Josefson–Nissenzweig property for $C_{p}$-spaces

Abstract

The famous Rosenthal–Lacey theorem asserts that for each infinite compact space K the Banach space C(K) admits a quotient isomorphic to Banach spaces c or $\ell _{2}$. The aim of the paper is to study a natural variant of this result for the space $C_{p}(X)$ of continuous real-valued maps on a Tychonoff space X with the pointwise topology. Following Josefson–Nissenzweig theorem for infinite-dimensional Banach spaces we introduce a corresponding property (called Josefson–Nissenzweig property, briefly, the JNP) for $C_{p}(X)$-spaces. We prove: for a Tychonoff space X the space $C_p(X)$ satisfies the JNP if and only if $C_p(X)$ has a quotient isomorphic to $c_{0}:=\{(x_n)_{n\in \mathbb N}\in \mathbb {R}^\mathbb {N}:x_n\rightarrow 0\}$ (with the product topology of $\mathbb {R}^\mathbb {N}$) if and only if $C_{p}(X)$ contains a complemented subspace isomorphic to $c_0$. The last statement provides a $C_{p}$-version of the Cembranos theorem stating that the Banach space C(K) is not a Grothendieck space if and only if C(K) contains a complemented copy of the Banach space $c_{0}$ with the sup-norm topology. For a pseudocompact space X the space $C_p(X)$ has the JNP if and only if $C_p(X)$ has a complemented metrizable infinite-dimensional subspace. An example of a compact space K without infinite convergent sequences with $C_{p}(K)$ containing a complemented subspace isomorphic to $c_{0}$ is given.

1 Introduction, motivations and two main problems

Let X be a Tychonoff space. By $C_{p}(X)$ we denote the space of real-valued continuous functions on X endowed with the pointwise topology.

We will need the following fact stating that each metrizable (linear) quotient $C_{p}(X)/Z$ of $C_p(X)$ by a closed vector subspace Z of $C_p(X)$ is separable. Indeed, this follows from the separability of metizable spaces of countable cellularity and the fact that $C_p(X)$ has countable cellularity, being a dense subspace of $\mathbb {R}^X$, see [2].

The classic Rosenthal–Lacey theorem, see [19, 23, 27], asserts that the Banach space C(K) of continuous real-valued maps on an infinite compact space K has a quotient isomorphic to Banach spaces c or $\ell _{2},$ or equivalently, there exists a continuous linear (and open; by the open mapping Banach theorem) map from C(K) onto c or $\ell _{2}$, see also a survey paper [14].

This theorem motivates the following natural question for spaces $C_{p}(X)$.

Problem 1

For which compact spaces K any of the following equivalent conditions holds:

• (1) The space $C_{p}(K)$ has an infinite dimensional metrizable quotient.

• (2) The space $C_{p}(K)$ has an infinite dimensional metrizable separable quotient.

• (3) The space $C_p(K)$ has a quotient isomorphic to a dense subspace of $\mathbb {R}^{\mathbb {N}}$.

In [21] it was shown that $C_{p}(K)$ has an infinite-dimensional separable quotient algebra if and only if K contains an infinite countable closed subset. Hence $C_{p}(\beta \mathbb {N})$ lacks infinite-dimensional separable quotient algebras. Nevertheless, as proved in [22, Theorem 4], the space $C_{p}(K)$ has infinite-dimensional separable quotient for any compact space K containing a copy of $\beta \mathbb {N}$.

Problem 1 has been already partially studied in [3], where we proved that for a Tychonoff space X the space $C_p(X)$ has an infinite-dimensional metrizable quotient if X either contains an infinite discrete $C^{*}$-embedded subspace or else X has a sequence $(K_n)_{n\in \mathbb {N}}$ of compact subsets such that for every n the space $K_n$ contains two disjoint topological copies of $K_{n+1}$. If fact, the first case (for example if compact X contains a copy of $\beta \mathbb {N}$) asserts that $C_{p}(X)$ has a quotient isomorphic to the subspace $\ell _\infty =\{(x_n)\in \mathbb {R}^{\mathbb {N}}:\sup _n |x_n|<\infty \}$ of $\mathbb {R}^{\mathbb {N}}$ or to the product $\mathbb {R}^{\mathbb {N}}$.

Consequently, this theorem reduces Problem 1 to the case when K is an Efimov space (i.e. K is an infinite compact space that contains neither a non-trivial convergent sequence nor a copy of $\beta \mathbb {N}$). Although, it is unknown if Efimov spaces exist in ZFC (see [7,8,9,10, 12, 13, 16, 18]) we showed in [22] that under $\lozenge$ for some Efimov spaces K the function space $C_{p}(K)$ has an infinite dimensional metrizable quotient.

By $c_{0}$ we mean the subspace $\{(x_n)_{n\in \mathbb N}\in \mathbb {R}^{\mathbb {N}}:x_n\rightarrow 0\}$ of $\mathbb {R}^{\mathbb {N}}$ endowed with the product topology. The term “the Banach space ${c}_{0}$” means the classic Banach space of null-sequences with the sup-norm topology.

It is known that the Banach space C(K) over an infinite compact K contains a copy of the Banach space $c_{0}$, see for example [6]. By a result of Cembranos, see [4, Theorem, page 74], the space C(K) is not a Grothendieck space if and only if C(K) contains a complemented copy of the Banach space $c_{0}$. Recall a Banach space E is a Grothendieck space if every weak$^{*}$ converging sequence in the dual $E^{*}$ weakly converges in $E^{*}$. It is well-known that if a compact space K contains a non-trivial converging sequence, C(K) is not a Grothendieck space; hence C(K) contains a complemented copy of the Banach space $c_{0}$. It is also easy to see that for every infinite compact space K the space $C_{p}(K)$ contains a closed copy of the space $c_{0}$ endowed with the product topology of $\mathbb {R}^{\mathbb {N}}$.

Cembranos theorem motivates the following next problem (connected with Problem 1).

Problem 2

Characterize those spaces $C_{p}(K)$ which contain a complemented copy of $c_{0}$ with the product topology of $\mathbb {R}^{\mathbb {N}}$.

2 The main results

For a Tychonoff space X and a point $x\in X$ let $\delta _x:C_p(X)\rightarrow \mathbb {R},\,\,\, \delta _x:f\mapsto f(x),$ be the Dirac measure concentrated at x. The linear hull $L_p(X)$ of the set $\{\delta _x:x\in X\}$ in $\mathbb {R}^{C_p(X)}$ can be identified with the dual space of $C_p(X)$. We refer also the reader to [15] for more information about the dual $L_p(X)$.

Elements of the space $L_p(X)$ will be called finitely supported sign-measures (or simply sign-measures) on X.

Each $\mu \in L_p(X)$ can be uniquely written as a linear combination of Dirac measures $\mu =\sum _{x\in F}\alpha _x\delta _x$ for some finite set $F\subset X$ and some non-zero real numbers $\alpha _x$. The set F is called the support of the sign-measure $\mu$ and is denoted by $\mathrm {supp}(\mu )$. The measure $\sum _{x\in F}|\alpha _x|\delta _x$ will be denoted by $|\mu |$ and the real number $\Vert \mu \Vert =\sum _{x\in F}|\alpha _x|$ coincides with the norm of $\mu$ (in the dual Banach space $C(\beta X)^{*}$).

The sign-measure $\mu =\sum _{x\in F}\alpha _x\delta _x$ determines the function $\mu :2^X\rightarrow \mathbb {R}$ defined on the power-set of X and assigning to each subset $A\subset X$ the real number $\sum _{x\in A\cap F}\alpha _x$. So, a finitely supported sign-measure will be considered both as a linear functional on $C_p(X)$ and an additive function on the power-set $2^X$.

The famous Josefson–Nissenzweig theorem asserts that for each infinite-dimensional Banach space E there exists a null sequence in the weak$^{*}$-topology of the topological dual $E^{*}$ of E and which is of norm one in the dual norm, see for example [6].

We propose the following corresponding property for spaces $C_{p}(X)$.

Definition 1

For a Tychonoff space X the space $C_{p}(X)$ satisfies the Josefson–Nissenzweig property (JNP in short) if there exists a sequence $(\mu _n)$ of finitely supported sign-measures on X such that $\Vert \mu _n\Vert =1$ for all $n\in \mathbb {N}$, and $\mu _n(f)\rightarrow _n 0$ for each $f\in C_p(X)$.

Concerning the JNP of function spaces $C_p(X)$ on compacta we have the following:

• (1) If a compact spaceKcontains a non-trivial convergent sequence, say$x_{n}\rightarrow x$, then$C_{p}(K)$satisfies the JNP. This is witnessed by the weak$^{*}$ null sequence $(\mu _n)$ of sign-measures $\mu _{n}=\frac{1}{2}(\delta _{x_{n}}-\delta _{x})$, $n\in \mathbb {N}$.

• (2) The space$C_{p}(\beta \mathbb {N})$does not satisfy the JNP. This follows directly from the Grothendieck theorem, see [5, Corollary 4.5.8].

• (3) There exists a compact spaceKcontaining a copy of$\beta \mathbb {N}$but without non-trivial convergent sequences such that$C_{p}(K)$satisfies the JNP, see Example 1 below.

If a compact space K contains a closed subset Z that is metrizable, then $C_p(K)$ has a complemented subspace isomorphic to $C_p(Z)$. Consequently, if compact K contains a non-trivial convergent sequence, then $C_{p}(K)$ has a complemented subspace isomorphic to $c_{0}$. However for every infinite compact K the space $C_{p}(K)$ contains a subspace isomorphic to $c_{0}$ but not necessary complemented in $C_{p}(K)$. Nevertheless, there exists a compact space K without infinite convergent sequences and such that $C_{p}(K)$ enjoys the JNP and so contains a complemented subspace isomorphic to $c_{0}$, as follows from Theorem 1 below.

It turns out that the Josefson–Nissenzweig property characterizes an interesting case related with Problem 1 and provides a complete solution to Problem 2.

Theorem 1

For a Tychonoff space X the following conditions are equivalent:

• (1) $C_{p}(X)$ satisfies the JNP;

• (2) $C_p(X)$ contains a complemented subspace isomorphic to $c_0$;

• (3) $C_p(X)$ has a quotient isomorphic to $c_0$;

• (4) $C_p(X)$ admits a linear continuous map onto $c_0.$ If the space X is pseudocompact, then the conditions (1)–(4) are equivalent to

• (5) $C_{p}(X)$ contains a complemented infinite-dimensional metrizable subspace;

• (6) $C_{p}(X)$ contains a complemented infinite-dimensional separable subspace;

• (7) $C_p(X)$ has an infinite-dimensional Polishable quotient.

We recall that a locally convex space X is Polishable if X admits a stronger Polish locally convex topology. Equivalently, Polishable locally convex spaces can be defined as images of separable Fréchet spaces under continuous linear maps. Clearly, the subspace $c_0$ of $\mathbb {R}^{\mathbb {N}}$ is Polishable.

A topological space X is pseudocompact if it is Tychonoff and each continuous real-valued function on X is bounded. It is known (see [3]) that a Tychonoff space X is not pseudocompact if and only if $C_{p}(X)$ contains a complemented copy of $\mathbb {R}^{\mathbb {N}}$. Combining this characterization with Theorem 1, we obtain another characterization related to Problem 1.

Corollary 1

For a Tychonoff space X the following conditions are equivalent:

• (1) $C_p(X)$ has an infinite-dimensional Polishable quotient;

• (2) $C_p(X)$ contains a complemeneted infinite-dimensional Polishable subspace;

• (3) $C_p(X)$ contains a complemented subspace isomorphic to $\mathbb {R}^{\mathbb {N}}$ or $c_0$;

Corollary 2

The space $C_{p}(\beta \mathbb {N})$

• (1) has a quotient isomorphic to $\ell _{\infty }$;

• (2) contains a subspace isomorphic to $c_{0}$;

• (3) does not admit a continuous linear map onto $c_0$;

• (4) has no Polishable infinite-dimensional quotients;

• (4) contains no complemented separable infinite-dimensional subspaces.

Indeed, the first claim follows from [3, Proposition], the others follow from Theorem 1 and the statement (2) after Definition 1.

In the final Sect. 5 we shall characterize Tychonoff spaces X whose function space $C_p(X)$ is Polishable and prove the following theorem.

Theorem 2

For a Tychonoff space X the following conditions are equivalent:

• (1) $C_p(X)$ is Polishable;

• (2) $C_k(X)$ is Polishable;

• (3) $C_k(X)$ is Polish;

• (4) X is a submetrizable hemicompact k-space.

In this theorem $C_k(X)$ denotes the space of continuous real-valued functions on X, endowed with the compact-open topology. It should be mentioned that a locally convex space is Polish if and only if it is a separable Fréchet space, by using, for example, the Birkhoff–Kakutani theorem [20, Theorem 9.1].

3 Proof of Theorem 1

We start with the following

Lemma 1

Let a Tychonoff space X be continuously embedded into a compact Hausdorff space K (i.e. there exists a continuous injection from X into K.) Let $(\mu _n)$ be a sequence of finitely supported sign-measures on X (and so, on K) such that

• (1) $\Vert \mu _n\Vert =1$ for all $n\in \mathbb {N}$, and

• (2) $\mu _n(f)\rightarrow _n 0$ for all $f\in C(K)$.

Then there exists an infinite subset $\Omega$ of $\mathbb {N}$ such that

• (a) the closed subspace $Z=\bigcap _{k\in \Omega }\{f\in C_p(X):\mu _k(f)=0\}$ of $C_p(X)$ is complemented in the subspace $L=\big \{f\in C_p(X):\lim _{k\in \Omega }\mu _k(f)=0\big \}$ of $C_p(X)$;

• (b) the quotient space L / Z is isomorphic to the subspace $c_0$ of $\mathbb {R}^\mathbb {N}$;

• (c) L contains a complemented subspace isomorphic to $c_0$;

• (d) the quotient space $C_p(X)/Z$ is infinite-dimensional and metrizable (and so, separable).

Proof

• (I) First we show that the set $M=\{\mu _n: n\in \mathbb {N}\}$ is not relatively weakly compact in the dual of the Banach space C(K). Indeed, assume on the contrary that the closure $\overline{M}$ of M in the weak topology of $C(K)^{*}$ is weakly compact. Applying the Eberlein–Šmulian theorem [1, Theorem 1.6.3], we conclude that $\overline{M}$ is weakly sequentially compact. Thus $(\mu _n)$ has a subsequence $(\mu _{k_n})$ that weakly converges to some element $\mu _0\in C(K)^{*}$. Taking into account that the sequence $(\mu _n)$ converges to zero in the weak$^{*}$ topology of $C(K)^{*}$, we conclude that $\mu _0=0$ and hence $(\mu _{k_n})$ is weakly convergent to zero in $C(K)^{*}$. Denote by W the countable set $\bigcup _{n\in \mathbb {N}}\mathrm {supp}(\mu _n)$. The measures $\mu _n, n\in \mathbb {N},$ can be considered as elements of the unit sphere of the Banach space $\ell _1(W)\subset C(K)^{*}$. By the Schur theorem [1, Theorem 2.3.6], the weakly convergent sequence $(\mu _{k_n})$ is convergent to zero in the norm topology of $\ell _1(W)$, which is not possible as $\Vert \mu _n\Vert =1$ for all $n\in \mathbb {N}$. Thus the set M is not relatively weakly compact in $C(K)^{*}$.

• (II) By the Grothendieck theorem [1, Theorem 5.3.2] there exist a number $\epsilon >0$, a sequence $(m_n) \subset \mathbb {N}$ and a sequence $(U_n)$ of pairwise disjoint open sets in K such that $|\mu _{m_n}(U_n)|>\epsilon$ for any $n\in \mathbb {N}$. Clearly, $\lim _{n\rightarrow \infty }\mu _k(U_n)=0$ for any $k\in \mathbb {N}$, since

  $$\begin{aligned} \sum _{n\in \mathbb {N}} |\mu _k|(U_n) = |\mu _k|\left( \bigcup _{n\in \mathbb {N}} U_n\right) \le |\mu _k|(K)=1. \end{aligned}$$

  Thus we can assume that the sequence $(m_n)$ is strictly increasing.

For some strictly increasing sequence $(n_k)\subset \mathbb {N}$ we have $U_{n_k}\cap \mathrm {supp}(\mu _{m_{n_i}})=\emptyset$ for all $k,i\in \mathbb {N}$ with $k>i$.

Put $\nu _k=\mu _{m_{n_k}}$ and $W_k=U_{n_k}$ for all $k\in \mathbb {N}.$ Then

• (A1) $\nu _k(f)\rightarrow _k 0$ for every $f\in C(K);$

• (A2) $|\nu _k(W_k)|>\epsilon$ for every $k\in \mathbb {N};$

• (A3) $|\nu _k|(W_n)=0$ for all $k,n\in \mathbb {N}$ with $k<n.$

• (III) By induction we shall construct a decreasing sequence $(N_k)$ of infinite subsets of $\mathbb {N}$ with $\min N_k< \min N_{k+1}$ for $k\in \mathbb {N}$ such that $|\nu _n|(W_m)\le \epsilon /3^k$ for every $k\in \mathbb {N}, m=\min N_k, n\in N_k$ and $n>m$. Let $N_0=\mathbb {N}.$ Assume that for some $k\in \mathbb {N}$ an infinite subset $N_{k-1}$ of $\mathbb {N}$ has been constructed. Let F be a finite subset of $N_{k-1}$ with $|F|>3^k/\epsilon$ and $\min F> \min N_{k-1}.$ For every $i\in F$ consider the set

  $$\begin{aligned} \Lambda _i=\{n\in N_{k-1}: |\nu _n|(W_i) \le \epsilon /3^k\}. \end{aligned}$$

  For every $n\in N_{k-1}$ we get $|\nu _n|(X)\ge \sum _{i\in F} |\nu _n|(W_i).$ Hence there exists $i\in F$ such that

  $$\begin{aligned} |\nu _n|(W_i)\le 1/|F| \le \epsilon /3^k. \end{aligned}$$

  Thus $N_{k-1}=\bigcup _{i\in F} \Lambda _i,$ so for some $m\in F$ the set $\Lambda _m$ is infinite. Put

  $$\begin{aligned} N_k=\{n\in \Lambda _m: n>m\}\cup \{m\}. \end{aligned}$$

  Then $\min N_{k-1}< \min F \le m=\min N_k$ and $|\nu _n|(W_m)\le \epsilon /3^k$ for $n\in N_k$ with $n>m.$

• (IV) Let $i_k= \min N_k, \lambda _k=\nu _{i_k}$ and $V_k=W_{i_k}$ for $k\in \mathbb {N}.$ Then

  • (B1) $\lambda _k(f)\rightarrow _k 0$ for every $f\in C(K);$

  • (B2) $|\lambda _k(V_k)|> \epsilon$ for every $k\in \mathbb {N}$;

  • (B3) $|\lambda _k|(V_l)=0$ and $|\lambda _l|(V_k) \le \epsilon /3^k$ for all $k,l\in \mathbb {N}$ with $k<l.$

Clearly, the set

$$\begin{aligned} \Omega =\{n\in \mathbb {N}:\mu _n=\lambda _k\; \text{ for } \text{ some }\; k\in \mathbb {N}\} \end{aligned}$$

is infinite. Put

$$\begin{aligned} Z=\bigcap _{n\in \mathbb {N}} \{f\in C_p(X): \lambda _n(f)=0\} \end{aligned}$$

and $L=\{f\in C_p(X): \lambda _n(f)\rightarrow _n 0\}.$ Clearly, Z and L are subspaces of $C_p(X)$ and Z is closed in L and in $C_p(X)$. The linear operator

$$\begin{aligned} S: L \rightarrow c_0,\; S:f\mapsto (\lambda _n(f))_{n}, \end{aligned}$$

is continuous and $\ker S=Z.$

We shall construct a linear continuous map $P:c_0 \rightarrow L$ such that $S\circ P$ is the identity map on $c_0$. For every $k\in \mathbb {N}$ there exists a continuous function $\varphi _k: K\rightarrow [-1,1]$ such that

$$\begin{aligned} \varphi _k(s)=\lambda _k(V_k)/|\lambda _k(V_k)| \end{aligned}$$

for $s\in V_k \cap \mathrm {supp}(\lambda _k)$ and $\varphi _k(s)=0$ for $s\in (K{\setminus } V_k).$ Then

$$\begin{aligned} \lambda _k(\varphi _k)=|\lambda _k(V_k)|>\epsilon , \end{aligned}$$

$\lambda _n(\varphi _k)=0$ for all $n,k \in \mathbb {N}$ with $n<k$ and

$$\begin{aligned} |\lambda _n(\varphi _k)| \le |\lambda _n|(V_k)\le \epsilon / 3^k \end{aligned}$$

for all $n,k \in \mathbb {N}$ with $n>k.$

• (V) Let $(x_n)\in c_0$. Define a sequence $(x'_n)\in \mathbb {R}^{\mathbb {N}}$ by the recursive formula

  $$\begin{aligned} x'_n:=\left[ x_n-\sum _{1\le k<n}x'_k \lambda _n(\varphi _k)\right] /\lambda _n(\varphi _n) \ \text{ for } \ n\in \mathbb {N}. \end{aligned}$$

We shall prove that $(x'_n)\in c_0$.

First we show that $\sup _n |x'_n|<\infty$. Since $(x_n)\in c_0$, there exists $m\in \mathbb {N}$ such that $\sup _{n\ge m}|x_n|<\epsilon .$ Put

$$\begin{aligned} M_n=\max \left\{ 2, \max _{1\le k<n} |x'_k|\right\} \end{aligned}$$

for $n\ge 2.$ For every $n>m$ we get

$$\begin{aligned} |x'_n|=|x_n-\sum _{1\le k<n}x'_k\lambda _n(\varphi _k)|/\lambda _n(\varphi _n)\le \left[ \epsilon +M_n\sum _{1\le k<n} \epsilon /3^k\right] /\epsilon \le 1+M_n/2\le M_n. \end{aligned}$$

Hence $M_{n+1}=\max \{M_n, |x'_n|\}\le M_n$ for $n>m.$ Thus $d:=\sup _n |x'_n|\le M_{m+1}<\infty .$

Now we show that $x'_n\rightarrow _n 0$. Given any $\delta >0$, find $v\in \mathbb {N}$ such that $d<3^v\delta$. Since $(x_n)\in c_0$ and $\lambda _n(f)\rightarrow _n 0$ for any $f\in C(K)$, there exists $m>v$ such that for every $n\ge m$

$$\begin{aligned} |x_n|<\delta \epsilon \ \text{ and } \ d\sum _{1\le k\le v} |\lambda _n(\varphi _k)|<\delta \epsilon . \end{aligned}$$

Then for $n\ge m$ we obtain

$$\begin{aligned} |x'_n|\le & {} \bigg [|x_n|+ \sum _{1\le k\le v} |x'_k|{\cdot }|\lambda _n(\varphi _k)| + \sum _{v< k<n} |x'_k|{\cdot }|\lambda _n(\varphi _k)|\bigg ]/\lambda _n(\varphi _n)\\\le & {} \bigg [\delta \epsilon +\sum _{1\le k\le v} d{\cdot }|\lambda _n(\varphi _k)| + \sum _{v< k<n} d{\cdot }|\lambda _n(\varphi _k)|\bigg ]/\epsilon<\delta + \delta \\&+\sum _{v< k<n} d/3^k< 2\delta +d/3^v<3\delta . \end{aligned}$$

Thus $(x'_n)\in c_0.$

Clearly, the operator

$$\begin{aligned} \Theta : c_0 \rightarrow c_0,\;\;\Theta :(x_n) \mapsto (x'_n), \end{aligned}$$

is linear and continuous. We prove that $\Theta$ is surjective. Let $(y_n)\in c_0$. Set $t=\sup _n |y_n|.$ Let

$$\begin{aligned} x_n=\sum _{k=1}^n \lambda _n (\varphi _k) y_k\ \text{ for }\ n\in \mathbb {N}. \end{aligned}$$

First we show that $(x_n)\in c_0.$ Given any $\delta >0$, find $v\in \mathbb {N}$ with $\epsilon t<\delta 3^v$. Clearly, there exists $m>v+2$ such that $|y_n|<\delta$ and $\sum _{k=1}^v t|\lambda _n(\varphi _k)|<\delta$ for $n\ge m.$ Then for every $n\ge m$ we obtain

$$\begin{aligned} |x_n|\le & {} \sum _{k=1}^n |\lambda _n(\varphi _k)|{\cdot }|y_k| \le \sum _{k=1}^v t{\cdot }|\lambda _n(\varphi _k)| + \sum _{v<k<n} t{\cdot }|\lambda _n(\varphi _k)| + |\lambda _n(\varphi _n)|{\cdot }|y_n|\\< & {} \delta + \sum _{v<k<n} t\epsilon /3^k + ||\lambda _n||{\cdot }|y_n|< \delta +t\epsilon /3^v +\delta <3\delta . \end{aligned}$$

Thus $(x_n)\in c_0.$ Clearly, $\Theta ((x_n)_{n\in \mathbb N})=(y_n)_{n\in \mathbb N};$ so $\Theta$ is surjective.

• (VI) The operator

  $$\begin{aligned} T: c_0 \rightarrow C_p(X),\;\; T:(x_n)\mapsto \sum _{n=1}^{\infty } x_n{\cdot }\varphi _n|X, \end{aligned}$$

is well-defined, linear and continuous, since the functions $\varphi _n, n\in \mathbb {N},$ have pairwise disjoint supports and $\varphi _n(X)\subset [-1,1], n\in \mathbb {N}.$ Thus the linear operator

$$\begin{aligned} \Phi =T\circ \Theta : c_0 \rightarrow C_p(X) \end{aligned}$$

is continuous.

Let $x=(x_k) \in c_0$ and $x'=(x'_k)=\Theta (x).$ Then

$$\begin{aligned} \Phi (x)=T(x')=\sum _{k=1}^{\infty } x'_k \varphi _k|X. \end{aligned}$$

Using (B3) and the definition of $\Theta$, we get for every $n\in \mathbb {N}$

$$\begin{aligned} \lambda _n(\Phi (x))= & {} \sum _{k=1}^{\infty } x'_k\lambda _n(\varphi _k)=x'_n\lambda _n(\varphi _n)+ \sum _{1\le k<n} x'_k \lambda _n(\varphi _k)\\= & {} \left( x_n-\sum _{1\le k<n} x'_k \lambda _n(\varphi _k)\right) +\sum _{1\le k<n} x'_k \lambda _n(\varphi _k)=x_n; \end{aligned}$$

so $\lambda _n(\Phi (x))\rightarrow _n 0.$ This implies that $\Phi (x)\in L$ and $S\circ \Phi (x)=x$ for every $x\in c_0.$ Therefore the operator $P:=\Phi \circ S{:}L\rightarrow L$ is a continuous linear projection with $\ker P=\ker S=Z.$ Thus the subspace Z is complemented in L. Since $S\circ \Phi$ is the identity map on $c_0$, the map $S: L\rightarrow c_0$ is open. Indeed, let U be a neighbourhood of zero in L;  then $V=\Phi ^{-1}(U)$ is a neighbourhood of zero in $c_0$ and

$$\begin{aligned} V=S\circ \Phi (V) \subset S(U). \end{aligned}$$

Thus the quotient space L / Z is topologically isomorphic to $c_0$ and $\Phi (c_0)$ is a complemented subspace of L, isomorphic to $c_0$. In particular, Z has infinite codimension in L and in $C_p(X).$

• (VII) Finally we prove that the quotient space $C_p(X)/Z$ is first countable and hence metrizable. Let

  $$\begin{aligned} U_n=\{f\in C_p(X): |f(x)|<1/n\;\text{ for } \text{ every }\;x\in \bigcup _{k=1}^n \mathrm {supp}(\lambda _k)\}, n\in \mathbb {N}. \end{aligned}$$

  The first countability of the quotient space $C_p(X)/Z$ will follow as soon as for every neighbourhood U of zero in $C_p(X)$ we find $n\in \mathbb {N}$ with $Z+U_n \subset Z+U.$ Clearly we can assume that

  $$\begin{aligned} U=\bigcap _{x\in F}\{f\in C_p(X): |f(x)|<\delta \} \end{aligned}$$

  for some finite subset F of X and some $\delta >0.$

By the continuity of the operator $\Phi : c_0 \rightarrow C_p(X),$ there exists $n\in \mathbb {N}$ such that for any $y=(y_k)\in c_0$ with

$$\begin{aligned} \max _{1\le k \le n}|y_k| \le 1/n \end{aligned}$$

we get $\Phi (y)\in \frac{1}{2}U$. Replacing n by a larger number, we can assume that $\frac{1}{n}<\frac{1}{2}\delta$ and

$$\begin{aligned} F\cap \bigcup _{k=1}^{\infty }\mathrm {supp}(\lambda _k) \subset \bigcup _{k=1}^n \mathrm {supp}(\lambda _k). \end{aligned}$$

Let $f\in U_n.$ Choose a function $h\in C_p(K)$ such that $h(x)=f(x)$ for every

$$\begin{aligned} x\in F{\setminus }\bigcup _{k=1}^{\infty }\mathrm {supp}(\lambda _k) \end{aligned}$$

and $h(x)=0$ for every $x\in \bigcup _{k=1}^n \mathrm {supp}(\lambda _k)$. Put $g=h|X.$ Then $g\in L$, since $\lambda _k (g)=\lambda _k (h)\rightarrow _k 0$. Put $y=S(g)$ and $\xi =\Phi (y).$ Since $g(x)=0$ for

$$\begin{aligned} x\in \bigcup _{k=1}^n \mathrm {supp}\lambda _k, \end{aligned}$$

we have $|\lambda _k(g)|=0<\frac{1}{n}$ for $1\le k \le n$, so $\max _{1\le k \le n} |y_k|< \frac{1}{n}$. Hence $\xi =\Phi (y) \in \frac{1}{2}U,$ so $\max _{x\in F} |\xi (x)|<\frac{1}{2}\delta .$ For $\varsigma =g-\xi$ we obtain

$$\begin{aligned} S(\varsigma )=S(g)-S\circ \Phi \circ S (g)= S(g)-S(g)=0, \end{aligned}$$

so $\varsigma \in Z.$ Moreover $f-\varsigma \in U.$ Indeed, we have

$$\begin{aligned} |f(x)-\varsigma (x)|=|f(x)-g(x)+\xi (x)|=|\xi (x)|<\delta \end{aligned}$$

for $x\in F{\setminus }\bigcup _{k=1}^{\infty }\mathrm {supp}(\lambda _k)$ and

$$\begin{aligned} |f(x)-\varsigma (x)|=|f(x)-g(x)+\xi (x)|\le |f(x)|+|g(x)|+|\xi (x)|<\tfrac{1}{n}+0+\tfrac{1}{2}\delta <\delta \end{aligned}$$

for every $x\in \bigcup _{k=1}^n \mathrm {supp}(\lambda _k)$. Thus $f=\varsigma + (f-\varsigma ) \in Z+U,$ so $U_n \subset Z+U.$ Hence $Z+U_n \subset Z+U$. $\square$

Lemma 2

Let X be a Tychonoff space. Each metrizable continuous image of $C_p(X)$ is separable.

Proof

It is well-known [11, 2.3.18] that the Tychonoff product $\mathbb {R}^{X}$ has countable cellularity, which means that $\mathbb {R}^{X}$ contains no uncountable family of pairwise disjoint non-empty open sets. Then the dense subspace $C_p(X)$ of $\mathbb {R}^{X}$ also has countable cellularity and so does any continuous image Y of $C_p(X)$. If Y is metrizable, then Y is separable according to Theorem 4.1.15 in [11]. $\square$

Lemma 3

Let X be a pseudocompact space. A closed linear subspace S of $C_p(X)$ is separable if and only if S is Polishable.

Proof

If S is Polishable, then S is separable, being a continuous image of a separable Fréchet locally convex space. Now assume that S is separable. Fix a countable dense subset $\{f_n\}_{n\in \mathbb N}$ in S and consider the continuous map

$$\begin{aligned} f:X\rightarrow \mathbb {R}^\mathbb {N},\;\;f:x\mapsto (f_n(x))_{n\in \mathbb {N}}. \end{aligned}$$

By the pseudocompactness of X and the metrizability of $\mathbb {R}^{\mathbb {N}}$, the image $M:=f(X)$ is a compact metrizable space. The continuous surjective map $f:X\rightarrow M$ induces an isomorphic embedding

$$\begin{aligned} C_pf:C_p(M)\rightarrow C_p(X),\;\;C_pf:\varphi \mapsto \varphi \circ f. \end{aligned}$$

So, we can identify the space $C_p(M)$ with its image $C_pf(C_p(M))$ in $C_p(X)$. We claim that $C_p(M)$ is closed in $C_p(X)$. Given any function $\varphi \in C_p(X){\setminus }C_p(M)$, we should find a neighborhood $O_\varphi \subset C_p(X)$ of $\varphi$, which is disjoint with $C_p(M)$.

We claim that there exist points $x,y\in X$ such that $f(x)=f(y)$ and $\varphi (x)\ne \varphi (y)$. In the opposite case, $\varphi =\psi \circ f$ for some bounded function $\psi :M\rightarrow \mathbb {R}$. Let us show that the function $\psi$ is continuous. Consider the continuous map

$$\begin{aligned} h:X\rightarrow M\times \mathbb {R},\,\,\, h:x\mapsto (f(x),\varphi (x)). \end{aligned}$$

The pseudocompactness of X implies that the image $h(X)\subset M\times \mathbb {R}$ is a compact closed subset of $M\times \mathbb {R}$. Let $\mathrm {pr}_M:h(X)\rightarrow M$ and $\mathrm {pr}_\mathbb {R}:h(X)\rightarrow \mathbb {R}$ be the coordinate projections. It follows that

$$\begin{aligned} \mathrm {pr}_\mathbb {R}\circ h=\varphi =\psi \circ f=\psi \circ \mathrm {pr}_M\circ h, \end{aligned}$$

which implies that $\mathrm {pr}_\mathbb {R}=\psi \circ \mathrm {pr}_M$. The map $\mathrm {pr}_M:h(X)\rightarrow M$ between the compact metrizable spaces h(X) and M is closed and hence is quotient. Then the continuity of the map $\mathrm {pr}_\mathbb {R}=\psi \circ \mathrm {pr}_M$ implies the continuity of $\psi$. Now we see that the function $\varphi =\psi \circ f$ belongs to the subspace $C_p(M)\subset C_p(X)$, which contradicts the choice of $\varphi$. This contradiction shows that $\varphi (x)\ne \varphi (y)$ for some points $x,y\in X$ with $f(x)=f(y)$. Then

$$\begin{aligned} O_\varphi :=\{\phi \in C_p(X):\phi (x)\ne \phi (y)\} \end{aligned}$$

is a required neighborhood of $\varphi$, disjoint with $C_p(M)$.

Therefore the subspace $C_p(M)$ of $C_p(X)$ is closed and hence $C_p(M)$ contains the closure S of the dense set $\{f_n\}_{n\in \mathbb N}$ in S. Since the space $C_p(M)$ is Polishable, so is its closed subspace S. $\square$

Now we are at the position to prove the main Theorem 1:

Proof of Theorem 1

First, for a Tychonoff space X we prove the equivalence of the conditions:

• (1) $C_{p}(X)$ satisfies the JNP;

• (2) $C_p(X)$ contains a complemented subspace isomorphic to $c_0$;

• (3) $C_{p}(X)$ has a quotient isomorphic to $c_0$;

• (4) $C_p(X)$ admits a continuous linear map onto $c_0.$

The implication $(1)\Rightarrow (2)$ follows from Lemma 1, applied to the Stone-Čech compactification $K=\beta X$ of X. The implications $(2)\Rightarrow (3)\Rightarrow (4)$ are trivial.

To prove the implication $(4)\Rightarrow (1)$ assume that there exists a continuous linear map T from $C_p(X)$ onto $c_0$. Let

$$\begin{aligned} \{e^{*}_n\}_{n\in \mathbb {N}}\subset c_0^{*} \end{aligned}$$

be the sequence of coordinate functionals. By definition of $c_0$, $e^{*}_n(y)\rightarrow _n 0$ for every $y\in c_0$. For every $n\in \mathbb {N}$ consider the linear continuous functional

$$\begin{aligned} \lambda _n\in C_p(X)^{*},\;\;\lambda _n(f)= e^{*}_n(Tf), \end{aligned}$$

which can be thought as a finitely supported sign-measure on X. It follows that $\lambda _n(f)=e^{*}_n(Tf)\rightarrow _n0$ for every $f\in C_p(X)$.

We shall show that the union $S=\bigcup _{n=1}^{\infty } S_n$ of supports $S_n=\mathrm {supp}(\lambda _n)$ of the sign-measures $\lambda _n$ is bounded in X in the sense that for any $\varphi \in C_p(X)$ the image $\varphi (S)$ is bounded in $\mathbb {R}$, since in the opposite case we get a function $\psi \in C_p(X)$ with $\lambda _n (\psi )\not \rightarrow 0.$ Indeed, suppose that for some $\varphi \in C_p(X)$ the image $\varphi (S)$ is unbounded in $\mathbb {R}$; without loss of generality we can assume that $\varphi$ is non-negative.

We can find inductively an increasing sequence $\{n_k\}_{k\in \mathbb {N}} \subset \mathbb {N}$ such that

$$\begin{aligned} \max \varphi (S_{n_k}) > 3+ \max \varphi (S_{n_i}) \end{aligned}$$

for all $k,i \in \mathbb {N}$ with $k>i$. For every $k\in \mathbb {N}$ choose $x_k\in S_{n_k}$ with $\varphi (x_k)= \max \varphi (S_{n_k})$. Then $\varphi (x_k) - \varphi (x_i)>3$ for all $k,i \in \mathbb {N}$ with $k>i$. Since the space X is Tychonoff, for every $k\in \mathbb {N}$ we can find an open neighborhood $U_k\subset \{x\in X:|\varphi (x)-\varphi (x_k)|<1\}$ of $x_k$ such that $U_k\cap \mathrm {supp}(\lambda _{n_k})=\{x_k\}$.

Observe that for any $k>i$ and any points $x\in U_k$ and $y\in U_i\cup \mathrm {supp}(\lambda _{n_i})$ we get

$$\begin{aligned} \varphi (x)-\varphi (y)\ge \varphi (x_k)-1-(\max \varphi (S_{n_i})+1)>3-2=1, \end{aligned}$$

which implies $U_k\cap \mathrm {supp}(\lambda _{n_i})=\emptyset$ and that the family $\big (\varphi (U_k)\big )_{k\in \mathbb {N}}$ is discrete in $\mathbb {R}$ and consequently $(U_k)_{k\in \mathbb {N}}$ is discrete in X.

Taking into account that $U_k\cap \mathrm {supp}(\lambda _{n_k})=\{x_k\}$, we can inductively construct a sequence $(\psi _k)_{k\in \mathbb {N}}$ of functions $\psi _k:X\rightarrow \mathbb {R}$ such that $\mathrm {supp}(\psi _k):=\{x\in X:\psi _i(x)\ne 0\}\subset U_k$ and

$$\begin{aligned} \lambda _{n_k}(\psi _k)>1+\sum _{i<k}|\lambda _{n_k}(\psi _i)| \end{aligned}$$

for all $k\in \mathbb {N}$.

Since the family $(U_k)_{k\in \mathbb {N}}$ is discrete, the function

$$\begin{aligned} \psi :X\rightarrow \mathbb {R}, \,\,\,\psi :x\mapsto \sum _{k=1}^\infty \psi _k(x) \end{aligned}$$

is well-defined and continuous. For every $k\in \mathbb {N}$ and $i>k$ we have $U_i\cap \mathrm {supp}(\lambda _{n_k})=\emptyset$ and hence

$$\begin{aligned} |\lambda _{n_k}(\psi )|=\big |\sum _{i\le k}\lambda _{n_k}(\psi _i)\big |\ge |\lambda _{n_k}(\psi _k)|-\sum _{i<k}|\lambda _{n_k}(\psi _i)|>1. \end{aligned}$$

But this contradicts $\lambda _n(\psi )\rightarrow _n 0$. This contradiction shows that the set $S=\bigcup _{k\in \mathbb {N}}\mathrm {supp}(\lambda _k)$ is bounded in X.

Next, we show that the set $B=\{(x_n)\in c_0:\forall n\in \mathbb {N}\;\; |x_n|\le \Vert \lambda _n\Vert \}$ is absorbing. Given any sequence $(x_n) \in c_0$, find $f\in C_p(X)$ with $Tf=(x_n)$. Put $\Vert f\Vert _S:= \sup \{|f(x)|: x\in S\}$ and observe that $|x_n|=|e^{*}_n(Tf)|=|\lambda _n (f)| \le \Vert \lambda _n\Vert {\cdot }\Vert f\Vert _S$, so $(x_n) \in \Vert f\Vert _S\cdot B$. Thus B is absorbing. Clearly, B is absolutely convex and closed in $c_0$, so it is a barrel in the Banach space $(c_0, \Vert \cdot \Vert _{\infty })$. Thus B is a neighbourhood of zero in $(c_0, \Vert \cdot \Vert _{\infty })$ and consequently, $\inf _n \Vert \lambda _n\Vert >0$. For every $n\in \mathbb {N}$ put

$$\begin{aligned} \mu _n=\frac{\lambda _n}{\Vert \lambda _n\Vert }\in C_p(X)^{*} \end{aligned}$$

and observe that the sequence $(\mu _n)$ witnesses that the function space $C_p(X)$ has the JNP.

Now assuming that the space X is pseudocompact, we shall prove that the conditions (1)–(4) are equivalent to

• (5) $C_p(X)$ contains a complemented infinite-dimensional metrizable subspace;

• (6) $C_{p}(X)$ contains a complemented infinite-dimensional separable subspace;

• (7) $C_p(X)$ has an infinite-dimensional Polishable quotient.

It suffices to prove the implications $(2)\Rightarrow (5)\Rightarrow (6)\Rightarrow (7)\Rightarrow (1)$. The implication $(2)\Rightarrow (5)$ is trivial and $(5)\Rightarrow (6)\Rightarrow (7)$ follow from Lemmas 2 and 3, respectively.

$(7) \Rightarrow (1)$ Assume that the space $C_p(X)$ contains a closed subspace Z of infinite codimension such that the quotient space $E:=C_p(X)/Z$ is Polishable. Denote by $\tau _p$ the quotient topology of $C_p(X)/Z$ and by $\tau _0\supset \tau _p$ a stronger separable Fréchet locally convex topology on E. Denote by $\tau _\infty$ the topology of the quotient Banach space C(X) / Z. Here C(X) is endowed with the sup-norm $\Vert f\Vert _\infty :=\sup _{x\in X}|f(x)|$ (which is well-defined as X is pseudocompact).

The identity maps between $(E, \tau _0)$ and $(E, \tau _{\infty })$ have closed graphs, since $\tau _p \subset \tau _0 \cap \tau _{\infty }.$ Using the Closed Graph Theorem we infer that the topologies $\tau _0$ and $\tau _{\infty }$ are equal. Let G be a countable subset of C(X) such that the set $\{g+Z: g\in G\}$ is dense in the Banach space C(X) / Z. Then the set

$$\begin{aligned} G+Z=\{g+z:g\in G,\; z\in Z\} \end{aligned}$$

is dense in C(X). Let $(g_n)_{n\in \mathbb N}$ be a linearly independent sequence in G such that its linear span $G_0$ has $G_0\cap Z=\{0\}$ and $G_0+Z=G+Z$. Let $f_1=g_1$ and $\nu _1 \in C_p(X)^{*}$ with $\nu _1|Z=0$ such that $\nu _1(f_1)=1.$

Assume that for some $n\in \mathbb {N}$ we have chosen

$$\begin{aligned} (f_1, \nu _1), \ldots , (f_n, \nu _n) \in C_p(X)\times C_p(X)^{*} \end{aligned}$$

such that

$$\begin{aligned} \text{ lin }\{f_1, \ldots , f_n\}=\text{ lin }\{g_1, \ldots , g_n\} \end{aligned}$$

and

$$\begin{aligned} \nu _j|Z=0,\;\; \nu _j(f_i)=\delta _{j,i}\ \quad \mathrm{for~all} \ \, i,j\in \{1, \ldots , n\}. \end{aligned}$$

Put

$$\begin{aligned} f_{n+1}=g_{n+1}-\sum _{i=1}^n \nu _i(g_{n+1})f_i. \end{aligned}$$

Then

$$\begin{aligned} \text{ lin }\{f_1, \ldots , f_{n+1}\}=\text{ lin }\{g_1, \ldots , g_{n+1}\} \end{aligned}$$

and $\nu _j(f_{n+1})=0$ for $1\le j \le n.$ Let $\nu _{n+1}\in C_p(X)^{*}$ with $\nu _{n+1}|Z=0$ such that $\nu _{n+1}(f_i)=0$ for $1\le i \le n$ and $\nu _{n+1}(f_{n+1})=1.$

Continuing on this way we can construct inductively a biorthogonal sequence $((f_n, \nu _n))_{n\in \mathbb N}$ in $C_p(X)\times C_p(X)^{*}$ such that $\text{ lin } \{f_n: n\in \mathbb {N}\}= \text{ lin } \{g_n: n\in \mathbb {N}\}$ and $\nu _n|Z=0$, $\nu _n(f_m)=\delta _{n,m}$ for all $n,m \in \mathbb {N}.$ Then $\text{ lin } \{f_n: n\in \mathbb {N}\} +Z$ is dense in $(C(X), \Vert .\Vert _{\infty })$. Let $\mu _n=\nu _n/\Vert \nu _n\Vert$ for $n\in \mathbb {N}.$ Then $\Vert \mu _n\Vert =1$ and $\mu _n(f_m)=0$ for all $n,m \in \mathbb {N}$ with $n\ne m.$

We prove that $\mu _n(f)\rightarrow _n 0$ for every $f\in C_p(X).$ Given any $f\in C(X)$ and $\varepsilon >0$, find $m\in \mathbb {N}$ and $g\in \text{ lin }\{f_1, \ldots , f_m\} +Z$ with $d(f,g)<\varepsilon ;$ clearly $d(f,g)=\Vert f-g\Vert _{\infty }.$ Then $\mu _n(g)=0$ for $n>m,$ so

$$\begin{aligned} |\mu _n(f)|=|\mu _n(f-g)| \le \Vert \mu _n\Vert {\cdot }\Vert f-g\Vert _{\infty }<\varepsilon \end{aligned}$$

for $n>m.$ Thus $\mu _n(f)\rightarrow _n 0$, which means that the space $C_p(X)$ has the JNP. $\square$

4 An example of Plebanek

In this section we describe the following example suggested to the authors by Grzegorz Plebanek [26].

Example 1

(Plebanek) There exists a compact Hausdorff space K such that

• (1) K contain no nontrivial converging sequences but contains a copy of $\beta \mathbb {N}$;

• (2) the function space $C_p(K)$ has the JNP.

We need some facts to present the construction of the space K. By definition, the asymptotic density of a subset $A\subset \mathbb N$ is the limit

$$\begin{aligned} d(A):=\lim _{n\rightarrow \infty }\frac{|A\cap [1,n]|}{n} \end{aligned}$$

if this limit exists. The family $\mathcal Z=\{A\subset \mathbb N:d(A)=0\}$ of sets of asymptotic density zero in $\mathbb N$ is an ideal on $\mathbb N$. Recall the following standard fact (here $A\subset ^{*} B$ means that $A{\setminus } B$ is finite).

Fact 1

For any countable subfamily $\mathcal C\subset \mathcal {Z}$ there is a set $B\in \mathcal {Z}$ such that $C\subset ^{*} B$ for all $C\in \mathcal C$.

Let $\mathfrak {A}=\big \{A\subset \mathbb N:d(A)\in \{0,1\}\big \}$ be the algebra of subsets of $\mathbb {N}$ generated by $\mathcal {Z}$. We now let K be the Stone space of the algebra $\mathfrak {A}$ so we treat elements of K as ultrafilters on $\mathfrak {A}$. There are three types of such $x\in K$:

• (1) $\{n\} \in x$ for some $n\in \mathbb {N}$; then $x=\{A\in \mathfrak {A}: n\in A\}$ is identified with n;

• (2) x contains no finite subsets of $\mathbb {N}$ but $Z\in x$ for some $Z\in \mathcal {Z}$;

• (3) $Z\notin x$ for every $Z\in \mathcal {Z}$; this defines the unique

  $$\begin{aligned} p=\{A\in \mathfrak {A}:d(A)=1\}\in K. \end{aligned}$$

To see that K is the required space it is enough to check the following two facts.

Fact 2

The space K contains no nontrivial converging sequence.

Proof

In fact we check that every infinite $X\subset K$ contains an infinite set Y such that $\overline{Y}$ is homeomorphic to $\beta \mathbb {N}$. Note first that for every $Z\in \mathcal {Z}$, the corresponding clopen set

$$\begin{aligned} \widehat{Z}=\{x\in K: Z\in x\}, \end{aligned}$$

is homeomorphic to $\beta \mathbb {N}$ because $\{A\in \mathfrak {A}: A\subset Z\}= 2^{Z}$.

For an infinite set $X\subset K$, we have two cases:

Case 1, $X\cap \mathbb {N}$ is infinite. There is an infinite $Z\subset X\cap \mathbb {N}$ having density zero. Then every subset of Z is in $\mathfrak {A}$, which implies that $\overline{Z}\cong \beta \mathbb {N}$ .

Case 2, $X\cap ({K{\setminus }\mathbb {N}})$ is infinite. Let us fix a sequence of different $x_n \in X\cap (K{\setminus }\mathbb {N})$ such that $x_n\ne p$ for every n. Then for every n we have $Z_n\in x_n$ for some $Z_n\in \mathcal {Z}$. Take $B\in \mathcal {Z}$ as in Fact 1. Then $B\in x_n$ because $x_n$ is a nonprincipial ultrafilter on $\mathfrak A$ so $A_n{\setminus }B\notin x_n$. Again, we conclude that $\overline{\{x_n: n\in \mathbb {N}\}}$ is $\beta \mathbb {N}$. $\square$

Fact 3

If $\nu _n=\frac{1}{n}\sum _{k\le n} \delta _k$ and $\mu _n=\frac{1}{2}(\nu _n - \delta _p)$ for $n \in \mathbb {N}$, then $\nu _n(f)\rightarrow _n \delta _p(f)$ and $\mu _n(f)\rightarrow _n 0$ for every $f\in C(K)$.

Proof

Observe $\nu _n(A)\rightarrow _n d(A)$ for every $A\in \mathfrak {A}$ since elements of $\mathfrak {A}$ have asymptotic density either 0 or 1. This means that, when we treat $\nu _n$ as measures on K then $\nu _n(V)$ converges to $\delta _p(V)$ for every clopen set $V\subset K$. This implies the assertion since every continuous function on K can be uniformly approximated by simple functions built from clopens. $\square$

5 Proof of Theorem 2

Let us recall that a topological space X is called

• submetrizable if X admits a continuous metric;

• hemicompact if X has a countable family $\mathcal K$ of compact sets such that each compact subset of X is contained in some compact set $K\in \mathcal {K}$;

• a k-space if a subset $F\subset X$ is closed if and only if for every compact subset $K\subset X$ the intersection $F\cap K$ is closed in K.

In order to prove Theorem 2, we should check the equivalence of the following conditions for every Tychonoff space X:

• (1) $C_k(X)$ is Polishable;

• (2) $C_p(X)$ is Polishable;

• (3) $C_k(X)$ is Polish;

• (4) X is a submetrizable hemicompact k-space.

The implication $(1)\Rightarrow (2)$ follows from the continuity of the identity map $C_k(X)\rightarrow C_p(X)$.

$(2)\Rightarrow (3)$ Assume that the space $C_p(X)$ is Polishable and fix a stronger Polish locally convex topology $\tau$ on $C_p(X)$. Let $C_\tau (X)$ denote the separable Fréchet space $(C_p(X),\tau )$. By $\tau _{k}$ denote the compact open topology of $C_{k}(X)$. Taking into account that the space $C_p(X)$ is a continuous image of the Polish space $C_\tau (X)$, we conclude that $C_p(X)$ has countable network and by [2, I.1.3], the space X has countable network and hence is Lindelöf. By the normality (and the Lindelöf property) of X, each closed bounded set in X is countably compact (and hence compact). So X is a $\mu$-space. By Theorem 10.1.20 in [25, Theorem 10.1.20] the function space $C_{k}(X)$ is barrelled. The continuity of the identity maps $C_k(X)\rightarrow C_p(X)$ and $C_\tau (X)\rightarrow C_p(X)$ implies that the identity map $C_k(X)\rightarrow C_\tau (X)$ has closed graph. Since $C_k(X)$ is barelled and $C_\tau (X)$ is Fréchet, we can apply the Closed Graph Theorem 4.1.10 in [25] and conclude that the identity map $C_k(X)\rightarrow C_\tau (X)$ is continuous.

Next, we show that the identity map $C_\tau (X)\rightarrow C_k(X)$ is continuous. Given any compact set $K\subset X$ and any $\varepsilon >0$ we have to find a neighborhood $U\subset C_\tau (X)$ of zero such that

$$\begin{aligned} U\subset \{f\in C(X):f(K)\subset (-\varepsilon ,\varepsilon )\}. \end{aligned}$$

The continuity of the restriction operator $R:C_p(X)\rightarrow C_p(K)$, $R:f\mapsto f{\upharpoonright }K$, and the continuity of the idenity map $C_\tau (X)\rightarrow C_p(X)$ imply that the restriction operator $R:C_\tau (X)\rightarrow C_p(K)$ is continuous and hence has closed graph. The continuity of the identity map $C_k(K)\rightarrow C_p(K)$ implies that R seen as an operator $R:C_\tau (X)\rightarrow C_k(K)$ still has closed graph. Since the spaces $C_\tau (X)$ and $C_k(K)$ are Fréchet, the Closed Graph Theorem 1.2.19 in [25] implies that the restriction operator $R:C_\tau (X)\rightarrow C_k(K)$ is continuous. So, there exists a neighborhood $U\subset C_\tau (X)$ of zero such that

$$\begin{aligned} R(U)\subset \{f\in C_k(K):f(K)\subset (-\varepsilon ,\varepsilon )\}. \end{aligned}$$

Then $U\subset \{f\in C(X):f(K)\subset (-\varepsilon ,\varepsilon )\}$ and we are done. Hence $\tau =\tau _{k}$ is a Polish locally convex topology as claimed.

The implication $(3)\Rightarrow (1)$ is trivial.

$(3)\Rightarrow (4)$ If the function space $C_k(X)$ is Polish, then by Theorem 4.2 in [24], X is a hemicompact k-space. Taking into account that the space $C_p(X)$ is a continuous image of the space $C_k(X)$, we conclude that $C_p(X)$ has countable network and by [2, I.1.3], the space X has countable network. By [17, 2.9], the space X is submetrizable.

$(4)\Rightarrow (3)$ If X is a submetrizable hemicompact k-space, then $X=\bigcup _{n\in \omega }X_n$ for some increasing sequence $(X_n)_{n\in \omega }$ of compact metrizable spaces such that each compact subset of X is contained in some compact set $X_n$. Then the function space $C_k(X)$ is Polish, being topologically isomorphic to the closed subspace

$$\begin{aligned} \left\{ (f_n)_{n\in \omega }\in \prod _{n\in \omega }C_k(X_n):\forall n\in \omega \;\;f_{n+1}{\upharpoonright }X_n=f_n\right\} \end{aligned}$$

of the countable product $\prod _{n=1}^\infty C_k(X_n)$ of separable Banach spaces. $\square$

The authors thank to the referee for his/her valuable comments and remarks.