Dominated positive type and positive commutant contractions

For $\eta \in \{\psi ,\phi -\psi ,\phi \}$ let $B_{\eta }(f,h)={\sum }_{x,y}f(x)\overline{h(y)}\eta (y^{-1}x)$ on finitely supported functions. By [F1] and Positive-type functions define the GNS pre-Hilbert form, these are Hermitian positive-semidefinite forms, and $B_{\phi }=B_{\psi }+B_{\phi -\psi }$. Thus $0\le B_{\psi }(f,f)\le B_{\phi }(f,f)$.

The form $B_{\psi }$ induces an inner product on the quotient by its null space, so Cauchy–Schwarz there gives $\mathrm{\mid }{B}_{\psi }(f,h){\mathrm{\mid }}^2\le B_{\psi }(f,f)B_{\psi }(h,h)$ for all finitely supported $f,h$, including null vectors.

Conversely, let $T$ satisfy the stated positive-contraction and commutant conditions, and put ${\psi }_T(g)=⟨{\pi }_{\phi }(g)T{\xi }_{\phi },{\xi }_{\phi }⟩$. For a finite list $g_1,\dots ,g_n$ and scalars $c_1,\dots ,c_n$, set $u={\sum }_j{c}_j{\pi }_{\phi }(g_j){\xi }_{\phi }$. Commutation and unitarity give ${\sum }_{i,j}\overline{c_i}{c}_j{\psi }_T(g_i^{-1}{g}_j)=⟨Tu,u⟩\ge 0.$ Thus ${\psi }_T$ is positive type; the same calculation with $I-T$ shows that $\phi -{\psi }_T$ is positive type. Strong continuity and the matrix coefficient definition make both functions continuous, so $0\le {\psi }_T\le \phi$.

Let $D={\mathrm{span}}_{\mathbb{C}}\{{\pi }_{\phi }(g){\xi }_{\phi }:g\in G\}$. Identify a finite sum of orbit vectors with the corresponding quotient class $[f]\in {\mathbb{C}}^{(G)}\mathrm{/}{N}_{\phi }$. Define $b_0([f],[h])=B_{\psi }(f,h)$. If $[f]=0$, then $B_{\phi }(f,f)=0$, so [F1] gives $B_{\psi }(f,f)=0$; by step 1.2, $B_{\psi }(f,h)=0$ for every $h$. Hermitian symmetry gives independence in the second variable as well. Hence $b_0$ is well-defined on $D$. If $\phi =0$, this also gives $B_{\psi }({\delta }_g,{\delta }_e)=\psi (g)=0$ for every $g$, and the quotient is zero.

For $u=[f],v=[h]\in D$, steps 1.1–1.2 give $0\le b_0(u,u)\le \mathrm{\parallel }u{\mathrm{\parallel }}^2,\mathrm{\mid }{b}_0(u,v)\mathrm{\mid }\le \mathrm{\parallel }u\mathrm{\parallel }\mathrm{\parallel }v\mathrm{\parallel }\mathrm{.}$ Since $D$ is dense in $H_{\phi }$, this bound extends $b_0$ uniquely to a continuous Hermitian sesquilinear form $b$ on $H_{\phi }\times H_{\phi }$, still satisfying $0\le b(u,u)\le \mathrm{\parallel }u{\mathrm{\parallel }}^2$.

Fix $u\in H_{\phi }$. The map $v\mapsto \overline{b(u,v)}$ is a bounded linear functional of norm at most $\mathrm{\parallel }u\mathrm{\parallel }$. By [F3] there is a unique $Tu$ with $\overline{b(u,v)}=⟨v,Tu⟩$, equivalently $b(u,v)=⟨Tu,v⟩$. Uniqueness of representing vectors and linearity of $b$ in its first argument show that $u\mapsto Tu$ is linear; the norm bound gives $\mathrm{\parallel }Tu\mathrm{\parallel }\le \mathrm{\parallel }u\mathrm{\parallel }$, so $T$ is bounded.

Hermitian symmetry gives $⟨Tu,v⟩=⟨u,Tv⟩$ for all $u,v$, so $T=T^{\ast }$ by uniqueness of the Hilbert adjoint. Also $⟨Tu,u⟩=b(u,u)\ge 0$ and $⟨(I-T)u,u⟩=\mathrm{\parallel }u{\mathrm{\parallel }}^2-b(u,u)\ge 0$. Thus $T$ and $I-T$ are positive.

For $k\in G$, simultaneous left translation leaves each kernel entry $\psi (h^{-1}g)$ unchanged, since $(kh{)}^{-1}(kg)=h^{-1}g$. Hence $b({\pi }_{\phi }(k)u,{\pi }_{\phi }(k)v)=b(u,v)$ first on $D$ and then on all of $H_{\phi }$ by continuity. Using $b(u,v)=⟨Tu,v⟩$ and unitarity, this identity gives $⟨{\pi }_{\phi }(k{)}^{-1}T{\pi }_{\phi }(k)u,v⟩=⟨Tu,v⟩$ for all $u,v$. Therefore $T{\pi }_{\phi }(k)={\pi }_{\phi }(k)T$, so $T\in {\pi }_{\phi }(G{)}^{\mathrm{\prime }}$.

On point masses, the form formula gives $b({\pi }_{\phi }(g){\xi }_{\phi },{\xi }_{\phi })=B_{\psi }({\delta }_g,{\delta }_e)=\psi (g)$. Since $T$ commutes with ${\pi }_{\phi }(g)$, this is $⟨{\pi }_{\phi }(g)T{\xi }_{\phi },{\xi }_{\phi }⟩$.

Suppose $S$ is another bounded operator in the commutant with the same coefficient. For all $g,h\in G$, commutation and unitarity give $⟨S{\pi }_{\phi }(g){\xi }_{\phi },{\pi }_{\phi }(h){\xi }_{\phi }⟩=⟨S{\pi }_{\phi }(h^{-1}g){\xi }_{\phi },{\xi }_{\phi }⟩=\psi (h^{-1}g)\mathrm{.}$ The same identity holds for $T$. The orbit vectors span the dense subspace $D$, so boundedness and continuity imply $⟨(S-T)u,v⟩=0$ for all $u,v\in H_{\phi }$, whence $S=T$. This proves uniqueness, including the zero space.

AC is used to infer Countable Choice for the GNS completion and bounded extensions and for the Riesz representation in step 3.1; it also satisfies the hypotheses of the adjoint and positive-operator definitions. The finite form calculations, extension from the specified dense orbit span, uniqueness, and converse matrix tests use no further choice.