# Is the free locally convex space $L(X)$ nuclear?

Arkady Leiderman  Department of Mathematics, Ben-Gurion University of the
Negev, Beer Sheva, P.O.B. 653, Israel<EMAIL_ADDRESS>and  Vladimir
Uspenskij  321 Morton Hall, Department of Mathematics, Ohio University,
Athens, Ohio 45701, USA<EMAIL_ADDRESS>Dedicated to María Jesús Chasco on
the occasion of her birthday

###### Abstract.

Given a class ${\mathcal{P}}$ of Banach spaces, a locally convex space (LCS)
$E$ is called multi-${\mathcal{P}}$ if $E$ can be isomorphically embedded into
a product of spaces that belong to ${\mathcal{P}}$. We investigate the
question whether the free locally convex space $L(X)$ is strongly nuclear,
nuclear, Schwartz, multi-Hilbert or multi-reflexive.

If $X$ is a Tychonoff space containing an infinite compact subset then, as it
follows from the results of [1], $L(X)$ is not nuclear. We prove that for such
$X$ the free LCS $L(X)$ has the stronger property of not being multi-Hilbert.
We deduce that if $X$ is a $k$-space, then the following properties are
equivalent: (1) $L(X)$ is strongly nuclear; (2) $L(X)$ is nuclear; (3) $L(X)$
is multi-Hilbert; (4) $X$ is countable and discrete. On the other hand, we
show that $L(X)$ is strongly nuclear for every projectively countable
$P$-space (in particular, for every Lindelöf $P$-space) $X$.

We observe that every Schwartz LCS is multi-reflexive. It is known that if $X$
is a $k_{\omega}$-space, then $L(X)$ is a Schwartz LCS [3], hence $L(X)$ is
multi-reflexive. We show that for any first-countable paracompact (in
particular, metrizable) space $X$ the converse is true, so $L(X)$ is multi-
reflexive if and only if $X$ is a $k_{\omega}$-space, equivalently, if $X$ is
a locally compact and $\sigma$-compact space.

Similarly, we show that for any first-countable paracompact space $X$ the free
abelian topological group $A(X)$ is a Schwartz group if and only if $X$ is a
locally compact space such that the set $X^{(1)}$ of all non-isolated points
of $X$ is $\sigma$-compact.

###### Key words and phrases:

Free locally convex space, nuclear locally convex space, Schwartz locally
convex space, Hilbert space, reflexive Banach space

###### 2010 Mathematics Subject Classification:

Primary 46A03; Secondary 46B25, 54D30

##  1\. Introduction

All topological spaces are assumed to be Tychonoff and all topological vector
spaces and groups are Hausdorff. All vector spaces are considered over the
field of real numbers $\mathbb{R}$. The abbreviation LCS means locally convex
space.

Every Tychonoff space $X$ can be considered as a subspace of its free locally
convex space $L(X)$ characterized by the following property: every continuous
mapping $f:X\to E$ to an LCS $E$ uniquely extends to a continuous linear
mapping $\widehat{f}:L(X)\to E$.

Similarly, one defines the free abelian group $A(X)$ over a Tychonoff space
$X$: $X$ is a subspace of generators of $A(X)$ and the topology of $A(X)$ is
characterized by the property that every continuous mapping $f:X\to G$ to an
abelian topological group $G$ uniquely extends to a continuous homomorphism
$\widehat{f}:A(X)\to G$. It is known that $A(X)$ naturally embeds into $L(X)$
([30], [32]).

Free topological groups were introduced in 1941 by A. A. Markov in the short
note [20]. The complete construction appeared 4 years later, in [21], where
certain basic properties of free topological group $F(X)$ and free abelian
topological group $A(X)$ over a Tychonoff space $X$ were established. In
particular, Markov answered in the negative Kolmogorov’s question as to
whether every topological group is a normal space. Several proofs of the
existence of free topological groups have been given, it seems that the
shortest one is due to S. Kakutani [19]. Another proof, due to M. I. Graev
[12], [13] is based on extending continuous pseudometrics from a Tychonoff
space $X$ to invariant continuous pseudometrics on $F(X)$ (or $A(X)$). Free
topological vector spaces were mentioned by A. A. Markov [20], without any
details. Later, in 1964, D. A. Raikov constructed the free locally convex
space $L(X)$ for a uniform space $X$ [24].

Free LCS constitute a very important subclass of locally convex spaces. It
suffices to mention that every locally convex space is a linear quotient of a
free LCS and that every Tychonoff space $X$ embeds as a _closed_ subspace into
$L(X)$. Analogous basic facts are valid for $F(X)$ and $A(X)$.

Given a class ${\mathcal{P}}$ of Banach spaces, an LCS $E$ is called
multi-${\mathcal{P}}$ if $E$ can be isomorphically embedded in a product of
spaces in ${\mathcal{P}}$. The following result was obtained in [33]: for
every Tychonoff space $X$ the free LCS $L(X)$ can be isomorphically embedded
in the product of Banach spaces of the form $l^{1}(\Gamma)$, in other words,
$L(X)$ is multi-${\mathcal{L}}^{1}$, where ${\mathcal{L}}^{1}$ is the class of
all Banach spaces of the form $l^{1}(\Gamma)$. It follows that, as a
topological group, $L(X)$ admits a topologically faithful unitary
representation, that is, $L(X)$ is isomorphic to a subgroup of the unitary
group.

Vladimir Pestov raised the following

######  Question 1.1.

When is $L(X)$ multi-Hilbert?

The question was motivated by the fact that $L(X)$ is nuclear, hence multi-
Hilbert, if $X$ is a countable discrete space [28, Ch. 3, Theorem 7.4]; in
that case $L(X)$ is the locally convex direct sum of countably many one-
dimensional spaces, and will be denoted by $\varphi$.

In Section 2 we prove that $L(X)$ is not multi-Hilbert whenever $X$ contains
an infinite compact subset. As a consequence we deduce that if $X$ is a
$k$-space, then the following properties are equivalent: (1) $L(X)$ is
strongly nuclear; (2) $L(X)$ is nuclear; (3) $L(X)$ is multi-Hilbert; (4) $X$
is countable and discrete. On the other hand, we show that $L(X)$ is strongly
nuclear (hence nuclear) for every projectively countable $P$-space (in
particular, for every Lindelöf $P$-space) $X$.

Recall the definition of nuclear maps and nuclear spaces.

######  Definition 1.2.

A linear map $u:E\to B$ from an LCS $E$ to a Banach space $B$ is called
nuclear if it can be written in the form

$u(x)=\sum_{n=1}^{\infty}\lambda_{n}f_{n}(x)y_{n},$

where $\sum|\lambda_{n}|<+\infty$, $(f_{n})$ is an equicontinuous sequence of
linear functionals on $E$, and $(y_{n})$ is a bounded sequence in $B$. An LCS
$E$ is called nuclear if any continuous linear map from $E$ to a Banach space
is nuclear.

The definition of nuclear LCS and the following important permanence results
about nuclear LCS are due to A. Grothendieck [15] (see also [28, Ch. 3, §7]).

  * •

Every subspace of a nuclear LCS is nuclear.

  * •

Every Hausdorff quotient space of a nuclear LCS is nuclear.

  * •

The product of an arbitrary family of nuclear LCS is nuclear.

  * •

The locally convex direct sum of a countable family of nuclear LCS is a
nuclear space.

  * •

Every nuclear LCS is multi-Hilbert.

Note also

  * •

The class of multi-Hilbert LCS is closed under Hausdorff quotients.

Lemma 2.4 below also should be attributed to A. Grothendieck. For the
definition of a nuclear group and a study of free abelian nuclear groups we
refer the reader to [1]. Note here only that the class of nuclear groups
contains all nuclear LCS.

Michael Megrelishvili asked whether $L(X)$ is multi-reflexive. It turns out
that this problem is tightly connected with the properties of the Schwartz
locally convex spaces. Schwartz LCS play an important role in analysis and its
applications. Remark that in several aspects the properties of Schwartz LCS
resemble the finite-dimensional Banach spaces, for instance, their bounded
subsets are precompact. The latter property for Banach spaces is equivalent to
finite-dimensionality. The notion of Schwartz LCS was introduced by A.
Grothendieck in [14]. In this paper we will use the following equivalent
versions:

######  Definition 1.3.

An LCS $E$ is called a Schwartz space if one of the following equivalent
conditions holds:

  * (a)

Every continuous linear map from $E$ to a Banach space is compact, that is,
sends a neighborhood of zero to a set with a compact closure;

  * (b)

For every neighborhood $U$ of zero in $E$, there exits another neighborhood
$V$ of zero such that for every $\lambda>0$ the set $V$ can be covered by
finitely many translates of $\lambda U$.

For other equivalent definitions of a Schwartz space see [16, Section 3.15].
Several relevant problems in the theory of Schwartz LCS were solved in [4],
[5], [25], [26], [29] (our list of references does not attempt to be
complete). Note that the class of all Schwartz LCS (as well as the classes of
all strongly nuclear, all nuclear, all multi-Hilbert and all multi-reflexive
LCS) constitutes a variety, i.e. is closed under the formation of subspaces
(not necessarily closed), Hausdorff quotients, arbitrary products and
isomorphic images (see [6]).

We observe that every Schwartz LCS is multi-reflexive. This follows, for
example, from [17, 17.2.9 and 21.1.1(c)]; we give a short proof of this
assertion in Theorem 3.3. It is known that $L(X)$ is a Schwartz space, hence
multi-relexive, for every $k_{\omega}$-space $X$ [3, Theorem 5.2]. In Section
3 we provide an alternative proof of this fact (Theorem 3.5).

Recall that a topological space $X$ satisfies the first axiom of countability
if each point of $X$ has a countable base of neighborhoods. In Section 3 we
give a complete answer to the question: when is $L(X)$ multi-reflexive? — for
paracompact spaces $X$ satisfying the first axiom of countability; namely, if
$X$ is such a space, then the following properties are equivalent: (1) $L(X)$
is a Schwartz space; (2) $L(X)$ is multi-reflexive; (3) $X$ is locally compact
and $\sigma$-compact (= $X$ is the union of countably many compact sets). In
particular, for a metrizable space $X$, $L(X)$ is multi-reflexive if and only
if $X$ is locally compact space and has a countable base. We also give a short
direct proof of the fact that $L(X)$ is not multi-reflexive if $X$ is the
space of irrational numbers (Example 3.15).

A group version of the concept of a Schwartz space is introduced and studied
in the paper [3] (see also [2]). In Section 4 we give a complete answer to the
question: when is the free abelian group $A(X)$ a Schwartz group? — for
paracompact spaces $X$ satisfying the first axiom of countability. In that
case, $A(X)$ is a Schwartz group if and only if $X$ is a locally compact space
such that the set $X^{(1)}$ of all non-isolated points of $X$ is
$\sigma$-compact.

Our notations are standard, the reader is advised to consult with the
monographs [8], [9], [28] for the notions which are not explicitly defined in
the text. In the end of the article we pose several open questions.

##  2\. Nuclear and multi-Hilbert $L(X)$

Let $X$ be a Tychonoff space containing an infinite compact subset $K$. Every
continuous pseudometric defined on the compact space $K$ extends to a
continuous pseudometric on $X$, therefore $L(K)$ can be identified by a linear
topological isomorphism with a subspace of $L(X)$ [32]. Since the free abelian
group $A(K)$ naturally embeds into $L(K)$, we observe that $A(K)$ is
isomorphic to a topological subgroup of $L(X)$. Hence, if $L(X)$ is a nuclear
LCS, then $A(K)$ is a nuclear group. However, the free abelian group $A(K)$ is
nuclear if and only if a compact space $K$ is finite [1, Theorem 7.3]. This
observation leads to

######  Proposition 2.1.

If a Tychonoff space $X$ contains an infinite compact subset, then $L(X)$ is
not nuclear.

Below we prove a significantly stronger result, the proof of which requires
more elaborate arguments from functional analysis.

######  Theorem 2.2.

If a Tychonoff space $X$ contains an infinite compact subset, then $L(X)$ is
not multi-Hilbert.

In order to prove Theorem 2.2 we need some preparations. Let us say that a
subset $C$ of a Banach space $E$ is an ellipsoid if $C$ is the image of a
closed ball in a Hilbert space $H$ under a bounded linear mapping $H\to E$.
Note that ellipsoids are weakly compact, hence closed in $E$.

######  Lemma 2.3.

There exists a Banach space $E$ and a countable sequence $S=(x_{n})$
converging to zero in $E$ such that $S$ is not contained in any ellipsoid in
$E$.

###### Proof.

Let $E$ be a Banach space without the approximation property. According to
[10, Theorem 1.2], there is a compact subset $K$ in $E$ such that for any
Banach space $F$ with a basis and any injective bounded operator $T:F\to E$
the image $T(F)$ does not contain $K$. Then $K$ is not contained in any
ellipsoid. Indeed, if $H$ is a Hilbert space and $A:H\to E$ is a bounded
linear operator, we can write $A$ as the composition $H\to\breve{H}\to E$,
where $\breve{H}$ is the quotient of $H$ by the kernel of the operator $A$.
Then $\breve{H}$ is a Hilbert space and the operator $T:\breve{H}\to E$ is
injective, hence $K$ is not contained in the image $T(\breve{H})=A(H)$.

Now we find a countable sequence $S=(x_{n})$ in $E$ converging to zero and
such that $K$ is contained in the closed convex hull of $S$. If an ellipsoid
contains $S$, then it also contains $K$, which is impossible. ∎

Note that in fact a sequence with properties stated in Lemma 2.3 can be found
in any Banach space $E$ not isomorphic to a Hilbert space. For the proof one
can use, for example, the following: in every Banach space which is non-
isomorphic to a Hilbert space there is a sequence $F_{n}$ of finite-
dimensional subspaces whose Banach-Mazur distances from the Euclidean spaces
of the same dimension go to infinity [18].

######  Lemma 2.4.

Let $L$ be a multi-Hilbert LCS. Then every continuous linear map $f:L\to E$
from $L$ to a Banach space $E$ can be represented as a composition $f=p\circ
g$, where $g:L\to H$ and $p:H\to E$ are two continuous linear maps and $H$ is
a Hilbert space.

###### Proof.

We assume that $L$ is isomorphically embedded into the product $\prod_{i\in
I}E_{i}$, where all $E_{i}$ are Hilbert spaces. Denote by $V$ the unit open
ball of $E$. Since $f$ is continuous, and by the definition of the product
topology, there is a finite face $E_{A}=\prod_{i\in A}E_{i}$ of the product
$\prod_{i\in I}E_{i}$, and a neighborhood $U$ of zero in $E_{A}$, such that
$f(p^{-1}(U))$ is contained in $V$. Here the set of indexes $A\subset I$ is
finite and $p:X\to E_{A}$ is the corresponding projection.

We claim that $\ker(p)\subset\ker(f)$. Indeed, if $x\in L$ and $p(x)=0$, then
$p(tx)=0$ for every scalar $t$, so $tx\in p^{-1}(U))$. This means that
$f(tx)=t(f(x))$ belongs to the same unit ball $V$ for every scalar $t$, which
implies that $f(x)=0$. So there is some map $g:p(X)\to E$ such that $f=g\circ
p$. Clearly, $g$ is a linear map. It is also bounded, because
$g(U)=f(p^{-1}(U))$ was assumed to be contained in $V$. The map $g$ is defined
on the vector subspace $p(X)$ of the Hilbert space $E_{A}$. (The LCS $E_{A}$
is isomorphic to a Hilbert space being a finite product of Hilbert spaces).
Further, the map $g$ can be extended by continuity to the closure of that
vector space in $E_{A}$. A closed vector subspace of a Hilbert space is
Hilbert itself, and we get the required factorization. ∎

######  Fact 2.5.

Let $S$ be a non-trivial countable convergent sequence. Then $L(S)$ is not
multi-Hilbert.

###### Proof.

On the contrary, suppose that $L(S)$ is multi-Hilbert. By Lemma 2.3 we define
a 1-to-1 continuous map $f:S\to E$ such that $f(S)$ is not contained in an
ellipsoid. Extend $f$ to a linear continuous map $\widehat{f}:L(S)\to E$. Now,
by Lemma 2.4, we can represent $\widehat{f}$ as a composition
$\widehat{f}=p\circ\widehat{g}$, where $\widehat{g}:L(S)\to H$ and $p:H\to E$
are two continuous linear maps, $\widehat{g}$ extends the map $g:S\to H$ and
$H$ is a Hilbert space. The following diagram illustrates our construction.

$\textstyle{H\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{p}$$\textstyle{S\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{f}$$\scriptstyle{g}$$\textstyle{E}$
$\textstyle{H\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{p}$$\textstyle{L(S)\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\scriptstyle{\widehat{f}}$$\scriptstyle{\widehat{g}}$$\textstyle{E}$

The compact set $g(S)\subset H$ is contained in a ball, hence $f(S)=p(g(S))$
is contained in an ellipsoid. The obtained contradiction finishes the proof. ∎

Proof of Theorem 2.2. Let $K$ be an infinite compact subset of $X$ and assume
that $L(X)$ is multi-Hilbert. As we have noted before $L(K)$ can be identified
with a linear subspace of $L(X)$, hence $L(K)$ is multi-Hilbert. $K$ is an
infinite compact space, therefore we can choose a continuous map
$\pi:K\to[0,1]$ with an infinite image. Let $M=\pi(K)$. We have that $\pi:K\to
M$ is a closed (hence, quotient) continuous surjective mapping. In this
situation $\pi$ lifts to a linear continuous quotient mapping $\widehat{\pi}$
from $L(K)$ onto $L(M)$. The latter fact can be proved analogously to [23,
Proposition 1.8]. Since the class of multi-Hilbert LCS is closed under
Hausdorff quotients we conclude that $L(M)$ is also multi-Hilbert. However,
$M$ is an infinite compact subset of the segment $[0,1]$, therefore $M$
contains a copy of a convergent sequence $S$. This would mean that $L(S)$ is
multi-Hilbert contradicting Fact 2.5. The obtained contradiction finishes the
proof of Theorem 2.2. $\Box$

Recall that a topological space $X$ is called a $k$-space whenever $F\subset
X$ is closed iff the intersection $F\cap K$ is closed in $K$ for every compact
$K\subset X$. If $X$ can be covered by countably many compact subsets $K_{n}$
such that $F\subset X$ is closed iff all intersections $F\cap K_{n}$ are
closed, then $X$ is said to be a $k_{\omega}$-space [11].

An LCS $E$ is called strongly nuclear [25, 1.4] if for every continuous
seminorm $q$ on $E$ there exist a rapidly decreasing sequence $(\lambda_{n})$
(this means that $\sum n^{k}|\lambda_{n}|<\infty$ for every $k$) and an
equicontinuous sequence $(a_{n})$ of linear functionals on $E$ such that
$q(x)\leq\sum|\lambda_{n}|\langle x,a_{n}\rangle|$ for all $x\in E$. It is
easy to verify that $L(\mathbb{N})=\varphi$ is strongly nuclear, where
$\mathbb{N}$ denotes the discrete space of natural numbers.

Following [6] denote by ${\mathcal{V}}(\varphi)$ the variety of LCS generated
by $\varphi$. Note that ${\mathcal{V}}(\varphi)$ is the (unique) second
smallest variety, in the sense that every variety properly containing
${\mathcal{V}}({\mathbb{R}})$ contains ${\mathcal{V}}(\varphi)$ [6].

######  Corollary 2.6.

Let $X$ be a $k$-space. The following conditions are equivalent:

  * (i)

$L(X)$ is strongly nuclear;

  * (ii)

$L(X)$ is nuclear;

  * (iii)

$L(X)$ is multi-Hilbert;

  * (iv)

$X$ is a countable discrete space.

###### Proof.

Only the implication (iii) $\Rightarrow$ (iv) is new and needs to be proved.
By Theorem 2.2, every compact subset of $X$ is finite, therefore the $k$-space
$X$ is discrete. Every multi-Hilbert space is multi-reflexive, and we will
show later that $L(X)$ is not multi-reflexive if $X$ is an uncountable
discrete space (Fact 3.9). ∎

Besides the requirement that $X$ should contain no infinite compact subspaces,
there is another necessary condition for $L(X)$ to be nuclear.

######  Theorem 2.7.

Let $L(X)$ be a nuclear LCS. Then every metrizable image of $X$ under a
continuous map must be separable.

###### Proof.

Let $f:X\to M$ be a continuous map onto a metrizable space $M$. Assume that
the topology of $M$ is generated by a metric $d$. Denote by $B=C_{b}(M)$ the
Banach space of all bounded continuous real-valued functions on $M$ equipped
with the supremum norm. It is known that the Kuratowski mapping isometrically
embeds $M$ into the Banach space $B$. Specifically, if $(M,d)$ is a metric
space, $a$ is a fixed point in $M$, then the mapping $\Phi:M\rightarrow
C_{b}(M)$ defined by

$\Phi(x)(y)=d(x,y)-d(a,y)\quad{\mbox{for all}}\quad x,y\in M,$

is an isometry.

Extend the map $f:X\to M$ to a linear continuous map $\widehat{f}:L(X)\to B$.
It follows from the definition of a nuclear map that the range of
$\widehat{f}$ is separable. Since $M$ is contained in that range, it is
separable, too. ∎

It follows from Theorem 2.8 below that there exist non-separable spaces $X$
such that $L(X)$ are strongly nuclear, hence nuclear. We say that a Tychonoff
space $X$ is projectively countable if every metrizable image of $X$ under a
continuous map is countable. Recall that a $P$-space is a topological space in
which the intersection of countably many open sets is open. To see that
Lindelöf $P$-spaces are projectively countable, observe that if $X$ is a
$P$-space, all points of $Y$ are $G_{\delta}$, and $f:X\to Y$ is a continuous
map, then the inverse images of points from $Y$ form a disjoint open cover of
$X$. It is known that $\omega$-modification of a topology of any scattered
Lindelöf space produces a Lindelöf $P$-space. A concrete example of an
uncountable Lindelöf $P$-space is the so-called one-point Lindelöfication
$X=\Gamma\cup\\{p\\}$ of an uncountable discrete space $\Gamma$. All open
neighborhoods of the point $p$ in X are of the form $A\cup\\{p\\}$, where
$A\subset\Gamma$ and $|\Gamma\setminus A|\leq\aleph_{0}$.

######  Theorem 2.8.

If $X$ is a projectively countable $P$-space (in particular, a Lindelöf
$P$-space), then $L(X)$ is strongly nuclear.

###### Proof.

Let $M$ be a metrizable space and $f:X\to M$ be a continuous map. Then $f(X)$
is countable. Every point $m\in M$ is a $G_{\delta}$-set in $M$, therefore
every fiber $f^{-1}(m)$ is open in the $P$-space $X$. We see that $X$ is the
disjoint countable union of open fibers $f^{-1}(m)$, $m\in f(X)$. It follows
that every continuous map $f:X\to M$ admits a factorization $X\to\mathbb{N}\to
M$, where $\mathbb{N}$ is the discrete space of natural numbers, hence any
linear continuous map $L(X)\to B$ to a Banach space $B$ admits a factorization
$L(X)\to L(\mathbb{N})\to B$. Using [25, Remark 3.4] we can conclude that
$L(X)$ can be isomorphically embedded into a power of the space $\varphi$.
This means that $L(X)$ belongs to ${\mathcal{V}}(\varphi)$ which is a sub-
variety of the variety of all strongly nuclear spaces [6], since
$L(\mathbb{N})=\varphi$ is strongly nuclear. ∎

It turns out that there are projectively countable $P$-spaces which are not
Lindelöf $P$-spaces. A Tychonoff space $X$ is called cellular-Lindelöf if for
every disjoint family ${\mathcal{U}}$ of open nonempty subsets of $X$ there is
a Lindelöf subspace $L$ of $X$ such that $L$ meets every member of the family
${\mathcal{U}}$. It is easy to see that every cellular-Lindelöf $P$-space is
projectively countable. Hence, Theorem 2.8 implies

######  Proposition 2.9.

If $X$ is a cellular-Lindelöf $P$-space, then $L(X)$ is a strongly nuclear
LCS.

Another quick consequence provides sufficient conditions for the free abelian
group $A(X)$ to be nuclear:

######  Corollary 2.10.

If $X$ is a cellular-Lindelöf $P$-space, then $A(X)$ is a nuclear topological
group.

See [31, Theorem 3.16] for an example of a cellular-Lindelöf $P$-space which
is not even weakly Lindelöf.

##  3\. Schwartz and Multi-reflexive $L(X)$

Recall that an LCS is multi-reflexive if it is isomorphic to a subspace of a
product of reflexive Banach spaces. For instance, every LCS endowed with its
weak topology is multi-Hilbert (hence multi-reflexive), since it embeds into a
product of the real lines. However, a reflexive LCS need not be multi-
reflexive [22, Remark 4.14 (a)].

Our proof of Theorem 2.2 is based on the fact that certain compact subsets of
Banach spaces are not contained in ellipsoids. However, every weakly compact
subset of a Banach space is contained in the image of the closed unit ball of
a reflexive space under a bounded linear mapping (see [9, Theorem 13.22]).
This motivates the following question (suggested to us by Michael
Megrelishvili):

######  Question 3.1.

When is $L(X)$ multi-reflexive?

We have seen that $L(X)$ is not multi-Hilbert if $X$ is an infinite compact
space (Theorem 2.2). Nevertheless, $L(X)$ is a Schwartz LCS for every compact
space $X$ [3]. We provide an alternative proof of this statement (Theorem 3.5)
and due to the fact that every Schwartz LCS is multi-reflexive (Theorem 3.3),
we conclude that $L(X)$ is multi-reflexive for every compact space $X$
(Theorem 3.6).

In general, $L(X)$ need not be Schwartz or multi-reflexive. We provide a
complete characterization of first-countable paracompact spaces $X$ such that
$L(X)$ is Schwartz or multi-reflexive (Theorem 3.13). In particular, if $X$ is
metrizable, then $L(X)$ is multi-reflexive iff $L(X)$ is a Schwartz LCS iff
$X$ is locally compact and has a countable base (Corollary 3.14).

######  Lemma 3.2.

If $L$ is a Schwartz LCS, then every continuous linear map $L\to B$ to a
Banach space $B$ admits a factorization $L\to B_{1}\to B$, where $B_{1}$ is a
Banach space, $L\to B_{1}$ is a continuous linear map and $B_{1}\to B$ is a
compact operator.

###### Proof.

Let $U$ be a balanced convex neighborhood of zero in $L$ such that the image
of $U$ in $B$ has a compact closure. Let $p$ be the gauge functional
corresponding to $U$. We can take for $B_{1}$ the completion
$\widetilde{L}_{U}$ of the normed space $L_{U}$ associated with the seminorm
$p$, as in [28, Ch. 3, §7]. ∎

######  Theorem 3.3.

Every Schwartz LCS is multi-reflexive.

###### Proof.

Let $L$ be a Schwartz LCS. We want to prove that continuous linear maps from
$L$ to reflexive Banach spaces generate the topology of $L$. Since maps of $L$
to all Banach spaces generate the topology of $L$, it suffices to prove that
every continuous linear map $L\to B$ to a Banach space admits a factorization
$L\to B_{1}\to B$, where $B_{1}$ is a reflexive Banach space. According to
Lemma 3.2, we can find a factorization $L\to B_{2}\to B$, where $B_{2}\to B$
is a compact operator between Banach spaces. Compact operators are weakly
compact, and in virtue of the Davis–Figiel–Johnson–Pełczyński theorem [9,
Theorem 13.33] weakly compact operators between Banach spaces admit a
factorization through reflexive spaces. Thus we have a required factorization
$L\to B_{2}\to B_{1}\to B$ for some reflexive Banach space $B_{1}$. ∎

Below we collect several basic facts about the topology of $L(X)$ and
associated Banach spaces. For a space $X$ we denote by $L_{0}(X)$ the
hyperplane

$L_{0}(X)=\left\\{\sum c_{i}x_{i}:\sum c_{i}=0,\ c_{i}\in{\mathbb{R}},\
x_{i}\in X\right\\}$

of $L(X)$. The topology of $L_{0}(X)$ is generated by the seminorms
$\overline{d}$ of the following form. For a continuous pseudometric $d$ on
$X$, the seminorm $\overline{d}$ on $L_{0}(X)$ is defined by

(3.1) $\overline{d}(v)=\inf\left\\{\sum|c_{i}|d(x_{i},y_{i}):v=\sum
c_{i}(x_{i}-y_{i}),\ c_{i}\in\mathbb{R},\ x_{i},y_{i}\in X\right\\}.$

Denote by $B_{d}$ the Banach space associated with the seminormed space
$(L_{0}(X),\overline{d})$, that is, the completion of the quotient by the
kernel of $\overline{d}$. The dual Banach space $B_{d}^{*}$ is naturally
isomorphic to the quotient $C_{d}/\mathbb{R}$ of the space of $d$-Lipschitz
functions on $X$ by the one-dimensional subspace of constant functions. The
seminorm on $C_{d}$ assigns to each $d$-Lipschitz function $f$ its Lipschitz
constant

(3.2) $\left\|f\right\|=\inf\left\\{k\geq 0:|f(x)-f(y)|\leq k\cdot
d(x,y)\,\mbox{for all}\,x,y\in X\right\\}.$

######  Lemma 3.4.

If $X$ is a compact space and $d$ is a continuous pseudometric on $X$, then
the natural operator $B_{\sqrt{d}}\to B_{d}$ is compact.

###### Proof.

We may assume that $d$ is a metric. Pick a point $x_{0}\in X$, and identify
the quotient $C_{d}/\mathbb{R}$ with the Banach space $\breve{C}_{d}=\\{f\in
C_{d}:f(x_{0})=0\\}$. A bounded linear operator between Banach spaces is
compact if and only if its dual is compact, by the Shauder theorem [9, Theorem
15.3], so it suffices to prove that the embedding
$\breve{C}_{d}\to\breve{C}_{\sqrt{d}}$ is compact. Take any sequence $(f_{n})$
in the unit ball of $\breve{C}_{d}$, that is, a sequence of $d$-non-expanding
functions such that $f_{n}(x_{0})=0$. By the Arzelà–Ascoli theorem, the
sequence $(f_{n})$ has a uniformly convergent susbsequence. To simplify the
notation, assume that the sequence $(f_{n})$ itself uniformly converges to a
limit $f$. We claim that $(f_{n})$ converges to $f$ in $\breve{C}_{\sqrt{d}}$
as well. Let $\epsilon>0$ be given. Put $g_{n}=f_{n}-f$. We must prove that
for $n$ large enough we have

(3.3) $|g_{n}(x)-g_{n}(y)|\leq\epsilon\sqrt{d(x,y)}$

for all $x,y\in X$. Note that each $g_{n}$ is 2-Lipschitz with respect to $d$.
If $x$ and $y$ are close to each other, namely, if $d(x,y)\leq\epsilon^{2}/4$,
then (3.3) holds:

$|g_{n}(x)-g_{n}(y)|\leq
2d(x,y)=2\sqrt{d(x,y)}\sqrt{d(x,y)}\leq\epsilon\sqrt{d(x,y)}.$

On the other hand, for pairs $x,y$ such that $d(x,y)>\epsilon^{2}/4$ the
inequality 3.3 holds for $n$ large enough because the sequence $(g_{n})$
uniformly converges to zero. ∎

######  Theorem 3.5.

If $X$ is a compact space, then $L(X)$ is a Schwartz LCS.

###### Proof.

Let $T:L(X)\to B$ be a continuous linear operator, where $B$ is a Banach
space. We want to prove that $T$ is compact. Let $d$ be the pseudometric on
$X$ induced by $T$ from the metric on $B$. We have a factorization $L(X)\to
B_{d}\to B$ of the operator $T$. By Lemma 3.4, the middle arrow in the
factorization $L(X)\to B_{\sqrt{d}}\to B_{d}\to B$ is a compact operator. It
follows that $T$ is a compact operator as well. ∎

Combining Theorems 3.3 and 3.5, we get the result we were aiming at.

######  Theorem 3.6.

If $X$ is a compact space, then $L(X)$ is multi-reflexive.

Theorems 3.5 and 3.6 can be readily generalized to the case of
$k_{\omega}$-spaces, similarly to [3]. Indeed, let
$X=\bigcup_{n\in{\mathbb{N}}}K_{n}$ be a $k_{\omega}$-space, where $(K_{n})$
is a sequence of compact subsets of $X$ witnessing the $k_{\omega}$-property.
Then $X$ is a quotient of the free sum $\bigoplus_{n\in{\mathbb{N}}}K_{n}$
[11], and $L(X)$ is a quotient of $L(\bigoplus_{n\in{\mathbb{N}}}K_{n})$ which
is isomorphic to the countable locally convex sum of $L(K_{n})$. Since
Schwartz spaces are preserved by quotients and by locally convex countable
sums [17, Proposition 21.1.7], we conclude that $L(X)$ is a Schwartz LCS.

In particular, we have established the following (which is a somewhat special
case of [3]):

######  Theorem 3.7.

If $X$ is a locally compact $\sigma$-compact space, then $L(X)$ is a Schwartz
LCS and hence (Theorem 3.3) multi-reflexive.

Our next aim is to show that for a wide and natural class of Tychonoff spaces
$X$ the converse of Theorem 3.7 is true: $L(X)$ is a Schwartz LCS if and only
if $X$ is a locally compact and $\sigma$-compact space.

######  Lemma 3.8.

If $L(X)$ is multi-reflexive, then for every Banach space $E$ and every
continuous map $f:X\to E$ the image $f(X)$ can be covered by countably many
weakly compact subspaces of $E$.

###### Proof.

If $L(X)$ is multi-reflexive, by literally the same arguments as we have used
in the proof of Lemma 2.4, we can represent $f$ as a composition $f=p\circ g$,
where $g:X\to F$ is a continuous map to a reflexive Banach space $F$ and
$p:F\to E$ is a bounded linear map. Since the reflexive Banach space $F$ is
the union of countably many weakly compact sets, the same is true for the set
$p(F)$ which is a subset of $E$ containing $f(X)$. ∎

######  Fact 3.9.

If $X$ is an uncountable discrete space, then $L(X)$ is not multi-reflexive.

###### Proof.

Consider the Banach space $E=l^{1}(\Gamma)$, where $\Gamma$ is a set of
indexes with the same cardinality as $X$. Let
$Y=\left\\{e_{\gamma}:\gamma\in\Gamma\right\\}$ be the canonical basis of $E$.
Let $f$ be any bijection from $X$ onto $Y$, then $f:X\to E$ is a continuous
map, because $Y$ is discrete in $E$. We claim that $Y=f(X)$ cannot be covered
by countably many weakly compact subspaces of $E$. This follows from the
easily verified fact that $Y$ is a closed discrete subset of $E$ endowed with
the weak topology (use the natural pairing
$\langle(a_{\gamma}),(b_{\gamma})\rangle=\sum a_{\gamma}b_{\gamma}$ between
$E=l^{1}(\Gamma)$ and $E^{\prime}=l^{\infty}(\Gamma)$, and consider 0–1
sequences in $E^{\prime}$). Alternatively, note that $l^{1}(\Gamma)$ has the
Schur property, that is, every weakly converging sequence in $l^{1}(\Gamma)$
converges in the norm topology, and then every weakly compact subset of
$l^{1}(\Gamma)$ is compact in the norm topology (see [9, Exercise 5.47]).
Hence, no infinite subset of the closed discrete set $Y$ can be covered by a
weakly compact subset of $E$, and the claim follows. ∎

The following known fact plays a crucial role in the sequel.

######  Theorem 3.10 ([32]).

Let $X$ be a paracompact (in particular, metrizable) space. Then for every
closed subspace $Y\subset X$, the natural map $L(Y)\to L(X)$ is a topological
embedding.

######  Corollary 3.11.

If $X$ is a non-separable metric space, then $L(X)$ is not multi-reflexive.

###### Proof.

$X$ contains an uncountable closed discrete subspace $Y$, so Fact 3.9 and
Theorem 3.10 apply. ∎

The metric fan $M$ is the metrizable space defined as follows. As a set, $M$
is the set ${\mathbb{N}}\times{\mathbb{N}}$ of pairs of natural numbers plus a
point $p$ “at infinity”; $p$ is the only non-isolated point of $M$, and a
basic neighborhood $U_{n}$ of $p$ consists of $p$ and all pairs $(a,b)$ such
that $b>n$. Thus, $M$ is the union of countably many sequences converging to
$p$. It is easy to verify that the space $M$ is not locally compact.

######  Fact 3.12.

Let $M$ be the metric fan defined above. Then the free LCS $L(M)$ is not
multi-reflexive.

###### Proof.

Consider the separable non-reflexive Banach space $E=l^{1}$, and let $(e_{n})$
be the canonical basis. The collection of all vectors of the form $e_{n}/k$,
$k=1,2,\dots$, together with the zero vector, is a subspace $A\subset E$
homeomorphic to $M$. Let $f:M\to A$ be a homeomorphism, $\widehat{f}:L(M)\to
E$ its linear extension. We want to prove that there is no factorization
$L(M)\to F\to E$ of $\widehat{f}$, where $F$ is a reflexive Banach space.
Suppose there is such a factorization. Let $\breve{A}$ be the image of $M$ in
$F$, and let $p$ be the non-isolated point of $\breve{A}$. Denote the arrow
$F\to E$ by $g$. We have continuous bijections $M\to\breve{A}\to A$ such that
their composition is a homeomorphism. It follows that the restriction
$g\restriction_{\breve{A}}:\breve{A}\to A$ is a homeomorphism. Let $U$ be the
closed unit ball in $F$. Then $(p+U)\cap\breve{A}$ is a neighborhood of $p$ in
$\breve{A}$. Since $g\restriction_{\breve{A}}:\breve{A}\to A$ is a
homeomorphism, $g((p+U)\cap\breve{A})$ is a neighborhood of $g(p)=0$ in $A$,
and so is the bigger set $g(U)\cap A$. Thus for $k$ large enough all the
vectors $e_{n}/k$ lie in the weakly compact set $g(U)$. Multiplying by $k$, we
conclude that all the vectors $e_{n}$ lie in a weakly compact subset of $E$.
That is a contradiction: we noted in the proof of Fact 3.9 that the set
$\\{e_{n}\\}$ is closed and discrete with respect to the weak topology of $E$.
∎

We now are ready to prove one of the main results of our paper.

######  Theorem 3.13.

Let $X$ be a first-countable paracompact space. The following conditions are
equivalent:

  * (i)

$L(X)$ is a multi-reflexive LCS;

  * (ii)

$L(X)$ is a Schwartz LCS;

  * (iii)

$X$ is locally compact and $\sigma$-compact.

###### Proof.

(iii) $\Rightarrow$ (ii): this is Theorem 3.7.

(ii) $\Rightarrow$ (i): this is Theorem 3.3.

(i) $\Rightarrow$ (iii): by Theorem 3.10 and Fact 3.12, $X$ does not contain a
closed copy of the metric fan $M$. A first-countable paracompact space is
locally compact if and only if it does not contain a closed subspace
homeomorphic to $M$ (by [7, Lemma 8.3]). Hence $X$ is locally compact. Being
locally compact and paracompact, $X$ can be represented as a disjoint union of
clopen $\sigma$-compact subspaces (by [8, Theorem 5.1.27]). Since $X$ does not
contain an uncountable closed discrete subspace (by Fact 3.9 and Theorem
3.10), the number of components in the representation of $X$ above is at most
countable and the proof is complete. ∎

######  Corollary 3.14.

Let $X$ be a metrizable space. The following conditions are equivalent:

  * (i)

$L(X)$ is a multi-reflexive LCS;

  * (ii)

$L(X)$ is a Schwartz LCS;

  * (iii)

$X$ is locally compact and has a countable base.

######  Example 3.15.

If $P$ is the space of irrational numbers, then as a straightforward
consequence of Theorem 3.13 we have that $L(P)$ is not multi-reflexive. We
note that an alternative proof of this fact readily follows from Lemma 3.8.

Indeed, consider any separable non-reflexive Banach space $E$. For example, we
can take the separable Banach space $E=l^{1}$. Then weakly compact subsets of
$E$ have empty interior, and it follows from the Baire Category Theorem that
$E$ is not the union of countably many weakly compact sets. On the other hand,
$E$, like any Polish space, is the image of $P$ under a continuous map. Now
apply Lemma 3.8.

##  4\. Schwartz $A(X)$

A concept of a Schwartz topological group which turned out to be coherent with
the concept of a Schwartz locally convex space has been introduced in [3].
Similarly to Schwartz LCS, the class of Schwartz groups enjoys analogous
permanence properties: it is closed with respect to subgroups, Hausdorff
quotients, products and local isomorphisms. Also, bounded subsets of locally
quasi-convex Schwartz groups are precompact [3].

A natural question arises: when is the free abelian topological group $A(X)$
Schwartz? Note that the authors of [3] found sufficient conditions on $X$
implying that $A(X)$ is a Schwartz group, which are identical to those
conditions on $X$ implying that $L(X)$ is a Schwartz LCS. In this section we
give a complete characterization of first-countable paracompact (in
particular, metrizable) spaces $X$ such that the free abelian topological
group $A(X)$ is a Schwartz group.

For a subset $U$ of an abelian group $G$ such that $0\in U$, and a natural
number $n$, we denote $U_{(n)}=\\{x\in G:kx\in U:k=1,2,\dots,n\\}$ [3]. The
next formulation is adjusted to Definition 1.3(b) of a Schwartz LCS.

######  Definition 4.1.

[3] Let $G$ be an abelian topological group. We say that $G$ is a Schwartz
group if for every neighborhood $U$ of zero in $G$ there exists another
neighborhood $V$ of zero in $G$ and a sequence $(F_{n})$ of finite subsets of
$G$ such that $V\subseteq F_{n}+U_{(n)}$ for every $n\in{\mathbb{N}}$.

Below we collect several basic facts about the topology of $A(X)$ and its
relations with the topology of $L(X)$ (for the details see [32]).

For a space $X$ we denote by $A_{0}(X)$ the open subgroup

$A_{0}(X)=\left\\{\sum c_{i}x_{i}:\sum c_{i}=0,\ c_{i}\in\mathbb{Z},\ x_{i}\in
X\right\\}$

of $A(X)$. If $d$ is a continuous pseudometric on $X$, $d$ has a canonical
translation-invariant extension over $A_{0}(X)$ defined by

(4.1) $\overline{d}(v,0)=\inf\left\\{\sum|c_{i}|d(x_{i},y_{i}):v=\sum
c_{i}(x_{i}-y_{i}),\ c_{i}\in\mathbb{Z},\ x_{i},y_{i}\in X\right\\}$

and $\overline{d}(u,v)=\overline{d}(u-v,0)$. We denote the seminorm
$v\mapsto\overline{d}(v,0)$ on $A_{0}(X)$ by the same symbol $\overline{d}$.
This seminorm is the restriction of the seminorm on $L_{0}(X)$ denoted by the
same symbol in the equality (3.1). In other words, in the definition above we
may replace the condition $c_{i}\in\mathbb{Z}$ by $c_{i}\in\mathbb{R}$ without
affecting the result [32]. The fact that in the equality (4.1) the minimum
over $c_{i}\in\mathbb{Z}$ is the same as the minimum over $c_{i}\in\mathbb{R}$
follows from the integrality property for the transportation problem. Let
$a_{1},\dots,a_{m}$ and $b_{1},\dots,b_{n}$ be non-negative integers. Consider
the compact convex set of all $(m\times n)$-matrices $(c_{ij})$ with real
entries $\geq 0$ such that the sum of all entries in the $i$-th row is $a_{i}$
and the sum in the $j$-th column is $b_{j}$ ($1\leq i\leq m$, $1\leq j\leq
n$). Then all extreme points of this convex set are matrices with integral
entries.

Since the seminorm $\overline{d}$ on $A_{0}(X)$ is the restriction of a
seminorm on the LCS $L_{0}(X)$, we immediately deduce the following assertion.

######  Lemma 4.2.

Let $d$ be a pseudometric on a set $X$, $\overline{d}$ the seminorm on
$A_{0}(X)$ given by the equality (4.1). For every integer $n$ and $t\in
A_{0}(X)$ we have $\overline{d}(nt)=|n|\overline{d}(t)$.

Denote by $X^{(1)}$ the set of all non-isolated points of $X$.

######  Theorem 4.3.

Let $X$ be a first-countable paracompact space. The following conditions are
equivalent:

  * (i)

$A(X)$ is a Schwartz group;

  * (ii)

$X$ is a locally compact space such that the set $X^{(1)}$ is
$\sigma$-compact.

###### Proof.

(ii) $\Rightarrow$ (i): $X^{(1)}$ is a Lindelöf space, therefore we can
represent $X$ as a disjoint sum $D\oplus Y$ of two clopen subsets $D$ and $Y$,
where $D$ consists of isolated points of $X$ and $Y$ is locally compact and
$\sigma$-compact. Further, the group $A(D)$ is discrete and the group $A(Y)$
is Schwartz, since $L(Y)$ is Schwartz, by Theorem 3.7. We conclude that $A(X)$
is a Schwartz group, in view of the known fact that $A(X)$ is isomorphic to
the product $A(D)\times A(Y)$.

(i) $\Rightarrow$ (ii): On the contrary, assume that $X$ is not locally
compact. Then, as we have observed earlier in the proof of Theorem 3.13, $X$
contains a closed copy of the metric fan $M$. By Theorem 3.10, this means that
the free abelian topological group $A(X)$ contains an isomorphic copy of the
free abelian topological group $A(M)$. It remains to prove the following

######  Fact 4.4.

Let $M$ be the metric fan. Then $A(M)$ is not a Schwartz group.

###### Proof.

The fan $M=({\mathbb{N}}\times{\mathbb{N}})\cup\\{p\\}$ can be represented as
a countable union of disjoint closed discrete layers
$\\{M_{k}:k\in{\mathbb{N}}\\}$, $M_{k}={\mathbb{N}}\times\\{k\\}$, and a
single non-isolated point $p$ such that for every choice $x_{k}\in M_{k}$ the
sequence $(x_{k})$ converges to $p$ in $M$. Consider the natural metric $d$ on
$M$ defined as follows: the distance $d(x,p)=1/k$ for every $x\in M_{k}$; the
distance $d(x,y)=2/k$ between any two distinct points $x,y\in M_{k}$; and
$d(x,y)=1/k+1/l$ if $x\in M_{k},y\in M_{l},k\neq l$. Let $U$ be the unit ball
in $A_{0}(M)$ with respect to the metric $\bar{d}$ on $A_{0}(M)$. If $V$ is
any neighborhood of zero, we can find a natural $k$ such that all points in
the $k$-th layer $M_{k}$ of the fan are in $p+V$. Since $d(x,y)=2/k$ for any
two distinct points $x,y$ of $M_{k}$, we conclude that $\bar{d}(nx,ny)=2n/k$
(by Lemma 4.2). It follows that for $n>k$ the difference $x-y$ does not belong
to $U_{(n)}-U_{(n)}$, so the balls $x+U_{(n)}$, as $x$ runs over $M_{k}$, are
pairwise disjoint. Therefore, $M_{k}$ cannot be covered by a set of the form
$F_{n}+U_{(n)}$, where $F_{n}$ is finite. The same is true for the translate
$M_{k}-p$ of $M_{k}$ and moreover for the larger set $V$. ∎

So, we obtained a contradiction assuming that $X$ is not locally compact.
Therefore, $X^{(1)}$ as a closed subset of $X$ is itself locally compact (and
paracompact). Then, as we have observed earlier in the proof of Theorem 3.13,
being locally compact and paracompact, $X^{(1)}$ can be represented as a
disjoint union of clopen (in $X^{(1)}$) $\sigma$-compact subspaces. Striving
for a contradiction, assume that the number of clopen components in this
representation is uncountable. Then we can find an uncountable closed discrete
subset $W\subset X^{(1)}\subset X$. Paracompact spaces are collectionwise
normal [8, Theorem 5.1.18], hence there exists a discrete family
$\\{U_{w}:w\in W\\}$ of open sets in $X$ such that $w\in U_{w}$ for every
$w\in W$. Every $w\in W$ is the limit of a convergent sequence $S_{w}\subset
U_{w}$. Put $Y=\bigcup_{w\in W}S_{w}$. Then $Y$ is closed in $X$ (as the union
of a discrete family of closed sets), and $Y$ is homeomorphic to a disjoint
sum of uncountably many convergent sequences. In order to get a contradiction
it now suffices to prove the following

######  Fact 4.5.

Let $Y$ be the disjoint sum of uncountably many convergent sequences. Then
$A(Y)$ is not a Schwartz group.

###### Proof.

Let $S=\\{1,1/2,1/3,\dots\\}\cup\\{0\\}$ be the standard convergent sequence
together with its limit, $W$ be an uncountable index set (regarded as a
discrete space), and $Y=S\times W$. Equip $Y$ with the natural metric $d$ such
that the distance $d$ between $(1/k,w^{\prime})$ and $(1/l,w^{\prime\prime})$
is $|1/k-1/l|$ if $w^{\prime}=w^{\prime\prime}$ and $1$ otherwise. Take
$U=\\{t\in A_{0}(Y):\overline{d}(t)<1\\}$. It suffices to prove that there do
not exist a neighborhood $V$ of zero in $A_{0}(Y)$ and finite sets $F_{n}$
such that $V\subset F_{n}+U_{(n)}$ for all $n=1,2,\dots$.

Assume the contrary: such $V$ and $F_{n}$’s exist. If $w\in W$ and
$\epsilon>0$, let us say that $V$ is $\epsilon$-fat in the direction $w$ if
$x-y\in V$ for all points $x=(1/k,w)$ and $y=(1/k^{\prime},w)$ such that
$1/k\leq\epsilon$ and $1/k^{\prime}\leq\epsilon$. For every $w\in W$ there
exists an integer $k\in{\mathbb{N}}$ such that $V$ is $1/k$-fat in the
direction $w$, so we can choose an infinite set $J\subset W$ and
$k\in{\mathbb{N}}$ such that $V$ is $1/k$-fat in the direction $w$ for every
$w\in J$.

For $w\in J$ consider the points $x_{w}=(1/k,w)$ and $y_{w}=(1/(k+1),w)$. Note
that the distance from $x_{w}$ to any other point in $Y$ is $\geq 1/k(k+1)$.
From the equality (4.1) for $\overline{d}$ we conclude that
$\overline{d}(t)\geq 1/k(k+1)$ for every $t\in A_{0}(Y)$ such that the support
of $t$ contains $x_{w}$. (The support $\text{supp\,}t$ of $t=\sum_{y\in
Y}n_{y}y$ is the set of all $y\in Y$ such that $n_{y}\neq 0$.) If $n\geq
k(k+1)$, we have $\overline{d}(nt)=n\overline{d}(t)\geq 1$ (by Lemma 4.2), so
$t\notin U_{(n)}$. Since $x_{w}-y_{w}\in V\subset F_{n}+U_{(n)}$ and the
support of any $t\in U_{(n)}$ does not contain $x_{w}$, it follows that there
is $t_{w}\in F_{n}$ such that $x_{w}$ belongs to the support of $t_{w}$. Since
this is true for every $w\in J$, we get a contradiction: the infinite set
$\\{x_{w}:w\in J\\}$ is covered by finitely many finite sets $\text{supp\,}t$,
$t\in F_{n}$. ∎

We now can finish the proof of Theorem 4.3: $X^{(1)}$ can be represented as at
most countable union of clopen $\sigma$-compact subspaces, and the proof is
complete. ∎

######  Corollary 4.6.

Let $X$ be a metrizable space. The following conditions are equivalent:

  * (i)

$A(X)$ is a Schwartz group;

  * (ii)

$X$ can be represented as a disjoint union of two clopen subsets $D$ and $Y$,
where $D$ consists of isolated points of $X$ and $Y$ is locally compact and
has a countable base.

The last part of this section is devoted to the following goal: find
sufficient conditions on $X$, more general than those in Theorem 3.7, which
imply that $L(X)$ is a Schwartz LCS (and $A(X)$ is a Schwartz group).

Recall that a Tychonoff space is pseudocompact if every continuous real-valued
function defined on $X$ is bounded. Let us say that a Tychonoff space $X$ is
$QS$-pseudocompact, if there is a countable sequence of pseudocompact spaces
$(X_{n})$ such that $X$ can be represented as a quotient of the free sum
$\bigoplus_{n\in{\mathbb{N}}}X_{n}$.

######  Theorem 4.7.

Let $X$ be a $QS$-pseudocompact space. Then $L(X)$ is a Schwartz LCS.

###### Proof.

By the argument used in the paragraph after Theorem 3.6, the general case can
be reduced to the case when $X$ is pseudocompact. In that case $L(X)$ is
isomorphic to a linear subspace of $L(\beta X)$, where $\beta X$ is the
Čech–Stone compactification of $X$ (see [30] or [32]). Since $L(\beta X)$ is
Schwartz, by Theorem 3.5, we conclude that $L(X)$ also is Schwartz. ∎

######  Corollary 4.8.

Let X be a free sum of a discrete space and a $QS$-pseudocompact space. Then
$A(X)$ is a Schwartz group.

######  Example 4.9.

Let $X$ be an ordered topological space $[0,\omega_{1})$ consisting of all
countable ordinals. Then $X$ is a countably compact (hence pseudocompact)
first-countable locally compact space. Further, $X$ is not $\sigma$-compact.
Nevertheless, by Theorem 4.7, $L(X)$ is a Schwartz LCS. There is no
contradiction with Theorem 3.13 because $X$ is not paracompact.

##  5\. Open problems

######  Problem 5.1.

Characterize all Tychonoff spaces $X$ such that $L(X)$ are nuclear / multi-
Hilbert.

######  Problem 5.2.

Characterize all Tychonoff spaces $X$ such that $L(X)$ are Schwartz / multi-
reflexive.

######  Problem 5.3.

Does there exist a Tychonoff space $X$ such that $L(X)$ is multi-Hilbert but
not nuclear?

######  Problem 5.4.

Does there exist a Tychonoff space $X$ such that $L(X)$ is multi-reflexive but
not Schwartz?

A more specific question is the following.

######  Problem 5.5.

Let $p$ be a point from the the remainder of the Stone-Čech compactification
$\beta\mathbb{N}\setminus\mathbb{N}$. Denote by $X=\mathbb{N}\cup\\{p\\}$. Is
$L(X)$ nuclear or multi-Hilbert?

Note that every Schwartz space can be isomorphically embedded in some
sufficiently high power of the Banach space $c_{0}$ [26]. There are examples
of Schwartz spaces which cannot be isomorphically embedded in any power of the
Banach space $l^{p}$ $(1<p<\infty)$ [4]. On the other hand, every nuclear
space can be isomorphically embedded in some sufficiently high power of every
Banach space [27]. These facts provide motivation for the following questions.

######  Problem 5.6.

Characterize all Tychonoff spaces $X$ such that $L(X)$ can be isomorphically
embedded in some sufficiently high power of the Banach space $c_{0}$.

######  Problem 5.7.

Characterize all Tychonoff spaces $X$ such that $L(X)$ can be isomorphically
embedded in some sufficiently high power of the Banach space $l^{p}$ $(1\leq
p<\infty)$.

Acknowledgements. We would like to express our gratitude to Vladimir Pestov
and Michael Megrelishvili for the most stimulating questions and discussions
on the topic. We thank Mikhail Ostrovskii for guiding us towards the paper
[18].

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