Dual system

Throughout, (X, Y, b) will be a pairing over 𝕂. The absolute polar or polar of a subset A of X is the set:[4]

${\displaystyle A^{\circ }:=\left\{y\in Y:\sup _{x\in A}\left|b(x,y)\right|\leq 1\right\}}$.

Dually, the absolute polar or polar of a subset B of Y is denoted by B° and defined by

${\displaystyle B^{\circ }:=\left\{x\in X:\sup _{y\in B}\left|b(x,y)\right|\leq 1\right\}}$.

In this case, the absolute polar of a subset B of Y is also called the absolute prepolar or prepolar of B and may be denoted by °B.

The polar B° is necessarily a convex set containing 0 ∈ Y where if B is balanced then so is B° and if B is a vector subspace of X then so too is B° a vector subspace of Y.[5]

If A ⊆ X then the bipolar of A, denoted by A°°, is the set °(A^⊥). Similarly, if B ⊆ Y then the bipolar of B is B°° := (°B)°.

If A is a vector subspace of X, then A° = A^⊥ and this is also equal to the real polar of A.

Given a pairing (X, Y, b), define a new pairing (Y, X, d) where d(y, x) := b(x, y) for all x ∈ X and all y ∈ Y.[2] For the purpose of explaining the following convention, the map d will be called the mirror of b (this is not standard terminology).

There is a repeating theme in duality theory, which is that any definition for a pairing (X, Y, b) has a corresponding dual definition for the pairing (Y, X, d).

Convention and Definition: Given any definition for a pairing (X, Y, b), one obtains a dual definition by applying it to the pairing (Y, X, d). This conventions also apply to theorems.

Convention: Adhering to common practice, unless clarity is needed, whenever a definition (or result) for a pairing (X, Y, b) is given then this article will omit mention of the corresponding dual definition (or result) but nevertheless use it.

For instance, if "X distinguishes points of Y" (resp, "S is a total subset of Y") is defined as above, then this convention immediately produces the dual definition of "Y distinguishes points of X" (resp, "S is a total subset of X").

This following notation is almost ubiquitous because it allows us to avoid having to assign a symbol to the mirror of b.

Convention and Notation: If a definition and its notation for a pairing (X, Y, b) depends on the order of X and Y (e.g. the definition of the Mackey topology 𝜏(X, Y, b) on X) then by switching the order of X and Y, then it is meant that definition applied to (Y, X, d) (e.g. 𝜏(Y, X, b) actually denotes the topology 𝜏(Y, X, d)).

For instance, once the weak topology on X is defined, which is denoted by 𝜎(X, Y, b), then this definition will automatically be applied to the pairing (Y, X, d) so as to obtain the definition of the weak topology on Y, where this topology will be denoted by 𝜎(Y, X, b) rather than 𝜎(Y, X, d).

Identification of (X, Y) with (Y, X)

Although it is technically incorrect and an abuse of notation, this article will also adhere to the following nearly ubiquitous convention:

Convention: This article will use the common practice of treating a pairing (X, Y, b) interchangeably with (Y, X, d) and also denoting (Y, X, d) by (Y, X, b).

Suppose that (X, Y, b) is a pairing, M is a vector subspace of X, and N is a vector subspace of Y. Then the restriction of (X, Y, b) to M × N is the pairing ${\displaystyle \left(M,N,b{\big \vert }_{M\times N}\right)}$. If (X, Y, b) is a duality then it's possible for a restrictions to fail to be a duality (e.g. if Y ≠ { 0 } and N = { 0 }).

Convention: This article will use the common practice of denoting the restriction ${\displaystyle \left(M,N,b{\big \vert }_{M\times N}\right)}$ by (M, N, b).

Suppose that X is a vector space and let X^# denote the algebraic dual space of X (i.e. the space of all linear functionals on X). There is a canonical duality (X, X^#, c) where c(x, x') = ⟨x, x'⟩ := x'(x), which is called the evaluation map or the natural or canonical bilinear functional on X × X^#. Note in particular that for any x' ∈ X^#, c(•, x') is just another way of denoting x'; i.e. c(•, x') = x'(•) = x'.

If N is a vector subspace of X^# then the restriction of (X, X^#, c) to X × N is called the canonical pairing where if this pairing is a duality then it is instead called the canonical duality. Clearly, X always distinguishes points of N so the canonical pairing is a dual system if and only if N separates points of X. The following notation is now nearly ubiquitous in duality theory.

Notation: The evaluation map will be denoted by ⟨x, x'⟩ := x'(x) (rather than by c) and ⟨X, N⟩ will be written rather than (X, N, c).

Assumption: As is common practice, if X is a vector space and N is a vector space of linear functionals on X, then unless stated otherwise, it will be assumed that they are associated with the canonical pairing ⟨X, N⟩.

If N is a vector subspace of X^# then X distinguishes points of N (or equivalently, (X, N, c) is a duality) if and only if N distinguishes points of X, or equivalently if N is total (i.e. n(x) = 0 for all n ∈ N implies x = 0).[2]

Suppose X is a topological vector space (TVS) with continuous dual space X'. Then the restriction of the canonical duality ${\displaystyle \left(X,X^{\#},c\right)}$ to X × X' defines a pairing ${\displaystyle \left(X,X^{\prime },c{\big \vert }_{X\times X^{\prime }}\right)}$ for which X separates points of X'. If X' separates points of X (which is true if, for instance, X is a Hausdorff locally convex space) then this pairing forms a duality.[3]

Assumption: As is commonly done, whenever X is a TVS then, unless indicated otherwise, it will be assumed without comment that it's associated with the canonical pairing ⟨X, X'⟩.

Polars and duals of TVSs

The following result shows that the continuous linear functionals on a TVS are exactly those linear functionals that are bounded on a neighborhood of the origin.

A pre-Hilbert space (H, ⟨ ⋅, ⋅ ⟩) is a dual pairing if and only if H is vector space over ℝ or H has dimension 0. Here it is assumed that the sesquilinear form ⟨ ⋅, ⋅ ⟩ is conjugate homogeneous in its second coordinate and homogeneous in its first coordinate.

• If (H, ⟨ ⋅, ⋅ ⟩) is a real Hilbert space then (H, H, ⟨ ⋅, ⋅ ⟩) forms a dual system.
• If (H, ⟨ ⋅, ⋅ ⟩) is a complex Hilbert space then(H, H, ⟨ ⋅, ⋅ ⟩) forms a dual system if and only if dim H = 0. If H is non-trivial then (H, H, ⟨ ⋅, ⋅ ⟩) doesn't even form pairing since the inner product is sesquilinear rather than bilinear.[2]

Suppose that (H, ⟨ ⋅, ⋅ ⟩) is a complex pre-Hilbert space with multiplication denoted by juxtaposition or by a dot  ⋅ . Define the map

• ⊥ • : ℂ × H → H       by       c ⊥ x  :=  c x,

where the right hand side uses the scalar multiplication of H. Let H denote the complex conjugate vector space of H, where H denotes the additive group of (H, +) (so vector addition in H is identical to vector addition in H) but with scalar multiplication in H being the map • ⊥ • rather than the scalar multiplication that H is endowed with.

The map b : H × H → ℂ defined by b (x, y) := ⟨ x, y ⟩ is linear in both coordinates[note 2] and so (H, H, ⟨ ⋅, ⋅ ⟩) forms a dual pairing.

• Suppose X = ℝ², Y = ℝ³, and ${\displaystyle b\left(\left(x_{1},y_{1}\right),\left(x_{1},y_{1},z_{1}\right)\right):=x_{1}x_{2}+y_{1}y_{2}}$ for all ${\displaystyle \left(x_{1},y_{1}\right)}$ ∈ X and ${\displaystyle \left(x_{1},y_{1},z_{1}\right)}$ ∈ Y. Then (X, Y, b) is a pairing such that X distinguishes points of Y, but Y does not distinguish points of X. Furthermore, X^⊥ := { y ∈ Y : X ⊥ y } = { (0, 0, z) : z ∈ ℝ }.
• Let 1 < p < ∞, X := L^p(μ), Y := L^q(μ) (where q is such that  1/p + 1/q = 1), and b(f, g) := ${\displaystyle \int fg\,\mathrm {d} \mu }$. Then (X, Y, b) is a dual system.
• Let X and Y be vector spaces over the same field 𝕂. Then the bilinear form ${\displaystyle b\left(x\otimes y,x^{*}\otimes y^{*}\right)=\left\langle x^{\prime },x\right\rangle \left\langle y^{\prime },y\right\rangle }$ places X × Y and ${\displaystyle X^{\#}\times Y^{\#}}$ in duality.[3]
• A sequence space X and its beta dual ${\displaystyle Y:=X^{\beta }}$ with the bilinear map defined as ${\displaystyle \langle x,y\rangle$ :=\sum _{i=1}^{\infty }x_{i}y_{i}} for x ∈ X, y ∈ ${\displaystyle X^{\beta }}$ forms a dual system.

The following theorem is of fundamental importance to duality theory because it completely characterizes the continuous dual space of (X, 𝜎(X, Y, b)).

If (X, Y, b) is a pairing then for any subset S of X:

• S^⊥ = (span S)^⊥ = (Cl_{𝜎(Y, X, b)}(span S))^⊥ = S^⊥⊥⊥ and this set is 𝜎(Y, X, b)-closed;[2]
• S ⊆ S^⊥⊥ = Cl_{𝜎(X, Y, b)}(span S);[2]
  • Thus if S is a 𝜎(X, Y, b)-closed vector subspace of X then S = S^⊥⊥.
• If (S_i)_{i ∈ I} is a family of 𝜎(X, Y, b)-closed vector subspaces of X then (∩_{i ∈ I} S_i)^⊥ = Cl_{𝜎(Y, X, b)}(span (∪_{i ∈ I} S^⊥
  _i)).[2]
• If (S_i)_{i ∈ I} is a family of subsets of X then (∪_{i ∈ I} S_i)^⊥ = ∩_{i ∈ I} S^⊥
  _i.[2]

If X is a normed space then under the canonical duality, S^⊥ is norm closed in X ' and S^⊥⊥ is norm closed in X.[2]

Subspaces

Suppose that M is a vector subspace of X and let (M, Y, b) denote the restriction of (X, Y, b) to M × Y. The weak topology 𝜎(M, Y, b) on M is identical to the subspace topology that M inherits from (X, 𝜎(X, Y, b)).

Also, (M, Y/M^⊥, b|_M) is a paired space (where Y/M^⊥ means Y/(M^⊥)) where b|_M : M × Y/M^⊥ → 𝕂 is defined by

(m, y + M^⊥) ↦ b(m, y).

The topology 𝜎(M, Y/M^⊥, b|_M) is equal to the subspace topology that M inherits from (X, 𝜎(X, Y, b)).[6] Furthermore, if (X, 𝜎(X, Y, b)) is a dual system then so is (M, Y/M^⊥, b|_M).[6]

Quotients

Suppose that M is a vector subspace of X. Then (X/M, M^⊥, b/M) is a paired space where b/M : X/M × M^⊥ → 𝕂 is defined by

(x + M, y) ↦ b(x, y).

The topology 𝜎(X/M, M^⊥) is identical to the usual quotient topology induced by (X, 𝜎(X, Y, b)) on X/M.[6]

If X is a locally convex space and if H is a subset of the continuous dual space X', then H is 𝜎(X', X)-bounded if and only if H ⊆ B° for some barrel B in X.[2]

The following results are important for defining polar topologies.

If (X, Y, b) is a pairing and A ⊆ X, then:[2]

1. The polar of A, A°, is a closed subset of (Y, 𝜎(Y, X, b)).
2. The polars of the following sets are identical: (a) A; (b) the convex hull of A; (c) the balanced hull of A; (d) the 𝜎(X, Y, b)-closure of A; (e) the 𝜎(X, Y, b)-closure of the convex balanced hull of A.
3. The bipolar theorem: The bipolar of A, A°°, is equal to the 𝜎(X, Y, b)-closure of the convex balanced hull of A.
   • The bipolar theorem in particular "is an indispensable tool in working with dualities."[5]
4. A is 𝜎(X, Y, b)-bounded if and only if A° is absorbing in Y.
5. If in addition Y distinguishes points of X then A is 𝜎(X, Y, b)-bounded if and only if it is 𝜎(X, Y, b)-totally bounded.

If (X, Y, b) is a pairing and 𝜏 is a locally convex topology on X that is consistent with duality, then a subset B of X is a barrel in (X, 𝜏) if and only if B is the polar of some 𝜎(Y, X, b)-bounded subset of Y.[7]

Let (X, Y, b) and (W, Z, c) be pairings over 𝕂 and let F : X → W be a linear map.

For all z ∈ Z, let c(F(•), z) : X → 𝕂 be the map defined by x ↦ c(F(x), z). It is said that F's transpose or adjoint is well-defined if the following conditions are satisfies:

1. X distinguishes points of Y (or equivalently, the map y ↦ b(•, y) from Y into the algebraic dual X^# is injective), and
2. c(F(•), Z) ⊆ b(•, Y), where c(F(•), Z) := { c(F(•), z) : z ∈ Z }.

In this case, for any z ∈ Z there exists (by condition 2) a unique (by condition 1) y ∈ Y such that c(F(•), z) = b(•, y), where this element of Y will be denoted by ^tF(z). This defines a linear map

^tF : Z → Y

called the transpose of adjoint of F with respect to (X, Y, b) and (W, Z, c) (this should not to be confused with the Hermitian adjoint). It is easy to see that the two conditions mentioned above (i.e. for "the transpose is well-defined") are also necessary for ^tF to be well-defined. For every z ∈ Z, the defining condition for ^tF(z) is

c(F(•), z) = b(•, ^tF(z)),

that is,

c(F(x), z) = b(x, ^tF(z))      for all x ∈ X.

By the conventions mentioned at the beginning of this article, this also defines the transpose of linear maps of the form Z → Y,[note 4] X → Z,[note 5] W → Y,[note 6] Y → W,[note 7] etc. (see footnote for details).

Properties of the transpose

Throughout, (X, Y, b) and (W, Z, c) be pairings over 𝕂 and F : X → W will be a linear map whose transpose ^tF : Z → Y is well-defined.

• ^tF : Z → Y is injective (i.e. ker ^tF = { 0 }) if and only if the range of F is dense in (W, 𝜎(W, Z, c)).[2]
• If in addition to ^tF being well-defined, the transpose of ^tF is also well-defined then ^{t t}F = F.
• Suppose (U, V, a) is a pairing over 𝕂 and E : U → X is a linear map whose transpose ^tE : Y → V is well-defined. Then the transpose of F ∘ E : U → W, which is ^t(F ∘ E) : Z → V, is well-defined and ^t(F ∘ E) = ^tE ∘ ^tF.
• If F : X → W is a vector space isomorphism then ^tF : Z → Y is bijective, the transpose of F⁻¹ : W → X, which is ^t(F⁻¹) : Y → Z, is well-defined, and ^t(F⁻¹) = (^tF)⁻¹.[2]
• Let S ⊆ X and let S° denotes the absolute polar of A, then:[2]
  1. [F(S)]° = (^tF)⁻¹(S°);
  2. if F(S) ⊆ T for some T ⊆ W, then ^tF(T°) ⊆ S°;
  3. if T ⊆ W is such that ^tF(T°) ⊆ S°, then F(S) ⊆ T°°;
  4. if T ⊆ W and S ⊆ X are weakly closed disks then ^tF(T°) ⊆ S° if and only if F(S) ⊆ T;
  5. ker ^tF = [Im F]° = [Im F]^⊥, where Im F := F(X) is the image of F.

These results hold when the real polar is used in place of the absolute polar.

If X and Y are normed spaces under their canonical dualities and if F : X → Y is a continuous linear map, then ||F|| = ||^tF||.[2]

A linear map F : X → W is weakly continuous (with respect to (X, Y, b) and (W, Z, c)) if F : (X, 𝜎(X, Y, b)) → (W, 𝜎(W, Z, c)) is continuous.

The following result shows that the existence of the transpose map is intimately tied to the weak topology.

Suppose that X is a vector space and that X^# is its the algebraic dual. Then every 𝜎(X, X^#)-bounded subset of X is contained in a finite dimensional vector subspace and every vector subspace of X is 𝜎(X, X^#)-closed.[2]

Call X 𝜎(X, Y, b)-complete or (if no ambiguity can arise) weakly-complete if (X, 𝜎(X, Y, b)) is a complete vector space. There exist Banach spaces that are not weakly-complete (despite being complete).[2]

If X is a vector space then under the canonical duality, (X^#, 𝜎(X^#, X)) is complete.[2] Conversely, if Z is a Hausdorff locally convex TVS with continuous dual space Z', then (Z, 𝜎(Z, Z')) is complete if and only if Z = (Z')^# (i.e. the map Z → (Z')^# defined by sending z ∈ Z to the evaluation map at z (i.e. z' ↦ z'(z)) is a bijection).[2]

In particular, if Y is a vector subspace of X^# such that Y separates points of X, then (Y, 𝜎(Y, X)) is complete if and only if Y = X^#.

If X distinguishes points of Y and if Z denotes the range of the injection y ↦ b(•, y) then Z is a vector subspace of the algebraic dual space of X and the pairing (X, Y, b) becomes canonically identified with the canonical pairing ⟨X, Z⟩ (where ⟨x, x'⟩ := x'(x) is the natural evaluation map). In particular, in this situation it will be assumed without loss of generality that Y is a vector subspace of X's algebraic dual and b is the evaluation map.

Convention: Often, whenever y ↦ b(•, y) is injective (especially when (X, Y, b) forms a dual pair) then it is common practice to assume without loss of generality that Y is a vector subspace of the algebraic dual space of X, that b is the natural evaluation map, and also denote Y by X'.

In a completely analogous manner, if Y distinguishes points of X then it is possible for X to be identified as a vector subspace of Y's algebraic dual space.[3]

In the spacial case where the dualities are the canonical dualities ⟨X, X^#⟩ and ⟨W, W^#⟩, the transpose of a linear map F : X → W is always well-defined. This transpose is called the algebraic adjoint of F and it will be denoted by F^#; that is, F^# = ^tF : W^# → X^#. In this case, for all w' ∈ W^#, F^#(w') := w' ∘ F[2][8] where the defining condition for F^#(w') is:

⟨x, F^#(w')⟩ = ⟨F(x), w'⟩     for all x ∈ X,

or equivalently, F^#(w')(x) = w'(F(x)) for all x ∈ X.

Examples

If X = Y = 𝕂ⁿ for some integer n, ℰ = {e₁, ..., e₁} is a basis for X with dual basis ℰ' = {e₁', ..., e₁' }, F : 𝕂ⁿ → 𝕂ⁿ is a linear operator, and the matrix representation of F with respect to ℰ is M := (f_i,j), then the transpose of M is the matrix representation with respect to ℰ' of F^#.

Suppose that ⟨X, Y⟩ and ⟨W, Z⟩ are canonical pairings (so Y ⊆ X^# and Z ⊆ W^#) that are dual systems and let F : X → W be a linear map. Then F : X → W is weakly continuous if and only if it satisfies any of the following equivalent conditions:[2]

1. F : (X, 𝜎(X, Y)) → (W, 𝜎(W, Z)) is continuous;
2. F^#(Z) ⊆ Y
3. the transpose of F, ^tF : Z → Y, with respect to ⟨X, Y⟩ and ⟨W, Z⟩ is well-defined.

If F is weakly continuous then ^tF : (Z, 𝜎(Z, W)) → (Y, 𝜎(Y, X)) will be continuous and furthermore, ^{t t}F = F.[8]

A map g : A → B between topological spaces is relatively open if g : A → Im g is an open mapping, where Im g is the range of g.[2]

Suppose that ⟨X, Y⟩ and ⟨W, Z⟩ are dual systems and F : X → W is a weakly continuous linear map. Then the following are equivalent:[2]

1. F : (X, 𝜎(X, Y)) → (W, 𝜎(W, Z)) is relatively open;
2. The range of ^tF is 𝜎(Y, X)-closed in Y;
3. Im ^tF = (ker F)^⊥

Furthermore,

• F : X → W is injective (resp. bijective) if and only if ^tF is surjective (resp. bijective);
• F : X → W is surjective if and only if ^tF : (Z, 𝜎(Z, W)) → (Y, 𝜎(Y, X)) is relatively open and injective.

The transpose of map between two TVSs is defined if and only if F is weakly continuous.

If F : X → Y is a linear map between two Hausdorff locally convex topological vector spaces then:[2]

• If F is continuous then it is weakly continuous and ^tF is both Mackey continuous and strongly continuous.
• If F is weakly continuous then it is both Mackey continuous and strongly continuous (defined below).
• If F is weakly continuous then it is continuous if and only if ^tF : Y' → X' maps equicontinuous subsets of Y' to equicontinuous subsets of X'.
• If X and Y are normed spaces then F is continuous if and only if it is weakly continuous, in which case ||F|| = ||^tF||.
• If F is continuous then F : X → Y is relatively open if and only if F is weakly relatively open (i.e. F : (X, 𝜎(X, X')) → (Y, 𝜎(Y, Y')) is relatively open) and every equicontinuous subsets of Im ^tF = ^tF(Y') is the image of some equicontinuous subsets of Y'.
• If F is continuous injection then F : X → Y is a TVS-embedding (or equivalently, a topological embedding) if and only if every equicontinuous subsets of X' is the image of some equicontinuous subsets of Y'.

Let X be a locally convex space with continuous dual space X' and let K ⊆ X'.[2]

1. If K is equicontinuous or 𝜎(X', X)-compact, and if D ⊆ X' is such that span D is dense in X, then the subspace topology that K inherits from (X', 𝜎(X', D)) is identical to the subspace topology that K inherits from (X', 𝜎(X', X)).
2. If X is separable and K is equicontinuous then K, when endowed with the subspace topology induced by (X', 𝜎(X', X)), is metrizable.
3. If X is separable and metrizable, then (X', 𝜎(X', X)) is separable.
4. If X is a normed space then X is separable if and only if the closed unit call the continuous dual space of X is metrizable when given the subspace topology induced by (X', 𝜎(X', X)).
5. If X is a normed space whose continuous dual space is separable (when given the usual norm topology), then X is separable.

The polar topology on Y determined by 𝒢 (and b) or the 𝒢-topology on Y is the unique topological vector space (TVS) topology on Y for which

${\displaystyle \left\{rG^{\circ }:G\in {\mathcal {G}},r>0\right\}}$

forms a subbasis of neighborhoods at the origin.[2] When Y is endowed with this 𝒢-topology then it is denoted by Y_𝒢. Every polar topology is necessarily locally convex.[2] When 𝒢 is a directed set with respect to subset inclusion (i.e. if for all G, H ∈ 𝒢 there exists some K ∈ 𝒢 such that G ∪ H ⊆ K) then this neighborhood subbasis at 0 actually forms a neighborhood basis at 0.[2]

The following table lists some of the more important polar topologies.

Notation: If 𝛥(Y, X, b) denotes a polar topology on Y then Y endowed with this topology will be denoted by Y_{𝛥(Y, X, b)}, Y_{𝛥(Y, X)} or simply Y_𝛥 (e.g. for 𝜎(X, Y, b) we'd have 𝛥 = 𝜎 so that Y_{σ(Y, X, b)}, Y_{σ(Y, X)} and Y_σ all denote Y with endowed with 𝜎(X, Y, b)).

${\displaystyle {\mathcal {G}}\subseteq {\mathcal {P}}\left(X\right)}$ ("topology of uniform convergence on ...")	Notation	Name ("topology of...")	Alternative name
finite subsets of X (or 𝜎(X, Y, b)-closed disked hulls of finite subsets of X)	𝜎(X, Y, b) s(X, Y, b)	pointwise/simple convergence	weak/weak* topology
𝜎(X, Y, b)-compact disks	τ(X, Y, b)		Mackey topology
𝜎(X, Y, b)-compact convex subsets	γ(X, Y, b)	compact convex convergence
𝜎(X, Y, b)-compact subsets (or balanced 𝜎(X, Y, b)-compact subsets)	c(X, Y, b)	compact convergence
𝜎(X, Y, b)-bounded subsets	b(X, Y, b) β(X, Y, b)	bounded convergence	strong topology Strongest polar topology

Continuity

A linear map F : X → W is Mackey continuous (with respect to (X, Y, b) and (W, Z, c)) if F : (X, τ(X, Y, b)) → (W, τ(W, Z, c)) is continuous.[2]

A linear map F : X → W is strongly continuous (with respect to (X, Y, b) and (W, Z, c)) if F : (X, β(X, Y, b)) → (W, β(W, Z, c)) is continuous.[2]

Bounded subsets

A subset of X is weakly bounded (resp. Mackey bounded, strongly bounded) if it is bounded in (X, 𝜎(X, Y, b)) (resp. bounded in (X, τ(X, Y, b)), bounded in (X, β(X, Y, b))).

If (X, Y, b) is a pairing over 𝕂 and 𝒯 is a vector topology on X then 𝒯 is a topology of the pairing and that it is compatible (or consistent) with the pairing (X, Y, b) if it is locally convex and if the continuous dual space of (X, 𝒯) = b(•, Y).[note 8] If X distinguishes points of Y then by identifying Y as a vector subspace of X's algebraic dual, the defining condition becomes: (X, 𝒯)' = Y.[2] Some authors (e.g. [Trèves 2006] and [Schaefer 1999]) require that a topology of a pair also be Hausdorff,[3][9] which it would have to be if Y distinguishes the points of X (which these authors assume).

The weak topology 𝜎(X, Y, b) is compatible with the pairing (X, Y, b) (as was shown in the Weak representation theorem) and it is in fact the weakest such topology. There is a strongest topology compatible with this pairing and that is the Mackey topology. If N is a normed space that is not reflexive then the usual norm topology on its continuous dual space is not compatible with the duality (N', N).[2]

The following is one of the most important theorems in duality theory.

It follows that the Mackey topology 𝜏(X, Y, b), which recall is the polar topology generated by all 𝜎(X, Y, b)-compact disks in Y, is the strongest locally convex topology on X that is compatible with the pairing (X, Y, b). A locally convex space whose given topology is identical to the Mackey topology is called a Mackey space. The following consequence of the above Mackey-Arens theorem is also called the Mackey-Arens theorem.

If X is a TVS (over ℝ or ℂ) then a half-space is a set of the form { x ∈ X : f(x) ≤ r} for some real r and some continuous real linear functional f on X.

The above theorem implies that the closed and convex subsets of a locally convex space depend entirely on the continuous dual space. Consequently, the closed and convex subsets are the same in any topology compatible with duality; that is, if 𝒯 and ℒ are any locally convex topologies on X with the same continuous dual spaces, then a convex subset of X is closed in the 𝒯 topology if and only if it is closed in the ℒ topology. This implies that the 𝒯-closure of any convex subset of X is equal to its ℒ-closure and that for any 𝒯-closed disk A in X, A = A°°.[2] In particular, if B is a subset of X then B is a barrel in (X, 𝒯) if and only if it is a barrel in (X, ℒ).[2]

The following theorem shows that barrels (i.e. closed absorbing disks) are exactly the polars of weakly bounded subsets.

If X is a topological vector space then:[2][11]

1. A closed absorbing and balanced subset B of X absorbs each convex compact subset of X (i.e. there exists a real r > 0 such that rB contains that set).
2. If X is Hausdorff and locally convex then every barrel in X absorbs every convex bounded complete subset of X.

All of this leads to Mackey's theorem, which is one of the central theorems in the theory of dual systems. In short, it states the bounded subsets are the same for any two Hausdorff locally convex topologies that are compatible with the same duality.

Space of finite sequences

Let X denote the space of all sequences of scalars r_• = (r_i)^∞
_i=1 such that r_i = 0 for all sufficiently large i. Let Y = X and define a bilinear map b : X × X → 𝕂 by

b(r_•, s_•) := ∑ r_is_i.

Then 𝜎(X, X, b) = τ(X, X, b).[2] Moreover, a subset T ⊆ X is 𝜎(X, X, b)-bounded (resp. β(X, X, b)-bounded) if and only if there exists a sequence m_• = (m_i)^∞
_i=1 of positive real numbers such that |t_i| ≤ m_i for all t_• = (t_i)^∞
_i=1 ∈ T and all indices i (resp. and m_• ∈ X).[2] It follows that there are weakly bounded (i.e. 𝜎(X, X, b)-bounded) subsets of X that are not strongly bounded (i.e. not β(X, X, b)-bounded).