Irreducible group vector functionals are extreme in the positive dual ball

$K$ is weak-* compact and convex. The unit ball $B:=\{\omega \in A^{\ast }:\mathrm{\parallel }\omega \mathrm{\parallel }\le 1\}$ is weak-* compact by [F5], and the positivity set $P:=\{\omega :\omega (a^{\ast }a)\ge 0\text{ for all }a\in A\}$ is an intersection of weak-* closed sets, because for fixed $a$ the map $\omega \mapsto \omega (a^{\ast }a)$ is evaluation at $a^{\ast }a$ and hence weak-* continuous; thus $K=B\cap P$ is a weak-* closed subset of a compact space, hence compact by [F5]. Convexity is immediate from the linearity of $\omega \mapsto \omega (a^{\ast }a)$ for each $a$: if ${\omega }_1,{\omega }_2\ge 0$ then $(t{\omega }_1+(1-t){\omega }_2)(a^{\ast }a)\ge 0$, and the norm bound is convex.

The canonical images $q(e_U)$ form a two-sided norm approximate identity for $A$, and $\pi (e_U)\eta \to \eta$ for every $\eta \in H_{\pi }$. For $f\in L^1(G)$ the contractivity of $q$ gives $\mathrm{\parallel }q(e_U)q(f)-q(f)\mathrm{\parallel }\le \mathrm{\parallel }{e}_U\ast f-f{\mathrm{\parallel }}_1\to 0$ and similarly on the right; given $x\in A$, choose $f$ with $\mathrm{\parallel }x-q(f)\mathrm{\parallel }<ϵ$; then $\mathrm{\parallel }q(e_U)x-x\mathrm{\parallel }\le \mathrm{\parallel }q(e_U)\mathrm{\parallel }\mathrm{\parallel }x-q(f)\mathrm{\parallel }+\mathrm{\parallel }q(e_U)q(f)-q(f)\mathrm{\parallel }+\mathrm{\parallel }q(f)-x\mathrm{\parallel }<2ϵ+o(1)$, and $\mathrm{\parallel }q(e_U)\mathrm{\parallel }\le \mathrm{\parallel }{e}_U{\mathrm{\parallel }}_1=1$, so the left convergence follows; the right convergence is identical. For $\pi (e_U)\eta ={\int }_G{e}_U(g)\pi (g)\eta dg$, nonnegativity, unit mass and $\mathrm{supp}{e}_U\subseteq U$ give $\mathrm{\parallel }\pi (e_U)\eta -\eta \mathrm{\parallel }\le {\int }_G{e}_U(g)\mathrm{\parallel }\pi (g)\eta -\eta \mathrm{\parallel }dg\le {\sup }_{g\in U}\mathrm{\parallel }\pi (g)\eta -\eta \mathrm{\parallel }$, which tends to $0$ as $U$ shrinks, by strong continuity of $\pi$ at $e$.

$\omega (e_U)\to 1$ and $\mathrm{\parallel }\omega \mathrm{\parallel }=1$. Since $\pi (e_U)\xi \to \xi$ by step 1.2, $\mathrm{\mid }\omega (e_U)-1\mathrm{\mid }=\mathrm{\mid }⟨\pi (e_U)\xi -\xi ,\xi ⟩\mathrm{\mid }\le \mathrm{\parallel }\pi (e_U)\xi -\xi \mathrm{\parallel }\to 0$, where $\omega :={\omega }_{\pi ,\xi }$. Thus $1={\lim }_U\omega (e_U)\le {\mathrm{lim\hspace{0.17em}sup}}_U\mathrm{\parallel }\omega \mathrm{\parallel }\mathrm{\parallel }{e}_U\mathrm{\parallel }=\mathrm{\parallel }\omega \mathrm{\parallel }$, using $\mathrm{\parallel }{e}_U{\mathrm{\parallel }}_1=1$ and $\mathrm{\parallel }q(e_U)\mathrm{\parallel }\le 1$; the reverse inequality $\mathrm{\parallel }\omega \mathrm{\parallel }\le 1$ holds because $\mathrm{\mid }\omega (a)\mathrm{\mid }=\mathrm{\mid }⟨\pi (a)\xi ,\xi ⟩\mathrm{\mid }\le \mathrm{\parallel }\pi (a)\mathrm{\parallel }\le \mathrm{\parallel }a\mathrm{\parallel }$ for all $a\in A$ by [F1].

The subspace $D:=\pi (A)\xi$ is dense in $H_{\pi }$. It is nonzero: $\pi (e_U)\xi \in D$ and $\pi (e_U)\xi \to \xi \mathrm{\ne }0$ by step 1.2, so $\xi \in \bar{D}$. It is $\pi (G)$-invariant: for $a\in A$ and $g\in G$, $\pi (g)\pi (a)\xi =\pi ({\tau }_{g}a)\xi \in D$ by [F2]. Hence $\bar{D}$ is a nonzero closed invariant subspace of $H_{\pi }$, so $\bar{D}=H_{\pi }$ by irreducibility [F8].

Domination lemma. If a positive functional $\psi \in A^{\ast }$ satisfies $0\le \psi \le c\omega$ for some $c\ge 0$, then $\psi =\lambda \omega$ for a unique $\lambda \in [0,c]$. Define $q(\pi (a)\xi ,\pi (b)\xi ):=\psi (b^{\ast }a)$ on $D\times D$. This is well defined: if $\pi (a)\xi =\pi (a^{\mathrm{\prime }})\xi$ and $d=a-a^{\mathrm{\prime }}$, then $\omega (d^{\ast }d)=\mathrm{\parallel }\pi (d)\xi {\mathrm{\parallel }}^2=0$, so $0\le \psi (d^{\ast }d)\le c\omega (d^{\ast }d)=0$ and Cauchy-Schwarz [F4] gives $\mathrm{\mid }\psi (b^{\ast }d){\mathrm{\mid }}^2\le \psi (d^{\ast }d)\psi (b^{\ast }b)=0$, hence $\psi (b^{\ast }a)=\psi (b^{\ast }{a}^{\mathrm{\prime }})$; conjugate symmetry and conjugate linearity in $b$ follow from [F4] and linearity of $\psi$. It is bounded: $\mathrm{\mid }q(\pi (a)\xi ,\pi (b)\xi ){\mathrm{\mid }}^2=\mathrm{\mid }\psi (b^{\ast }a){\mathrm{\mid }}^2\le \psi (a^{\ast }a)\psi (b^{\ast }b)\le c^2\omega (a^{\ast }a)\omega (b^{\ast }b)=c^2\mathrm{\parallel }\pi (a)\xi {\mathrm{\parallel }}^2\mathrm{\parallel }\pi (b)\xi {\mathrm{\parallel }}^2$ by [F4], and $q(u,u)=\psi (a^{\ast }a)\ge 0$. Since $D$ is dense by step 2.2, $q$ extends uniquely to a bounded sesquilinear form on $H_{\pi }$ with $0\le q(u,u)\le c\mathrm{\parallel }u{\mathrm{\parallel }}^2$, and [F6] provides $T\in B(H_{\pi })$ with $q(u,v)=⟨Tu,v⟩$, $0\le T\le cI$. For $d,a,b\in A$, $q(\pi (d)\pi (a)\xi ,\pi (b)\xi )=\psi (b^{\ast }da)=q(\pi (a)\xi ,\pi (d^{\ast })\pi (b)\xi )$, that is $⟨T\pi (d)\pi (a)\xi ,\pi (b)\xi ⟩=⟨\pi (d)T\pi (a)\xi ,\pi (b)\xi ⟩$; as $\{\pi (b)\xi :b\in A\}=D$ is dense, $T$ commutes with every $\pi (d)$, and then with every $\pi (g)$: for each $U$, it commutes with $\pi (L_g{e}_U)=\pi (g)\pi (e_U)$ by [F2], and these operators converge strongly to $\pi (g)$ by step 1.2. Bounded $T$ commutes with this strong limit. Schur's lemma [F7] gives $T=\lambda I$ with $\lambda \in [0,c]$. Finally, for $a\in A$ one has $\psi (e_U^{\ast }a)=q(\pi (a)\xi ,\pi (e_U)\xi )=⟨T\pi (a)\xi ,\pi (e_U)\xi ⟩\to ⟨\lambda \pi (a)\xi ,\xi ⟩=\lambda \omega (a)$, because $e_U^{\ast }a\to a$ in norm (the two-sided approximate identity of step 1.2 applied to $a^{\ast }$ and adjunction) and $\pi (e_U)\xi \to \xi$ by step 1.2; hence $\psi =\lambda \omega$.

The functional $\omega$ is extreme in $K$. Let $0<t<1$ and $\omega =t{\psi }_1+(1-t){\psi }_2$ with ${\psi }_1,{\psi }_2\in K$. Then $0\le {\psi }_1\le \omega \mathrm{/}t$ and $0\le {\psi }_2\le \omega \mathrm{/}(1-t)$, so step 3.1 gives ${\psi }_j={\lambda }_j\omega$ with ${\lambda }_j\ge 0$. Since $\mathrm{\parallel }\omega \mathrm{\parallel }=1$ by step 2.1 and $\mathrm{\parallel }{\psi }_j\mathrm{\parallel }\le 1$, ${\lambda }_j=\mathrm{\parallel }{\psi }_j\mathrm{\parallel }\le 1$. Substituting into the convex decomposition gives $(t{\lambda }_1+(1-t){\lambda }_2)\omega =\omega$, and $\omega \mathrm{\ne }0$, so $t{\lambda }_1+(1-t){\lambda }_2=1$; with $0\le {\lambda }_j\le 1$ this forces ${\lambda }_1={\lambda }_2=1$. Hence ${\psi }_1={\psi }_2=\omega$, and $\omega$ is extreme in $K$.

The Axiom of Choice is spent through the ultrafilter lemma in Banach-Alaoglu and is inherited from the maximal-norm completion and Schur's lemma; the domination and convexity arguments are choice-free (The Axiom of Choice).