Proof, [Theorem 3.9, Theorem 3.11, Fol16]. (1, Well-Defined): For each $x \in G$, $\pi_{0}(x)$ is unitary, so for any $f \in L^{1}(G; \complex)$ and $\xi, \eta \in H$,

and the representation is well-defined.

(1, Homomorphism): For any $f, g \in L^{1}(G; \complex)$ and $\xi, \eta \in H$, since $\pi_{0}$ is a unitary representation,

(1, *-Homomorphism): For any $f \in L^{1}(G; \complex)$ and $\xi, \eta \in H$, since $\pi_{0}$ is unitary,

(1, Non-Degenerate): For any $\xi \in H \setminus \bracsn{0}$, since $\pi_{0}$ is strong-operator continuous, there exists $V \in \cn_{G}(1)$ such that $\norm{\pi_0(x)\xi - \xi}_{H} < \norm{\xi}_{H}$ for all $x \in V$. In which case, if $f = |V|^{-1}\one_{V}$^[1], then for any $\eta \in H$,

so $\norm{\pi(f)\xi - \xi}_{H} < \norm{\xi}_{H}$, and $\pi(f)\xi \ne 0$.

(3): Using Proposition 31.4.10, let $\angles{\phi_\alpha}_{\alpha \in A}$ be an approximate identity in $L^{1}(G; \complex)$. For each $x \in G$ and $f \in L^{1}(G; \complex)$, $L_{x}\phi_{\alpha} * f \to L_{x} f$ in $L^{1}(G; \complex)$ implies that $\angles{\pi(L_x\phi_\alpha)}_{\alpha \in A}$ converges pointwise on

Given that $\pi$ is non-degenerate, $K$ is dense in $H$. As $\bracsn{\pi(L_x\phi_\alpha)|\alpha \in A}$ is equicontinuous and $H$ is complete, the Arzelà-Ascoli Theorem implies that there exists $\pi_{0}(x) \in \ol{B_{B(H)}(0, 1)}$ such that $\pi(L_{x}\phi_{\alpha}) \to \pi_{0}(x)$ in the strong operator topology. In particular, $\pi_{0}(1) = \text{Id}$.

For each $f \in L^{1}(G; \complex)$, $\xi \in H$, and $x, y \in G$,

By density of $K$ in $H$, $\pi_{0}(xy) = \pi_{0}(x)\pi_{0}(y)$, so $\pi_{0}$ is a homomorphism. For any $x \in G$, $\text{Id}= \pi_{0}(1) = \pi_{0}(x)\pi_{0}(x^{-1})$, so $\pi_{0}(x)$ is isometric and invertible, and hence unitary.

Finally, for any $f \in L^{1}(G; \complex)$ and $\xi, \eta \in H$, by the Dominated Convergence Theorem,

$\square$