Integral linear operator

{{Short description|Mathematical function}}

In mathematical analysis, an integral linear operator is a linear operator T given by integration; i.e.,

:(Tf)(x) = \int f(y) K(x, y) \, dy

where K(x, y) is called an integration kernel.

More generally, an integral bilinear form is a bilinear functional that belongs to the continuous dual space of X \widehat{\otimes}_{\epsilon} Y, 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

{{See also|Injective tensor product|Projective tensor product}}

Let X and Y be locally convex TVSs, let X \otimes_{\pi} Y denote the projective tensor product, X \widehat{\otimes}_{\pi} Y denote its completion, let X \otimes_{\epsilon} Y denote the injective tensor product, and X \widehat{\otimes}_{\epsilon} Y denote its completion.

Suppose that \operatorname{In} : X \otimes_{\epsilon} Y \to X \widehat{\otimes}_{\epsilon} Y denotes the TVS-embedding of X \otimes_{\epsilon} Y into its completion and let {}^{t}\operatorname{In} : \left( X \widehat{\otimes}_{\epsilon} Y \right)^{\prime}_b \to \left( X \otimes_{\epsilon} Y \right)^{\prime}_b be its transpose, which is a vector space-isomorphism. This identifies the continuous dual space of X \otimes_{\epsilon} Y as being identical to the continuous dual space of X \widehat{\otimes}_{\epsilon} Y.

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

{{math theorem | name = Theorem{{sfn | Schaefer|Wolff| 1999 | p=168}}{{sfn | Trèves | 2006 | pp=500-502}} | math_statement =

The dual {{math|J(X, Y)}} of X \widehat{\otimes}_{\epsilon} Y consists of exactly of the continuous bilinear forms {{mvar|u}} on X \times Y of the form

: u(x,y) = \int_{S \times T} \langle x, x'\rangle \langle y, y' \rangle\; d \mu\!\left( x', y' \right),

where {{mvar|S}} and {{mvar|T}} are respectively some weakly closed and equicontinuous (hence weakly compact) subsets of the duals X^{\prime} and Y^{\prime}, and \mu is a (necessarily bounded) positive Radon measure on the (compact) set S \times T.

}}

There is also a closely related formulation {{sfn | Grothendieck | 1955 | pp=124-126}} of the theorem above that can also be used to explain the terminology integral bilinear form: a continuous bilinear form u on the product X\times Y of locally convex spaces is integral if and only if there is a compact topological space \Omega equipped with a (necessarily bounded) positive Radon measure \mu and continuous linear maps \alpha and \beta from X and Y to the Banach space L^{\infty}(\Omega,\mu) such that

:u(x,y) = \langle\alpha(x),\beta(y)\rangle = \int_{\Omega}\alpha(x)\beta(y)\;d\mu,

i.e., the form u can be realised by integrating (essentially bounded) functions on a compact space.

Integral linear maps

A continuous linear map \kappa : X \to Y' is called integral if its associated bilinear form is an integral bilinear form, where this form is defined by (x, y) \in X \times Y \mapsto (\kappa x)(y).{{sfn | Schaefer|Wolff| 1999 | p=169}} It follows that an integral map \kappa : X \to Y' is of the form:{{sfn | Schaefer|Wolff| 1999 | p=169}}

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

for suitable weakly closed and equicontinuous subsets S and T of X' and Y', respectively, and some positive Radon measure \mu of total mass ≤ 1.

The above integral is the weak integral, so the equality holds if and only if for every y \in Y, \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).

Given a linear map \Lambda : X \to Y, one can define a canonical bilinear form B_{\Lambda} \in Bi\left(X, Y' \right), called the associated bilinear form on X \times Y', by B_{\Lambda}\left( x, y' \right) := \left( y' \circ \Lambda \right) \left( x \right).

A continuous map \Lambda : X \to Y is called integral if its associated bilinear form is an integral bilinear form.{{sfn | Trèves | 2006 | pp=502-505}} An integral map \Lambda: X \to Y is of the form, for every x \in X and y' \in Y':

: \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)

for suitable weakly closed and equicontinuous aubsets A' and B of X' and Y, respectively, and some positive Radon measure \mu of total mass \leq 1.

= Relation to Hilbert spaces =

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

Proposition:{{sfn | Trèves | 2006 | pp=506-508}} Suppose that u : X \to Y is an integral map between locally convex TVS with Y Hausdorff and complete. There exists a Hilbert space H and two continuous linear mappings \alpha : X \to H and \beta : H \to Y such that u = \beta \circ \alpha.

Furthermore, every integral operator between two Hilbert spaces is nuclear.{{sfn | Trèves | 2006 | pp=506-508}} 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.{{sfn | Trèves | 2006 | pp=502-505}} An important partial converse is that every integral operator between two Hilbert spaces is nuclear.{{sfn | Trèves | 2006 | pp=506-508}}

Suppose that A, B, C, and D are Hausdorff locally convex TVSs and that \alpha : A \to B, \beta : B \to C, and \gamma: C \to D are all continuous linear operators. If \beta : B \to C is an integral operator then so is the composition \gamma \circ \beta \circ \alpha : A \to D.{{sfn | Trèves | 2006 | pp=506-508}}

If u : X \to Y is a continuous linear operator between two normed space then u : X \to Y is integral if and only if {}^{t}u : Y' \to X' is integral.{{sfn | Trèves | 2006 | pp=505}}

Suppose that u : X \to Y is a continuous linear map between locally convex TVSs.

If u : X \to Y is integral then so is its transpose {}^{t}u : Y^{\prime}_b \to X^{\prime}_b.{{sfn | Trèves | 2006 | pp=502-505}} Now suppose that the transpose {}^{t}u : Y^{\prime}_b \to X^{\prime}_b of the continuous linear map u : X \to Y is integral. Then u : X \to Y is integral if the canonical injections \operatorname{In}_X : X \to X (defined by x \mapsto value at {{mvar|x}}) and \operatorname{In}_Y : Y \to Y are TVS-embeddings (which happens if, for instance, X and Y^{\prime}_b are barreled or metrizable).{{sfn | Trèves | 2006 | pp=502-505}}

= Properties =

Suppose that A, B, C, and D are Hausdorff locally convex TVSs with B and D complete. If \alpha : A \to B, \beta : B \to C, and \gamma: C \to D are all integral linear maps then their composition \gamma \circ \beta \circ \alpha : A \to D is nuclear.{{sfn | Trèves | 2006 | pp=506-508}}

Thus, in particular, if {{mvar|X}} is an infinite-dimensional Fréchet space then a continuous linear surjection u : X \to X cannot be an integral operator.

See also

References

{{Reflist}}

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