Weak containment implies kernel inclusion

For every compactly supported continuous $f$ and unit vector $\xi \in H_{\pi }$: $\mathrm{\mid }⟨\pi (f)\xi ,\xi ⟩\mathrm{\mid }\le \mathrm{\parallel }\rho (f)\mathrm{\parallel }$. Indeed, let $Q\supseteq \mathrm{supp}f\cup \{e\}$ be compact and let $ϵ>0$; [F1] provides ${\eta }_1,\dots ,{\eta }_n$ with ${\sup }_Q\mathrm{\mid }⟨\pi (g)\xi ,\xi ⟩-{\sum }_j⟨\rho (g){\eta }_j,{\eta }_j⟩\mathrm{\mid }<ϵ$, and integrating against $f$ gives $\mathrm{\mid }⟨\pi (f)\xi ,\xi ⟩-{\sum }_j⟨\rho (f){\eta }_j,{\eta }_j⟩\mathrm{\mid }\le ϵ\mathrm{\parallel }f{\mathrm{\parallel }}_1$ by [F2]. Evaluating the same coefficient comparison at $e\in Q$ gives $\mathrm{\mid }\mathrm{\parallel }\xi {\mathrm{\parallel }}^2-{\sum }_j\mathrm{\parallel }{\eta }_j{\mathrm{\parallel }}^2\mathrm{\mid }<ϵ$, so ${\sum }_j\mathrm{\parallel }{\eta }_j{\mathrm{\parallel }}^2\le 1+ϵ$, and hence $\mathrm{\mid }{\sum }_j⟨\rho (f){\eta }_j,{\eta }_j⟩\mathrm{\mid }\le (1+ϵ)\mathrm{\parallel }\rho (f)\mathrm{\parallel }$. Therefore $\mathrm{\mid }⟨\pi (f)\xi ,\xi ⟩\mathrm{\mid }\le (1+ϵ)\mathrm{\parallel }\rho (f)\mathrm{\parallel }+ϵ\mathrm{\parallel }f{\mathrm{\parallel }}_1$ for every $ϵ>0$, and letting $ϵ\to 0$ gives the claim.

For every $f\in C_c(G)$: $\mathrm{\parallel }\pi (f)\mathrm{\parallel }\le \mathrm{\parallel }\rho (f)\mathrm{\parallel }$. Indeed, $\pi (f^{\ast }\ast f)$ is self-adjoint with $\pi (f^{\ast }\ast f)=\pi (f{)}^{\ast }\pi (f)$, so [F3] gives $\mathrm{\parallel }\pi (f){\mathrm{\parallel }}^2=\mathrm{\parallel }\pi (f^{\ast }\ast f)\mathrm{\parallel }={\sup }_{\mathrm{\parallel }\xi \mathrm{\parallel }=1}\mathrm{\mid }⟨\pi (f^{\ast }\ast f)\xi ,\xi ⟩\mathrm{\mid }\le \mathrm{\parallel }\rho (f^{\ast }\ast f)\mathrm{\parallel }=\mathrm{\parallel }\rho (f){\mathrm{\parallel }}^2$ by step 1.1 and the multiplicativity of the integrated forms.

The inequality holds for all $f\in L^1(G)$ by density of $C_c(G)$: both $f\mapsto \mathrm{\parallel }\pi (f)\mathrm{\parallel }$ and $f\mapsto \mathrm{\parallel }\rho (f)\mathrm{\parallel }$ are continuous in the $L^1$ norm (the integrated forms are contractive), and the set where the inequality holds is closed in $L^1(G)$.

The inequality extends to $C^{\ast }(G)$: for $a\in C^{\ast }(G)$ and $f_n\in L^1(G)$ with $\mathrm{\parallel }a-f_n{\mathrm{\parallel }}_{C^{\ast }}\to 0$ (using density of the image of $L^1(G)$ in $C^{\ast }(G)$), continuity of the extended representations gives $\mathrm{\parallel }\pi (a)\mathrm{\parallel }=\lim \mathrm{\parallel }\pi (f_n)\mathrm{\parallel }\le \lim \mathrm{\parallel }\rho (f_n)\mathrm{\parallel }=\mathrm{\parallel }\rho (a)\mathrm{\parallel }$ by step 3.1; in particular $a\in \ker \rho$ implies $\pi (a)=0$, that is $\ker \rho \subseteq \ker \pi$.

The Axiom of Choice is inherited from the completion and quadratic-form suppliers; the coefficient, integration and density arguments use no further choice (The Axiom of Choice).