Integral linear operator

Let X and Y be locally convex TVSs, let ${\displaystyle X\otimes _{\pi }Y}$ denote the projective tensor product, ${\displaystyle X{\widehat {\otimes }}_{\pi }Y}$ denote its completion, let ${\displaystyle X\otimes _{\epsilon }Y}$ denote the injective tensor product, and ${\displaystyle X{\widehat {\otimes }}_{\epsilon }Y}$ denote its completion. Suppose that ${\displaystyle \operatorname {In} :X\otimes _{\epsilon }Y\to X{\widehat {\otimes }}_{\epsilon }Y}$ denotes the TVS-embedding of ${\displaystyle X\otimes _{\epsilon }Y}$ into its completion and let ${\displaystyle {}^{t}\operatorname {In}$ :\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 ${\displaystyle X\otimes _{\epsilon }Y}$ as being identical to the continuous dual space of ${\displaystyle X{\widehat {\otimes }}_{\epsilon }Y}$.

Let ${\displaystyle \operatorname {Id} :X\otimes _{\pi }Y\to X\otimes _{\epsilon }Y}$ denote the identity map and ${\displaystyle {}^{t}\operatorname {Id}$ :\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 ${\displaystyle \left(X\otimes _{\pi }Y\right)^{\prime }}$ is canonically identified with ${\displaystyle B(X,Y)}$, the space of continuous bilinear maps on ${\displaystyle X\times Y}$. In this way, the continuous dual space of ${\displaystyle X\otimes _{\epsilon }Y}$ can be canonically identified as a subvector space of ${\displaystyle B(X,Y)}$, denoted by ${\displaystyle J(X,Y)}$. The elements of ${\displaystyle J(X,Y)}$ are called integral (bilinear) forms on ${\displaystyle X\times Y}$. The following theorem justifies the word integral.

Theorem[1][2] The dual J(X, Y) of ${\displaystyle X{\widehat {\otimes }}_{\epsilon }Y}$ consists of exactly those continuous bilinear forms v on ${\displaystyle X\times Y}$ that can be represented in the form of a map

${\displaystyle b\in B(X,Y)\mapsto v(b)=\int _{S\times T}b{\big \vert }_{S\times T}\left(x^{\prime },y^{\prime }\right)\operatorname {d} \mu \left(x^{\prime },y^{\prime }\right)}$

where S and T are some closed, equicontinuous subsets of ${\displaystyle X_{\sigma }^{\prime }}$ and ${\displaystyle Y_{\sigma }^{\prime }}$, respectively, and ${\displaystyle \mu }$ is a positive Radon measure on the compact set ${\displaystyle S\times T}$ with total mass ${\displaystyle \leq 1}$. Furthermore, if A is an equicontinuous subset of J(X, Y) then the elements ${\displaystyle v\in A}$ can be represented with ${\displaystyle S\times T}$ fixed and ${\displaystyle \mu }$ running through a norm bounded subset of the space of Radon measures on ${\displaystyle S\times T}$.

A continuous linear map ${\displaystyle \kappa :X\to Y^{\prime }}$ is called integral if its associated bilinear form is an integral bilinear form, where this form is defined by ${\displaystyle (x,y)\in X\times Y\mapsto (\kappa x)(y)}$.[3] It follows that an integral map ${\displaystyle \kappa :X\to Y^{\prime }}$ is of the form:[3]

${\displaystyle x\in X\mapsto \kappa (x)=\int _{S\times T}\left\langle x^{\prime },x\right\rangle y^{\prime }\operatorname {d} \mu \left(x^{\prime },y^{\prime }\right)}$

for suitable weakly closed and equicontinuous subsets S and T of ${\displaystyle X^{\prime }}$ and ${\displaystyle Y^{\prime }}$, respectively, and some positive Radon measure ${\displaystyle \mu }$ of total mass ≤ 1. The above integral is the weak integral, so the equality holds if and only if for every ${\displaystyle y\in Y}$, ${\displaystyle \left\langle \kappa (x),y\right\rangle =\int _{S\times T}\left\langle x^{\prime },x\right\rangle \left\langle y^{\prime },y\right\rangle \operatorname {d} \mu \left(x^{\prime },y^{\prime }\right)}$.

Given a linear map ${\displaystyle \Lambda :X\to Y}$, one can define a canonical bilinear form ${\displaystyle B_{\Lambda }\in Bi\left(X,Y^{\prime }\right)}$, called the associated bilinear form on ${\displaystyle X\times Y^{\prime }}$, by ${\displaystyle B_{\Lambda }\left(x,y^{\prime }\right):=\left(y^{\prime }\circ \Lambda \right)\left(x\right)}$. A continuous map ${\displaystyle \Lambda :X\to Y}$ is called integral if its associated bilinear form is an integral bilinear form.[4] An integral map ${\displaystyle \Lambda :X\to Y}$ is of the form, for every ${\displaystyle x\in X}$ and ${\displaystyle y^{\prime }\in Y^{\prime }}$:

${\displaystyle \left\langle y^{\prime },\Lambda (x)\right\rangle =\int _{A^{\prime }\times B^{\prime \prime }}\left\langle x^{\prime },x\right\rangle \left\langle y^{\prime \prime },y^{\prime }\right\rangle \operatorname {d} \mu \left(x^{\prime },y^{\prime \prime }\right)}$

for suitable weakly closed and equicontinuous aubsets ${\displaystyle A^{\prime }}$ and ${\displaystyle B^{\prime \prime }}$ of ${\displaystyle X^{\prime }}$ and ${\displaystyle Y^{\prime \prime }}$, respectively, and some positive Radon measure ${\displaystyle \mu }$ of total mass ${\displaystyle \leq 1}$.

• Auxiliary normed spaces
• Final topology
• Injective tensor product
• Nuclear operators
• Nuclear spaces
• Projective tensor product
• Topological tensor product

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• Nuclear space at ncatlab