Dual system

Given a pairing (X, Y, b) we can define a new pairing (Y, X, ${\displaystyle {\hat {b}}}$) where ${\displaystyle {\hat {b}}}$(y, x) := b(x, y).[3]

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, ${\displaystyle {\hat {b}}}$).

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, ${\displaystyle {\hat {b}}}$). If the definition depends on the order of X and Y (e.g. the definition of "the weak topology 𝜎(X, Y) defined on X by Y") then if we switch the order of X and Y then we mean that definition applied to (Y, X, ${\displaystyle {\hat {b}}}$) (e.g. this gives us the definition of "the weak topology 𝜎(Y, X) defined on Y by X").

For instance, if we define "X distinguishes points of Y" (resp, "S is a total subset of Y") as above, then we immediately obtain the dual definition of "Y distinguishes points of X" (resp, "S is a total subset of X"). Once we define, for instance, 𝜎(X, Y) then we will automatically assume that 𝜎(Y, X) has been defined without mentioning the analogous definition. The same applies to many theorems.

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

Although it is technically incorrect and an abuse of notation, we will also adhere to the following.

Convention: This article will use the common practice of treating a pairing (X, Y, b) interchangeably with (Y, X, ${\displaystyle {\hat {b}}}$) and also denoting (Y, X, ${\displaystyle {\hat {b}}}$) by (Y, X, b).

Suppose that (X, Y, b) is a pairing. 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 ${\displaystyle B^{\circ }}$ 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 ${\displaystyle {}^{\circ }B}$.

The polar ${\displaystyle B^{\circ }}$ is necessarily a convex set containing 0 ∈ Y where if B is balanced then so is ${\displaystyle B^{\circ }}$ and if B is a vector subspace of X then so too is ${\displaystyle B^{\circ }}$ a vector subspace of Y.[5]

If A ⊆ X then the bipolar of A, denoted by A^∘∘, is A^∘∘ := ${\displaystyle {}^{\circ }}$(A^⊥). Similarly, if B ⊆ Y then the bipolar of B is B^∘∘ := ${\displaystyle \left({}^{\circ }B\right)^{\circ }}$.

Competing definitions

Some authors define "polar sets" differently. For instance, some authors do not include absolute values around b(x, y) and/or use the real part of b (i.e. Re b(x, y)) in place of b(x, y). We now briefly discuss how these various definitions relate to one another.

If b is real-valued and A is symmetric (i.e. A = -A) then ${\displaystyle A^{\circ }=\left\{y\in Y:\sup _{x\in A}b(x,y)\leq 1\right\}}$, where some authors use the right hand side to define the polar set. If b is complex-valued and ${\displaystyle e^{ir}}$A ⊆ A for all real r then ${\displaystyle A^{\circ }=\left\{y\in Y:\sup _{x\in A}\operatorname {Re} b(x,y)\leq 1\right\}=\left\{y\in Y:\sup _{x\in A}\left|\operatorname {Re} b(x,y)\right|\leq 1\right\}}$,[3] where some authors use one of the latter two sets on the right hand side to define the polar set. Thus for all the common definitions of the polar set A^∘ to agree, it suffices that sA ⊆ A for all scalars s of unit length (i.e. all scalars s such that |s| = 1). In particular, all definitions of the polar of A agree when A is a balanced set.

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 ${\displaystyle \left(M,N,b{\big \vert }_{M\times N}\right)}$. Note that 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: We 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 ${\displaystyle X^{\#}}$ denote the algebraic dual space of X (i.e. the space of all linear functionals on X). There is a canonical duality ${\displaystyle \left(X,X^{\#},c\right)}$ where ${\displaystyle c(x,x^{\prime })=\langle x,x^{\prime }\rangle =x^{\prime }(x)}$ is evaluation, which is called the natural or canonical bilinear functional on X × ${\displaystyle X^{\#}}$. This duality is often denoted by ${\displaystyle \left\langle x,x^{\prime }\right\rangle$ :=x^{\prime }\left(x\right)} and it is the canonical duality.

If N is a vector subspace of ${\displaystyle X^{\#}}$ then X distinguishes points of N so (X, M, c) is a duality if and only if N distinguishes points of X, or equivalently if M is total (i.e. n(x) = 0 for all n ∈ N implies x = 0).[3]

Suppose X is a topological vector space (TVS) with continuous dual space ${\displaystyle X^{\prime }}$. Then the restriction of ${\displaystyle \left(X,X^{\#},c\right)}$ to X × ${\displaystyle X^{\prime }}$ defines a pairing ${\displaystyle \left(X,X^{\prime },c{\big \vert }_{X\times X^{\prime }}\right)}$ for which X separates points of ${\displaystyle X^{\prime }}$. If ${\displaystyle X^{\prime }}$ separates points of X (which is true if, for instance, X is a Hausdorff locally convex space) then this pairing forms a duality.[1]

• If (H, ${\displaystyle \left\langle \cdot ,\cdot \right\rangle }$) is a complex Hilbert space then ${\displaystyle \left(H,H,\left\langle \cdot ,\cdot \right\rangle \right)}$ forms a dual system if and only if dim H = 0. If H is non-trivial then ${\displaystyle \left(H,H,\left\langle \cdot ,\cdot \right\rangle \right)}$ doesn't even form pairing since the inner product is sesquilinear rather than bilinear.[3]
• If (H, ${\displaystyle \left\langle \cdot ,\cdot \right\rangle }$) is a real Hilbert space then ${\displaystyle \left(H,H,\left\langle \cdot ,\cdot \right\rangle \right)}$ forms a dual system.
• 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.[1]
• 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.

Theorem (Weak representation theorem):[3] Let (X, Y, b) be a pairing over the field 𝕂. Then the continuous dual space of (X, 𝜎(X, Y, b)) is ${\displaystyle \left\{b\left(\bullet ,y\right):y\in Y\right\}}$, where b( •, y) : X → 𝕂 is defined by x ↦ b(x, y). Furthermore,

1. If f is a continuous linear functional on (X, 𝜎(X, Y, b)) then there exists some y ∈ Y such that f = b( •, y); y is unique if and only if X distinguishes points of Y.
2. The continuous dual space of (X, 𝜎(X, Y, b)) may be identified with Y/X^⊥, where X^⊥ := { y ∈ Y : b(x, y) = 0 for all x ∈ X }.
   • Note that this is true regardless of whether or not X distinguishes points of Y or Y distinguishes points of X.

The following results are important for defining polar topologies. The bipolar theorem in particular "is an indispensable tool in working with dualities."[5]

Theorem: Let (X, Y, b) be a pairing and A ⊆ X. Then

1. The polar of A, A^∘, is a closed subset of (Y, 𝜎(Y, X, b)).[3]
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]
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.[3]
4. A is 𝜎(X, Y, b)-bounded if and only if A^∘ is absorbing in Y.[3]
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.[3]

Given a dual system ${\displaystyle \langle X,Y\rangle }$, since Y is identified as a subspace of the algebraic dual of X, we can define on X the weak topology induced by Y, denoted by 𝜎(X, Y), as being the weakest TVS topology making all of the linear functionals in Y continuous. X endowed with this topology is denoted by X_{𝜎(X, Y)} or X_𝜎. If ℬ is a Hamel basis of Y then the topology 𝜎(X, Y) is generated by the seminorms ${\displaystyle x\mapsto |\langle x,b\rangle |}$ as b ranges over ℬ.[5] The continuous dual space of X_𝜎 is Y, meaning that if f is a linear form on X then f is continuous as a map on X_𝜎 if and only if there exists some y ∈ Y such that ${\displaystyle f(x)=\langle x,y\rangle }$ for every x ∈ X.[5] If X is an infinite dimensional TVS then every weakly open neighborhood of the origin contains an infinite-dimensional vector subspace;[3] indeed, this vector subspace even has finite codimension in X.

Similarly, since X is identified as a subspace of the algebraic dual of Y, we can define on Y the weak topology induced by X, denoted by 𝜎(Y, X), as being the weakest TVS topology making all of the linear functionals in X continuous. Y endowed with this topology is denoted by Y_𝜎. The continuous dual space of Y_𝜎 is X, meaning that if f is a linear form on Y then f is continuous as a map on Y_𝜎 if and only if there exists some x ∈ X such that ${\displaystyle f(y)=\langle x,y\rangle }$ for every y ∈ Y.[5] If X is a TVS and if Y is the continuous dual space of a X, then the weak topology 𝜎(Y, X) is called the weak-* topology.

If Y₁ ⊆ Y then the topology 𝜎(X, Y₁) is coarser than 𝜎(X, Y) and furthermore, 𝜎(X, Y₁) is strictly coarser than 𝜎(X, Y) if and only if Y₁ ≠ Y.[5] In addition, the canonical bilinear form of ${\displaystyle \langle X,Y\rangle }$ places X and Y₁ in duality if and only if Y₁ is 𝜎(Y, X)-dense in Y.[5] So for instance, if X is a vector space and Y₁ is a subspace of the algebraic dual ${\displaystyle X^{\#}}$ then ${\displaystyle \left\langle X,X^{\#}\right\rangle }$ induces a duality between X and Y₁ if and only if Y₁ is ${\displaystyle \sigma \left(X^{\#},X\right)}$-dense in ${\displaystyle X^{\#}}$.[5]

Note that if X is a TVS, then since X always distinguishes points on ${\displaystyle X^{\prime }}$, ${\displaystyle \sigma \left(X^{\prime },X\right)}$ is always Hausdorff.[3]

If X is any locally convex TVS, then the family of all barrels in X and the family of all subsets of ${\displaystyle X^{\prime }}$ that are convex, balanced, closed, and bounded in ${\displaystyle X_{\sigma }^{\prime }}$, correspond to each other by polarity (with respect to ${\displaystyle \langle X,X^{\prime }\rangle }$).[5] It follows that a locally convex TVS X is barreled if and only if each bounded subset of ${\displaystyle X_{\sigma }^{\prime }}$ is equicontinuous.[5]

Theorem Suppose that X is a separable TVS. Then every closed equicontinuous subset of ${\displaystyle X_{\sigma }^{\prime }}$ is a compact metrizable space (under the subspace topology). If in addition X is metrizable then ${\displaystyle X_{\sigma }^{\prime }}$ is separable.[5]

Here are some corollaries to the bipolar theorem:

• For any subset S of X, ${\displaystyle S^{\circ \circ \circ }\subseteq S^{\circ }}$.[5]
• If ${\displaystyle \left(S_{\alpha }\right)_{\alpha \in A}}$ is a family of 𝜎(X, Y)-closed subsets of X containing 0 ∈ X, then the polar of ${\displaystyle \cap _{\alpha \in A}S_{\alpha }}$ is the closed convex hull of ${\displaystyle \cup _{\alpha \in A}\left(S_{\alpha }^{\circ }\right)}$.[5]
• If X is a locally convex TVS then the polars (taken with respect to ${\displaystyle \left\langle X,X^{\prime }\right\rangle }$) of any 0-neighborhood base forms a fundamental family of equicontinuous subsets of ${\displaystyle X^{\prime }}$ (i.e. given any bounded subset ${\displaystyle B^{\prime }}$ of ${\displaystyle X_{\sigma }^{\prime }}$, there exists a neighborhood S of 0 in X such that ${\displaystyle B^{\prime }\subseteq S^{\circ }}$).[5]
  • Conversely, if X is a locally convex TVS then the polars (taken with respect to ${\displaystyle \langle X,X^{\prime }\rangle }$) of any fundamental family of equicontinuous subsets of ${\displaystyle X^{\prime }}$ form a neighborhood base of the origin in X.[5]
• Let X be a TVS with a topology 𝜏. Then 𝜏 is a locally convex TVS topology if and only if 𝜏 is the topology of uniform convergence on the equicontinuous subsets of ${\displaystyle X^{\prime }}$.[5] The last two results explain why equicontinuous subsets of the continuous dual space play such a prominent role in the modern theory of functional analysis: because equicontinuous subsets encapsulate all information about the locally convex space X's original topology.