelfand (aka Fourier) transform then gives you the isomorphism. – 2017-03-01
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    I am following the book "C* algebras and Finite Dimensional Approximations" by Brown and Ozawa. This is example 2.5.1 on Page-43. This is what it is written there: "More generally, for every abelian group $\Gamma$, Pontryagin duality gives an identification of $C_{\lambda}^{*}(\Gamma)$ with $C(\hat{\Gamma})$, the continuous functions on the dual group" – 2017-03-01

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