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Integral linear operator

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Mathematical function

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 ^ 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.

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

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Let X and Y be locally convex TVSs, let X p Y {\displaystyle X\otimes _{\pi }Y} denote the projective tensor product, X ^ p Y {\displaystyle X{\widehat {\otimes }}_{\pi }Y} denote its completion, let X Y {\displaystyle X\otimes _{\epsilon }Y} denote the injective tensor product, and X ^ Y {\displaystyle X{\widehat {\otimes }}_{\epsilon }Y} denote its completion. Suppose that In : X Y - X ^ Y {\displaystyle \operatorname {In} :X\otimes _{\epsilon }Y\to X{\widehat {\otimes }}_{\epsilon }Y} denotes the TVS-embedding of X Y {\displaystyle X\otimes _{\epsilon }Y} into its completion and let t In : ( X ^ Y ) b ' - ( X 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 Y {\displaystyle X\otimes _{\epsilon }Y} as being identical to the continuous dual space of X ^ Y {\displaystyle X{\widehat {\otimes }}_{\epsilon }Y} .

Let Id : X p Y - X Y {\displaystyle \operatorname {Id} :X\otimes _{\pi }Y\to X\otimes _{\epsilon }Y} denote the identity map and t Id : ( X Y ) b ' - ( X p 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 p 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 x Y {\displaystyle X\times Y} . In this way, the continuous dual space of X 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 x Y {\displaystyle X\times Y} . The following theorem justifies the word integral.

Theorem[1][2]--The dual J(X, Y) of X ^ Y {\displaystyle X{\widehat {\otimes }}_{\epsilon }Y} consists of exactly of the continuous bilinear forms u on X x Y {\displaystyle X\times Y} of the form

u ( x , y ) = S x T < x , x ' > < y , y ' > d m ( 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 m {\displaystyle \mu } is a (necessarily bounded) positive Radon measure on the (compact) set S x T {\displaystyle S\times T} .

There is also a closely related formulation [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 x Y {\displaystyle X\times Y} of locally convex spaces is integral if and only if there is a compact topological space O {\displaystyle \Omega } equipped with a (necessarily bounded) positive Radon measure m {\displaystyle \mu } and continuous linear maps a {\displaystyle \alpha } and b {\displaystyle \beta } from X {\displaystyle X} and Y {\displaystyle Y} to the Banach space L ( O , m ) {\displaystyle L^{\infty }(\Omega ,\mu )} such that

u ( x , y ) = < a ( x ) , b ( y ) > = O a ( x ) b ( y ) d m {\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

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A continuous linear map k : 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 x Y - ( k x ) ( y ) {\displaystyle (x,y)\in X\times Y\mapsto (\kappa x)(y)} .[4] It follows that an integral map k : X - Y ' {\displaystyle \kappa :X\to Y'} is of the form:[4]

x X - k ( x ) = S x T < x ' , x > y ' d m ( 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 m {\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} , < k ( x ) , y > = S x T < x ' , x > < y ' , y > d m ( 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 L : X - Y {\displaystyle \Lambda :X\to Y} , one can define a canonical bilinear form B L B i ( X , Y ' ) {\displaystyle B_{\Lambda }\in Bi\left(X,Y'\right)} , called the associated bilinear form on X x Y ' {\displaystyle X\times Y'} , by B L ( x , y ' ) := ( y ' L ) ( x ) {\displaystyle B_{\Lambda }\left(x,y'\right):=\left(y'\circ \Lambda \right)\left(x\right)} . A continuous map L : X - Y {\displaystyle \Lambda :X\to Y} is called integral if its associated bilinear form is an integral bilinear form.[5] An integral map L : 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 ' , L ( x ) > = A ' x B '' < x ' , x > < y '' , y ' > d m ( 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 m {\displaystyle \mu } of total mass <= 1 {\displaystyle \leq 1} .

Relation to Hilbert spaces

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

Proposition:[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 a : X - H {\displaystyle \alpha :X\to H} and b : H - Y {\displaystyle \beta :H\to Y} such that u = b a {\displaystyle u=\beta \circ \alpha } .

Furthermore, every integral operator between two Hilbert spaces is nuclear.[6] Thus a continuous linear operator between two Hilbert spaces is nuclear if and only if it is integral.

Sufficient conditions

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Every nuclear map is integral.[5] An important partial converse is that every integral operator between two Hilbert spaces is nuclear.[6]

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

Properties

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Suppose that A, B, C, and D are Hausdorff locally convex TVSs with B and D complete. If a : A - B {\displaystyle \alpha :A\to B} , b : B - C {\displaystyle \beta :B\to C} , and g : C - D {\displaystyle \gamma :C\to D} are all integral linear maps then their composition g b a : A - D {\displaystyle \gamma \circ \beta \circ \alpha :A\to D} is nuclear.[6] Thus, in particular, if X is an infinite-dimensional Frechet space then a continuous linear surjection u : X - X {\displaystyle u:X\to X} cannot be an integral operator.

See also

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References

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  1. ^ Schaefer & Wolff 1999, p. 168.
  2. ^ Treves 2006, pp. 500-502.
  3. ^ Grothendieck 1955, pp. 124-126.
  4. ^ a b Schaefer & Wolff 1999, p. 169.
  5. ^ a b c d Treves 2006, pp. 502-505.
  6. ^ a b c d e Treves 2006, pp. 506-508.
  7. ^ Treves 2006, pp. 505.

Bibliography

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