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The weak topology $\sigma (X,X^{\ast })$ is the coarsest topology making every $f\in X^{\ast }$ continuous. Its basic neighbourhoods impose finitely many inequalities $|f_j(x-x_0)|<\epsilon$. The weak-star topology $\sigma (X^{\ast },X)$ is the coarsest topology making each evaluation $f\mapsto f(x)$ continuous, for $x\in X$.

The canonical embedding into the bidual isIt is linear and $\Vert J_{X}x\Vert \le \Vert x\Vert$. For $x\ne 0$, the Hahn-Banach theorem supplies a norming functional $f$ with $\Vert f\Vert =1$ and $f(x)=\Vert x\Vert$, proving equality. Thus $J_X$ is an isometric embedding. The coordinates defining the weak-star topology on $X^{\ast \ast }$, restricted to $J_X(X)$, are precisely $J_{X}x\mapsto f(x)$ for $f\in X^{\ast }$. These are exactly the coordinates defining the weak topology on $X$, so $J_X$ is also a topological embedding for these topologies.

A normed vector space is a reflexive space if $J_X$ is onto $X^{\ast \ast }$. Such a space is automatically a Banach space, since it is isometric to the complete continuous dual space $X^{\ast \ast }$. The Banach-Alaoglu theorem states that the closed unit ball of a continuous dual space is compact for its weak-star topology. The Goldstine theorem states that $J_X(B_X)$ is dense in the weak-star topology in $B_{X^{\ast \ast }}$. ConsequentlyIf $J_X$ is onto, its weak-to-weak-star homeomorphism carries $B_X$ onto the Banach-Alaoglu theorem's compact ball $B_{X^{\ast \ast }}$. Conversely, if $B_X$ is weakly compact, $J_X(B_X)$ is compact and hence closed in the Hausdorff weak-star topology. Its density from the Goldstine theorem makes it all of $B_{X^{\ast \ast }}$, and scaling gives surjectivity of $J_X$.

Now suppose $X$ is separable and nonzero. Choose a countable dense set $(u_n)$ in its unit sphere and, using the Hahn-Banach theorem, choose norming functionals $f_n$ with $\Vert f_n\Vert =1$ and $f_n(u_n)=1$. These continuous linear functionals separate points: for $x\ne 0$, choose $\Vert u_n-x/\Vert x\Vert \Vert <1/2$ to get $f_n(x)>\Vert x\Vert /2$. DefineThe series converges uniformly, separates distinct points, and obeys the triangle inequality because $\min (c,a+b)\le \min (c,a)+\min (c,b)$. Thus $d$ is a metric. Its topology is coarser than the weak topology: a finite set of coordinate controls makes the finite initial sum small, while the remaining sum is at most ${\sum }_{n>N}{2}^{-n}=2^{-N}$. On a weakly compact set $K$, the identity $(K,\text{weak})\to (K,d)$ is therefore a continuous bijection from a compact space into a Hausdorff space, hence a homeomorphism. This proves that $K$ is metrizable in its weak topology, an instance of countable separating family metrizes a weakly compact set.

If $X$ is also a reflexive space, its closed unit ball is weakly compact and therefore metrizable by this construction. Compactness in a metric space implies sequential compactness, so every sequence in $B_X$ has a weakly convergent subsequence. Scaling proves the claim for every bounded sequence.

For the converse, assume every bounded sequence in $X$ has a weakly convergent subsequence. Use the same metric $d$ on $B_X$. A weak limit of points of $B_X$ remains in $B_X$: $|f(x)|\le 1$ for every $\Vert f\Vert \le 1$, and the Hahn-Banach theorem recovers $\Vert x\Vert \le 1$. Thus every sequence in $B_X$ has a subsequence convergent in $d$ to a point of $B_X$, so $(B_X,d)$ is sequentially compact and hence compact.

We must still recover the entire weak topology, rather than just the selected coordinates. Suppose $x_n\to x$ in $d$ but not weakly. There is a neighbourhood in the weak topology $U$ of $x$ and a subsequence outside $U$. By hypothesis that subsequence has a further subsequence converging weakly to $y\in B_X$. It then converges in $d$ to $y$, so uniqueness of metric limits gives $y=x$, contradicting eventual membership in $U$. Hence convergence in $d$ implies weak convergence. Every $f\in X^{\ast }$ is therefore sequentially continuous on $(B_X,d)$ and hence continuous, since the domain is a metric space. The weak topology and $d$-topology on $B_X$ coincide. The ball is weakly compact, so the criterion above provesThe zero space satisfies all statements directly. The converse did not assume completeness, nor use the full Eberlein-Šmulian theorem.

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.