The domain carries $\sigma (Z^{\ast \ast },Z^{\ast })$, and the target carries $\sigma (X^{\ast },X)$. For each $x\in X\subseteq Z^{\ast }$,which is one of the defining weak-star continuous evaluations on $Z^{\ast \ast }$. Thus $R$ is weak-star continuous.

Restriction cannot increase a functional's norm, so $R(B_{Z^{\ast \ast }})\subseteq B_{X^{\ast }}$. Conversely, any $f\in B_{X^{\ast }}$ is a bounded linear functional on the subspace $X$ of $Z^{\ast }$. The Hahn-Banach theorem extends it to a functional $\theta \in (Z^{\ast }{)}^{\ast }=Z^{\ast \ast }$ with the same norm. Then $R\theta =f$ and $\Vert \theta \Vert \le 1$. ConsequentlyThis proves exact surjectivity on closed balls, not just density of the image.

A reflexive Banach space is one for which $J_X:X\to X^{\ast \ast }$ is onto. If $X$ is reflexive, every $\theta \in X^{\ast \ast }$ has the form $\theta (f)=f(x)$ for some $x\in X$. Thus the functionals defining $\sigma (X^{\ast },X^{\ast \ast })$ and $\sigma (X^{\ast },X)$ are precisely the same, proving equality of the topologies.

Conversely, assume those topologies coincide. For any $\theta \in X^{\ast \ast }$, the linear functional $f\mapsto \theta (f)$ on $X^{\ast }$ is then weak-star continuous. Its continuity at zero gives finitely many $x_1,\dots ,x_m\in X$ and $ϵ>0$ such thatIf all these evaluations are zero, arbitrary scalar multiples of $f$ still satisfy the inequalities, forcing $\theta (f)=0$. Hence $\theta$ factors through the finite-dimensional evaluation map $f\mapsto (f(x_1),\dots ,f(x_m))$. A linear functional on its image extends to ${\mathbb{R}}^m$, so for suitable scalars $a_j$,Thus $\theta$ lies in $J_X(X)$. Since it was arbitrary, $X$ is reflexive. We have proved

The weak topology is coarser than the norm topology, so norm convergence implies weak convergence, andFor the reverse inclusion, let $x\notin K={\overline{C}}^{\Vert \cdot \Vert }$. The set $K$ is closed and convex. The Hahn-Banach separation theorem supplies $f\in X^{\ast }$ and $a\in \mathbb{R}$ such that $f(x)>a\ge {\sup }_{y\in K}f(y)$. The set $\{z:f(z)>a\}$ is weakly open, contains $x$ and misses $C$. Hence $x$ is outside the weak closure as well. If $C$ is empty the equality is immediate. ThereforeThe separating functional explains the role of convexity: an arbitrary norm-closed subset need not be weakly closed.

The weak topology $\sigma (X,X^{\ast })$ is the coarsest topology on $X$ making every continuous linear functional $f\in X^{\ast }$ continuous. A basic neighborhood of $x_0$ is specified by finitely many inequalities $|f_j(x-x_0)|<ϵ$.

The weak-star topology $\sigma (X^{\ast },X)$ is the coarsest topology on $X^{\ast }$ making every evaluation $f\mapsto f(x)$, for $x\in X$, continuous. Its basic neighborhoods use finitely many such evaluations. Nets converge in these topologies exactly when all the corresponding scalar evaluations converge.

Let $J_X:X\to X^{\ast \ast }$ be the canonical isometric embedding, $J_{X}x(f)=f(x)$. Goldstine theorem statesHere the balls are closed unit balls. Thus any bidual unit vector can be approximated in finitely many dual evaluations by unit vectors from $X$; this does not claim norm density or sequential density in complete generality.

For a self-adjoint C-star element $a$, the following are three equivalent characterizations of a Positive element of a C-star algebra:If $a=b^2$ with $b=b^{\ast }$, the spectral mapping theorem gives nonnegative spectrum. Conversely, the continuous functional calculus applied to $\sqrt{\lambda }$ on $\sigma (a)$ gives a self-adjoint C-star element $b$ with $b^2=a$. For real $\sigma (a)$,Thus the third condition means $\sigma (a)\subseteq [0,2t]$, while the first implies the second by taking the interval $[0,\Vert a\Vert ]$. The second implies the third with $t=\Vert a\Vert$ if $a\ne 0$; if $a=0$, any $t>0$ works. This proves every equivalence.

The positive cone of a C-star algebra contains zero and is closed under multiplication by nonnegative real scalars. To verify addition without assuming that positive elements commute, put $r=\Vert a\Vert$, $s=\Vert b\Vert$ for $a,b\ge 0$. ThenIf $r+s>0$, the third characterization makes $a+b$ positive; otherwise both are zero. Hence it is a convex cone. If $a$ and $-a$ are positive, $\sigma (a)=\{0\}$ and $a=0$ by the self-adjoint spectral norm formula; therefore the cone is pointed and defines a partial order on $A_{\mathrm{sa}}$ by $h\le k$ if $k-h\in A_{+}$. It is generating because the continuous functional calculus givesFor $a\ge 0$, the spectral mapping theorem yieldsFinally, let $a\in M_2(\mathbb{C})$ be Hermitian, and write $\tau =\mathrm{tr}a$, $d=\det a$. Its two eigenvalues are real. Both are nonnegative exactly when their sum $\tau$ and product $d$ are nonnegative. The corresponding sum and product for $I-a$ are $2-\tau$ and $1-\tau +d$. Therefore the necessary and sufficient conditions for a positive contraction areIndeed, nonnegative sum and product force two real numbers to be nonnegative, including the zero-product cases. Applying this to $a$ and $I-a$ places both eigenvalues in $[0,1]$.

A unital C-star algebra is a complex Banach algebra $A$ with a multiplicative identity $1_A$ and a conjugate-linear involution $a\mapsto a^{\ast }$ satisfying $(ab{)}^{\ast }=b^{\ast }{a}^{\ast }$ and the C-star identity $\Vert a^{\ast }a\Vert =\Vert a{\Vert }^2$. A C-star isomorphism $\Phi :A\to B$ is a bijective complex-linear map preserving products and the involution. It carries $1_A$ to $1_B$ and is automatically isometric.

For a commutative unital C-star algebra, let $\Omega$ be its maximal ideal space, equivalently the space of Gelfand characters, with the weak-star topology. A Gelfand character has $\chi (1)=1$ and $\Vert \chi \Vert =1$. The equations $\chi (ab)=\chi (a)\chi (b)$ and $\chi (1)=1$ define a closed in the weak-star topology subset of the dual unit ball. The Banach-Alaoglu theorem makes $\Omega$ a compact Hausdorff space. Maximal ideals give characters by the Gelfand-Mazur theorem, and the Banach algebra spectrum satisfies $\sigma (a)=\{\chi (a):\chi \in \Omega \}$.

The Gelfand transform is $\hat{a}(\chi )=\chi (a)$. It preserves products, constants and the involution; the latter uses the fact that characters of a C-star algebra respect the involution. Every element of a commutative C-star algebra is normal, so the C-star identity and the spectral radius formula giveThus its image is a closed self-adjoint subalgebra of $C(\Omega )$, contains the constants, and separates characters. The complex Stone-Weierstrass theorem makes that image dense and therefore all of $C(\Omega )$. This is the Commutative Gelfand--Naimark theorem.

For the convergent sequence space $c$, coordinatewise multiplication satisfies $\Vert \alpha \beta {\Vert }_{\infty }\le \Vert \alpha {\Vert }_{\infty }\Vert \beta {\Vert }_{\infty }$, and coordinatewise complex conjugation is an involution. The constant sequence $1$ is the identity, andIt is complete: a uniform limit of convergent sequences is convergent, so $c$ is a closed linear subspace of the Banach space ${\ell }_{\infty }$. Hence it is a commutative unital C-star algebra.

TakeEvery finite point is isolated, and the neighbourhoods of $\infty$ contain all but finitely many integers. This is a compact Hausdorff space: an open cover has a set containing $\infty$, leaving only finitely many points to cover. DefineContinuity at $\infty$ is exactly convergence of the sequence. Thus $\Phi$ is onto $C(\Omega )$, and pointwise operations show it is a C-star isomorphism with $\Vert \Phi \alpha {\Vert }_{\infty }=\Vert \alpha {\Vert }_{\infty }$. Explicitly, these are all Gelfand characters: each $\chi (e_n)$ is $0$ or $1$ and at most one is $1$. If $\chi (e_n)=1$, multiplication by $e_n$ gives $\chi (\alpha )={\alpha }_n$. If all vanish, density of finitely supported sequences in l-p in $c_0$ gives $\chi {|}_{c_0}=0$, so $\chi (\alpha )=\lim {\alpha }_n$.

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.