States and positive functionals on a C star algebra

Let $A$ be a complex C*-algebra (C star algebra) and recall that positivity in $A$ is the algebraic condition $a=b^{\ast }b$, without spectral hypotheses (Self-adjoint positive unitary and normal elements). A linear functional $\omega :A\to \mathbb{C}$ is positive if $\omega (a^{\ast }a)\ge 0\text{for every }a\in A,$ and a state if in addition $\mathrm{\parallel }\omega \mathrm{\parallel }=1$, the norm being the operator norm of the bounded linear functional on $A$.

• The positive functionals form a convex cone. If $\omega ,\phi$ are positive, $s,t\ge 0$ and $a\in A$, then $(s\omega +t\phi )(a^{\ast }a)=s\omega (a^{\ast }a)+t\phi (a^{\ast }a)\ge 0$. In particular the positive functionals of norm at most one form a convex set, since $\mathrm{\parallel }s\omega +t\phi \mathrm{\parallel }\le s+t$ whenever $s,t\ge 0$. Being bounded with norm one is part of the definition of a state, not a consequence claimed here.
• Cauchy–Schwarz. Every positive functional satisfies $\mathrm{\mid }\omega (b^{\ast }a){\mathrm{\mid }}^2\le \omega (a^{\ast }a)\omega (b^{\ast }b)(a,b\in A)\mathrm{.}$ Indeed, for all $z\in \mathbb{C}$ linearity gives $\omega ((a+zb{)}^{\ast }(a+zb))=\omega (a^{\ast }a)+z\omega (a^{\ast }b)+\bar{z}\omega (b^{\ast }a)+\mathrm{\mid }z{\mathrm{\mid }}^2\omega (b^{\ast }b)\ge 0.$ The left side is a nonnegative real number for every $z$. Taking $z=1$ and $z=i$ and using that both resulting values are real shows $\omega (a^{\ast }b)+\omega (b^{\ast }a)\in \mathbb{R}$ and $i(\omega (a^{\ast }b)-\omega (b^{\ast }a))\in \mathbb{R}$; hence $\omega (b^{\ast }a)=\overline{\omega (a^{\ast }b)}$. Writing $p=\omega (a^{\ast }a)\ge 0$, $q=\omega (b^{\ast }b)\ge 0$ and $u=\omega (a^{\ast }b)$, the displayed inequality reads $p+2\mathrm{Re}(zu)+\mathrm{\mid }z{\mathrm{\mid }}^{2}q\ge 0$ for all $z$. If $q>0$, insert $z=-\bar{u}\mathrm{/}q$ to obtain $p-\mathrm{\mid }u{\mathrm{\mid }}^2\mathrm{/}q\ge 0$, that is $\mathrm{\mid }u{\mathrm{\mid }}^2\le pq$; if $q=0$, the same inequality forces $\mathrm{Re}(zu)\ge -p\mathrm{/}2$ for all $z$, which is impossible unless $u=0$, and then $\mathrm{\mid }u{\mathrm{\mid }}^2=0\le pq$. Since $\mathrm{\mid }\omega (b^{\ast }a)\mathrm{\mid }=\mathrm{\mid }\overline{\omega (a^{\ast }b)}\mathrm{\mid }=\mathrm{\mid }u\mathrm{\mid }$, this is the stated inequality.
• Continuity and the norm formula. Assume AC for the calculus and approximate-unit suppliers (The Axiom of Choice, Positive calculus and order estimates in a C star algebra, Positive contractive approximate units for C star algebras and ideals). Write $A^{+}$ here for the positive cone, not the unitization, and put $M=\sup \{\omega (b):b\in A^{+},\mathrm{\parallel }b\mathrm{\parallel }\le 1\}$. This supremum is finite: otherwise choose positive contractions $b_n$ with $\omega (b_n)>n{2}^n$. The norm-convergent series $b={\sum }_{n\ge 1}{2}^{-n}{b}_n$ is positive, and $b-2^{-n}{b}_n$ is positive because the positive cone is closed. Positivity would give $\omega (b)\ge 2^{-n}\omega (b_n)>n$ for every $n$, a contradiction. Every self-adjoint $h$ is $h_{+}-h_{-}$ with $h_{\pm }\ge 0$ and $\mathrm{\parallel }{h}_{\pm }\mathrm{\parallel }\le \mathrm{\parallel }h\mathrm{\parallel }$ by the calculus. Thus $\omega (h)$ is real and $\mathrm{\mid }\omega (h)\mathrm{\mid }\le M\mathrm{\parallel }h\mathrm{\parallel }$. Decomposing $x=h+ik$, with $\mathrm{\parallel }h\mathrm{\parallel },\mathrm{\parallel }k\mathrm{\parallel }\le \mathrm{\parallel }x\mathrm{\parallel }$, gives $\mathrm{\mid }\omega (x)\mathrm{\mid }\le 2M\mathrm{\parallel }x\mathrm{\parallel }$, proving continuity. For a positive contractive approximate unit $u_{\lambda }$, Cauchy–Schwarz gives $\mathrm{\mid }\omega (u_{\lambda }x){\mathrm{\mid }}^2\le \omega (x^{\ast }x)\omega (u_{\lambda }^2)\le M^2\mathrm{\parallel }x{\mathrm{\parallel }}^2$. Passing to $u_{\lambda }x\to x$ gives $\mathrm{\parallel }\omega \mathrm{\parallel }\le M$; the reverse inequality follows from the definition of the operator norm. Hence $\mathrm{\parallel }\omega \mathrm{\parallel }=M$. A nonzero positive functional therefore becomes a state upon division by its norm. The zero algebra has no state.