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.

An element $a$ is a Normal element of a C-star algebra if $a^{\ast }a=a{a}^{\ast }$. It is a Unitary element of a C-star algebra if $a^{\ast }a=a{a}^{\ast }=1$. Its Banach algebra spectrum iswhere $A^{\times }$ is the group of invertible elements of a Banach algebra.

For normal $a$, the continuous functional calculus is the unique unital isometric star-isomorphismwhich sends the coordinate function $z\mapsto z$ to $a$. It preserves sums, products, complex scalars and the involution: ${\Phi }_a(\overline{f})={\Phi }_a(f{)}^{\ast }$. In particular $\Vert f(a)\Vert ={\max }_{z\in {\sigma }_A(a)}|f(z)|$. The spectral mapping theorem in this setting statesThe common unit and spectral permanence for C-star subalgebras permit computing this spectrum in $A$ itself.

For unitary $u$, the C-star identity gives $\Vert u\Vert =\Vert u^{\ast }\Vert =1$. If $|z|>1$, $u-z1=-z(1-u/z)$ has a multiplicative inverse by the Neumann series. If $|z|<1$, write $u-z1=u(1-z{u}^{\ast })$ and apply the same series. Therefore ${\sigma }_A(u)$ lies on the unit circle.

For $|t|=1$, the denominator $1+\overline{\lambda }t$ is nonzero because $|\lambda |<1$. The function is consequently continuous on the spectrum, andThus $|f(t)|=1$ throughout ${\sigma }_A(u)$, so $\overline{f}f=f\overline{f}=1$ there. Applying the star-homomorphism ${\Phi }_u$ provesHence $f(u)$ is unitary. Explicitly, the same calculus giveswith the denominator's inverse also supplied by a norm-convergent Neumann series. This is the Möbius transform of a unitary element. The conjugation on $\lambda$ in the denominator is essential to the unit-modulus identity.

A unital Banach star-algebra is a complex unital Banach algebra with a continuous conjugate-linear algebra involution satisfyingThe algebra norm is complete and submultiplicative, with the unit conventionally normalized to norm one. Many definitions additionally require the involution to be isometric, $\Vert a^{\ast }\Vert =\Vert a\Vert$; the function algebra here satisfies this stronger convention as well. A Banach star-algebra does not by definition require the extra C-star identity $\Vert a^{\ast }a\Vert =\Vert a{\Vert }^2$.

Here is the lattice proof of the Stone-Weierstrass theorem in its complex self-adjoint form. Let $B$ be the uniform closure of the given unital star-subalgebra in $C(\Omega )$, and let $B_{\mathbb{R}}$ be its real-valued part. Closure preserves multiplication and complex conjugation, so $B$ is still a unital star-algebra. Real and imaginary parts lie in $B_{\mathbb{R}}$, and separation by the complex algebra implies separation by its real-valued part.

For $h\in B_{\mathbb{R}}$, polynomial approximation of absolute value on a bounded interval gives $|h|\in B_{\mathbb{R}}$. For example, this polynomial approximation follows from the binomial series for the square root, whose coefficients give uniform convergence on the relevant closed interval. Thusalso lie in $B_{\mathbb{R}}$. It is therefore a lattice of functions.

To see how the lattice gives density, fix a real continuous target $f$ and a tolerance $ϵ>0$. For any two distinct points $x,y$, separation and constants provide $h_{xy}\in B_{\mathbb{R}}$ with $h_{xy}(x)=f(x)$ and $h_{xy}(y)=f(y)$, by an affine rescaling of a separating function. For $x=y$, use the constant $f(x)$.

Fix $x$. The sets where $h_{xy}>f-ϵ$ cover $\Omega$ as $y$ varies. A finite subcover and a pointwise maximum give $g_x\in B_{\mathbb{R}}$ with $g_x>f-ϵ$ everywhere and $g_x(x)=f(x)$. The neighborhoods where $g_x<f+ϵ$ now cover $\Omega$ as $x$ varies. A second finite subcover and a pointwise minimum give $h\in B_{\mathbb{R}}$ satisfyingThis is lattice approximation from two-point approximation. Using half the desired tolerance gives strict uniform error below the desired bound. Since $B_{\mathbb{R}}$ is closed, it contains every real continuous function. Splitting a complex function into real and imaginary parts proves $\boxed{B=C(\Omega )}$.

The annulus test. For the closed annulus, the family holomorphic in the interior is not dense in the full continuous-function algebra. A uniform limit on the annulus is locally uniform in its interior, so locally uniform convergence of holomorphic functions makes that limit holomorphic there. Thus the proposed family is closed in $C(\Omega )$. It is proper because $z\mapsto \overline{z}$ is continuous but not holomorphic on the annulus interior.

There is also an explicit continuous functional detecting the obstruction. Choose a small closed disk of radius $r$ contained in the annulus interior and let $L(h)$ be the integral of $h(z)dz$ around its positively oriented boundary. The Cauchy integral theorem gives $L(h)=0$ for every member of the holomorphic family, while parametrizing the circle gives $L(\overline{z})=2\pi i{r}^2\ne 0$. Since $|L(h)|\le 2\pi r\Vert h{\Vert }_{\infty }$, no uniform approximation of $\overline{z}$ by that family is possible.