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The continuous dual space of ${\ell }^1$ is ${\ell }^{\infty }$ under the complex-linear coordinate pairing. Its subspace $c_0$ is norm closed: uniform approximation by a sequence with a small tail leaves a small tail. Nevertheless its preannihilator in ${\ell }^1$ is zero, because the coordinate unit vectors belong to $c_0$ and test every coordinate. Its double continuous annihilator is therefore all of ${\ell }^{\infty }$. It is weak-star dense directly as well: the finite truncations of any bounded sequence lie in $c_0$ and converge against every ${\ell }^1$ test by absolute summability. This demonstrates why norm closure cannot replace closure in the weak-star topology in the dual-side annihilator theorem.

A bilinear form $B:X\times X\to \mathbb{R}$ on a real normed vector space is bounded when one constant $0\le C<\infty$ satisfiesThis is an upper bound on the magnitude of every pairing. It says nothing about the sign of $B(v,v)$ and does not require a symmetric bilinear form. The chosen norm is part of the hypothesis. Boundedness is equivalent to joint continuity, as shown by the boundedness criterion for a bilinear form.^[ref]

For example, on ${\mathbb{R}}^2$ with the Euclidean norm, the bilinear formsboth satisfy the upper bound with $C=1$, by the Cauchy-Schwarz inequality. Yet their quadratic forms are $B_{-}(v,v)=x^2-y^2$ and $B_0(v,v)=x^2$: the former takes negative values and the latter vanishes at the nonzero vector $(0,1)$. Thus neither is a coercive bilinear form. The latter is also a degenerate bilinear form, whereas $B_{-}$ is nondegenerate. Boundedness controls size, not positivity or loss of directions.

Every bilinear form on a real or complex finite-dimensional vector space equipped with any norm is bounded. In coordinates, write $B(u,v)=u^{T}Av$ using its matrix of a bilinear form. The Cauchy-Schwarz inequality gives $|u^{T}Av|\le \Vert A{\Vert }_2\Vert u{\Vert }_2\Vert v{\Vert }_2$, with $\Vert A{\Vert }_2$ the spectral norm; finite-dimensional equivalence of norms transfers this to the chosen norm. The conclusion concerns each fixed norm and fixed bilinear form: it does not supply a single constant for an arbitrary family of matrices or changing dimensions.

On an infinite-dimensional vector space, the upper bound is an additional condition. Equip the finitely supported sequences $c_{00}$ with the norm inherited from the l2 sequence space and setEach pairing is a finite sum, so this is a well-defined bilinear form. For the basis vector $e_n$, however, $\Vert e_n{\Vert }_2=1$ and $B(e_n,e_n)=n$; no finite $C$ works. It is even a coercive bilinear form with $\alpha =1$, since $\sum n{v}_n^2\ge \sum v_n^2$. Consequently coercivity alone does not imply boundedness on a normed vector space. This example uses $c_{00}$, which is not a Hilbert space; it does not define this pairing on all of the l2 sequence space.

A bounded bilinear form defines a bounded linear operator $T:X\to X^{\prime }$ by $(Tu)(v)=B(u,v)$, where $X^{\prime }$ is the continuous dual space. Its operator norm is the smallest admissible $C$:This converts an upper bound for pairings into an upper bound for a linear map taking an unknown vector to a linear functional.

The closed unit ball of a normed vector space is the set of vectors of norm at most one. In a continuous dual space it is compact in the weak-star topology by the Banach-Alaoglu theorem.

Suppose a linear map $S:c_0\to Y$ has compact image of the closed unit ball. It is bounded, and each $g\in Y^{\ast }$ attains its maximum on this compact image. Thus $S^{\ast }g$ attains its norm, so norm attainment on the vanishing sequence space implies $S^{\ast }(Y^{\ast })\subseteq c_{00}$. The Baire category theorem applied to the inverse images of the first $n$ coordinate spans makes this adjoint finite rank: these are closed linear subspaces covering the complete continuous dual space $Y^{\ast }$, so one has interior and is all of $Y^{\ast }$. Finally the Hahn-Banach theorem identifies $\ker S$ as the common kernel of $S^{\ast }(Y^{\ast })$, proving finite rank of $S$. Ordinary compact operators require only relative compactness of the ball image, so they need not have finite rank.

The continuous dual ${\mathcal{E}}^{\prime }(X)$ of $\mathcal{E}(X)$ is the space of compactly supported distributions on $X$. Its weak topology is pointwise convergence on $\mathcal{E}(X)$.

The continuous dual space $X^{\ast }$ of a topological vector space is the vector space of its continuous linear functionals. Unlike the algebraic dual, it requires that a small change in an input produce a small change in its scalar measurement. For a normed vector space, this is equivalent to a finite boundfor all vectors $x$ and some constant $C\ge 0$. The least such constant is the operator norm of $\phi$.

This is the continuity-restricted subspace of the algebraic dual space; in infinite dimension the two can differ.

For the linear functional $\phi (x,y)=2x-y$ on ${\mathbb{R}}^2$ with its Euclidean norm, the Cauchy-Schwarz inequality givesEquality holds on the unit vector $(2,-1)/\sqrt{5}$, so its operator norm is exactly $\sqrt{5}$. The same estimate controls measurement error: $|\phi (v+h)-\phi (v)|\le \sqrt{5}\Vert h{\Vert }_2$. The linear functional diagram shows which changes are entirely invisible to this measurement; its kernel of a linear map consists of the changes with zero response. All linear functionals on a finite-dimensional real or complex normed vector space are continuous, but discontinuous functional detected by vanishing sequence tails shows why the restriction matters in infinite dimension.

A continuous linear functional is a continuous linear map from a topological vector space to its scalar field. The collection of all such functionals is the continuous dual space.

The algebraic double dual of a vector space is the dual space of its algebraic dual. An element assigns a scalar linearly to each linear functional on the original space. The bidual evaluation map sends $v$ to the rule $\ell \mapsto \ell (v)$. It is a canonical linear isomorphism in finite dimension, unlike an identification of $V$ with $V^{\ast }$, which generally requires an additional choice. For spaces with topology, using the continuous dual space instead defines the topological bidual; that is a distinct construction in infinite dimension.

For $1\le p<\infty$ and conjugate exponent $q$, every bounded linear functional on $L^p$ over a sigma-finite measure space has the formfor a unique $\omega \in L^q$, and its operator norm is $\Vert \omega {\Vert }_q$. At the endpoint $p=1$, this says $(L^1{)}^{\ast }=L^{\infty }$.

This identifies the continuous dual space of an Lp space; it is not an alternative definition of the entire Lp-space topic.

The dual space of a vector space $V$ over a field $F$ is the vector space of all linear functionals $\phi :V\to F$. Think of a vector as an input and a linear functional as a scalar-valued measurement of that input. For $V={\mathbb{R}}^2$, the rules $(x,y)\mapsto x$, $(x,y)\mapsto y$, and $(x,y)\mapsto 2x-y$ are different elements of the dual space. None needs a notion of length or angle.

Its operations combine measurements pointwise:The dual basis explains the finite-dimensional structure: specifying a linear functional's values on a basis determines its value on every vector. Thus an $n$-dimensional vector space has an $n$-dimensional dual space. The spaces have the same dimension, but their elements play different roles: an input and a rule that measures an input. ^[ref].

This separation is useful when changing coordinates. The dual pairing $\phi (v)$ must stay the same while the component lists for $v$ and $\phi$ change in opposite ways. Covector transformation under a change of basis makes the compensation explicit. An inner product on a finite-dimensional real vector space gives one way to represent each linear functional by a vector, but that representation uses the chosen inner product; the dual space itself does not require one.

Here dual space means the algebraic dual: all linear functionals. For a topological vector space, the continuous dual space keeps only the continuous linear functionals. In infinite dimension, this can be a proper vector subspace; discontinuous functional detected by vanishing sequence tails gives an example. The distinction matters when a scalar-valued measurement must respond continuously to small input changes.

A linear functional is a linear map from a vector space $V$ to its scalar field $F$:Its output is one scalar, rather than an arbitrary vector in another space. In ${\mathbb{R}}^2$, $\phi (x,y)=2x-y$ gives $\phi (3,1)=5$, $\phi (1,2)=0$, and $\phi (4,3)=5$: adding the two inputs adds their measured values.

The level sets $\phi (v)=c$ visualize what the measurement retains and discards. For this example they are parallel straight lines $y=2x-c$. Moving parallel to $(1,2)$ changes the input but leaves $\phi$ unchanged. The zero level set is its kernel of a linear map, while the other level sets are translates of that vector subspace.

Inputs need not be geometric arrows. On the vector space of polynomials of polynomial degree at most two, evaluating at zero is a linear functional, as is taking the definite integral from zero to one. For $p(t)=a+bt+c{t}^2$, they give $p(0)=a$ and ${\int }_0^{1}p(t)dt=a+b/2+c/3$. Thus polynomial evaluation and integration are two different measurements of the same polynomial. Coordinate extraction is another measurement, developed under dual basis.

A linear functional is defined by linearity, not by continuity. In finite-dimensional real or complex normed vector spaces it is automatically continuous; in infinite dimension it need not be. The continuous dual space adds this requirement when small changes in the input should give small changes in the output.

A locally convex space is a topological vector space with a neighbourhood basis at zero consisting of convex sets. Equivalently, its topology is a seminorm topology generated by some family $\mathcal{P}$ of seminorms. We allow non-Hausdorff spaces here; a separating family of seminorms gives precisely the Hausdorff case. Every normed vector space is locally convex, but smooth convergence generally requires a whole family of measurements.

The geometric and measurement descriptions fit together: seminorm sublevel sets are convex, while the Minkowski functional of an open, convex zero neighbourhood invariant under all scalar multiplications with $|\lambda |\le 1$ is a continuous seminorm. Such neighbourhoods can be chosen as a neighbourhood basis. ^[ref].

Local convexity makes Hahn-Banach theorem and separation of a point and an open convex set available. It turns geometric distinctions into continuous linear functionals, connecting convex sets to the continuous dual space. For a seminorm topology generated by $\mathcal{P}$, a linear functional $\ell$ is continuous exactly when there are $p_1,\dots ,p_m\in \mathcal{P}$ and $C>0$ withThus one continuous linear functional needs only finitely many of the available measurements. For example, point evaluation of a derivative on the space of smooth functions on a compact interval is controlled by a suitable derivative seminorm.

For a real Banach space $X$, a map $A:X\to X^{\ast }$ has operator monotonicity when $⟨A(u)-A(v),u-v⟩\ge 0$ for all $u,v\in X$. This is the monotone operator property, with the pairing between $X$ and its continuous dual space $X^{\ast }$. In a Hilbert space one can identify that pairing with the inner product.

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.

The continuous dual space $X^{\prime }$ consists of all bounded complex-linear functionals $f:X\to \mathbb{C}$, with pointwise vector operations and norm $\Vert f\Vert ={\sup }_{\Vert x\Vert \le 1}|f(x)|$. This is finite by boundedness. The scalar norm identity and triangle inequality follow under the supremum; if the norm is zero, scaling every nonzero vector into the unit ball shows that $f$ vanishes. Hence $X^{\prime }$ is a normed complex vector space.

For the dual of the convergent sequence space, take $\beta \in {\ell }^1$. The displayed formula is a bounded linear functional, since the series is absolutely convergent andIt is linear in $\beta$, so $\Vert \varphi (\beta )\Vert \le \Vert \beta {\Vert }_1$. For equality fix $N$, choose ${\alpha }_j$ for $j\le N$ of modulus at most one so that ${\alpha }_j{\beta }_{j+1}=|{\beta }_{j+1}|$, and make all later terms equal a constant $c$ of modulus at most one with $c{\beta }_1=|{\beta }_1|$. The choices for a zero coefficient can be arbitrary. This sequence lies in the unit ball of $c$ and converges to $c$. The reverse triangle inequality givesLetting $N\to \infty$ proves $\Vert \varphi (\beta )\Vert =\Vert \beta {\Vert }_1$, including injectivity.

For surjectivity let $F\in c^{\prime }$. By the permitted description of $c_0^{\prime }$, there is $\gamma \in {\ell }^1$ with $F(z)={\sum }_j{\gamma }_j{z}_j$ on $c_0$. Every $\alpha \in c$ decomposes as $\alpha -{\alpha }_{\infty }1\in c_0$ plus ${\alpha }_{\infty }1$. ConsequentlyTake ${\beta }_{j+1}={\gamma }_j$ and ${\beta }_1=F(1)-{\sum }_j{\gamma }_j$. This is an ${\ell }^1$ sequence and $F=\varphi (\beta )$, proving the stated isometric linear isomorphism onto $c^{\prime }$. No complex conjugation is inserted, since these are complex-linear dual functionals.

Equip the continuous dual space $X^{\prime }$ with the operator normContinuity of a linear functional makes this finite; homogeneity, definiteness and the triangle inequality follow directly from the supremum formula.

For completeness of the dual space, let $(f_n)$ be Cauchy in this norm. For each $x\in X$, $|f_n(x)-f_m(x)|\le \Vert f_n-f_m\Vert \Vert x\Vert$, so $(f_n(x))$ is Cauchy in $\mathbb{R}$. Define $f(x)={\lim }_n{f}_n(x)$. Passing to the limit in the linearity identities makes $f$ linear. Since a Cauchy sequence is bounded, ${\sup }_n\Vert f_n\Vert \le C$ and $|f(x)|\le C\Vert x\Vert$, so $f\in X^{\prime }$.

Given $\epsilon >0$, choose $N$ with $\Vert f_n-f_m\Vert \le \epsilon$ for $n,m\ge N$. Let $m\to \infty$ pointwise; then $|f_n(x)-f(x)|\le \epsilon \Vert x\Vert$ and therefore $\Vert f_n-f\Vert \le \epsilon$. Thus $X^{\prime }$ is a Banach space. This proof in fact works even if $X$ itself is not complete.

The continuous dual space $X^{\prime }$ is the vector space of bounded scalar-valued linear functionals, with operator normThis makes it a normed space, and in fact a Banach space whether or not $X$ is complete.

The norm-preserving Hahn-Banach theorem states that a bounded linear functional $f_0$ on any linear subspace $M\subseteq X$ extends to a bounded linear functional $f$ on all of $X$ with $f{|}_M=f_0$ and $\Vert f\Vert =\Vert f_0\Vert$. The subspace need not be closed. This holds over both the real and complex scalar fields, using the complex version when necessary.

For the spaces in part (b)(i) and fixed $f\in Y$, define $A:X\to X^{\ast }$ byFor each $u$, this is a linear functional in $v$, and the estimates in part (b)(i) showHence $A(u)$ belongs to the continuous dual space $X^{\ast }$, and $A$ sends bounded sets to bounded sets. The weak formulation is exactlyThe forcing term is included in $A$ so that its right side is zero, as requested.

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.

Since $W$ is proper, $T$ is a nonzero linear functional. Choose $e\in X$ with $Te=1$, by rescaling any vector outside its kernel. For every $x\in X$,If $ae\in W$, applying $T$ gives $a=0$. Consequently this is a direct sum, and the quotient vector space $X/W$ has dimension one. This computes the codimension of a vector subspace:

To prove boundedness without assuming completeness, use the linear functional with closed kernel criterion directly. Since $W$ is a closed set and $e\notin W$, the distanceis positive. If $Tx\ne 0$, then $e-x/(Tx)\in W$, soFor $Tx=0$ the resulting estimate is immediate. ThereforeHere $X^{\ast }$ denotes the continuous dual space.