Past exam of the functional Analysis and Banach and C*-Algebras course of the University of Oxford / 2010 / B / 6

Let $c=x(1_A)\ge 0$. Positivity on $a^{\ast }a$ implies positivity on every Positive element of a C-star algebra, by taking its positive square root. For a self-adjoint C-star element $h$, the positive cone of a C-star algebra givesBoth $x(\Vert h\Vert 1_A+h)$ and $x(\Vert h\Vert 1_A-h)$ are nonnegative real numbers. Their difference shows $x(h)$ is real, and the same inequalities give $|x(h)|\le c\Vert h\Vert$.

For an arbitrary $a\in A$, write $a=h+ik$ with $h=(a+a^{\ast })/2$ and $k=(a-a^{\ast })/(2i)$ self-adjoint. Since the involution is isometric, $\Vert h\Vert ,\Vert k\Vert \le \Vert a\Vert$. HenceThus every positive functional on a C-star algebra is bounded, with the preliminary bound $\Vert x\Vert \le 2x(1_A)$. This proof does not presuppose the Cauchy–Schwarz inequality for positive C-star functionals. It also gives $x(a^{\ast })=\overline{x(a)}$, since $x(h)$ and $x(k)$ are real.

Set $A_0=x(a^{\ast }a)$, $B_0=x(b^{\ast }b)$ and $c=x(a^{\ast }b)$. The preceding reality argument gives $x(b^{\ast }a)=\bar{c}$. Positivity therefore implies, for every $\lambda \in \mathbb{C}$,If $A_0>0$, choose $\lambda =-c/A_0$ to obtain $0\le B_0-|c{|}^2/A_0$. If $A_0=0$ and $c\ne 0$, choosing $\lambda =-tc$ and letting $t\to \infty$ contradicts positivity. Thus $c=0$ in that case too. This proves the Cauchy–Schwarz inequality for positive C-star functionals:Apply it with $a=1_A$, $b=z$. Since $0\le z^{\ast }z\le \Vert z{\Vert }^2{1}_A$, positivity givesThus $\Vert x\Vert \le x(1_A)$, while evaluation on $1_A$ gives the reverse inequality. ConsequentlyThe zero functional is covered as well. In the trivial zero algebra all assertions are immediate, and its state space is empty.

For a nonzero unital C-star algebra, the set $S_A$ of state on a C-star algebra functionals is convex, since both positivity and evaluation $x(1_A)=1$ are preserved by convex combinations. The norm identity places it in the closed dual unit ball. Suppose a state satisfies $x=ty+(1-t)z$, with $0<t<1$ and $\Vert y\Vert ,\Vert z\Vert \le 1$. Evaluating at the unit and taking real parts givesSince both weights are strictly positive, both real parts equal $1$. Also $|y(1_A)|,|z(1_A)|\le 1$, so $y(1_A)=z(1_A)=1$. HenceThe stipulated converse to the norm identity makes $y,z$ positive, so both belong to $S_A$. Therefore $S_A$ is a face of a convex set, namely the closed dual unit ball. More precisely it is the exposed face of the dual ball defined by the unit, since it is cut out by $\mathrm{Re}x(1_A)=1$.