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For a state on a C-star algebra $x$ on a unital algebra $A$, put $⟨[a],[b]⟩=x(b^{\ast }a)$ on $A/N_x$, with the inner product linear in its first argument. The Cauchy–Schwarz inequality for positive C-star functionals makes the null vectors pair to zero with every vector. Moreover $c^{\ast }c\le \Vert c{\Vert }^{2}1$ gives $x((ca{)}^{\ast }(ca))\le \Vert c{\Vert }^{2}x(a^{\ast }a)$, so $N_x$ is a left ideal and $[a]\mapsto [ca]$ is well-defined and bounded by $\Vert c\Vert$. Completion gives a Hilbert space $H_x$ and a star-representation ${\pi }_x(c)[a]=[ca]$. The adjoint identity follows from $x(b^{\ast }ca)=⟨[a],{\pi }_x(c^{\ast })[b]⟩$. The unit vector ${\xi }_x=[1]$ is cyclic and recovers the state as $x(a)=⟨{\pi }_x(a){\xi }_x,{\xi }_x⟩$.

A state on a C-star algebra is a positive linear functional of norm one. In the unital case this is equivalent to positivity and $x(1)=1$. A unital star-homomorphism $\pi :A\to \mathcal{B}(K)$ is a cyclic representation of a C-star algebra if it has a vector $\xi$ with $\overline{\pi (A)\xi }=K$. For the representation associated with a state, the cyclic vector is taken to have norm one.

Use inner products linear in the first argument, as in the displayed formula for the projection in the source. In the Gelfand–Naimark–Segal construction, setThe Cauchy–Schwarz inequality for positive C-star functionals makes this inner product well-defined on $A/N_x$. Indeed a null vector pairs to zero with every vector. The set $N_x$ is a left ideal: $c^{\ast }c\le \Vert c{\Vert }^{2}1$ impliesso $a\in N_x$ implies $ca\in N_x$. The same estimate shows that left multiplication induces a bounded operatorComplete the quotient pre-Hilbert space to $H_x$, and extend these operators by continuity. The formula for the inner product gives ${\pi }_x(c{)}^{\ast }={\pi }_x(c^{\ast })$, while multiplication and the unit are preserved. Put ${\xi }_x=[1]$. Then $\Vert {\xi }_x\Vert =1$, the vectors ${\pi }_x(a){\xi }_x=[a]$ are dense in $H_x$, andThis gives the requested cyclic representation for an arbitrary state.

For the specified unit vector $\eta$, the functional $x_{\eta }$ is complex-linear and obeys $|x_{\eta }(a)|\le \Vert a\eta \Vert \Vert \eta \Vert \le \Vert a\Vert$. MoreoverThus it is positive and has norm exactly one, proving that $x_{\eta }$ is a vector state on a C-star algebra.

The rank-one projection satisfiesIf $a\eta =0$, this is zero for every $\zeta$, so $a{p}_{\eta }=0$. Conversely, evaluation at $\zeta =\eta$ gives $(a{p}_{\eta })\eta =a\eta$. HenceIn particular $N_{x_{\eta }}=\{a:a\eta =0\}=\{a:a{p}_{\eta }=0\}$.

The concrete GNS space is the completion of $A/N_{x_{\eta }}$ withDefine $U_0[a]=a\eta$. The description of the null ideal proves it is well-defined, and the displayed equality proves it is a linear isometry with range $A\eta$. It therefore extends to an isometry $U:H_{x_{\eta }}\to \overline{A\eta }$. The extended range is closed, since its domain is complete, and contains the dense set $A\eta$. Thus $U$ is onto the required closure.

In fact here $A=\mathcal{B}(H)$, so $A\eta =H$: given any $\zeta \in H$, the bounded rank-one operator $b(v)=⟨v,\eta ⟩\zeta$ sends $\eta$ to $\zeta$. Thus $U$ is a unitary map onto all of $H$, and $U{\xi }_{x_{\eta }}=\eta$.

Finally, on the dense quotient vectors,Both sides are bounded operators, so this identity extends to all of $H_{x_{\eta }}$. Multiplication by the inverse unitary map givesThis is the GNS representation of a vector state, with the vector state, null ideal, cyclic vector and intertwining isometry all made explicit.

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 $\mathrm{Re}x(1_A)=1$.

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.

Apply the correct Cauchy–Schwarz inequality for positive C-star functionals to $x^{\ast }$ and $1$. Its left side is $|f(x){|}^2$, and its diagonal values are $f(x{x}^{\ast })$ and $f(1)$. Thus the printed conclusion of this part is valid:This does not depend on the false mixed-order version in the preceding part.

The inequality printed in this part is false for general positive functionals. In $A=M_2(\mathbb{C})$ take the state on a C-star algebra $f(a)=a_{22}$ and $x=y=E_{12}$. It is positive since $f(z^{\ast }z)=\Vert z{e}_2{\Vert }^2\ge 0$, and it has norm one. ButThus the literal requested inequality would say $1\le 0$.

The correct Cauchy–Schwarz inequality for positive C-star functionals keeps the orders consistent:To prove it, set $a=f(x^{\ast }x)$, $d=f(y^{\ast }y)$ and $b=f(x^{\ast }y)$. By part (i) and positivity,If $d>0$, choose $t=-\overline{b}/d$ to obtain $|b{|}^2\le ad$. If $d=0$, arbitrary choices of $t$ force $b=0$, and the inequality again follows. Replacing $x,y$ by $x^{\ast },y^{\ast }$ also yields the valid alternativeEither replacing the printed right-hand factors by $f(x^{\ast }x),f(y^{\ast }y)$, or replacing the left-hand product by $x{y}^{\ast }$, repairs the statement. No tracial hypothesis is supplied, so it cannot be silently used to exchange those products.

The functional's stipulated positivity initially means $f(z^{\ast }z)\ge 0$, without assuming continuity. For a Hermitian element of a C-star algebra $h$, the identityexpresses it as a real linear combination of two adjoint squares. Their functional values are real, so $f(h)$ is real. Write $x=h+ik$ with $h=(x+x^{\ast })/2$ and $k=(x-x^{\ast })/(2i)$ Hermitian. Complex linearity gives $f(x)=f(h)+if(k)$ and $f(x^{\ast })=f(h)-if(k)$. ThereforeThis is the Hermitian symmetry used in the Cauchy–Schwarz inequality for positive C-star functionals.

A complex linear functional $\tau$ on a C-star algebra is positive when $\tau (a)\ge 0$ for every Positive element of a C-star algebra $a$. Such a linear functional is automatically bounded and satisfies the Cauchy–Schwarz inequality for positive C-star functionals:For a unital C-star algebra, positivity gives $\Vert \tau \Vert =\tau (1)$. Conversely, a bounded complex linear functional with $\Vert \tau \Vert =\tau (1)$ is positive. The unit assumption belongs to these formulas; it is not needed for the definition of positivity.

A concrete example on complex matrices is $\tau (a)=\mathrm{tr}(\rho a)$ with $\rho$ a positive semidefinite matrix. It is a state on a C-star algebra when $\mathrm{tr}\rho =1$. The positivity follows from $\mathrm{tr}(\rho b^{\ast }b)=\mathrm{tr}((b{\rho }^{1/2}{)}^{\ast }b{\rho }^{1/2})\ge 0$.