Faithfulness of a given operator family on von Neumann tensor product Let $G$ be a locally compact (Hausdorff) group and $H$ a Hilbert space. Given $\xi,\eta\in C_c(G)$, we denote by $\kappa_{\xi,\eta}$ the normal (i.e. ultraweakly continuous) linear functional on $B(L^2(G))$ given by $T\mapsto \langle T\xi,\eta\rangle$ and let $\kappa_{\xi,\eta}\overline{\otimes}\text{id}_{B(H)}\colon B(L^2(G))\overline{\otimes}B(H)\to B(H)$ be the associated ultraweakly continuous and completely positive slice map.
I want to conclude the following for an element $x\in B(L^2(G))\overline{\otimes}B(H)$:
$$x=0 \Leftrightarrow \kappa_{\xi,\eta}\overline{\otimes}\text{id}_{B(H)}(x)=0 \quad \text{for every} \ \xi,\eta\in C_c(G) $$
Apparently, this should be "clear enough". All hints are appreciated.
 A: We have the identification
$$
B(L^2(G))\overline{\otimes} B(H)=B(L^2(G)\otimes H)\cong B(L^2(G;H)),
$$
where the last isomorphism is spatial, i.e., $L^2(G)\otimes H\cong L^2(G;H)$ via $\phi\otimes\zeta\mapsto \phi(\cdot)\zeta$. Under this identification, $\kappa_{\phi,\psi}\otimes \mathrm{id}$ acts as
$$
\kappa_{\phi,\psi}\otimes\mathrm{id}(x)\zeta=\int_G x(\phi(\cdot)\zeta)(g)\overline {\psi(g)}\,dg.
$$
Indeed, it suffices to show this for $x=y\otimes z$ with $y\in B(L^2(G))$ and $z\in B(H)$. In this case we have (keeping the identifications in mind)
$$
\kappa_{\phi,\psi}\otimes\mathrm{id}(y\otimes z)\zeta=\int y\phi(g)\overline{\psi(g)}\,dg\cdot z\zeta=\int (y\otimes z)(\phi(\cdot)\zeta)(g)\overline{\psi(g)}\,dg
$$
as desired.
Now if $\int x(\phi(\cdot)\zeta)\overline{\psi}=0$ for all $\psi\in C_c(G)$, then $x((\phi(\cdot)\zeta)=0$. Functions of the form $\phi(\cdot)\zeta$ however form a total subset of $L^2(G;H)$, so that $x=0$.
A: Let $s,t \in H$ and note that
$$0 = \kappa_{s,t}((\kappa_{\xi, \eta}\overline{\otimes} \operatorname{id})(x))= (\kappa_{\xi, \eta}\overline{\otimes}\kappa_{s,t})(x)= \kappa_{\xi \otimes s, \eta \otimes t}(x)= \langle x(\xi\otimes s), \eta \otimes t\rangle.$$
Since $C_c(G)$ is norm-dense in $L^2(G)$, it easily follows that $\langle x\mu, \mu'\rangle = 0$ for all $\mu, \mu' \in L^2(G)\otimes H.$ This implies $x=0$.
