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

See also: Injective tensor product and Projective 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 $y\in Y$, ${\textstyle \left\langle \kappa (x),y\right\rangle =\int _{S\times T}\left\langle x',x\right\rangle \left\langle y',y\right\rangle \mathrm {d} \mu \!\left(x',y'\right)}$.

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.

See also

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

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External links

• Nuclear space at ncatlab