32.4 Positive-Definite Functions

Proof.

$\square$

Proof, [Proposition 3.35, Fol16]. (1) $\Rightarrow$ (2): Using Proposition 31.4.10, let $\angles{\psi_\alpha}_{\alpha \in A}\subset L^{1}(G; \complex)$ be an approximate identity such that for all $U \in \cn_{G}(1)$, there exists $\alpha_{0} \in A$ with $\supp{\psi_\alpha}\subset U$ for all $\alpha \ge \alpha_{0}$.

For each $\alpha \in A$, $\seqf{\lambda_j}\subset \complex$, and $\seqf{x_j}\subset G$, let $f_{\alpha} = \sum_{j = 1}^{n} \lambda_{j} L_{x_j}\psi_{\alpha}$, then by continuity of $\phi$,

(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 Proposition 31.1.2. 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

As

and the above holds for all $\eps > 0$, $\int_{G} (f^{*} * f) \phi \ge 0$. By density of $C_{c}(G; \complex)$ in $L^{1}(G; \complex)$ and continuity of the convolution product, $\int_{G} (g^{*} * g) \phi \ge 0$ for all $g \in L^{1}(G; \complex)$.$\square$

Proof, [Proposition 3.14, Fol16]. For any $f \in L^{1}(G; \complex)$,

$\square$

Proof. (1): Let $f \in L^{1}(G; \complex)$, then by Lemma 32.4.1,

(2): Let $\pi: G \to B(L^{2}(G; \complex))$ be the left regular representation of $G$, then

By (1) and Lemma 32.4.3, $f * \tilde f$ is positive-definite.$\square$

Proof. For each $h \in C_{c}(G; \complex)$, denote $\tilde h(x) = \ol{h(x^{-1})}$, then by polarisation,

Since $C_{c}(G; \complex) \subset L^{2}(G; \complex)$, (2) of Lemma 32.4.4 implies that $(f + i^{k} \tilde g) * (f + i^{k} \tilde g)^{\sim}$ is positive-definite.$\square$

Proof, [Lemma 3.30, Fol16]. Let $(H_{\phi}, \pi_{\phi}, \xi_{\phi})$ be the GNS triple associated with $\phi$, then by Corollary 32.3.4, $\phi(x) = \dpn{\pi_\phi(x)\xi_\phi, \xi_\phi}{H_\phi}$ for all $x \in G$. Since $\norm{\phi}_{u} = \phi(1_{G}) = 1$, $\norm{\xi_\phi}_{H_\phi}= 1$. As such,

$\square$

Proof. ($\supset$): 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 $\phi, \psi \in S(G)$,

($\subset$): Let $\phi \in S(G)$ and $V \in \cn_{G}(1_{G})$ be relatively compact, then for each $x \in G$,

by Lemma 32.4.6 and the Cauchy-Schwarz inequality. As such, for any $\phi, \psi \in S(G)$ and $x \in G$,

For any $K \subset G$ compact,

Let $\eps > 0$. By Corollary 32.3.4, $\phi$ 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 Arzelà-Ascoli Theorem, 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

for all $\psi \in W$. Therefore for any $\psi \in W$, $\sup_{x \in K}|\phi(x) - \psi(x)| \le 2\eps$.$\square$

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

$\square$

Proof, [Theorem 3.34, Fol16]. Let $x, y \in G$ and suppose that for any irreducible unitary representation $(H, \pi)$ of $G$, $\pi(x) = \pi(y)$. Let $\phi \in S(G)$ be an extreme point of $S(G)$ and $(H_{\phi}, \pi_{\phi}, \xi_{\phi})$ be its GNS triple, then $(H_{\phi}, \pi_{\phi})$ is irreducible by Proposition 32.3.5. In particular, by Corollary 32.3.4,

By (2) of Theorem 32.4.8, the convex hull of extreme points of $S(G)$ is weak*-dense in $S(G)$. By Theorem 32.4.7, 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 Lemma 32.4.5, $(f * g)(x) = (f * g)(y)$ for all $f, g \in C_{c}(G; \complex)$. Using Proposition 31.4.10, 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 $g \in C_{c}(G; \complex)$.

Should $x \ne y$, then Urysohn’s Lemma implies that there exists $g \in C_{c}(G; \complex)$ with $g(x) \ne g(y)$. Since this is not the case, $x = y$.$\square$