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Integral linear operator

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

( T f ) ( x ) = ∫ ∫ f ( y ) K ( x , y ) d y {\displaystyle (Tf)(x)=\int f(y)K(x,y)\,dy}

where K ( x , y ) {\displaystyle 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 ⊗ ⊗ ^ ^ ϵ ϵ Y {\displaystyle 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.

Contents

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

• Integral linear maps

• Relation to Hilbert spaces

• Sufficient conditions

• Properties

• See also

• References

• Bibliography

• External links

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

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

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

Theoremcite-ref-footnoteschaeferwolff1999168-1-0[1]cite-ref-footnotetr-ves2006500-502-2-0[2]—The dual J(X, Y) of X ⊗ ⊗ ^ ^ ϵ ϵ Y {\displaystyle X{\widehat {\otimes }}_{\epsilon }Y} consists of exactly of the continuous bilinear forms u on X × × Y {\displaystyle X\times Y} of the form

u ( x , y ) = ∫ ∫ S × × T ⟨ ⟨ x , x ′ ⟩ ⟩ ⟨ ⟨ y , y ′ ⟩ ⟩ d μ μ ( x ′ , y ′ ) , {\displaystyle u(x,y)=\int _{S\times T}\langle x,x'\rangle \langle y,y'\rangle \;d\mu \!\left(x',y'\right),}

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

There is also a closely related formulation cite-ref-footnotegrothendieck1955124-126-3-0[3] of the theorem above that can also be used to explain the terminology integral bilinear form: a continuous bilinear form u {\displaystyle u} on the product X × × Y {\displaystyle X\times Y} of locally convex spaces is integral if and only if there is a compact topological space Ω Ω {\displaystyle \Omega } equipped with a (necessarily bounded) positive Radon measure μ μ {\displaystyle \mu } and continuous linear maps α α {\displaystyle \alpha } and β β {\displaystyle \beta } from X {\displaystyle X} and Y {\displaystyle Y} to the Banach space L ∞ ∞ ( Ω Ω , μ μ ) {\displaystyle L^{\infty }(\Omega ,\mu )} such that

u ( x , y ) = ⟨ ⟨ α α ( x ) , β β ( y ) ⟩ ⟩ = ∫ ∫ Ω Ω α α ( x ) β β ( y ) d μ μ {\displaystyle u(x,y)=\langle \alpha (x),\beta (y)\rangle =\int _{\Omega }\alpha (x)\beta (y)\;d\mu } ,

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

Integral linear maps

A continuous linear map κ κ : X → → Y ′ {\displaystyle \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 ) ∈ ∈ X × × Y ↦ ↦ ( κ κ x ) ( y ) {\displaystyle (x,y)\in X\times Y\mapsto (\kappa x)(y)} .cite-ref-footnoteschaeferwolff1999169-4-0[4] It follows that an integral map κ κ : X → → Y ′ {\displaystyle \kappa :X\to Y'} is of the form:cite-ref-footnoteschaeferwolff1999169-4-1[4]

x ∈ ∈ X ↦ ↦ κ κ ( x ) = ∫ ∫ S × × T ⟨ x ′ , x ⟩ y ′ d μ μ ( x ′ , y ′ ) {\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)}

for suitable weakly closed and equicontinuous subsets S and T of X ′ {\displaystyle X'} and Y ′ {\displaystyle Y'} , respectively, and some positive Radon measure μ μ {\displaystyle \mu } of total mass ≤ 1. The above integral is the weak integral, so the equality holds if and only if for every y ∈ ∈ Y {\displaystyle y\in Y} , ⟨ κ κ ( x ) , y ⟩ = ∫ ∫ S × × T ⟨ x ′ , x ⟩ ⟨ y ′ , y ⟩ d μ μ ( x ′ , y ′ ) {\textstyle \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 Λ Λ : X → → Y {\displaystyle \Lambda :X\to Y} , one can define a canonical bilinear form B Λ Λ ∈ ∈ B i ( X , Y ′ ) {\displaystyle B_{\Lambda }\in Bi\left(X,Y'\right)} , called the associated bilinear form on X × × Y ′ {\displaystyle X\times Y'} , by B Λ Λ ( x , y ′ ) := ( y ′ ∘ ∘ Λ Λ ) ( x ) {\displaystyle B_{\Lambda }\left(x,y'\right):=\left(y'\circ \Lambda \right)\left(x\right)} . A continuous map Λ Λ : X → → Y {\displaystyle \Lambda :X\to Y} is called integral if its associated bilinear form is an integral bilinear form.cite-ref-footnotetr-ves2006502-505-5-0[5] An integral map Λ Λ : X → → Y {\displaystyle \Lambda :X\to Y} is of the form, for every x ∈ ∈ X {\displaystyle x\in X} and y ′ ∈ ∈ Y ′ {\displaystyle y'\in Y'} :

⟨ y ′ , Λ Λ ( x ) ⟩ = ∫ ∫ A ′ × × B ″ ⟨ x ′ , x ⟩ ⟨ y ″ , y ′ ⟩ d μ μ ( x ′ , y ″ ) {\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)}

for suitable weakly closed and equicontinuous aubsets A ′ {\displaystyle A'} and B ″ {\displaystyle B''} of X ′ {\displaystyle X'} and Y ″ {\displaystyle Y''} , respectively, and some positive Radon measure μ μ {\displaystyle \mu } of total mass ≤ ≤ 1 {\displaystyle \leq 1} .

Relation to Hilbert spaces

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

Proposition:cite-ref-footnotetr-ves2006506-508-6-0[6] Suppose that u : X → → Y {\displaystyle 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 α α : X → → H {\displaystyle \alpha :X\to H} and β β : H → → Y {\displaystyle \beta :H\to Y} such that u = β β ∘ ∘ α α {\displaystyle u=\beta \circ \alpha } .

Furthermore, every integral operator between two Hilbert spaces is nuclear.cite-ref-footnotetr-ves2006506-508-6-1[6] 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.cite-ref-footnotetr-ves2006502-505-5-1[5] An important partial converse is that every integral operator between two Hilbert spaces is nuclear.cite-ref-footnotetr-ves2006506-508-6-2[6]

Suppose that A, B, C, and D are Hausdorff locally convex TVSs and that α α : A → → B {\displaystyle \alpha :A\to B} , β β : B → → C {\displaystyle \beta :B\to C} , and γ γ : C → → D {\displaystyle \gamma :C\to D} are all continuous linear operators. If β β : B → → C {\displaystyle \beta :B\to C} is an integral operator then so is the composition γ γ ∘ ∘ β β ∘ ∘ α α : A → → D {\displaystyle \gamma \circ \beta \circ \alpha :A\to D} .cite-ref-footnotetr-ves2006506-508-6-3[6]

If u : X → → Y {\displaystyle u:X\to Y} is a continuous linear operator between two normed space then u : X → → Y {\displaystyle u:X\to Y} is integral if and only if t u : Y ′ → → X ′ {\displaystyle {}^{t}u:Y'\to X'} is integral.cite-ref-footnotetr-ves2006505-7-0[7]

Suppose that u : X → → Y {\displaystyle u:X\to Y} is a continuous linear map between locally convex TVSs. If u : X → → Y {\displaystyle u:X\to Y} is integral then so is its transpose t u : Y b ′ ′ → → X b ′ ′ {\displaystyle {}^{t}u:Y_{b}^{\prime }\to X_{b}^{\prime }} .cite-ref-footnotetr-ves2006502-505-5-2[5] Now suppose that the transpose t u : Y b ′ ′ → → X b ′ ′ {\displaystyle {}^{t}u:Y_{b}^{\prime }\to X_{b}^{\prime }} of the continuous linear map u : X → → Y {\displaystyle u:X\to Y} is integral. Then u : X → → Y {\displaystyle u:X\to Y} is integral if the canonical injections In X : X → → X ″ {\displaystyle \operatorname {In} _{X}:X\to X''} (defined by x ↦ ↦ {\displaystyle x\mapsto } value at x) and In Y : Y → → Y ″ {\displaystyle \operatorname {In} _{Y}:Y\to Y''} are TVS-embeddings (which happens if, for instance, X {\displaystyle X} and Y b ′ ′ {\displaystyle Y_{b}^{\prime }} are barreled or metrizable).cite-ref-footnotetr-ves2006502-505-5-3[5]

Properties

Suppose that A, B, C, and D are Hausdorff locally convex TVSs with B and D complete. If α α : A → → B {\displaystyle \alpha :A\to B} , β β : B → → C {\displaystyle \beta :B\to C} , and γ γ : C → → D {\displaystyle \gamma :C\to D} are all integral linear maps then their composition γ γ ∘ ∘ β β ∘ ∘ α α : A → → D {\displaystyle \gamma \circ \beta \circ \alpha :A\to D} is nuclear.cite-ref-footnotetr-ves2006506-508-6-4[6] Thus, in particular, if X is an infinite-dimensional Fréchet space then a continuous linear surjection u : X → → X {\displaystyle u:X\to X} cannot be an integral operator.