For a bounded linear functional on a unital C*-algebra, the state criterion isEquivalently it is positive and takes the unit to one.

In the Gelfand–Naimark–Segal construction, set $N_x=\{b:x(b^{\ast }b)=0\}$. This is a left ideal. Give $A/N_x$ the inner product $⟨[b],[c]⟩=x(c^{\ast }b)$, linear in the first variable, and complete it to a Hilbert space $H_x$. Left multiplication gives the bounded *-representation ${\pi }_x(a)[b]=[ab]$. The vector $\xi =[1_A]$ is cyclic and satisfies $x(a)=⟨{\pi }_x(a)\xi ,\xi ⟩$.

The Hermitian matrix $\rho$ has eigenvalues ${\lambda }_{\pm }=1/2\pm |\alpha |$ and trace one. In an orthonormal eigenbasis $v_{+},v_{-}$,If $|\alpha |\le 1/2$, both weights are nonnegative and sum to one, so this functional is positive and normalized, hence a state. If $|\alpha |>1/2$, evaluating on the positive rank-one projection onto the negative-eigenvalue eigenspace gives a negative value, so it is not a state. ThereforeTo obtain the representation dimension, cyclicity of trace givesThus $A/N_{x_{\rho }}$ identifies isometrically with the matrix space $\{b{\rho }^{1/2}:b\in M_2(\mathbb{C})\}$, with left multiplication as the representation. If $r=\mathrm{rank}\rho$, this space consists, after a unitary change of basis, of arbitrary first $r$ columns and zero remaining columns. Its complex dimension is $2r$. It is already complete because it is finite-dimensional. Rank is one at $|\alpha |=1/2$ and two at $|\alpha |<1/2$, proving

A character of an algebra is a nonzero multiplicative complex-linear map $\chi :A\to \mathbb{C}$. A maximal proper ideal is a proper ideal contained in no larger proper ideal. For a character, $\chi (1)=1$ and $\chi (a)\in \sigma (a)$: otherwise $a-\chi (a)1$ would be invertible although its character value is zero. The spectral bound gives $|\chi (a)|\le \Vert a\Vert$, so the character is continuous. It is onto, and its kernel is a closed maximal ideal because $A/\ker \chi \cong \mathbb{C}$.

Conversely let $M$ be maximal. Its closure is an ideal and remains proper: otherwise an element of $M$ would lie within distance one of the unit, hence would be invertible by the Neumann series, forcing $1\in M$. Maximality then gives $\overline{M}=M$. The Banach algebra $A/M$ is a field. For any $b$ in this quotient choose $\lambda \in \sigma (b)$, using nonemptiness of the complex spectrum. In a field only zero is noninvertible, so $b-\lambda 1=0$. Hence every quotient element is a scalar multiple of the unit. The quotient map therefore determines a character ${\chi }_M$, uniquely normalized by ${\chi }_M(1)=1$. These constructions are inverse:For the convolution algebra, write $e_j$ for the sequence supported at $j$ with value one. Then $e_j{e}_k=e_{j+k}$, $e_0=1$, and $e_1{e}_{-1}=1$. If $z=\chi (e_1)$, continuity and $\Vert e_{\pm 1}{\Vert }_1=1$ imply $|z|\le 1$ and $|z^{-1}|\le 1$. Thus $|z|=1$, so uniquely $z=e^{it}$ for $0\le t<2\pi$, and $\chi (e_j)=e^{ijt}$ for every integer $j$. Absolute convergence in ${\ell }^1$ now givesConversely each displayed map is bounded, nonzero, and multiplicative, since the absolutely convergent double sum can be rearranged along $k+l=j$. Thus these are all characters, and all maximal proper ideals areThey are distinct because the character is uniquely recovered from its kernel and its value on the unit.

First suppose $x$ is a positive linear functional, without assuming boundedness. Every Hermitian C-star element $h$ is a difference of positive elements, since $\Vert h\Vert 1\pm h\ge 0$. Thus $x(h)$ is real. For a Hermitian C-star element $h$ of norm at most one, $-1\le h\le 1$ gives $|x(h)|\le x(1)$.

For arbitrary $a$, choose a scalar $\zeta$ of modulus one so that $\zeta x(a)=|x(a)|$. The real part of $\zeta a$ is a Hermitian C-star element and has norm at most $\Vert a\Vert$. Since $x$ is real on Hermitian C-star elements, it respects real parts, givingThis proves boundedness and $\Vert x\Vert \le x(1)$. Evaluation at the unit gives the reverse inequality, so positivity implies $\Vert x\Vert =x(1)$.

Conversely let $x$ be bounded with $x(1)=\Vert x\Vert =c\ge 0$. If $c=0$, $x=0$ is positive. Otherwise normalize to $c=1$. For a Hermitian C-star element $h$ and real $t$, the C-star identity givesWriting $x(h)=a+ib$ and squaring yields $-2tb+t^2(a^2+b^2)\le t^2\Vert h{\Vert }^2$. Taking $t\to 0$ through positive and negative values forces $b=0$. Hence $x$ is real on Hermitian C-star elements. If $0\le a\le 1$, then $1-a$ has norm at most one, so $x(1-a)\le 1$ and $x(a)\ge 0$. Scaling proves nonnegativity on every positive element, in particular every $b^{\ast }b$. Therefore

A Unitary element of a C-star algebra satisfies $u^{\ast }u=u{u}^{\ast }=1$. A projection in a C-star algebra satisfies $p=p^{\ast }=p^2$.

If $q=u^{\ast }u$ is a projection in a C-star algebra, the C-star identity gives $\Vert u(1-q){\Vert }^2=\Vert (1-q)q(1-q)\Vert =0$. Hence $u=uq=u{u}^{\ast }u$. Conversely $u{u}^{\ast }u=u$ implies $(u^{\ast }u{)}^2=u^{\ast }u$, and $u^{\ast }u$ is a Hermitian C-star element, so it is a projection in a C-star algebra. Taking adjoints of the triple-product equality gives the corresponding equality for $u^{\ast }$, provingFor nonzero $u$, let $p=u{u}^{\ast }$ and $q=u^{\ast }u$. These are projections, $u=pu=uq$, and $\Vert u\Vert =1$. The rectangular corner $pAq$ is closed because it is the range of the bounded idempotent map $a\mapsto paq$.

Define $\Theta :pAq\to qAq$ by $\Theta (a)=u^{\ast }a$. It lands in the corner, and its inverse is $b\mapsto ub$. The identities $u{u}^{\ast }a=pa=a$ and $u^{\ast }ub=qb=b$ verify both compositions. Both maps are contractive because $\Vert u\Vert =\Vert u^{\ast }\Vert =1$, so $\Theta$ is an isometry. For the proposed multiplication and involution,Thus the operations correspond exactly to the ordinary operations in the closed C-star algebra $qAq$. This proves associativity, closure, all involution identities and the C*-norm identity at once. Moreover $\Theta (u)=q$, so the unit of $A_u$ is $u$. If $u$ is unitary, $p=q=1$, and the map $a\mapsto u^{\ast }a$ is an onto isometric *-isomorphism from $A_u$ to $A$.

The spectrum of an element is ${\sigma }_A(a)=\{\lambda \in \mathbb{C}:a-\lambda 1_A\text{ is not invertible in }A\}$. A Normal element of a C-star algebra satisfies $a^{\ast }a=a{a}^{\ast }$.

For a Hermitian C-star element $h$, the C-star identity gives $\Vert h^2\Vert =\Vert h{\Vert }^2$. For normal $a$, commutation givesAll powers remain normal, so induction yields $\Vert a^{2^k}\Vert =\Vert a{\Vert }^{2^k}$. Taking the spectral-radius limit along these powers givesLet $\phi :A\to B$ be a C-star homomorphism, without assuming continuity or preservation of the ambient unit. Put $p=\phi (1_A)$, a projection in a C-star algebra. If $p=0$, then $\phi =0$. Otherwise its image lies in the unital corner $pBp$ and it is unital as a map into that corner. An inverse in $A$ maps to an inverse in the corner, so ${\sigma }_{pBp}(\phi (b))\subseteq {\sigma }_A(b)$. Apply this to the normal element $b=a^{\ast }a$ and use the C-star identity in both algebras:Thus every C-star homomorphism is norm non-increasing, including nonunital maps.

For the real lattice version of the Stone-Weierstrass theorem, let $f\in C_{\mathbb{R}}(\Omega )$ and $\epsilon >0$. For any $x,y\in \Omega$, separation of points and constants give $h_{xy}\in L$ matching $f$ at both $x$ and $y$: take an affine combination of a separating function, and for $x=y$ take a constant. Fix $x$. The open sets where $h_{xy}>f-\epsilon$ cover $\Omega$ as $y$ varies. Compactness gives finitely many such functions; their maximum $g_x$ belongs to the lattice $L$, satisfies $g_x>f-\epsilon$ everywhere, and has $g_x(x)=f(x)$. The open sets where $g_x<f+\epsilon$ now cover $\Omega$ as $x$ varies. Choose finitely many and take their minimum $h\in L$. Then $f-\epsilon <h<f+\epsilon$ everywhere. Thus $L$ is uniformly dense in the real continuous functions.

For the complex *-algebra, its real subalgebra $A_{\mathbb{R}}$ of Hermitian C-star elements contains the real and imaginary parts of every member. It separates points, because at least one of those parts distinguishes any pair distinguished by a complex function. Its uniform closure is a unital real subalgebra and is a lattice by the allowed result. The real theorem makes this closure all of $C_{\mathbb{R}}(\Omega )$. Approximating the real and imaginary parts separately proves $A$ is uniformly dense in $C(\Omega )$.

On $[0,1]$, the algebra of polynomials in $t^2$ is unital, closed under conjugation, and separates points because $t\mapsto t^2$ is injective there. Choose $p(t^2)$ within $\epsilon /3$ of $a(t)$, and correct its endpoints byThis is still a polynomial in $t^2$ and has exactly the requested endpoint values. The correction has absolute value at most $\epsilon /3$, because it is a convex combination of two errors of that size. Therefore

The open mapping theorem says that a bounded surjective linear map $T:E\to F$ between Banach spaces is open. Here is a proof using only Baire and completeness. Surjectivity gives $F={\bigcup }_{n\ge 1}T(n{\overline{B}}_E)$, hence also a cover by their closures. Baire makes one such closure contain a ball $B(y_0,r)$. Taking differences gives $B(0,r)\subseteq \overline{T(2n{\overline{B}}_E)}$. Scaling yields some $\delta >0$ withIf $\Vert y\Vert <\delta /2$, recursively use scaled versions of this inclusion to choose $x_k$ with $\Vert x_k\Vert \le 2^{-k}$ and residual $\Vert y-T{\sum }_{j=1}^k{x}_j\Vert <\delta 2^{-k-1}$. The sum converges in $E$ to an element of norm at most one, whose image is $y$. Thus $B(0,\delta /2)\subseteq T({\overline{B}}_E)$, and scaling gives $B(0,\delta /4)\subseteq T(\frac{1}{2}{\overline{B}}_E)\subseteq T(B_E)$. Translating and scaling proves openness on every open set. This proof does not use the closed graph or inverse mapping theorem.

Apply it to the evaluation map $E:X\to {\ell }^{\infty }$, $Ef=(f(z_n){)}_n$. It is bounded with norm at most one and onto by hypothesis. Hence its image of the unit ball contains a ball about zero, yielding a constant $c$ and, for every $\alpha$, an interpolant $g$ withFor completeness, scale a ball of radius $\eta$ inside $E(B_X)$ by $2\Vert \alpha {\Vert }_{\infty }/\eta$ to obtain $c=2/\eta$; zero data use $g=0$.

For each $n$, interpolate the data which are one at index $n$ and zero elsewhere, obtaining $g_n$ of norm at most $c$. The supplied factorization property gives $g_n=k_n{h}_n$ with $\Vert k_n{\Vert }_{\infty }=\Vert g_n{\Vert }_{\infty }\le c$. Evaluation at $z_n$ gives $1=|k_n(z_n)h_n(z_n)|\le c|h_n(z_n)|$, soThe three enumerated properties of $h_n$ are supplied hypotheses for this argument, rather than additional requests to prove them.