Oopsies daisies.

src/harmonic/rep/pos.tex

@@ -181,23 +181,23 @@
 \label{theorem:lcg-state-extreme}
     Let $G$ be a locally compact group, then:
     \begin{enumerate}
-        \item The extreme points of $P_1(G)$ is the extreme points of $S(G)$ and $0$. 
+        \item The extreme points of $P_0(G)$ is the extreme points of $S(G)$ and $0$. 
         \item $S(G)$ is contained in the closed convex hull of its extreme points\footnote{In general, $L^1(G; \complex)$ is not unital and $S(G)$ is not closed.}. 
     \end{enumerate}
 \end{theorem}
 \begin{proof}[Proof, {{\cite[Lemma 3.26, Theorem 3.27]{FollandHarmonic}}}. ]
-    (1): Let $\phi \in S(G)$, $\psi_1, \psi_2 \in P_1(G)$, and $t \in (0, 1)$ such that $\phi = (1 - t)\psi_1 + t\psi_2$, then
+    (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
     \[
     1 = \norm{\phi}_{L^1(G; \complex)^*} = (1 - t)\norm{\psi_1}_{L^1(G; \complex)^*} + t\norm{\psi_2}_{L^1(G; \complex)^*} \le 1
     \]
 
-    so $\norm{\psi_2}_{L^1(G; \complex)^*} = \norm{\psi_2}_{L^1(G; \complex)^*} = 1$, and $S(G)$ is an extreme subset of $P_1(G)$. By \autoref{lemma:extreme-subset-extreme-point}, every extreme point of $S(G)$ is an extreme point of $P_1(G)$. 
+    so $\norm{\psi_2}_{L^1(G; \complex)^*} = \norm{\psi_2}_{L^1(G; \complex)^*} = 1$, and $S(G)$ is an extreme subset of $P_0(G)$. By \autoref{lemma:extreme-subset-extreme-point}, every extreme point of $S(G)$ is an extreme point of $P_0(G)$. 
 
-    On the other hand, every extreme point of $P_1(G)$ in $S(G)$ is an extreme point of $S(G)$. 
+    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 $P_1(G)$. To this end, let $\phi, \psi \in P_1(G)$ 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 \autoref{corollary:group-l1-positive-function}, $0 = \phi(1_G) = \norm{\phi}_u$ and $0 = \psi(1_G) = \norm{\psi}_u$ so $0$ is an extreme point of $P_1(G)$. For any $\phi \in P_1(G) \setminus (S(G) \cup \bracsn{0})$, $\phi/\phi(1_G) \in S(G)$, and $\phi = 0 + \phi(1_G) \cdot \phi/\phi(1_G)$. Therefore $0$ is the only other extreme point of $P_1(G)$. 
+    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 \autoref{corollary:group-l1-positive-function}, $0 = \phi(1_G) = \norm{\phi}_u$ 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 $\phi = 0 + \phi(1_G) \cdot \phi/\phi(1_G)$. 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 \autoref{proposition:semicontinuous-properties}, $\norm{\cdot}_{L^1(G; \complex)^*}$ is lower semicontinuous on $L^1(G; \complex)^*$ with respect to the weak topology, so $\bracsn{\psi \in P_1(G)|\ \norm{\psi}_{L^1(G; \complex)^*} > 1 - \eps}$ is a weak*-neighbourhood of $\phi$. 
+    (2): Let $\phi \in S(G)$ and $\eps \in (0, 1)$. By definition of the operator norm and (4) of \autoref{proposition:semicontinuous-properties}, $\norm{\cdot}_{L^1(G; \complex)^*}$ 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 \hyperref[Krein-Milman Theorem]{theorem:krein-milman}, there exists $\seqf{\phi_j} \subset S(G)$ 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
     \[