Injective tensor product

• ${\displaystyle \sigma \left(X,X^{\prime }\right)}$ denotes the coarsest topology on ${\displaystyle X}$ making every map in ${\displaystyle X^{\prime }}$ continuous and ${\displaystyle X_{\sigma \left(X,X^{\prime }\right)}}$ or ${\displaystyle X_{\sigma }}$ denotes ${\displaystyle X}$ endowed with this topology.
• ${\displaystyle \sigma \left(X^{\prime },X\right)}$ denotes weak-* topology on ${\displaystyle X^{\prime }}$ and ${\displaystyle X_{\sigma \left(X^{\prime },X\right)}}$ or ${\displaystyle X_{\sigma }^{\prime }}$ denotes ${\displaystyle X^{\prime }}$ endowed with this topology.
  • Note that every ${\displaystyle x_{0}\in X}$ induces a map ${\displaystyle X^{\prime }\to \mathbb {R} }$ defined by ${\displaystyle \lambda \mapsto \lambda \left(x_{0}\right).}$ ${\displaystyle \sigma \left(X^{\prime },X\right)}$ is the coarsest topology on X′ making all such maps continuous.
• ${\displaystyle b\left(X,X^{\prime }\right)}$ denotes the topology of bounded convergence on ${\displaystyle X}$ and ${\displaystyle X_{b\left(X,X^{\prime }\right)}}$ or ${\displaystyle X_{b}}$ denotes ${\displaystyle X}$ endowed with this topology.
• ${\displaystyle b\left(X^{\prime },X\right)}$ denotes the topology of bounded convergence on ${\displaystyle X^{\prime }}$ or the strong dual topology on ${\displaystyle X^{\prime }}$ and ${\displaystyle X_{b\left(X^{\prime },X\right)}}$ or ${\displaystyle X_{b}^{\prime }}$ denotes ${\displaystyle X^{\prime }}$ endowed with this topology.
  • As usual, if ${\displaystyle X^{\prime }}$ is considered as a topological vector space but it has not been made clear what topology it is endowed with, then the topology will be assumed to be ${\displaystyle b\left(X^{\prime },X\right).}$
• ${\displaystyle \tau \left(X,X^{\prime }\right)}$ denotes the Mackey topology on ${\displaystyle X}$ or the topology of uniform convergence on the convex balanced weakly compact subsets of ${\displaystyle X^{\prime }}$ and ${\displaystyle X_{\tau \left(X,X^{\prime }\right)}}$ or ${\displaystyle X_{\tau }}$ denotes ${\displaystyle X}$ endowed with this topology. ${\displaystyle \tau (X,X^{\prime })}$ is the finest locally convex TVS topology on ${\displaystyle X}$ whose continuous dual space is equal to ${\displaystyle X^{\prime }.}$
• ${\displaystyle \tau \left(X^{\prime },X\right)}$ denotes the Mackey topology on ${\displaystyle X^{\prime }}$ or the topology of uniform convergence on the convex balanced weakly compact subsets of ${\displaystyle X}$ and ${\displaystyle X_{\tau \left(X^{\prime },X\right)}}$ or ${\displaystyle X_{\tau }^{\prime }}$ denotes ${\displaystyle X}$ endowed with this topology.
  • Note that ${\displaystyle \tau \left(X^{\prime },X\right)\subseteq b\left(X^{\prime },X\right)\subseteq \tau \left(X^{\prime },X^{\prime \prime }\right).}$[1]
• ${\displaystyle \varepsilon \left(X,X^{\prime }\right)}$ denotes the topology of uniform convergence on equicontinuous subsets of ${\displaystyle X^{\prime }}$ and ${\displaystyle X_{\varepsilon \left(X,X^{\prime }\right)}}$ or ${\displaystyle X_{\varepsilon }}$ denotes ${\displaystyle X}$ endowed with this topology.
  • If ${\displaystyle H}$ is a set of linear mappings ${\displaystyle X\to Y}$ then ${\displaystyle H}$ is equicontinuous if and only if it is equicontinuous at the origin; that is, if and only if for every neighborhood ${\displaystyle V}$ of the origin in ${\displaystyle Y,}$ there exists a neighborhood ${\displaystyle U}$ of the origin in ${\displaystyle X}$ such that ${\displaystyle \lambda (U)\subseteq V}$ for every ${\displaystyle \lambda \in H.}$
• A set ${\displaystyle H}$ of linear maps from ${\displaystyle X}$ to ${\displaystyle Y}$ is called equicontinuous if for every neighborhood ${\displaystyle V}$ of the origin in ${\displaystyle Y,}$ there exists a neighborhood ${\displaystyle U}$ of the origin in ${\displaystyle X}$ such that ${\displaystyle h(U)\subseteq V}$ for all ${\displaystyle h\in H.}$[2]

Although the question of whether or not one vector space is a tensor product of two other vector spaces is a purely algebraic one (that is, the answer does not depend on the topologies of ${\displaystyle X}$ or ${\displaystyle Y}$), nevertheless the vector space ${\displaystyle B\left(X_{\sigma }^{\prime },Y_{\sigma }^{\prime }\right)}$ of continuous bilinear functionals is always a tensor product of ${\displaystyle X}$ and ${\displaystyle Y,}$ as is now described.

For every ${\displaystyle (x,y)\in X\times Y}$ we now define a bilinear form, denoted by the symbol ${\displaystyle x\otimes y,}$ from ${\displaystyle X^{\prime }\times Y^{\prime }}$ into the underlying field (that is, ${\displaystyle x\otimes y:X^{\prime }\times Y^{\prime }\to \mathbb {C} }$) by

$${\displaystyle (x\otimes y)\left(x^{\prime },y^{\prime }\right):=x^{\prime }(x)y^{\prime }(y).}$$

This induces a canonical map

$${\displaystyle \cdot \otimes \cdot :X\times Y\to {\mathcal {B}}\left(X_{\sigma }^{\prime },Y_{\sigma }^{\prime }\right)}$$

defined by sending ${\displaystyle (x,y)\in X\times Y}$ to the bilinear form ${\displaystyle x\otimes y.}$ The span of the range of this map is ${\displaystyle B\left(X_{\sigma }^{\prime },Y_{\sigma }^{\prime }\right).}$ The following theorem may be used to verify that ${\displaystyle B\left(X_{\sigma }^{\prime },Y_{\sigma }^{\prime }\right)}$ together with the above map ${\displaystyle \,\otimes \,}$ is a tensor product of ${\displaystyle X}$ and ${\displaystyle Y.}$

Henceforth, all topological vector spaces considered will be assumed to be locally convex. If ${\displaystyle Z}$ is any locally convex topological vector space, then for any equicontinuous subsets ${\displaystyle G\subseteq X^{\prime }}$ and ${\displaystyle H\subseteq Y^{\prime },}$ and any neighborhood ${\displaystyle N}$ in ${\displaystyle Z,}$ define

$${\displaystyle {\mathcal {U}}(G,H,N)=\left\{b\in {\mathcal {B}}\left(X_{b}^{\prime },Y_{b}^{\prime };Z\right)~:~b(G,H)\subseteq N\right\}}$$

Every set ${\displaystyle b(G,H)}$ is bounded, which is necessary and sufficient for the collection of all such ${\displaystyle {\mathcal {U}}(G,H,N)}$ to form a locally convex TVS topology on ${\displaystyle {\mathcal {B}}\left(X_{b}^{\prime },Y_{b}^{\prime };Z\right)}$ called the ${\displaystyle \varepsilon }$-topology. The inclusions

$${\displaystyle {\mathcal {U}}(G,H,N)~\subseteq ~B\left(X_{\sigma }^{\prime },Y_{\sigma }^{\prime };Z\right)~\subseteq ~{\mathcal {B}}\left(X_{b}^{\prime },Y_{b}^{\prime };Z\right).}$$

always hold and whenever any one of these vector spaces is endowed with the ${\displaystyle \varepsilon }$-topology then this will be indicated by placing ${\displaystyle \varepsilon }$ as a subscript before the opening parenthesis. For example, ${\displaystyle {\mathcal {B}}\left(X_{b}^{\prime },Y_{b}^{\prime };Z\right)}$ endowed with the ${\displaystyle \varepsilon }$-topology will be denoted by ${\displaystyle {\mathcal {B}}_{\varepsilon }\left(X_{b}^{\prime },Y_{b}^{\prime };Z\right).}$

In particular, when ${\displaystyle Z}$ is the underlying scalar field then since ${\displaystyle B\left(X_{\sigma }^{\prime },Y_{\sigma }^{\prime }\right)=X\otimes Y,}$ the topological vector space ${\displaystyle B_{\varepsilon }\left(X_{\sigma }^{\prime },Y_{\sigma }^{\prime }\right)}$ will be denoted by ${\displaystyle X\otimes _{\varepsilon }Y,}$ which is called the injective tensor product of ${\displaystyle X}$ and ${\displaystyle Y.}$ This TVS is not necessarily complete so its completion will be denoted by ${\displaystyle X{\widehat {\otimes }}_{\varepsilon }Y.}$ The space ${\displaystyle {\mathcal {B}}\left(X_{\sigma }^{\prime },Y_{\sigma }^{\prime }\right)}$ is complete if and only if both ${\displaystyle X}$ and ${\displaystyle Y}$ are complete, in which case the completion of ${\displaystyle B\left(X_{\sigma }^{\prime },Y_{\sigma }^{\prime }\right)}$ is a subvector space, denoted by ${\displaystyle X{\widehat {\otimes }}_{\varepsilon }Y,}$ of ${\displaystyle {\mathcal {B}}\left(X_{\sigma }^{\prime },Y_{\sigma }^{\prime }\right).}$ If ${\displaystyle X}$ and ${\displaystyle Y}$ are normed then so is ${\displaystyle {\mathcal {B}}_{\varepsilon }\left(X_{\sigma }^{\prime },Y_{\sigma }^{\prime }\right).}$ And ${\displaystyle {\mathcal {B}}_{\varepsilon }\left(X_{\sigma }^{\prime },Y_{\sigma }^{\prime }\right)}$ is a Banach space if and only if both ${\displaystyle X}$ and ${\displaystyle Y}$ are Banach spaces.[4]

One reason for converging on equicontinuous subsets (of all possibilities) is the following important fact:

A set of continuous linear functionals ${\displaystyle H}$ on a TVS ${\displaystyle X}$[note 1] is equicontinuous if and only if it is contained in the polar of some neighborhood ${\displaystyle U}$ of ${\displaystyle 0}$ in ${\displaystyle X}$; that is, ${\displaystyle H\subseteq U^{\circ }.}$

A TVS's topology is completely determined by the open neighborhoods of the origin. This fact together with the bipolar theorem means that via the operation of taking the polar of a subset, the collection of all equicontinuous subsets of ${\displaystyle X^{\prime }}$ "encodes" all information about ${\displaystyle X}$'s given topology. Specifically, distinct LCTVS topologies on ${\displaystyle X}$ produce distinct collections of equicontinuous subsets and conversely, given any such collection of equicontinuous sets, the TVS's original topology can be recovered by taking the polar of every (equicontinuous) set in the collection. Thus through this identification, uniform convergence on the collection of equicontinuous subsets is essentially uniform convergence on the very topology of the TVS; this allows one to directly relate the injective topology with the given topologies of ${\displaystyle X}$ and ${\displaystyle Y.}$ Furthermore, the topology of a locally convex Hausdorff space ${\displaystyle X}$ is identical to the topology of uniform convergence on the equicontinuous subsets of ${\displaystyle X^{\prime }.}$[5]

For this reason, the article now lists some properties of equicontinuous sets that are relevant for dealing with the injective tensor product. Throughout ${\displaystyle X}$ and ${\displaystyle Y}$ are arbitrary TVSs and ${\displaystyle H}$ is a collection of linear maps from ${\displaystyle X}$ into ${\displaystyle Y.}$

• If ${\displaystyle H\subseteq L(X;Y)}$ is equicontinuous then the subspace topologies that ${\displaystyle H}$ inherits from the following topologies on ${\displaystyle L(X;Y)}$ are identical:[6]
  1. the topology of precompact convergence;
  2. the topology of compact convergence;
  3. the topology of pointwise convergence;
  4. the topology of pointwise convergence on a given dense subset of ${\displaystyle X.}$
• An equicontinuous set ${\displaystyle H\subseteq L(X;Y)}$ is bounded in the topology of bounded convergence (that is, bounded in ${\displaystyle L_{b}(X;Y)}$).[6] So in particular, ${\displaystyle H}$ will also bounded in every TVS topology that is coarser than the topology of bounded convergence.
• If ${\displaystyle X}$ is a barrelled space and ${\displaystyle Y}$ is locally convex then for any subset ${\displaystyle H\subseteq L(X;Y),}$ the following are equivalent:
  1. ${\displaystyle H}$ is equicontinuous;
  2. ${\displaystyle H}$ is bounded in the topology of pointwise convergence (that is, bounded in ${\displaystyle L_{\sigma }(X;Y)}$);
  3. ${\displaystyle H}$ is bounded in the topology of bounded convergence (that is, bounded in ${\displaystyle L_{b}(X;Y)}$).

In particular, to show that a set ${\displaystyle H}$ is equicontinuous it suffices to show that it is bounded in the topology of pointwise converge.[7]

• If ${\displaystyle X}$ is a Baire space then any subset ${\displaystyle H\subseteq L(X;Y)}$ that is bounded in ${\displaystyle L_{\sigma }(X;Y)}$ is necessarily equicontinuous.[7]
• If ${\displaystyle X}$ is separable, ${\displaystyle Y}$ is metrizable, and ${\displaystyle D}$ is a dense subset of ${\displaystyle X,}$ then the topology of pointwise convergence on ${\displaystyle D}$ makes ${\displaystyle L(X;Y)}$ metrizable so that in particular, the subspace topology that any equicontinuous subset ${\displaystyle H\subseteq L(X;Y)}$ inherits from ${\displaystyle L_{\sigma }(X;Y)}$ is metrizable.[6]

For equicontinuous subsets of the continuous dual space ${\displaystyle X^{\prime }}$ (where ${\displaystyle Y}$ is now the underlying scalar field of ${\displaystyle X}$), the following hold:

• The weak closure of an equicontinuous set of linear functionals on ${\displaystyle X}$ is a compact subspace of ${\displaystyle X_{\sigma }^{\prime }.}$[6]
• If ${\displaystyle X}$ is separable then every weakly closed equicontinuous subset of ${\displaystyle X_{\sigma }^{\prime }}$ is a metrizable compact space when it is given the weak topology (that is, the subspace topology inherited from ${\displaystyle X_{\sigma }^{\prime }}$).[6]
• If ${\displaystyle X}$ is a normable space then a subset ${\displaystyle H\subseteq X^{\prime }}$ is equicontinuous if and only if it is strongly bounded (that is, bounded in ${\displaystyle X_{b}^{\prime }}$).[6]
• If ${\displaystyle X}$ is a barrelled space then for any subset ${\displaystyle H\subseteq X^{\prime },}$ the following are equivalent:[7]
  1. ${\displaystyle H}$ is equicontinuous;
  2. ${\displaystyle H}$ is relatively compact in the weak dual topology;
  3. ${\displaystyle H}$ is weakly bounded;
  4. ${\displaystyle H}$ is strongly bounded.

We mention some additional important basic properties relevant to the injective tensor product:

• Suppose that ${\displaystyle B:X_{1}\times X_{2}\to Y}$ is a bilinear map where ${\displaystyle X_{1}}$ is a Fréchet space, ${\displaystyle X_{2}}$ is metrizable, and ${\displaystyle Y}$ is locally convex. If ${\displaystyle B}$ is separately continuous then it is continuous.[8]

The set equality ${\displaystyle L\left(X_{\sigma }^{\prime };Y_{\sigma }\right)=L\left(X_{\tau }^{\prime };Y\right)}$ always holds; that is, if ${\displaystyle u:X^{\prime }\to Y}$ is a linear map, then ${\displaystyle u:X_{\sigma \left(X^{\prime },X\right)}^{\prime }\to Y_{\sigma \left(Y,Y^{\prime }\right)}}$ is continuous if and only if ${\displaystyle u:X_{\tau \left(X^{\prime },X\right)}^{\prime }\to Y}$ is continuous, where here ${\displaystyle Y}$ has its original topology.[9]

There also exists a canonical vector space isomorphism[9]

$${\displaystyle J:{\mathcal {B}}\left(X_{\sigma \left(X^{\prime },X\right)}^{\prime },Y_{\sigma \left(Y^{\prime },Y\right)}^{\prime }\right)\to L\left(X_{\sigma \left(X^{\prime },X\right)}^{\prime };Y_{\sigma \left(Y,Y^{\prime }\right)}\right).}$$

To define it, for every separately continuous bilinear form ${\displaystyle B}$ defined on ${\displaystyle X_{\sigma \left(X^{\prime },X\right)}^{\prime }\times Y_{\sigma \left(Y^{\prime },Y\right)}^{\prime }}$ and every ${\displaystyle x^{\prime }\in X^{\prime },}$ let ${\displaystyle B_{x^{\prime }}\in \left(Y_{\sigma }^{\prime }\right)^{\prime }}$ be defined by

$${\displaystyle B_{x^{\prime }}\left(y^{\prime }\right):=B\left(x^{\prime },y^{\prime }\right).}$$

Because ${\displaystyle \left(Y_{\sigma }^{\prime }\right)^{\prime }}$ is canonically vector space-isomorphic to ${\displaystyle Y}$ (via the canonical map ${\displaystyle y\mapsto }$ value at ${\displaystyle y}$), ${\displaystyle B_{x^{\prime }}}$ will be identified as an element of ${\displaystyle Y,}$ which will be denoted by ${\displaystyle {\tilde {B}}_{x^{\prime }}\in Y.}$ This defines a map ${\displaystyle {\tilde {B}}:X^{\prime }\to Y}$ given by ${\displaystyle x^{\prime }\mapsto {\tilde {B}}_{x^{\prime }}}$ and so the canonical isomorphism is of course defined by ${\displaystyle J(B):={\tilde {B}}.}$

When ${\displaystyle L\left(X_{\sigma }^{\sigma };Y_{\sigma }\right)}$ is given the topology of uniform convergence on equicontinous subsets of ${\displaystyle X^{\prime },}$ the canonical map becomes a TVS-isomorphism[9]

$${\displaystyle J:{\mathcal {B}}_{\varepsilon }\left(X_{\sigma }^{\prime },Y_{\sigma }^{\prime }\right)\to L_{\varepsilon }\left(X_{\tau }^{\prime };Y\right).}$$

In particular, ${\displaystyle X\otimes _{\varepsilon }Y=B_{\varepsilon }\left(X_{\sigma }^{\prime },Y_{\sigma }^{\prime }\right)}$ can be canonically TVS-embedded into ${\displaystyle L_{\varepsilon }\left(X_{\tau }^{\prime };Y\right)}$; furthermore the image in ${\displaystyle L\left(X_{\sigma }^{\prime };Y_{\sigma }\right)}$ of ${\displaystyle X\otimes _{\varepsilon }Y=B_{\varepsilon }\left(X_{\sigma }^{\prime },Y_{\sigma }^{\prime }\right)}$ under the canonical map ${\displaystyle J}$ consists exactly of the space of continuous linear maps ${\displaystyle X_{\sigma \left(X^{\prime },X\right)}^{\prime }\to Y}$ whose image is finite dimensional.[4]

The inclusion ${\displaystyle L\left(X_{\tau }^{\prime };Y\right)\subseteq L\left(X_{b}^{\prime };Y\right)}$ always holds. If ${\displaystyle X}$ is normed then ${\displaystyle L_{\varepsilon }\left(X_{\tau }^{\prime };Y\right)}$ is in fact a topological vector subspace of ${\displaystyle L_{b}\left(X_{b}^{\prime };Y\right).}$ And if in addition ${\displaystyle Y}$ is Banach then so is ${\displaystyle L_{b}\left(X_{b}^{\prime };Y\right)}$ (even if ${\displaystyle X}$ is not complete).[4]

Suppose that

$${\displaystyle \operatorname {In} :X\otimes _{\varepsilon }Y\to X{\widehat {\otimes }}_{\varepsilon }Y}$$

denotes the TVS-embedding of ${\displaystyle X\otimes _{\varepsilon }Y}$ into its completion and let

$${\displaystyle {}^{t}\operatorname {In}$$ :\left(X{\widehat {\otimes }}_{\varepsilon }Y\right)_{b}^{\prime }\to \left(X\otimes _{\varepsilon }Y\right)_{b}^{\prime }}

be its transpose, which is a vector space-isomorphism. This identifies the continuous dual space of ${\displaystyle X\otimes _{\varepsilon }Y}$ as being identical to the continuous dual space of ${\displaystyle X{\widehat {\otimes }}_{\varepsilon }Y.}$

The identity map

$${\displaystyle \operatorname {Id} _{X\otimes Y}:X\otimes _{\pi }Y\to X\otimes _{\varepsilon }Y}$$

is continuous (by definition of the π-topology) so there exists a unique continuous linear extension

$${\displaystyle {\hat {I}}:X{\widehat {\otimes }}_{\pi }Y\to X{\widehat {\otimes }}_{\varepsilon }Y.}$$

If ${\displaystyle X}$ and ${\displaystyle Y}$ are Hilbert spaces then ${\displaystyle {\hat {I}}:X{\widehat {\otimes }}_{\pi }Y\to X{\widehat {\otimes }}_{\varepsilon }Y}$ is injective and the dual of ${\displaystyle X{\widehat {\otimes }}_{\varepsilon }Y}$ is canonically isometrically isomorphic to the vector space ${\displaystyle L^{1}\left(X;Y^{\prime }\right)}$ of nuclear operators from ${\displaystyle X}$ into ${\displaystyle Y}$ (with the trace norm).

There is a canonical map

$${\displaystyle K:X\otimes Y\to L\left(X^{\prime };Y\right)}$$

that sends ${\displaystyle z=\sum _{i=1}^{n}x_{i}\otimes y_{i}}$ to the linear map ${\displaystyle K(z):X^{\prime }\to Y}$ defined by

$${\displaystyle K(z)\left(x^{\prime }\right):=\sum _{i=1}^{n}x^{\prime }(x_{i})y_{i}\in Y,}$$

where it may be shown that the definition of ${\displaystyle K(z):X\to Y}$ does not depend on the particular choice of representation ${\textstyle \sum _{i=1}^{n}x_{i}\otimes y_{i}}$ of ${\displaystyle z.}$ The map

$${\displaystyle K:X\otimes _{\varepsilon }Y\to L_{b}\left(X_{b}^{\prime };Y\right)}$$

is continuous and when ${\displaystyle L_{b}\left(X_{b}^{\prime };Y\right)}$ is complete, it has a continuous extension

$${\displaystyle {\hat {K}}:X{\widehat {\otimes }}_{\varepsilon }Y\to L_{b}\left(X_{b}^{\prime };Y\right).}$$

When ${\displaystyle X}$ and ${\displaystyle Y}$ are Hilbert spaces then ${\displaystyle {\hat {K}}:X{\widehat {\otimes }}_{\varepsilon }Y\to L_{b}\left(X_{b}^{\prime };Y\right)}$ is a TVS-embedding and isometry (when the spaces are given their usual norms) whose range is the space of all compact linear operators from ${\displaystyle X}$ into ${\displaystyle Y}$ (which is a closed vector subspace of ${\displaystyle L_{b}\left(X^{\prime };Y\right).}$ Hence ${\displaystyle X{\widehat {\otimes }}_{\varepsilon }Y}$ is identical to space of compact operators from ${\displaystyle X^{\prime }}$ into ${\displaystyle Y}$ (note the prime on ${\displaystyle X}$). The space of compact linear operators between any two Banach spaces (which includes Hilbert spaces) ${\displaystyle X}$ and ${\displaystyle Y}$ is a closed subset of ${\displaystyle L_{b}(X;Y).}$[17]

Furthermore, the canonical map ${\displaystyle X{\widehat {\otimes }}_{\pi }Y\to X{\widehat {\otimes }}_{\varepsilon }Y}$ is injective when ${\displaystyle X}$ and ${\displaystyle Y}$ are Hilbert spaces. [17]

Denote the identity map by

$${\displaystyle \operatorname {Id} :X\otimes _{\pi }Y\to X\otimes _{\varepsilon }Y}$$

and let

$${\displaystyle {}^{t}\operatorname {Id}$$ :\left(X\otimes _{\varepsilon }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 ${\displaystyle X\otimes _{\varepsilon }Y}$ 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[18][19] — The dual ${\displaystyle J(X,Y)}$ of ${\displaystyle X{\widehat {\otimes }}_{\varepsilon }Y}$ 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 ${\displaystyle S}$ and ${\displaystyle T}$ 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 ${\displaystyle A}$ is an equicontinuous subset of ${\displaystyle J(X,Y)}$ 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 ${\displaystyle S\times T.}$

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

$${\displaystyle B_{\Lambda }\left(x,y^{\prime }\right):=\left(y^{\prime }\circ \Lambda \right)(x).}$$

A continuous map ${\displaystyle \Lambda :X\to Y}$ is called integral if its associated bilinear form is an integral bilinear form.[20] 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 subsets ${\displaystyle A^{\prime }}$ 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 ${\displaystyle \leq 1.}$

There is a canonical map ${\displaystyle K:X^{\prime }\otimes Y\to L(X;Y)}$ that sends ${\textstyle z=\sum _{i=1}^{n}x_{i}^{\prime }\otimes y_{i}}$ to the linear map ${\displaystyle K(z):X\to Y}$ defined by ${\textstyle K(z)(x):=\sum _{i=1}^{n}x_{i}^{\prime }(x)y_{i}\in Y,}$ where it may be shown that the definition of ${\displaystyle K(z):X\to Y}$ does not depend on the particular choice of representation ${\textstyle \sum _{i=1}^{n}x_{i}^{\prime }\otimes y_{i}}$ of ${\displaystyle z.}$

Throughout this section we fix some arbitrary (possibly uncountable) set ${\displaystyle A,}$ a TVS ${\displaystyle X,}$ and we let ${\displaystyle {\mathcal {F}}(A)}$ be the directed set of all finite subsets of ${\displaystyle A}$ directed by inclusion ${\displaystyle \subseteq .}$

Let ${\displaystyle \left(x_{\alpha }\right)_{\alpha \in A}}$ be a family of elements in a TVS ${\displaystyle X}$ and for every finite subset ${\displaystyle H\subseteq A,}$ let ${\textstyle x_{H}:=\sum _{i\in H}x_{i}.}$ We call ${\displaystyle \left(x_{\alpha }\right)_{\alpha \in A}}$ summable in ${\displaystyle X}$ if the limit ${\textstyle \lim _{H\in {\mathcal {F}}(A)}x_{H}}$ of the net ${\displaystyle \left(x_{H}\right)_{H\in {\mathcal {F}}(A)}}$ converges in ${\displaystyle X}$ to some element (any such element is called its sum). The set of all such summable families is a vector subspace of ${\displaystyle X^{A}}$ denoted by ${\displaystyle S.}$

We now define a topology on ${\displaystyle S}$ in a very natural way. This topology turns out to be the injective topology taken from ${\displaystyle l^{1}(A){\widehat {\otimes }}_{\varepsilon }X}$ and transferred to ${\displaystyle S}$ via a canonical vector space isomorphism (the obvious one). This is a common occurrence when studying the injective and projective tensor products of function/sequence spaces and TVSs: the "natural way" in which one would define (from scratch) a topology on such a tensor product is frequently equivalent to the injective or projective tensor product topology.

Let ${\displaystyle {\mathfrak {U}}}$ denote a base of convex balanced neighborhoods of 0 in ${\displaystyle X}$ and for each ${\displaystyle U\in {\mathfrak {U}},}$ let ${\displaystyle \mu _{U}:X\to \mathbb {R} }$ denote its Minkowski functional. For any such ${\displaystyle U}$ and any ${\displaystyle x=\left(x_{\alpha }\right)_{\alpha \in A}\in S,}$ let

$${\displaystyle q_{U}(x):=\sup _{x^{\prime }\in U^{\circ }}\sum _{\alpha \in A}\left|\left\langle x^{\prime },x_{\alpha }\right\rangle \right|}$$

where ${\displaystyle q_{U}}$ defines a seminorm on ${\displaystyle S.}$ The family of seminorms ${\displaystyle \left\{q_{U}:U\in {\mathfrak {U}}\right\}}$ generates a topology making ${\displaystyle S}$ into a locally convex space. The vector space ${\displaystyle S}$ endowed with this topology will be denoted by ${\displaystyle l^{1}(A,X).}$[21] The special case where ${\displaystyle X}$ is the scalar field will be denoted by ${\displaystyle l^{1}(A).}$

There is a canonical embedding of vector spaces ${\displaystyle l^{1}(A)\otimes X\to l^{1}(A,E)}$ defined by linearizing the bilinear map ${\displaystyle l^{1}(A)\times X\to l^{1}(A,E)}$ defined by ${\displaystyle \left(\left(r_{\alpha }\right)_{\alpha \in A},x\right)\mapsto \left(r_{\alpha }x\right)_{\alpha \in A}.}$[21]