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On tensor fractions and tensor products in the category of stereotype spaces
S. S. Akbarov School of Applied Mathematics, Tikhonov Moscow Institute of Electronics and Mathematics of the National Research University Higher School of Economics, Moscow, Russia
Abstract:
We prove two identities that connect some natural tensor products in the category $\operatorname{LCS}$ of locally convex spaces with tensor products in the category $\operatorname{Ste}$ of stereotype spaces. In particular, we give sufficient conditions under which the identity
$$
X^\vartriangle\odot Y^\vartriangle\cong (X^\vartriangle\cdot Y^\vartriangle)^\vartriangle\cong (X\cdot Y)^\vartriangle
$$
holds, where $\odot$ is the injective tensor product in the category $\operatorname{Ste}$, $\cdot\,$ is the primary tensor product in $\operatorname{LCS}$, and $\vartriangle$ is the pseudosaturation operation in $\operatorname{LCS}$. The study of relations of this type is justified by the fact that they turn out to be important instruments for constructing duality theory based on the notion of an envelope. In particular, they are used in the construction of the duality theory for the class of (not necessarily Abelian) countable discrete groups.
Bibliography: 15 titles.
Keywords:
stereotype space, pseudosaturation.
Received: 24.09.2020 and 08.12.2021
§ 1. Introduction In [1]–[5] this author described the properties of the class $\operatorname{Ste}$ of locally convex spaces named stereotype spaces and defined by the condition1[x]1In Köthe’s book [6] spaces of this type are called polar reflexive.
$$
\begin{equation*}
X\cong (X^\star)^\star,
\end{equation*}
\notag
$$
where each star $\star$ means the dual space of functionals endowed with the topology of uniform convergence on totally bounded sets. As was noticed, it has a series of remarkable properties, in particular, - • $\operatorname{Ste}$ is very wide, since it contains all quasicomplete barreled spaces (in particular, all Fréchet spaces, and thus, all Banach spaces),
- • $\operatorname{Ste}$ forms a pre-Abelian complete and co-complete category with linear continuous maps as morphisms,
- • $\operatorname{Ste}$ is an autodual category with respect to the functor $X\mapsto X^\star$, and
- • $\operatorname{Ste}$ has three natural bifunctors $\circledast$, $\odot$, $\oslash$ with the following properties:
$$
\begin{equation}
\mathbb C\circledast X \cong X \cong X\circledast \mathbb C, \qquad \mathbb C\odot X \cong X \cong X\odot \mathbb C,
\end{equation}
\tag{1.1}
$$
$$
\begin{equation}
X\circledast Y \cong Y\circledast X, \qquad X\odot Y \cong Y\odot X,
\end{equation}
\tag{1.2}
$$
$$
\begin{equation}
(X\circledast Y)\circledast Z \cong X\circledast (Y\circledast Z), \qquad (X\odot Y)\odot Z \cong X\odot (Y\odot Z),
\end{equation}
\tag{1.3}
$$
$$
\begin{equation}
Z\oslash (Y\circledast X)\cong (Z\oslash Y)\oslash X, \qquad (X\odot Y)\oslash Z \cong X\odot (Y\oslash Z),
\end{equation}
\tag{1.4}
$$
$$
\begin{equation}
(X\circledast Y)^\star \cong Y^\star \odot X^\star, \qquad (X\odot Y)^\star \cong Y^\star \circledast X^\star,
\end{equation}
\tag{1.5}
$$
$$
\begin{equation}
X\circledast Y\cong (X^\star\oslash Y)^\star, \qquad X\odot Y\cong Y\oslash (X^\star).
\end{equation}
\tag{1.6}
$$
Identities (1.1)–(1.3) in this list mean that $\operatorname{Ste}$ is a symmetric monoidal category with respect to the bifunctors $\circledast$ and $\odot$, and this justifies the name of stereotype tensor products for them ($\circledast$, the projective tensor product, and $\odot$, the injective). On the other hand, identities (1.4) justify the notification and the name stereotype tensor fraction for the bifunctor $\oslash$ (since among arithmetic operations the fraction is the most natural analogy for the operation $\oslash$ in the description of the properties of the products $\circledast$ and $\odot$, presented in (1.4)). At the same time, the left identity in (1.4) means that with respect to $\circledast$ the monoidal category $\operatorname{Ste}$ is closed with the inner hom-functor $\oslash$. In further studies the author (and later Kuznetsova and Aristov) found numerous applications of stereotype spaces in functional analysis and in geometry, in particular, in generalizations of Pontryagin duality to noncommutative groups. For example, from the results of [4] and [7] it follows that Pontryagin duality extends from the category of finite Abelian groups to the category of affine algebraic groups by the functor $G\mapsto {\mathcal O}^\star(G)$ of taking the group algebra of analytic functionals. This generalization can be expressed by the following functorial diagram: where $\mathfrak{e}$ denotes the natural embedding of categories, $G^\bullet$ the Pontryagin dual group, $\unicode{8224}$ the composition of the functor $H\mapsto \widehat{H}$ of the Arens-Michael envelope (see the definition in [5] or in [8]) and the functor $X\mapsto X^\star$ of taking the dual stereotype space,
$$
\begin{equation*}
H^\unicode{8224}:=( \widehat{H} )^\star,
\end{equation*}
\notag
$$
and by Hopf algebras reflexive with respect to the Arens-Michael envelope, we mean Hopf algebras in the monoidal category $(\operatorname{Ste},\circledast)$ that satisfy the identity
$$
\begin{equation*}
H\cong \big(H^\unicode{8224}\big)^\unicode{8224}
\end{equation*}
\notag
$$
(for details, see [8]). The commutativity of the diagram (1.7) is understood as an isomorphism of two functors going from the lower left corner to the upper right one:
$$
\begin{equation}
{\mathcal O}^\star(G)^\unicode{8224}\cong {\mathcal O}^\star(G^\bullet).
\end{equation}
\tag{1.8}
$$
After the author’s work [4], where the duality with respect to an envelope was first described and applied to complex Lie groups, the same idea (with the replacement of the Arens-Michael envelope by other envelopes) was used to generalize Pontryagin duality to various other classes of groups. In particular, Kuznetsova [9] suggested a duality theory for (not necessarily commutative) Moore groups (later her results were specified in [10]–[12]), and, besides this, the present author himself described in [10] and [11] a duality theory for differential geometry that works at least for the real Lie groups of the form $\mathbb{R}^n\times K\times D$, where $K$ is a compact Lie group, and $D$ is a discrete Moore group. These observations have not yet taken shape in general theories with natural boundaries, since it is obvious that all these results can be generalized to broader classes of groups, and it is not clear where it would be logical to stop in these studies. Within the framework of this program, in [8] this author considered the question of whether it is possible, by analogy with diagram (1.7), to generalize Pontryagin duality from the category of finite Abelian groups to the category of countable discrete groups. In [8] it was answered positively, with the describing diagram where the holomorphic reflexivity of Hopf algebras is understood as reflexivity with respect to one of the stereotype variants of the Arens-Michael envelope (for details, see [8]). The proof of this fact, however, as it turns out, requires a deep immersion into the foundations of the theory of stereotype spaces with the establishment of a series of technical statements, which, due to their volume and ideological isolation, should be presented as a separate text. This article is such a text. Here we derive two formulae expressing the injective stereotype tensor product $\odot$ and the tensor fraction $\oslash$, mentioned above, in terms of some auxiliary constructions in the theory of topological vector spaces. The first formula expresses $\odot$ in terms of one of the variants of tensor product in the category of locally convex spaces, which we denote by the dot $\cdot$ and call a primary tensor product. Formally $X\cdot Y$ is defined as the space of linear continuous mappings $\varphi\colon X^\star\to Y$ with the topology of uniform convergence on the polars $U^\circ$ of neighbourhoods of zero in $X$, and for it we prove the formula
$$
\begin{equation}
X^\vartriangle\odot Y^\vartriangle\cong (X^\vartriangle\cdot Y^\vartriangle)^\vartriangle\cong (X\cdot Y)^\vartriangle,
\end{equation}
\tag{1.10}
$$
which holds for all pseudocomplete locally convex spaces $X$ and $Y$ (without the assumption that one of them possesses the classical approximation property). And the second formula completes the formal definition of the tensor fraction $\oslash$ via spaces of operators. Recall that in [3] the space $\oslash$ was defined in two steps. First the space $Y:X$ (the primary tensor product) was defined, which was the space of linear continuous mappings $\varphi\colon X\to Y$ with the topology of uniform convergence on totally bounded sets. After that the space $Y\oslash X$ was defined as the pseudosaturation of $Y:X$:
$$
\begin{equation}
Y\oslash X=(Y:X)^\vartriangle.
\end{equation}
\tag{1.11}
$$
And here is another formula connecting $\oslash$ with $:$, and proved in our article:
$$
\begin{equation}
Y^\vartriangle\oslash X^\triangledown\cong(Y^\vartriangle:X^\triangledown)^\vartriangle\cong (Y:X)^\vartriangle
\end{equation}
\tag{1.12}
$$
($\triangledown$ is the operation of pseudocompletion in the category of locally convex spaces (see [3], § 1.3), and $\vartriangle$ is again the operation of pseudosaturation). The importance of formula (1.10) is that it allows us to study the space $X^\vartriangle\odot Y^\vartriangle$ without the detailed study of the pseudosaturations $X^\vartriangle$ and $Y^\vartriangle$. For instance, in [8] such formulae are used in the proof of identities like
$$
\begin{equation}
{\mathcal F}(M)^\vartriangle\odot {\mathcal F}(N)^\vartriangle\cong {\mathcal F}(M\times N)^\vartriangle,
\end{equation}
\tag{1.13}
$$
where $M$ and $N$ are manifolds of certain type, and ${\mathcal F}(\,{\dots}\,)$ is a certain type of (topological vector) functional space on these manifolds. A typical situation in the theory of stereotype spaces is when we can easily describe the topology of a given space, in this case, of ${\mathcal F}(\,{\dots}\,)$, but the topology of its pseudosaturation ${\mathcal F}(\,{\dots}\,)^\vartriangle$ looks extremely complicated, so that for its description it is usually impossible to find anything simpler than the very definition of the pseudosaturation. And formula (1.13) gives a description of the stereotype tensor product without the necessity of analyzing the topologies of the spaces ${\mathcal F}(\,{\dots}\,)^\vartriangle$. Similarly, formula (1.12) allows us to reduce the description of the space $Y^\vartriangle\oslash X^\triangledown$ to the description of the space $Y:X$ without the necessity of studying the spaces $Y^\vartriangle$ and $X^\triangledown$ (however, we give formula (1.12) for the future, since it is not used in [8] directly). These are the questions the present work is devoted to. Acknowledgements The author thanks O. Yu. Aristov, J. Wengenroth and the reviewers for useful advice and consultations.
§ 2. Pseudosaturation of the primary fraction $Y:X$ Everywhere in the text we use the terminology and notation of [3]. We consider locally convex spaces over the field $\mathbb{C}$. For each locally convex space $X$ the symbol ${\mathcal U}(X)$ denotes the set of all neighbourhoods of zero in $X$, ${\mathcal S}(X)$ the set of all totally bounded subsets in $X$, and ${\mathcal D}(X)$ the set of all capacious2[x]2A set $D\subseteq X$ is said to be capacious if for each totally bounded set $S\subseteq X$ there is a finite set $F\subseteq X$ such that $S\subseteq D+F$. sets in $X$. The term operator will be used for linear continuous mappings $\varphi\colon X\to Y$ of locally convex spaces. We start with the proof of formula (1.12), the last one in the list of the introduction. 2.1. The bilinear form $(g,x)\mapsto g:x$ Recall that in [3], § 5.6, a bilinear mapping $\beta\colon X\times Y\to Z$ was said to be continuous if 1) for each neighbourhood of zero $W\subseteq Z$ and each totally bounded set $S\subseteq X$ there is a neighbourhood of zero $V\subseteq Y$ such that
$$
\begin{equation*}
\beta(S,V)\subseteq W;
\end{equation*}
\notag
$$
2) for each neighbourhood of zero $W\subseteq Z$ and each totally bounded set $T\subseteq Y$ there is a neighbourhood of zero $U\subseteq X$ such that
$$
\begin{equation*}
\beta(U,T)\subseteq W.
\end{equation*}
\notag
$$
Recall again that for any locally convex spaces $X$ and $Y$ the symbol $Y:X$ denotes the space of linear continuous mappings $\varphi\colon X\to Y$ with the topology of uniform convergence on totally bounded sets (see [3], § 5.1). We call the space $Y:X$ the primary tensor fraction, having in mind formula (2.23), which we prove below and according to which the application of the operations $\vartriangle$ and $\triangledown$ to $Y:X$ gives the stereotype tensor fraction $Y\oslash X$. If $A\subseteq X$ and $B\subseteq Y$, then the symbol $B:A$ denotes the subset of $Y:X$ consisting of mappings that turn $A$ into $B$:
$$
\begin{equation}
\varphi\in B\colon A \quad\Longleftrightarrow\quad \varphi\in Y:X\ \& \ \varphi(A)\subseteq B
\end{equation}
\tag{2.1}
$$
(see [3], § 5.4). For each $x\in X$ and $g\in Y^\star$ we define the functional
$$
\begin{equation}
(g:x)\colon (Y:X)\to\mathbb C \quad\bigm|\quad (g:x)(\varphi)=g(\varphi(x)), \qquad \varphi\in Y:X.
\end{equation}
\tag{2.2}
$$
Theorem 1. Suppose $X$ and $Y$ are locally convex spaces and let $X$ be pseudosaturated. Then the bilinear mapping
$$
\begin{equation*}
(g,x)\in Y^\star\times X\mapsto g\colon x\in (Y:X)^\star
\end{equation*}
\notag
$$
is continuous (in the above sense). Proof. 1. Take $g_i\to 0$ and $S\in{\mathcal S}(X)$. Then for an arbitrary totally bounded set $\varPhi\subseteq Y:X$ the set $\varPhi(S)$ is totally bounded in $Y$ by Theorem 5.1 in [3]. So the net $g_i$ tends to zero uniformly on $\varPhi(S)$ (by the definition of the topology in $Y^\star$):
$$
\begin{equation*}
(g_i:x)(\varphi)=g_i(\varphi(x)) \overset{\mathbb C}{\underset{i\to\infty}{\underset{\varphi\in\varPhi,\,x\in S}{\rightrightarrows}}}0.
\end{equation*}
\notag
$$
This is true for each totally bounded set $\varPhi\subseteq Y:X$, hence
$$
\begin{equation*}
g_i\colon x \overset{Y:X}{\underset{i\to\infty}{\underset{x\in S}{\rightrightarrows}}}0.
\end{equation*}
\notag
$$
2. Take $G\in {\mathcal S}(Y^\star)$ and $x_i\to 0$. Then for an arbitrary totally bounded set $\varPhi\subseteq Y:X$ the set $G\circ\varPhi$ is totally bounded in $X^\star$ by Theorem 5.1 in [3]. Since $X$ is pseudosaturated, the set $G\circ\varPhi$ is equicontinuous on $X$. As a corollary,
$$
\begin{equation*}
(g:x_i)(\varphi)=g(\varphi(x_i)) \overset{\mathbb C}{\underset{i\to\infty}{\underset{\varphi\in\varPhi,\,g\in G}{\rightrightarrows}}}0.
\end{equation*}
\notag
$$
This is true for each totally bounded set $\varPhi\subseteq Y:X$, hence
$$
\begin{equation*}
g\colon x_i \overset{Y:X}{\underset{i\to\infty}{\underset{g\in G}{\rightrightarrows}}}0.
\end{equation*}
\notag
$$
Corollary 1. Suppose $X$ and $Y$ are locally convex spaces, and $X$ is pseudosaturated. If $S\subseteq X$ is totally bounded and $g\in Y^\star$, then $g:S\subseteq (Y:X)^\star$ is totally bounded. Dually, if $x\in X$ and $G\subseteq Y^\star$ is totally bounded, then $G:x\subseteq (Y:X)^\star$ is totally bounded. 2.2. Pseudosaturation of $Y:X$ Let $\{X^\lambda;\,\lambda\in\mathbf{Ord}\}$ be an injective series of the space $X$, and $\{Y_\mu;\,\mu\in\mathbf{Ord}\}$ be a projective series of the space $Y$, and let
$$
\begin{equation*}
\vee_\lambda^X\colon X\to X^\lambda\quad\text{and} \quad \wedge_\mu^Y\colon Y_\mu\to Y
\end{equation*}
\notag
$$
be the corresponding natural mappings. Consider the analogue of the diagram (5.1) in [3], It defines a mapping
$$
\begin{equation}
(\wedge_\mu^Y:\vee_\lambda^X)\colon (Y_\mu:X^\lambda) \to (Y:X).
\end{equation}
\tag{2.4}
$$
Lemma 1. If $X$ is pseudosaturated and $Y$ is pseudocomplete, then the mapping (2.4) is a bijection, so that the spaces $Y_\mu:X^\lambda$ and $Y:X$ can be identified as sets (while they can have different topologies as locally convex spaces):
$$
\begin{equation}
Y_\mu:X^\lambda=Y:X.
\end{equation}
\tag{2.5}
$$
Proof. The mapping $\vee_\lambda^X\colon X\to X^\lambda$ is an epimorphism. On the other hand $\wedge_\mu^Y$: $Y_\mu\to Y$ is an injection. This implies that the mapping
$$
\begin{equation*}
\psi\mapsto \wedge_\mu^Y\circ\psi\circ (\wedge_\lambda^X)^\star
\end{equation*}
\notag
$$
is an injection. We have to verify that it is a surjection. Indeed, let $\varphi\colon X\to Y$ be an arbitrary operator. We construct subsequently two operators $\varphi_\mu$ and $\psi$ such that the following diagram is commutative: Since $X$ is pseudosaturated, the operator $\varphi$ is uniquely extended to an operator $\varphi^\vartriangle\colon X\to Y^\vartriangle$. Its composition with the natural mapping $Y^\vartriangle\to Y_\mu$ is an operator that extends $\varphi$ to an operator $\varphi_\mu\colon X\to Y_\mu$.
Further, since $Y$ is pseudocomplete, $Y_\mu$ is again pseudocomplete (by Proposition 3.16 in [3]). Hence $\varphi_\mu\colon X\to Y_\mu$ is extended to an operator $\chi\colon X^\triangledown\to Y_\mu$. Its composition with the natural embedding $X^\lambda\to X^\triangledown$ is the operator $\psi\colon X^\lambda\to Y_\mu$.
The lemma is proved. Lemma 2. Let $X$ be pseudosaturated and $Y$ be pseudocomplete. If $\varPhi\in Y:X$ is totally bounded, then for any $\lambda,\mu\in\mathbf{Ord}$ its representation in the space $Y_\mu:X^\lambda$ (by the bijection (2.5)) is again a totally bounded set. Proof. 1. First consider the case $\lambda=0$. If $\varPhi\subseteq Y:X$ is totally bounded, then by Theorem 5.1 in [3] this means that $\varPhi$ is equicontinuous and uniformly totally bounded on each totally bounded set $S\subseteq X$. As a corollary, the image $\varPhi(S)$ is totally bounded in $Y$. The space $Y_\mu$ is a strengthening of the topology of $Y$, but the class of totally bounded sets in $Y$ and the topology on them are not changed. As a corollary, $\varPhi(S)$ is totally bounded in the space $Y_\mu$ as well, and moreover, with the same topology as the one induced from $Y$. We can conclude that the set $\varPhi$, being considered as the set of operators from $X$ into $Y_\mu$, is equicontinuous and uniformly totally bounded on $S$. And this is true for each totally bounded set $S\subseteq X$. Hence (again by Theorem 5.1 in [3]) $\varPhi$ is totally bounded in $Y_\mu:X$.
2. Thus, we have understood that if $\varPhi\subseteq Y:X$ is totally bounded in $Y:X$, then it is totally bounded in $Y_\mu:X$ for each $\mu\in\mathbf{Ord}$. If we take a big enough ordinal $\mu\in\mathbf{Ord}$, then we obtain that $\varPhi\subseteq Y:X$ is totally bounded in $Y^\vartriangle:X$. From this, by Lemma 5.10 in [3] we obtain that $\varPhi\subseteq Y:X$ is totally bounded in the space $Y^{\vartriangle\triangledown}:X^\triangledown$.
Since, by assumption, $Y$ is pseudocomplete, by Proposition 3.16 in [3], the space $Y^{\vartriangle}$ is also pseudocomplete, and as a corollary,
$$
\begin{equation*}
Y^{\vartriangle\triangledown}=Y^{\vartriangle}.
\end{equation*}
\notag
$$
We obtain that the set $\varPhi\subseteq Y:X$ is totally bounded in $Y^\vartriangle:X^\triangledown$.
Now, if we take arbitrary $\lambda,\mu\in{\bf Ord}$ and denote by
$$
\begin{equation*}
\sigma\colon X^\lambda\to X^\triangledown \quad\text{and}\quad \pi\colon Y_\mu\gets Y^\vartriangle
\end{equation*}
\notag
$$
the natural mappings, we obtain a linear continuous mapping
$$
\begin{equation*}
(\pi:\sigma)\colon (Y^\vartriangle:X^\triangledown) \to (Y_\mu:X^\lambda).
\end{equation*}
\notag
$$
It turns the totally bounded set $\varPhi\subseteq Y^\vartriangle:X^\triangledown$ into the totally bounded set ${\varPhi\subseteq Y_\mu:X^\lambda}$.
The lemma is proved. Lemma 3. Let $X$ be pseudosaturated and $Y$ be pseudocomplete. If $S\in {\mathcal S}(X^\lambda)$ and $V\in {\mathcal U}(Y_\mu)$, then $V:S\in {\mathcal D}(Y:X)$.3[x]3The notation $V:S$ was introduced in (2.1). Proof. Let $\varPhi\subseteq Y:X$ be a totally bounded set. By Lemma 2, $\varPhi$ is totally bounded in the space $Y_\mu:X^\lambda$ as well. Since $V:S$ is a neighbourhood of zero in the space $Y_\mu:X^\lambda$, there is a finite set $A\subseteq Y_\mu:X^\lambda$ such that
$$
\begin{equation}
\varPhi\subseteq V:S+A.
\end{equation}
\tag{2.6}
$$
By Lemma 1, all these sets can be considered as subsets of $Y:X$. We obtain the following: for each totally bounded set $\varPhi$ in $Y:X$ there is a finite set $A$ in $Y:X$ such that (2.6) holds. This means that the set $V:S$ is capacious in $Y:X$.
The lemma is proved. Lemma 4. If $X$ is pseudosaturated and $Y$ is pseudocomplete, then for any ordinals $\lambda,\mu\in\mathbf{Ord}$ (i) the spaces $Y_\mu:X^\lambda$ and $Y:X$ become isomorphic after pseudosaturation:
$$
\begin{equation}
(Y_\mu:X^\lambda)^\vartriangle=(Y:X)^\vartriangle;
\end{equation}
\tag{2.7}
$$
(ii) for each $x\in X^\lambda$ and $g\in (Y_\mu)^\star$ the functional
$$
\begin{equation*}
(g:x)\colon (Y_\mu:X^\lambda)\to\mathbb C \quad\bigm|\quad (g:x)(\varphi)=g(\varphi(x)), \qquad \varphi\in Y_\mu:X^\lambda,
\end{equation*}
\notag
$$
is continuous on the space $(Y:X)^\vartriangle$, that is, there is a unique functional $h\in (({Y:X})^\vartriangle)^\star$ such that the following diagram is commutative: Proof. We have to organize double induction on $\lambda$ and $\mu$. For $\lambda=\mu=0$ equality (2.8) becomes trivial:
$$
\begin{equation*}
(Y:X)^\vartriangle=(Y:X)^\vartriangle,
\end{equation*}
\notag
$$
and diagram (2.8) turns into the diagram where we just have to put $h=g\colon x\mathbin{\circ}{\vartriangle}_{X\cdot Y}$.
Suppose we have already proved assertions (i)$^\circ$ and (ii)$^\circ$ for all ordinals $\iota<\lambda$ and $\varkappa<\mu$, where $\lambda$ and $\mu$ are two given ordinals:
(i)$^\circ$ the spaces $Y_\varkappa:X^\iota$ and $Y:X$ become isomorphic after pseudosaturation:
$$
\begin{equation}
(Y_\varkappa:X^\iota)^\vartriangle=(Y:X)^\vartriangle;
\end{equation}
\tag{2.9}
$$
(ii)$^\circ$ for each $x\in X^\iota$ and $g\in (Y_\varkappa)^\star$ the functional
$$
\begin{equation*}
(g:x)\colon (Y_\varkappa:X^\iota)\to\mathbb C \quad\bigm|\quad (g:x)(\varphi)=g(\varphi(x)), \qquad \varphi\in Y_\varkappa:X^\iota,
\end{equation*}
\notag
$$
is continuous on the space $(Y:X)^\vartriangle$, that is, there exists a unique functional $h\in ((Y:X)^\vartriangle)^\star$ such that the following diagram is commutative:
Let us show that then (i) and (ii) hold for the ordinals $\lambda$ and $\mu$.
Step 1. First we fix an arbitrary ordinal $\varkappa$ such that $\varkappa<\mu$, and show that (i)$^\circ$ and (ii)$^\circ$ hold after the substitution $\iota=\lambda$. We have to consider two cases here.
a) Suppose $\lambda$ is an isolated ordinal, that is,
$$
\begin{equation*}
\lambda=\iota+1
\end{equation*}
\notag
$$
for some ordinal $\iota<\lambda$. We show that then (ii)$^\circ$ holds for $\lambda$ (substituted instead of $\iota$). Take
$$
\begin{equation*}
x\in X^\lambda=X^{\iota+1}=(X^\iota)^\vee \quad\text{and}\quad g\in(Y_\varkappa)^\star.
\end{equation*}
\notag
$$
Then $x$ is the limit of some totally bounded net $\{x_i;\,i\in I\}\subseteq X^\iota$:
$$
\begin{equation}
x_i\overset{(X^\iota)^\vee}{\underset{i\to\infty}{\longrightarrow}}x.
\end{equation}
\tag{2.11}
$$
Since the set $\{x_i\}\subseteq X^\iota$ is totally bounded, by Corollary 1 the set
$$
\begin{equation*}
\{g:x_i;\,i\in I\}
\end{equation*}
\notag
$$
is totally bounded in $(Y_\varkappa:X^\iota)^\star$. On the other hand, from Theorem 1 we obtain that $\{g:x_i;\, i\in I\}$ is a Cauchy net. Thus, $\{g:x_i;\, i\in I\}$ is a totally bounded Cauchy net in $(Y_\varkappa:X^\iota)^\star$, and as a corollary, it converges in the enveloping pseudocomplete space $(Y_\varkappa:X^\iota)^{\star\triangledown}$:
$$
\begin{equation*}
(Y_\varkappa:X^\iota)^\star\subseteq (Y_\varkappa:X^\iota)^{\star\triangledown} =(Y_\varkappa:X^\iota)^{\vartriangle\star} \stackrel{(2.9)}{=} (Y:X)^{\vartriangle\star}
\end{equation*}
\notag
$$
(the first equality follows from [ 3], Theorem 3.14). In other words, there is a functional $h\in (Y:X)^{\vartriangle\star}$ such that
$$
\begin{equation}
g\colon x_i\overset{(Y:X)^{\vartriangle\star}}{\underset{i\to\infty}{\longrightarrow}}h.
\end{equation}
\tag{2.12}
$$
On the other hand, from the chain of equalities
$$
\begin{equation*}
(X^\iota)^\vee=X^{\iota+1}=X^\lambda
\end{equation*}
\notag
$$
we obtain that (2.11) is equivalent to the relation
$$
\begin{equation}
x_i\overset{X^\lambda}{\underset{i\to\infty}{\longrightarrow}} x,
\end{equation}
\tag{2.13}
$$
which by Theorem 1 implies that
$$
\begin{equation}
g\colon x_i\overset{(Y_\varkappa:X^\lambda)^{\star}}{\underset{i\to\infty}{\longrightarrow}} g:x.
\end{equation}
\tag{2.14}
$$
Together (2.12) and (2.14) mean, in particular, that the functionals $g:x$ and $h$ coincide at each operator
$$
\begin{equation*}
\varphi\in Y_\varkappa:X^\lambda
\end{equation*}
\notag
$$
(by Lemma 1, the spaces $Y_\varkappa:X^\lambda$ and $Y:X$ coincide as sets, and the difference can be only in topologies):
$$
\begin{equation*}
(g:x)(\varphi)=h(\varphi).
\end{equation*}
\notag
$$
This proves (ii)$^\circ$ (for $\lambda=\iota+1$ substituted instead of $\iota$).
Further we prove (i)$^\circ$ (again for $\lambda=\iota+1$ instead of $\iota$). The map
$$
\begin{equation*}
(Y_\varkappa:X^\lambda)^\vartriangle\to (Y:X)^\vartriangle
\end{equation*}
\notag
$$
is also continuous. We need to verify that its inverse map
$$
\begin{equation*}
(Y:X)^\vartriangle\to (Y_\varkappa:X^\lambda)^\vartriangle
\end{equation*}
\notag
$$
is continuous. Since the space $(Y:X)^\vartriangle$ is pseudosaturated, it is sufficient to prove the continuity of the map
$$
\begin{equation*}
(Y:X)^\vartriangle\to (Y_\varkappa:X^\lambda).
\end{equation*}
\notag
$$
Take a basic neighbourhood of zero in $Y_\varkappa:X^\lambda$, that is, a set $V:S$, where $V$ is a closed convex balanced neighbourhood of zero in $Y_\varkappa$, and $S$ is a totally bounded set in $X^\lambda$. By Lemma 3, the set $V:S$ is capacious in $Y:X$. Hence if we prove that it is closed in $(Y:X)^\vartriangle$, this will mean that it is a neighbourhood of zero in $(Y:X)^\vartriangle$.
This becomes obvious if we represent $V:S$ as the polar of the system of functionals of the form $\{g:x | \,g\in V^\circ,\,x\in S\}$:
$$
\begin{equation}
V:S=\{g:x;\,g\in V^\circ,\,x\in S\}^\circ.
\end{equation}
\tag{2.15}
$$
Indeed,
$$
\begin{equation}
\begin{aligned} \, \notag \varphi\in V:S \quad&\Longleftrightarrow\quad \varphi(S)\subseteq V \quad\Longleftrightarrow\quad \forall\, x\in S \quad \varphi(x)\in V \\ &\Longleftrightarrow\quad \forall\, g\in V^\circ, \quad \forall\, x\in S \quad |\underbrace{g(\varphi(x))}_{\substack{\| \\ (g:x)(\varphi)}}|\leqslant 1 \notag\\ &\Longleftrightarrow\quad \varphi\in \{g:x;\,g\in V^\circ,\, x\in S\}^\circ. \end{aligned}
\end{equation}
\tag{2.16}
$$
Now we must notice that in formula (2.15) we have $x\in S$, where $S$ is a totally bounded set in $X^\lambda$, and $g\in V^\circ$, where $V$ is a neighbourhood of zero in $Y_\varkappa$. Hence $x\in X^\lambda$ and $g\in (Y_\varkappa)^\star$. From this, by (ii)$^\circ$ (which we have already proved for $\lambda=\iota+1$) we have that the $g:x$ are continuous functionals on the space $(Y:X)^\vartriangle$. Thus $V:S$ is the polar of a system of continuous functionals on $(Y:X)^\vartriangle$, and therefore $V:S$ is a closed set in $(Y:X)^\vartriangle$. In addition, it is capacious, hence it is a neighbourhood of zero in the (pseudosaturated) space $(Y:X)^\vartriangle$. We have proved (i)$^\circ$ (for $\lambda=\iota+1$ substituted instead of $\iota$).
b) Now we consider the case when $\lambda$ is a limit ordinal, that is,
$$
\begin{equation*}
\lambda\ne \iota+1
\end{equation*}
\notag
$$
for all $\iota<\lambda$. Here we must again start with (ii)$^\circ$. Take $x\in X^\lambda$ and $g\in (Y_\varkappa)^\star$. Then
$$
\begin{equation*}
x\in X^\lambda=\bigcup_{\iota<\lambda} X^\iota
\end{equation*}
\notag
$$
(the equality follows from formula (1.19) in [ 3]), hence there exists an ordinal ${\iota<\lambda}$ such that $x\in X^\iota$. We obtain $x\in X^\iota$ and $g\in (Y_\varkappa)^\star$, hence by induction assumption (ii)$^{\circ}$, the functional $g:x$ must be continuous on the space $(Y:X)^\vartriangle$. This proves (ii)$^\circ$ (in the case of the limit ordinal $\lambda$ substituted instead of $\iota<\lambda$).
Now we can prove (i)$^\circ$. Here we repeat the reasoning in a): we have to prove the continuity of the mapping
$$
\begin{equation*}
(Y:X)^\vartriangle\to (Y_\varkappa:X^\lambda).
\end{equation*}
\notag
$$
We take a basic neighbourhood of zero in $Y_\varkappa:X^\lambda$, that is, a set $V:S$, where $S$ is totally bounded in $X^\lambda$ and $V$ is a closed convex balanced neighbourhood of zero in $Y_\varkappa$. Then we prove (2.15), and we notice that in this equality $x\mkern-3mu \in\mkern-3mu X^\lambda$ and ${g\mkern-3mu \in\mkern-3mu (Y_\varkappa)^\star}$. After that, by the already proved (ii)$^\circ$ (for $\lambda$ substituted instead of ${\iota<\lambda}$) we obtain that the $g:x$ are continuous functionals on the space $(Y:X)^\vartriangle$. Therefore, $V:S$ is a closed set in $(Y:X)^\vartriangle$. In addition, it is capacious by Lemma 3, hence it is a neighbourhood of zero in the (pseudosaturated) space $(Y:X)^\vartriangle$. Thus we have proved (i)$^\circ$ (in the case of the limit ordinal $\lambda$ substituted instead of $\iota<\lambda$).
Step 2. We have shown that, given $\varkappa<\mu$, conditions (i)$^\circ$ and (ii)$^\circ$ hold if we replace there $\iota<\lambda$ by $\iota=\lambda$. Now let us take $\iota=\lambda$, and show that (i)$^\circ$ and (ii)$^\circ$ (with $\iota=\lambda$) remain true if we replace there $\varkappa<\mu$ by $\varkappa=\mu$. Again, we have to consider two cases.
a) First we assume that $\mu$ is an isolated ordinal, that is,
$$
\begin{equation*}
\mu=\varkappa+1
\end{equation*}
\notag
$$
for some ordinal $\varkappa<\mu$. Let us show that condition (ii)$^\circ$ also holds for $\mu$ (substituted instead of $\varkappa$). Take
$$
\begin{equation*}
g\in (Y_\mu)^\star=(Y_{\varkappa+1})^\star=((Y_\varkappa)^\wedge)^\star =((Y_\varkappa)^\star)^\vee
\end{equation*}
\notag
$$
(the third equality follows from [ 3], Theorem 3.10) and
$$
\begin{equation*}
x\in X^\lambda.
\end{equation*}
\notag
$$
Then $g$ is the limit of some totally bounded net $\{g_i;\,i\in I\}\subseteq (Y_\varkappa)^\star$:
$$
\begin{equation}
g_i\overset{((Y_\varkappa)^\star)^\vee}{\underset{i\to\infty}{\longrightarrow}}g.
\end{equation}
\tag{2.17}
$$
Since the set $\{g_i\}\subseteq (Y_\varkappa)^\star$ is totally bounded, by Corollary 1 the set
$$
\begin{equation*}
\{g_i:x;\,i\in I\}
\end{equation*}
\notag
$$
is totally bounded in $(Y_\varkappa:X^\lambda)^\star$. On the other hand, Theorem 1 implies that $\{g_i:x;\,i\in I\}$ is a Cauchy net. Thus, we obtain that $\{g_i:x;\,i\in I\}$ is a totally bounded Cauchy net in the space $(Y_\varkappa:X^\lambda)^\star$, hence it converges in the enveloping pseudocomplete space $(Y_\varkappa:X^\lambda)^{\star\triangledown}$:
$$
\begin{equation*}
(Y_\varkappa:X^\lambda)^\star\subseteq (Y_\varkappa:X^\lambda)^{\star\triangledown}= (Y_\varkappa:X^\lambda)^{\vartriangle\star} \stackrel{(2.9)}{=}(Y:X)^{\vartriangle\star}
\end{equation*}
\notag
$$
(the first equality follows from Theorem 3.14 in [ 3]). That is, there exists a functional $h\in (Y:X)^{\vartriangle\star}$ such that
$$
\begin{equation}
g_i\colon x\overset{(Y:X)^{\vartriangle\star}}{\underset{i\to\infty}{\longrightarrow}}h.
\end{equation}
\tag{2.18}
$$
On the other hand, from the chain of equalities
$$
\begin{equation*}
((Y_\varkappa)^\star)^\vee= ((Y_\varkappa)^\wedge)^\star=(Y_{\varkappa+1})^\star=(Y_\mu)^\star
\end{equation*}
\notag
$$
(the first equality follows from Theorem 3.10 in [ 3]) we see that the relation (2.17) is equivalent to the relation
$$
\begin{equation}
g_i\overset{(Y_\mu)^\star}{\underset{i\to\infty}{\longrightarrow}} g,
\end{equation}
\tag{2.19}
$$
from which by Theorem 1 it follows that
$$
\begin{equation}
g_i\colon x\overset{(Y_\mu:X^\lambda)^{\star}}{\underset{i\to\infty}{\longrightarrow}} g:x.
\end{equation}
\tag{2.20}
$$
Together (2.18) and (2.20) mean, in particular, that the functionals $g:x$ and $h$ coincide at each operator
$$
\begin{equation*}
\varphi\in Y_\mu:X^\lambda
\end{equation*}
\notag
$$
(by Lemma 1, the spaces $Y_\mu:X^\lambda$ and $Y:X$ coincide as sets, but the difference can be in their topologies):
$$
\begin{equation*}
(g:x)(\varphi)=h(\varphi).
\end{equation*}
\notag
$$
This proves (ii)$^\circ$ (for $\mu=\varkappa+1$ substituted instead of $\varkappa$).
Now we prove (i)$^\circ$ (again for $\mu=\varkappa+1$ substituted instead of $\varkappa$). The mapping
$$
\begin{equation*}
(Y_\mu:X^\lambda)^\vartriangle\to (Y:X)^\vartriangle
\end{equation*}
\notag
$$
is continuous. We need to prove that its inverse mapping
$$
\begin{equation*}
(Y:X)^\vartriangle\to (Y_\mu:X^\lambda)^\vartriangle
\end{equation*}
\notag
$$
is also continuous. Since the space $(Y:X)^\vartriangle$ is pseudosaturated, it is sufficient to prove the continuity of the mapping
$$
\begin{equation*}
(Y:X)^\vartriangle\to (Y_\mu:X^\lambda).
\end{equation*}
\notag
$$
Consider a basic neighbourhood of zero in $Y_\mu:X^\lambda$, that is, a set $V:S$, where $V$ is a closed convex balanced neighbourhood of zero in $Y_\mu$ and $S$ is a totally bounded set in $X^\lambda$. By Lemma 3, the set $V:S$ is capacious in $Y:X$. Hence if we prove that it is closed in $(Y:X)^\vartriangle$, this will mean that it is a neighbourhood of zero in $(Y:X)^\vartriangle$.
This becomes obvious if we represent $V:S$ as the polar of the system of functionals of the form $\{g:x;\,g\in V^\circ,\,x\in S\}$:
$$
\begin{equation}
V:S=\{g:x;\,g\in V^\circ,\,x\in S\}^\circ,
\end{equation}
\tag{2.21}
$$
which is proved by the same chain of implications (2.16) as before. Next we notice that in formula (2.21) $x\in S$, where $S$ is a totally bounded set in $X^\lambda$, and ${g\in V^\circ}$, where $V$ is a neighbourhood of zero in $Y_\mu$. Hence $x\in X^\lambda$, and $g\in (Y_\mu)^\star$. From this, by (ii)$^\circ$ (which we have already proved for $\mu=\varkappa+1$) the $g:x$ are continuous functionals on the space $(Y:X)^\vartriangle$. We see that $V:S$ is the polar of a system of continuous functionals on $(Y:X)^\vartriangle$, hence $V:S$ is a closed set in $(Y:X)^\vartriangle$. In addition, it is capacious, hence it is a neighbourhood of zero in the (pseudosaturated) space $(Y:X)^\vartriangle$. We proved (i)$^\circ$ (for $\mu=\varkappa+1$ substituted instead of $\varkappa$).
b) Now consider the case where $\mu$ is a limit ordinal, that is,
$$
\begin{equation*}
\mu\ne \varkappa+1
\end{equation*}
\notag
$$
for all $\varkappa<\mu$. Here we start with (ii)$^\circ$ again. Take $x\in X^\lambda$ and $g\in (Y_\mu)^\star$. Then
$$
\begin{equation*}
g\in (Y_\mu)^\star =(Y^\star)^\mu =\bigcup_{\varkappa<\mu} (Y^\star)^\varkappa=\bigcup_{\varkappa<\mu}(Y_\varkappa)^\star
\end{equation*}
\notag
$$
(the first and third equalities follow from Theorem 3.12 in [ 3], and the second follows from [ 3], (1.19)), so there is an ordinal $\varkappa<\mu$ such that $g\in (Y^\star)^\varkappa=(Y_\varkappa)^\star$. We have $x\in X^\iota$ and $g\in (Y_\varkappa)^\star$, hence by the induction assumption in (ii)$^\circ$ the functional $g:x$ must be continuous on the space $(Y:X)^\vartriangle$. This proves (ii)$^\circ$ (for the limit ordinal $\mu$ substituted instead of $\varkappa$).
We can now prove (i)$^\circ$. We repeat here the reasoning of a): we need to prove the continuity of the mapping
$$
\begin{equation*}
(Y:X)^\vartriangle\to (Y_\mu:X^\lambda).
\end{equation*}
\notag
$$
We take a basic neighbourhood of zero in $Y_\mu:X^\lambda$, that is, a set $V:S$, where $S$ is a totally bounded set in $X^\lambda$, and $V$ is a closed convex balanced neighbourhood of zero in $Y_\mu$. Then we prove (2.21), and we notice that there $x\in X^\lambda$ and $g\in (Y_\mu)^\star$. After that, by (ii)$^\circ$ (which we have already proved for the limit ordinal $\mu$ substituted instead of $\varkappa$), we obtain that $g:x$ is a continuous functional on the space $(Y:X)^\vartriangle$. Hence $V:S$ is a closed set in $(Y:X)^\vartriangle$. In addition, it is capacious by Lemma 3, so it is a neighbourhood of zero in the (pseudosaturated) space $(Y:X)^\vartriangle$. We have proved (i)$^\circ$ (in the case of the limit ordinal $\mu$ substituted instead of $\varkappa$).
Lemma 4 is proved. For arbitrary locally convex spaces $X$ and $Y$ the mappings
$$
\begin{equation*}
\triangledown_X\colon X\to X^\triangledown \quad\text{and}\quad \vartriangle_Y\colon Y^\vartriangle\to Y
\end{equation*}
\notag
$$
define the mapping
$$
\begin{equation}
(\vartriangle_Y:\triangledown_X)\colon (Y^\vartriangle:X^\triangledown) \to (Y:X)
\end{equation}
\tag{2.22}
$$
(which certainly coincides with the mapping (2.4) for some $\lambda$ and $\mu$). Theorem 2. Let $X$ be pseudosaturated and $Y$ be pseudocomplete. Then the pseudosaturation of the mapping (2.22) is an isomorphism of locally convex (and stereotype) spaces:
$$
\begin{equation}
Y^\vartriangle\oslash X^\triangledown\cong (Y^\vartriangle:X^\triangledown)^\vartriangle\cong (Y:X)^\vartriangle.
\end{equation}
\tag{2.23}
$$
Proof. The second equality follows from (2.7) if we choose ordinals $\lambda$ and $\mu$ such that $X^\lambda=X^\triangledown$ and $Y_\mu=Y^\vartriangle$. And the first is just the definition (1.11) of the tensor fraction $\oslash$ (here the spaces $X^\triangledown$ and $Y^\vartriangle$ are stereotype by Propositions 3.17 and 3.16 in [3]).
The theorem is proved.
§ 3. Pseudocompletion of the primary tensor product $X\cdot Y$3.1. Primary tensor product $X\cdot Y$ The construction of the primary tensor product $X\cdot Y$ that we define here coincides with the so-called $\varepsilon$-product introduced by Bierstedt in [13],
$$
\begin{equation*}
X\cdot Y=X\varepsilon Y,
\end{equation*}
\notag
$$
and in the case when $X$ is complete, with the $\varepsilon$-product in Jarchow’s book [14]. We use the term ‘primary tensor product’ having in mind formula (3.29) that we prove below, from which it follows that $X\cdot Y$ plays the same role for the injective stereotype tensor product $X\odot Y$ as the primary fraction $Y:X$ plays for the stereotype tensor fraction $Y\oslash X$ in (2.23): applying $\vartriangle$ to $X\cdot Y$ we obtain $X\odot Y$. Let $X$ and $Y$ be locally convex spaces. By the primary tensor product $X\cdot Y$ we mean the locally convex space of linear continuous mappings $\varphi\colon X^\star\to Y$, endowed with the topology of uniform convergence on the polars of neighbourhoods of zero $U\subseteq X$:
$$
\begin{equation}
\varphi_i\overset{X\cdot Y}{\underset{i\to\infty}{\longrightarrow}}\varphi \quad\Longleftrightarrow\quad \forall\, U\in{\mathcal U}(X) \quad \varphi_i(f)\overset{Y}{\underset{f\in U^\circ}{\underset{i\to\infty}{\rightrightarrows}}}\varphi(f).
\end{equation}
\tag{3.1}
$$
It will be useful to choose a letter for this topology; let it be $\xi$. Then the space $X\cdot Y$ can be presented by the formula
$$
\begin{equation}
X\cdot Y=Y\underset{\xi}{:} X^\star
\end{equation}
\tag{3.2}
$$
(the index $\xi$ means convergence in the topology $\xi$). Obviously, there is a bijective linear continuous mapping
$$
\begin{equation*}
Y\colon X\to Y\underset{\xi}{:} X^\star=X\cdot Y
\end{equation*}
\notag
$$
(which however is not an isomorphism if $X$ is not pseudosaturated). If $A\subseteq X$ and $B\subseteq Y$, then the symbol $A\cdot B$ denotes the set
$$
\begin{equation*}
A\cdot B=B:(A^\circ)
\end{equation*}
\notag
$$
(where the colon ‘ : ’ was defined in (2.1)). If $\alpha\colon X'\to X$ and $\beta\colon Y'\to Y$ are operators, then the diagram defines a mapping
$$
\begin{equation*}
\varphi\cdot\chi\colon X'\cdot Y'\to X\cdot Y.
\end{equation*}
\notag
$$
Proposition 1. For any operators $\alpha:X'\to X$ and $\beta:Y'\to Y$ the mapping
$$
\begin{equation*}
\varphi\cdot\chi\colon X'\cdot Y'\to X\cdot Y
\end{equation*}
\notag
$$
is continuous. Proof. The continuity of $\alpha$ implies that the dual operator $\alpha^\star\colon X^\star\to (X')^\star$ turns the polar of an arbitrary neighbourhood of zero $U\subseteq X$ into the polar of some neighbourhood of zero $U'$, namely, $U'=\alpha^{-1}(U)$: As a corollary the convergence $\psi_i\to \psi$ in $X'\cdot Y'$
$$
\begin{equation*}
\forall\, U'\in{\mathcal U}(X') \quad \psi_i(g)\overset{Y'}{\underset{g\in (U')^\circ}{\underset{i\to\infty}{\rightrightarrows}}}\psi(g)
\end{equation*}
\notag
$$
implies the convergence $(\alpha\cdot\beta)(\psi_i)\to (\alpha\cdot\beta)(\psi)$ in $X\cdot Y$:
$$
\begin{equation*}
\forall\, U\in{\mathcal U}(X) \quad (\alpha\cdot\beta)(\psi_i)(f)=\beta\bigl(\psi_i(\alpha^\star(f))\bigr) \overset{Y}{\underset{f\in U^\circ}{\underset{i\to\infty}{\rightrightarrows}}} \beta\bigl(\psi(\alpha^\star(f))\bigr) =(\alpha\cdot\beta)(\psi)(f).
\end{equation*}
\notag
$$
The proposition is proved. Suppose further that $X$ and $Y$ are locally convex spaces, and $X$ is pseudocomplete. For each operator $\varphi\colon X^\star\to Y$ we can consider the dual operator $\varphi\colon {Y^\star\to X^{\star\star}}$, and after that form the composition with the operator $i_X^{-1}\colon {X^{\star\star}\,{\to}\, X}$ (which due to Corollary 2.13 in [3] exists and is continuous, since $X$ is pseudocomplete). Let us denote this composition by $\omega_{X,Y}(\varphi)$: We have the mapping
$$
\begin{equation}
\omega_{X,Y}\colon X\cdot Y=Y\underset{\xi}{:}X^\star\to X\underset{\xi}{:}Y^\star=Y\cdot X \quad\bigm|\quad \omega_{X,Y}(\varphi)=i_X^{-1}\circ\varphi^\star.
\end{equation}
\tag{3.5}
$$
Theorem 3. If locally convex spaces $X$ and $Y$ are pseudocomplete, then the mapping (3.5) establishes an isomorphism of locally convex spaces
$$
\begin{equation}
X\cdot Y\cong Y\cdot X.
\end{equation}
\tag{3.6}
$$
Proof. 1. First we must notice the following identity:
$$
\begin{equation}
f(\omega_{X,Y}(\varphi)(g))=g(\varphi(f)), \qquad f\in X^\star, \quad g\in Y^\star, \quad \varphi\in Y:(X^\star).
\end{equation}
\tag{3.7}
$$
For a proof we put
$$
\begin{equation*}
x=\omega_{X,Y}(\varphi)(g)=i_X^{-1}(\varphi^\star(g))=i_X^{-1}(g\circ\varphi)\in X.
\end{equation*}
\notag
$$
Then we have
$$
\begin{equation*}
f(\underbrace{\omega_{X,Y}(\varphi)(g)}_{\substack{\| \\ x}})=f(x)=i_X(x)(f) =i_X(i_X^{-1}(g\circ\varphi))(f)=(g\circ\varphi)(f)=g(\varphi(f)).
\end{equation*}
\notag
$$
2. When (3.7) is proved, we must verify that the mapping $\omega_{X,Y}$ is bijective. This follows from the identity
$$
\begin{equation}
\omega_{Y,X}(\omega_{X,Y}(\varphi))=\varphi, \qquad \varphi\in X\cdot Y=Y\underset{\xi}{:}X^\star.
\end{equation}
\tag{3.8}
$$
Indeed, for each $f\in X^\star$ and $g\in Y^\star$ we have
$$
\begin{equation*}
g(\omega_{Y,X}(\omega_{X,Y}(\varphi))(f)) \stackrel{(3.7)}{=}f(\omega_{X,Y}(\varphi)(g)) \stackrel{(3.7)}{=}g(\varphi(f)).
\end{equation*}
\notag
$$
3. Finally, let us prove the continuity of $\omega_{X,Y}$ in both directions. Indeed, if $U:(V^\circ)$ is a basic neighbourhood of zero in $X\underset{\xi}{:}Y^\star=Y\cdot X$ (where $U\subseteq X$ and $V\subseteq Y$ are closed convex balanced neighbourhood of zero), then $V:(U^\circ)$ is a basic neighbourhood of zero in $Y\underset{\xi}{:}X^\star=X\cdot Y$, and they turn into each other under the mappings $\omega_{X,Y}$ and $\omega_{X,Y}^{-1}$:
$$
\begin{equation*}
\begin{aligned} \, \omega_{X,Y}(\varphi)\in U:(V^\circ) \quad&\Longleftrightarrow\quad \omega_{X,Y}(\varphi)(V^\circ)\subseteq U \\ &\Longleftrightarrow \quad \forall\, f\in U^\circ, \quad \forall\, g\in V^\circ \quad |\underbrace{f(\omega_{X,Y}(\varphi)(g))}_{ \substack{\|(3.7) \\ g(\varphi(f))}}|\leqslant 1 \\ &\Longleftrightarrow \ \ \forall\, f\in U^\circ, \ \ \forall\, g\in V^\circ \ \ |g(\varphi(f))|\leqslant 1 \\ &\Longleftrightarrow\ \ \varphi(U^\circ)\subseteq V \ \ \Longleftrightarrow \ \ \varphi\,{\in}\,V:(U^\circ). \end{aligned}
\end{equation*}
\notag
$$
Theorem 3 is proved. Let us go back to the diagram (3.3). In the special case when $\alpha$ and $\beta$ are functionals
$$
\begin{equation*}
f\colon X\to\mathbb C \quad\text{and}\quad g\colon Y\to\mathbb C,
\end{equation*}
\notag
$$
the mapping $\alpha\cdot\beta$ can be naturally understood as the functional
$$
\begin{equation*}
f\cdot g\colon X\cdot Y\to\mathbb C,
\end{equation*}
\notag
$$
acting by the formula
$$
\begin{equation}
(f\cdot g)(\varphi)=g(\varphi(f)), \qquad \varphi\colon X^\star\to Y.
\end{equation}
\tag{3.9}
$$
If $F\subseteq X^\star$ and $G\subseteq Y^\star$ are two sets of functionals, then the symbol $F\cdot G$ denotes the set
$$
\begin{equation*}
F\cdot G=\{f\cdot g;\,f\in F,\, g\in G\}.
\end{equation*}
\notag
$$
In particular, if $f\in X^\star$ and $G\subseteq Y^\star$, then
$$
\begin{equation*}
f\cdot G=\{f\cdot g;\,g\in G\},
\end{equation*}
\notag
$$
and if $F\subseteq X^\star$ and $g\in Y^\star$, then
$$
\begin{equation*}
F\cdot g=\{f\cdot g;\,f\in F\}.
\end{equation*}
\notag
$$
Theorem 4. For any pseudocomplete locally convex spaces $X$ and $Y$ the mapping
$$
\begin{equation*}
(f,g)\in X^\star\times Y^\star\mapsto f\cdot g\in (X\cdot Y)^\star
\end{equation*}
\notag
$$
is separately continuous. Proof. Let $\varPhi\subseteq X\cdot Y=Y\underset{\xi}{:}X^\star$ be a totally bounded set, and let $f\in X^\star$. By the definition of the topology $\xi$, $\varPhi$ is equicontinuous and uniformly totally bounded on the polar $U^\circ$ of an arbitrary neighbourhood of zero $U\subseteq X$ (we use here a variant of Theorem 5.1 from [3]). In particular, if we take the neighbourhood $U=\{f\}^\circ$, then the set of operators $\varPhi$ turns its polar
$$
\begin{equation*}
U^\circ=\{f\}^{\circ\circ}=\{\lambda\in\mathbb C\colon |\lambda|\leqslant 1\}\cdot f
\end{equation*}
\notag
$$
into a totally bounded set in $Y$:
$$
\begin{equation*}
\varPhi(U^\circ)=\{\lambda\in\mathbb C\colon |\lambda|\leqslant 1\}\cdot \varPhi(f)\subseteq Y.
\end{equation*}
\notag
$$
As a corollary, the polar of this set
$$
\begin{equation*}
V=\{\lambda\in\mathbb C\colon |\lambda|\leqslant 1\}\cdot \varPhi(f)^\circ\subseteq Y^\star
\end{equation*}
\notag
$$
is a neighbourhood of zero in $Y^\star$. For each $g\in V$ we have
$$
\begin{equation*}
\sup_{\varphi\in\varPhi}|(f\cdot g)(\varphi)|=\sup_{\varphi\in\varPhi}|g(\varphi(f))|\leqslant 1,
\end{equation*}
\notag
$$
that is, $f\cdot g\in\varPhi^\circ$. In other words,
$$
\begin{equation}
f\cdot V\subseteq \varPhi^\circ.
\end{equation}
\tag{3.10}
$$
We can say that for each basic neighbourhood of zero $\varPhi^\circ$ in $(X\cdot Y)^\star$ (where $\varPhi$ is a totally bounded set in $X\cdot Y$) and for any vector $f\in X^\star$ there is a neighbourhood of zero $V\subseteq Y^\star$ such that (3.10) holds. By (3.6) the same is true if we transpose $X$ and $Y$.
The theorem is proved. Corollary 2. If $F\subseteq X^\star$ is totally bounded, and $g\in Y^\star$, then $F\cdot g\subseteq (X\cdot Y)^\star$ is totally bounded. Dually, if $f\in X^\star$ and $G\subseteq Y^\star$ is totally bounded, then ${f\cdot G\subseteq (X\cdot Y)^\star}$ is totally bounded. Proposition 2. Let $X$ and $Y$ be complete locally convex spaces, with $Y$ having the (classical) approximation property. Then the primary tensor product $X\cdot Y$ is isomorphic to the injective tensor product
$$
\begin{equation}
X\cdot Y\cong X\widetilde{\otimes}_{\varepsilon} Y.
\end{equation}
\tag{3.11}
$$
If, in addition, $Y$ is nuclear, then $X\cdot Y$ is isomorphic to the projective tensor product
$$
\begin{equation}
X\cdot Y\cong X\widetilde{\otimes}_{\varepsilon} Y\cong X\widetilde{\otimes}_{\pi} Y.
\end{equation}
\tag{3.12}
$$
The first part of this proposition was stated in an article of Bierstedt (see [13], Satz 3.9(3)), and it can be deduced from proposition 16.1.5 in Jarchow’s book [14] (and we have to use a proposition from Köthe’s book (see [6], 21.9, (7)), where he states that the topologies that Bierstedt and Jarchow use on the space $X'$ in the definition of $X\varepsilon Y$ coincide). The second part follows from the characterization of nuclear spaces by the actions of the injective and projective tensor products on them; see [15], § 7.3.3. 3.2. Pseudosaturation of $X\cdot Y$ Let $\{X_\lambda;\,\lambda\in\mathbf{Ord}\}$ and $\{Y_\mu;\, \mu\in\mathbf{Ord}\}$ be projective series of the spaces $X$ and $Y$ (see [3], § 1.4, (b)) and
$$
\begin{equation*}
\wedge_\lambda^X\colon X_\lambda\to X \quad\text{and}\quad \wedge_\mu^Y\colon Y_\mu\to Y
\end{equation*}
\notag
$$
be the corresponding natural mappings. If we consider the analogue of the diagram (3.3), then we obtain the mapping
$$
\begin{equation}
\wedge_\lambda^X\cdot\,\wedge_\mu^Y\colon X_\lambda\cdot Y_\mu\to X\cdot Y,
\end{equation}
\tag{3.14}
$$
which is continuous by Proposition 1. Lemma 5. If $X$ and $Y$ are pseudocomplete, then (3.14) is a bijection, hence the spaces $X_\lambda\cdot Y_\mu$ and $X\cdot Y$ can be identified as sets (with the possible difference in topologies):
$$
\begin{equation}
X_\lambda\cdot Y_\mu=X\cdot Y.
\end{equation}
\tag{3.15}
$$
Proof. The mapping $\wedge_\lambda^X\colon X_\lambda\to X$ is an injection, hence that dual mapping $(\wedge_\lambda^X)^\star$: ${(X_\lambda)^\star\gets X^\star}$ is an epimorphism. On the other hand $\wedge_\mu^Y\colon Y_\mu\to Y$ is an injection. This implies that the mapping
$$
\begin{equation*}
\psi\mapsto \wedge_\mu^Y\circ\psi\circ (\wedge_\lambda^X)^\star
\end{equation*}
\notag
$$
is an injection. We only have to verify that it is a surjection. Indeed, let $\varphi\colon X^\star\to Y$ be an arbitrary operator. We construct subsequently two operators $\varphi_\mu$ and $\psi$, such that the following diagram is commutative: Since $X$ is pseudocomplete, the space $X^\star$ is pseudosaturated. This means that $\varphi$ is uniquely lifted to an operator $\varphi^\vartriangle\colon X^\star\to Y^\vartriangle$. If we consider its composition with the natural mapping $Y^\vartriangle\to Y_\mu$, we obtain a lifting of $\varphi$ to $\varphi_\mu\colon X^\star\to Y_\mu$.
Further, since $Y$ is pseudocomplete, $Y_\mu$ is also pseudocomplete. Hence the operator $\varphi_\mu\colon X^\star\to Y_\mu$ extends to some operator $\chi\colon (X^\star)^\triangledown\to Y_\mu$. If we consider its composition with the natural embedding $(X^\star)^\lambda\to (X^\star)^\triangledown$, then we obtain an operator $\psi\colon (X^\star)^\lambda\to Y_\mu$. The equality
$$
\begin{equation*}
(X^\star)^\lambda=(X_\lambda)^\star
\end{equation*}
\notag
$$
(see [ 3], (3.11)) allows us to present $\psi$ as an operator $\psi\colon (X_\lambda)^\star\to Y_\mu$.
The lemma is proved. Lemma 6. Let $X$ and $Y$ be pseudocomplete. If $\varPhi\in X\cdot Y$ is a totally bounded set, then for any $\lambda,\mu\in\mathbf{Ord}$ its representation in the space $X_\lambda\cdot Y_\mu$ (by the bijection (3.15)) is also a totally bounded set. Proof. 1. First consider the case when $\lambda=0$. If $\varPhi\subseteq Y\underset{\xi}{:}X^\star$ is a totally bounded set, then this means that $\varPhi$ is equicontinuous and uniformly totally bounded on the polar $U^\circ$ of every neighbourhood of zero $U\subseteq X$ (we use here a variant of Theorem 5.1 in [3]). As a corollary, the image $\varPhi(U^\circ)$ is a totally bounded subset in $Y$. The space $Y_\mu$ is a strengthening of the topology on $Y$ such that the class of totally bounded sets and the topology on these sets are not changed. Hence $\varPhi(U^\circ)$ is totally bounded in the space $Y_\mu$ as well, and moreover, with the same induced topology as the one induced from $Y$. We can conclude from this that $\varPhi$, considered as a set of operators from $X^\star$ to $Y_\mu$, is equicontinuous and uniformly totally bounded on $U^\circ$. And this is true for each neighbourhood of zero $U\subseteq X$. Hence $\varPhi$ is totally bounded in $Y_\mu\underset{\xi}{:}X^\star=X\cdot Y_\mu$.
2. Thus we have understood that if $\varPhi\subseteq X\cdot Y$ is totally bounded in $X\cdot Y$, then it is totally bounded in $X\cdot Y_\mu$ for each $\mu\in\mathbf{Ord}$. By Theorem 3 this means that $\varPhi$ is totally bounded in the space $Y_\mu\cdot X\cong X\cdot Y_\mu$. Again, what we have already proved allows us to conclude that $\varPhi$ is totally bounded in the space $Y_\mu\cdot X_\lambda$ for each ${\lambda\in\mathbf{Ord}}$. Applying Theorem 3 again, we can say that $\varPhi$ is totally bounded in the space $X_\lambda\cdot Y_\mu\cong Y_\mu\cdot X_\lambda$.
The lemma is proved. Lemma 7. Let $X$ and $Y$ be pseudocomplete. If $U\in {\mathcal U}(X_\lambda)$ and $V\in {\mathcal U}(Y_\mu)$, then $U\cdot V\in {\mathcal D}(X\cdot Y)$. Proof. Let $\varPhi\subseteq X\cdot Y=Y\underset{\xi}{:}X^\star$ be a totally bounded set. By Lemma 6, $\varPhi$ is totally bounded in the space $X_\lambda\cdot Y_\mu=Y_\mu\underset{\xi}{:}(X_\lambda)^\star$. Since $U\cdot V=U:(V^\circ)$ is a neighbourhood of zero in $X_\lambda\cdot Y_\mu=Y_\mu\underset{\xi}{:}(X_\lambda)^\star$, there is a finite set $A\subseteq X_\lambda\cdot Y_\mu=Y_\mu\underset{\xi}{:}(X_\lambda)^\star$ such that
$$
\begin{equation}
\varPhi\subseteq U\cdot V+A.
\end{equation}
\tag{3.16}
$$
But by Lemma 5 we can assume that all these sets lie in $X\cdot Y$. We obtain the following: for any totally bounded set $\varPhi$ in ${X\cdot Y}$ there exists a finite set $A$ in $X\cdot Y$ such that (3.16) holds. This means that the set $U\cdot V$ is capacious in the space ${X\cdot Y}$.
The lemma is proved. Lemma 8. For any pseudocomplete locally convex spaces $X$ and $Y$ and any ordinals $\lambda,\mu\in\mathbf{Ord}$ (i) the spaces $X_\lambda\cdot Y_\mu$ and $X\cdot Y$ become isomorphic after pseudosaturation:
$$
\begin{equation}
(X_\lambda\cdot Y_\mu)^\vartriangle=(X\cdot Y)^\vartriangle;
\end{equation}
\tag{3.17}
$$
(ii) for any functionals $f\in (X_\lambda)^\star$ and $g\in (Y_\mu)^\star$ the functional $f\cdot g\colon {X_\lambda\cdot Y_\mu\to\mathbb{C}}$ is continuous on the space $(X\cdot Y)^\vartriangle$, that is, there exists a unique functional $h\in ((X\cdot Y)^\vartriangle)^\star$ such that the following diagram is commutative: Proof. Step 1. We should notice first that it is sufficient to prove this only in the case when $\mu=0$. Indeed, suppose we have proved the following propositions:
(i)$^\circ$ the spaces $X_\lambda\cdot Y$ and $X\cdot Y$ become isomorphic after pseudosaturation:
$$
\begin{equation}
(X_\lambda\cdot Y)^\vartriangle=(X\cdot Y)^\vartriangle;
\end{equation}
\tag{3.19}
$$
(ii)$^\circ$ for any functionals $f\in (X_\lambda)^\star$ and $g\in Y^\star$ the functional $f\cdot g\colon {X_\lambda\cdot Y\to\mathbb{C}}$ is continuous on the space $(X\cdot Y)^\vartriangle$, that is, there exists a unique functional ${h\in ((X\cdot Y)^\vartriangle)^\star}$, such that the following diagram is commutative:
Then, first, (3.17) holds, since
$$
\begin{equation*}
(X_\lambda\cdot Y_\mu)^\vartriangle \stackrel{(3.19)}{=}(X\cdot Y_\mu)^\vartriangle\stackrel{(3.6)}{=} (Y_\mu\cdot X)^\vartriangle\stackrel{(3.19)}{=} (Y\cdot X)^\vartriangle\stackrel{(3.6)}{=}(X\cdot Y)^\vartriangle,
\end{equation*}
\notag
$$
and, second, for all functionals $f\in (X_\lambda)^\star$ and $g\in (Y_\mu)^\star$ the functional
$$
\begin{equation*}
f\cdot g\ \circ\vartriangle_{X_\lambda\cdot Y_\mu}
\end{equation*}
\notag
$$
will be continuous on the space $(X_\lambda\cdot Y_\mu)^\vartriangle$, and therefore on the space $(X\cdot Y)^\vartriangle$ (which is isomorphic to it by (3.17)), and this functional on $(X\cdot Y)^\vartriangle$ is exactly the functional $h$ constructed in (3.18). Its uniqueness also follows from (3.17).
Step 2. Thus, we have understood that it is sufficient to prove the weaker propositions (i)$^\circ$ and (ii)$^\circ$. They are proved by induction by ordinals $\lambda\in\mathbf{Ord}$.
First, for $\lambda=0$ the equality (3.17) becomes trivial
$$
\begin{equation*}
(X\cdot Y)^\vartriangle=(X\cdot Y)^\vartriangle,
\end{equation*}
\notag
$$
and (3.20) turns into the diagram where we just have to put $h=f\cdot g\circ\vartriangle_{X\cdot Y}$.
Further, suppose we have proved (i)$^\circ$ and (ii)$^\circ$ for all ordinals $\iota<\lambda$, where $\lambda$ is a given ordinal, that is,
(i)$^{\circ\circ}$ the spaces $X_\iota\cdot Y$ and $X\cdot Y$ become isomorphic after pseudosaturation:
$$
\begin{equation}
(X_\iota\cdot Y)^\vartriangle=(X\cdot Y)^\vartriangle;
\end{equation}
\tag{3.21}
$$
(ii)$^{\circ\circ}$ for any functionals $f\in (X_\iota)^\star$ and $g\in Y^\star$ the functional $f\cdot g\colon {X_\iota\cdot Y\to\mathbb{C}}$ is continuous on the space $(X\cdot Y)^\vartriangle$, that is, there exists a unique functional ${h\in ((X\cdot Y)^\vartriangle)^\star}$, such that the following diagram is commutative:
Let us show that (i)$^{\circ\circ}$ and (ii)$^{\circ\circ}$ hold for the ordinal $\lambda$ substituted instead of $\iota$. We have to consider two cases.
a) Suppose that $\lambda$ is an isolated ordinal, that is,
$$
\begin{equation*}
\lambda=\iota+1
\end{equation*}
\notag
$$
for some $\iota<\lambda$. First we show first that (ii)$^\circ$ holds for $\iota=\lambda$. Take
$$
\begin{equation*}
f\in (X_\lambda)^\star=(X_{\iota+1})^\star=((X_\iota)^\wedge)^\star= ((X_\iota)^\star)^\vee
\end{equation*}
\notag
$$
(the third equality follows from Theorem 3.10 in [ 3]) and
$$
\begin{equation*}
g\in Y^\star.
\end{equation*}
\notag
$$
Then $f$ is the limit of a totally bounded net $\{f_i;\,i\in I\}\subseteq (X_\iota)^\star$:
$$
\begin{equation}
f_i\overset{((X_\iota)^\star)^\vee}{\underset{i\to\infty}{\longrightarrow}}f.
\end{equation}
\tag{3.23}
$$
Since the set $\{f_i\}\subseteq (X_\iota)^\star$ is totally bounded, by Corollary 2, the set
$$
\begin{equation*}
\{f_i\cdot g;\,i\in I\}
\end{equation*}
\notag
$$
is totally bounded in $(X_\iota\cdot Y)^\star$. On the other hand, Theorem 4 implies that $\{{f_i\cdot g};\,{i\in I}\}$ is a Cauchy net. Thus, we see that $\{f_i\cdot g;\,i\in I\}$ is a totally bounded Cauchy net in the space $(X_\iota\cdot Y)^\star$, hence it converges in the enveloping space $(X_\iota\cdot Y)^{\star\triangledown}$:
$$
\begin{equation*}
(X_\iota\cdot Y)^\star\subseteq (X_\iota\cdot Y)^{\star\triangledown}= (X_\iota\cdot Y)^{\vartriangle\star} \stackrel{(3.21)}{=} (X\cdot Y)^{\vartriangle\star}
\end{equation*}
\notag
$$
(the first equality follows from Theorem 3.14 in [ 3]). In other words there is a functional $h\in (X\cdot Y)^{\vartriangle\star}$ such that
$$
\begin{equation}
f_i\cdot g\overset{(X\cdot Y)^{\vartriangle\star}}{\underset{i\to\infty}{\longrightarrow}}h.
\end{equation}
\tag{3.24}
$$
On the other hand, the chain of equalities
$$
\begin{equation*}
((X_\iota)^\star)^\vee= ((X_\iota)^\wedge)^\star=(X_{\iota+1})^\star= (X_\lambda)^\star
\end{equation*}
\notag
$$
(the first equality follows from Theorem 3.10 in [ 3]) implies that (3.23) is equivalent to the relation
$$
\begin{equation}
f_i\overset{(X_\lambda)^\star}{\underset{i\to\infty}{\longrightarrow}}f,
\end{equation}
\tag{3.25}
$$
which by Theorem 4 implies that
$$
\begin{equation}
f_i\cdot g\overset{(X_\lambda\cdot Y)^{\star}}{\underset{i\to\infty}{\longrightarrow}} f\cdot g.
\end{equation}
\tag{3.26}
$$
Together (3.24) and (3.26) mean, in particular, that the functionals $f\cdot g$ and $h$ coincide on each operator
$$
\begin{equation*}
\varphi\in X_\lambda\cdot Y=Y\underset{\xi}{:} (X_\iota)^\star=Y\underset{\xi}{:} X^\star=X_\lambda\cdot Y
\end{equation*}
\notag
$$
(by Lemma 5, the spaces $Y\underset{\xi}{:} (X_\iota)^\star$ and $Y\underset{\xi}{:} X^\star$ coincide as sets):
$$
\begin{equation*}
(f\cdot g)(\varphi)=h(\varphi).
\end{equation*}
\notag
$$
This proves (ii)$^\circ$ (for $\lambda=\iota+1$).
Now we prove (i)$^\circ$. The mapping
$$
\begin{equation*}
(X_\lambda\cdot Y)^\vartriangle=(Y\underset{\xi}{:}(X_\lambda)^\star)^\vartriangle\to (Y\underset{\xi}{:}X^\star)^\vartriangle=(X\cdot Y)^\vartriangle
\end{equation*}
\notag
$$
is always continuous. We have to prove that the inverse mapping
$$
\begin{equation*}
(X\cdot Y)^\vartriangle=(Y\underset{\xi}{:}X^\star)^\vartriangle\to (Y\underset{\xi}{:}(X_\lambda)^\star)^\vartriangle=(X_\lambda\cdot Y)^\vartriangle
\end{equation*}
\notag
$$
is continuous. Since the space $(X\cdot Y)^\vartriangle$ is pseudosaturated, it is sufficient to prove the continuity of the mapping
$$
\begin{equation*}
(X\cdot Y)^\vartriangle=(Y\underset{\xi}{:}X^\star)^\vartriangle\to Y\underset{\xi}{:}(X_\lambda)^\star=X_\lambda\cdot Y.
\end{equation*}
\notag
$$
Consider the basic neighbourhood of zero in $X_\lambda\cdot Y$, that is, a set $U\cdot V$, where $U$ is a closed convex balanced neighbourhood of zero in $X_\lambda$ and $V$ is a closed convex balanced neighbourhood of zero in $Y$. By Lemma 7 the set $U\cdot V$ is capacious in ${X\cdot Y}$. Hence if we prove that it is closed in $(X\cdot Y)^\vartriangle$, this will mean that it is a neighbourhood of zero in $(X\cdot Y)^\vartriangle$.
This becomes obvious if we represent $U\cdot V=V:U^\circ$ as the polar of the system of functionals of the form $\{f\cdot g;\,f\in U^\circ,\,g\in V^\circ\}$:
$$
\begin{equation}
U\cdot V=V:U^\circ=\{f\cdot g;\,f\in U^\circ,\, g\in V^\circ\}^\circ.
\end{equation}
\tag{3.27}
$$
Indeed,
$$
\begin{equation*}
\begin{aligned} \, &\varphi\in U\cdot V=V:U^\circ \quad\Longleftrightarrow\quad \varphi(U^\circ)\subseteq V \quad\Longleftrightarrow\quad \forall\, f\in U^\circ \quad \varphi(f)\in V \\ &\Longleftrightarrow\ \ \forall\, g\in V^\circ, \quad \forall\, f\in U^\circ \quad |\underbrace{g(\varphi(f))}_{\substack{\| \\ (f\cdot g)(\varphi)}}|\leqslant 1 \ \ \Longleftrightarrow\ \ \varphi\in \{f\cdot g;\,f\in U^\circ,\,g\in V^\circ\}^\circ. \end{aligned}
\end{equation*}
\notag
$$
Now we have to notice that in formula (3.27) we have $f\in U^\circ$, where $U$ is a neighbourhood of zero in $X_\lambda$ and $g\in V^\circ$, where $V$ is a neighbourhood of zero in $Y$. Hence $f\in (X_\lambda)^\star$ and $g\in Y^\star$. Thus, by (ii)$^\circ$ (which has already been proved for $\lambda=\iota+1$) the $f\cdot g$ are continuous functionals on the space $(X\cdot Y)^\vartriangle$. We see that $U\cdot V$ is a polar of some system of continuous functionals on $(X\cdot Y)^\vartriangle$, hence $U\cdot V$ is a closed set in $(X\cdot Y)^\vartriangle$. In addition, it is capacious, so it is a neighbourhood of zero in the (pseudosaturated) space $(X\cdot Y)^\vartriangle$. We have proved (i)$^\circ$ (for $\lambda=\iota+1$).
b) Now we consider the case when $\lambda$ is a limit ordinal, that is,
$$
\begin{equation*}
\lambda\ne \iota+1
\end{equation*}
\notag
$$
for all $\iota<\lambda$. Again, first we have to prove (ii)$^\circ$. Take $f\in (X_\lambda)^\star$ and $g\in Y^\star$. Then
$$
\begin{equation*}
f\in (X_\lambda)^\star=(X^\star)^\lambda= \bigcup_{\iota<\lambda}(X^\star)^\iota
\end{equation*}
\notag
$$
(the first equality follows from Theorem 3.12 in [ 3], and the second from [ 3], (1.32)), hence there is an ordinal $\iota<\lambda$ such that
$$
\begin{equation*}
f\in (X^\star)^\iota=(X_\iota)^\star
\end{equation*}
\notag
$$
(the equality follows from Theorem 3.12 in [ 3]). We see that $f\in (X_\iota)^\star$ and $g\in Y^\star$, hence by the induction assumption in (ii)$^{\circ\circ}$ the functional $f\cdot g$ must be continuous on the space $(X\cdot Y)^\vartriangle$. This proves (ii)$^\circ$ (in the case of the limit ordinal $\lambda$).
Now we can prove (i)$^\circ$. We repeat here the reasoning of a): we have to prove the continuity of the mapping
$$
\begin{equation*}
(X\cdot Y)^\vartriangle=(Y\underset{\xi}{:}X^\star)^\vartriangle\to Y\underset{\xi}{:}(X_\lambda)^\star=X_\lambda\cdot Y.
\end{equation*}
\notag
$$
We take a basic neighbourhood of zero in $X_\lambda\cdot Y$, that is, a set $U\cdot V$, where $U$ is a neighbourhood of zero in $X_\lambda$ and $V$ is a neighbourhood of zero in $Y$. Then we prove equality (3.27), and we notice that in it $f\in (X_\lambda)^\star$, and $g\in Y^\star$. After that, by the already proved (ii)$^\circ$ (for the limit ordinal $\lambda$) we have that $f\cdot g$ are continuous functionals on the space $(X\cdot Y)^\vartriangle$. Hence $U\cdot V$ is a closed set in $(X\cdot Y)^\vartriangle$. In addition, it is capacious by Lemma 7, so it is a neighbourhood of zero in the (pseudosaturated) space $(X\cdot Y)^\vartriangle$. We have proved (i)$^\circ$ (for the case of the limit $\lambda$).
Lemma 8 is proved. For arbitrary locally convex spaces $X$ and $Y$ the mappings
$$
\begin{equation*}
\vartriangle_X\colon X^\vartriangle\to X \quad\text{and}\quad \vartriangle_Y\colon Y^\vartriangle\to Y
\end{equation*}
\notag
$$
define the mapping
$$
\begin{equation}
(\vartriangle_X\cdot \vartriangle_Y)\colon (X^\vartriangle\cdot Y^\vartriangle) \to (X\cdot Y)
\end{equation}
\tag{3.28}
$$
(which coincide with (3.14) for some $\lambda$ and $\mu$). Theorem 5. For each pseudocomplete locally convex spaces $X$ and $Y$ the pseudosaturation of the mapping (3.28) is an isomorphism of locally convex (and stereotype) spaces:
$$
\begin{equation}
X^\vartriangle\odot Y^\vartriangle\cong (X^\vartriangle\cdot Y^\vartriangle)^\vartriangle\cong (X\cdot Y)^\vartriangle.
\end{equation}
\tag{3.29}
$$
Proof. The second isomorphism follows from (3.17) if we choose ordinals $\lambda$ and $\mu$ such that $X_\lambda=X^\vartriangle$ and $Y_\mu=Y^\vartriangle$. And the first isomorphism uses the fact that the pseudosaturation $X^\vartriangle$ of a pseudocomplete space $X$ is always pseudocomplete (and thus, stereotype); see [3], Proposition 3.16. From this we have that each totally bounded set $S$ in the space $(X^\vartriangle)^\star$ is contained in the polar $U^\circ$ of some neighbourhood of zero $U\subseteq X^\vartriangle$. As a corollary, in the space of operators ${Y^\vartriangle:(X^\vartriangle)^\star}$ the topology of uniform convergence on totally bounded sets coincides with the topology of uniform convergence on the polars $U^\circ$ of neighbourhoods of zero ${U\subseteq X^\vartriangle}$:
$$
\begin{equation*}
Y^\vartriangle:(X^\vartriangle)^\star=Y^\vartriangle\underset{\xi}{:}(X^\vartriangle)^\star.
\end{equation*}
\notag
$$
This implies the following chain of equalities, which proves the first isomorphism in (3.29):
$$
\begin{equation*}
X^\vartriangle\odot Y^\vartriangle=Y^\vartriangle\oslash (X^\vartriangle)^\star= (Y^\vartriangle:(X^\vartriangle)^\star)^\vartriangle= (Y^\vartriangle\underset{\xi}{:}(X^\vartriangle)^\star)^\vartriangle \stackrel{(3.2)}{=}(X^\vartriangle\cdot Y^\vartriangle)^\vartriangle.
\end{equation*}
\notag
$$
The theorem is proved.
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Citation:
S. S. Akbarov, “On tensor fractions and tensor products in the category of stereotype spaces”, Sb. Math., 213:5 (2022), 579–603
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