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The smallest closed convex set containing a set $C$, equivalently the closure of its convex hull in the specified topology. The topology must be stated: the norm topology, weak topology and weak-star topology need not give the same closure for an arbitrary set.

For a separated dual pair, the continuous linear functionals for its weak topology are exactly evaluations by members of the paired space. Continuity bounds a functional on a neighbourhood defined by finitely many evaluations; scalar rescaling makes it vanish on their common kernel. It then factors through their finite-dimensional evaluation map and is a finite linear combination of them.

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.

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.

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.

Assume instead that $T$ sends weakly compact sets to norm-compact sets. If $x_n\rightharpoonup 0$, then $K=\{0,x_1,x_2,\dots \}$ is weakly compact: an open cover has a member containing zero, which contains all but finitely many terms of the convergent sequence. Hence $T(K)$ is norm compact. A bounded linear operator is weak-to-weak continuous, because each $g\in F^{\ast }$ pulls back to $g\circ T\in E^{\ast }$. Thus $T{x}_n\rightharpoonup 0$. If $\Vert T{x}_n\Vert$ did not tend to zero, a subsequence with norms bounded below by some $\epsilon >0$ would have a norm-convergent further subsequence with limit $y$. That further subsequence also converges weakly to both $y$ and zero; uniqueness of limits in the Hausdorff weak topology forces $y=0$, contradicting the lower bound. Therefore $\boxed{\Vert T{x}_n\Vert \to 0}$, proving the equivalence and the characterization of completely continuous operators.

Now consider the inclusion of the space of continuous functions on a compact space $C(\mathbb{T})$ into L2 space for normalized Lebesgue measure. It is a bounded linear operator, since $\Vert f{\Vert }_2\le \Vert f{\Vert }_{\infty }$. For a weakly null sequence $(f_n)$ in $C(\mathbb{T})$, point evaluations are continuous linear functionals, so $f_n(t)\to 0$ for every $t$. The Uniform boundedness principle, applied to the canonical embeddings $f_n\in C(\mathbb{T}{)}^{\ast \ast }$, gives a uniform bound $\Vert f_n{\Vert }_{\infty }\le M$: each dual functional is bounded on the weakly convergent sequence. The dominated convergence theorem now yieldsThus $j_2$ satisfies both equivalent conditions.

It is nevertheless not a compact operator. Over complex scalars the functions $f_n(t)=e^{int}$ lie in the unit ball of $C(\mathbb{T})$ and form an orthonormal sequence in $L^2(\mu )$, soThere is no norm-convergent subsequence. Over real scalars use $\cos (nt)$, whose distinct images have distance $1$. Complete continuity controls weakly compact sets, rather than all bounded sets.

The weak topology $\sigma (E,E^{\ast })$ is the coarsest topology making every continuous linear functional in $E^{\ast }$ continuous. A basic neighborhood of $x$ imposes finitely many inequalities $|f_j(z-x)|<\epsilon$, and a sequence converges weakly precisely when each $f(x_n)$ converges to $f(x)$. The Hahn-Banach theorem makes the dual separate points, so the weak topology is Hausdorff.

Here is a proof of the requested weak subsequence property, the compact-to-sequential direction of the Eberlein-Šmulian theorem. Let $Y$ be the closed linear span of the countable set $\{x_n:n\ge 1\}$. It is a separable Banach space. The Hahn-Banach theorem separates points outside a closed vector subspace from that subspace, so $Y$ is weakly closed in $E$. Therefore $C=K\cap Y$ is weakly compact.

Choose a countable dense set $(u_j)$ in the unit sphere of $Y$, and choose norming functionals $f_j\in Y^{\ast }$ with $\Vert f_j\Vert =1$ and $f_j(u_j)=1$. They separate points: for $y\ne 0$, a sufficiently close $u_j$ to $y/\Vert y\Vert$ gives $|f_j(y)|>0$. This is a countable norming family. Extend the functionals to $E$ by the Hahn-Banach theorem. The evaluation mapis weakly continuous and injective. Its domain is compact and its target Hausdorff, so it is a homeomorphism onto its image: images of closed subsets are compact, hence closed. A countable product of real or complex lines is metrizable, for example by ${\sum }_j{2}^{-j}\min (1,|a_j-b_j|)$. Thus $C$ has a compact metrizable weak topology. Compact metric spaces are sequentially compact, giving a subsequence with $\boxed{x_{n_j}\rightharpoonup x\in C\subseteq K}$. If $Y=\{0\}$ the result is immediate. We did not assert that a separable Banach space has a separable dual or that its weak topology is generally metrizable.

The topology needed here is the weak-star topology $\sigma (E^{\ast },E)$, meaning pointwise convergence on $E$. With the literal weak topology $\sigma (E^{\ast },E^{\ast \ast })$, the claimed compactness is false. For example, $E=c_0$ is separable and $E^{\ast }={\ell }^1$. Its unit vectors $e_n$ have no weak cluster point: every coordinate functional forces a cluster point to have all coordinates zero, while the bounded functional $x\mapsto {\sum }_j{x}_j$ has value one throughout. Thus the dual unit ball is not weakly compact. We prove the intended weak-star assertion in full.

Choose a countable dense set $(x_k)$ in the unit ball of $E$. On $E_1^{\ast }$ defineEach term is at most $2\cdot 2^{-k}$, so the series converges. The triangle inequality is immediate, and $d=0$ forces equality on the dense set and hence on all of $E$. It is therefore a metric. Convergence in this metric is equivalent to convergence on every $x_k$: one direction follows from any individual term, and the other by bounding the series tail uniformly. Because $\Vert \varphi \Vert ,\Vert \psi \Vert \le 1$, approximation of any unit vector by $x_k$ then gives convergence on all of $E$. The same finite-tail and approximation argument for neighborhoods proves equality with the weak-star topology, not just agreement of its convergent sequences.

Given a sequence $({\varphi }_n)$ in this ball, diagonal selection gives a subsequence converging at every $x_k$, since each coordinate lies in a compact disk. Uniform norm bounds then make it pointwise Cauchy at every $x\in E$. Define $\varphi (x)$ as the limit. Linearity passes to the limit, and $|\varphi (x)|\le \Vert x\Vert$, so $\varphi \in E_1^{\ast }$. The subsequence converges to it in the metric. Thus this metric space is sequentially compact, hence compact, proving

The relevant Banach-Alaoglu theorem says that for $1<p<\infty$ the closed bounded balls of $L^p({\mathbb{R}}^n)$ are compact in the weak topology. In particular, every bounded sequence has a weakly convergent subsequence whose limit has norm no larger than the common bound. Applying the preceding duality of Lp spaces also identifies this as weak-star compactness of $L^p=(L^q{)}^{\ast }$; its weak and weak-star test spaces are both $L^q$.

Here is a direct proof of both compactness and the subsequence assertion. Let $\Vert f_j{\Vert }_p\le M$ and choose a countable dense set $\{g_m\}$ in the separable Lp space $L^q$. Each scalar sequence $\int f_j{g}_m$ is bounded. Successive subsequence selection followed by the diagonal subsequence argument gives $f_{j_k}$ for which these pairings converge for every $m$. For arbitrary $g\in L^q$,First approximate $g$ by $g_m$ and then take $k,l$ large. This proves convergence of every pairing, and the limit $\Lambda (g)={\lim }_k\int f_{j_k}g$ is linear with $|\Lambda (g)|\le M\Vert g{\Vert }_q$. Apply the already proved duality of Lp spaces with $q$ in place of $p$: $\Lambda (g)=\int fg$ for an $f\in L^p$ with $\Vert f{\Vert }_p\le M$. ThusTo obtain compactness rather than only a sequential statement, on the radius-$M$ ball defineDensity and Hölder's inequality show that this is a metric space distance and that its topology is exactly the weak topology on this ball: every remaining test pairing is uniformly approximated there by a dense-set pairing. The subsequence argument proves sequential compactness for this metric. A sequentially compact metric space is compact: otherwise either an infinite separated sequence contradicts total boundedness or a Cauchy sequence without a limit contradicts sequential compactness; completeness and total boundedness give compactness by successive finite coverings. This completes the proof. The same argument is the Sequential Banach-Alaoglu theorem for a separable predual.

True. In the norm topology on the unit ball, the set $\{x\in B:\Vert x\Vert <1/2\}$ is a neighbourhood of zero. If the relative weak topology agrees with it, there are finitely many continuous linear functionals $f_1,\dots ,f_m$ and positive ${ϵ}_j$ such thatIf ${\bigcap }_j\ker f_j$ contained a nonzero vector, rescaling it to have norm $3/4$ would put it in the left-hand set but not the right-hand set. Therefore ${\bigcap }_j\ker f_j=\{0\}$. The linear mapis injective into a finite-dimensional scalar space, soThis proof works for either the open or closed unit-ball convention and does not require completeness.

Work over $\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$. A Fréchet space is a Hausdorff, complete, metrizable locally convex space. Its topology is generated by a countable separating family of seminorms $(p_n{)}_{n\ge 1}$. A neighbourhood base at zero consists of finite intersectionsReplacing $p_n$ by ${\max }_{j\le n}{p}_j$ makes the family increasing. One compatible translation-invariant metric isCompleteness means that a sequence Cauchy for all these seminorms converges in the space. The separating condition ensures that $d(x,y)=0$ only when $x=y$.

Let $E^{\prime }$ denote the continuous dual space for the original topology $\tau$. The weak topology $\sigma (E,E^{\prime })$ is the coarsest topology making all $f\in E^{\prime }$ continuous. Its basic zero-neighbourhoods areEquivalently, it is generated by the seminorms $x\mapsto |f(x)|$.

To prove the Hausdorff property, fix $z\ne 0$. Some continuous seminorm $p$ has $p(z)>0$. On the one-dimensional span of $z$, define $f_0(az)=ap(z)$. Then $|f_0(az)|=p(az)$. The real or complex Hahn-Banach theorem extends this to $f:E\to \mathbb{F}$ with $|f(x)|\le p(x)$. Thus $f$ is originally continuous and $f(z)\ne 0$. This proves continuous-dual point separation. For distinct $x,y$, inverse images under such an $f$ of disjoint small scalar neighbourhoods of $f(x)$ and $f(y)$ are disjoint weak neighbourhoods. Hence the weak topology is Hausdorff.

If a linear functional is originally continuous, it belongs to $E^{\prime }$ and is weakly continuous by definition. Conversely, $\sigma (E,E^{\prime })$ is coarser than $\tau$, so every weakly continuous scalar function is originally continuous. In particular,This special case of the continuous dual of a weak topology needs no completeness argument.

For the linear map $T:E_1\to E_2$, original continuity implies that $g\circ T\in E_1^{\prime }$ whenever $g\in E_2^{\prime }$. Each such composition is weakly continuous on $E_1$, so the defining property of the target weak topology makes $T$ weak-to-weak continuous.

Conversely, suppose $T$ is weak-to-weak continuous. Every $g\circ T$, $g\in E_2^{\prime }$, is then weakly continuous, and hence continuous for ${\tau }_1$. We show that the graph of a linear operator $\Gamma (T)$ is closed in the original product topology. If $x_n\to x$ in $E_1$ and $T{x}_n\to y$ in $E_2$, thenBy continuous-dual point separation on $E_2$, this forces $y=Tx$. The product is metrizable, so sequential closedness is closedness. The closed graph theorem for Fréchet spaces now gives original continuity of $T$.

The relevant version of that theorem does use completeness of both spaces: a closed graph is a Fréchet space, its first-coordinate projection is a continuous linear bijection onto $E_1$, and the open mapping theorem for Fréchet spaces gives a continuous inverse. Composing with the second-coordinate projection gives $T$. ThusEquality of continuous scalar functionals does not imply equality of the two topologies; the closed graph theorem for Fréchet spaces is the extra step that makes this conclusion valid.

A separated dual pair consists of real vector spaces $E,F$ and a bilinear form $⟨x,y⟩$ that separates both variables: each nonzero $x\in E$ pairs nontrivially with some $y\in F$, and conversely. Thus $F$ embeds in the algebraic dual of $E$, and $E$ embeds in the algebraic dual of $F$.

The weak topology $\sigma (E,F)$ is the coarsest topology making all maps $x\mapsto ⟨x,y⟩$, $y\in F$, continuous. It makes $E$ a Hausdorff locally convex space, with topology generated by the seminorms $p_y(x)=|⟨x,y⟩|$. A neighbourhood base at zero consists of finite intersectionsSeparation ensures that the common zero set of all these seminorms is just zero, giving Hausdorffness.

We determine the continuous dual of a weak topology. Suppose a linear functional $\ell$ is continuous. A basic neighbourhood as above is contained in $\{x:|\ell (x)|<1\}$. If $x$ is in the common kernel of the finitely many evaluations, every scalar multiple of $x$ lies in that neighbourhood, so $\ell (x)=0$. Hence $\ell$ factors through the image ofExtend the resulting linear functional on $T(E)$ to ${\mathbb{R}}^m$. It has the form $z\mapsto {\sum }_j{a}_j{z}_j$, giving $\ell (x)=⟨x,{\sum }_j{a}_j{y}_j⟩$. Conversely every such evaluation is continuous by the definition of the weak topology. Since the pairing separates $F$, this identification is injective, andFor the remaining parts every topological assertion about $E^{\prime }$ or its subsets uses the stipulated weak-star topology $\sigma (E^{\prime },E)$.

Use the assumed separability of the Hilbert space to choose a countable Hilbertian basis $(e_j)$. For $\Vert f_n\Vert \le M$, every scalar sequence $⟨f_n,e_j⟩$ is bounded. Repeated extraction followed by the diagonal subsequence argument gives one subsequence with all these coordinates converging, say to $a_j$. The Bessel inequality gives ${\sum }_{j\le N}|a_j{|}^2\le M^2$ for every $N$, so $g={\sum }_j{a}_j{e}_j$ exists and $\Vert g\Vert \le M$.

For $h\in H$, let $h_N$ be its finite basis truncation. Coordinate convergence gives $⟨f_{\phi (n)}-g,h_N⟩\to 0$, while the Cauchy-Schwarz inequality bounds the remaining inner product by $2M\Vert h-h_N\Vert$, uniformly in $n$. First choose $N$ large, then $n$ large. ThusThis proves the asserted sequential conclusion. To obtain actual compactness for the weak topology, note that on a bounded ball it is induced by the metricFinite coordinates and the uniformly small tail show that $d$ gives coordinate convergence; uniform boundedness of the ball extends this to every inner product as above. Hence this metric induces exactly the restricted weak topology. In a metric space, sequential compactness implies compactness. The closed unit ball is therefore weakly compact, and the limit remains inside it.

The weak topology on a normed vector space $X$ is $\sigma (X,X^{\ast })$, the coarsest topology making every bounded linear functional continuous. A neighbourhood base at $x$ is given by finitely many inequalities $|f_j(y-x)|<\epsilon$, with $f_j\in X^{\ast }$. In the complex case, separation uses real parts of bounded linear functionals.

Mazur theorem states that the weak closure of a convex set equals its closure in the norm topology. The weak topology is coarser than the norm topology, so the norm closure is contained in the weak closure. Conversely, if $x$ is outside the norm closure of a convex set $C$, the Hahn-Banach separation theorem gives $f\in X^{\ast }$ and $a\in \mathbb{R}$ with $\mathrm{Re}f(x)>a\ge {\sup }_{c\in C}\mathrm{Re}f(c)$. The corresponding weak topology neighbourhood of $x$ misses $C$, so $x$ is outside its weak closure. This proves Mazur theorem, including the empty-set case. In particular, if $x_n$ converges weakly to $x$, then $x$ is in the weak closure of every tail and therefore in the norm closure of its convex hull. Choosing a finite convex combination of the $n$th tail within $1/n$ of $x$ proves the usual Mazur lemma formulation as well.

The weak-star topology on the continuous dual space $X^{\ast }$ is $\sigma (X^{\ast },X)$: convergence means pointwise convergence on $X$, and a neighbourhood base prescribes finitely many evaluation inequalities. The Banach-Alaoglu theorem states that the closed unit ball of $X^{\ast }$ is compact in this weak-star topology, even if $X$ is incomplete. Embed this closed unit ball intowhere $\mathbb{K}=\mathbb{R}$ or $\mathbb{C}$. Each factor is compact, so $P$ is compact by the Tychonoff theorem. Inside $P$, the equations $a_{x+y}=a_x+a_y$ and $a_{\lambda x}=\lambda a_x$ define a closed set. Every such point defines a linear functional satisfying $|a_x|\le \Vert x\Vert$, hence belongs to the closed unit ball of $X^{\ast }$. Thus this image is closed in $P$. The product topology on it is exactly the weak-star topology, proving Banach-Alaoglu theorem. Evaluations also separate its points, so the weak-star topology is Hausdorff.

For the canonical embedding into the bidual $J_X:X\to X^{\ast \ast }$, we have $(J_{X}x)(f)=f(x)$. Restricting all evaluations at $f\in X^{\ast }$ therefore gives exactly $\sigma (X,X^{\ast })$. Thus the induced subspace topology is the weak topology on $X$.

Now identify $X$ with $J_{X}X$. Let $C$ be a bounded convex set, write $D$ for its norm closure in $X$, and let $K$ be its weak-star topology closure in $X^{\ast \ast }$. Boundedness places $K$ in a multiple of the closed unit ball of $X^{\ast \ast }$, so $K$ is compact by Banach-Alaoglu theorem. The preceding subspace topology identification and Mazur theorem giveIf $D$ is a weakly compact set, its image in the Hausdorff weak-star topology is compact and therefore closed. It contains $C$, so $K\subseteq D\subseteq X$. Conversely, if $K\subseteq X$, the displayed identity gives $K=D$, and its compactness is precisely weak compactness in $X$. Hence $\boxed{D\text{ is weakly compact}⟺K\subseteq X}$. This argument also covers $C=\varnothing$.

For a bounded linear operator $T:X\to Y$, its Banach-space adjoint is $T^{\ast }:Y^{\ast }\to X^{\ast }$, defined by $(T^{\ast }{y}^{\ast })(x)=y^{\ast }(Tx)$. Evaluation at any fixed $x$ is thus evaluation at $Tx$ after applying $T^{\ast }$. Each is continuous in the relevant weak-star topology, proving that $T^{\ast }$ is weak-star continuous. Applying the same result to $T^{\ast }$ shows that $T^{\ast \ast }:X^{\ast \ast }\to Y^{\ast \ast }$ is weak-star continuous. Direct evaluation gives

Use $B_X$ for the closed unit ball; using the open ball gives the same norm closure of $T(B_X)$. Goldstine theorem says that $J_X{B}_X$ is weak-star dense in $B_{X^{\ast \ast }}$. Put $K=T^{\ast \ast }(B_{X^{\ast \ast }})$. It is compact and closed in the weak-star topology by Banach-Alaoglu theorem and the established continuity. It contains $J_{Y}T(B_X)$. Conversely, Goldstine theorem gives, for every $x^{\ast \ast }\in B_{X^{\ast \ast }}$, a net $(J_X{x}_{\alpha })$ from $J_X{B}_X$ converging weak-star to $x^{\ast \ast }$; its image converges weak-star to $T^{\ast \ast }{x}^{\ast \ast }$. ConsequentlyApply the preceding bounded convex set criterion in $Y$, and then scale the closed unit ball. We obtain the bidual characterization of weakly compact operators:The canonical embedding into the bidual on the right specifies exactly which copy of $Y$ is intended.