Proof, [Lemma 3.26, Theorem 3.27, Fol16]. (1): Let $\phi \in S(G)$, $\psi_{1}, \psi_{2} \in P_{0}(G)$, and $t \in (0, 1)$ such that $\phi = (1 - t)\psi_{1} + t\psi_{2}$, then

so $\norm{\psi_1}_{L^1(G; \complex)^*}= \norm{\psi_2}_{L^1(G; \complex)^*}= 1$, and $S(G)$ is an extreme subset of $P_{0}(G)$. By Lemma 13.3.4, every extreme point of $S(G)$ is an extreme point of $P_{0}(G)$.

On the other hand, every extreme point of $P_{0}(G)$ in $S(G)$ is an extreme point of $S(G)$.

It remains to show that $0$ is the only other extreme point of $P_{0}(G)$. To this end, let $\phi, \psi \in P_{0}(G)$ and $t \in (0, 1)$ such that $0 = (1 - t)\phi + t\psi$, then $0 = (1 - t)\phi(1_{G}) + t\psi(1_{G})$. By Corollary 32.3.4, $0 = \phi(1_{G}) = \norm{\phi}_{u}$ and $0 = \psi(1_{G}) = \norm{\psi}_{u}$ so $0$ is an extreme point of $P_{0}(G)$. For any $\phi \in P_{0}(G) \setminus (S(G) \cup \bracsn{0})$, $\phi/\phi(1_{G}) \in S(G)$, and $\phi = (1 - \phi(1_{G})) \cdot 0 + \phi(1_{G}) \cdot \phi/\phi(1_{G})$. Therefore $0$ is the only other extreme point of $P_{0}(G)$.

(2): Let $\phi \in S(G)$ and $\eps \in (0, 1)$. By definition of the operator norm and (4) of Proposition 5.22.3, $\norm{\cdot}_{L^1(G; \complex)^*}$ is lower semicontinuous on $L^{1}(G; \complex)^{*}$ with respect to the weak* topology, so $\bracsn{\psi \in P_0(G)|\ \norm{\psi}_{L^1(G; \complex)^*} > 1 - \eps}$ is a weak*-neighbourhood of $\phi$.

Let $\cf \subset L^{1}(G; \complex)$ be finite, then by (1) and the Krein-Milman Theorem, there exist extreme points $\seqf{\phi_j}\subset S(G)$ and $\bracsn{\lambda_j}_{0}^{n} \subset [0, 1]$ such that $\sum_{j = 0}^{n} \lambda_{j} = 1$, $\lambda_{0} < \eps/\max_{f \in \cf}\norm{f}_{L^1(G; \complex)}$, and

In which case, $\frac{1}{1 - \lambda_{0}}\sum_{j = 1}^{n} \lambda_{j} = 1$ and

so

$\square$