Integral linear operator
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 subvector space 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 v 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
Suppose that is a continuous linear map between locally convex TVSs.
- Every nuclear map is integral.[4]
- An important partial converse is that every integral operator between two Hilbert spaces is nuclear.[5]
- If is integral then so is its transpose .[4]
- 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, X and are barreled or metrizable).[4]
- If X and Y are normable spaces then is integral if and only if is integral.[6]
- 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]
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
- Schaefer 1999, p. 168.
- Treves 2006, pp. 500-502.
- Schaefer 1999, p. 169.
- Treves 2006, pp. 502-505.
- Treves 2006, pp. 506-508.
- Treves 2006, pp. 505.
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