Injective Tensor Product

In mathematics, the injective tensor product of two topological vector spaces (TVSs) was introduced by Alexander Grothendieck and was used by him to define nuclear spaces. An injective tensor product is in general not necessarily complete, so its completion is called the . Injective tensor products have applications outside of nuclear spaces. In particular, as described below, up to TVS-isomorphism, many TVSs that are defined for real or complex valued functions, for instance, the Schwartz space or the space of continuously differentiable functions, can be immediately extended to functions valued in a Hausdorff locally convex TVS $Y$ with any need to extend definitions (such as "differentiable at a point") from real/complex-valued functions to $Y$-valued functions.

Preliminaries and notation

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 $\operatorname L = L(X)$ has the subspace topology induced by $Y$
** If $S$ 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: $X \to X / \ker L \overset\rightarrow \operatorname L \to Y$ 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. $B(X, Y)$).
* The set of separately continuous bilinear maps $X \times Y \to Z$ (that is, continuous in each variable when the other variable is fixed) will be denoted by $\mathcal(X, Y; Z)$ where if $Z$ is the scalar field then we may instead write $\mathcal(X, Y).$
* We will denote the continuous dual space of $X$ by $X^$ or $X^$ 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 (for example, $x^$ denotes an element of $X^$ and not, say, a derivative and the variables $x$ and $x^$ need not be related in any way).

Notation for topologies

* $\sigma\left(X, X^\right)$ denotes the coarsest topology on $X$ making every map in $X^$ continuous and $X_$ or $X_\sigma$ denotes $X$ endowed with this topology.
* $\sigma\left(X^, X\right)$ denotes weak-* topology on $X^$ and $X_$ or $X^_\sigma$ denotes $X^$ endowed with this topology.
** Note that 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 $b\left(X^, X\right).$
* $\tau\left(X, X^\right)$ denotes the Mackey topology on $X$ or the topology of uniform convergence on the convex balanced weakly compact subsets of $X^\prime$ and $X_$ or $X_\tau$ denotes $X$ endowed with this topology. $\tau(X, X^)$ is the finest locally convex TVS topology on $X$ whose continuous dual space is equal to $X^.$
* $\tau\left(X^, X\right)$ denotes the Mackey topology on ''$X^$'' or the topology of uniform convergence on the convex balanced weakly compact subsets of $X$ and $X_$ or $X^_\tau$ denotes $X$ endowed with this topology.
** Note that $\tau\left(X^, X\right) \subseteq b\left(X^, X\right) \subseteq \tau \left(X^, X^\right).$
* $\varepsilon\left(X, X^\right)$ denotes the topology of uniform convergence on equicontinuous subsets of $X^$ and $X_$ or $X_\varepsilon$ denotes $X$ endowed with this topology.
** If $H$ is a set of linear mappings $X \to Y$ then $H$ is equicontinuous if and only if it is equicontinuous at the origin; that is, if and only if for every neighborhood $V$ of the origin in $Y,$ there exists a neighborhood $U$ of the origin in $X$ such that $\lambda(U) \subseteq V$ for every $\lambda \in H.$
* A set $H$ of linear maps from $X$ to $Y$ is called equicontinuous if for every neighborhood $V$ of the origin in $Y,$ there exists a neighborhood $U$ of the origin in $X$ such that $h(U) \subseteq V$ for all $h \in H.$

Definition

Throughout let $X$ and $Y$ be topological vector spaces with continuous dual spaces $X^$ and $Y^.$ Note that almost all results described are independent of whether these vector spaces are over $\R$ or $\Complex$ but to simplify the exposition we will assume that they are over the field $\Complex.$

Continuous bilinear maps as a tensor product

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 $X$ or $Y$), nevertheless the vector space $B\left(X_\sigma^, Y_\sigma^\right)$ of continuous bilinear functionals is always a tensor product of $X$ and $Y,$ as is now described.

For every $(x, y) \in X \times Y$ we now define a bilinear form, denoted by the symbol $x \otimes y,$ from $X^ \times Y^$ into the underlying field (that is, $x \otimes y : X^ \times Y^ \to \Complex$) by
$(x \otimes y) \left(x^, y^\right) := x^(x) y^(y).$
This induces a canonical map
$\cdot \otimes \cdot : X \times Y \to \mathcal\left(X_\sigma^, Y_\sigma^\right)$
defined by sending $(x, y) \in X \times Y$ to the bilinear form $x \otimes y.$
The span of the range of this map is $B\left(X_\sigma^, Y_\sigma^\right).$
The following theorem may be used to verify that $B\left(X_\sigma^, Y_\sigma^\right)$ together with the above map $\,\otimes\,$ is a tensor product of $X$ and $Y.$

Topology

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

Every set $b(G, H)$ is bounded, which is necessary and sufficient for the collection of all such $\mathcal(G, H, N)$ to form a locally convex TVS topology on $\mathcal\left(X^_b, Y^_b; Z\right)$ called the $\varepsilon$-topology.
The inclusions
$\mathcal(G, H, N) ~\subseteq~ B\left(X^_, Y^_\sigma; Z\right)~\subseteq~ \mathcal\left(X^_b, Y^_b; Z\right).$
always hold and whenever any one of these vector spaces is endowed with the $\varepsilon$-topology then this will be indicated by placing $\varepsilon$ as a subscript before the opening parenthesis. For example, $\mathcal\left(X^_b, Y^_b; Z\right)$ endowed with the $\varepsilon$-topology will be denoted by $\mathcal_\varepsilon\left(X^_b, Y^_b; Z\right).$

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

Equicontinuous sets

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

:A set of continuous linear functionals $H$ on a TVS $X$This is true even if $X$ is not assumed to be Hausdorff or locally convex. is equicontinuous if and only if it is contained in the polar of some neighborhood $U$ of $0$ in $X$; that is, $H \subseteq U^.$

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 $X^$ "encodes" all information about $X$'s given topology. Specifically, distinct LCTVS topologies on $X$ 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 $X$ and $Y.$
Furthermore, the topology of a locally convex Hausdorff space $X$ is identical to the topology of uniform convergence on the equicontinuous subsets of $X^.$

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

* If $H \subseteq L(X; Y)$ is equicontinuous then the subspace topologies that $H$ inherits from the following topologies on $L(X; Y)$ are identical:
*#the topology of precompact convergence;
*#the topology of compact convergence;
*#the topology of pointwise convergence;
*#the topology of pointwise convergence on a given dense subset of $X.$
* An equicontinuous set $H \subseteq L(X; Y)$ is bounded in the topology of bounded convergence (that is, bounded in $L_b(X; Y)$). So in particular, $H$ will also bounded in every TVS topology that is coarser than the topology of bounded convergence.
* If $X$ is a barrelled space and $Y$ is locally convex then for any subset $H \subseteq L(X; Y),$ the following are equivalent:
*#$H$ is equicontinuous;
*#$H$ is bounded in the topology of pointwise convergence (that is, bounded in $L_(X; Y)$);
*#$H$ is bounded in the topology of bounded convergence (that is, bounded in $L_b(X; Y)$).
In particular, to show that a set $H$ is equicontinuous it suffices to show that it is bounded in the topology of pointwise converge.
* If $X$ is a Baire space then any subset $H \subseteq L(X; Y)$ that is bounded in $L_\sigma(X; Y)$ is necessarily equicontinuous.
* If $X$ is separable, $Y$ is metrizable, and $D$ is a dense subset of $X,$ then the topology of pointwise convergence on $D$ makes $L(X; Y)$ metrizable so that in particular, the subspace topology that any equicontinuous subset $H \subseteq L(X; Y)$ inherits from $L_\sigma(X; Y)$ is metrizable.

For equicontinuous subsets of the continuous dual space $X^$ (where $Y$ is now the underlying scalar field of $X$), the following hold:
* The weak closure of an equicontinuous set of linear functionals on $X$ is a compact subspace of $X_\sigma^.$
* If $X$ is separable then every weakly closed equicontinuous subset of $X_\sigma^$ is a metrizable compact space when it is given the weak topology (that is, the subspace topology inherited from $X_\sigma^$).
* If $X$ is a normable space then a subset $H \subseteq X^$ is equicontinuous if and only if it is strongly bounded (that is, bounded in $X_b^$).
* If $X$ is a barrelled space then for any subset $H \subseteq X^,$ the following are equivalent:
*#$H$ is equicontinuous;
*#$H$ is relatively compact in the weak dual topology;
*#$H$ is weakly bounded;
*#$H$ is strongly bounded.

We mention some additional important basic properties relevant to the injective tensor product:
* Suppose that $B : X_1 \times X_2 \to Y$ is a bilinear map where $X_1$ is a Fréchet space, $X_2$ is metrizable, and $Y$ is locally convex. If $B$ is separately continuous then it is continuous.

Canonical identification of separately continuous bilinear maps with linear maps

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

There also exists a canonical vector space isomorphism
$J : \mathcal\left(X^_, Y^_\right) \to L\left(X^_; Y_\right).$
To define it, for every separately continuous bilinear form $B$ defined on $X^_ \times Y^_$ and every $x^ \in X^,$ let $B_ \in \left(Y_\sigma^\right)^$ be defined by
$B_\left(y^\right) := B\left(x^, y^\right).$
Because $\left(Y_\sigma^\right)^$ is canonically vector space-isomorphic to $Y$ (via the canonical map $y \mapsto$ value at $y$), $B_$ will be identified as an element of $Y,$ which will be denoted by $\tilde_ \in Y.$
This defines a map $\tilde : X^ \to Y$ given by $x^ \mapsto \tilde_$ and so the canonical isomorphism is of course defined by $J(B) := \tilde.$

When $L\left(X^\sigma_\sigma; Y_\sigma\right)$ is given the topology of uniform convergence on equicontinous subsets of $X^,$ the canonical map becomes a TVS-isomorphism
$J : \mathcal_\varepsilon\left(X^_\sigma, Y^_\sigma\right) \to L_\varepsilon\left(X^_\tau; Y\right).$
In particular, $X \otimes_\varepsilon Y = B_\varepsilon\left(X^_\sigma, Y^_\sigma\right)$ can be canonically TVS-embedded into $L_\varepsilon\left(X^_\tau; Y\right)$; furthermore the image in $L\left(X^_\sigma; Y_\sigma\right)$ of $X \otimes_\varepsilon Y = B_\varepsilon\left(X^_\sigma, Y^_\sigma\right)$ under the canonical map $J$ consists exactly of the space of continuous linear maps $X^_ \to Y$ whose image is finite dimensional.

The inclusion $L\left(X^_\tau; Y\right) \subseteq L\left(X^_b; Y\right)$ always holds. If $X$ is normed then $L_\varepsilon\left(X^_\tau; Y\right)$ is in fact a topological vector subspace of $L_b\left(X^_b; Y\right).$ And if in addition $Y$ is Banach then so is $L_b\left(X^_b; Y\right)$ (even if $X$ is not complete).

Properties

The canonical map $\cdot \otimes \cdot : X \times Y \to \mathcal\left(X_\sigma^, Y_\sigma^\right)$ is always continuous and the ε-topology is always finer than the π-topology and coarser than the inductive topology (which is the finest locally convex TVS topology making $X \times Y \to X \otimes Y$ separately continuous).
The space $X \otimes_\varepsilon Y$ is Hausdorff if and only if both $X$ and $Y$ are Hausdorff.

If $X$ and $Y$ are normed then $X \otimes_\varepsilon Y$ is normable in which case for all $\theta \in X \otimes Y,$ $\, \theta\, _\varepsilon \leq \, \theta\, _.$

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 $u \otimes v : X_1 \otimes_\varepsilon X_2 \to Y_1 \otimes_\varepsilon Y_2.$ Moreover:
* If $u$ and $v$ are both TVS-embeddings then so is $u \widehat_\varepsilon v : X_1 \widehat_\varepsilon X_2 \to Y_1 \widehat_\varepsilon Y_2.$
* If $X_1$ (resp. $Y_1$) is a linear subspace of $X_2$ (resp. $Y_2$) then $X_1 \otimes_\varepsilon Y_1$ is canonically isomorphic to a linear subspace of $X_2 \otimes_\varepsilon Y_2$ and $X_1 \widehat_\varepsilon Y_1$ is canonically isomorphic to a linear subspace of $X_2 \widehat_\varepsilon Y_2.$
* There are examples of $u$ and $v$ such that both $u$ and $v$ are surjective homomorphisms but $u \widehat_\varepsilon v : X_1 \widehat_\varepsilon X_2 \to Y_1 \widehat_\varepsilon Y_2$ is a homomorphism.
* If all four spaces are normed then $\, u \otimes v\, _\varepsilon = \, u\, \, v\, .$

Relation to projective tensor product and nuclear spaces

The strongest locally convex topology on $B\left(X^_\sigma, Y^_\sigma\right) = X \otimes Y$ making the canonical map $\cdot \otimes \cdot : X \times Y \to B\left(X_\sigma^, Y_\sigma^\right)$ (defined by sending $(x, y) \in X \times Y$ to the bilinear form $x \otimes y$) continuous is called the projective topology or the $\pi$-topology. When $B\left(X^_\sigma, Y^_\sigma\right) = X \otimes Y$ is endowed with this topology then it will be denoted by $X \otimes_ Y$ and called the projective tensor product of $X$ and $Y.$

The following definition was used by Grothendieck to define nuclear spaces.

Definition 0: Let $X$ be a locally convex topological vector space. Then $X$ is nuclear if for any locally convex space $Y,$ the canonical vector space embedding $X \otimes_\pi Y \to \mathcal_\varepsilon\left(X^_\sigma, Y^_\sigma\right)$ is an embedding of TVSs whose image is dense in the codomain.

Canonical identifications of bilinear and linear maps

In this section we describe canonical identifications between spaces of bilinear and linear maps. These identifications will be used to define important subspaces and topologies (particularly those that relate to nuclear operators and nuclear spaces).

Dual spaces of the injective tensor product and its completion

Suppose that
$\operatorname : X \otimes_\varepsilon Y \to X \widehat_\varepsilon Y$
denotes the TVS-embedding of $X \otimes_\varepsilon Y$ into its completion and let
$^t \operatorname : \left(X \widehat_\varepsilon Y\right)^_b \to \left(X \otimes_\varepsilon Y\right)^_b$
be its transpose, which is a vector space-isomorphism. This identifies the continuous dual space of $X \otimes_\varepsilon Y$ as being identical to the continuous dual space of $X \widehat_\varepsilon Y.$

The identity map
$\operatorname_ : X \otimes_ Y \to X \otimes_\varepsilon Y$
is continuous (by definition of the π-topology) so there exists a unique continuous linear extension
$\hat : X \widehat_ Y \to X \widehat_\varepsilon Y.$
If $X$ and $Y$ are Hilbert spaces then $\hat : X \widehat_ Y \to X \widehat_\varepsilon Y$ is injective and the dual of $X \widehat_\varepsilon Y$ is canonically isometrically isomorphic to the vector space $L^1\left(X; Y^\right)$ of nuclear operators from $X$ into $Y$ (with the trace norm).

Injective tensor product of Hilbert spaces

There is a canonical map
$K : X \otimes Y \to L\left(X^; Y\right)$
that sends $z = \sum_^n x_i \otimes y_i$ to the linear map $K(z) : X^ \to Y$ defined by
$K(z)\left(x^\right) := \sum_^n x^(x_i) y_i \in Y,$
where it may be shown that the definition of $K(z) : X \to Y$ does not depend on the particular choice of representation $\sum_^n x_i \otimes y_i$ of $z.$ The map
$K : X \otimes_\varepsilon Y \to L_b\left(X^_b; Y\right)$
is continuous and when $L_b\left(X^_b; Y\right)$ is complete, it has a continuous extension
$\hat : X \widehat_\varepsilon Y \to L_b\left(X^_b; Y\right).$

When $X$ and $Y$ are Hilbert spaces then $\hat : X \widehat_\varepsilon Y \to L_b\left(X^_b; Y\right)$ 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 $X$ into $Y$ (which is a closed vector subspace of $L_b\left(X^; Y\right).$ Hence $X \widehat_\varepsilon Y$ is identical to space of compact operators from $X^$ into $Y$ (note the prime on $X$). The space of compact linear operators between any two Banach spaces (which includes Hilbert spaces) $X$ and $Y$ is a closed subset of $L_b(X; Y).$

Furthermore, the canonical map $X \widehat_ Y \to X \widehat_\varepsilon Y$ is injective when $X$ and $Y$ are Hilbert spaces.

Integral forms and operators

Integral bilinear forms

Denote the identity map by
$\operatorname : X \otimes_ Y \to X \otimes_\varepsilon Y$
and let
$^\operatorname : \left(X \otimes_\varepsilon Y\right)^_b \to \left(X \otimes_ Y\right)^_b$
denote its transpose, which is a continuous injection. Recall that $\left(X \otimes_ Y\right)^$ is canonically identified with $B(X, Y),$ the space of continuous bilinear maps on $X \times Y.$ In this way, the continuous dual space of $X \otimes_\varepsilon Y$ can be canonically identified as a subvector space of $B(X, Y),$ denoted by $J(X, Y).$ The elements of $J(X, Y)$ are called integral (bilinear) forms on $X \times Y.$ The following theorem justifies the word .

Integral linear operators

Given a linear map $\Lambda : X \to Y,$ one can define a canonical bilinear form $B_ \in Bi\left(X, Y^\right),$ called the associated bilinear form on $X \times Y^,$ by
$B_\left(x, y^\right) := \left( y^ \circ \Lambda\right)(x).$
A continuous map $\Lambda : X \to Y$ is called integral if its associated bilinear form is an integral bilinear form. An integral map $\Lambda: X \to Y$ is of the form, for every $x \in X$ and $y^ \in Y^:$
$\left\langle y^, \Lambda(x)\right\rangle = \int_ \left\langle x^, x\right\rangle \left\langle y^, y^\right\rangle \operatorname \mu \left(x^, y^\right)$
for suitable weakly closed and equicontinuous subsets $A^$ and $B^$ of $X^$ and $Y^,$ respectively, and some positive Radon measure $\mu$ of total mass $\leq 1.$

Canonical map into ''L''(''X''; ''Y'')

There is a canonical map $K : X^ \otimes Y \to L(X; Y)$ that sends $z = \sum_^n x_i^ \otimes y_i$ to the linear map $K(z) : X \to Y$ defined by $K(z)(x) := \sum_^n x_i^(x) y_i \in Y,$ where it may be shown that the definition of $K(z) : X \to Y$ does not depend on the particular choice of representation $\sum_^n x_i^ \otimes y_i$ of $z.$

Examples

Space of 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 $H \subseteq A,$ 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). The set of all such summable families is a vector subspace of $X^$ denoted by $S.$

We now define a topology on $S$ in a very natural way. This topology turns out to be the injective topology taken from $l^1(A) \widehat_\varepsilon X$ and transferred to $S$ 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 $\mathfrak$ denote a base of convex balanced neighborhoods of 0 in $X$ and for each $U \in \mathfrak,$ let $\mu_U : X \to \R$ denote its Minkowski functional. For any such $U$ and any $x = \left(x_\right)_ \in S,$ let
$q_U(x) := \sup_ \sum_ \left, \left\langle x^, x_\right\rangle\$
where $q_U$ defines a seminorm on $S.$ The family of seminorms $\left\$ generates a topology making $S$ into a locally convex space. The vector space $S$ endowed with this topology will be denoted by $l^1(A, X).$ The special case where $X$ is the scalar field will be denoted by $l^1(A).$

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

Space of continuously differentiable vector-valued functions

Throughout, let $\Omega$ be an open subset of $\R^n,$ where $n \geq 1$ is an integer and let $Y$ be a locally convex topological vector space (TVS).

Definition Suppose $p^0 = \left(p^0_1, \ldots, p^0_n\right) \in \Omega$ and $f : \operatorname f \to Y$ is a function such that $p^0 \in \operatorname f$ with $p^0$ a limit point of $\operatorname f.$ Say that $f$ is differentiable at $p^0$ if there exist $n$ vectors $e_1, \ldots, e_n$ in $Y,$ called the partial derivatives of $f$, such that
$\lim_ \frac = 0 \text Y$
where $p = \left(p_1, \ldots, p_n\right).$

One may naturally extend the notion of to $Y$-valued functions defined on $\Omega.$
For any $k = 0, 1, \ldots, \infty,$ let $C^k(\Omega; Y)$ denote the vector space of all $C^k$ $Y$-valued maps defined on $\Omega$ and let $C_c^k(\Omega; Y)$ denote the vector subspace of $C^k(\Omega; Y)$ consisting of all maps in $C^k(\Omega; Y)$ that have compact support.

One may then define topologies on $C^k(\Omega; Y)$ and $C_c^k(\Omega; Y)$ in the same manner as the topologies on $C^k(\Omega)$ and $C_c^k(\Omega)$ are defined for the space of distributions and test functions (see the article: Differentiable vector-valued functions from Euclidean space).
All of this work in extending the definition of differentiability and various topologies turns out to be exactly equivalent to simply taking the completed injective tensor product:

Spaces of continuous maps from a compact space

If $Y$ is a normed space and if $K$ is a compact set, then the $\varepsilon$-norm on $C(K) \otimes Y$ is equal to $\, f \, _\varepsilon = \sup_ \, f(x)\, .$
If $H$ and $K$ are two compact spaces, then $C(H \times K) \cong C(H) \widehat_\varepsilon C(K),$ where this canonical map is an isomorphism of Banach spaces.

Spaces of sequences converging to 0

If $Y$ is a normed space, then let $l_(Y)$ denote the space of all sequences $\left(y_i\right)_^$ in $Y$ that converge to the origin and give this space the norm $\left\, \left(y_i\right)_^\right\, := \sup_ \left\, y_i\right\, .$
Let $l_$ denote $l_\left(\Complex\right).$
Then for any Banach space $Y,$ $l_ \widehat_\varepsilon Y$ is canonically isometrically isomorphic to $l_(Y).$

Schwartz space of functions

We will now generalize the Schwartz space to functions valued in a TVS.
Let $\mathcal\left(\R^n; Y\right)$ be the space of all $f \in C^\left(\R^n; Y\right)$ such that for all pairs of polynomials $P$ and $Q$ in $n$ variables, $\left\$ is a bounded subset of $Y.$
To generalize the topology of the Schwartz space to $\mathcal\left(\R^n; Y\right),$ we give $\mathcal\left(\R^n; Y\right)$ the topology of uniform convergence over $\R^n$ of the functions $P(x) Q\left(\partial / \partial x\right) f(x),$ as $P$ and $Q$ vary over all possible pairs of polynomials in $n$ variables.

See also

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Notes

References

Bibliography

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External links

Nuclear space at ncatlab

{{TopologicalTensorProductsAndNuclearSpaces

Functional analysis