Irreducible representations separate arbitrary C star algebras

Let $a=x^{\ast }x$. By [F1], $a$ is positive and $\mathrm{\parallel }a\mathrm{\parallel }=\mathrm{\parallel }x{\mathrm{\parallel }}^2>0$. Choose the unital C*-algebra $D$ of [F2], and set $C=C^{\ast }(1,a)\subseteq D$.

The positive calculus of Positive calculus and order estimates in a C star algebra gives $b=a^{1\mathrm{/}2}\in C^{\ast }(1,a)=C$ with $b=b^{\ast }$ and $b^{\ast }b=a$. Thus the Gelfand transform sends $a$ to the nonnegative continuous function $\widehat{a}(\chi )=\mathrm{\mid }\chi (b){\mathrm{\mid }}^2$ on the nonempty compact space $\mathrm{\Delta }(C)$. Its supremum norm is $\mathrm{\parallel }a\mathrm{\parallel }$ by [F4]. By [F5], its maximum is attained at some character ${\chi }_0\in \mathrm{\Delta }(C)$ and equals $\mathrm{\parallel }a\mathrm{\parallel }$.

The character ${\chi }_0$ has norm one and takes $1$ to $1$. Extend it by [F6] to $F\in D^{\ast }$ with $\mathrm{\parallel }F\mathrm{\parallel }=1$, $F(1)=1$, and $F(a)=\mathrm{\parallel }a\mathrm{\parallel }$.

Define $S(D)=\{\psi \in D^{\ast }:\mathrm{\parallel }\psi \mathrm{\parallel }\le 1,\psi (1)=1,\psi (b^{\ast }b)\ge 0\text{ for every }b\in D\}$. Every member is a state: the unit-ball bound and $\psi (1)=1$ give norm one, and the remaining condition is positivity; conversely every state belongs to this set by [F9]. It is convex because its normalization and positivity constraints are preserved by convex combinations, which remain in the unit ball. Within that ball its normalization and positivity conditions are intersections of closed evaluation constraints, so [F7] and the closed-subset compactness clause there make $S(D)$ weak-star compact. The weak-star topology is locally convex and Hausdorff by [F8].

If $c=c^{\ast }\in D$ and $t\in \mathbb{R}$, then $\mathrm{\mid }1+itF(c){\mathrm{\mid }}^2=\mathrm{\mid }F(1+itc){\mathrm{\mid }}^2\le \mathrm{\parallel }1+itc{\mathrm{\parallel }}^2=\mathrm{\parallel }1+t^2{c}^2\mathrm{\parallel }\le 1+t^2\mathrm{\parallel }c{\mathrm{\parallel }}^2$. Expanding the left side and letting $t$ tend to zero through positive and negative values forces $\mathrm{Im}F(c)=0$. If $0\le b\le 1$, then [F3] gives $\mathrm{\parallel }1-b\mathrm{\parallel }\le 1$; since $F(b)$ is real, $\mathrm{\mid }1-F(b)\mathrm{\mid }\le 1$ implies $F(b)\ge 0$. Scaling proves $F(b)\ge 0$ for every positive $b$, so $F$ is a state and $F(a)=\mathrm{\parallel }a\mathrm{\parallel }$.

Let $K=\{\psi \in S(D):\psi (a)=\mathrm{\parallel }a\mathrm{\parallel }\}$. It is nonempty by step 3.1 and is weak-star closed, hence compact. For every $\psi \in S(D)$, positivity and [F3], together with $\psi (1)=1$, give $0\le \psi (a)\le \mathrm{\parallel }a\mathrm{\parallel }$; therefore $K$ is convex and is a face of $S(D)$, since a proper convex combination can attain the upper bound only if both terms attain it.

By [F12], $K$ has an extreme point $\phi$. Because $K$ is a face of $S(D)$, any convex decomposition of $\phi$ in $S(D)$ has both terms in $K$; extremality in $K$ then makes both terms equal $\phi$. Thus $\phi$ is an extreme state of $D$, and $\phi (a)=\mathrm{\parallel }a\mathrm{\parallel }$.

Define $N=\{b\in D:\phi (b^{\ast }b)=0\}$. Cauchy–Schwarz [F9] makes $N$ a linear subspace, makes $⟨[b],[c]⟩=\phi (c^{\ast }b)$ independent of representatives, and makes it positive definite on $D\mathrm{/}N$; this form is linear in its first variable. For $d,b\in D$, [F3] gives $0\le \phi (b^{\ast }{d}^{\ast }db)\le \mathrm{\parallel }d{\mathrm{\parallel }}^2\phi (b^{\ast }b)$, so $N$ is a left ideal and left multiplication is bounded on the quotient with norm at most $\mathrm{\parallel }d\mathrm{\parallel }$. By [F10], the inner-product space $D\mathrm{/}N$ has a Hilbert completion $H$ under AC.

Define $\pi (d)[b]=[db]$ on $D\mathrm{/}N$ and extend it continuously to $H$. Then $\mathrm{\parallel }\pi (d)\mathrm{\parallel }\le \mathrm{\parallel }d\mathrm{\parallel }$, $\pi (de)=\pi (d)\pi (e)$, $\pi (1)=I_H$, and $\pi (d^{\ast })=\pi (d{)}^{\ast }$: the product and unit identities hold on quotient classes, and $⟨\pi (d)[b],[c]⟩=\phi (c^{\ast }db)=⟨[b],\pi (d^{\ast })[c]⟩$ gives the adjoint identity. The Hilbert adjoint has the properties used here by [F11].

The vector $\xi =[1]$ has norm one, the span of $\pi (D)\xi$ is dense by construction, and $\phi (d)=⟨\pi (d)\xi ,\xi ⟩$ for every $d\in D$. In particular, $\mathrm{\parallel }\pi (x)\xi {\mathrm{\parallel }}^2=\phi (x^{\ast }x)=\mathrm{\parallel }x{\mathrm{\parallel }}^2>0$. Thus $H\mathrm{\ne }\{0\}$ and $\pi (x)\mathrm{\ne }0$.

Suppose a nonzero proper closed subspace $M\subset H$ is invariant under every $\pi (d)$. Since $D$ is star-closed, $M^{\perp }$ is invariant too: for $v\in M^{\perp }$, $m\in M$, and $d\in D$, $⟨\pi (d)v,m⟩=⟨v,\pi (d^{\ast })m⟩=0$. By [F10] its orthogonal projection $E$ is a nonzero proper self-adjoint idempotent commuting with every $\pi (d)$. Cyclicity implies $t:=\mathrm{\parallel }E\xi {\mathrm{\parallel }}^2$ satisfies $0<t<1$, since either $E\xi =0$ or $(I-E)\xi =0$ would make $E=0$ or $E=I$ on the dense cyclic span.

Put ${\phi }_1(d)=t^{-1}⟨\pi (d)E\xi ,E\xi ⟩$ and ${\phi }_2(d)=(1-t{)}^{-1}⟨\pi (d)(I-E)\xi ,(I-E)\xi ⟩$. These vector functionals are positive; they have norm one because they take $1$ to $1$ and are bounded by one using $\mathrm{\parallel }\pi (d)\mathrm{\parallel }\le \mathrm{\parallel }d\mathrm{\parallel }$. Since $M$ and $M^{\perp }$ are invariant, the cross terms vanish and $\phi =t{\phi }_1+(1-t){\phi }_2$. The extremality of $\phi$ from step 5.1 gives ${\phi }_1={\phi }_2=\phi$.

For $b,c\in D$, commutation of $E$ with $\pi (D)$ and the orthogonality of $M$ and $M^{\perp }$ give $⟨E\pi (b)\xi ,\pi (c)\xi ⟩=⟨\pi (b)E\xi ,\pi (c)E\xi ⟩=t\phi (c^{\ast }b)=t⟨\pi (b)\xi ,\pi (c)\xi ⟩$. The vectors $\pi (b)\xi$ span a dense subspace, so continuity implies $E=t{I}_H$. This contradicts $E^2=E$ and $0<t<1$. Thus $\pi$ is irreducible.

If $A$ is unital, take $\sigma =\pi$. If $A$ is nonunital, its closed ideal property in [F2] makes $H_0=\stackrel{}{\mathrm{span}}\pi (A)H$ a reducing subspace for $\pi (D)$: for $d\in D$ and $a\in A$, both $da$ and $d^{\ast }a$ lie in $A$, so $\pi (d)$ and its adjoint preserve the dense spanning set. It is nonzero because $\pi (x)\xi \mathrm{\ne }0$. Irreducibility gives $H_0=H$, so $\sigma :=\pi {\mathrm{\mid }}_A$ is nondegenerate. Any closed $A$-invariant subspace is invariant under $\pi (D)=\pi (A)+\mathbb{C}{I}_H$, hence $\sigma$ is irreducible as well. In either case $\sigma$ is nonzero and $\sigma (x)\mathrm{\ne }0$.

We have constructed the required representation for each nonzero $x\in A$, so an element in the intersection of all the stated kernels must be zero. If $A=\{0\}$, there is no nonzero $x$, and the empty intersection inside $A$ is $A=\{0\}$, as stated.