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A map from an open subset of $\mathbb{C}$ to a complex Banach space is holomorphic when the displayed limit exists in the Banach-space norm at every point. A norm-convergent power series is holomorphic inside its radius of convergence: on every smaller disk the series of derivatives converges uniformly by the geometric coefficient bound, allowing differentiation term by term. Applying a continuous linear functional produces an ordinary scalar holomorphic function. Scalar Cauchy identities therefore imply the corresponding vector-integral identities when bounded linear functionals separate points, as they do by the Hahn-Banach theorem.

For a probability measure whose identity map is strongly measurable with integrable norm, its barycenter is the Bochner integral of that map. It is characterized by $\phi (b(\mu ))=\int \phi (x)d\mu (x)$ for every continuous linear functional. For a measure on a weakly compact set in a separable Banach space, its barycenter belongs to the norm-closed convex hull of that set.

Let $\mathcal{G}$ be the first Gaussian chaos of a centered Gaussian random variable in a Banach space. By Fernique's theorem, $Sg=\mathbb{E}[Xg]$ is a Bochner integral in $B$. The map $S$ is injective, because $Sg=0$ implies $\mathbb{E}[g\ell (X)]=0$ for every continuous linear functional $\ell$, hence $g=0$. Its range becomes a Hilbert space under $⟨Sg,Sv{⟩}_H=\mathbb{E}[gv]$. The embedding in $B$ is continuous by Cauchy-Schwarz inequality. Its reproducing property is $⟨h,S\ell (X){⟩}_H=\ell (h)$. Thus this is the Reproducing kernel Hilbert space of the random variable, including degenerate laws. For $h=Sg$, the scalar coordinate $\hat{h}(X)=g$ has variance $\Vert h{\Vert }_H^2$; it is generally a measurable linear coordinate rather than a continuous functional on $B$.

The continuous dual space of a Hausdorff locally convex space separates points: for every nonzero $x$ there is a continuous linear functional $f$ with $f(x)\ne 0$.

A separating family of continuous seminorms supplies $p(x)>0$. In the real case, define $g(tx)=tp(x)$ on the span of $x$ and extend it under $p$ by the Hahn-Banach theorem. The bound $|g(y)|\le p(y)$ gives continuity. In a complex vector space, apply the real argument to the underlying real vector space, then use $f(y)=g(y)-ig(iy)$ to obtain a complex-linear functional with nonzero real part at $x$. The Hausdorff condition is essential: a nontrivial vector invisible to every continuous seminorm is also invisible to every continuous linear functional.

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 dual norm measures how strongly a bounded linear functional can respond to a unit-sized input. If $p$ is a bounded linear functional on a normed vector space $X$, write $⟨p,x⟩=p(x)$ and defineIt is the operator norm of $p$ when its scalar output is measured by absolute value. On ${\mathbb{R}}^n$, represent $p(x)=a^{T}x$ using a coefficient vector $a$. The unit ball is symmetric, so maximizing $|a^{T}x|$ gives the same value as maximizing $a^{T}x$: negate an input whenever its output is negative.^[ref]

For an original two-coordinate example, fix $a=(2,-1{)}^T$, so $p(x)=2x_1-x_2$. The input norm changes the allowable unit ball and hence the dual norm:Under the L1 norm, $|2x_1-x_2|\le 2(|x_1|+|x_2|)\le 2$, and $(1,0{)}^T$ attains the bound. Under the Euclidean norm, the Cauchy-Schwarz inequality gives $|a^{T}x|\le \Vert a{\Vert }_2\Vert x{\Vert }_2\le \sqrt{5}$, with equality at the displayed normalized vector. Under the maximum norm, each coordinate can reach unit absolute value, so $|2x_1-x_2|\le 2+1=3$; $(1,-1{)}^T$ attains it. Thus the dual norms of these three vector norms are respectively the maximum norm, the Euclidean norm, and the L1 norm. For $1\le p\le \infty$, Lp norms on a finite coordinate space pair with their Holder conjugate exponents.^[ref]

This geometry explains the usefulness of a dual norm: it translates a bound on inputs into the sharp bound on a linear output. The support function of the unit ball gives its value in the coefficient direction $a$. Changing the input budget changes the set of possible inputs; the three different answers do not conflict. In finite dimensions a closed unit ball is compact, so a continuous linear functional achieves its maximum absolute value. In an infinite-dimensional normed vector space, the supremum remains the definition and need not be achieved.

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 Gaussian measure on a real topological vector space is a Borel probability measure whose image under every continuous linear functional is a one-dimensional normal distribution.

A random variable $X$ in a real separable Banach space $B$ is Gaussian when $\ell (X)$ has a normal distribution for every continuous linear functional $\ell \in B^{\ast }$. It is centered when all these expected values vanish. Its probability law is a Gaussian measure. Degenerate normal distributions are allowed.

Let $X$ be a compact metric space and $\tau :X\to X$ a continuous map. Its composition operator on the Banach space $C(X)$ is $Tf=f\circ \tau$. A continuous linear functional $\mu$ is invariant when $\mu (Tf)=\mu (f)$ for every $f\in C(X)$.

To construct a positive normalized invariant linear functional, fix $x\in X$ and average point evaluations along its orbit:These linear functionals have operator norm one. The Banach-Alaoglu theorem, together with separability of $C(X)$, gives a subsequence convergent in the weak-star topology. Its limit $\mu$ is positive and satisfies $\mu (1)=1$. Finally,so the limit is invariant. Continuity of $\tau$ ensures $Tf\in C(X)$, which is needed when passing to the limit in the weak-star topology.

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.

The continuous linear functional $f_a(x)=\sum a_n{x}_n$ on the real space of sequences converging to zero corresponds to $a\in {\ell }_1$ and has operator norm $\sum |a_n|$. Equality $f_a(x)=\Vert f_a\Vert$ on the closed unit ball forces $x_n=\mathrm{sgn}(a_n)$ on the support of $a$, since every term $|a_n|-a_n{x}_n$ is nonnegative. Such an $x$ can tend to zero only if that support is finite. Conversely, the finite sign sequence attains equality for every finitely supported sequence $a$.

For a nonzero vector in a real or complex normed vector space, a norming functional is a continuous linear functional satisfying the displayed conditions. The Hahn-Banach theorem extends the functional sending $av$ to $a\Vert v\Vert$ on its one-dimensional span. Consequently $\Vert w\Vert \ge \mathrm{Re}\ell (w)$, with equality at $v$, giving a supporting linear lower bound for the norm. At zero the zero functional can be used as a supporting functional.

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.

Put $\rho =r-\sqrt{r^2-1}$. Since $r>1$, $0<\rho <1$ and ${\rho }^2-2r\rho +1=0$. The nonzero square-summable sequence ${\beta }_j={\rho }^{j-1}$ defines a continuous linear functional on ${\ell }_2$ byThe Cauchy-Schwarz inequality proves continuity. For every $k$,Hence $f$ vanishes on the closed linear span $M$, but $f(e_1)=1$. ThereforeMore precisely, any vector in the orthogonal complement of $M$ satisfies the recurrence ${\beta }_{k+2}-2r{\beta }_{k+1}+{\beta }_k=0$. Its roots are $r\pm \sqrt{r^2-1}$, one less than one and the other greater than one. Square summability eliminates the growing solution, so $M^{\perp }$ is exactly the span of $({\rho }^{j-1})$. Thus $M=\ker f$ is a closed hyperplane of codimension one. This is annihilator of a second-difference span in l2.

A distribution is a continuous linear functional on the space of test functions $\mathcal{D}(\mathbb{R})=C_c^{\infty }(\mathbb{R})$. A precise useful form of continuity is: for every compact set $K\subset \mathbb{R}$ there exist an integer $m\ge 0$ and a constant $C_K$ such thatEquivalently, if a sequence of test functions has supports in one fixed compact set and every derivative tends uniformly to zero, then its images under $T$ tend to zero. Keeping a common compact support is part of the test function topology; mere pointwise convergence is insufficient.

The real Hahn-Banach separation theorem says that if $C$ is a nonempty open convex set in a real normed vector space and $y\notin C$, there is a nonzero continuous linear functional $g$ such that $g(y)\ge {\sup }_{z\in C}g(z)$.

Put $\delta =1-1/M\in (0,1)$. The prescribed set is open because it is a union of translates of an open ball, and it is convex because both summands are convex. If $\Vert y\Vert <1$ and $y\notin C$, Hahn-Banach separation gives a nonzero $g\in Y^{\ast }$ withThe displayed supremum is a sum of independent suprema; the symmetry of the unit balls identifies them with the respective operator norms. But $g(y)\le \Vert g\Vert \Vert y\Vert <\Vert g\Vert$, a contradiction. Hence $y=Tx+r$ for some $\Vert x\Vert \le 1$ and $\Vert r\Vert <\delta$.

To obtain the strict bound on $x$ required in the approximation hypothesis, choose $t<1$ sufficiently close to $1$. Then $\Vert tx\Vert <1$ andThus the approximation condition holds with exactly $\delta =1-1/M$. When $X$ is a Banach space, the preceding successive approximation for quotient operators proves $T$ is a quotient operator. Combining this with the Banach-space adjoint estimate proves

The compact set $S(B_{c_0})$ is bounded, so $S$ is a bounded linear operator. For each $g\in Y^{\ast }$, the continuous linear functional $g$ attains its maximum on that compact set. Symmetry of $S(B_{c_0})$ implies that maximum is $\Vert S^{\ast }g\Vert$, so $S^{\ast }g$ attains its norm on $B_{c_0}$. By norm attainment on the vanishing sequence space,Regard the Banach-space adjoint $S^{\ast }$ as a bounded linear operator from the Banach space $Y^{\ast }$ into $c_{00}$ with its inherited ${\ell }_1$ norm. The completeness of the dual space holds even if $Y$ is incomplete. The preceding Baire category theorem argument shows that $S^{\ast }$ has finite rank.

If $S^{\ast }(Y^{\ast })=\mathrm{span}\{f_1,\dots ,f_r\}$, set $N={\bigcap }_{j=1}^r\ker f_j$. For $x\in N$ and every $g\in Y^{\ast }$, $g(Sx)=(S^{\ast }g)(x)=0$. The Hahn-Banach theorem implies that continuous linear functionals separate points of $Y$, so $Sx=0$. Thus $N\subseteq \ker S$, andThis proves compact closed-ball image forces finite rank on c0.

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.

The weak topology $\sigma (X,X^{\ast })$ is the coarsest topology on $X$ making every continuous linear functional $f\in X^{\ast }$ continuous. A basic neighborhood of $x_0$ is specified by finitely many inequalities $|f_j(x-x_0)|<ϵ$.

The weak-star topology $\sigma (X^{\ast },X)$ is the coarsest topology on $X^{\ast }$ making every evaluation $f\mapsto f(x)$, for $x\in X$, continuous. Its basic neighborhoods use finitely many such evaluations. Nets converge in these topologies exactly when all the corresponding scalar evaluations converge.

Let $J_X:X\to X^{\ast \ast }$ be the canonical isometric embedding, $J_{X}x(f)=f(x)$. Goldstine theorem statesHere the balls are closed unit balls. Thus any bidual unit vector can be approximated in finitely many dual evaluations by unit vectors from $X$; this does not claim norm density or sequential density in complete generality.