I was reading Ozawa's book and in Chapter 5 they discuss $C^* $-algebras of locally compact Hausforff étale groupoids. I have a question regarding the reduced $C^* $-algebra construction.
It is first defined by considering the $* $-algebra $C_c(G)$ and endow it with a $C_0(G^{(0)})$-valued inner product defined for $f,g \in C_c(G)$ and $x \in G^{(0)}$ by $$ \langle f,g \rangle(x) = \sum\limits_{\gamma \in G_x}\overline{f(\gamma)}g(\gamma). $$ Thus we obtain the Hilbert module $L^2(G)$ completing $C_c(G)$ for this inner product. Now we define the left regular representation $\lambda: C_c(G) \xrightarrow{} B(L^2(G))$ by $$ \lambda(f)g = f* g, \quad \text{ for } f,g \in C_c(G), $$ and endow $C_c(G)$ with the norm $\|f\|_\lambda := \|\lambda(f)\|.$ Completing for this norm, we obtain the reduced groupoid $C^* $-algebra, denoted by $C_\lambda^* (G)$.
However, there is a footnote saying that we can consider a Hilbert space instead of module by letting $\mu$ be a regular Borel measure with full support on $G^{(0)}$ and defining an inner product on $C_c(G)$ by $$ \langle f,g \rangle_\mu := \int_{G^{(0)}} \langle f,g \rangle(x) d\mu, $$ obtaining the Hilber space $L^2(G,\mu)$ through completion. We can then consider the same regular representation on $B(L^2(G,\mu))$ and complete $C_c(G)$ for the norm $\|f\|_{\lambda,\mu} := \|\lambda(f) \|_{B(L^2(G,\mu))}$ and lets denote this by $C_{\lambda,\mu}^* (G)$.
My question is how do these norms relate.
The first one is $$ \|f\|_\lambda^2 = \|\lambda(f)\|_{B(L^2(G))}^2 = \sup\limits_{\|g\|_{L^2(G)}=1} \|f * g \|_{L^2(G)}^2 = \sup\limits_{\|g\|_{L^2(G)}=1} \| \langle f * g, f * g \rangle \|_\infty. $$ while the second, $$ \| f \|_{\lambda,\mu}^2 = \| \lambda(f) \|_{B(L^2(G,\mu))}^2 = \sup\limits_{\| g \|_{L^2(G,\mu)} = 1} \| f * g \|_{L^2(G,\mu)}^2 = \sup\limits_{\| g \|_{L^2(G,\mu)}=1} \int_{G^{(0)}} \langle f * g, f * g \rangle(x) d\mu . $$ We can see that in case 1 the norm is obtained by taking taking the infinity norm of $\langle f* g,f* g \rangle$ but on the second we take the $L^1$ norm of the same function.
Because of this I'm having troubles believing these form the same $C^* $-algebra, did I misunderstand something?
To distinguish the two regular representations, I will write $\lambda$ for the representation of $C_c(G)$ on $\mathbb{B}(L^2(G))$ and $\lambda_\mu$ for the representation of $C_c(G)$ on $\mathbb{B}(L^2(G,\mu))$.
The same footnote mentioned in your post states that there is an isomorphism between the Hilbert spaces $L^2(G,\mu)$ and $L^2(G)\otimes_{C_0(G^{(0)})}L^2(G^{(0)},\mu)$. In fact, there is a unitary \begin{align*} U:L^2(G)\otimes_{C_0(G^{(0)})}L^2(G^{(0)},\mu)\to L^2(G,\mu) \end{align*} such that for all $\xi\in C_c(G)\subset L^2(G)$, $\eta\in C_c(G^{(0)})\subset L^2(G^{(0)},\mu)$, and $\gamma\in G$, one has $U(\xi\otimes\eta)(\gamma)=\xi(\gamma)\eta(s(\gamma))$. Now fix $f\in C_c(G)$. If $\xi\in C_c(G)$, $\eta\in C_c(G^{(0)})$, and $\gamma\in G$, then \begin{align*} \left[\lambda_\mu(f)U(\xi\otimes\eta)\right](\gamma) &=\sum_{\alpha\beta=\gamma}f(\alpha)U(\xi\otimes\eta)(\beta) \\&=\left[\sum_{\alpha\beta=\gamma}f(\alpha)\xi(\beta)\right]\eta(s(\gamma)) \\&=[\lambda(f)\xi](\gamma)\eta(s(\gamma)) \\&=U([\lambda(f)\xi]\otimes\eta)(\gamma) \\&=[U(\lambda(f)\otimes1)(\xi\otimes\eta)](\gamma). \end{align*} It follows that $\lambda(f)\otimes 1=U^*\lambda_\mu(f)U$, and thus \begin{align*} \|\lambda(f)\|_{\mathbb{B}(L^2(G))} =\|\lambda(f)\otimes 1\|_{\mathbb{B}(L^2(G)\otimes_{C_0(G^{(0)})}L^2(G^{(0)},\mu))} =\|\lambda_\mu(f)\|_{\mathbb{B}(L^2(G,\mu))}. \end{align*}