On the Mackey problem for free locally convex spaces

TL;DR

This paper establishes that the free locally convex space over a Tychonoff space is a Mackey group or space if and only if the underlying space is discrete, clarifying the conditions for Mackey properties.

Contribution

It provides a complete characterization of when free locally convex spaces are Mackey groups or spaces based on the discreteness of the underlying space.

Findings

Free locally convex space is a Mackey group iff the base space is discrete.

Free locally convex space is a Mackey space iff the base space is discrete.

Discreteness of the base space is necessary and sufficient for Mackey properties.

Abstract

We show that the free locally convex space $L(X)$ over a Tychonoff space $X$ is a Mackey group iff $L(X)$ is a Mackey space iff $X$ is discrete.

Equations

$$\Vert \chi \Vert :=|a_1|+\cdots +|a_n|,\text{mbox}and\mathrm{supp}(\chi ):=\{x_1,\dots ,x_n\}.$$

$$t_n(x):=\max \left\{h(x)-\frac{1}{3n+1},0\right\}\cdot \max \left\{0,\frac{1}{3n-1}-h(x)\right\}$$

$$A_n:=\sup \{t_n(x):x\in X\}.$$

$$U_n:=h^{-1}\left(\frac{1}{3n+1},\frac{1}{3n-1}\right)\text{mbox}and{g}_n(x):=\frac{2}{A_n}\cdot t_n(x),$$

$$b_{n+1}<a_n<b_n,[b_{n+1},a_n]\cap \overline{h(W)}=\varnothing \text{mbox}and(a_n,b_n)\cap \overline{h(W)}\ne \varnothing ,\forall n\in \mathbb{N}.$$

$$c_n:=\frac{1}{2}(b_{n+1}+a_n)\text{mbox}and{d}_n:=\frac{1}{2}(b_n+a_{n-1}),\forall n\in \mathbb{N}.$$

$$r_{n}\big{(}[0,c_{n}]\cup[d_{n},1]\big{)}=\{0\}\mbox{ and }r_{n}\big{(}[a_{n},b_{n}]\big{)}=\{1\},$$

$$U_{n}:=h^{-1}(c_{n},d_{n})\;\mbox{ and }\;g_{n}(x):=r_{n}\big{(}h(x)\big{)},\quad x\in X.$$

$$R(\mu):=\big{(}\mu(g_{i})\big{)},\quad\mu\in M_{c}(X).$$

$$T(\mu):=\big{(}\mu,R(\mu)\big{)},\;\;\forall\mu\in M_{c}(X).$$

$$\chi=\big{(}F,(\lambda_{i})\big{)},\mbox{ where }F\in(M_{c}(X),\tau_{e})^{\prime}\mbox{ and }(\lambda_{i})\in\ell_{1}(\kappa).$$

$$\chi (\mu )=\mu (F)+\sum _{i\in \kappa }{\lambda }_i\cdot \mu (g_i)=\mu \left(F+\sum _{i\in \kappa }{\lambda }_i{g}_i\right)=\mu (G),\forall \mu \in M_c(X).$$

$$S:=\{{\delta }_x:\text{mbox}thereisani\in \kappa \text{mbox}suchthat{g}_i(x)\ge 1\}\subseteq M_c(X).$$

$$[K;\epsilon ]:=\{\mu \in M_c(X):|\mu (f)|<\epsilon \forall f\in K\}$$

$$|f(x)-f(z)|<\epsilon ,\forall f\in K,\forall x\in U.$$

$$\big{|}(\delta_{x_{i_{0}}}-\delta_{z})(f)\big{|}=|f(x_{i_{0}})-f(z)|<\varepsilon,\quad\forall f\in K.$$

$$\big{|}\delta_{x}(g_{j})-\delta_{z}(g_{j})\big{|}=|g_{j}(x)-g_{j}(z)|=g_{j}(x)\geq 1>1/2.$$

Full text

1. Introduction

Let $(E,\tau)$ be a locally convex space (lcs for short). A locally convex vector topology $\nu$ on $E$ is called compatible with $\tau$ if the spaces $(E,\tau)$ and $(E,\nu)$ have the same topological dual space. The classical Mackey–Arens theorem states that for every lcs $(E,\tau)$ there exists the finest locally convex vector space topology $\mu$ on $E$ compatible with $\tau$. The topology $\mu$ is called the Mackey toplogy on $E$ associated with $\tau$, and if $\mu=\tau$, the space $E$ is called a Mackey space.

An analogous notion in the class of locally quasi-convex (lqc for short) abelian groups was introduced in [3]. For an abelian topological group $(G,\tau)$ we denote by $\widehat{G}$ the group of all continuous characters of $(G,\tau)$ (for all relevant definitions see the next section). Two topologies $\mu$ and $\nu$ on an abelian group $G$ are said to be *compatible * if $\widehat{(G,\mu)}=\widehat{(G,\nu)}$. Following [3], an lqc abelian group $(G,\mu)$ is called a Mackey group if for every lqc group topology $\nu$ on $G$ compatible with $\tau$ it follows that $\nu\leq\mu$.

Not every Mackey lcs is a Mackey group. In [6] we show that the space $C_{p}(X)$, which is a Mackey space for every Tychonoff space $X$, is a Mackey group if and only if it is barrelled. In particular, this result shows that there are even metrizable lcs which are not Mackey groups that gives a negative answer to a question posed in [4]. Only very recently, answering Question 4.4 of [5], Außenhofer [1] and the author [7] independently have shown that there are lqc groups which do not admit a Mackey group topology. For historical remarks, references and open questions we referee the reader to [5, 12]. In Question 4.3 of [5], we ask: For which Tychonoff spaces $X$ the free lcs $L(X)$ is a Mackey space or a Mackey group? Below we give a complete answer to this question.

Theorem 1.1**.**

For a Tychonoff space $X$, the following assertions are equivalent:

1. (i)

$L(X)$* is a Mackey group;* 2. (ii)

$L(X)$* is a Mackey space;* 3. (iii)

$X$* is discrete.*

In particular, Theorem 1.1 essentially strengthen Theorem 6.4 of [9] which states that $L(X)$ is quasibarrelled if and only of $X$ is discrete.

2. Proof of Theorem 1.1

We start from some necessary definitions and notations. Let $X$ be a Tychonoff space. The space $X$ is called a $k_{\mathbb{R}}$-space if every real-valued function on $X$ which is continuous on every compact subset of $X$ is continuous on $X$. A subset $A$ of $X$ is called functionally bounded in $X$ if every continuous real-valued function on $X$ is bounded on $A$, and $X$ is a $\mu$-space if every functionally bounded subset of $X$ has compact closure. The Dieudonné completion $\mu X$ of $X$ is always a $\mu$-space.

Denote by $\mathbb{S}$ the unit circle group and set $\mathbb{S}_{+}:=\{z\in\mathbb{S}:\ {\rm Re}(z)\geq 0\}$. Let $G$ be an abelian topological group. A character $\chi\in\widehat{G}$ is a continuous homomorphism from $G$ into $\mathbb{S}$. A subset $A$ of $G$ is called quasi-convex if for every $g\in G\setminus A$ there exists $\chi\in\widehat{G}$ such that $\chi(g)\notin\mathbb{S}_{+}$ and $\chi(A)\subseteq\mathbb{S}_{+}$. The group $G$ is called locally quasi-convex if it admits a neighborhood base at the neutral element [math] consisting of quasi-convex sets. Every real locally convex space is a locally quasi-convex group by Proposition 2.4 of [2].

Following [11], the free locally convex space $L(X)$ on a Tychonoff space $X$ is a pair consisting of a locally convex space $L(X)$ and a continuous map $i:X\to L(X)$ such that every continuous map $f$ from $X$ to a locally convex space $E$ gives rise to a unique continuous linear operator ${\bar{f}}:L(X)\to E$ with $f={\bar{f}}\circ i$. The free locally convex space $L(X)$ always exists and is essentially unique. The set $X$ forms a Hamel basis for $L(X)$ and the map $i$ is a topological embedding, see [13, 14].

Let $X$ be a Tychonoff space. For $\chi=a_{1}x_{1}+\cdots+a_{n}x_{n}\in L(X)$ with distinct $x_{1},\dots,x_{n}\in X$ and nonzero $a_{1},\dots,a_{n}\in\mathbb{R}$, we set

[TABLE]

For an lcs $E$, we denote by $E^{\prime}$ the topological dual space of $E$. For a cardinal number $\kappa$, the classical Banach space $c_{0}(\kappa)$ consists of all bounded functions $g:\kappa\to\mathbb{R}$ such that the set $\{i\in\kappa:|g(i)|\geq\varepsilon\}$ is finite for every $\varepsilon>0$ and is endowed with the supremum norm $\|\cdot\|_{\infty}$.

We denote by $C_{k}(X)$ the space $C(X)$ of all real-valued continuous functions on $X$ endowed with the compact-open topology $\tau_{k}$. The support of a function $f\in C(X)$ is denoted by $\mathrm{supp}(f)$. Denote by $M_{c}(X)$ the space of all real regular Borel measures on $X$ with compact support. It is well-known that the dual space of $C_{k}(X)$ is $M_{c}(X)$, see [10, Proposition 7.6.4]. For every $x\in X$, we denote by $\delta_{x}\in M_{c}(X)$ the evaluation map (Dirac measure), i.e. $\delta_{x}(f):=f(x)$ for every $f\in C(X)$. Denote by $\tau_{e}$ the polar topology on $M_{c}(X)$ defined by the family of all equicontinuous pointwise bounded subsets of $C(X)$. We shall use the following deep result of Uspenskiĭ [14].

Theorem 2.1** ([14]).**

Let $X$ be a Tychonoff space and let $\mu X$ be the Dieudonné completion of $X$. Then the completion $\overline{L(X)}$ of $L(X)$ is topologically isomorphic to \big{(}M_{c}(\mu X),\tau_{e}\big{)}.

We need also the following corollary of Theorem 2.1 noticed in [8].

Corollary 2.2** ([8]).**

Let $X$ be a $\mu$-space. Then the topology $\tau_{e}$ on $M_{c}(X)$ is compatible with the duality $(C_{k}(X),M_{c}(X))$.

Proof.

It is well-known that $L(X)^{\prime}=C(X)$, see [13]. Now Theorem 2.1 implies $(M_{c}(X),\tau_{e})^{\prime}=L(X)^{\prime}=C(X)$. ∎

The next lemma follows from Proposition 2.5 of [5], we give its proof for the sake of completeness of the paper.

Lemma 2.3**.**

If a real lcs $(E,\tau)$ is a Mackey group, then it is a Mackey space.

Proof.

Let $\nu$ be a locally convex vector topology on $E$ compatible with $\tau$. Applying Proposition 2.3 of [2] we obtain $\widehat{(E,\nu)}=\widehat{(E,\tau)}$. Hence $\nu$ is a locally quasi-convex group topology (see Proposition 2.4 of [2]) compatible with $\tau$. Therefore $\nu\leq\tau$ since $(E,\tau)$ is a Mackey group. Thus $(E,\tau)$ is a Mackey space. ∎

We need the following characterization of non-discrete Tychonoff spaces.

Proposition 2.4**.**

A Tychonoff space $X$ is not discrete if and only if there exist an infinite cardinal $\kappa$, a point $z\in X$, a family $\{g_{i}\}_{i\in\kappa}$ of continuous functions from $X$ to $[0,2]$ and a family $\{U_{i}\}_{i\in\kappa}$ of open subsets of $X$ such that

1. (i)

$\mathrm{supp}(g_{i})\subseteq U_{i}$* for every $i\in\kappa$;* 2. (ii)

$U_{i}\cap U_{j}=\emptyset$* for all distinct $i,j\in\kappa$;* 3. (iii)

$z\not\in U_{i}$* for every $i\in\kappa$ and z\in\mathrm{cl}\big{(}\bigcup_{i\in\kappa}\{x\in X:g_{i}(x)\geq 1\}\big{)}.*

Proof.

The sufficiency follows from (i)-(iii) which cannot hold simultaneously for discrete spaces. To prove the necessity we consider two cases.

Case 1. There is a continuous function $h:X\to[0,1]$ such that the set $L:=\{x\in X:h(x)>0\}$ is not closed. So there is a $z\in\mathrm{cl}(L)$ such that $h(z)=0$. We distinguish between two subcases.

Subcase 1.1. For every neighborhood $W$ of $z$, the closure $\overline{h(W)}$ of $h(W)$ contains an interval of the form $[0,\varepsilon)$ for some $\varepsilon>0$. For every $n\in\mathbb{N}$, set

[TABLE]

and

[TABLE]

Let $m$ be the least natural number such that $A_{n}>0$ for every $n>m$. For every $n>m$, set

[TABLE]

Then, for every $n>m$, we have $g_{n}(X)\subseteq[0,2]$, $\mathrm{supp}(g_{n})\subseteq U_{n}$ and $z\not\in U_{n}$. Clearly, (i) and (ii) are fulfilled and $z\not\in U_{n}$ for every $n>m$. So to check that the sequences $\{U_{n}:n>m\}$ and $\{g_{n}:n>m\}$ and the point $z$ satisfy (i)-(iii) we have to show that z\in\mathrm{cl}\big{(}\bigcup_{n>m}g_{n}^{-1}\big{(}[1,2]\big{)}\big{)}.

Fix arbitrarily a neighborhood $W$ of $z$ in $X$. Then, by assumption, $\overline{h(W)}$ contains $[0,\varepsilon)$ for some $\varepsilon\in(0,1)$. Therefore, if $n_{0}>(1+3\varepsilon)/(3\varepsilon)$ there is a $y\in W$ such that $t_{n_{0}}(y)\geq(1/2)A_{n_{0}}$, and hence $g_{n_{0}}(y)\geq 1$. Thus g_{n_{0}}^{-1}\big{(}[1,2]\big{)}\cap W is not empty and hence z\in\mathrm{cl}\big{(}\bigcup_{n>m}g_{n}^{-1}\big{(}[1,2]\big{)}\big{)}.

Subcase 1.2. There is a neighborhood $W$ of $z$ such that the closure $\overline{h(W)}$ of $h(W)$ does not contain an interval of the form $[0,\varepsilon)$. Then there exist sequences $\{a_{n}\}_{n\in\mathbb{N}}$ and $\{b_{n}\}_{n\in\mathbb{N}}$ in $(0,1)$ converging to zero such that

[TABLE]

Set $a_{0}:=1$ and

[TABLE]

Then $c_{n}<a_{n}<b_{n}<d_{n}<1$. For every $n\in\mathbb{N}$, let $r_{n}(x)$ be the piecewise linear continuous function from $[0,1]$ to $[0,1]$ such that

[TABLE]

and set

[TABLE]

By construction, the sequences $\{U_{n}:n\in\mathbb{N}\}$ and $\{g_{n}:n\in\mathbb{N}\}$ and the point $z$ satisfy (i) and (ii) and $z\not\in U_{n}$ for every $n\in\mathbb{N}$. Let us show that every neighborhood $U$ of $z$ contains elements of \bigcup_{n\in\mathbb{N}}g_{n}^{-1}\big{(}\{1\}\big{)}. We can assume that $U\subseteq W$. Since $[b_{n+1},a_{n}]\cap\overline{h(W)}=\emptyset$ we obtain that $h(W)\subseteq\{0\}\cup\bigcup_{n\in\mathbb{N}}(a_{n},b_{n})$. Therefore, if $y\in U$ and $n_{0}\in\mathbb{N}$ is such that $h(y)\in(a_{n_{0}},b_{n_{0}})$ (such a $y$ exists because $z\in\mathrm{cl}(L)$), then $g_{n_{0}}(y)=1$.

Case 2. For every continuous function $h:X\to[0,1]$ the set $\{x\in X:h(x)>0\}$ is closed.

We claim that $X$ has a neighborhood base containing closed-and-open sets. Indeed, since $X$ is Tychonoff, for every point $x\in X$ and each open neighborhood $U$ of $x$ there is a continuous function $h:X\to[0,1]$ such that $h(x)=1$ and $h(X\setminus U)=\{0\}$. It remains to note that, by assumption, the open neighborhood h^{-1}\big{(}(0,1]\big{)}\subseteq U of $x$ also is closed.

Now, by the assumption of the proposition, there is a non-isolated point $z\in X$. By the Zorn lemma, there exists a maximal (under inclusion) family $\mathcal{U}=\{U_{i}:i\in\kappa\}$ of pairwise disjoint closed-and-open sets such that $z\not\in U_{i}$ for every $i\in\kappa$. The maximality of $\mathcal{U}$ and the claim imply that z\in\mathrm{cl}\big{(}\bigcup\mathcal{U}\big{)}. For every $i\in\kappa$, let $g_{i}$ be the characteristic function of $U_{i}$. Clearly, the families $\mathcal{U}$ and $\{g_{i}:i\in\kappa\}$ and the point $z$ satisfy conditions (i)-(iii). ∎

The following proposition is crucial for the proof of Theorem 1.1.

Proposition 2.5**.**

Let $X$ be a Dieudonné complete space. If \big{(}M_{c}(X),\tau_{e}\big{)} is a Mackey space, then $X$ is discrete.

Proof.

Suppose for a contradiction that $X$ is not discrete. Then, by Proposition 2.4, there exist an infinite cardinal $\kappa$, a point $z\in X$, a family $\{g_{i}\}_{i\in\kappa}$ of continuous functions from $X$ to $[0,2]$ and a family $\{U_{i}\}_{i\in\kappa}$ of open subsets of $X$ satisfying (i)-(iii) of that proposition. Define a map $R:M_{c}(X)\to c_{0}(\kappa)$ by

[TABLE]

Claim 1. The map $R$ is well-defined.

Indeed, let $\mu\in M_{c}(X)$ be a positive measure. Since $\mu$ is finite and $\sigma$-additive, the condition (ii) of Proposition 2.4 implies that for every $\varepsilon>0$ the number of indices $i\in\kappa$ for which $\mu(U_{i})\geq\varepsilon$ is finite. Now the claim follows from the inclusion $\mathrm{supp}(g_{i})\subseteq U_{i}$ (see (i)) and the inequalities $0\leq\mu(g_{i})\leq 2\mu(U_{i})$.

Consider a map T:M_{c}(X)\to\big{(}M_{c}(X),\tau_{e}\big{)}\times c_{0}(\kappa) defined by

[TABLE]

The map $T$ is well-defined by Claim 1. Denote by $\mathcal{T}$ the locally convex vector topology on $M_{c}(X)$ induced from the product \big{(}M_{c}(X),\tau_{e}\big{)}\times c_{0}(\kappa).

Claim 2. The topology $\mathcal{T}$ is compatible with $\tau_{e}$.

First we note that for every (\lambda_{i})\in\big{(}c_{0}(\kappa)\big{)}^{\prime}=\ell_{1}(\kappa), the function $\sum_{i}\lambda_{i}g_{i}$ belongs to $C(X)$. The Hahn–Banach extension theorem implies that every $\chi\in(M_{c}(X),\mathcal{T})^{\prime}$ has the form

[TABLE]

By Corollary 2.2, we have $F\in C(X)$ and hence $G:=F+\sum_{i\in\kappa}\lambda_{i}f_{i}\in C(X)$. Therefore

[TABLE]

Applying Corollary 2.2 once again we obtain $\chi=G\in(M_{c}(X),\tau_{e})^{\prime}$ as desired.

Claim 3. We claim that $\tau_{e}<\mathcal{T}$.

Indeed, it is clear that $\tau_{e}\leq\mathcal{T}$. Set

[TABLE]

To show that $\tau_{e}\not=\mathcal{T}$, we shall prove that (1) $\delta_{z}\in\mathrm{cl}_{\tau_{e}}(S)$, and (2) $\delta_{z}\not\in\mathrm{cl}_{\mathcal{T}}(S).$

To prove that $\delta_{z}\in\mathrm{cl}_{\tau_{e}}(S)$, fix arbitrarily a standard neighborhood

[TABLE]

of zero in $(M_{c}(X),\tau_{e})$, where $K$ is a pointwise bounded equicontinuous subset of $C(X)$ and $\varepsilon>0$. Choose a neighborhood $U$ of $z$ such that

[TABLE]

By (iii) of Proposition 2.4, take an $i_{0}\in\kappa$ and $x_{i_{0}}\in U$ such that $g_{i_{0}}(x_{i_{0}})\geq 1$. Then $\delta_{x_{i_{0}}}\in S$ and

[TABLE]

Thus $\delta_{x_{i_{0}}}\in\delta_{z}+[K;\varepsilon]$ and hence $\delta_{z}\in\mathrm{cl}_{\tau_{e}}(S)$.

To show that $\delta_{z}\not\in\mathrm{cl}_{\mathcal{T}}(S)$, consider the neighborhood $W:=M_{c}(X)\times U$ of zero in $\mathcal{T}$, where $U=\{g\in c_{0}(\kappa):\|g\|_{\infty}\leq 1/2\}$. Fix arbitrarily $\delta_{x}\in S$ and choose $j\in\kappa$ such that $g_{j}(x)\geq 1$. Then the $j$th coordinate $\delta_{x}(g_{j})$ of $R(\delta_{x})$ satisfies the following (in the last equality we use (i) and (ii) of Proposition 2.4)

[TABLE]

Therefore $R(\delta_{x})-R(\delta_{z})\not\in U$ and hence $\delta_{x}-\delta_{z}\not\in W$. As $x$ was arbitrary we obtain $\delta_{z}\not\in\mathrm{cl}_{\mathcal{T}}(S)$.

Finally, Claims 2 and 3 imply that \big{(}M_{c}(X),\tau_{e}\big{)} is not a Mackey space. This contradiction shows that $X$ must be discrete. ∎

Theorem 1.1 follows from the next more general result.

Theorem 2.6**.**

For a Tychonoff space $X$ the following assertions are equivalent:

1. (i)

$L(X)$* is a Mackey group;* 2. (ii)

$L(X)$* is a Mackey space;* 3. (iii)

\big{(}M_{c}(\mu X),\tau_{e}\big{)}* is a Mackey group;* 4. (iv)

\big{(}M_{c}(\mu X),\tau_{e}\big{)}* is a Mackey space;* 5. (v)

$X$* is discrete.*

Proof.

(i)$\Rightarrow$(ii) and (iii)$\Rightarrow$(iv) follow from Lemma 2.3.

(ii)$\Rightarrow$(iv) It is well known that the completion of a Mackey space is a Mackey space, see Proposition 8.5.8 of [10]. Therefore, by Theorem 2.1, the space \big{(}M_{c}(\mu X),\tau_{e}\big{)} is a Mackey space.

(iv)$\Rightarrow$(v) By Proposition 2.5, $\mu X$ is discrete. Thus $X$ is discrete as well.

(v)$\Rightarrow$(i),(iii) Since $X$ is discrete Theorem 2.1 implies L(X)=\big{(}M_{c}(X),\tau_{e}\big{)}. Therefore $L(X)$ is a barrelled space by [9, Theorem 6.4], and hence $L(X)$ is a Mackey group by [3, Theorem 4.2]. ∎