# Integral linear operator

In mathematical analysis, an integral linear operator is a linear operator T given by integration; i.e.,
    $(Tf)(x)=\int f(y)K(x,y)\,dy$
where $K(x,y)$ is called an integration kernel.
More generally, an integral bilinear form is a bilinear functional that belongs to the continuous dual space of $X{\widehat {\otimes }}_{\epsilon }Y$, 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

Let X and Y be locally convex TVSs, let $X\otimes _{\pi }Y$ denote the projective tensor product, $X{\widehat {\otimes }}_{\pi }Y$ denote its completion, let $X\otimes _{\epsilon }Y$ denote the injective tensor product, and $X{\widehat {\otimes }}_{\epsilon }Y$ denote its completion. Suppose that $\operatorname {In} :X\otimes _{\epsilon }Y\to X{\widehat {\otimes }}_{\epsilon }Y$ denotes the TVS-embedding of $X\otimes _{\epsilon }Y$ into its completion and let $^{t}\operatorname {In} :\left(X{\widehat {\otimes }}_{\epsilon }Y\right)_{b}^{\prime }\to \left(X\otimes _{\epsilon }Y\right)_{b}^{\prime }$ be its transpose, which is a vector space-isomorphism. This identifies the continuous dual space of $X\otimes _{\epsilon }Y$ as being identical to the continuous dual space of $X{\widehat {\otimes }}_{\epsilon }Y$.
Let $\operatorname {Id} :X\otimes _{\pi }Y\to X\otimes _{\epsilon }Y$ denote the identity map and $^{t}\operatorname {Id} :\left(X\otimes _{\epsilon }Y\right)_{b}^{\prime }\to \left(X\otimes _{\pi }Y\right)_{b}^{\prime }$ denote its transpose, which is a continuous injection. Recall that $\left(X\otimes _{\pi }Y\right)^{\prime }$ is canonically identified with $B(X,Y)$, the space of continuous bilinear maps on $X\times Y$. In this way, the continuous dual space of $X\otimes _{\epsilon }Y$ can be canonically identified as a vector subspace of $B(X,Y)$, denoted by $J(X,Y)$. The elements of $J(X,Y)$ are called integral (bilinear) forms on $X\times Y$. The following theorem justifies the word integral.
Theorem — The dual J(X, Y) of $X{\widehat {\otimes }}_{\epsilon }Y$ consists of exactly of the continuous bilinear forms u on $X\times Y$ of the form
    $u(x,y)=\int _{S\times T}\langle x,x'\rangle \langle y,y'\rangle \;d\mu \!\left(x',y'\right),$
where S and T are respectively some weakly closed and equicontinuous (hence weakly compact) subsets of the duals $X^{\prime }$ and $Y^{\prime }$, and $\mu$ is a (necessarily bounded) positive Radon measure on the (compact) set $S\times T$.
There is also a closely related formulation of the theorem above that can also be used to explain the terminology integral bilinear form: a continuous bilinear form $u$ on the product $X\times Y$ of locally convex spaces is integral if and only if there is a compact topological space $\Omega$ equipped with a (necessarily bounded) positive Radon measure $\mu$ and continuous linear maps $\alpha$ and $\beta$ from $X$ and $Y$ to the Banach space $L^{\infty }(\Omega ,\mu )$ such that
    $u(x,y)=\langle \alpha (x),\beta (y)\rangle =\int _{\Omega }\alpha (x)\beta (y)\;d\mu$,
i.e., the form $u$ can be realised by integrating (essentially bounded) functions on a compact space.

## Integral linear maps

A continuous linear map $\kappa :X\to Y'$ is called integral if its associated bilinear form is an integral bilinear form, where this form is defined by $(x,y)\in X\times Y\mapsto (\kappa x)(y)$. It follows that an integral map $\kappa :X\to Y'$ is of the form:
    $x\in X\mapsto \kappa (x)=\int _{S\times T}\left\langle x',x\right\rangle y'\mathrm {d} \mu \!\left(x',y'\right)$
for suitable weakly closed and equicontinuous subsets S and T of $X'$ and $Y'$, respectively, and some positive Radon measure $\mu$ of total mass ≤ 1. The above integral is the weak integral, so the equality holds if and only if for every $y\in Y$, ⟨κ(x), y⟩=∫$_{S×T}$⟨x$^′$, x⟩⟨y$^′$, y⟩dμ(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 $\Lambda :X\to Y$, one can define a canonical bilinear form $B_{\Lambda }\in Bi\left(X,Y'\right)$, called the associated bilinear form on $X\times Y'$, by $B_{\Lambda }\left(x,y'\right):=\left(y'\circ \Lambda \right)\left(x\right)$. A continuous map $\Lambda :X\to Y$ is called integral if its associated bilinear form is an integral bilinear form. An integral map $\Lambda :X\to Y$ is of the form, for every $x\in X$ and $y'\in Y'$:
    $\left\langle y',\Lambda (x)\right\rangle =\int _{A'\times B''}\left\langle x',x\right\rangle \left\langle y'',y'\right\rangle \mathrm {d} \mu \!\left(x',y''\right)$
for suitable weakly closed and equicontinuous aubsets $A'$ and $B''$ of $X'$ and $Y''$, respectively, and some positive Radon measure $\mu$ of total mass $\leq 1$.

### Relation to Hilbert spaces

The following result shows that integral maps "factor through" Hilbert spaces.
Proposition: Suppose that $u:X\to Y$ is an integral map between locally convex TVS with Y Hausdorff and complete. There exists a Hilbert space H and two continuous linear mappings $\alpha :X\to H$ and $\beta :H\to Y$ such that $u=\beta \circ \alpha$.
Furthermore, every integral operator between two Hilbert spaces is nuclear. 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. An important partial converse is that every integral operator between two Hilbert spaces is nuclear.
Suppose that A, B, C, and D are Hausdorff locally convex TVSs and that $\alpha :A\to B$, $\beta :B\to C$, and $\gamma :C\to D$ are all continuous linear operators. If $\beta :B\to C$ is an integral operator then so is the composition $\gamma \circ \beta \circ \alpha :A\to D$.
If $u:X\to Y$ is a continuous linear operator between two normed space then $u:X\to Y$ is integral if and only if $^{t}u:Y'\to X'$ is integral.
Suppose that $u:X\to Y$ is a continuous linear map between locally convex TVSs.
If $u:X\to Y$ is integral then so is its transpose $^{t}u:Y_{b}^{\prime }\to X_{b}^{\prime }$. Now suppose that the transpose $^{t}u:Y_{b}^{\prime }\to X_{b}^{\prime }$ of the continuous linear map $u:X\to Y$ is integral. Then $u:X\to Y$ is integral if the canonical injections $\operatorname {In} _{X}:X\to X''$ (defined by $x\mapsto$ value at x) and $\operatorname {In} _{Y}:Y\to Y''$ are TVS-embeddings (which happens if, for instance, $X$ and $Y_{b}^{\prime }$ are barreled or metrizable).

### Properties

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