About Fourier transform The reduced C*-algebra of $\Gamma$, denoted $C^{*}_{\lambda}(\Gamma)$, is  the completion of $\mathbb{C}(\Gamma)$ with respect to the norm
$$\|x\|_{r}=\|\lambda(x)\|_{\mathbb{B}(l^{2}(\Gamma))},$$
The $\Gamma$ is a discrete group and $\lambda$ is the left regular representation on $\Gamma.$
Example 2.5.1. If $\Gamma=\mathbb{Z}$, then $C_{\lambda}^{\ast}(\Gamma)=C(\mathbb{T})$, the contunuous functions on the circle. Indeed, the Fourier transform identifies $l^{2}(\mathbb{Z})$ with $L^{2}(\mathbb{T}, Lebesgue)$ and one checks that this unitary takes  $C_{\lambda}^{*}(\mathbb{Z})$ to (continuous) multiplication operators.
I am not familiar with Fourier transform, so my question is how to check the Forier transform is a unitary here, and why it takes $C_{\lambda}^{*}(\mathbb{Z})$ to multiplication operators? 
 A: Here you start with $\ell^2(\Gamma)=\ell^2(\mathbb Z)$. The left regular representation $\lambda$ maps every $n\in\mathbb Z$ into $B(\ell^2(\mathbb Z))$ by $\lambda(n)f(m)=f(m-n)$.
The Fourier transform $\ell^2(\mathbb Z)\to L^2(\mathbb T)$ in this context is the map $U:f\mapsto \sum_{n\in\mathbb Z}f(n)\,e^{int}$.
Now, using that $\{e^{int}\}_n$ is an orthonormal basis, 
$$
\langle U\lambda(n)f,U\lambda(n)f\rangle_{\ell^2(\mathbb Z)}=\langle \sum_{m\in\mathbb Z}f(m-n)\,e^{int},\sum_{m\in\mathbb Z}f(m-n)\,e^{int}\rangle_{L^2(\mathbb T)}\\
=\sum_{m\in\mathbb Z}|f(m-n)|^2=\sum_{m\in\mathbb Z}|\lambda(n)f(m)|^2=\langle\lambda(n)f,\lambda(n)f\rangle.
$$
This shows that $U$ is isometric, and it is easy to check that it is surjective: so it is a unitary. 
As $C^*_\lambda(\mathbb Z)$ is generated by $\lambda(1)$, we have that $UC^*_\lambda(\mathbb Z)U^*$ is generated by $U\lambda(1)U^*$. This is $e^{-it}$ since
\begin{align}
\langle U\lambda(1)U^*\sum_{m\in\mathbb Z}c_m\,e^{imt},\sum_{m\in\mathbb Z}c_m\,e^{imt}\rangle
&=\langle\lambda(1)(c_m),(c_m)\rangle=\langle (c_{m-1}),(c_m)\rangle\\ \ \\
&=\sum_{m\in\mathbb Z} c_{m-1}\overline{c_m}\\ \ \\
&=\langle e^{-it}\,\sum_{m\in\mathbb Z}c_m\,e^{imt},\sum_{m\in\mathbb Z}c_m\,e^{imt}\rangle.
\end{align}
So $UC^*_\lambda(\mathbb Z)U^*$ is the closure of the trigonometric polynomials, that is $C(\mathbb T)$. 
