Projective tensor product

• σ(X, X′) denotes the coarsest topology on X making every map in X′ continuous and ${\displaystyle X_{\sigma \left(X,X^{\prime }\right)}}$ or ${\displaystyle X_{\sigma }}$ denotes X endowed with this topology.
• σ(X′, X) denotes weak-* topology on X* and ${\displaystyle X_{\sigma \left(X^{\prime },X\right)}}$ or ${\displaystyle X_{\sigma }^{\prime }}$ denotes X′ endowed with this topology.
  • Note that every ${\displaystyle x_{0}\in X}$ induces a map ${\displaystyle X^{\prime }\to \mathbb {R} }$ defined by ${\displaystyle \lambda \mapsto \lambda \left(x_{0}\right)}$. σ(X′, X) is the coarsest topology on X′ making all such maps continuous.
• b(X, X′) denotes the topology of bounded convergence on X and ${\displaystyle X_{b\left(X,X^{\prime }\right)}}$ or ${\displaystyle X_{b}}$ denotes X endowed with this topology.
• b(X′, X) denotes the topology of bounded convergence on X′ or the strong dual topology on X′ and ${\displaystyle X_{b\left(X^{\prime },X\right)}}$ or ${\displaystyle X_{b}^{\prime }}$ denotes X′ endowed with this topology.
  • As usual, if X* 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 b(X′, X).

Let X and Y be vector spaces (no topology is needed yet) and let Bi(X, Y) be the space of all bilinear maps defined on ${\displaystyle X\times Y}$ and going into the underlying scalar field.

For every ${\displaystyle (x,y)\in X\times Y}$ define a canonical bilinear form by ${\displaystyle \chi _{(x,y)}}$ with domain Bi(X, Y) by ${\displaystyle \chi _{(x,y)}(u):=u(x,y)}$ for every ${\displaystyle u\in Bi(X,Y)}$. This induces a canonical map ${\displaystyle \chi :X\times Y\to Bi(X,Y)^{\#}}$ defined by ${\displaystyle \chi (x,y)=\chi _{(x,y)}}$, where ${\displaystyle Bi(X,Y)^{\#}}$ denotes the algebraic dual of Bi(X, Y). If we denote the span of the range of 𝜒 by X ⊗ Y then X ⊗ Y together with 𝜒 forms a tensor product of X and Y (where x ⊗ y = 𝜒(x, y)). This gives us a canonical tensor product of X and Y.

If Z is any other vector space then the mapping ${\displaystyle Li(X\otimes Y;Z)\to Bi(X,Y;Z)}$ given by ${\displaystyle u\mapsto u\circ \chi }$ is an isomorphism of vector spaces. In particular, this allows us to identify the algebraic dual of X ⊗ Y with the space of bilinear forms on ${\displaystyle X\times Y}$.[4] Moreover, if X and Y are locally convex topological vector spaces (TVSs) and if X ⊗ Y is given the 𝜋-topology then for every locally convex TVS Z, this map restricts to a vector space isomorphism ${\displaystyle L(X\otimes _{\pi }Y;Z)\to B(X,Y;Z)}$ from the space of continuous linear mappings onto the space of continuous bilinear mappings.[5] In particular, the continuous dual of X ⊗ Y can be canonically identified with the space B(X, Y) of continuous bilinear forms on ${\displaystyle X\times Y}$; furthermore, under this identification the equicontinuous subsets of B(X, Y) are the same as the equicontinuous subsets of ${\displaystyle (X\otimes _{\pi }Y)^{\prime }}$.[5]

Throughout we will let X and Y be locally convex topological vector spaces (local convexity allows us to define useful topologies[5]). If p is a seminorm on X then ${\displaystyle C_{p}:=\{x\in X:p(x)\leq 1\}}$ will be its closed unit ball.

If p is a seminorm on X and q is a seminorm on Y then we can define the tensor product of p and q to be the map p ⊗ q defined on X ⊗ Y by

${\displaystyle (p\otimes q)(b):=\inf _{r>0,b\in rW}r}$

where W is the balanced convex hull of ${\displaystyle C_{p}\otimes C_{q}=\left\{x\otimes y:p(x)\leq 1,q(y)\leq 1\right\}}$. Given b in X ⊗ Y, this can also be expressed as[6]

${\displaystyle (p\otimes q)(b):=\inf \sum _{i}p(x_{i})q(y_{i})}$

where the infimum is taken over all finite sequences ${\displaystyle x_{1},\ldots ,x_{n}\in X}$ and ${\displaystyle y_{1},\ldots ,y_{n}\in Y}$ (of the same length) such that ${\displaystyle b=x_{1}\otimes y_{1}+\cdots +x_{n}\otimes y_{n}}$ (recall that it may not be possible to express b as a simple tensor). If b = x⊗y then we have

${\displaystyle (p\otimes q)(x\otimes y)=p(x)q(y)}$.

The seminorm p ⊗ q is a norm if and only if both p and q are norms.[7]

If the topology of X (resp. Y) is given by the family of seminorms ${\displaystyle \left(p_{\alpha }\right)}$ (resp. ${\displaystyle \left(q_{\beta }\right)}$) then ${\displaystyle X\otimes _{\pi }Y}$ is a locally convex space whose topology if given by the family of all possible tensor products of the two families (i.e. by ${\displaystyle \left(p_{\alpha }\otimes q_{\beta }\right)}$). In particular, if X and Y are seminormed spaces with seminorms p and q, respectively, then ${\displaystyle X\otimes _{\pi }Y}$ is a seminormable space whose topology is defined by the seminorm p ⊗ q.[8] If (X, p) and (Y, q) are normed spaces then ${\displaystyle \left(X\otimes Y,p\otimes q\right)}$ is also a normed space, called the projective tensor product of (X, p) and (Y, q), where the topology induced by p ⊗ q is the same as the π-topology.[8]

If W is a convex subset of ${\displaystyle X\otimes Y}$ then W is a neighborhood of 0 in ${\displaystyle X\otimes _{\pi }Y}$ if and only if the preimage of W under the map ${\displaystyle (x,y)\mapsto x\otimes y}$ is a neighborhood of 0; equivalent, if and only if there exist open subsets ${\displaystyle U\subseteq X}$ and ${\displaystyle V\subseteq Y}$ such that this preimage contains ${\displaystyle U\otimes V:=\{u\otimes v:u\in U,v\in V\}}$.[9] It follows that if ${\displaystyle \left(U_{\alpha }\right)}$ and ${\displaystyle \left(V_{\beta }\right)}$ are neighborhood bases of the origin in X and Y, respectively, then the set of convex hulls of all possible set ${\displaystyle U_{\alpha }\otimes V_{\beta }}$ form a neighborhood basis of the origin in ${\displaystyle X\otimes _{\pi }Y}$.

If 𝜏 is a locally convex TVS topology on X ⊗ Y (X ⊗ Y with this topology will be denoted by ${\displaystyle X\otimes _{\tau }Y}$), then 𝜏 is equal to the π-topology if and only if it has the following property:[10]

For every locally convex TVS Z, if I is the canonical map from the space of all bilinear mappings of the form ${\displaystyle X\times Y\to Z}$, going into the space of all linear mappings of ${\displaystyle X\otimes Y\to Z}$, then when the domain of I is restricted to ${\displaystyle B\left(X,Y;Z\right)}$ then the range of this restriction is the space ${\displaystyle L\left(X\otimes _{\tau }Y;Z\right)}$ of continuous linear operators ${\displaystyle X\otimes _{\tau }Y\to Z}$.

In particular, the continuous dual space of ${\displaystyle X\otimes _{\pi }Y}$ is canonically isomorphic to the space ${\displaystyle B(X,Y)}$, the space of continuous bilinear forms on ${\displaystyle X\times Y}$.

Note that the canonical vector space isomorphism ${\displaystyle I:B\left(X_{\sigma }^{\prime },Y_{\sigma }^{\prime };Z\right)\to L\left(X\otimes _{\pi }Y;Z\right)}$ preserves equicontinuous subsets. Since ${\displaystyle B(X,Y)}$ is canonically isomorphic to the continuous dual of ${\displaystyle X\otimes _{\pi }Y}$, place on X ⊗ Y the topology of uniform convergence on equicontinuous subsets of ${\displaystyle \left(X\otimes _{\pi }Y\right)^{\prime }}$; this topology is identical to the π-topology.[10]

Let X and Y be locally convex TVSs.

• If both X and Y are Hausdorff (resp. locally convex, metrizable, semi-metrizable, normable, semi-normable) then so is ${\displaystyle X\otimes _{\pi }Y}$.

Recall that in a Hausdorff locally convex space X, a sequence ${\displaystyle \left(x_{i}\right)_{i=1}^{\infty }}$ in X is absolutely convergent if ${\displaystyle \sum _{i=1}^{\infty }p\left(x_{i}\right)<\infty }$ for every continuous seminorm p on X.[13] We write ${\displaystyle x=\sum _{i=1}^{\infty }x_{i}}$ if the sequence of partial sums ${\displaystyle \left(\sum _{i=1}^{n}x_{i}\right)_{n=1}^{\infty }}$ converges to x in X.[13]

The following fundamental result in the theory of topological tensor products is due to Alexander Grothendieck.[14]

Theorem Let X and Y be metrizable locally convex TVSs and let ${\displaystyle z\in X{\widehat {\otimes }}_{\pi }Y}$. Then z is the sum of an absolutely convergent series

${\displaystyle z=\sum _{i=1}^{\infty }\lambda _{i}x_{i}\otimes y_{i}}$

where ${\displaystyle \sum _{i=1}^{\infty }|\lambda _{i}|<\infty }$, and ${\displaystyle \left(x_{i}\right)_{i=1}^{\infty }}$ and ${\displaystyle \left(y_{i}\right)_{i=1}^{\infty }}$ are null sequences in X and Y, respectively.

The next theorem shows that it's possible to make the representation of z independent of the sequences ${\displaystyle \left(x_{i}\right)_{i=1}^{\infty }}$ and ${\displaystyle \left(y_{i}\right)_{i=1}^{\infty }}$.

Theorem[15] Let X and Y be Frechet spaces and let U (resp. V) be a balanced open neighborhood of the origin in X (resp. in Y). Let ${\displaystyle K_{0}}$ be a compact subset of the convex balanced hull of ${\displaystyle U\otimes V:=\{u\otimes v:u\in U,v\in V\}}$. There exists a compact subset ${\displaystyle K_{1}}$ of the unit ball in ${\displaystyle l^{1}}$ and sequences ${\displaystyle \left(x_{i}\right)_{i=1}^{\infty }}$ and ${\displaystyle \left(y_{i}\right)_{i=1}^{\infty }}$ contained in U and V, respectively, converging to the origin such that for every ${\displaystyle z\in K_{0}}$ there exists some ${\displaystyle \left(\lambda _{i}\right)_{i=1}^{\infty }\in K_{1}}$ such that

${\displaystyle z=\sum _{i=1}^{\infty }\lambda _{i}x_{i}\otimes y_{i}}$.

Let ${\displaystyle {\mathfrak {B}}_{X}}$ and ${\displaystyle {\mathfrak {B}}_{Y}}$ denote the families of all bounded subsets of X and Y, respectively. Since the continuous dual space of ${\displaystyle X{\widehat {\otimes }}_{\pi }Y}$ is the space of continuous bilinear forms ${\displaystyle B(X,Y)}$, we can place on ${\displaystyle B(X,Y)}$ the topology of uniform convergence on sets in ${\displaystyle {\mathfrak {B}}_{X}\times {\mathfrak {B}}_{Y}}$, which is also called the topology of bi-bounded convergence. This topology is coarser than the strong topology ${\displaystyle b\left(B(X,Y),X{\widehat {\otimes }}_{\pi }Y\right)}$, and in (Grothendieck 1955) harv error: no target: CITEREFGrothendieck1955 (help), Alexander Grothendieck was interested in when these two topologies were identical. This question is equivalent to the questions: Given a bounded subset ${\displaystyle B\subseteq X{\widehat {\otimes }}_{\pi }}$, do there exist bounded subsets ${\displaystyle B_{1}\subseteq X}$ and ${\displaystyle B_{2}\subseteq Y}$ such that B is a subset of the closed convex hull of ${\displaystyle B_{1}\otimes B_{2}:=\{b_{1}\otimes b_{2}:b_{1}\in B_{1},b_{2}\in B_{2}\}}$?

Grothendieck proved that these topologies are equal when X and Y are both Banach spaces or both are DF-spaces (a class of spaces introduced by Grothendieck[16]). They are also equal when both spaces are Fréchet with one of them being nuclear.[12]

Given a locally convex TVS X, ${\displaystyle X^{\prime }}$ is assumed to have the strong topology (so ${\displaystyle X^{\prime }=X_{b}^{\prime }}$) and unless stated otherwise, the same is true of the bidual ${\displaystyle X^{\prime \prime }}$ (so ${\displaystyle X^{\prime \prime }=\left(X_{b}^{\prime }\right)_{b}^{\prime }}$. Alexander Grothendieck characterized the strong dual and bidual for certain situations:

Theorem (Grothendieck):[17] Let N and Y be locally convex TVSs with N nuclear. Assume that both N and Y are Fréchet spaces or else that they are both DF-spaces. Then:

1. The strong dual of ${\displaystyle N{\widehat {\otimes }}_{\pi }Y}$ can be identified with ${\displaystyle N_{b}^{\prime }{\widehat {\otimes }}_{\pi }Y_{b}^{\prime }}$;
2. The dibual of ${\displaystyle N{\widehat {\otimes }}_{\pi }Y}$ can be identified with ${\displaystyle N{\widehat {\otimes }}_{\pi }Y^{\prime \prime }}$;
3. If in addition Y is reflexive then ${\displaystyle N{\widehat {\otimes }}_{\pi }Y}$ (and hence ${\displaystyle N_{b}^{\prime }{\widehat {\otimes }}_{\pi }Y_{b}^{\prime }}$) is a reflexive space;
4. Every separately continuous bilinear form on ${\displaystyle N_{b}^{\prime }\times Y_{b}^{\prime }}$ is continuous;
5. The strong dual of ${\displaystyle L_{b}\left(X_{b}^{\prime },Y\right)}$ can be identified with ${\displaystyle N_{b}^{\prime }{\widehat {\otimes }}_{\pi }Y_{b}^{\prime }}$, so in particular if Y is reflexive then so is ${\displaystyle L_{b}\left(X_{b}^{\prime },Y\right)}$.

• For all normed spaces ${\displaystyle \left(Z,\|\cdot \|\right)}$, the canonical vector space isomorphism of ${\displaystyle B(X,Y;Z)}$ onto ${\displaystyle L\left(X\otimes _{\pi }Y;Z\right)}$ is an isometry.[26]
• Suppose that ${\displaystyle \|\cdot \|}$ is a norm on ${\displaystyle X\otimes Y}$ and let the TVS topology that it induces on ${\displaystyle X\otimes Y}$ be denoted by ${\displaystyle \alpha }$. If the canonical linear map of ${\displaystyle B(X,Y)}$ into ${\displaystyle \left(X\otimes Y\right)^{\#}}$, which is the algebraic dual of ${\displaystyle X\otimes Y}$, is an isometry of ${\displaystyle B(X,Y)}$ onto ${\displaystyle \left(X\otimes _{\alpha }Y\right)^{\prime }}$, then ${\displaystyle \|\cdot \|=\|\cdot \|_{\pi }}$.[26]

• In general, the projective tensor product does not respect subspaces (e.g. if ${\displaystyle Z}$ is a vector subspace of ${\displaystyle X}$ then the TVS ${\displaystyle Z\otimes _{\pi }Y}$ has in general a coarser topology than the subspace topology inherited from ${\displaystyle X\otimes _{\pi }Y}$).[27]
• Suppose that ${\displaystyle E}$ and ${\displaystyle F}$ are complemented subspaces of X and Y, respectively. Then ${\displaystyle E\otimes F}$ is a complemented subvector space of ${\displaystyle X\otimes _{\pi }Y}$ and the projective norm on ${\displaystyle E\otimes _{\pi }F}$ is equivalent to the projective norm on ${\displaystyle X\otimes _{\pi }Y}$ restricted to the subspace ${\displaystyle E\otimes F}$; Furthermore, if E and F are complemented by projections of norm 1, then ${\displaystyle E\otimes _{\pi }F}$ is complemented by a projection of norm 1.[27]
• If ${\displaystyle I:X\otimes _{\pi }Y\to Z}$ is an isometric embedding into a Banach space Z, then its unique continuous extension ${\displaystyle I:X{\widehat {\otimes }}_{\pi }Y\to Z}$ is also an isometric embedding.
• If ${\displaystyle \alpha :W\to X}$ and ${\displaystyle \beta :Y\to Z}$ are quotient operators between Banach spaces, then so is ${\displaystyle \alpha {\widehat {\otimes }}_{\pi }\beta :W{\widehat {\otimes }}_{\pi }Y\to X{\widehat {\otimes }}_{\pi }Z}$.[28]
  • Recall a continuous linear operator ${\displaystyle \beta :Y\to Z}$ between normed spaces is a quotient operator if it is surjective and it maps the open unit ball of Y into the open unit ball of Z, or equivalently if for all ${\displaystyle z\in Z}$, ${\displaystyle \|z\|=\inf _{y\in \beta ^{-1}\left(z\right)}\|y\|}$. [28]
• Let E and F be vector subspaces of the Banach spaces X and Y, respectively. Then ${\displaystyle E{\widehat {\otimes }}_{\pi }F}$ is a TVS-subspace of ${\displaystyle X{\widehat {\otimes }}_{\pi }Y}$ if and only if every bounded bilinear form on ${\displaystyle E\times F}$ extends to a continuous bilinear form on ${\displaystyle X\times Y}$ with the same norm.[29]

Assuming that X and Y are Banach spaces over the field ${\displaystyle \mathbb {F} }$, one may define a dual system between ${\displaystyle X{\widehat {\otimes }}_{\pi }Y}$ and ${\displaystyle L_{b}\left(X;Y^{\prime }\right)}$ with the duality map ${\displaystyle \left\langle \cdot ,\cdot \right\rangle :L_{b}\left(X;Y^{\prime }\right)\times \left(X{\widehat {\otimes }}_{\pi }Y\right)\to \mathbb {F} }$ defined by ${\displaystyle \left\langle u,z\right\rangle$ :=\operatorname {Tr} \left(\left(u\,{\widehat {\otimes }}_{\pi }\operatorname {Id} _{Y}\right)\left(z\right)\right)} , where ${\displaystyle \operatorname {Id} _{Y}:Y\to Y}$ is the identity map and ${\displaystyle u\,{\widehat {\otimes }}_{\pi }\operatorname {Id} _{Y}:X{\widehat {\otimes }}_{\pi }Y\to Y^{\prime }{\widehat {\otimes }}_{\pi }Y}$ is the unique continuous extension of the continuous map ${\displaystyle u\,\otimes _{\pi }\operatorname {Id} _{Y}:X\otimes _{\pi }Y\to Y^{\prime }\otimes _{\pi }Y}$. If we write ${\displaystyle z=\sum _{i=1}^{\infty }\lambda _{i}x_{i}\otimes y_{i}}$ with ${\displaystyle \sum _{i=1}^{\infty }\left|\lambda _{i}\right|<\infty }$ and the sequences ${\displaystyle \left(x_{i}\right)_{i=1}^{\infty }}$ and ${\displaystyle \left(y_{i}\right)_{i=1}^{\infty }}$ each converging to zero, then we have

${\displaystyle \left\langle u,z\right\rangle =\sum _{i=1}^{\infty }\lambda _{i}\left\langle u\left(x_{i}\right),y_{i}\right\rangle }$.[31]

Assuming that X and Y are Banach spaces, then the map ${\displaystyle I:X_{b}^{\prime }\otimes _{\pi }Y\to L_{b}(X;Y)}$ has norm ${\displaystyle 1}$ so it has a continuous extension to a map ${\displaystyle {\hat {I}}:X_{b}^{\prime }{\widehat {\otimes }}_{\pi }Y\to L_{b}(X;Y)}$, where it is known that this map is not necessarily injective.[32] The range of this map is denoted by ${\displaystyle L^{1}(X;Y)}$ and its elements are called nuclear operators.[33] ${\displaystyle L^{1}(X;Y)}$ is TVS-isomorphic to ${\displaystyle \left(X_{b}^{\prime }{\widehat {\otimes }}_{\pi }Y\right)/\operatorname {ker} {\hat {I}}}$ and the norm on this quotient space, when transferred to elements of ${\displaystyle L^{1}(X;Y)}$ via the induced map ${\displaystyle {\hat {I}}:\left(X_{b}^{\prime }{\widehat {\otimes }}_{\pi }Y\right)/\operatorname {ker} {\hat {I}}\to L^{1}(X;Y)}$, is called the trace-norm and is denoted by ${\displaystyle \|\cdot \|_{\operatorname {Tr} }}$.

Suppose that U is a convex balanced closed neighborhood of the origin in X and B is a convex balanced bounded Banach disk in Y with both X and Y locally convex spaces. Let ${\displaystyle p_{U}(x)=\inf _{r>0,x\in rU}r}$ and let ${\displaystyle \pi :X\to X/p_{U}^{-1}(0)}$ be the canonical projection. One can define the auxiliary Banach space ${\displaystyle {\hat {X}}_{U}}$ with the canonical map ${\displaystyle {\hat {\pi }}_{U}:X\to {\hat {X}}_{U}}$ whose image, ${\displaystyle X/p_{U}^{-1}(0)}$, is dense in ${\displaystyle {\hat {X}}_{U}}$ as well as the auxiliary space ${\displaystyle F_{B}=\operatorname {span} B}$ normed by ${\displaystyle p_{B}(y)=\inf _{r>0,y\in rB}r}$ and with a canonical map ${\displaystyle \iota :F_{B}\to F}$ being the (continuous) canonical injection. Given any continuous linear map ${\displaystyle T:{\hat {X}}_{U}\to Y_{B}}$ one obtains through composition the continuous linear map ${\displaystyle {\hat {\pi }}_{U}\circ T\circ \iota :X\to Y}$; thus we have an injection ${\displaystyle L\left({\hat {X}}_{U};Y_{B}\right)\to L(X;Y)}$ and we henceforth use this map to identify ${\displaystyle L\left({\hat {X}}_{U};Y_{B}\right)}$ as a subspace of ${\displaystyle L(X;Y)}$.[33]

Definition: Let X and Y be Hausdorff locally convex spaces. The union of all ${\displaystyle L^{1}\left({\hat {X}}_{U};Y_{B}\right)}$ as U ranges over all closed convex balanced neighborhoods of the origin in X and B ranges over all bounded Banach disks in Y, is denoted by ${\displaystyle L^{1}(X;Y)}$ and its elements are call nuclear mappings of X into Y.[33]

When X and Y are Banach spaces, then this new definition of nuclear mapping is consistent with the original one given for the special case where X and Y are Banach spaces.

Every nuclear operator is an integral operator but the converse is not necessarily true. However, every integral operator between Hilbert spaces is nuclear.[34]

Theorem:[35] Let X and Y be Hilbert spaces and endow ${\displaystyle L^{1}\left(X;Y\right)}$ (the space of nuclear linear operators) with the trace-norm. When the space of compact linear operators ${\displaystyle X\to Y}$ is equipped with the operator norm (induced by the usual norm on ${\displaystyle L_{b}(X;Y)}$) then its (strong) dual is ${\displaystyle L^{1}\left(X;Y\right)}$ (with the trace-norm) and its bidual is the space ${\displaystyle L\left(X;Y\right)}$ of all continuous linear operators ${\displaystyle X\to Y}$.

Assuming that X and Y are Banach spaces, then the map ${\displaystyle J:X_{b}^{\prime }\otimes _{\pi }Y_{b}^{\prime }\to {\mathcal {B}}_{b}(X,Y)}$ has norm ${\displaystyle 1}$ so it has a continuous extension to a map ${\displaystyle {\hat {J}}:X_{b}^{\prime }{\widehat {\otimes }}_{\pi }Y_{b}^{\prime }\to {\mathcal {B}}_{b}(X,Y)}$. The range of this map is denoted by ${\displaystyle B^{1}(X,Y)}$ and its elements are called nuclear bilinear forms.[36] ${\displaystyle B^{1}(X,Y)}$ is TVS-isomorphic to ${\displaystyle \left(X_{b}^{\prime }{\widehat {\otimes }}_{\pi }Y_{b}^{\prime }\right)/\operatorname {ker} {\hat {J}}}$ and the norm on this quotient space, when transferred to elements of ${\displaystyle B^{1}(X,Y)}$ via the induced map ${\displaystyle {\hat {J}}:\left(X_{b}^{\prime }{\widehat {\otimes }}_{\pi }Y_{b}^{\prime }\right)/\operatorname {ker} {\hat {J}}\to B^{1}(X,Y)}$, is called the nuclear-norm and is denoted by ${\displaystyle \|\cdot \|_{\operatorname {N} }}$.

Suppose that X and Y are Banach spaces and that ${\displaystyle N}$ is a continuous bilinear from on ${\displaystyle X\times Y}$.

• The following are equivalent:

1. ${\displaystyle N}$ is nuclear.
2. There exist bounded sequences ${\displaystyle \left(x_{i}^{\prime }\right)_{i=1}^{\infty }}$ in ${\displaystyle X_{b}^{\prime }}$ and ${\displaystyle \left(y_{i}^{\prime }\right)_{i=1}^{\infty }}$ in ${\displaystyle Y_{b}^{\prime }}$ such that ${\displaystyle \sum _{i=1}^{\infty }\|x_{i}^{\prime }\|\|y_{i}^{\prime }\|<\infty }$ and ${\displaystyle N}$ is equal to the mapping:[37] ${\displaystyle N(x,y)=\sum _{i=1}^{\infty }x_{i}^{\prime }(x)y_{i}^{\prime }(y)}$ for all ${\displaystyle (x,y)\in X\times Y}$.