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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.

Even a bounded infinite input need not have a closed convex hull. In the real line, let $S=\{1/n:n\ge 1\}$. Every finite convex combination of these positive points is positive, so zero is absent. Conversely, any $x\in (0,1]$ lies between some $1/n$ and $1$, and is their convex combination. HenceThe closed convex hull adds the missing limit point zero. For the finite eight-point example in convex hull, the convex hull is already closed, so taking the closed convex hull adds nothing.

A convex set that is also a closed subset in the specified topology. In a Hilbert space with its norm topology, every nonempty closed convex set has a unique nearest point to every ambient vector, by the Hilbert projection theorem.

In a normed vector space, the closed linear span of $S$ is the closure of its linear span in the norm topology. It includes limits of finite linear combinations. In a Hilbert space, an infinite orthonormal basis spans in this closed sense, allowing norm-convergent series in addition to finite sums.

A vector subspace closed in the norm topology of a Hilbert space is complete in the inherited norm and inner product. Its orthogonal complement gives a decomposition of the ambient Hilbert space into an orthogonal direct sum. This closedness permits orthogonal projection and ensures limits used in range arguments remain in the subspace. Every finite-dimensional vector subspace is closed.

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 every finite $d\ge 1$, a subset $K\subseteq {\mathbb{R}}^d$ of Euclidean space is compact if and only if it is a closed set and a bounded set. Here closed means closed in the ambient Euclidean space, not merely in $K$ itself, and the bounded set condition uses the Euclidean norm. The empty set is included: it is a closed set, a bounded set, and compact. ^[ref]

This makes an apparently infinite test practical. To verify compactness of a closed interval $[a,b]$ with finite endpoints, we need not examine every open cover: it is closed in the real line and a bounded set. By contrast, the open interval $(0,1)$ is a bounded set but is not closed in the real line, as the missing-endpoint figure in compact space illustrates. The whole real line is closed but is not a bounded set, so it too fails compactness. These examples show why both hypotheses are needed.

The same criterion holds in any real or complex finite-dimensional vector space equipped with a norm: all norms there are equivalent norms, so they give the same open sets, closed sets, and bounded sets. Since compactness depends only on the topology, choosing a different norm does not change the conclusion. Without an appropriate topology, a bare vector space alone has no notion of compactness.

The restriction to a finite-dimensional vector space is essential. In the l2 sequence space, an infinite-dimensional vector space that is also a Hilbert space, let $e_n$ be the vectors of its standard orthonormal basis. All lie in the closed unit ball $B=\{x:\Vert x\Vert \le 1\}$, which is a closed set and a bounded set in the norm topology. For distinct indices, their inner product is zero, givingThus distinct terms remain distance $\sqrt{2}$ apart. No subsequence is a Cauchy sequence, hence none is a convergent sequence. The equivalence of compactness and sequential compactness in a metric space proves that this closed unit ball is not compact. The conclusion concerns the norm topology.

A short proof outline in Euclidean space explains the role of a finite-dimensional vector space. For a sequence in a bounded set, apply the Bolzano-Weierstrass theorem to each of the Cartesian coordinates in turn, retaining a subsequence at each of the finitely many steps. The final subsequence is a convergent sequence in all Cartesian coordinates, and a closed set contains its limit of a sequence. This yields sequential compactness, hence compactness. Conversely, a compact subset of a metric space is a closed set and a bounded set. ^[ref]

A normed vector space is a vector space equipped with a norm, whose induced metric is $d(x,y)=\Vert x-y\Vert$. The norm must be positive on nonzero vectors, scale by the absolute value of a scalar, and satisfy the triangle inequality. These conditions make distance compatible with addition and scaling.

On $C([0,1])$, both the supremum norm $\Vert f{\Vert }_{\infty }={\max }_{0\le x\le 1}|f(x)|$ and the L2 norm $\Vert f{\Vert }_2=({\int }_0^1|f(x){|}^{2}dx{)}^{1/2}$ are genuine norms. A nonzero continuous function stays nonzero on an interval of positive length, so its squared integral cannot vanish. For $f(x)=x$, they give $1$ and $1/\sqrt{3}$, respectively. Always $\Vert f{\Vert }_2\le \Vert f{\Vert }_{\infty }$, but the shrinking peaks below show that no uniform reverse bound exists. Unlike the finite-dimensional equivalence of norms, infinite-dimensional vector spaces can support inequivalent norm topologies.

A point $x$ of a convex set $K$ is an extreme point if $x=(1-t)a+tb$ with $a,b\in K$ and $0<t<1$ forces $a=b=x$. It is enough to test midpoints: a nontrivial interior point of a segment is the midpoint of a smaller nontrivial segment.

Here is a proof of the Krein-Milman theorem. The empty-set case is immediate, so take $K\ne \varnothing$. A face of a convex set is a convex subset $F\subseteq K$ such that an interior convex combination lying in $F$ has both endpoints in $F$. Consider nonempty compact faces. They include $K$. Every inclusion chain has nonempty intersection, because its closed subsets of the compact set $K$ have the finite intersection property. The intersection remains a compact face. The Zorn lemma, ordered by reverse inclusion, therefore supplies a minimal nonempty compact face $F$.

If $F$ contains two distinct points, the Hausdorff locally convex hypotheses and the allowed separation theorem supply a continuous real linear functional $\ell$ taking different values there. Its maximizer set in $F$ is nonempty and compact. It is convex, and if a convex combination attains the maximum, both endpoint values attain it too; hence it is a face of $F$. A face of a face is a face of $K$, by applying the defining endpoint property twice. This maximizer face is proper because $\ell$ is nonconstant, contradicting minimality. Thus $F$ is a singleton, whose point is extreme in $K$. The same minimal compact face argument inside any nonempty compact face shows that each such face contains an extreme point of $K$.

Let $H=\overline{\mathrm{conv}}(\mathrm{ext}K)$. Since $K$ is compact, hence closed in the Hausdorff space, and convex, $H\subseteq K$. It is nonempty, closed and convex. If $p\in K\setminus H$, the allowed strict separation theorem supplies a continuous linear functional withThe maximizer face of $\ell$ on $K$ contains an extreme point $e$ by the preceding argument. Then $e\in H$ but $\ell (e)={\max }_K\ell \ge \ell (p)>{\sup }_H\ell$, a contradiction. ThereforeAll closures refer to the given locally convex topology, and compactness is what justified both maxima and chain intersections.

For the real l-infinity sequence space, if $|x_j|<1$ for any coordinate, choose $0<ϵ\le 1-|x_j|$. The two distinct sequences $x\pm ϵe_j$ belong to the closed unit ball and have midpoint $x$, so $x$ is not extreme. Conversely, if every $x_j$ is one or minus one, writing $x=(1-t)a+tb$ for unit-ball sequences forces $a_j=b_j=x_j$ at every coordinate, because an endpoint of $[-1,1]$ cannot be a nontrivial convex average of its points. ThusFor the space of sequences converging to zero, every unit-ball sequence has some coordinate with $|x_j|<1$, since its coordinates tend to zero. The same opposite single-coordinate perturbations still belong to $c_0$, givingThese are extreme points of real sequence-space unit balls. There is no conflict with Krein-Milman: the norm-closed unit ball of $c_0$ is not compact in the norm topology, as the coordinate vectors have mutual distance one.

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.

A separable Hilbert space has a countable subset dense in its norm topology.

The printed assertion about all locally square-integrable functions is false. The proposed average is not even finite for every such function: for $f(x)=x$,It also fails positive definiteness. The nonzero function $f=1_{[0,1]}$ hasConsequently this formula cannot define an inner product, much less a Hilbert space, on $L_{\mathrm{loc}}^2(\mathbb{R})$.

A precise version of the intended nonseparability argument uses the mean-square completion of trigonometric polynomials. Start with the real vector space $V$ of finite linear combinations of $1$, $\cos (\lambda x)$, and $\sin (\lambda x)$, with arbitrary $\lambda >0$. Product-to-sum identities show that all the proposed cross averages exist. Distinct frequencies are orthogonal, each sine and cosine has squared norm one, and the constant function has squared norm two. Thus, after collecting equal frequencies,This is positive definite on $V$. Its Hilbert space completion $H$ contains the uncountable orthonormal set $\{\cos (\lambda x):\lambda >0\}$. The distance between two distinct members is $\sqrt{2}$. Their open balls of radius $1/2$ are pairwise disjoint, and a dense subset must meet each one. A countable dense subset is therefore impossible: this corrected completed space is nonseparable. Completion is an essential additional construction; it does not validate the printed claim about all of $L_{\mathrm{loc}}^2$.

Throughout this question the Banach spaces and Hilbert spaces are complex. Separability is an additional standing hypothesis, not part of the general definition of a Hilbert space. Here and below unit ball means the closed unit ball. An open unit ball in positive finite dimension is not compact. In finite dimension, equivalence of norms identifies the closed unit ball with a closed bounded subset of Euclidean space, so the Heine-Borel theorem gives compactness in the norm topology.

In an infinite-dimensional Hilbert space, the Gram-Schmidt process constructs an infinite orthonormal sequence $(e_n)$. Put $f_n=e_n/\sqrt{2}$. Then $\Vert f_n\Vert =1/\sqrt{2}\le 1$ and, for $m\ne n$, $\Vert f_n-f_m{\Vert }^2=(1+1)/2=1$. No subsequence is a Cauchy sequence, contradicting sequential compactness. Thus the closed unit ball is norm compact exactly in finite dimension.

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.

For a convex subset of a Banach space, weak closure equals norm closure. The Mazur theorem states that for every convex set $C\subseteq X$,Since every bounded linear functional is norm-continuous, the weak topology is weaker than the norm topology, giving ${\overline{C}}^{\Vert \cdot \Vert }\subseteq {\overline{C}}^w$.

For the reverse inclusion, use this form of the Hahn-Banach separation theorem: if $D$ is a nonempty norm-closed convex set in a real normed vector space and $x\notin D$, there is a continuous real-linear functional $F$ such thatApply it to $D={\overline{C}}^{\Vert \cdot \Vert }$. In a complex Banach space, use the underlying real space; every continuous real-linear $F$ is the real part of the complex bounded linear functional $f(v)=F(v)-iF(iv)$. The separating strict inequality gives a neighbourhood open for the weak topology of $x$ disjoint from $D$, so $x\notin {\overline{C}}^w$. The empty set case is immediate. This proves the Mazur theorem.

A weakly null sequence has disjoint convex blocks converging to zero in norm. If $x_n\stackrel{w}{\to }0$, then for every starting index $m$,The Mazur theorem places $0$ in the norm closure of that tail convex hull. Thus a finite convex combination of vectors from any tail can have norm less than any prescribed $\epsilon >0$.

Choose the convex blocks recursively. After selecting the previous terminal index $q_{n-1}$, let $p_n>q_{n-1}$, and choose a finite convex combination from $\{x_i:i\ge p_n\}$ with norm less than $2^{-n}$. Choose $q_n$ beyond its largest used index, padding the intervening coefficients and the final coefficient with zero. This ensures the printed strict condition $p_n<q_n$ as well as $q_n<p_{n+1}$. Setting unused coefficients to zero yieldsTherefore $u_n\to 0$ in norm. The recursive tail selection is what makes these convex blocks, rather than merely unrelated convex combinations.

Next suppose $X^{\ast }$ is separable and $(x_n)$ is bounded. Use the canonical embedding into the bidual $J_X:X\to X^{\ast \ast }$, where $J_{X}x(f)=f(x)$. If $\Vert x_n\Vert \le M$, then $J_X{x}_n\in M{B}_{X^{\ast \ast }}$. The Banach-Alaoglu theorem makes this ball compact in $\sigma (X^{\ast \ast },X^{\ast })$, and weak-star metrizability of the dual ball makes it metrizable because its predual $X^{\ast }$ is separable. A compact space that is a metric space is sequentially compact, so some subsequence $(y_n)$ satisfiesThe limit need not belong to $J_{X}X$; it is precisely the use of the bidual space that supplies compactness without assuming reflexivity.

For a convex block $u_n={\sum }_{i=p_n}^{q_n}{a}_i{y}_i$, every $f\in X^{\ast }$ satisfiesBoth $f(y_n)$ and $f(u_n)$ tend to $\varphi (f)$, soThis is the useful convex-block cancellation of a weak-star limit; it does not require the limit to be a vector in $X$.

The quotient sequence has approximate lifts in $3B_X$ that are weakly null after passing to a subsequence. Let $q:X\to Z=X/Y$ be the quotient map, with the quotient norm. Since $\Vert z_n\Vert \le 1$, choose $v_n\in X$ such thatThis uses the infimum defining the quotient norm and does not assume that it is attained. By the preceding compactness argument, pass to a subsequence $y_n=v_{k_n}$ with a common weak-star limit in $X^{\ast \ast }$. Its quotient images $q(y_n)=z_{k_n}$ form a weakly null sequence.

Apply the convex block construction to $(q(y_n))$ in the quotient Banach space. Using the same coefficients on $(y_n)$ gives convex blocksSet $x_n=y_n-u_n$. The convex-block cancellation of a weak-star limit proves $x_n\stackrel{w}{\to }0$, whileThus the approximate weakly null lifting through a quotient is achieved with the required bound:

Weak-star-to-weak continuity of the adjoint implies its weak compactness. By the Banach-Alaoglu theorem, $B_{Y^{\ast }}$ is compact for the weak-star topology. Under the assumed continuous map $T^{\ast }:(Y^{\ast },w^{\ast })\to (X^{\ast },w)$, its image $T^{\ast }(B_{Y^{\ast }})$ is weakly compact in $X^{\ast }$.

This image is closed for the weak topology, and hence for the norm topology, because the weak topology is Hausdorff and weaker than the norm topology. Thus its norm closure is the same weakly compact set. By the definition of a weakly compact operator, $T^{\ast }$ is weakly compact, proving $(iii)⟹(iv)$.

The bidual range condition implies both weak compactness and the stronger continuity of $T^{\ast }$. First assume $T^{\ast \ast }(X^{\ast \ast })\subseteq J_{Y}Y$. The Banach-Alaoglu theorem makes $B_{X^{\ast \ast }}$ weak-star compact, and $T^{\ast \ast }$ is weak-star continuous. Hence its image is compact for the weak-star topology in $Y^{\ast \ast }$. Because it lies in $J_{Y}Y$, it corresponds to a weakly compact subset $C\subseteq Y$ under $J_Y^{-1}$.

The identity $T^{\ast \ast }{J}_X=J_{Y}T$ gives $T(B_X)\subseteq C$. A weakly compact set in a Hausdorff space is closed for the weak topology, so $C$ is also norm-closed and contains the norm closure of $T(B_X)$. That closure is a convex set closed in the norm topology, hence weakly closed by the Mazur theorem. It is therefore a weakly closed subset of the weakly compact $C$, and is a weakly compact set. This proves $(ii)⟹(i)$.

For $(ii)⟹(iii)$, fix $F\in X^{\ast \ast }$. By the range assumption there is $y_F\in Y$ with $T^{\ast \ast }F=J_Y{y}_F$. For every $y^{\ast }\in Y^{\ast }$,The right side is a defining weak-star continuous evaluation on $Y^{\ast }$. Since the weak topology on $X^{\ast }$ is determined by all $F\in X^{\ast \ast }$, every scalar coordinate of $T^{\ast }$ for that topology is weak-star continuous. Therefore

Weak compactness of the adjoint forces weak compactness of the original operator. Assume $T^{\ast }:Y^{\ast }\to X^{\ast }$ is weakly compact. Apply the already proved $(i)⟹(ii)$ to this operator, rather than assuming the adjoint equivalence in advance. It givesTake $\psi \in Y^{\ast \ast \ast }$ annihilating $J_{Y}Y$, so $\psi (J_{Y}y)=0$ for every $y\in Y$. Write $T^{\ast \ast \ast }\psi =J_{X^{\ast }}f$ for some $f\in X^{\ast }$. For every $x\in X$,Hence $f=0$ and $T^{\ast \ast \ast }\psi =0$. It follows thatfor every $\psi$ annihilating $J_{Y}Y$.

The vector subspace $J_{Y}Y$ is closed in the norm topology in $Y^{\ast \ast }$ because $Y$ is complete and $J_Y$ is an isometry. The Hahn-Banach theorem says that any point outside a closed linear subspace can be separated from it by a bounded linear functional vanishing on that subspace. Applying this in $Y^{\ast \ast }$ shows that an element annihilated by all such $\psi \in Y^{\ast \ast \ast }$ must belong to $J_{Y}Y$. Therefore $T^{\ast \ast }F\in J_{Y}Y$ for every $F$, proving $(iv)⟹(ii)$. The reverse implications above now establish all four equivalences, including weak compactness of an operator and its adjoint:

Finally, if $X$ is a reflexive Banach space, every $F\in X^{\ast \ast }$ is $J_{X}x$ for some $x$, and $T^{\ast \ast }F=J_{Y}Tx\in J_{Y}Y$. If $Y$ is a reflexive Banach space, $Y^{\ast \ast }=J_{Y}Y$, so the same range condition holds automatically. In either case, $(ii)⟹(i)$ proves

The weak-star topology $\sigma (X^{\ast },X)$ is generated by the seminorms $p_x(g)=|g(x)|$, $x\in X$. Thus every evaluation is continuous, so $f_n\stackrel{w^{\ast }}{\to }f$ implies $f_n(x)\to f(x)$ for each $x$. Conversely, a basic neighborhood of $f$ imposes finitely many conditions $|(g-f)(x_j)|<\epsilon$. Pointwise convergence eventually satisfies all of them simultaneously. HenceThe Uniform boundedness principle, applied on the Banach space $X$, gives $M={\sup }_n\Vert f_n\Vert <\infty$. If $x_n\to x$ in the norm topology, thenTherefore $\boxed{f_n(x_n)\to f(x)}$. Completeness of $X$ is used in this uniform boundedness step.

If $C\subseteq X^{\ast }$ is norm compact, the identity map from $(C,\Vert \cdot \Vert )$ to $(C,w^{\ast })$ is a continuous bijection. The target is Hausdorff, since distinct functionals differ at some $x\in X$. The compact-to-Hausdorff continuous bijection theorem gives

The Banach-Alaoglu theorem states that the closed dual unit ball $B_{X^{\ast }}$ is compact in the weak-star topology; in fact $X$ need only be a normed vector space. To prove it, work over $\mathbb{F}=\mathbb{R}$ or $\mathbb{C}$ and formEach $D_x$ is compact, so $Q$ is compact by the Tychonoff theorem. Embed $B_{X^{\ast }}$ by $g\mapsto (g(x){)}_{x\in X}$. Its image consists exactly of those families $(z_x)$ satisfyingThese are closed conditions in the product topology. Conversely, any family satisfying them defines a linear functional bounded by $|z_x|\le \Vert x\Vert$, hence an element of $B_{X^{\ast }}$. The image is therefore closed and compact. The induced product topology is precisely the weak-star topology, proving the theorem. This is a compactness proof for arbitrary nets, with no separability assumption.

For the dual map of bounded linear operator $T^{\ast }:Y^{\ast }\to X^{\ast }$, each $x\in X$ satisfiesEvaluation at the fixed vector $Tx$ is weak-star continuous, so $\boxed{T^{\ast }\text{ is weak-star-to-weak-star continuous}}$.

Suppose $T$ is a compact operator. By the allowed Schauder theorem for compact operators, $T^{\ast }$ is compact. If $S\subseteq Y^{\ast }$ is bounded, the norm closure $C=\overline{T^{\ast }(S)}$ is norm compact. We already proved that its norm and weak-star topologies coincide. The weak-star continuous map $T^{\ast }{|}_S$ into $C$ is consequently norm continuous. Conversely, if $T^{\ast }$ is weak-star-to-norm continuous on bounded subsets, its restriction to $B_{Y^{\ast }}$ maps a weak-star compact set, by Banach-Alaoglu theorem, to a norm compact set. Thus $T^{\ast }$ is a compact operator, and the Schauder theorem for compact operators gives

Let $X$ have basis $e_1,\dots ,e_d$, and write $x={\sum }_j{a}_j{e}_j$. On the compact coordinate sphere ${\sum }_j|a_j|=1$, the continuous positive functionhas a positive minimum $m$. Homogeneity and the triangle inequality therefore give constants $m,M>0$ such thatIn particular, each coordinate functional ${\varphi }_j(x)=a_j$ belongs to the continuous dual space $X^{\ast }$.

Every member of $X^{\ast }$ is norm-continuous, so the weak topology is coarser than the norm topology. Conversely, given $x_0\in X$ and $\epsilon >0$, the basic weak neighbourhoodsatisfies $U\subseteq \{x:\Vert x-x_0\Vert <\epsilon \}$. Thus every norm-open set is weakly open, proving that the two topologies coincide. This is the finite-dimensional weak and norm topologies coincide theorem.

Suppose $a^k\rightharpoonup a$ in ${\ell }^1$ but $a^k$ does not converge to $a$ in the norm topology. There are $\epsilon >0$ and a subsequence $(a^{m_k})$ such thatSet $x^k=a^{m_k}-a$. Thenwhich proves the first assertion.

We now use a gliding hump argument. Put $b_0=0$. Having chosen $b_{n-1}$ and $k_{n-1}$, coordinatewise convergence $x_i^k\to 0$ lets us choose $k_n>k_{n-1}$ so thatFor this fixed element of ${\ell }^1$, choose $b_n>b_{n-1}$ so far out thatDefine one sequence $y$ byThe blocks partition the positive integers and $|y_i|\le 1$, so $y\in {\ell }^{\infty }$. On the $n$th assigned block, the signs agree; outside it, use $|y_i|\le 1$. The duality of l1 and l infinity givesBut $x^{k_n}\rightharpoonup 0$ requires $⟨x^{k_n},y⟩\to 0$ for this fixed $y\in ({\ell }^1{)}^{\ast }={\ell }^{\infty }$, a contradiction. Therefore every weakly convergent sequence in ${\ell }^1$ converges in norm. This is the Schur property of l1.

A Banach space has the Schur property when every weakly convergent sequence converges with respect to the norm topology.