Integral linear operator

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Short description: Mathematical function

An integral bilinear form is a bilinear functional that belongs to the continuous dual space of [math]\displaystyle{ X \widehat{\otimes}_{\epsilon} Y }[/math], the injective tensor product of the locally convex topological vector spaces (TVSs) X and Y. An integral linear operator is a continuous linear operator that arises in a canonical way from an integral bilinear form.

These maps play an important role in the theory of nuclear spaces and nuclear maps.

Definition - Integral forms as the dual of the injective tensor product

Let X and Y be locally convex TVSs, let [math]\displaystyle{ X \otimes_{\pi} Y }[/math] denote the projective tensor product, [math]\displaystyle{ X \widehat{\otimes}_{\pi} Y }[/math] denote its completion, let [math]\displaystyle{ X \otimes_{\epsilon} Y }[/math] denote the injective tensor product, and [math]\displaystyle{ X \widehat{\otimes}_{\epsilon} Y }[/math] denote its completion. Suppose that [math]\displaystyle{ \operatorname{In} : X \otimes_{\epsilon} Y \to X \widehat{\otimes}_{\epsilon} Y }[/math] denotes the TVS-embedding of [math]\displaystyle{ X \otimes_{\epsilon} Y }[/math] into its completion and let [math]\displaystyle{ {}^{t}\operatorname{In} : \left( X \widehat{\otimes}_{\epsilon} Y \right)^{\prime}_b \to \left( X \otimes_{\epsilon} Y \right)^{\prime}_b }[/math] be its transpose, which is a vector space-isomorphism. This identifies the continuous dual space of [math]\displaystyle{ X \otimes_{\epsilon} Y }[/math] as being identical to the continuous dual space of [math]\displaystyle{ X \widehat{\otimes}_{\epsilon} Y }[/math].

Let [math]\displaystyle{ \operatorname{Id} : X \otimes_{\pi} Y \to X \otimes_{\epsilon} Y }[/math] denote the identity map and [math]\displaystyle{ {}^{t}\operatorname{Id} : \left( X \otimes_{\epsilon} Y \right)^{\prime}_b \to \left( X \otimes_{\pi} Y \right)^{\prime}_b }[/math] denote its transpose, which is a continuous injection. Recall that [math]\displaystyle{ \left( X \otimes_{\pi} Y \right)^{\prime} }[/math] is canonically identified with [math]\displaystyle{ B(X, Y) }[/math], the space of continuous bilinear maps on [math]\displaystyle{ X \times Y }[/math]. In this way, the continuous dual space of [math]\displaystyle{ X \otimes_{\epsilon} Y }[/math] can be canonically identified as a vector subspace of [math]\displaystyle{ B(X, Y) }[/math], denoted by [math]\displaystyle{ J(X, Y) }[/math]. The elements of [math]\displaystyle{ J(X, Y) }[/math] are called integral (bilinear) forms on [math]\displaystyle{ X \times Y }[/math]. The following theorem justifies the word integral.

Theorem[1][2] — The dual J(X, Y) of [math]\displaystyle{ X \widehat{\otimes}_{\epsilon} Y }[/math] consists of exactly those continuous bilinear forms c on [math]\displaystyle{ X \times Y }[/math] that can be represented in the form of a map

[math]\displaystyle{ b \in B(X, Y) \mapsto v(b) = \int_{S \times T} b\big\vert_{S \times T} \left( x', y' \right) \mathrm{d} \mu\!\left( x', y' \right) }[/math]

where S and T are some closed, equicontinuous subsets of [math]\displaystyle{ X^{\prime}_{\sigma} }[/math] and [math]\displaystyle{ Y^{\prime}_{\sigma} }[/math], respectively, and [math]\displaystyle{ \mu }[/math] is a positive Radon measure on the compact set [math]\displaystyle{ S \times T }[/math] with total mass [math]\displaystyle{ \leq 1. }[/math] Furthermore, if A is an equicontinuous subset of J(X, Y) then the elements [math]\displaystyle{ v \in A }[/math] can be represented with [math]\displaystyle{ S \times T }[/math] fixed and [math]\displaystyle{ \mu }[/math] running through a norm bounded subset of the space of Radon measures on [math]\displaystyle{ S \times T. }[/math]

Integral linear maps

A continuous linear map [math]\displaystyle{ \kappa : X \to Y' }[/math] is called integral if its associated bilinear form is an integral bilinear form, where this form is defined by [math]\displaystyle{ (x, y) \in X \times Y \mapsto (\kappa x)(y) }[/math].[3] It follows that an integral map [math]\displaystyle{ \kappa : X \to Y' }[/math] is of the form:[3]

[math]\displaystyle{ x \in X \mapsto \kappa(x) = \int_{S \times T} \left\langle x', x \right\rangle y' \mathrm{d} \mu\! \left( x', y' \right) }[/math]

for suitable weakly closed and equicontinuous subsets S and T of [math]\displaystyle{ X' }[/math] and [math]\displaystyle{ Y' }[/math], respectively, and some positive Radon measure [math]\displaystyle{ \mu }[/math] of total mass ≤ 1. The above integral is the weak integral, so the equality holds if and only if for every [math]\displaystyle{ y \in Y }[/math], [math]\displaystyle{ \left\langle \kappa(x), y \right\rangle = \int_{S \times T} \left\langle x', x \right\rangle \left\langle y', y \right\rangle \mathrm{d} \mu\! \left( x', y' \right) }[/math].

Given a linear map [math]\displaystyle{ \Lambda : X \to Y }[/math], one can define a canonical bilinear form [math]\displaystyle{ B_{\Lambda} \in Bi\left(X, Y' \right) }[/math], called the associated bilinear form on [math]\displaystyle{ X \times Y' }[/math], by [math]\displaystyle{ B_{\Lambda}\left( x, y' \right) := \left( y' \circ \Lambda \right) \left( x \right) }[/math]. A continuous map [math]\displaystyle{ \Lambda : X \to Y }[/math] is called integral if its associated bilinear form is an integral bilinear form.[4] An integral map [math]\displaystyle{ \Lambda: X \to Y }[/math] is of the form, for every [math]\displaystyle{ x \in X }[/math] and [math]\displaystyle{ y' \in Y' }[/math]:

[math]\displaystyle{ \left\langle y', \Lambda(x) \right\rangle = \int_{A' \times B''} \left\langle x', x \right\rangle \left\langle y'', y' \right\rangle \mathrm{d} \mu\! \left( x', y'' \right) }[/math]

for suitable weakly closed and equicontinuous aubsets [math]\displaystyle{ A' }[/math] and [math]\displaystyle{ B'' }[/math] of [math]\displaystyle{ X' }[/math] and [math]\displaystyle{ Y'' }[/math], respectively, and some positive Radon measure [math]\displaystyle{ \mu }[/math] of total mass [math]\displaystyle{ \leq 1 }[/math].

Relation to Hilbert spaces

The following result shows that integral maps "factor through" Hilbert spaces.

Proposition:[5] Suppose that [math]\displaystyle{ u : X \to Y }[/math] is an integral map between locally convex TVS with Y Hausdorff and complete. There exists a Hilbert space H and two continuous linear mappings [math]\displaystyle{ \alpha : X \to H }[/math] and [math]\displaystyle{ \beta : H \to Y }[/math] such that [math]\displaystyle{ u = \beta \circ \alpha }[/math].

Furthermore, every integral operator between two Hilbert spaces is nuclear.[5] Thus a continuous linear operator between two Hilbert spaces is nuclear if and only if it is integral.

Sufficient conditions

Every nuclear map is integral.[4] An important partial converse is that every integral operator between two Hilbert spaces is nuclear.[5]

Suppose that A, B, C, and D are Hausdorff locally convex TVSs and that [math]\displaystyle{ \alpha : A \to B }[/math], [math]\displaystyle{ \beta : B \to C }[/math], and [math]\displaystyle{ \gamma: C \to D }[/math] are all continuous linear operators. If [math]\displaystyle{ \beta : B \to C }[/math] is an integral operator then so is the composition [math]\displaystyle{ \gamma \circ \beta \circ \alpha : A \to D }[/math].[5]

If [math]\displaystyle{ u : X \to Y }[/math] is a continuous linear operator between two normed space then [math]\displaystyle{ u : X \to Y }[/math] is integral if and only if [math]\displaystyle{ {}^{t}u : Y' \to X' }[/math] is integral.[6]

Suppose that [math]\displaystyle{ u : X \to Y }[/math] is a continuous linear map between locally convex TVSs. If [math]\displaystyle{ u : X \to Y }[/math] is integral then so is its transpose [math]\displaystyle{ {}^{t}u : Y^{\prime}_b \to X^{\prime}_b }[/math].[4] Now suppose that the transpose [math]\displaystyle{ {}^{t}u : Y^{\prime}_b \to X^{\prime}_b }[/math] of the continuous linear map [math]\displaystyle{ u : X \to Y }[/math] is integral. Then [math]\displaystyle{ u : X \to Y }[/math] is integral if the canonical injections [math]\displaystyle{ \operatorname{In}_X : X \to X'' }[/math] (defined by [math]\displaystyle{ x \mapsto }[/math] value at x) and [math]\displaystyle{ \operatorname{In}_Y : Y \to Y'' }[/math] are TVS-embeddings (which happens if, for instance, [math]\displaystyle{ X }[/math] and [math]\displaystyle{ Y^{\prime}_b }[/math] are barreled or metrizable).[4]

Properties

Suppose that A, B, C, and D are Hausdorff locally convex TVSs with B and D complete. If [math]\displaystyle{ \alpha : A \to B }[/math], [math]\displaystyle{ \beta : B \to C }[/math], and [math]\displaystyle{ \gamma: C \to D }[/math] are all integral linear maps then their composition [math]\displaystyle{ \gamma \circ \beta \circ \alpha : A \to D }[/math] is nuclear.[5] Thus, in particular, if X is an infinite-dimensional Fréchet space then a continuous linear surjection [math]\displaystyle{ u : X \to X }[/math] cannot be an integral operator.

See also

References

  1. Schaefer & Wolff 1999, p. 168.
  2. Trèves 2006, pp. 500-502.
  3. 3.0 3.1 Schaefer & Wolff 1999, p. 169.
  4. 4.0 4.1 4.2 4.3 Trèves 2006, pp. 502-505.
  5. 5.0 5.1 5.2 5.3 5.4 Trèves 2006, pp. 506-508.
  6. Trèves 2006, pp. 505.

Bibliography

External links