An integral bilinear form is a bilinear functional that belongs to the continuous dual space of , 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 denote the projective tensor product, denote its completion, let denote the injective tensor product, and denote its completion. Suppose that denotes the TVS-embedding of into its completion and let be its transpose, which is a vector space-isomorphism. This identifies the continuous dual space of as being identical to the continuous dual space of .

Let denote the identity map and denote its transpose, which is a continuous injection. Recall that is canonically identified with , the space of continuous bilinear maps on . In this way, the continuous dual space of can be canonically identified as a vector subspace of , denoted by . The elements of are called integral (bilinear) forms on . The following theorem justifies the word integral.

Theorem[1][2]  The dual J(X, Y) of consists of exactly those continuous bilinear forms c on that can be represented in the form of a map

where S and T are some closed, equicontinuous subsets of and , respectively, and is a positive Radon measure on the compact set with total mass Furthermore, if A is an equicontinuous subset of J(X, Y) then the elements can be represented with fixed and running through a norm bounded subset of the space of Radon measures on

Integral linear maps

A continuous linear map is called integral if its associated bilinear form is an integral bilinear form, where this form is defined by .[3] It follows that an integral map is of the form:[3]

for suitable weakly closed and equicontinuous subsets S and T of and , respectively, and some positive Radon measure of total mass ≤ 1. The above integral is the weak integral, so the equality holds if and only if for every , .

Given a linear map , one can define a canonical bilinear form , called the associated bilinear form on , by . A continuous map is called integral if its associated bilinear form is an integral bilinear form.[4] An integral map is of the form, for every and :

for suitable weakly closed and equicontinuous aubsets and of and , respectively, and some positive Radon measure of total mass .

Relation to Hilbert spaces

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

Proposition:[5] Suppose that is an integral map between locally convex TVS with Y Hausdorff and complete. There exists a Hilbert space H and two continuous linear mappings and such that .

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 , , and are all continuous linear operators. If is an integral operator then so is the composition .[5]

If is a continuous linear operator between two normed space then is integral if and only if is integral.[6]

Suppose that is a continuous linear map between locally convex TVSs. If is integral then so is its transpose .[4] Now suppose that the transpose of the continuous linear map is integral. Then is integral if the canonical injections (defined by value at x) and are TVS-embeddings (which happens if, for instance, and are barreled or metrizable).[4]

Properties

Suppose that A, B, C, and D are Hausdorff locally convex TVSs with B and D complete. If , , and are all integral linear maps then their composition is nuclear.[5] Thus, in particular, if X is an infinite-dimensional Fréchet space then a continuous linear surjection cannot be an integral operator.

See also

References

Bibliography

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