Show that $\lambda^{\phi}: \Gamma \to \mathbb{B}(\cal{l}^2_{\phi}(\Gamma)$ is an unitary representation Suppose that $\phi$ is a positive definite function on $\Gamma$. Show that $\lambda^{\phi}: \Gamma \to \mathbb{B}(\cal{l}^2_{\phi}(\Gamma)$ is the unitary representation given by $\lambda_s^{\phi}(\hat{f})=\hat{s.f}$, where $s.f(t)=f(s^{-1}t)$, for all $t \in \Gamma$. 
Here $\Gamma$ is a discrete group and $\phi: \Gamma \to \mathbb{C}$ is said to be positive definite if the matrix $[\phi(s^{-1}t]_{s,t \in F} \in \mathbb{M}_F(\mathbb{C})$ is positive for every finite set $F \subset \Gamma$. 
Now Let $C_c(\Gamma)$ be the finitely supported functions on $\Gamma$. Define a sesquilinear form $C_c(\Gamma) \times C_c(\Gamma) \to \mathbb{C}$ by $$\langle f,g\rangle_{\phi}=\sum_{s,t \in \gamma}\phi(s^{-1}t)f(t)\bar{g(s)}$$
This form is positive semidefinite: For $f \in C_c(\Gamma)$ has support $F$, then $$\langle f, f \rangle_{\phi}=\sum_{s,t \in \Gamma}\phi(s^{-1}t)f(t)\bar{f(s)}=\langle[\phi(s^{-1}t)]_{s,t \in F}(f) ,f \rangle =\langle[\phi(s^{-1}t)]^{\frac{1}{2}}_{s,t \in F}(f) ,[\phi(s^{-1}t)]^{\frac{1}{2}}f \rangle \ge 0$$
Let $N=\{f \in C_c(\Gamma):\langle f,f\rangle_{\phi}=0\}$. Then Let $\cal{l}^2_{\phi}(\Gamma)$ be the Hilbert completion of $C_c(\Gamma)/N$. This is all the process of creation of the space. Now I try to show that $\lambda_s^{\phi}:\cal{l}^2_{\phi}(\Gamma) \to \cal{l}^2_{\phi}(\Gamma)$ is unitary for every $s \in \Gamma$. I claim that $(\lambda_s^{\phi})^*:\cal{l}^2_{\phi}(\Gamma) \to \cal{l}^2_{\phi}(\Gamma)$ is nothing but $\lambda_{s^*}^{\phi}$, where $\lambda_{s^*}^{\phi}:l^2_{\phi}(\Gamma)\to \cal{l}^2_{\phi}(\Gamma)$ is defined by $\lambda_{s^*}^{\phi}(f)=\hat {s^{-1}.f}$. But what throws me up is the little footnote at the end of the page of the book which says that "It isn't hard, just tedious, to check that this is really a unitary representation". (I am following the book "C*-algebras and Finite-dimensional Approximations"by Brown and Ozawa. This  is on Page-45) Am I missing something? What else do I need to check?
Thanks for the help!!
 A: We need to show that for all $s\in \Gamma$ the operator $\lambda_s^{\varphi}$ is unitary.
Let $f,g \in C_c(\Gamma)$ be given, then $\  \hat{f}, \hat{g} \in l^2_{\varphi}(\Gamma)$ and:
$<\lambda_s^{\varphi}(\hat{f}), \lambda_s^{\varphi}(\hat{g})>_{\varphi} = <\hat{s.f}, \hat{s.g}>_{\varphi}= \sum_{x,y\in \Gamma}\varphi(x^{-1}y)f(s^{-1}x)\overline{g(s^{-1}y)}=\sum_{sx,sy\in \Gamma}\varphi((sx)^{-1}sy)f(s^{-1}(sx))\overline{g(s^{-1}(sy))} =\sum_{sx,sy\in \Gamma}\varphi(x^{-1}y)f(x)\overline{g(y)}=<\hat{f},\hat{g}>_{\varphi}$
By density and continuity $\lambda_s^{\varphi}$ is an isometric map.
Now, we just need to verify that $\lambda_s^{\varphi}$ is surjective.
It is easy to check that $\lambda_s^{\varphi}(\hat{s^{-1}.f})=\hat{f}$ for all $f \in C_c(\Gamma)$ and using density (and the fact that the map is isometric) the result follows.
One more remark regarding this construction:
If we take $\varphi: \Gamma \to \Bbb{C}$ to be $\varphi(e)=1$ and $\varphi(g)=0$ for all $g\neq e$, then $\varphi$ is a positive-definite function since the matrix $[\varphi(s^{-1}t)]_{s,t\in F}$ is the identity matrix for every finite subset $F\subseteq \Gamma$.
Moreover, for $f,g\in C_c(\Gamma)$ we get $<f,g>_{\varphi}=\sum_{t\in \Gamma} f(t)\overline{g(t)}$, so this form is exactly the inner-product of $l^2(\Gamma)$. So, $\hat{f}=f$ for all $f\in C_c(\Gamma)$ and the completion of $C_c(\Gamma)$ w.r.t. this inner product is just $l^2(\Gamma)$.
Note that $\lambda_{\varphi}:\Gamma\to B(l^2(\Gamma))$ given by $\lambda_s^{\varphi}(f)=s.f$ is the left-regular representation.
