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Integral linear operator

In mathematical analysis, anintegral linear operator is a linear operatorT given by integration; i.e.,

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

whereK(x,y){\displaystyle K(x,y)} is called an integration kernel.

More generally, anintegral bilinear form is abilinear functional that belongs to the continuous dual space ofX^ϵY{\displaystyle X{\widehat {\otimes }}_{\epsilon }Y}, theinjective tensor product of the locally convextopological vector spaces (TVSs)X andY. Anintegral 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 ofnuclear spaces andnuclear maps.

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

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LetX andY be locally convex TVSs, letXπY{\displaystyle X\otimes _{\pi }Y}  denote theprojective tensor product,X^πY{\displaystyle X{\widehat {\otimes }}_{\pi }Y}  denote its completion, letXϵY{\displaystyle X\otimes _{\epsilon }Y}  denote theinjective tensor product, andX^ϵY{\displaystyle X{\widehat {\otimes }}_{\epsilon }Y}  denote its completion. Suppose thatIn:XϵYX^ϵY{\displaystyle \operatorname {In} :X\otimes _{\epsilon }Y\to X{\widehat {\otimes }}_{\epsilon }Y}  denotes the TVS-embedding ofXϵY{\displaystyle X\otimes _{\epsilon }Y}  into its completion and lettIn:(X^ϵY)b(XϵY)b{\displaystyle {}^{t}\operatorname {In} :\left(X{\widehat {\otimes }}_{\epsilon }Y\right)_{b}^{\prime }\to \left(X\otimes _{\epsilon }Y\right)_{b}^{\prime }}  be itstranspose, which is a vector space-isomorphism. This identifies the continuous dual space ofXϵY{\displaystyle X\otimes _{\epsilon }Y}  as being identical to the continuous dual space ofX^ϵY{\displaystyle X{\widehat {\otimes }}_{\epsilon }Y} .

LetId:XπYXϵY{\displaystyle \operatorname {Id} :X\otimes _{\pi }Y\to X\otimes _{\epsilon }Y}  denote the identity map andtId:(XϵY)b(XπY)b{\displaystyle {}^{t}\operatorname {Id} :\left(X\otimes _{\epsilon }Y\right)_{b}^{\prime }\to \left(X\otimes _{\pi }Y\right)_{b}^{\prime }}  denote itstranspose, which is a continuous injection. Recall that(XπY){\displaystyle \left(X\otimes _{\pi }Y\right)^{\prime }}  is canonically identified withB(X,Y){\displaystyle B(X,Y)} , the space of continuous bilinear maps onX×Y{\displaystyle X\times Y} . In this way, the continuous dual space ofXϵY{\displaystyle X\otimes _{\epsilon }Y}  can be canonically identified as a vector subspace ofB(X,Y){\displaystyle B(X,Y)} , denoted byJ(X,Y){\displaystyle J(X,Y)} . The elements ofJ(X,Y){\displaystyle J(X,Y)}  are calledintegral (bilinear) forms onX×Y{\displaystyle X\times Y} . The following theorem justifies the wordintegral.

Theorem[1][2]The dualJ(X,Y) ofX^ϵY{\displaystyle X{\widehat {\otimes }}_{\epsilon }Y}  consists of exactly of the continuous bilinear formsu onX×Y{\displaystyle X\times Y}  of the form

u(x,y)=S×Tx,xy,ydμ(x,y),{\displaystyle u(x,y)=\int _{S\times T}\langle x,x'\rangle \langle y,y'\rangle \;d\mu \!\left(x',y'\right),} 

whereS andT are respectively some weakly closed and equicontinuous (hence weakly compact) subsets of the dualsX{\displaystyle X^{\prime }}  andY{\displaystyle Y^{\prime }} , andμ{\displaystyle \mu }  is a (necessarily bounded) positiveRadon measure on the (compact) setS×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 terminologyintegral bilinear form: a continuous bilinear formu{\displaystyle u}  on the productX×Y{\displaystyle X\times Y}  of locally convex spaces is integral if and only if there is acompact topological spaceΩ{\displaystyle \Omega }  equipped with a (necessarily bounded) positive Radon measureμ{\displaystyle \mu }  and continuous linear mapsα{\displaystyle \alpha }  andβ{\displaystyle \beta }  fromX{\displaystyle X}  andY{\displaystyle Y}  to the Banach spaceL(Ω,μ){\displaystyle L^{\infty }(\Omega ,\mu )}  such that

u(x,y)=α(x),β(y)=Ωα(x)β(y)dμ{\displaystyle u(x,y)=\langle \alpha (x),\beta (y)\rangle =\int _{\Omega }\alpha (x)\beta (y)\;d\mu } ,

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

Integral linear maps

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A continuous linear mapκ:XY{\displaystyle \kappa :X\to Y'}  is calledintegral if its associated bilinear form is an integral bilinear form, where this form is defined by(x,y)X×Y(κx)(y){\displaystyle (x,y)\in X\times Y\mapsto (\kappa x)(y)} .[4] It follows that an integral mapκ:XY{\displaystyle \kappa :X\to Y'}  is of the form:[4]

xXκ(x)=S×Tx,xydμ(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 subsetsS andT ofX{\displaystyle X'}  andY{\displaystyle Y'} , respectively, and some positive Radon measureμ{\displaystyle \mu }  of total mass ≤ 1. The above integral is theweak integral, so the equality holds if and only if for everyyY{\displaystyle y\in Y} ,κ(x),y=S×Tx,xy,ydμ(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 mapΛ:XY{\displaystyle \Lambda :X\to Y} , one can define a canonical bilinear formBΛBi(X,Y){\displaystyle B_{\Lambda }\in Bi\left(X,Y'\right)} , called theassociated bilinear form onX×Y{\displaystyle X\times Y'} , byBΛ(x,y):=(yΛ)(x){\displaystyle B_{\Lambda }\left(x,y'\right):=\left(y'\circ \Lambda \right)\left(x\right)} . A continuous mapΛ:XY{\displaystyle \Lambda :X\to Y}  is calledintegral if its associated bilinear form is an integral bilinear form.[5] An integral mapΛ:XY{\displaystyle \Lambda :X\to Y}  is of the form, for everyxX{\displaystyle x\in X}  andyY{\displaystyle y'\in Y'} :

y,Λ(x)=A×Bx,xy,ydμ(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 aubsetsA{\displaystyle A'}  andB{\displaystyle B''}  ofX{\displaystyle X'}  andY{\displaystyle Y''} , respectively, and some positive Radon measureμ{\displaystyle \mu }  of total mass1{\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 thatu:XY{\displaystyle u:X\to Y}  is an integral map between locally convex TVS withY Hausdorff and complete. There exists a Hilbert spaceH and two continuous linear mappingsα:XH{\displaystyle \alpha :X\to H}  andβ:HY{\displaystyle \beta :H\to Y}  such thatu=βα{\displaystyle u=\beta \circ \alpha } .

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

Sufficient conditions

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Everynuclear map is integral.[5] An important partial converse is that every integral operator between twoHilbert spaces isnuclear.[6]

Suppose thatA,B,C, andD are Hausdorff locally convex TVSs and thatα:AB{\displaystyle \alpha :A\to B} ,β:BC{\displaystyle \beta :B\to C} , andγ:CD{\displaystyle \gamma :C\to D}  are all continuous linear operators. Ifβ:BC{\displaystyle \beta :B\to C}  is an integral operator then so is the compositionγβα:AD{\displaystyle \gamma \circ \beta \circ \alpha :A\to D} .[6]

Ifu:XY{\displaystyle u:X\to Y}  is a continuous linear operator between two normed space thenu:XY{\displaystyle u:X\to Y}  is integral if and only iftu:YX{\displaystyle {}^{t}u:Y'\to X'}  is integral.[7]

Suppose thatu:XY{\displaystyle u:X\to Y}  is a continuous linear map between locally convex TVSs. Ifu:XY{\displaystyle u:X\to Y}  is integral then so is itstransposetu:YbXb{\displaystyle {}^{t}u:Y_{b}^{\prime }\to X_{b}^{\prime }} .[5] Now suppose that the transposetu:YbXb{\displaystyle {}^{t}u:Y_{b}^{\prime }\to X_{b}^{\prime }}  of the continuous linear mapu:XY{\displaystyle u:X\to Y}  is integral. Thenu:XY{\displaystyle u:X\to Y}  is integral if the canonical injectionsInX:XX{\displaystyle \operatorname {In} _{X}:X\to X''}  (defined byx{\displaystyle x\mapsto }  value atx) andInY:YY{\displaystyle \operatorname {In} _{Y}:Y\to Y''}  areTVS-embeddings (which happens if, for instance,X{\displaystyle X}  andYb{\displaystyle Y_{b}^{\prime }}  are barreled or metrizable).[5]

Properties

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Suppose thatA,B,C, andD are Hausdorff locally convex TVSs withB andDcomplete. Ifα:AB{\displaystyle \alpha :A\to B} ,β:BC{\displaystyle \beta :B\to C} , andγ:CD{\displaystyle \gamma :C\to D}  are all integral linear maps then their compositionγβα:AD{\displaystyle \gamma \circ \beta \circ \alpha :A\to D}  isnuclear.[6] Thus, in particular, ifX is an infinite-dimensionalFréchet space then a continuous linear surjectionu:XX{\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. ^Trèves 2006, pp. 500–502.
  3. ^Grothendieck 1955, pp. 124–126.
  4. ^abSchaefer & Wolff 1999, p. 169.
  5. ^abcdTrèves 2006, pp. 502–505.
  6. ^abcdeTrèves 2006, pp. 506–508.
  7. ^Trèves 2006, pp. 505.

Bibliography

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External links

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