RACSAM https://doi.org/10.1007/s13398-019-00667-8 ORIGINAL PAPER
Josefson–Nissenzweig property for Cp-spaces
T. Banakh1,2 · J. K¸akol3,4 · W. Sliwa´ 5
Received: 6 March 2019 / Accepted: 2 April 2019 © The Author(s) 2019
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 2.Theaimof 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 N N quotient isomorphic to c0 := {(xn)n∈N ∈ R : xn → 0} (with the product topology of R )if and only if C p(X) contains a complemented subspace isomorphic to c0. 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 c0 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 c0 is given.
Keywords The separable quotient problem · Spaces of continuous functions · Quotient spaces · The Josefson–Nissenzweig theorem · Efimov space
To the memory of our Friend Professor Mati Rubin.
J. K¸akol supported by GACRˇ Project 16-34860L and RVO: 67985840.
B W. Sliwa´ [email protected]; [email protected] T. Banakh [email protected] J. K¸akol [email protected]
1 Ivan Franko National University of Lviv, Lviv, Ukraine 2 Jan Kochanowski University in Kielce, Kielce, Poland 3 Faculty of Mathematics and Informatics, A. Mickiewicz University, 61-614 Poznan, Poland 4 Institute of Mathematics Czech Academy of Sciences, Prague, Czech Republic 5 Faculty of Mathematics and Natural Sciences, University of Rzeszów, 35-310 Rzeszow, Poland 123 T. Banakh et al.
Mathematics Subject Classification 46E10 · 54C35
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 RX ,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 2, or equivalently, there exists a continuous linear (and open; by the open mapping Banach theorem) map from C(K ) onto c or 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. N (3) The space C p(K ) has a quotient isomorphic to a dense subspace of R .
In [21]itwasshownthatC p(K ) has an infinite-dimensional separable quotient algebra if and only if K contains an infinite countable closed subset. Hence C p(βN) 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 acopyofβN. 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 (Kn)n∈N of compact subsets such that for every n the space Kn contains two disjoint topological copies of Kn+1. If fact, the first case (for example if compact X contains a copy of βN) asserts that C p(X) ={( ) ∈ RN : | | < ∞} RN has a quotient isomorphic to the subspace ∞ xn supn xn of or to the product RN. 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 βN). Although, it is unknown if Efimov spaces exist in ZFC (see [7–10,12,13,16,18]) we showed in [22] that under ♦ for some Efimov spaces K the function space C p(K ) has an infinite dimensional metrizable quotient. N N By c0 we mean the subspace {(xn)n∈N ∈ R : xn → 0} of R endowed with the product topology. The term “the Banach space c0” 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 c0, 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 c0. 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 c0. It is also easy to 123 Josefson–Nissenzweig property for Cp-spaces see that for every infinite compact space K the space C p(K ) contains a closed copy of the N space c0 endowed with the product topology of R . 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 c0 with the product topology of RN.
2 The main results
For a Tychonoff space X and a point x ∈ X let δx : C p(X) → R,δx : f → f (x), be the C (X) Dirac measure concentrated at x. The linear hull L p(X) of the set {δx : x ∈ X} in R p 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)onX. μ ∈ ( ) Each L p X can be uniquely written as a linear combination of Dirac measures μ = α δ ⊂ α x∈F x x for some finite set F X and some non-zero real numbers x .Theset F is called the support of the sign-measure μ and is denoted by supp(μ). The measure |α |δ |μ| μ= |α | x∈F x x will be denoted by and the real number x∈F x coincides with ∗ the norm of μ (in the dual Banach space C(β X) ). μ = α δ μ : X → R The sign-measure x∈F x x determines the function 2 defined on ⊂ α the power-set of X and assigning to each subset A X the real number x∈A∩F 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 2X . 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 (μn) of finitely supported sign-measures on X such that μn=1foralln ∈ N,andμn( f ) →n 0 for each f ∈ C p(X).
Concerning the JNP of function spaces C p(X) on compacta we have the following:
(1) If a compact space K contains a non-trivial convergent sequence, say xn → x, then ∗ C p(K ) satisfies the JNP. This is witnessed by the weak null sequence (μn) of sign- μ = 1 (δ − δ ) ∈ N measures n 2 xn x , n . (2) The space C p(βN) does not satisfy the JNP. This follows directly from the Grothendieck theorem, see [5, Corollary 4.5.8]. (3) There exists a compact space K containing a copy of βN 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 c0. However for every infinite compact K the space C p(K ) contains a subspace isomorphic to c0 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 c0, as follows from Theorem 1 below. 123 T. Banakh et al.
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 c0; (3) C p(X) has a quotient isomorphic to c0; (4) C p(X) admits a linear continuous map onto c0. 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 N of separable Fréchet spaces under continuous linear maps. Clearly, the subspace c0 of R 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 N if and only if C p(X) contains a complemented copy of R . 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; N (3) C p(X) contains a complemented subspace isomorphic to R or c0;
Corollary 2 The space C p(βN)
(1) has a quotient isomorphic to ∞; (2) contains a subspace isomorphic to c0; (3) does not admit a continuous linear map onto c0; (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) Ck(X) is Polishable; (3) Ck(X) is Polish; (4) X is a submetrizable hemicompact k-space.
In this theorem Ck(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]. 123 Josefson–Nissenzweig property for Cp-spaces
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 (μn) be a sequence of finitely supported sign-measures on X (and so, on K ) such that
(1) μn=1 for all n ∈ N, and (2) μn( f ) →n 0 for all f ∈ C(K ). Then there exists an infinite subset of N such that = { ∈ ( ) : μ ( ) = } ( ) (a) the closed subspace Z k∈ f C p X k f 0 of C p X is complemented in the subspace L = f ∈ C p(X) : limk∈ μk( f ) = 0 of C p(X); N (b) the quotient space L/Z is isomorphic to the subspace c0 of R ; (c) L contains a complemented subspace isomorphic to c0; (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 ={μn : n ∈ N} is not relatively weakly compact in the dual of the Banach space C(K ).Indeed,assumeonthecontrarythattheclosureM 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 M is weakly sequentially compact. Thus (μ ) (μ ) μ ∈ ( )∗ n has a subsequence kn that weakly converges to some element 0 C K . ∗ Taking into account that the sequence (μn) converges to zero in the weak topology ( )∗ μ = (μ ) of C K , we conclude that 0 0 and hence kn is weakly convergent to zero in ( )∗ (μ ) μ , ∈ N, C K . Denote by W the countable set n∈N supp n . The measures n n can ∗ be considered as elements of the unit sphere of the Banach space 1(W) ⊂ C(K ) . (μ ) By the Schur theorem [1, Theorem 2.3.6], the weakly convergent sequence kn is convergent to zero in the norm topology of 1(W), which is not possible as μn=1 for all n ∈ N.ThusthesetM is not relatively weakly compact in C(K )∗. (II) By the Grothendieck theorem [1, Theorem 5.3.2] there exist a number >0, a sequence (mn) ⊂ N and a sequence (Un) of pairwise disjoint open sets in K such that |μm (Un)| >for any n ∈ N. Clearly, limn→∞ μk (Un) = 0foranyk ∈ N,since n |μk |(Un) =|μk| Un ≤|μk |(K ) = 1. n∈N n∈N
Thus we can assume that the sequence (mn) is strictly increasing. For some strictly increasing sequence (n ) ⊂ N we have U ∩ supp(μ ) =∅for all k nk mni k, i ∈ N with k > i. Put ν = μ and W = U for all k ∈ N. Then k mnk k nk (A1) νk( f ) →k 0forevery f ∈ C(K ); (A2) |νk(Wk )| >for every k ∈ N; (A3) |νk|(Wn) = 0forallk, n ∈ N with k < n.
(III) By induction we shall construct a decreasing sequence (Nk) of infinite subsets of N k with min Nk < min Nk+1 for k ∈ N such that |νn|(Wm) ≤ /3 for every k ∈ N, m = min Nk, n ∈ Nk and n > m.LetN0 = N. Assume that for some k ∈ N an infinite subset k Nk−1 of N has been constructed. Let F be a finite subset of Nk−1 with |F| > 3 / and min F > min Nk−1. For every i ∈ F consider the set k i ={n ∈ Nk−1 :|νn|(Wi ) ≤ /3 }. 123 T. Banakh et al. ∈ |ν |( ) ≥ |ν |( ). ∈ For every n Nk−1 we get n X i∈F n Wi Hence there exists i F such that k |νn|(Wi ) ≤ 1/|F|≤/3 . = , ∈ Thus Nk−1 i∈F i so for some m F the set m is infinite. Put
Nk ={n ∈ m : n > m}∪{m}. k Then min Nk−1 < min F ≤ m = min Nk and |νn|(Wm) ≤ /3 for n ∈ Nk with n > m. = ,λ = ν = ∈ N. (IV) Let ik min Nk k ik and Vk Wik for k Then
(B1) λk ( f ) →k 0forevery f ∈ C(K ); (B2) |λk(Vk)| >for every k ∈ N; k (B3) |λk|(Vl ) = 0and|λl |(Vk) ≤ /3 for all k, l ∈ N with k < l. Clearly, the set
={n ∈ N : μn = λk for some k ∈ N} is infinite. Put
Z = { f ∈ C p(X) : λn( f ) = 0} n∈N and L ={f ∈ C p(X) : λn( f ) →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
S : L → c0, S : f → (λn( f ))n, is continuous and ker S = Z. We shall construct a linear continuous map P : c0 → L such that S ◦ P is the identity map on c0.Foreveryk ∈ N there exists a continuous function ϕk : K →[−1, 1] such that
ϕk(s) = λk (Vk)/|λk(Vk)| for s ∈ Vk ∩ supp(λk ) and ϕk(s) = 0fors ∈ (K \Vk). Then
λk(ϕk ) =|λk(Vk)| >,
λn(ϕk ) = 0foralln, k ∈ N with n < k and k |λn(ϕk )|≤|λn|(Vk) ≤ /3 for all n, k ∈ N with n > k. ( ) ∈ ( ) ∈ RN (V) Let xn c0. Define a sequence xn by the recursive formula ⎡ ⎤ := ⎣ − λ (ϕ )⎦ /λ (ϕ ) ∈ N. xn xn xk n k n n for n 1≤k Thus (xn) ∈ c0. Clearly, ((xn)n∈N) = (yn)n∈N; so is surjective. (VI) The operator ∞ T : c0 → C p(X), T : (xn) → xn·ϕn|X, n=1 is well-defined, linear and continuous, since the functions ϕn, n ∈ N, have pairwise disjoint supports and ϕn(X) ⊂[−1, 1], n ∈ N. Thus the linear operator = T ◦ : c0 → C p(X) 123 T. Banakh et al. is continuous. = ( ) ∈ = ( ) = ( ). Let x xk c0 and x xk x Then ∞ ( ) = ( ) = ϕ | . x T x xk k X k=1 Using (B3) and the definition of ,wegetforeveryn ∈ N ∞ λ ( ( )) = λ (ϕ ) = λ (ϕ ) + λ (ϕ ) n x xk n k xn n n xk n k = ≤ < ⎛k 1 ⎞ 1 k n = ⎝ − λ (ϕ )⎠ + λ (ϕ ) = ; xn xk n k xk n k xn 1≤k Thus the quotient space L/Z is topologically isomorphic to c0 and (c0) is a complemented subspace of L, isomorphic to c0. 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 metriz- able. Let n Un ={f ∈ C p(X) :|f (x)| < 1/n for every x ∈ supp(λk)}, n ∈ N. k=1 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 ∈ N with Z + Un ⊂ Z + U. Clearly we can assume that U = { f ∈ C p(X) :|f (x)| <δ} x∈F for some finite subset F of X and some δ>0. By the continuity of the operator : c0 → C p(X), there exists n ∈ N such that for any y = (yk ) ∈ c0 with max |yk|≤1/n 1≤k≤n ( ) ∈ 1 1 < 1 δ we get y 2 U. Replacing n by a larger number, we can assume that n 2 and ∞ n F ∩ supp(λk) ⊂ supp(λk). k=1 k=1 Let f ∈ Un. Choose a function h ∈ C p(K ) such that h(x) = f (x) for every ∞ x ∈ F\ supp(λk ) k=1 123 Josefson–Nissenzweig property for Cp-spaces ( ) = ∈ n (λ ) = | . ∈ λ ( ) = and h x 0foreveryx k=1 supp k .Putg h X Then g L,since k g λk(h) →k 0. Put y = S(g) and ξ = (y). Since g(x) = 0for n x ∈ suppλk, k=1 |λ ( )|= < 1 ≤ ≤ | | < 1 ξ = ( ) ∈ 1 , we have k g 0 n for 1 k n,somax1≤k≤n yk n . Hence y 2 U |ξ( )| < 1 δ. ς = − ξ so maxx∈F x 2 For g we obtain S(ς) = S(g) − S ◦ ◦ S(g) = S(g) − S(g) = 0, so ς ∈ Z. Moreover f − ς ∈ U. Indeed, we have | f (x) − ς(x)|=|f (x) − g(x) + ξ(x)|=|ξ(x)| <δ ∈ \ ∞ (λ ) for x F k=1 supp k and | ( ) − ς( )|=| ( ) − ( ) + ξ( )|≤| ( )|+| ( )|+|ξ( )| < 1 + + 1 δ<δ f x x f x g x x f x g x x n 0 2 ∈ n (λ ) = ς + ( − ς) ∈ + , ⊂ + . for every x k=1 supp k . Thus f f Z U so Un Z U Hence Z + Un ⊂ Z + U. Lemma 2 Let X be a Tychonoff space. Each metrizable continuous image of C p(X) is sep- arable. Proof It is well-known [11, 2.3.18] that the Tychonoff product RX has countable cellularity, which means that RX contains no uncountable family of pairwise disjoint non-empty open X sets. Then the dense subspace C p(X) of R also has countable cellularity and so does any continuous image Y of C p(X).IfY is metrizable, then Y is separable according to Theorem 4.1.15 in [11]. 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 { fn}n∈N in S and consider the continuous map N f : X → R , f : x → ( fn(x))n∈N. By the pseudocompactness of X and the metrizability of RN,theimageM := f (X) is a compact metrizable space. The continuous surjective map f : X → M induces an isomorphic embedding C p f : C p(M) → C p(X), C p f : ϕ → ϕ ◦ f . So, we can identify the space C p(M) with its image C p f (C p(M)) in C p(X).Weclaim that C p(M) is closed in C p(X). Given any function ϕ ∈ C p(X)\C p(M), we should find a neighborhood Oϕ ⊂ C p(X) of ϕ, which is disjoint with C p(M). We claim that there exist points x, y ∈ X such that f (x) = f (y) and ϕ(x) = ϕ(y).In the opposite case, ϕ = ψ ◦ f for some bounded function ψ : M → R. Let us show that the function ψ is continuous. Consider the continuous map h : X → M × R, h : x → ( f (x), ϕ(x)). 123 T. Banakh et al. The pseudocompactness of X implies that the image h(X) ⊂ M × R is a compact closed × R : ( ) → : ( ) → R subset of M .LetprM h X M and prR h X be the coordinate projections. It follows that ◦ = ϕ = ψ ◦ = ψ ◦ ◦ , prR h f prM h = ψ ◦ : ( ) → which implies that prR prM .ThemapprM h X M between the compact metrizable spaces h(X) and M is closed and hence is quotient. Then the continuity of the map = ψ ◦ ψ ϕ = ψ ◦ prR prM implies the continuity of . Now we see that the function f belongs to the subspace C p(M) ⊂ C p(X), which contradicts the choice of ϕ. This contradiction shows that ϕ(x) = ϕ(y) for some points x, y ∈ X with f (x) = f (y).Then Oϕ := {φ ∈ C p(X) : φ(x) = φ(y)} is a required neighborhood of ϕ, 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 { fn}n∈N in S. Since the space C p(M) is Polishable, so is its closed subspace S. 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 c0; (3) C p(X) has a quotient isomorphic to c0; (4) C p(X) admits a continuous linear map onto c0. The implication (1) ⇒ (2) follows from Lemma 1, applied to the Stone-Cechˇ compacti- fication K = β X of X. The implications (2) ⇒ (3) ⇒ (4) are trivial. To prove the implication (4) ⇒ (1) assume that there exists a continuous linear map T from C p(X) onto c0.Let { ∗} ⊂ ∗ en n∈N c0 ∗( ) → ∈ be the sequence of coordinate functionals. By definition of c0, en y n 0foreveryy c0. For every n ∈ N consider the linear continuous functional λ ∈ ( )∗,λ( ) = ∗( ), n C p X n f en Tf which can be thought as a finitely supported sign-measure on X. It follows that λn( f ) = ∗( ) → ∈ ( ) en Tf n 0forevery f C p X . = ∞ = (λ ) We shall show that the union S n=1 Sn of supports Sn supp n of the sign- measures λn is bounded in X in the sense that for any ϕ ∈ C p(X) the image ϕ(S) is bounded in R, since in the opposite case we get a function ψ ∈ C p(X) with λn(ψ) → 0. Indeed, suppose that for some ϕ ∈ C p(X) the image ϕ(S) is unbounded in R; without loss of generality we can assume that ϕ is non-negative. We can find inductively an increasing sequence {nk}k∈N ⊂ N such that ϕ( )> + ϕ( ) max Snk 3 max Sni , ∈ N > ∈ N ∈ ϕ( ) = ϕ( ) for all k i with k i.Foreveryk choose xk Snk with xk max Snk . Then ϕ(xk) − ϕ(xi )>3forallk, i ∈ N with k > i. Since the space X is Tychonoff, for every k ∈ N we can find an open neighborhood Uk ⊂{x ∈ X :|ϕ(x) − ϕ(xk)| < 1} of xk ∩ (λ ) ={ } such that Uk supp nk xk . 123 Josefson–Nissenzweig property for Cp-spaces > ∈ ∈ ∪ (λ ) Observe that for any k i and any points x Uk and y Ui supp ni we get ϕ( ) − ϕ( ) ≥ ϕ( ) − − ( ϕ( ) + )> − = , x y xk 1 max Sni 1 3 2 1 ∩ (λ ) =∅ ϕ( ) R which implies Uk supp ni and that the family Uk k∈N is discrete in and consequently (Uk)k∈N is discrete in X. ∩ (λ ) ={ } Taking into account that Uk supp nk xk , we can inductively construct a sequence (ψk )k∈N of functions ψk : X → R such that supp(ψk ) := {x ∈ X : ψi (x) = 0}⊂Uk and λ (ψ )> + |λ (ψ )| nk k 1 nk i i λn ∗ μn = ∈ C p(X) λn and observe that the sequence (μn) 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) ⇒ (5) ⇒ (6) ⇒ (7) ⇒ (1). The implication (2) ⇒ (5) is trivial and (5) ⇒ (6) ⇒ (7) follow from Lemmas 2 and 3, respectively. (7) ⇒ (1) Assume that the space C p(X) contains a closed subspace Z of infinite codi- mension such that the quotient space E := C p(X)/Z is Polishable. Denote by τp the quotient topology of C p(X)/Z and by τ0 ⊃ τp a stronger separable Fréchet locally convex topology on E. Denote by τ∞ the topology of the quotient Banach space C(X)/Z.HereC(X) is endowed := | ( )| with the sup-norm f ∞ supx∈X f x (which is well-defined as X is pseudocompact). The identity maps between (E,τ0) and (E,τ∞) have closed graphs, since τp ⊂ τ0 ∩ τ∞. Using the Closed Graph Theorem we infer that the topologies τ0 and τ∞ are equal. Let G be 123 T. Banakh et al. a countable subset of C(X) such that the set {g + Z : g ∈ G} is dense in the Banach space C(X)/Z. Then the set G + Z ={g + z : g ∈ G, z ∈ Z} is dense in C(X).Let(gn)n∈N be a linearly independent sequence in G such that its linear ∗ span G0 has G0 ∩ Z ={0} and G0 + Z = G + Z.Let f1 = g1 and ν1 ∈ C p(X) with ν1|Z = 0 such that ν1( f1) = 1. Assume that for some n ∈ N we have chosen ∗ ( f1,ν1),...,(fn,νn) ∈ C p(X) × C p(X) such that lin{ f1,..., fn}=lin{g1,...,gn} and ν j |Z = 0,νj ( fi ) = δ j,i for all i, j ∈{1,...,n}. Put n fn+1 = gn+1 − νi (gn+1) fi . i=1 Then lin{ f1,..., fn+1}=lin{g1,...,gn+1} ∗ and ν j ( fn+1) = 0for1 ≤ j ≤ n. Let νn+1 ∈ C p(X) with νn+1|Z = 0 such that νn+1( fi ) = 0for1≤ i ≤ n and νn+1( fn+1) = 1. Continuing on this way we can construct inductively a biorthogonal sequence (( fn,νn))n∈N ∗ in C p(X) × C p(X) such that lin{ fn : n ∈ N}=lin{gn : n ∈ N} and νn|Z = 0, νn( fm) = δn,m for all n, m ∈ N. Then lin{ fn : n ∈ N}+Z is dense in (C(X), .∞). Let μn = νn/νn for n ∈ N. Then μn=1andμn( fm) = 0foralln, m ∈ N with n = m. We prove that μn( f ) →n 0forevery f ∈ C p(X). Given any f ∈ C(X) and ε>0, find m ∈ N and g ∈ lin{ f1,..., fm}+Z with d( f , g)<ε; clearly d( f , g) =f − g∞. Then μn(g) = 0forn > m, so |μn( f )|=|μn( f − g)|≤μn· f − g∞ <ε for n > m. Thus μn( f ) →n 0, which means that the space C p(X) has the JNP. 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 βN; (2) the function space C p(K ) has the JNP. 123 Josefson–Nissenzweig property for Cp-spaces Weneed some facts to present the construction of the space K . By definition, the asymptotic density of a subset A ⊂ N is the limit |A ∩[1, n]| d(A) := lim n→∞ n if this limit exists. The family Z ={A ⊂ N : d(A) = 0} of sets of asymptotic density zero in N is an ideal on N. Recall the following standard fact (here A ⊂∗ B means that A\B is finite). Fact 1 For any countable subfamily C ⊂ Z there is a set B ∈ Z such that C ⊂∗ B for all C ∈ C. Let A = A ⊂ N : d(A) ∈{0, 1} be the algebra of subsets of N generated by Z.We now let K be the Stone space of the algebra A so we treat elements of K as ultrafilters on A. There are three types of such x ∈ K : (1) {n}∈x for some n ∈ N;thenx ={A ∈ A : n ∈ A} is identified with n; (2) x contains no finite subsets of N but Z ∈ x for some Z ∈ Z; (3) Z ∈/ x for every Z ∈ Z; this defines the unique p ={A ∈ A : d(A) = 1}∈K . 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 ⊂ K contains an infinite set Y such that Y is homeomorphic to βN. Note first that for every Z ∈ Z, the corresponding clopen set Z ={x ∈ K : Z ∈ x}, is homeomorphic to βN because {A ∈ A : A ⊂ Z}=2Z . For an infinite set X ⊂ K , we have two cases: Case 1, X ∩ N is infinite. There is an infinite Z ⊂ X ∩ N having density zero. Then every subset of Z is in A, which implies that Z =∼ βN . Case 2, X ∩ (K \N) is infinite. Let us fix a sequence of different xn ∈ X ∩ (K \N) such that xn = p for every n. Then for every n we have Zn ∈ xn for some Zn ∈ Z.TakeB ∈ Z as in Fact 1.ThenB ∈ xn because xn is a nonprincipial ultrafilter on A so An\B ∈/ xn.Again, we conclude that {xn : n ∈ N} is βN. ν = 1 δ μ = 1 (ν − δ ) ∈ N ν ( ) → δ ( ) Fact 3 If n n k≤n k and n 2 n p for n ,then n f n p f and μn( f ) →n 0 for every f ∈ C(K ). Proof Observe νn(A) →n d(A) for every A ∈ A since elements of A have asymptotic density either 0 or 1. This means that, when we treat νn as measures on K then νn(V ) converges to δp(V ) for every clopen set V ⊂ K . This implies the assertion since every continuous function on K can be uniformly approximated by simple functions built from clopens. 5 Proof of Theorem 2 Let us recall that a topological space X is called • submetrizable if X admits a continuous metric; 123 T. Banakh et al. • hemicompact if X has a countable family K of compact sets such that each compact subset of X is contained in some compact set K ∈ K; • a k-space if a subset F ⊂ X is closed if and only if for every compact subset K ⊂ X the intersection F ∩ 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) Ck(X) is Polishable; (2) C p(X) is Polishable; (3) Ck(X) is Polish; (4) X is a submetrizable hemicompact k-space. The implication (1) ⇒ (2) follows from the continuity of the identity map Ck(X) → C p(X). (2) ⇒ (3) Assume that the space C p(X) is Polishable and fix a stronger Polish locally convex topology τ on C p(X).LetCτ (X) denote the separable Fréchet space (C p(X), τ).By τk denote the compact open topology of Ck(X). Taking into account that the space C p(X) is a continuous image of the Polish space Cτ (X), we conclude that C p(X) has countable network andby[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 μ-space. By Theorem 10.1.20 in [25, Theorem 10.1.20] the function space Ck(X) is barrelled. The continuity of the identity maps Ck(X) → C p(X) and Cτ (X) → C p(X) implies that the identity map Ck(X) → Cτ (X) has closed graph. Since Ck(X) is barelled and Cτ (X) is Fréchet, we can apply the Closed Graph Theorem 4.1.10 in [25] and conclude that the identity map Ck(X) → Cτ (X) is continuous. Next, we show that the identity map Cτ (X) → Ck(X) is continuous. Given any compact set K ⊂ X and any ε>0 we have to find a neighborhood U ⊂ Cτ (X) of zero such that U ⊂{f ∈ C(X) : f (K ) ⊂ (−ε, ε)}. The continuity of the restriction operator R : C p(X) → C p(K ), R : f → f K ,and the continuity of the idenity map Cτ (X) → C p(X) imply that the restriction operator R : Cτ (X) → C p(K ) is continuous and hence has closed graph. The continuity of the identity map Ck(K ) → C p(K ) implies that R seen as an operator R : Cτ (X) → Ck(K ) still has closed graph. Since the spaces Cτ (X) and Ck(K ) are Fréchet, the Closed Graph Theorem 1.2.19 in [25] implies that the restriction operator R : Cτ (X) → Ck(K ) is continuous. So, there exists a neighborhood U ⊂ Cτ (X) of zero such that R(U) ⊂{f ∈ Ck(K ) : f (K ) ⊂ (−ε, ε)}. Then U ⊂{f ∈ C(X) : f (K ) ⊂ (−ε, ε)} and we are done. Hence τ = τk is a Polish locally convex topology as claimed. The implication (3) ⇒ (1) is trivial. (3) ⇒ (4) If the function space Ck(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 Ck(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 3 If X is a submetrizable hemicompact k-space, then X n∈ω Xn for some increasing sequence (Xn)n∈ω of compact metrizable spaces such that each compact subset of X is contained in some compact set Xn. Then the function space Ck(X) is Polish, being topologically isomorphic to the closed subspace 123 Josefson–Nissenzweig property for Cp-spaces ( fn)n∈ω ∈ Ck(Xn) :∀n ∈ ω fn+1Xn = fn n∈ω ∞ ( ) of the countable product n=1 Ck Xn of separable Banach spaces. The authors thank to the referee for his/her valuable comments and remarks. 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