Review.

src/harmonic/rep/gns.tex

@@ -57,7 +57,7 @@
     
     (2): By the \hyperref[Cauchy-Schwarz inequality]{proposition:cauchy-schwarz}, for any $f, g \in L^1(G; \complex)$, 
     \[
-    \dpn{f, g}{\phi} \le \dpn{f, f}{\phi}^{1/2} \cdot \dpn{g, g}{\phi}^{1/2}
+    |\dpn{f, g}{\phi}| \le \dpn{f, f}{\phi}^{1/2} \cdot \dpn{g, g}{\phi}^{1/2}
     \]
 
     In particular, if $g \in N_\phi$, then

src/harmonic/rep/pos.tex

@@ -42,7 +42,7 @@
     \end{align*}
     
 
-    (2) $\Rightarrow$ (1): Let $f \in C_c(G; \complex)$, then $F(x, y) = f(x)\ol{f(y)}\phi(y^{-1}x)$ is uniformly continuous by \autoref{proposition:lcg-cc-uc}. As such, there exists a partition $K = \bigsqcup_{j = 1}^n U_j$ and $\seqf{x_j} \subset G$ such that $|F(x_i, x_j) - F(x, y)| \le \eps$ for all $(x, y) \in U_i \times U_j$ and $1 \le i, j \le n$. Thus
+    (2) $\Rightarrow$ (1): Let $f \in C_c(G; \complex)$ and $K = \supp{f}$, then $F(x, y) = f(x)\ol{f(y)}\phi(y^{-1}x)$ is uniformly continuous by \autoref{proposition:lcg-cc-uc}. As such, there exists a partition $K = \bigsqcup_{j = 1}^n U_j$ and $\seqf{x_j} \subset G$ such that $|F(x_i, x_j) - F(x, y)| \le \eps$ for all $(x, y) \in U_i \times U_j$ and $1 \le i, j \le n$. Thus
     \[
     \abs{\int_G (f^* * f) \phi - \sum_{i, j = 1}^n F(x_i, x_j) \cdot |U_i| \cdot |U_j|} \le \eps |K|^2
     \]
@@ -99,10 +99,10 @@
 \begin{proof}
     For each $h \in C_c(G; \complex)$, denote $\tilde h(x) = \ol{h(x^{-1})}$, then by \hyperref[polarisation]{proposition:polarisation-complex},
     \[
-        f * \tilde g = \frac{1}{4}\sum_{k = 0}^3 i^k (f + i^k g) * (f + i^k g)^\sim
+        f * g = \frac{1}{4}\sum_{k = 0}^3 i^k (f + i^k \tilde g) * (f + i^k \tilde g)^\sim
     \]
     
-    Since $C_c(G; \complex) \subset L^2(G; \complex)$, (2) of \autoref{lemma:positive-definite-from-representation} implies that $(f + i^k g) * (f + i^k g)^\sim$ is positive-definite. 
+    Since $C_c(G; \complex) \subset L^2(G; \complex)$, (2) of \autoref{lemma:positive-definite-from-representation} implies that $(f + i^k \tilde g) * (f + i^k \tilde g)^\sim$ is positive-definite. 
 \end{proof}
 
 
@@ -132,11 +132,11 @@
     Let $G$ be a locally compact group, then the topology of uniform convergence on compact subsets of $G$ on $S(G)$ and the weak* topology on $S(G)$ as a subset of $L^1(G; \complex)^*$ coincide. 
 \end{theorem}
 \begin{proof}
-    ($\supset$): Let $\seqf{f_j} \subset L^1(G; \complex)$. By inner regularity of the Haar measure and hence $|f_j(x)|dx$, there exists $K \subset G$ compact such that $\int_{G \setminus K}|f_j| < \eps$ for all $1 \le j \le n$. In which case, for any $\phi, \psi \in \mathcal{F}$, 
+    ($\supset$): Let $\seqf{f_j} \subset L^1(G; \complex)$. By inner regularity of the Haar measure and hence of $|f_j(x)|dx$, there exists $K \subset G$ compact such that $\int_{G \setminus K}|f_j| < \eps$ for all $1 \le j \le n$. In which case, for any $\phi, \psi \in S(G)$, 
     \begin{align*}
         \max_{1 \le j \le n}\abs{\int_{G} f_j (\phi - \psi)} &\le \max_{1 \le j \le n}\int_G|f_j(\phi - \psi)| \\
         &\le \max_{1 \le j \le n}\braks{\int_{K} |f_j(\phi - \psi)| + \int_{G \setminus K}|f_j(\phi - \psi)|} \\
-        &\le \sup_{x \in K}|\phi(x) - \psi(x)|\max_{1 \le j \le n}\norm{f_j}_{L^1(G; \complex)} + \eps
+        &\le \sup_{x \in K}|\phi(x) - \psi(x)|\max_{1 \le j \le n}\norm{f_j}_{L^1(G; \complex)} + 2\eps
     \end{align*}
 
     ($\subset$): Let $\phi \in S(G)$ and $V \in \cn_G(1_G)$ be relatively compact, then for each $x \in G$, 
@@ -150,23 +150,23 @@
     \begin{align*}
         |\phi(x) - \psi(x)| &\le \abs{\phi(x) - \frac{\one_V}{|V|} * \phi(x)} + \frac{1}{|V|}\abs{\one_V * (\phi - \psi)(x)}\\
         &+ \abs{\psi(x) - \frac{\one_V}{|V|} * \psi(x)} \\
-        &\le |V|^{1/2}\braks{\int_V 2 - 2\text{Re}(\phi(y)) dy}^{1/2} + \abs{\angles{\frac{L_x\one_V}{|V|}, \phi - \psi}_{L^1(G; \complex)}} \\
+        &\le |V|^{-1/2}\braks{\int_V 2 - 2\text{Re}(\phi(y)) dy}^{1/2} + \abs{\angles{\frac{L_x\one_V}{|V|}, \phi - \psi}_{L^1(G; \complex)}} \\
-        &+ |V|^{1/2}\braks{\int_V 2 - 2\text{Re}(\psi(y)) dy}^{1/2} \\
+        &+ |V|^{-1/2}\braks{\int_V 2 - 2\text{Re}(\psi(y)) dy}^{1/2} \\
-        &\le 2|V|^{1/2}\braks{\int_V 2 - 2\text{Re}(\phi(y)) dy}^{1/2}+ |2\dpn{\one_V, \phi - \psi}{L^1(G; \complex)}|^{1/2} \\
+        &\le 2|V|^{-1/2}\braks{\int_V 2 - 2\text{Re}(\phi(y)) dy}^{1/2}+ 2|V|^{-1/2}|\dpn{\one_V, \phi - \psi}{L^1(G; \complex)}|^{1/2} \\
         &+ \abs{\angles{\frac{L_x\one_V}{|V|}, \phi - \psi}_{L^1(G; \complex)}} \\
-        &\le 4|V|\sup_{y \in V}|\phi(y) - 1|^{1/2}+ |2\dpn{\one_V, \phi - \psi}{L^1(G; \complex)}|^{1/2}\\
+        &\le 4\sup_{y \in V}|\phi(y) - 1|^{1/2}+ |2\dpn{\one_V, \phi - \psi}{L^1(G; \complex)}|^{1/2}\\
         &+ \abs{\angles{\frac{L_x\one_V}{|V|}, \phi - \psi}_{L^1(G; \complex)}} \\
     \end{align*}
 
     For any $K \subset G$ compact,
     \begin{align*}
-    \sup_{x \in K}|\phi(x) - \psi(x)| &\le 4|V|\sup_{y \in V}|\phi(y) - 1|^{1/2}+ |2\dpn{\one_V, \phi - \psi}{L^1(G; \complex)}|^{1/2} \\
+    \sup_{x \in K}|\phi(x) - \psi(x)| &\le 4\sup_{y \in V}|\phi(y) - 1|^{1/2}+ |2\dpn{\one_V, \phi - \psi}{L^1(G; \complex)}|^{1/2} \\
     &+\sup_{x \in K}\abs{\angles{\frac{L_x\one_V}{|V|}, \phi - \psi}_{L^1(G; \complex)}}
     \end{align*}
 
-    Let $\eps > 0$. By \autoref{corollary:group-l1-positive-function}, $\phi$ is continuous with $\phi(1_G) = 1$, so there exists a relatively compact neighbourhood $V \in \cn_G(1_G)$ with $4|V|\sup_{y \in V}|\phi(y) - 1|^{1/2} < \eps$. 
+    Let $\eps > 0$. By \autoref{corollary:group-l1-positive-function}, $\phi$ is continuous with $\phi(1_G) = 1$, so there exists a relatively compact neighbourhood $V \in \cn_G(1_G)$ with $4\sup_{y \in V}|\phi(y) - 1|^{1/2} < \eps$. 
     
-    Since $S(G)$ is a bounded subset of $L^1(G; \complex)^*$, it is equicontinuous. By the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}, the weak* topology and the topology of uniform convergence on compact subsets of $L^1(G; \complex)$ coincide on $S(G)$. In particular, given that $K$ is compact, $\bracsn{L_x\one_V|x \in K}$ is compact in $L^1(G; \complex)$, so there exists $W \in \cn_{S(G)}(\phi)$ such that
+    Since $S(G)$ is a bounded subset of $L^1(G; \complex)^*$, it is equicontinuous. By the \hyperref[Arzelà-Ascoli Theorem]{theorem:arzela-ascoli}, the weak* topology and the topology of uniform convergence on compact subsets of $G$ coincide on $S(G)$. In particular, given that $K$ is compact, $\bracsn{L_x\one_V|x \in K}$ is compact in $L^1(G; \complex)$, so there exists $W \in \cn_{S(G)}(\phi)$ such that
     \[
     |2\dpn{\one_V, \phi - \psi}{L^1(G; \complex)}|^{1/2} +\sup_{x \in K}\abs{\angles{\frac{L_x\one_V}{|V|}, \phi - \psi}_{L^1(G; \complex)}} \le \eps
     \]
@@ -188,18 +188,18 @@
 \begin{proof}[Proof, {{\cite[Lemma 3.26, Theorem 3.27]{FollandHarmonic}}}. ]
     (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
     \[
-    1 = \norm{\phi}_{L^1(G; \complex)^*} = (1 - t)\norm{\psi_1}_{L^1(G; \complex)^*} + t\norm{\psi_2}_{L^1(G; \complex)^*} \le 1
+    1 = \norm{\phi}_{L^1(G; \complex)^*} \le (1 - t)\norm{\psi_1}_{L^1(G; \complex)^*} + t\norm{\psi_2}_{L^1(G; \complex)^*} \le 1
     \]
 
     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 \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 $P_0(G)$ 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 \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 $P_0(G)$. 
+    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 $\phi = (1 - \phi(1_G)) \cdot 0 + \phi(1_G) \cdot \phi/\phi(1_G)$. 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 \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
+    Let $\cf \subset L^1(G; \complex)$ be finite, then by (1) and the \hyperref[Krein-Milman Theorem]{theorem:krein-milman}, 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
     \[
     \max_{f \in \cf}\angles{f, \phi - \sum_{j = 1}^n \lambda_j \phi_j}_{L^1(G; \complex)} < \eps
     \]
@@ -227,7 +227,7 @@
     \dpn{\pi_\phi(y)\xi_\phi, \xi_\phi}{H_\phi} = \phi(y)
     \]
 
-    By (2) of \autoref{theorem:lcg-state-extreme}, the extreme points of $S(G)$ are weak*-dense in $S(G)$. By \autoref{theorem:group-bc-weak-topology}, the weak* topology and the topology of uniform convergence on compact subsets of $G$ coincide on $S(G)$. As such, $\psi(x) = \psi(y)$ for all $\psi \in S(G)$. By linearity, $\psi(x) = \psi(y)$ for all $\psi \in P(G)$ as well. 
+    By (2) of \autoref{theorem:lcg-state-extreme}, the convex hull of extreme points of $S(G)$ is weak*-dense in $S(G)$. By \autoref{theorem:group-bc-weak-topology}, the weak* topology and the topology of uniform convergence on compact subsets of $G$ coincide on $S(G)$. As such, $\psi(x) = \psi(y)$ for all $\psi \in S(G)$. By linearity, $\psi(x) = \psi(y)$ for all $\psi \in P(G)$ as well. 
 
     By linearity and \autoref{lemma:positive-definite-cc-decompose}, $(f * g)(x) = (f * g)(y)$ for all $f, g \in C_c(G; \complex)$. Using \autoref{proposition:group-algebra-approximate-identity}, let $\angles{\psi_U}_{U \in \cn_G(1_G)} \subset C_c(G; \complex)$ be an approximate identity for $L^1(G; \complex)$ such that $\supp{\psi_U} \subset U$ for all $U \in \cn_G(1_G)$, then for any $g \in C_c(G; \complex)$, $\psi_U * g \to g$ uniformly. Therefore $g(x) = g(y)$ for all $g \in C_c(G; \complex)$.