In mathematical analysis, an integral linear operator is a linear operator T given by integration; i.e.,

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

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

More generally, an integral bilinear form is a bilinear functional that belongs to the continuous dual space of ${\displaystyle X{\widehat {\otimes }}_{\epsilon }Y}$, 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.

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 vector subspace 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.

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 ${\displaystyle u}$ on the product ${\displaystyle X\times Y}$ of locally convex spaces is integral if and only if there is a compact topological space ${\displaystyle \Omega }$ equipped with a (necessarily bounded) positive Radon measure ${\displaystyle \mu }$ and continuous linear maps ${\displaystyle \alpha }$ and ${\displaystyle \beta }$ from ${\displaystyle X}$ and ${\displaystyle Y}$ to the Banach space ${\displaystyle L^{\infty }(\Omega ,\mu )}$ such that

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

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

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

${\displaystyle x\in X\mapsto \kappa (x)=\int _{S\times T}\left\langle x',x\right\rangle y'\mathrm {d} \mu \!\left(x',y'\right)}$

for suitable weakly closed and equicontinuous subsets S and T of ${\displaystyle X'}$ and ${\displaystyle Y'}$, 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 ${\displaystyle y\in Y}$, ${\textstyle \left\langle \kappa (x),y\right\rangle =\int _{S\times T}\left\langle x',x\right\rangle \left\langle y',y\right\rangle \mathrm {d} \mu \!\left(x',y'\right)}$.

Given a linear map ${\displaystyle \Lambda :X\to Y}$, one can define a canonical bilinear form ${\displaystyle B_{\Lambda }\in Bi\left(X,Y'\right)}$, called the associated bilinear form on ${\displaystyle X\times Y'}$, by ${\displaystyle B_{\Lambda }\left(x,y'\right):=\left(y'\circ \Lambda \right)\left(x\right)}$. A continuous map ${\displaystyle \Lambda :X\to Y}$ is called integral if its associated bilinear form is an integral bilinear form.^[5] An integral map ${\displaystyle \Lambda :X\to Y}$ is of the form, for every ${\displaystyle x\in X}$ and ${\displaystyle y'\in Y'}$:

${\displaystyle \left\langle y',\Lambda (x)\right\rangle =\int _{A'\times B''}\left\langle x',x\right\rangle \left\langle y'',y'\right\rangle \mathrm {d} \mu \!\left(x',y''\right)}$

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

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

Proposition:^[6] Suppose that ${\displaystyle u:X\to Y}$ is an integral map between locally convex TVS with Y Hausdorff and complete. There exists a Hilbert space H and two continuous linear mappings ${\displaystyle \alpha :X\to H}$ and ${\displaystyle \beta :H\to Y}$ such that ${\displaystyle u=\beta \circ \alpha }$.

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.

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 ${\displaystyle \alpha :A\to B}$, ${\displaystyle \beta :B\to C}$, and ${\displaystyle \gamma :C\to D}$ are all continuous linear operators. If ${\displaystyle \beta :B\to C}$ is an integral operator then so is the composition ${\displaystyle \gamma \circ \beta \circ \alpha :A\to D}$.^[6]

If ${\displaystyle u:X\to Y}$ is a continuous linear operator between two normed space then ${\displaystyle u:X\to Y}$ is integral if and only if ${\displaystyle {}^{t}u:Y'\to X'}$ is integral.^[7]

Suppose that ${\displaystyle u:X\to Y}$ is a continuous linear map between locally convex TVSs. If ${\displaystyle u:X\to Y}$ is integral then so is its transpose ${\displaystyle {}^{t}u:Y_{b}^{\prime }\to X_{b}^{\prime }}$.^[5] Now suppose that the transpose ${\displaystyle {}^{t}u:Y_{b}^{\prime }\to X_{b}^{\prime }}$ of the continuous linear map ${\displaystyle u:X\to Y}$ is integral. Then ${\displaystyle u:X\to Y}$ is integral if the canonical injections ${\displaystyle \operatorname {In} _{X}:X\to X''}$ (defined by ${\displaystyle x\mapsto }$ value at x) and ${\displaystyle \operatorname {In} _{Y}:Y\to Y''}$ are TVS-embeddings (which happens if, for instance, ${\displaystyle X}$ and ${\displaystyle Y_{b}^{\prime }}$ are barreled or metrizable).^[5]

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

• 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