Weak topology

The weak topology $\sigma (X,X^{\ast })$ on a normed space $X$ is the coarsest topology making every member of $X^{\ast }$ continuous. A net converges weakly exactly when every bounded linear functional converges on it.

For Fréchet spaces, the two continuity conditions for a linear map are equivalent. Original continuity makes every target continuous linear functional compose to a source continuous linear functional, proving weak continuity. Conversely, weak continuity makes these compositions continuous in the original source topology. They separate target points, so limits in the graph of a linear operator remain in the graph. The closed graph theorem for Fréchet spaces finishes the proof.

Without the hypotheses, equal continuous dual spaces do not imply equal topologies. The identity from an infinite-dimensional Hilbert space with its weak topology to its norm topology is weak-to-weak continuous but not continuous in those original topologies.

If a countable family in $X^{\ast }$ separates points of $X$, the displayed metric induces the weak topology on every weakly compact set $K$. The coordinate map into the countable product of scalar lines is continuous and injective; a continuous injection from a compact space into a Hausdorff space is a homeomorphism onto its image. Equivalently, uniform convergence of the metric series makes the identity from weak $K$ to metric $K$ continuous. Compactness is essential: point separation alone need not generate the weak topology on an arbitrary bounded set.

A subset $D$ of a normed vector space is weakly bounded when ${\sup }_{x\in D}|f(x)|<\infty$ for every $f\in X^{\ast }$. It is norm bounded: its canonical embedding into the bidual is a pointwise bounded family of functionals on the complete continuous dual space, so the Uniform boundedness principle applies. The converse is immediate. A weakly compact set is weakly bounded because each scalar evaluation has compact image.

The weak topology and norm topology of a separable Banach space generate the same Borel sigma-algebra. A countable norming family makes every norm ball weakly Borel measurable, and separability makes every norm-open set a countable union of such balls. The reverse inclusion follows because the weak topology is coarser.

A sequence $x_n$ converges weakly to $x$ when $f(x_n)\to f(x)$ for every $f$ in the continuous dual space $X^{\ast }$.

The nonnegative Lp space functions $f_j$ have $L^1$ norm one. Any weak limit tested against indicators supported outside the origin must vanish almost everywhere. Testing against the constant bounded function one would nevertheless require integral one. This contradiction applies to every subnet whose indices tend to infinity, so the $L^1$ unit ball is not weakly compact. The abstract Banach-Alaoglu theorem remains valid: weak compactness of the $L^1$ unit ball is a different assertion from weak-star compactness of a dual ball.

A normed vector space has the Radon-Riesz property when weak convergence $u_k\rightharpoonup u$ together with $\Vert u_k\Vert \to \Vert u\Vert$ implies strong convergence $u_k\to u$. The Hilbert-space case is the Radon-Riesz theorem. The condition prevents norm from remaining in weakly invisible directions.

On a finite-dimensional vector space equipped with a norm, the weak topology and norm topology coincide. The weak topology is no finer because every element of the dual is norm-continuous. Conversely, finitely many coordinate functionals in a basis control the norm, so every sufficiently small basic weak neighbourhood lies in a prescribed norm ball.

Every weakly convergent sequence in a normed vector space is a bounded sequence. Apply the Uniform boundedness principle on the Banach space $X^{\ast }$ to the evaluation maps $J{x}_n:f\mapsto f(x_n)$. They are pointwise bounded, while the canonical embedding into the bidual gives $\Vert J{x}_n\Vert =\Vert x_n\Vert$.

Let $X^{\ast }$ have a norm-dense sequence $({\varphi }_m)$, and let $B\subset X$ be norm-bounded. Thenis a metric on $B$ that induces its weak topology. Density lets convergence against the ${\varphi }_m$ extend uniformly on $B-B$ to convergence against every $\varphi \in X^{\ast }$.

If the relative weak topology on the unit ball of a normed vector space is first countable at zero, choose a countable collection of finite families of continuous linear functionals defining a local base. For any $f\in E^{\prime }$ and $ϵ>0$, one family $F$ has common kernel $M$ on which $\Vert f{|}_M\Vert \le ϵ$. Extend that restriction to $h\in E^{\prime }$ with $\Vert h\Vert \le ϵ$ using the Hahn-Banach theorem. Then $f-h$ vanishes on $M$, so it belongs to $\mathrm{span}F$ by finite-dimensional factorization. Consequently the countable union of these families has norm-dense linear span in $E^{\prime }$. The converse is the metrization of the weak topology on a bounded set construction from a norm-dense sequence in $E^{\prime }$. In particular, separability of $E$ alone does not suffice: ${\ell }^1$ has nonseparable dual ${\ell }^{\infty }$.

The weak closure ${\overline{K}}^w$ of $K$ is its closure in the weak topology. Thus $x\in {\overline{K}}^w$ exactly when every weak neighbourhood of $x$ meets $K$.

A sequence $(x_n)$ is weakly null when $f(x_n)\to 0$ for every member $f$ of the continuous dual space. It may remain bounded away from zero in norm.

For $1<p<\infty$, identify $({\ell }^p{)}^{\ast }$ with ${\ell }^q$. If $y=(y_j)\in {\ell }^q$, thenThus the standard unit vectors converge weakly to zero in ${\ell }^p$, although $\Vert e_n{\Vert }_p=1$ for every $n$.

Let $1<p<\infty$. Every norm-bounded sequence in ${\ell }^p$ whose terms have pairwise disjoint supports converges weakly to zero. Indeed, the average of $N$ distinct terms has norm $O(N^{1/p-1})$, so Mazur theorem gives weak convergence through its convex-combination criterion.

A sequence $(x_n)$ is weakly Cauchy when $(f(x_n))$ is a Cauchy sequence for every continuous linear functional $f$. Its consecutive differences form a weakly null sequence.

In an infinite-dimensional normed space, the weak closure of the unit sphere is the closed unit ball. Every basic weak neighbourhood imposes only finitely many linear conditions, whose common kernel contains a nonzero direction that can move any interior point onto the sphere.

The weak-star topology $\sigma (X^{\ast },X)$ is pointwise convergence of functionals on $X$. If $X$ is nonreflexive, it is strictly weaker than the weak topology $\sigma (X^{\ast },X^{\ast \ast })$ on $X^{\ast }$.

If $T=S^{\ast }$ for a bounded operator on a separable Banach space, and $C$ is a nonempty weak-star closed convex subset of the dual unit ball with $T(C)\subseteq C$, then $T$ has a fixed point in $C$. The Cesàro averages of the orbit remain in $C$, and $T$ minus the identity sends those averages to vectors of norm at most $2/n$. Weak-star compactness supplies a cluster point, and the adjoint action is weak-star continuous. The corresponding assertion with only weak closedness is false for the right shift on the probability simplex in ${\ell }^1$, the dual of $c_0$.

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.

For an infinite-dimensional normed space, the weak-star open unit ball of its dual is not a Baire space. The smaller closed norm balls are weak-star closed and have empty relative interior: a nonzero functional annihilating any finitely many tested vectors can move the norm beyond the smaller radius while remaining below one. Their countable union is the open norm ball. Thus this topology is not topologically complete, regardless of separability.

A countable local base at zero would give countably many finite sets of evaluation vectors. Any other evaluation must lie in the span of one of those finite sets: otherwise a bounded functional vanishing on that set but not on the evaluation vector, supplied by Hahn-Banach theorem, contradicts inclusion of neighbourhoods after scalar rescaling. Thus the predual has countable Hamel dimension, impossible for an infinite-dimensional Banach space by the Baire category theorem. This concerns the entire dual, unlike weak-star metrizability of the dual ball.

A linear functional on $X^{\ast }$ is continuous for the weak-star topology exactly when it is evaluation at an element of $X$. Continuity makes it vanish on the common kernel of finitely many evaluations, so it is a linear combination of those evaluations. The canonical embedding into the bidual identifies this dual with $X$.

In $L^{\infty }(\mathbb{R})=(L^1(\mathbb{R}){)}^{\ast }$, the sequence $f_n(x)=\sin (nx)$ converges weak-star to zero by the Riemann-Lebesgue lemma, while $f_n^2$ converges weak-star to the constant $1/2$. Nonlinear pointwise operations therefore need not preserve weak-star limits.

If $X$ is separable, the weak-star topology on $B_{X^{\ast }}$ is metrizable. For a dense sequence $(x_n)$ in $B_X$, a compatible metric is obtained by summing bounded multiples of $2^{-n}|f(x_n)-g(x_n)|$.

The dual unit ball $B_{X^{\ast }}$ is weak-star metrizable if and only if $X$ is separable. For the converse direction, Banach-Alaoglu theorem makes a metrizable dual ball a compact metric space, and the evaluation map embeds $X$ isometrically into the separable space $C(B_{X^{\ast }})$.

A Banach space $X$ is a Grothendieck space when every weak-star convergent sequence in $X^{\ast }$ is weakly convergent. Every reflexive Banach space has this property, and every separable Grothendieck space is reflexive.

For a weak-star compact set $K\subseteq X^{\ast }$, its Szlenk derivation at scale $\epsilon >0$ isIt removes points having a relatively weak-star open neighbourhood of norm diameter at most $\epsilon$.

Transfinite iteration of this operation defines the Szlenk index; one derivation step is not itself the index.

If $X^{\ast }$ is separable and $K$ is a nonempty weak-star closed subset of $B_{X^{\ast }}$, then $K_{\epsilon }^{\prime }$ is a proper subset of $K$ for every $\epsilon >0$. Metrizability produces sequences witnessing membership in the derivative, while a Baire-category argument applied to a universal weakly null sequence yields the strict inclusion.

A subset of a Banach space is weakly compact when it is compact in the weak topology; it is relatively weakly compact when its weak closure is weakly compact.

A bounded sequence in the space of continuous functions on a compact space that has a discontinuous pointwise limit cannot lie in a weakly compact set. Compactness would give a weakly convergent subnet, and continuous point-evaluation functionals would identify its continuous limit with the prescribed pointwise limit. No assumption of sequential compactness is needed.

Coordinate evaluations on the l-infinity sequence space form a countable separating family, so a weakly compact set $K$ is metrizable by countable separating family metrizes a weakly compact set. Choose a countable weakly dense subset $D$. Then $K$ lies in the weak closure of $\mathrm{conv}D$, which equals its norm closure by Mazur theorem. Rational convex combinations make that norm closure separable, and every subset of a separable metric space is separable. The weakly dense subset itself need not be norm dense in a nonconvex $K$.

Every weakly compact set in a normed vector space is norm bounded, even if the space is incomplete. Its canonical embedding into the bidual is pointwise bounded on the continuous dual space; the Uniform boundedness principle on that Banach dual gives the norm bound.

A set is weakly sequentially compact when every sequence in it has a subsequence that converges weakly to a point of the set.

For a subset of a Banach space, relative weak compactness is equivalent to every sequence having a weakly convergent subsequence. In particular, weak compactness and weak sequential compactness coincide.

The closed convex hull of a weakly compact subset of a Banach space is weakly compact.

The other standard formulation says that a convex subset $C$ of a Banach-space dual is weak-star closed if its intersection with every norm-closed ball is weak-star closed.

The Szlenk index measures how many transfinite Szlenk derivations are needed to empty the dual unit ball at each positive norm-diameter scale. At a successor ordinal take another derivation; at a limit ordinal intersect the previous sets. Taking the supremum of the emptying ordinals over the positive scales gives an ordinal invariant of the Banach space.