projective tensor product

The strongest locally convex topological vector space (TVS) topology on $X \otimes Y,$ the tensor product of two locally convex TVSs, making the canonical map $\cdot \otimes \cdot : X \times Y \to X \otimes Y$ (defined by sending $(x, y) \in X \times Y$ to $x \otimes y$) continuous is called the projective topology or the π-topology. When $X \otimes Y$ is endowed with this topology then it is denoted by $X \otimes_ Y$ and called the projective tensor product of $X$ and $Y.$

Preliminaries

Throughout let $X, Y,$ and $Z$ be topological vector spaces and $L : X \to Y$ be a linear map.

* $L : X \to Y$ is a topological homomorphism or homomorphism, if it is linear, continuous, and $L : X \to \operatorname L$ is an open map, where $\operatorname L,$ the image of $L,$ has the subspace topology induced by $Y.$
** If $S \subseteq X$ is a subspace of $X$ then both the quotient map $X \to X / S$ and the canonical injection $S \to X$ are homomorphisms. In particular, any linear map $L : X \to Y$ can be canonically decomposed as follows: $X \to X / \operatorname L \overset \operatorname L \to Y$ where $L_0(x + \ker L) := L(x)$ defines a bijection.
* The set of continuous linear maps $X \to Z$ (resp. continuous bilinear maps $X \times Y \to Z$) will be denoted by $L(X; Z)$ (resp. $B(X, Y; Z)$) where if $Z$ is the scalar field then we may instead write $L(X)$ (resp. $B(X, Y)$).
* We will denote the continuous dual space of $X$ by $X^$ and the algebraic dual space (which is the vector space of all linear functionals on $X,$ whether continuous or not) by $X^.$
** To increase the clarity of the exposition, we use the common convention of writing elements of $X^$ with a prime following the symbol (e.g. $x^$ denotes an element of $X^$ and not, say, a derivative and the variables $x$ and $x^$ need not be related in any way).
* A linear map $L : H \to H$ from a Hilbert space into itself is called positive if $\langle L(x), x \rangle \geq 0$ for every $x \in H.$ In this case, there is a unique positive map $r : H \to H,$ called the square-root of $L,$ such that $L = r \circ r.$
** If $L : H_1 \to H_2$ is any continuous linear map between Hilbert spaces, then $L^* \circ L$ is always positive. Now let $R : H \to H$ denote its positive square-root, which is called the absolute value of $L.$ Define $U : H_1 \to H_2$ first on $\operatorname R$ by setting $U(x) = L(x)$ for $x = R \left(x_1\right) \in \operatorname R$ and extending $U$ continuously to $\overline,$ and then define $U$ on $\operatorname R$ by setting $U(x) = 0$ for $x \in \operatorname R$ and extend this map linearly to all of $H_1.$ The map $U\big\vert_ : \operatorname R \to \operatorname L$ is a surjective isometry and $L = U \circ R.$
* A linear map $\Lambda : X \to Y$ is called compact or completely continuous if there is a neighborhood $U$ of the origin in $X$ such that $\Lambda(U)$ is precompact in $Y.$
** In a Hilbert space, positive compact linear operators, say $L : H \to H$ have a simple spectral decomposition discovered at the beginning of the 20th century by Fredholm and F. Riesz:
::There is a sequence of positive numbers, decreasing and either finite or else converging to 0, $r_1 > r_2 > \cdots > r_k > \cdots$ and a sequence of nonzero finite dimensional subspaces $V_i$ of $H$ ($i = 1, 2, \ldots$) with the following properties: (1) the subspaces $V_i$ are pairwise orthogonal; (2) for every $i$ and every $x \in V_i,$ $L(x) = r_i x$; and (3) the orthogonal of the subspace spanned by $\cup_i V_i$ is equal to the kernel of $L.$

Notation for topologies

* $\sigma\left(X, X^\right)$ denotes the coarsest topology on $X$ making every map in $X^$ continuous and $X_$ or $X_$ denotes $X$ endowed with this topology.
* $\sigma\left(X^, X\right)$ denotes weak-* topology on $X^$ and $X_$ or $X^_$ denotes $X^$ endowed with this topology.
** Every $x_0 \in X$ induces a map $X^ \to \R$ defined by $\lambda \mapsto \lambda \left(x_0\right).$ $\sigma\left(X^, X\right)$ is the coarsest topology on $X^$ making all such maps continuous.
* $b\left(X, X^\right)$ denotes the topology of bounded convergence on $X$ and $X_$ or $X_b$ denotes $X$ endowed with this topology.
* $b\left(X^, X\right)$ denotes the topology of bounded convergence on $X^$ or the strong dual topology on $X^$ and $X_$ or $X^_b$ 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 $b\left(X^, X\right).$

A canonical tensor product as a subspace of the dual of Bi(X, Y)

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

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

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

The projective tensor product

Tensor product of seminorms

Throughout we will let $X$ and $Y$ be locally convex topological vector spaces (local convexity allows us to define useful topologies).
If $p$ is a seminorm on $X$ then $C_p := \$ 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 $p \otimes q$ defined on $X \otimes Y$ by
$(p \otimes q)(b) := \inf_ r$
where $W$ is the balanced convex hull of $C_p \otimes C_q = \left\.$
Given $b$ in $X \otimes Y,$ this can also be expressed as
$(p \otimes q)(b) := \inf \sum_i p(x_i) q(y_i)$
where the infimum is taken over all finite sequences $x_1, \ldots, x_n \in X$ and $y_1, \ldots, y_n \in Y$ (of the same length) such that $b = x_1 \otimes y_1 + \cdots + x_n \otimes y_n$ (recall that it may not be possible to express $b$ as a simple tensor).
If $b = x \otimes y$ then we have
$(p \otimes q)(x \otimes y) = p(x) q(y).$
The seminorm $p \otimes q$ is a norm if and only if both $p$ and $q$ are norms.

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

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

Universal property

If $\tau$ is a locally convex TVS topology on $X \otimes Y$ ($X \otimes Y$ with this topology will be denoted by $X \otimes_ Y$), then $\tau$ is equal to the π-topology if and only if it has the following property:
:For every locally convex TVS $Z,$ if $I$ is the canonical map from the space of all bilinear mappings of the form $X \times Y \to Z,$ going into the space of all linear mappings of $X \otimes Y \to Z,$ then when the domain of $I$ is restricted to $B\left(X, Y; Z\right)$ then the range of this restriction is the space $L\left(X \otimes_ Y; Z\right)$ of continuous linear operators $X \otimes_ Y \to Z.$

In particular, the continuous dual space of $X \otimes_ Y$ is canonically isomorphic to the space $B(X, Y),$ the space of continuous bilinear forms on $X \times Y.$

The π-topology

Note that the canonical vector space isomorphism $I : B\left(X_^, Y_^; Z\right) \to L\left(X \otimes_ Y; Z\right)$ preserves equicontinuous subsets. Since $B(X, Y)$ is canonically isomorphic to the continuous dual of $X \otimes_ Y,$ place on $X \otimes Y$ the topology of uniform convergence on equicontinuous subsets of $\left(X \otimes_ Y\right)^$; this topology is identical to the π-topology.

Preserved properties

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 $X \otimes_ Y.$

Completion

In general, the space $X \otimes_ Y$ is not complete, even if both $X$ and $Y$ are complete (in fact, if $X$ and $Y$ are both infinite-dimensional Banach spaces then $X \otimes_ Y$ is necessarily complete). However, $X \otimes_ Y$ can always be linearly embedded as a dense vector subspace of some complete locally convex TVS, which is generally denoted by $X \widehat_ Y,$ via a linear topological embedding. Explicitly, this means that there is a continuous linear injection $\operatorname : X \otimes_ Y \to X \widehat_ Y$ whose image is dense in $X \widehat_ Y$ and that is a TVS-isomorphism onto its image. Using this map, $X \otimes_ Y$ is identified as a subspace of $X \widehat_ Y.$

The continuous dual space of $X \widehat_ Y$ is the same as that of $X \otimes_ Y,$ namely the space of continuous bilinear forms $B(X, Y).$:

Any continuous map on $X \otimes_ Y$ can be extended to a unique continuous map on $X \widehat_ Y.$ In particular, if $u : X_1 \to Y_1$ and $v : X_2 \to Y_2$ are continuous linear maps between locally convex spaces then their tensor product $u \otimes v : X_1 \otimes_ X_2 \to Y_1 \otimes_ Y_2 \subseteq Y_1 \widehat_ Y_2,$ which is necessarily continuous, can be extended to a unique continuous linear function $u \widehat_ v : X_1 \widehat_ X_2 \to Y_1 \widehat_ Y_2,$ which may also be denoted by $u \widehat v$ if no ambiguity would arise.

Note that if $X$ and $Y$ are metrizable then so are $X \otimes_ Y$ and $X \widehat_ Y,$ where in particular $X \widehat_ Y$ will be an F-space.

Grothendieck's representation of elements of $X \widehat_ Y$

In a Hausdorff locally convex space $X,$ a sequence $\left(x_i\right)_^$ in $X$ is absolutely convergent if $\sum_^ p \left(x_i\right) < \infty$ for every continuous seminorm $p$ on $X.$ We write $x = \sum_^ x_i$ if the sequence of partial sums $\left(\sum_^n x_i\right)_^$ converges to $x$ in $X.$

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

The next theorem shows that it is possible to make the representation of $z$ independent of the sequences $\left(x_i\right)_^$ and $\left(y_i\right)_^.$

Topology of bi-bounded convergence

Let $\mathfrak_X$ and $\mathfrak_Y$ denote the families of all bounded subsets of $X$ and $Y,$ respectively. Since the continuous dual space of $X \widehat_ Y$ is the space of continuous bilinear forms $B(X, Y),$ we can place on $B(X, Y)$ the topology of uniform convergence on sets in $\mathfrak_X \times \mathfrak_Y,$ which is also called the topology of bi-bounded convergence. This topology is coarser than the strong topology $b\left(B(X, Y), X \widehat_ Y\right),$ and in , Alexander Grothendieck was interested in when these two topologies were identical.
This question is equivalent to the questions: Given a bounded subset $B \subseteq X \widehat_ Y,$ do there exist bounded subsets $B_1 \subseteq X$ and $B_2 \subseteq Y$ such that $B$ is a subset of the closed convex hull of $B_1 \otimes 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). They are also equal when both spaces are Fréchet with one of them being nuclear.

Strong dual and bidual

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

Properties

* $X \otimes_ Y$ is Hausdorff if and only if both $X$ and $Y$ are Hausdorff.
* Suppose that $u : X_1 \to Y_1$ and $v : X_2 \to Y_2$ are two linear maps between locally convex spaces. If both $u$ and $v$ are continuous then so is their tensor product $u \otimes v : X_1 \otimes_ X_2 \to Y_1 \otimes_ Y_2.$
** $u \otimes v : X_1 \otimes_ X_2 \to Y_1 \otimes_ Y_2$ has a unique continuous extension to $X_1 \widehat_ X_2$ denoted by $u \widehat v : X_1 \widehat_ X_2 \to Y_1 \widehat_ Y_2.$
** If in addition both $u$ and $v$ are TVS-homomorphisms and the image of each map is dense in its codomain, then $u \widehat_ v : X_1 \widehat_ X_2 \to Y_1 \widehat_ Y_2$ is a homomorphism whose image is dense in $Y_1 \widehat_ Y_2$; if $X_1$ and $Y_1$ are both metrizable then this image is equal to all of $Y_1 \widehat_ Y_2.$
** There are examples of $u$ and $v$ such that both $u$ and $v$ are surjective homomorphisms but $u \widehat_ v : X_1 \widehat_ X_2 \to Y_1 \widehat_ Y_2$ is surjective.
** There are examples of $u$ and $v$ such that both $u$ and $v$ are TVS-embeddings but $u \widehat_ v : X_1 \widehat_ X_2 \to Y_1 \widehat_ Y_2$ is a TVS-embedding. In order for $u \widehat_ v$ to be a TVS-embedding, it is necessary and sufficient to additionally show that every equicontinuous subset of $B\left(X_1, X_2\right)$ is the image under $^\left(u \widehat_ v\right)$ of an equicontinuous subset of $B\left(Y_1, Y_2\right).$
** If all four spaces are normed then $\, u \otimes v \, _ = \, u \, \, v \, .$
* The π-topology is finer than the ε-topology (since the canonical bilinear map $X \times Y \to X \otimes_ Y$ is continuous).
* If $X$ and $Y$ are Frechet spaces then $X \otimes_ Y$ is barelled.
* If $Y$ and $\left(X_\right)$ are locally convex spaces then the canonical map $\left(\prod_ X_\right) \widehat_ Y \to \prod_ \left(X_ \widehat_ Y\right)$ is a TVS-isomorphism.
* If $X$ and $Y$ are Frechet spaces and $Z$ is a complete Hausdorff locally convex space, then the canonical vector space isomorphism $I : B(X, Y; Z) \to L\left(X \widehat_ Y; Z\right)$ becomes a homeomorphism when these spaces are given the topologies of uniform convergence on products of compact sets and, for the second one, the topology of compact convergence (i.e. $I : B_(X, Y; Z) \to L_\left(X \widehat_ Y; Z\right)$ is a TVS-isomorphism).
* Suppose $X$ and $Y$ are Frechet spaces. Every compact subset of $X \widehat_ Y$ is contained in the closed convex balanced hull of the tensor product if a compact subset of $X$ and a compact subset of $Y.$
* If $X$ and $Y$ are nuclear then $X \otimes_ Y$ and $X \widehat_ Y$ are nuclear.

Projective norm

Suppose now that $\left(X, \, \,\cdot\,\, \right)$ and $\left(Y, \, \,\cdot\,\, \right)$ are normed spaces. Then $X \otimes_ Y$ is a normable space with a canonical norm denoted by $\, \,\cdot\,\, _.$
The $\pi$-norm is defined on $X \otimes Y$ by
$\, b\, _ := \inf_ r$
where $W$ is the balanced convex hull of $C_p \otimes C_q = \left\.$
Given $b$ in $X \otimes Y,$ this can also be expressed as
$\, b\, _ := \inf \sum_i \, x_i \, \, y_i \,$
where the infimum is taken over all finite sequences $x_1, \ldots, x_n \in X$ and $y_1, \ldots, y_n \in Y$ (of the same length) such that $b = x_1 \otimes y_1 + \cdots + x_n \otimes y_n.$
If $b$ is in $X \widehat_ Y$ then
$\, b\, _ := \inf \sum_i \, x_i \, \, y_i \,$
where the infimum is taken over all (finite or infinite) sequences $x_1, \ldots, \in X$ and $y_1, \ldots, \in Y$ (of the same length) such that $b = x_1 \otimes y_1 + \cdots.$ Also,
$\, b\, _ := \inf \sum_i, \lambda_i,$
where the infimum is taken over all sequences $\left(x_i\right)$ in $X$ and $\left(y_i\right)$ in $Y$ and scalars $\lambda_1, \cdots$ (of the same length) such that $b = \lambda_1 x_1 \otimes y_1 + \cdots,$ $\, x_i \, = \, y_i \, = 1,$ and $\sum_, \lambda_i, < \infty.$ Also,
$\, b\, _ := \inf \sum_i, \lambda_i, \, x_i \, \, y_i \,$
where the infimum is taken over all sequences $\left(x_i\right)$ in $X$ and $\left(y_i\right)$ in $Y$ and scalars $\lambda_1, \cdots$ (of the same length) such that $b = \lambda_1 x_1 \otimes y_1 + \cdots,$ $\left(x_i\right)$ and $\left(y_i\right)$ converge to the origin, and $\sum_, \lambda_i, < \infty.$

If $X$ and $Y$ are Banach spaces then the closed unit ball of $X \widehat_ Y$ is the closed convex hull of the tensor product of the closed unit ball in $X$ with that of $Y.$

Properties

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

Preserved properties

* In general, the projective tensor product does not respect subspaces (e.g. if $Z$ is a vector subspace of $X$ then the TVS $Z \otimes_ Y$ has in general a coarser topology than the subspace topology inherited from $X \otimes_ Y$).
* Suppose that $E$ and $F$ are complemented subspaces of $X$ and $Y,$ respectively. Then $E \otimes F$ is a complemented subvector space of $X \otimes_ Y$ and the projective norm on $E \otimes_ F$ is equivalent to the projective norm on $X \otimes_ Y$ restricted to the subspace $E \otimes F$; Furthermore, if $X$ and $F$ are complemented by projections of norm 1, then $E \otimes_ F$ is complemented by a projection of norm 1.
* If $I : X \otimes_ Y \to Z$ is an isometric embedding into a Banach space $Z,$ then its unique continuous extension $I : X \widehat_ Y \to Z$ is also an isometric embedding.
* If $\alpha : W \to X$ and $\beta : Y \to Z$ are quotient operators between Banach spaces, then so is $\alpha \widehat_ \beta : W \widehat_ Y \to X \widehat_ Z.$
** A continuous linear operator $\beta : Y \to Z$ 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 $z \in Z,$ $\, z \, = \inf_ \, y\, .$
* Let $X$ and $F$ be vector subspaces of the Banach spaces $X$ and $Y,$ respectively. Then $E \widehat_ F$ is a TVS-subspace of $X \widehat_ Y$ if and only if every bounded bilinear form on $E \times F$ extends to a continuous bilinear form on $X \times Y$ with the same norm.

Trace form

Suppose that $X$ is a locally convex spaces. There is a bilinear form on $X \times X^$ defined by $\left(x, x^\right) \mapsto x^(x),$ which when $X$ is a Banach space has norm equal to 1. This bilinear form corresponds to a linear form on $X \otimes X^$ given by mapping $z := \sum_^n x_i \otimes x^_$ to $\sum_^n x^_i\left(x_i\right)$ (where of course this value is in fact independent of the representation $\sum_^n x_i \otimes x^_$ of $z$ chosen). Letting $X^$ have its strong dual topology, we can continuously extend this linear map to a map $\operatorname : X \widehat_ X^_b \to \Complex$ (assuming that the vector spaces have scalar field $\Complex$) called the trace of $X.$
This name originates from the fact that if we write $z = \sum_^n z_ e_i \otimes e_j^$ where $e_j^\left(e_i\right) = 1$ if $i = j$ and 0 otherwise, then $\operatorname(z) = \sum_^n z_.$

Duality with L(X; Y')

Assuming that $X$ and $Y$ are Banach spaces over the field $\mathbb,$ one may define a dual system between $X \widehat_ Y$ and $L_b\left(X; Y^\right)$ with the duality map
$\left\langle \cdot, \cdot \right\rangle : L_b\left(X; Y^\right) \times \left(X \widehat_ Y\right) \to \mathbb$
defined by
$\langle u, z \rangle := \operatorname\left(\left(u \, \widehat_ \operatorname_Y\right) (z)\right),$
where $\operatorname_Y : Y \to Y$ is the identity map and
$u \, \widehat_ \operatorname_Y : X \widehat_ Y \to Y^ \widehat_ Y$
is the unique continuous extension of the continuous map
$u \, \otimes_ \operatorname_Y : X \otimes_ Y \to Y^ \otimes_ Y.$
If we write $z = \sum_^ \lambda_ x_i \otimes y_i$ with $\sum_^ \left, \lambda_i \ < \infty$ and the sequences $\left(x_i\right)_^$ and $\left(y_i\right)_^$ each converging to zero, then we have$\left\langle u, z \right\rangle = \sum_^ \lambda_i \left\langle u\left(x_i\right), y_i \right\rangle.$

Nuclear operators

There is a canonical vector space embedding $I : X^ \otimes Y \to L(X; Y)$ defined by sending $z := \sum_^n x_i^ \otimes y_i$ to the map
$x \mapsto \sum_i^n x_i^(x) y_i$
where it can be shown that this value is independent of the representation of $z$ chosen.

Nuclear operators between Banach spaces

Assuming that $X$ and $Y$ are Banach spaces, then the map $I : X^_b \otimes_ Y \to L_b(X; Y)$ has norm $1$ so it has a continuous extension to a map $\hat : X^_b \widehat_ Y \to L_b(X; Y),$ where it is known that this map is not necessarily injective. The range of this map is denoted by $L^1(X; Y)$ and its elements are called nuclear operators. $L^1(X; Y)$ is TVS-isomorphic to $\left(X^_b \widehat_ Y\right) / \operatorname \hat$ and the norm on this quotient space, when transferred to elements of $L^1(X; Y)$ via the induced map $\hat : \left(X^_b \widehat_ Y\right) / \operatorname \hat \to L^1(X; Y),$ is called the trace-norm and is denoted by $\, \,\cdot\,\, _.$

Nuclear operators between locally convex spaces

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 $p_U(x) = \inf_ r$ and let $\pi : X \to X/p_U^(0)$ be the canonical projection. One can define the auxiliary Banach space $\hat_U$ with the canonical map $\hat_U : X \to \hat_U$ whose image, $X/p_U^(0),$ is dense in $\hat_U$ as well as the auxiliary space $F_B = \operatorname B$ normed by $p_B(y) = \inf_ r$ and with a canonical map $\iota : F_B \to F$ being the (continuous) canonical injection.
Given any continuous linear map $T : \hat_U \to Y_B$ one obtains through composition the continuous linear map $\hat_U \circ T \circ \iota : X \to Y$; thus we have an injection $L \left(\hat_U; Y_B\right) \to L(X; Y)$ and we henceforth use this map to identify $L \left(\hat_U; Y_B\right)$ as a subspace of $L(X; Y).$

Let $X$ and $Y$ be Hausdorff locally convex spaces. The union of all $L^1\left(\hat_U; Y_B\right)$ 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 $L^1(X; Y)$ and its elements are call nuclear mappings of $X$ into $Y.$

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.

Nuclear operators between Hilbert spaces

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

Nuclear bilinear forms

There is a canonical vector space embedding $J : X^ \otimes Y^ \to \mathcal(X, Y)$ defined by sending $z := \sum_^n x_i^ \otimes y_i^$ to the map
$(x, y) \mapsto \sum_i^n x_i^(x) y_i^(y)$
where it can be shown that this value is independent of the representation of $z$ chosen.

Nuclear bilinear forms on Banach spaces

Assuming that $X$ and $Y$ are Banach spaces, then the map $J : X^_b \otimes_ Y^_b \to \mathcal_b(X, Y)$ has norm $1$ so it has a continuous extension to a map $\hat : X^_b \widehat_ Y^_b \to \mathcal_b(X, Y).$ The range of this map is denoted by $B^1(X, Y)$ and its elements are called nuclear bilinear forms. $B^1(X, Y)$ is TVS-isomorphic to $\left(X^_b \widehat_ Y^_b\right) / \operatorname \hat$ and the norm on this quotient space, when transferred to elements of $B^1(X, Y)$ via the induced map $\hat : \left(X^_b \widehat_ Y^_b\right) / \operatorname \hat \to B^1(X, Y),$ is called the nuclear-norm and is denoted by $\, \,\cdot\,\, _.$

Suppose that $X$ and $Y$ are Banach spaces and that $N$ is a continuous bilinear from on $X \times Y.$
* The following are equivalent:
# $N$ is nuclear.
# There exist bounded sequences $\left(x_i^\right)_^$ in $X^_b$ and $\left(y_i^\right)_^$ in $Y^_b$ such that $\sum_^ \, x_i^ \, \, y_i^ \, < \infty$ and $N$ is equal to the mapping: $N(x, y) = \sum_^ x^_i(x) y^_i(y)$ for all $(x, y) \in X \times Y.$
* In this case we call $\sum_^ x^_i \otimes y^_i$ a nuclear representation of $N.$

The nuclear norm of $N$ is:
$\, N\, _ = \inf \left\.$
Note that $\, N\, \leq \, N\, _.$

Examples

Space of absolutely summable families

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

Let $\left(x_\right)_$ be a family of elements in a TVS $X$ and for every finite subset $H$ of $A,$ let $x_H := \sum_ x_i.$ We call $\left(x_\right)_$ summable in $X$ if the limit $\lim_ x_$ of the net $\left(x_H\right)_$ converges in $X$ to some element (any such element is called its sum). We call $\left(x_\right)_$ absolutely summable if it is summable and if for every continuous seminorm $p$ on $X,$ the family $\left(p \left(x_\right)\right)_$ is summable in $\R.$ The set of all such absolutely summable families is a vector subspace of $X^$ denoted by $S_a.$

Note that if $X$ is a metrizable locally convex space then at most countably many terms in an absolutely summable family are non-0.
A metrizable locally convex space is nuclear if and only if every summable sequence is absolutely summable. It follows that a normable space in which every summable sequence is absolutely summable, is necessarily finite dimensional.

We now define a topology on $S_a$ in a very natural way. This topology turns out to be the projective topology taken from $\ell^1(A) \widehat_ X$ and transferred to $S_a$ 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 projective or injective tensor product topology.

Let $\mathfrak$ denote a base of convex balanced neighborhoods of the origin in $X$ and for each $U \in \mathfrak,$ let $\mu_U : X \to \R$ denote its Minkowski functional. For any such $U$ and any $x = \left(x_\right)_ \in S_a,$ let $p_U(x) := \sum_ \mu_U\left(x_\right)$ where $p_U$ defines a seminorm on $S_a.$ The family of seminorms $\$ generates a topology making $S_a$ into a locally convex space. The vector space $S_a$ endowed with this topology will be denoted by $\ell^1$, X The special case where $X$ is the scalar field will be denoted by \ell^1

There is a canonical embedding of vector spaces \ell^1(A) \otimes X \to \ell^1 , E/math> defined by linearizing the bilinear map \ell^1(A) \times X \to \ell^1 , E/math> defined by $\left(\left(r_\right)_, x\right) \mapsto \left(r_ x\right)_.$

See also

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References

Bibliography

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External links

Nuclear space at ncatlab

{{Functional Analysis

Functional analysis