Topological properties of strict $(LF)$-spaces and strong duals of Montel strict $(LF)$-spaces

TL;DR

This paper characterizes when strict $(LF)$-spaces and their strong duals are Ascoli spaces, revealing that only certain classical spaces like Fréchet or $ ext{ extphi}$ qualify, impacting the understanding of test functions and distributions.

Contribution

It provides a complete characterization of Ascoli properties for strict $(LF)$-spaces and their duals, extending previous results and clarifying the structure of these spaces.

Findings

Strict $(LF)$-spaces are Ascoli iff they are Fréchet or $ ext{ extphi}$.

The strong dual of a Montel strict $(LF)$-space is Ascoli iff it is a Fréchet–Montel space or $ ext{ extphi}$.

Spaces of test functions and distributions are not Ascoli, strengthening prior results.

Abstract

Following [2], a Tychonoff space $X$ is Ascoli if every compact subset of $C_k(X)$ is equicontinuous. By the classical Ascoli theorem every $k$-space is Ascoli. We show that a strict $(LF)$-space $E$ is Ascoli iff $E$ is a Fr\'{e}chet space or $E=\phi$. We prove that the strong dual $E'_\beta$ of a Montel strict $(LF)$-space $E$ is an Ascoli space iff one of the following assertions holds: (i) $E$ is a Fr\'{e}chet--Montel space, so $E'_\beta$ is a sequential non-Fr\'{e}chet--Urysohn space, or (ii) $E=\phi$, so $E'_\beta= \mathbb{R}^\omega$. Consequently, the space $\mathcal{D}(\Omega)$ of test functions and the space of distributions $\mathcal{D}'(\Omega)$ are not Ascoli that strengthens results of Shirai [20] and Dudley [5], respectively.

Equations

$$\genfrac{}{}{0ex}{}{\text{mbox}Fr\acute{e}chet--}{\text{mbox}Urysohn}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

$$\underset{n\in \omega }{⨁}{G}_n:=\left\{(g_n{)}_{n\in \omega }\in \prod _{n\in \omega }{G}_n:g_n=e_n\text{mbox}foralmostalln\right\}.$$

$$\prod _{n\in \omega }{U}_n:=\left\{(g_n{)}_{n\in \omega }\in \prod _{n\in \omega }{G}_n:g_n\in U_n\text{mbox}foralln\in \omega \right\}.$$

$$B_{n,i}:=\{x\in X_i:{\rho }_i(x,e)<2^{-n}\},n\in \omega .$$

$${\mathcal{W}}_m:=\{W_{n,i}:n\in \omega ,i\le m\}$$

$${\mathcal{V}}_k:=\{U_{n,i}\cap X_k:n\in \omega ,1\le i\le k\}$$

$$a_{n,k}:=x_{n,k-1}{y}_{n,k}\text{mbox}and{U}_{n,k}:=a_{n,k}{V}_{n,k}.$$

$$\lim_{n}\rho_{k}\big{(}U_{n,i}\cap X_{k},X_{0})=\lim_{n}\rho_{k}\big{(}y_{n,i}V_{n,i}\cap X_{k},X_{0})\leq\lim_{n}\rho_{k}\big{(}y_{n,i}B_{n,k},e)=0.$$

$$u_n=x\cdot t_n=x_{n,i-1}{y}_{n,i}{z}_n.$$

$$x_{n,i-1}=x\cdot\left(t_{n}z_{n}^{-1}y_{n,i}^{-1}\right)\in\big{(}xU_{x}^{3}\cap X_{0}\big{)}\cap W_{n,i-1}$$

$$A:=\{a_{i,k}:i\in \omega ,k\ge 1\},\mathcal{U}:=\{U_{i,k}:i\in \omega ,k\ge 1\}$$

$$a_{k,n}:=\frac{1}{k}{\eta }_n+k{\xi }_n\text{mbox}and{U}_{k,n}:=a_{k,n}+V_{k,n}.$$

$$A:=\{a_{k,n}:k,n\ge 1\},\mathcal{U}:=\{U_{k,n}:k,n\ge 1\}\text{mbox}andz:=0.$$

$$I:=\{(k,n)\in \mathbb{N}\times \mathbb{N}:C\cap U_{k,n}\ne \varnothing \}.$$

$$S_n^{\ast }\left(\frac{1}{k}{\eta }_n\right)\in {\tilde{W}}_n,\forall k\ge l.$$

$$S_n^{\ast }\left(a_{s,n}+y_s\right)-S_n^{\ast }\left(a_{t,n}+y_t\right)\notin {\tilde{W}}_n$$

$$(t-s)S_n^{\ast }({\xi }_n)=\zeta -S_n^{\ast }\left(\frac{1}{s}{\eta }_n\right)+S_n^{\ast }\left(\frac{1}{t}{\eta }_n\right)-S_n^{\ast }(y_s)+S_n^{\ast }(y_t)\in 5{\tilde{W}}_n$$

$$|a_{k,n+1}(x)|=\left|\left(\frac{1}{k}{\eta }_{n+1}+k{\xi }_{n+1}\right)(x)\right|=\left|\left(\frac{1}{k}{\eta }_{n+1}\right)(x)\right|\le 1.$$

Full text

1. Introduction.

The class of strict $(LF)$-spaces was intensively studied in the classic paper of Dieudonné and Schwartz [3]. It turns out that many of strict $(LF)$-spaces, in particular a lot of linear spaces considered in Schwartz’s theory of distributions [18], are not metrizable. Even the simplest $\aleph_{0}$-dimensional strict $(LF)$-space $\varphi$, the inductive limit of the sequence $\{\mathbb{R}^{n}\}_{n\in\omega}$, is not metrizable. Nyikos [16] showed that $\varphi$ is a sequential non-Fréchet–Urysohn space (all relevant definitions are given in the next section). On the other hand, Shirai [20] proved the space $\mathcal{D}(\Omega)$ of test functions over an open subset $\Omega$ of $\mathbb{R}^{n}$, which is one of the most famous example of strict $(LF)$-spaces, is not sequential. These results motivate the study of sequential properties of strict $(LF)$-spaces and more generally of $(LM)$-spaces. Sequential properties of $(LM)$-spaces were studied by Dudley in [4]. Webb [22] and Ka̧kol and Saxon [14] proved the following remarkable result:

Theorem 1.1** ([14, 22]).**

An $(LM)$-space $E$ is a $k$-space if and only if $E$ is sequential if and only if $E$ is metrizable or is a Montel $(DF)$-space.

Being motivated by the Ascoli theorem we introduced in [2] a new class of topological spaces, namely, the class of Ascoli spaces. A Tychonoff space $X$ is Ascoli if and only if every compact subset of $C_{k}(X)$ is equicontinuous, where $C_{k}(X)$ is the space $C(X)$ of all real-valued continuous functions on $X$ endowed with the compact-open topology. By Ascoli’s theorem [6], every $k$-space is an Ascoli space. So we have the following diagram

[TABLE]

and none of these implications is reversible. The Ascoli property for function spaces and Banach spaces and their closed unit balls with the weak topology has been studied recently in [1, 7, 8, 9, 10, 12]. Taking into account Theorem 1.1 it is natural to consider the following question: Which $(LM)$-spaces are Ascoli? For strict $(LF)$-spaces we obtain a complete answer.

Theorem 1.2**.**

A strict $(LF)$-space $E$ is Ascoli if and only if $E$ is a Fréchet space or $E=\varphi$.

In particular, $\mathcal{D}(\Omega)$ is not an Ascoli space that strengthens Shirai’s result.

Antedating the Nyikos result, Yoshinaga [23] showed that every strong dual of a Fréchet–Schwartz space is sequential. Webb [22] extended this result to strong duals of Fréchet–Montel spaces (equivalently, to Montel $(DF)$-spaces, see Theorem 1.1). Let us recall that a locally convex space $E$ is called semi-Montel if every bounded subset of $E$ is relatively compact, and $E$ is a Montel space if it is a barrelled semi-Montel space. Since every Montel space is reflexive, these results motivate the following problem: Characterize strong duals of Montel spaces which are Ascoli. Note that this problem is quite natural for Montel spaces. Indeed, if $E$ is a Montel space, then: (1) each compact subset of the strong dual space $E^{\prime}_{\beta}$ of $E$ is equicontinuous, and (2) the strong topology $\beta(E^{\prime},E)$ coincides with the compact-open topology on $E^{\prime}$ and therefore $E^{\prime}_{\beta}$ is a closed subspace of $C_{k}(E)$. So the above problem can be reformulated in a more general form as follows: Let $E$ be a locally convex space and $\tau_{k}$ be the compact-open topology on the dual space $E^{\prime}$. When the equicontinuity of the compact subsets of the space $(E^{\prime},\tau_{k})$ implies the equicontinuity of the compact subsets of its ‘functional envelope’ $C_{k}(E)$? Proposition 3.6 below gives a partial answer to this question and plays a crucial role in the proof of our second main result, see Theorem 1.3.

By Theorem 1.1, the strong dual $E^{\prime}_{\beta}$ of an infinite-dimensional Fréchet–Montel space $E$ is sequential, and Webb [22] also has shown that $E^{\prime}_{\beta}$ is not a Fréchet–Urysohn space (below we generalize these results, see Proposition 2.3). However, it seems a little known about topological properties of strong duals of proper Montel strict $(LF)$-spaces. To the best of our knowledge only one nontrivial result is known: Dudley in [5] has shown that the strong dual $\mathcal{D}^{\prime}(\Omega)$ of $\mathcal{D}(\Omega)$, the space of distributions, is not sequential. In the next theorem we eliminate this gap and strengthen Dudley’s result.

Theorem 1.3**.**

Let $E$ be a Montel strict $(LF)$-space. Then the strong dual $E^{\prime}_{\beta}$ of $E$ is an Ascoli space if and only if one of the following assertions holds: (i) $E$ is a Fréchet–Montel space, so $E^{\prime}_{\beta}$ is a sequential non-Fréchet–Urysohn $\mathcal{MK}_{\omega}$-space, or (ii) $E=\varphi$, so $E^{\prime}_{\beta}=\mathbb{R}^{\omega}$.

Consequently, the space of distributions $\mathcal{D}^{\prime}(\Omega)$ is not Ascoli. For another topological properties of $\mathcal{D}^{\prime}(\Omega)$ see [11].

2. Definitions and auxiliary results

All topological spaces in the article are assumed to be Hausdorff. Let us recall some basic definitions. A topological space $X$ is called

• $\bullet$

Fréchet-Urysohn if for any cluster point $a\in X$ of a subset $A\subseteq X$ there is a sequence $\{a_{n}\}_{n\in\mathbb{N}}\subseteq A$ which converges to $a$;

• $\bullet$

sequential if for each non-closed subset $A\subseteq X$ there is a sequence $\{a_{n}\}_{n\in\mathbb{N}}\subseteq A$ converging to some point $a\in\bar{A}\setminus A$;

• $\bullet$

a $k$-space if for each non-closed subset $A\subseteq X$ there is a compact subset $K\subseteq X$ such that $A\cap K$ is not closed in $K$.

Let $\{(X_{n},\tau_{n})\}_{n\in\omega}$ be a sequence of topological spaces such that $X_{n}\subseteq X_{n+1}$ and $\tau_{n+1}|_{X_{n}}=\tau_{n}$ for all $n\in\omega$. The union $X=\bigcup_{n\in\omega}X_{n}$ with the weak topology $\tau$ (i.e., $U\in\tau$ if and only if $U\cap X_{n}\in\tau_{n}$ for every $n\in\omega$) is called the inductive limit of the sequence $\{(X_{n},\tau_{n})\}_{n\in\omega}$ and it is denoted by $(X,\tau)=\underrightarrow{\lim}(X_{n},\tau_{n})$. If $X_{n}$ is closed in $X_{n+1}$ for every $n\in\omega$, then, clearly, $X_{n}$ is a closed subspace of $X$. A topological space $X$ is called a $k_{\omega}$-space (an $\mathcal{MK}_{\omega}$-space) if it is the inductive limit of an increasing sequence $\{C_{n}\}_{n\in\omega}$ of its (respectively, metrizable) compact subsets. So $\varphi=\underrightarrow{\lim}\,\mathbb{R}^{n}$ is an $\mathcal{MK}_{\omega}$-space. In [21, Lemma 9.3] Steenrod proved the following useful result.

Proposition 2.1**.**

If $K$ is a compact subset of $(X,\tau)=\underrightarrow{\lim}(X_{n},\tau_{n})$, then $K\subseteq X_{n}$ for some $n\in\mathbb{N}$.

In what follows we shall omit $\tau_{n}$ and write simply $X=\underrightarrow{\lim}X_{n}$.

Proposition 2.2**.**

If a topological group $(G,\tau)$ is an $\mathcal{MK}_{\omega}$-space, then it is either a locally compact metrizable group or is a sequential non-Fréchet–Urysohn space.

Proof.

It is well-known that any $\mathcal{MK}_{\omega}$-space is sequential. Assume that $G$ is metrizable. Then $G$ is locally compact by Lemma 4.3 of [17]. If $G$ is not metrizable, then it is a sequential non-Fréchet–Urysohn space by Theorem 2.4 of [19]. ∎

We denote by $A^{\circ}$ the polar of a subset $A$ of a locally convex space $E$. Proposition 2.2 and Alaoglu’s theorem and the Banach–Dieudonné theorem imply the following example of $k_{\omega}$-spaces.

Proposition 2.3**.**

Let $E$ be a metrizable infinite-dimensional locally convex space. Then $(E^{\prime},\tau_{k})$ is a $k_{\omega}$-space. If in addition $E$ is separable, then $(E^{\prime},\tau_{k})$ is a sequential non-Fréchet–Urysohn $\mathcal{MK}_{\omega}$-space.

Proof.

Let $\{U_{n}\}_{n\in\omega}$ be a decreasing base of absolutely convex neighborhoods of zero. Then $E^{\prime}=\bigcup_{n\in\omega}U_{n}^{\circ}$, where $U_{n}^{\circ}$ is $\sigma(E^{\prime},E)$-compact by the Alaoglu theorem. Hence $U^{\circ}$ is $\tau_{k}$-compact by Proposition 3.9.8 of [13]. By (3) of [15, §21.10], $\tau_{k}$ coincides with the precompact topology $\tau_{pc}$ on $E^{\prime}$. As every equicontinuous subset of $E^{\prime}$ is contained in some $U_{n}^{\circ}$, (4) of [15, §21.10] implies that $(E^{\prime},\tau_{k})=\underrightarrow{\lim}\,U_{n}^{\circ}$. Thus $(E^{\prime},\tau_{k})$ is a $k_{\omega}$-space.

If $E$ is separable, then $(E^{\prime},\tau_{k})$ admits a weaker metrizable locally convex vector topology. So every $\tau_{k}$-compact subset of $E^{\prime}$ is metrizable, and hence $(E^{\prime},\tau_{k})$ is an $\mathcal{MK}_{\omega}$-space. Since $E$ is infinite-dimensional, Proposition 2.2 implies that $(E^{\prime},\tau_{k})$ is a sequential non-Fréchet–Urysohn $\mathcal{MK}_{\omega}$-space. ∎

An important case of inductive limits of sequences of topological groups is the direct sum a sequence of topological groups endowed with the box topology. Let $\{(G_{n},\tau_{n})\}_{n\in\omega}$ be a sequence of topological groups and $\mathcal{N}(G_{n})$ a basis of open neighborhoods at the identity in $G_{n}$, for each $n\in\omega$. The direct sum of $G_{n}$ is denoted by

[TABLE]

For each $n\in\omega$, fix $U_{n}\in\mathcal{N}(G_{n})$ and put

[TABLE]

Then the sets of the form \big{(}\bigoplus_{n\in\omega}G_{n}\big{)}\cap\prod_{n\in\omega}U_{n}, where $U_{n}\in\mathcal{N}(G_{n})$ for every $n\in\omega$, form a basis of open neighborhoods at the identity of a topological group topology ${\mathcal{T}}_{b}$ on $\bigoplus_{n\in\omega}G_{n}$ that is called the box topology. Set $\widetilde{G}_{n}:=\bigoplus_{i\leq n}G_{i},\,n\in\omega$. Then \big{(}\bigoplus_{n\in\omega}G_{n},\mathcal{T}_{b}\big{)}=\underrightarrow{\lim}\,\widetilde{G}_{n}.

Recall that a locally convex space $E$ is a strict $(LF)$-space if $E$ is the inductive limit of a sequence $\{(E_{n},\tau_{n})\}_{n\in\omega}$ of Fréchet spaces considered as topological spaces, i.e., $\tau_{n+1}|_{E_{n}}=\tau_{n}$ holds for all $n\in\omega$, see [15, §19.4]. A strict $(LF)$-space is proper if a sequence $\{(E_{n},\tau_{n})\}_{n\in\omega}$ can be taken such that $E_{n}\subsetneq E_{n+1}$ for every $n\in\omega$.

We shall use also the following proposition to show that a space is not Ascoli.

Proposition 2.4** ([12]).**

Assume $X$ admits a family $\mathcal{U}=\{U_{i}:i\in I\}$ of open subsets of $X$, a subset $A=\{a_{i}:i\in I\}\subset X$ and a point $z\in X$ such that

1. (i)

$a_{i}\in U_{i}$* for every $i\in I$;* 2. (ii)

\big{|}\{i\in I:C\cap U_{i}\not=\emptyset\}\big{|}<\infty* for each compact subset $C$ of $X$;* 3. (iii)

$z$* is a cluster point of $A$.*

Then $X$ is not an Ascoli space.

3. Proofs

The following proposition plays a crucial role in the proof of Theorem 1.2.

Proposition 3.1**.**

Let $X=\underrightarrow{\lim}\,X_{n}$ be the inductive limit of a sequence $\{X_{n}\}_{n\in\omega}$ of metrizable groups such that $X_{n}$ is a closed non-open subgroup of $X_{n+1}$ for every $n\in\omega$. If $X$ is an Ascoli space, then all the $X_{n}$ are locally compact.

Proof.

Suppose for a contradiction that there is $X_{i}$, say $X_{0}$, which is not locally compact. For every $i\in\omega$ we denote by $\rho_{i}$ a left invariant metric on $X_{i}$ and set

[TABLE]

Step 1. Consider the open base of neighborhoods $\{B_{n,0}\}_{n\in\omega}$ of the unit $e$ of $X_{0}$, so $\overline{B_{n+1,0}}\subseteq B_{n,0}$. Then there is a strictly increasing sequence $\{n_{k}\}_{k\in\omega}$ such that $n_{k+1}>n_{k}+1$ and for every $k\in\omega$, the set $\overline{B_{n_{k},0}}\setminus B_{n_{k}+1,0}$ is not compact. (Indeed, otherwise, there is $n_{0}$ such that $\overline{B_{n,0}}\setminus B_{n+1,0}$ is compact for all $n\geq n_{0}$. Since $B_{n,0}$ converges to $e$, we obtain that $\overline{B_{n_{0},0}}$ is compact. So $X_{0}$ is locally compact, a contradiction.)

Set $P_{k}:=\overline{B_{n_{k},0}}\setminus B_{n_{k}+1,0}$. Then $P_{k}$ is metrizable and non-compact, and hence $P_{k}$ is not pseudocompact. By [6, Theorem 3.10.22] there exists a locally finite collection $\{W_{n,k}\}_{n\in\omega}$ of nonempty open subsets of $P_{k}$. We may assume in addition that every $W_{n,k}\subseteq\mathrm{Int}(P_{k})$. Note that the family

[TABLE]

is also locally finite in $X_{0}$ for every $m\in\omega$. For every $n,k\in\omega$ pick arbitrarily a point $x_{n,k}\in W_{n,k}$.

Step 2. We claim that for every $k\geq 1$ there are

1. (a)

a one-to-one sequence $\{y_{n,k}\}_{n\in\omega}$ in $X_{k}\setminus X_{k-1}$ converging to the unit $e\in X$; 2. (b)

for every $n\in\omega$, an open neighborhood $U_{n,k}$ of $a_{n,k}:=x_{n,k-1}\cdot y_{n,k}$ in $X$;

such that

1. (c)

$U_{n,k}\cap X_{k-1}=\emptyset$ for every $n\in\omega$; 2. (d)

the family

[TABLE]

is locally finite in $X_{k}$.

Indeed, for every $k\geq 1$, let $\{y_{n,k}\}_{n\in\omega}$ be an arbitrary one-to-one sequence in $X_{k}\setminus X_{k-1}$ converging to $e$ (such a sequence exists because $X_{k-1}$ is not open in $X_{k}$ by assumption). For every $k\geq 1$ and each $n\in\omega$ choose an open symmetric neighborhood $V_{n,k}$ of $e$ in $X$ such that

1. $(\alpha)$

$V_{n,k}^{3}\cap X_{i}\subseteq B_{n,i}$ for every $0\leq i\leq n$; 2. $(\beta)$

\big{(}y_{n,k}\cdot V_{n,k}^{3}\big{)}\cap X_{k-1}=\emptyset (recall that $X_{k-1}$ is closed in $X$).

For every $k\geq 1$ and each $n\in\omega$ set

[TABLE]

Clearly, (a) and (b) hold. Also (c) holds since if $U_{n,k}\cap X_{k-1}\not=\emptyset$ for some $k\geq 1$ and $n\in\omega$, then $x_{n,k-1}y_{n,k}v\in X_{k-1}$ for some $v\in V_{n,k}$. So $y_{n,k}v\in x_{n,k-1}^{-1}X_{k-1}=X_{k-1}$ that contradicts $(\beta)$.

Let us check (d): $\mathcal{V}_{k}$ is locally finite in $X_{k}$ for every $k\geq 1$. Fix $x\in X_{k}$ and consider the two possible cases.

Case 1. Let $x\in X_{k}\setminus X_{0}$. So $\rho_{k}(x,X_{0})>0$ as $X_{0}$ is closed in $X_{k}$. For every $1\leq i\leq k$, since $y_{n,i}\to e$ in $X$ and $x_{n,i-1}\in X_{0}$, the condition $(\alpha)$ implies (note that $V_{n,i}\cap X_{k}\subseteq B_{n,k}$ for every $1\leq i\leq k<n$)

[TABLE]

Hence there is an open neighborhood $U_{x}$ of $x$ in $X$ such that $U_{x}\cap X_{k}$ intersects only with a finite subfamily of $\mathcal{V}_{k}$.

Case 2. Let $x\in X_{0}$. Choose an open symmetric neighborhood $U_{x}$ of $e$ in $X$ such that $xU_{x}^{3}\cap X_{0}$ intersects only with a finite subfamily of $\mathcal{W}_{k}$. We claim that $xU_{x}\cap X_{k}$ intersects only a finite subfamily of $\mathcal{V}_{k}$. Indeed, assuming the converse we can find $1\leq i\leq k$ such that $(xU_{x}\cap X_{k})\cap U_{n,i}\not=\emptyset$ for every $n\in I$, where $I$ is an infinite subset of $\omega$. Then for every $n\in I$ there are $u_{n}\in X_{k}$, $t_{n}\in U_{x}$ and $z_{n}\in V_{n,i}$ such that

[TABLE]

Note that $z_{n}=y_{n,i}^{-1}x_{n,i-1}^{-1}u_{n}\in V_{n,i}\cap X_{k}$ belongs to $U_{x}\cap X_{k}$ for all sufficiently large $n\in I$ by $(\alpha)$, and also $y_{n,i}\in U_{x}\cap X_{k}$ for all sufficiently large $n\in I$ because $y_{n,i}\to e$. So

[TABLE]

for all sufficiently large $n\in I$. But this contradicts the choice of $U_{x}$.

Cases 1 and 2 show that $\mathcal{V}_{k}$ is locally finite in $X_{k}$.

Step 3. Let us show that the families

[TABLE]

and $z:=e$ satisfy (i)-(iii) of Proposition 2.4. Indeed, (i) is clear. To check (ii) let $K$ be a compact subset of $X$. By Proposition 2.1, there is $m\in\omega$ such that $K\subseteq X_{m}$. So (c) implies that if $K\cap U_{n,i}\not=\emptyset$, then $i\leq m$, and hence $U_{n,i}\cap X_{m}\in\mathcal{V}_{m}$. Since the family $\mathcal{V}_{m}$ is locally finite in $X_{m}$, we obtain that $K$ intersects only a finite subfamily of $\mathcal{U}$ that proves (ii).

To prove (iii) let $V$ be an open neighborhood of $e$. Take an open neighborhood $U$ of $e$ such that $U^{2}\subseteq V$, and choose $k_{0}\in\omega$ such that $W_{i,k_{0}}\subseteq X_{0}\cap U$ for every $i\in\omega$. So $x_{i,k_{0}}\in U$ for every $i\in\omega$. Since $\lim_{i}y_{i,k_{0}+1}=e$ we obtain that $a_{i,k_{0}+1}=x_{i,k_{0}}y_{i,k_{0}+1}\in U\cdot U\subseteq V$ for all sufficiently large $i$. Thus $e\in\overline{A}$ and (iii) holds. Finally, Proposition 2.4 implies that the group $X$ is not Ascoli which is a desired contradiction. ∎

Theorem 3.2**.**

Let $X=\underrightarrow{\lim}\,X_{n}$ be the inductive limit of a sequence $\{X_{n}\}_{n\in\omega}$ of metrizable groups such that $X_{n}$ is a closed subgroup of $X_{n+1}$ for every $n\in\omega$. Then the following assertions are equivalent:

1. (i)

$X$* is an Ascoli space;* 2. (ii)

one of the following assertions holds:

1. $(ii)_{1}$

there is $m\in\omega$ such that $X_{n}$ is open in $X_{n+1}$ for every $n\geq m$, so $X$ is metrizable; 2. $(ii)_{2}$

all the $X_{n}$ are locally compact, so $X$ contains an open $\mathcal{MK}_{\omega}$-subspace and hence $X$ is a sequential non-Fréchet–Urysohn space.

Proof.

(i)$\Rightarrow$(ii) If there is $m\in\omega$ such that $X_{n}$ is open in $X_{n+1}$ for every $n\geq m$, then $X_{m}$ is an open subgroup of $X$. Thus $X$ is metrizable. Assume that for infinitely many $n\in\omega$ the group $X_{n}$ is not open in $X_{n+1}$. Without loss of generality we can assume that $X_{n}$ is not open in $X_{n+1}$ for all $n\in\omega$. Since $X$ is Ascoli, Proposition 3.1 implies that all the $X_{n}$ are locally compact. Let $Y_{n}$ be an open $\sigma$-compact subgroup of $X_{n}$. We can assume that $Y_{n}\subseteq Y_{n+1}$ for every $n\in\omega$. As all the $Y_{n}$ are metrizable, the group $Y:=\underrightarrow{\lim}\,Y_{n}$ is an $\mathcal{MK}_{\omega}$-space, and hence $Y$ is a sequential non-Fréchet–Urysohn space by Proposition 2.2. Clearly, $Y$ is an open subgroup of $X$. Thus $X$ is also a sequential non-Fréchet–Urysohn space.

(ii)$\Rightarrow$(i) follows from the Ascoli theorem [6]. ∎

Now we prove Theorem 1.2.

Proof of Theorem 1.2.

For every $n\in\omega$ the space $E_{n}$ is closed in $E_{n+1}$ as a complete subspace of a complete metrizable space. Taking into account that a locally convex space $E$ is locally compact if and only if $E$ is finite dimensional, the assertion follows from Theorem 3.2. ∎

It is mentioned in [16, Footnote 2] that van Douwen has shown the following: if even one of the factors in the direct sum $X=(\bigoplus_{n\in\omega}X_{n},\mathcal{T}_{b})$ of a sequence $\{X_{n}\}_{n\in\omega}$ of metrizable groups with the box topology $\mathcal{T}_{b}$ is not locally compact and infinitely many of the $X_{n}$ are not discrete, then $X$ is not sequential. The next corollary generalizes this result.

Corollary 3.3**.**

Let $\{X_{n}\}_{n\in\omega}$ be a sequence of metrizable groups such that infinitely many of the $X_{n}$ are not discrete and let $X=(\bigoplus_{n\in\omega}X_{n},\mathcal{T}_{b})$ be the direct sum endowed with the box topology $\mathcal{T}_{b}$. Then $X$ is an Ascoli space if and only if all the $X_{n}$ are locally compact. In this case $X$ has an open subgroup which is a sequential non-Fréchet–Urysohn $\mathcal{MK}_{\omega}$-space, and hence $X$ is also a sequential non-Fréchet–Urysohn space.

Corollary 3.4**.**

Let $E=\bigoplus_{n\in\omega}E_{n}$ be the direct locally convex sum of a sequence $\{E_{n}\}_{n\in\omega}$ of nontrivial metrizable locally convex spaces. Then $E$ is Ascoli if and only if $E=\varphi$.

Proof.

Taking into account that a locally convex space $E$ is locally compact if and only if $E$ is finite dimensional, the assertion follows from Corollary 3.3. ∎

In particular, the space $\mathbb{R}^{\omega}\times\varphi$ is not Ascoli, and hence the product of a metrizable space and a sequential space can be not Ascoli.

To prove Theorem 1.3 we need two assertions.

Let $E$ be a locally convex space and $E^{\prime}$ be its dual. Denote by $\tau_{k}$ and $\tau_{pc}$ the compact-open topology and the precompact-open topology on $E^{\prime}$, respectively.

Lemma 3.5**.**

Let $E$ be an infinite-dimensional barrelled space and let $\mathcal{T}$ be a locally convex topology on the dual space $E^{\prime}$ such that $\sigma(E^{\prime},E)\leq\mathcal{T}\leq\tau_{pc}$ and the space $(E^{\prime},\tau_{\mathcal{T}})$ is barrelled. Then for every neighborhood $U$ of zero in $E$ there is an absolutely convex neighborhood $V\subseteq U$ of zero in $E$ and a nonzero $\chi\in V^{\circ}$ such that $\mathbb{R}\chi\cap U^{\circ}=\{0\}$.

Proof.

By the Alaoglu theorem, the polar $U^{\circ}$ is $\sigma(E^{\prime},E)$-compact, and hence $U^{\circ}$ is a $\tau_{pc}$-compact disc by Proposition 3.9.8 of [13]. Therefore the $\mathcal{T}$-compact set $U^{\circ}$ is not absorbing (otherwise, $U^{\circ}$ would be a compact neighborhood of zero in $(E^{\prime},\mathcal{T})$ by the barrelledness of $(E^{\prime},\mathcal{T})$, and hence $E$ is finite-dimensional). So there is $\chi\in E^{\prime}$ such that $\mathbb{R}\chi\cap U^{\circ}=\{0\}$. It remains to choose a neighborhood $V\subseteq U$ of zero such that $\chi\in V^{\circ}$. ∎

For a subspace $H$ of a locally convex space $E$ we set $H^{\perp}:=\{\chi\in E^{\prime}:\chi|_{H}=0\}$.

Proposition 3.6**.**

Let $E=\underrightarrow{\lim}\,E_{n}$ be a strict inductive limit of a sequence $\{(E_{n},\tau_{n})\}_{n\in\omega}$ of locally convex spaces such that $E_{n}$ is a closed proper subspace of $E_{n+1}$ for every $n\in\omega$. Assume that $E_{0}$ is an infinite-dimensional barrelled metrizable space such that $(E^{\prime}_{0},\tau_{k})$ is also barrelled. Then $(E^{\prime},\tau_{k})$ is not an Ascoli space.

Proof.

We shall apply Proposition 2.4. For every $n\in\omega$, we denote by $S_{n}:E_{n}\to E$ the canonical embedding and observe that the adjoint map $S^{\ast}_{n}$ is weak* continuous and $\tau_{k}$-continuous.

Step 1. The basic construction. By Lemma 3.5, choose a decreasing base $\{O_{n}\}_{n\in\omega}$ of absolutely convex neighborhoods of zero in $E_{0}$ such that for every $n\geq 1$ there is a nonzero $\chi_{n}\in O_{n}^{\circ}$ satisfying $\mathbb{R}\chi_{n}\cap O_{n-1}^{\circ}=\{0\}$. For every $n,k\geq 1$ choose a $\sigma(E^{\prime}_{0},E_{0})$-open neighborhood $W_{k,n}$ of zero in $E^{\prime}_{0}$ such that

1. (a)

$\left(\frac{1}{k}\chi_{n}+W_{k,n}\right)\cap O_{n-1}^{\circ}=\emptyset$.

For every $n\geq 1$, choose $\eta_{n},\xi_{n}\in E^{\prime}$ such that

1. (b)

$S_{0}^{\ast}(\eta_{n})=\chi_{n}$; 2. (c)

$\xi_{n}\in E_{n-1}^{\perp}\setminus E_{n}^{\perp}$ (recall that $E_{n}/E_{n-1}$ is nontrivial and Hausdorff by the assumption of the proposition);

and now choose an absolutely convex $\sigma(E^{\prime}_{n},E_{n})$-neighborhood $\widetilde{W}_{n}$ of zero in $E^{\prime}_{n}$ such that

1. (d)

$S_{n}^{\ast}(\xi_{n})\not\in 5\widetilde{W}_{n}$.

For every $n,k\geq 1$, choose a $\tau_{k}$-open neighborhood $V_{k,n}$ of zero in $E^{\prime}$ such that

1. (e)

$S_{0}^{\ast}(V_{k,n})\subseteq W_{k,n}$, 2. (f)

$S_{n}^{\ast}(V_{k,n})\subseteq\widetilde{W}_{n}$,

and set

[TABLE]

Define

[TABLE]

Step 2. We show that $A$, $\mathcal{U}$ and $z$ satisfy (i)-(iii) of Proposition 2.4. (i) is clear. To prove (ii), let $C$ be a compact subset of $(E^{\prime},\tau_{k})$ and set

[TABLE]

We have to show that $I$ is finite. Since $E_{0}$ is barrelled, the $\sigma(E^{\prime}_{0},E_{0})$-compact subset $S_{0}^{\ast}(C)$ of $E^{\prime}_{0}$ is equicontinuous. So there is $m\geq 1$ such that $S_{0}^{\ast}(C)\subseteq O_{m}^{\circ}$. By (a)-(c) and (e) we obtain that $n\leq m$ for every $(k,n)\in I$. Now for a fixed $n$, $1\leq n\leq m$, take $l\geq 1$ such that

[TABLE]

We claim that

[TABLE]

for every $t>s\geq l$ and $y_{s}\in V_{k,n}$ and $y_{t}\in V_{t,n}$. Indeed, denote by $\zeta$ the element in the left side and suppose for a contradiction that $\zeta\in\widetilde{W}_{n}$. Then (f) and (3.1) imply

[TABLE]

and therefore $S_{n}^{\ast}(\xi_{n})\in\frac{5}{t-s}\widetilde{W}_{n}$ that contradicts (d) (recall that $\widetilde{W}_{n}$ is absolutely convex). Since the compact set $S_{n}^{\ast}(C)$ cannot contain an infinite uniformly discrete subset, this means that the set $\{k:(k,n)\in I\}$ is finite. Thus $I$ is finite and (ii) holds true.

To check (iii), let $U=K^{\circ}$ be a basic $\tau_{k}$-neighborhood of $z$, where $K$ is a compact subset of $E$. By Proposition 2.1, there is $n\in\omega$ such that $K\subseteq E_{n}$ (we identify $E_{n}$ with its image $S_{n}(E_{n})$ in $E$). Take $k\geq 1$ sufficiently large such that $\frac{1}{k}\eta_{n+1}\in U$. Then, for every $x\in K$, (c) implies

[TABLE]

Therefore $a_{k,n+1}\in U$ and hence $z$ belongs to the $\tau_{k}$-closure of $A$. Finally, Proposition 2.4 implies that $(E^{\prime},\tau_{k})$ is not an Ascoli space. ∎

Now we are ready to prove Theorem 1.3.

Proof of Theorem 1.3.

The ‘if’ part is trivial. Assume that $E=\underrightarrow{\lim}\,E_{n}$ is an infinite-dimensional Montel strict $(LF)$-space such that the strong dual $E^{\prime}_{\beta}$ of $E$ is an Ascoli space. Note that for every $n\in\omega$ the space $E_{n}$ is closed in $E_{n+1}$ as a complete subspace of a complete metrizable space, and the strong topology $\beta(E^{\prime},E)$ coincides with the compact-open topology $\tau_{k}$ on $E^{\prime}$ as $E$ is Montel.

If $E$ is not proper and $E=E_{m}$ for some $m\in\omega$, then $E$ is a Fréchet–Montel space. Therefore $E^{\prime}_{\beta}$ is a sequential non-Fréchet–Urysohn $\mathcal{MK}_{\omega}$-space by Proposition 2.3.

Assume that $E$ is proper. If all $E_{n}$ are finite-dimensional, then $E=\varphi$. If $E_{n}$ is infinite-dimensional for some $n\in\omega$, say $E_{0}$, then $E_{0}$ is a Fréchet–Montel space. Therefore its strong dual $(E^{\prime}_{0},\beta(E^{\prime}_{0},E_{0}))$ is also a Montel space and hence is barrelled. Since $\tau_{k}=\beta(E^{\prime}_{0},E_{0})$, the space $E^{\prime}_{\beta}$ is not Ascoli by Proposition 3.6. ∎