projective tensor product

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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 $\left(x, y\right) \in X \times Y$ 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 $L_0\left(x + \ker L\right) := L\left(x\right)$ 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 $L\left(X; Z\right)$ (resp. $B\left(X, Y; Z\right)$) where if $Z$ is the scalar field then we may instead write $L\left(X\right)$ (resp. $B\left(X, Y\right)$). * 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 $\langle L\left(x\right), X \rangle \geq 0$ 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 $U\left(x\right) = L\left(x\right)$ for $x = R \left\left(x_1\right\right) \in \operatorname R$ and extending $U$ continuously to $\overline,$ and then define $U$ on $\operatorname R$ by setting $U\left(x\right) = 0$ 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 $\Lambda\left(U\right)$ is Totally bounded space#In topological groups, 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 $x \in V_i,$ $L\left(x\right) = r_i x$; and (3) the orthogonal of the subspace spanned by $\cup_i V_i$ is equal to the kernel of $L.$

## Notation for topologies

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

## 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 $\operatorname\left(X, Y\right)$ be the space of all bilinear maps defined on $X \times Y$ and going into the underlying scalar field. For every $\left(x, y\right) \in X \times Y$ define a canonical bilinear form by $\chi_$ with domain $\operatorname\left(X, Y\right)$ by $\chi_\left(u\right) := u\left(x, y\right)$ for every $u \in Bi\left(X, Y\right).$ This induces a canonical map $\chi : X \times Y \to \operatorname\left(X, Y\right)^$ defined by $\chi\left(x, y\right) = \chi_,$ where $\operatorname\left(X, Y\right)^$ denotes the algebraic dual of $\operatorname\left(X, Y\right).$ 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 $x \otimes y = \chi\left(x, y\right)$). This gives us a canonical tensor product of $X$ and $Y.$ If $Z$ is any other vector space then the mapping $Li\left(X \otimes Y; Z\right) \to \operatorname\left(X, Y; Z\right)$ 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 $L\left(X \otimes_ Y; Z\right) \to B\left(X, Y; Z\right)$ 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 $\operatorname\left(X, Y\right)$ of continuous bilinear forms on $X \times Y$; furthermore, under this identification the equicontinuous subsets of $\operatorname\left(X, Y\right)$ are the same as the equicontinuous subsets of ''$\left(X \otimes_ Y\right)^.$

# 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 $\left\left(p_\right\right)$ (resp. $\left\left(q_\right\right)$) 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 $\left\left(p_ \otimes q_\right\right)$). 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 $\left(X, p\right)$ and $\left(Y, q\right)$ are normed spaces then $\left\left(X \otimes Y, p \otimes q\right\right)$ is also a normed space, called the projective tensor product of $\left(X, p\right)$ and $\left(Y, q\right),$ 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 $\left(x, y\right) \mapsto x \otimes y$ 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 $\left\left(U_\right\right)$ and $\left\left(V_\right\right)$ 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 $B\left\left(X, Y; Z\right\right)$ then the range of this restriction is the space $L\left\left(X \otimes_ Y; Z\right\right)$ 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 $B\left(X, Y\right),$ the space of continuous bilinear forms on $X \times Y.$

## The π-topology

Note that the canonical vector space isomorphism $I : B\left\left(X_^, Y_^; Z\right\right) \to L\left\left(X \otimes_ Y; Z\right\right)$ preserves equicontinuous subsets. Since $B\left(X, Y\right)$ 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 $\left\left(X \otimes_ Y\right\right)^$; 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 topological vector space, metrizable, semi-metrizable, Normable space, 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 injective function, 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 $B\left(X, Y\right).$: 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 $\left\left(x_i\right\right)_^$ in $X$ is absolutely convergent if $\sum_^ p \left\left(x_i\right\right) < \infty$ for every continuous seminorm $p$ on $X.$ We write $x = \sum_^ x_i$ if the sequence of partial sums $\left\left(\sum_^n x_i\right\right)_^$ 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 $\left\left(x_i\right\right)_^$ and $\left\left(y_i\right\right)_^.$

## 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 $B\left(X, Y\right),$ we can place on $B\left(X, Y\right)$ 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 $b\left\left(B\left(X, Y\right), X \widehat_ Y\right\right),$ 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 $X^ = \left\left(X^_b\right\right)^_b.$ 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 $B\left\left(X_1, X_2\right\right)$ is the image under $^\left\left(u \widehat_ v\right\right)$ of an equicontinuous subset of $B\left\left(Y_1, Y_2\right\right).$ ** If all four spaces are normed then $\, u \otimes v \, _ = \, u \, \, v \, .$ * The π-topology is finer than the Injective tensor product, 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 $\left\left(X_\right\right)$ are locally convex spaces then the canonical map $\left\left(\prod_ X_\right\right) \widehat_ Y \to \prod_ \left\left(X_ \widehat_ Y\right\right)$ 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 $I : B\left(X, Y; Z\right) \to L\left\left(X \widehat_ Y; Z\right\right)$ 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. $I : B_\left(X, Y; Z\right) \to L_\left\left(X \widehat_ Y; Z\right\right)$ 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 $\left\left(X, \, \,\cdot\,\, \right\right)$ and $\left\left(Y, \, \,\cdot\,\, \right\right)$ 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 $\left\left(x_i\right\right)$ in $X$ and $\left\left(y_i\right\right)$ 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 $\left\left(x_i\right\right)$ in $X$ and $\left\left(y_i\right\right)$ in $Y$ and scalars $\lambda_1, \cdots$ (of the same length) such that $b = \lambda_1 x_1 \otimes y_1 + \cdots,$ $\left\left(x_i\right\right)$ and $\left\left(y_i\right\right)$ 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 $\left\left(Z, \, \,\cdot\,\, \right\right),$ the canonical vector space isomorphism of $B\left(X, Y; Z\right)$ onto $L\left\left(X \otimes_ Y; Z\right\right)$ 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 $B\left(X, Y\right)$ into $\left\left(X \otimes Y\right\right)^,$ which is the algebraic dual of $X \otimes Y,$ is an isometry of $B\left(X, Y\right)$ onto $\left\left(X \otimes_ Y\right\right)^,$ 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 $\left\left(x, x^\right\right) \mapsto x^\left(x\right),$ 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 $\sum_^n x^_i\left\left(x_i\right\right)$ (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 $e_j^\left\left(e_i\right\right) = 1$ if $i = j$ and 0 otherwise, then $\operatorname\left(z\right) = \sum_^n z_.$

## 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 $L_b\left\left(X; Y^\right\right)$ with the duality map $\left\langle \cdot, \cdot \right\rangle : L_b\left\left(X; Y^\right\right) \times \left\left(X \widehat_ Y\right\right) \to \mathbb$ defined by $\langle u, z \rangle := \operatorname\left\left(\left\left(u \, \widehat_ \operatorname_Y\right\right) \left(z\right)\right\right),$ 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 $\left\left(x_i\right\right)_^$ and $\left\left(y_i\right\right)_^$ 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 $I : X^ \otimes Y \to L\left(X; Y\right)$ 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 $I : X^_b \otimes_ Y \to L_b\left(X; Y\right)$ has norm $1$ so it has a continuous extension to a map $\hat : X^_b \widehat_ Y \to L_b\left(X; Y\right),$ where it is known that this map is not necessarily injective. The range of this map is denoted by $L^1\left(X; Y\right)$ and its elements are called nuclear operators. $L^1\left(X; Y\right)$ is TVS-isomorphic to $\left\left(X^_b \widehat_ Y\right\right) / \operatorname \hat$ and the norm on this quotient space, when transferred to elements of $L^1\left(X; Y\right)$ via the induced map $\hat : \left\left(X^_b \widehat_ Y\right\right) / \operatorname \hat \to L^1\left(X; Y\right),$ 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 $p_U\left(x\right) = \inf_ r$ and let $\pi : X \to X/p_U^\left(0\right)$ be the canonical projection. One can define the Auxiliary normed spaces, auxiliary Banach space $\hat_U$ with the canonical map $\hat_U : X \to \hat_U$ whose image, $X/p_U^\left(0\right),$ is dense in $\hat_U$ as well as the auxiliary space $F_B = \operatorname B$ normed by $p_B\left(y\right) = \inf_ r$ 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 $L \left\left(\hat_U; Y_B\right\right) \to L\left(X; Y\right)$ and we henceforth use this map to identify $L \left\left(\hat_U; Y_B\right\right)$ as a subspace of $L\left(X; Y\right).$ Let $X$ and $Y$ be Hausdorff locally convex spaces. The union of all $L^1\left\left(\hat_U; Y_B\right\right)$ 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 $L^1\left(X; Y\right)$ 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 $J : X^ \otimes Y^ \to \mathcal\left(X, Y\right)$ 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 $J : X^_b \otimes_ Y^_b \to \mathcal_b\left(X, Y\right)$ has norm $1$ so it has a continuous extension to a map $\hat : X^_b \widehat_ Y^_b \to \mathcal_b\left(X, Y\right).$ The range of this map is denoted by $B^1\left(X, Y\right)$ and its elements are called nuclear bilinear forms. $B^1\left(X, Y\right)$ is TVS-isomorphic to $\left\left(X^_b \widehat_ Y^_b\right\right) / \operatorname \hat$ and the norm on this quotient space, when transferred to elements of $B^1\left(X, Y\right)$ via the induced map $\hat : \left\left(X^_b \widehat_ Y^_b\right\right) / \operatorname \hat \to B^1\left(X, Y\right),$ 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 $\left\left(x_i^\right\right)_^$ in $X^_b$ and $\left\left(y_i^\right\right)_^$ in $Y^_b$ such that $\sum_^ \, x_i^ \, \, y_i^ \, < \infty$ and $N$ is equal to the mapping: $N\left(x, y\right) = \sum_^ x^_i\left(x\right) y^_i\left(y\right)$ for all $\left(x, y\right) \in X \times Y.$ * 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 $\mathcal\left(A\right)$ be the directed set of all finite subsets of $A$ directed by inclusion $\subseteq.$ Let $\left\left(x_\right\right)_$ 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 $\left\left(x_\right\right)_$ summable in $X$ if the limit $\lim_ x_$ of the Net (mathematics), net $\left\left(x_H\right\right)_$ converges in $X$ to some element (any such element is called its sum). We call $\left\left(x_\right\right)_$ absolutely summable if it is summable and if for every continuous seminorm $p$ on $X,$ the family $\left\left(p \left\left(x_\right\right)\right\right)_$ 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 space, 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 $\ell^1\left(A\right) \widehat_ X$ 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 $x = \left\left(x_\right\right)_ \in S_a,$ let $p_U\left(x\right) := \sum_ \mu_U\left\left(x_\right\right)$ 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 $\ell^1\left[A, X\right].$ The special case where $X$ is the scalar field will be denoted by $\ell^1\left[A\right].$ There is a canonical embedding of vector spaces $\ell^1\left(A\right) \otimes X \to \ell^1\left[A, E\right]$ defined by linearizing the bilinear map $\ell^1\left(A\right) \times X \to \ell^1\left[A, E\right]$ defined by $\left\left(\left\left(r_\right\right)_, x\right\right) \mapsto \left\left(r_ x\right\right)_.$

# See also

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# Bibliography

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# External links

Nuclear space at ncatlab
{{Functional Analysis Functional analysis