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For a unital Banach algebra, $\phi (a)$ belongs to ${\sigma }_A(a)$: otherwise $a-\phi (a)1$ would be invertible, while applying $\phi$ to its inverse identity would give $0=1$. Thereforeso every character of an algebra is continuous and has norm one. The nonunital case follows by extending the character to the unitization of an algebra.

The Gelfand topology on ${\Phi }_A$ is the weak-star topology inherited from $A^{\ast }$: a net ${\phi }_i$ converges to $\phi$ exactly when ${\phi }_i(a)\to \phi (a)$ for every $a\in A$. If $A$ is unital, ${\Phi }_A$ lies in the weak-star compact dual unit ball by the Banach-Alaoglu theorem. The equationsdefine a weak-star closed subset, so ${\Phi }_A$ is compact.

LetChoose positive ${\epsilon }_n$ with ${\prod }_n(1+{\epsilon }_n)<\infty$. We construct $x_n\in K$ inductively. Once $E_n=\mathrm{span}\{x_1,\dots ,x_n\}$ has been chosen, compactness of its unit sphere gives finitely many members of $X^{\ast }$ that almost norm every vector of $E_n$. Because $0\in {\overline{K}}^w$, the next $x_{n+1}$ may be chosen so that all those functionals are as small on it as required. Choosing the error relative to $d$ givesfor all scalars $a_1,\dots ,a_{n+1}$. Indeed, if the last coefficient could threaten this estimate, $\Vert x_{n+1}\Vert \ge d$ first bounds that coefficient by a fixed multiple of the norm of the preceding sum; the selected norming functional then gives the displayed inequality. Iteration and the finite product bound satisfy the standard basis criterion, so $(x_n)$ is a basic sequence contained in $K$.

The same proof works for any Hausdorff locally convex vector topology $\tau$ weaker than the weak topology: on each finite-dimensional $E_n$, the $\tau$-continuous linear functionals still norm the space, and $0\in {\overline{K}}^{\tau }$ supplies the next point. A canonical strictly weaker example arises on $X^{\ast }$ when $X$ is not reflexive: the weak-star topology $\sigma (X^{\ast },X)$ is then strictly weaker than the weak topology $\sigma (X^{\ast },X^{\ast \ast })$.

We next prove the Eberlein-Šmulian theorem. If the weak closure of a bounded set $K$ is weakly compact, take any sequence in $K$ and let $Y$ be its closed linear span. The relevant weak closure lies in the separable space $Y$. A countable weak-star dense subset of the dual unit ball separates points of this compact set, so its weak topology is metrizable. Compact metrizability gives a weakly convergent subsequence.

For the converse, suppose $K$ is not relatively weakly compact. In the canonical embedding into $X^{\ast \ast }$, chooseThe Hahn-Banach theorem gives $x^{\ast \ast \ast }\in X^{\ast \ast \ast }$ that vanishes on $J_X(X)$ but satisfies $x^{\ast \ast \ast }(x^{\ast \ast })=1$. Alternating Goldstine approximation with the fact that $x^{\ast \ast }$ lies in the weak-star closure of $J_{X}K$ constructs bounded $x_n^{\ast }\in X^{\ast }$ and $x_n\in K$ such that, up to errors tending to zero,If a subsequence $x_{n_j}$ converged weakly to $x$, then for each fixed $i$ the second relation would give $x_i^{\ast }(x)=1$. A weak-star cluster point $x^{\ast }$ of the bounded sequence $(x_i^{\ast })$ satisfies $x^{\ast }(x_{n_j})=0$ by the first relation, and hence $x^{\ast }(x)=0$ by weak convergence; but passing to the same cluster point in $x_i^{\ast }(x)=1$ gives $x^{\ast }(x)=1$. This contradiction produces a sequence in $K$ with no weakly convergent subsequence. Relative weak compactness is therefore equivalent to the subsequence condition.

Finally suppose $K$ is weakly sequentially compact and $x\in {\overline{K}}^w$. If $x$ lies in the norm closure of $K$, a norm-convergent sequence suffices. Otherwise apply the first part to $K-x$ and obtain a basic sequence $(y_n)$. A subsequence converges weakly by hypothesis. Its weak limit lies in the closed span of the basic sequence, while every coordinate functional is eventually zero on that subsequence. The limit is consequently zero, so the corresponding sequence from $K$ converges weakly to $x$.

This also shows that a weakly sequentially compact $K$ is weakly closed: any point of its weak closure is the limit of a sequence in $K$, and a weakly convergent subsequence has its limit in $K$. The relative form of the Eberlein-Šmulian theorem then makes $K$ weakly compact. The reverse implication follows from the first direction of that theorem. Thus weak compactness and weak sequential compactness are equivalent.

Weak convergence $u_j\rightharpoonup u$ means $F(u_j)\to F(u)$ for every $F\in X^{\ast }$. Weak-star convergence $T_j\stackrel{\ast }{\rightharpoonup }T$ means $T_j(x)\to T(x)$ for every $x\in X$, viewing $X$ as the predual. The Sequential Banach-Alaoglu theorem for a separable predual says that every norm-bounded sequence in $X^{\ast }$ has a weak-star convergent subsequence. Separability is what turns the bounded weak-star compactness statement into this sequential form.

For the weak-star limit of periodic rescalings, apply it with $X=L^1(\mathbb{R})$, which is separable, and $X^{\ast }=L^{\infty }(\mathbb{R})$. The rescaled functions have uniform essential-supremum bound $\Vert \theta {\Vert }_{\infty }$. Let $q=\theta -\bar{\theta }$ and $F(t)={\int }_0^{t}q(s)ds$. Since $q$ is one-periodic with zero mean, $F$ is a bounded one-periodic absolutely continuous primitive. Thus for every finite interval,Every weak-star subsequential limit $h$ therefore satisfies ${\int }_a^b(h-\bar{\theta })=0$ for all $a<b$. To apply the supplied uniqueness fact without assuming global integrability, fix $R>0$ and set $v=(h-\bar{\theta })1_{[-R,R]}$. This is in $L^1(\mathbb{R})$, and its integral over any interval is zero, since intersecting with $[-R,R]$ again gives an interval. Hence $v=0$ almost everywhere. Taking integer $R\to \infty$ yields $h=\bar{\theta }$ almost everywhere on the whole line.

Finally, if the full sequence did not converge weak-star to this constant, some $g\in L^1$ and some subsequence would have its pairing with $\theta (jx)-\bar{\theta }$ bounded away from zero. Sequential compactness gives a further weak-star convergent subsequence, whose limit must be the constant just identified, a contradiction. ThereforeThis is averaging against integrable test functions, not a claim of pointwise or strong norm convergence.

For a complex measure, its total variation measure isIts norm is $\Vert \mu \Vert =|\mu |(\mathbb{T})$. For Radon measures on the compact circle this is also ${\sup }_{\Vert f{\Vert }_{\infty }\le 1,f\in C(\mathbb{T})}|\int fd\mu |$. The stated convergence is weak-star convergence against continuous tests:It is the weak-star topology from the identification of these measures with the dual of $C(\mathbb{T})$, rather than convergence of their values on every Borel set.

For an absolutely p-summing operator $T:E\to F$, there is a probability measure $\mu$ on the dual unit ball with its weak-star topology such that $\Vert Tx\Vert \le {\pi }_p(T)(\int |\varphi (x){|}^{p}d\mu (\varphi ){)}^{1/p}$. Thus $T$ factors through the closed span of the evaluation functions in $L^p(\mu )$, with the final operator having norm at most ${\pi }_p(T)$. Conversely this factorization implies the summing inequality by summing under the integral. Existence of $\mu$ follows from finite-dimensional Hahn-Banach separation theorem: failure for finitely many vectors would contradict the summing inequality for a nonnegative weighted family. Compactness of probability measures then combines the finite constraints. The final operator need only be defined on the indicated closed subspace, not on all of $L^p(\mu )$.

A pure state is an extreme point of the state space. The Krein-Milman theorem guarantees pure states, and every positive element attains its norm at some pure state.

Every continuous self-map of a nonempty compact convex subset of a locally convex space has a fixed point. Compactness is in the locally convex topology, which may be weaker than a norm topology. This version permits fixed-point arguments on a weak-star topology compact set of linear functionals. See Hartman's fixed-point chapter, Theorem 0.2.

A topological space is separable when it has a countable dense subset. The topology is part of the assertion: a dual may be separable in its weak-star topology and not in its norm topology. A compact metric space is separable by taking the union of finite nets at scales tending to zero.

If a normed vector space $X$ is separable, every bounded sequence in $X^{\ast }$ has a subsequence that converges pointwise on $X$. Equivalently, the closed unit ball of $X^{\ast }$ is sequentially compact in the weak-star topology. A diagonal subsequence argument on a countable dense subset proves the result.

A state is a positive functional of norm one. The state space $S(A)$ is a weak-star compact convex subset of $A^{\ast }$.

For a real function in the $L^{\infty }$ unit ball, push uniform measure on $t\in [0,1]$ forward by the displayed sign-valued functions. In the weak-star topology this is a Borel probability concentrated on the extreme points, and Fubini's theorem shows that its barycenter is $f$. Weak-star compactness and separability of the predual $L^1$ place the example within Choquet theorem.

A bounded linear operator between Banach spaces is weakly compact exactly when its Banach-space adjoint is weakly compact. Equivalently, that adjoint is continuous from its domain's weak-star topology to its range's weak topology. For the reverse direction, the bidual condition applied to $T^{\ast }$ forces $T^{\ast \ast \ast }$ to annihilate every functional on $Y^{\ast \ast }$ vanishing on $J_{Y}Y$; the Hahn-Banach theorem then places $T^{\ast \ast }(X^{\ast \ast })$ inside $J_{Y}Y$.

A bounded measurable one-periodic function rescaled to frequency $j$ converges weak-star in $L^{\infty }(\mathbb{R})$ to its period average. Subtract the average and take a bounded periodic primitive $F$. Its integral on $[a,b]$ after rescaling is $[F(jb)-F(ja)]/j$, which tends to zero. Sequential Banach-Alaoglu with separable predual $L^1$ then identifies every cluster point by interval integrals; uniqueness of all cluster points gives full weak-star convergence. Localizing to bounded intervals avoids falsely assuming that a bounded cluster function is globally integrable.

The continuous dual of a separable Banach space is separable in its weak-star topology, even when it is not norm separable. Banach-Alaoglu theorem and weak-star metrizability of the dual ball make the dual unit ball a separable compact metric space. The countable union of positive integer multiples of a countable dense subset of that ball is dense in the whole dual. This does not imply global weak-star metrizability.

The same Banach space can acquire incomparable weak-star topologies from different isometric preduals. Pair ${\ell }^1$ with $c_0$ by $\sum {\alpha }_n{\beta }_n$ and with the space $c$ of convergent sequences by ${\alpha }_1{\gamma }_{\infty }+{\sum }_{n\ge 2}{\alpha }_n{\gamma }_{n-1}$. The functional $\alpha \mapsto {\alpha }_1$ is continuous only for the first topology: $e_1-e_n\to 0$ for the second but its values remain one. Conversely $\alpha \mapsto \sum {\alpha }_n$ is continuous only for the second: $e_n\to 0$ for the first but its values remain one. More generally, comparable weak-star topologies arising from isometric preduals agree on the common closed unit ball, since its identity map is a continuous bijection from a compact space to a Hausdorff space by the Banach-Alaoglu theorem.