Josefson-Nissenzweig property for $C_p$-spaces

The famous Rosenthal-Lacey theorem asserts that for each infinite compact space $K$ the Banach space $C(K)$ admits a quotient which is either a copy of $c_{0}$ or $\ell_{2}$. The aim of the paper is to study a natural variant of this result for the space $C_{p}(K)$ of continuous real-valued maps on $K$ with the pointwise topology. Following famous Josefson-Nissenzweig theorem for infinite-dimensional Banach spaces we introduce a corresponding property (called Josefson-Nissenzweig property, briefly, the JNP) for $C_{p}$-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 $c_{0}$ (with the product topology of $\mathbb R^\mathbb{N}$) if and only if $C_{p}(X)$ contains a complemented subspace, isomorphic to $c_0$. 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. This applies to show that for a Tychonoff space $X$ the space $C_p(X)$ has a complemented subspace isomorphic to $\mathbb R^{\mathbb N}$ or $c_0$ if and only if $X$ is not pseudocompact or $C_p(X)$ has the JNP. The space $C_{p}(\beta\mathbb{N})$ contains a subspace isomorphic to $c_0$ and admits a quotient isomorphic to $\ell_{\infty}$ but fails to have a quotient isomorphic to $c_{0}$. An example of a compact space $K$ without infinite convergent sequences with $C_{p}(K)$ containing a complemented subspace isomorphic to $c_{0}$ is constructed.


Introduction and the main problem
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 simple observation 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 R X , see [2].
The classic Rosenthal-Lacey theorem, see [24], [16], and [20], asserts that the Banach space C(K) of continuous real-valued maps on an infinite compact space K has a quotient isomorphic to c 0 or ℓ 2 , or equivalently, there exists a continuous linear (and open; by the open mapping Banach theorem) map from C(K) onto c 0 or ℓ 2 .
This theorem motivates the following natural question for spaces C p (X). The space C p (K) has an infinite dimensional metrizable quotient. (4) The space C p (K) has an infinite dimensional metrizable separable quotient. (5) The space C p (K) has a quotient isomorphic to a dense subspace of R N .
Note that there is a continuous linear map from a real topological vector space E onto a dense subspace of R N if and only if the continuous dual E ′ is infinite dimensional. Thus when K is infinite, (1) and (2) hold provided we delete "open" in both cases. When we retain "open" and delete "metrizable" in (2), the question is unsolved and more general: For every infinite compact set K, does C p (K) admit an infinite dimensional separable quotient?
In [18] 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 (βN) lacks infinite-dimensional separable quotient algebras. Nevertheless, as proved in [19,Theorem 4], the space C p (K) has infinite-dimensional separable quotient for any compact space K containing a copy of βN.
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∈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 βN) asserts that C p (X) has a quotient isomorphic to the subspace ℓ ∞ = {(x n ) ∈ R N : sup n |x n | < ∞} of R N or to the product R 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 βN). Although, it is unknown if Efimov spaces exist in ZFC (see [6], [7], [8], [9], [11], [12], [13], [15]) we showed in [19] that under ♦ for some Efimov spaces K the function space C p (K) has an infinite dimensional metrizable quotient.
In this paper c 0 means the subspace {(x n ) n∈N ∈ R N : x n → n 0} of R N endowed with the product topology.

The main results
For a Tychonoff space X and a point x ∈ X let δ x : C p (X) → R, δ x : f → f (x), be the Dirac measure concentrated at x. The linear hull L p (X) of the set {δ x : x ∈ X} in R Cp(X) can be identified with the dual space of C p (X).
Elements of the space L p (X) will be called finitely supported sign-measures (or simply signmeasures) on X.
Each µ ∈ L p (X) can be uniquely written as a linear combination of Dirac measures µ = x∈F α x δ x for some finite set F ⊂ X and some non-zero real numbers α x . The set F is called the support of the sign-measure µ and is denoted by supp(µ). The measure x∈F |α x |δ x will be denoted by |µ| and the real number µ = x∈F |α x | coincides with the norm of µ (in the dual Banach space C(βX) * ).
The sign-measure µ = x∈F α x δ x determines the function µ : 2 X → R defined on the power-set of X and assigning to each subset A ⊂ X the real number x∈A∩F α 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 [5].
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 (µ n ) of finitely supported sign-measures on X such that µ n = 1 for all n ∈ N, and µ n (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 x n → x, then C p (K) satisfies the JNP. This is witnessed by the weak * null sequence (µ n ) of signmeasures µ n := 1 2 (δ xn − δ x ), n ∈ N. (2) The space C p (βN) does not satisfy the JNP. This follows directly from the Grothendieck theorem, see [4,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. Consequently, if compact K contains an infinite convergent sequence x n → x, then C p (K) satisfies the JNP with C p (Z) complemented in C p (K) and isomorphic to c 0 , where Z := {x} ∪ {x n } n∈N . 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 (hence contains complemented subspaces 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.
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) contains a quotient isomorphic to c 0 . If the space X is pseudocompact, then the conditions (1)-(3) are equivalent to (4) C p (X) contains a complemented infinite-dimensional metrizable subspace; (5) C p (X) contains a complemented infinite-dimensional separable subspace; (6) C p (X) has an infinite-dimensional Polishable quotient.
We recall that a locally convex space X is Polishable if X admits a stronger separable Frécht (= complete metrizable) 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 R 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 R 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 R N or c 0 ; (4) X is not pseudocompact or C p (X) has the JNP.
Indeed, the first claim follows from [3,Proposition], the others follow from Theorem 1 and the statement (1) after Definition 1.
In the final Section 5 we shall characterize Tychonoff spaces 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: ( 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 [17, Theorem 9.1].

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. Let (µ n ) be a sequence of finitely supported sign-measures on X (and so, on K) such that (1) µ n = 1 for all n ∈ N, and (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 = {µ n : n ∈ N} in not relatively weakly compact in the dual of the Banach space C(K). Indeed, assume by contrary that the closure 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 M is weakly sequentially compact. Thus (µ n ) has a subsequence (µ kn ) that weakly converges to some element µ 0 ∈ C(K) * . Taking into account that the sequence (µ n ) converges to zero in the weak * topology of C(K) * , we conclude that µ 0 = 0 and hence (µ kn ) is weakly convergent to zero in C(K) * . Denote by S the countable set n∈N supp(µ n ). The measures µ n , n ∈ N, can be considered as elements of the unit sphere of the Banach space ℓ 1 (S) ⊂ C(K) * . By the Schur theorem [1, Theorem 2.3.6], the weakly convergent sequence (µ kn ) is convergent to zero in the norm topology of ℓ 1 (S), which is not possible as µ n = 1 for all n ∈ 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 ǫ > 0, a sequence (m n ) ⊂ N and a sequence (U n ) of pairwise disjoint open sets in K such that |µ mn (U n )| > ǫ for any n ∈ N. Clearly, lim n→∞ µ k (U n ) = 0 for any k ∈ N, since Thus we can assume that the sequence (m n ) is strictly increasing.
For some strictly increasing sequence (n k ) (A3) |ν k |(W n ) = 0 for all k, n ∈ N with k < n.
(III) By induction we shall construct a decreasing sequences (N k ) of infinite subsets of N with min N k < min N k+1 for k ∈ N such that |ν n |(W m ) ≤ ǫ/3 k for every k ∈ N, m = min N k , n ∈ N k and n > m. Let N 0 = N. Assume that for some k ∈ N an infinite subset N k−1 of N has been constructed. Let F be a finite subset of N k−1 with |F | > 3 k /ǫ and min F > min N k−1 . For every i ∈ F consider the set Clearly, the set and L = {f ∈ C p (X) : λ n (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 is continuous and ker S = Z. We shall construct a linear continuous map P : c 0 → L such that S • P is the identity map on c 0 . For every k ∈ N there exists a continuous function Then for n ≥ m we obtain Thus (x ′ n ) ∈ c 0 . Clearly, the operator Θ : c 0 → c 0 , Θ : (x n ) → (x ′ n ), is linear and continuous. We prove that Θ is surjective. Let (y n ) ∈ c 0 . Set t = sup n |y n |. Let x n = n k=1 λ n (ϕ k )y k for n ∈ N.
(VI) The operator is well-defined, linear and continuous, since the functions ϕ n , n ∈ N, have pairwise disjoint supports and ϕ n (X) ⊂ [−1, 1], n ∈ N. Thus the linear operator Using (B3) and the definition of Θ, we get for every n ∈ N Thus the quotient space L/Z is topologically isomorphic to c 0 and Φ(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 U n = {f ∈ C p (X) : |f (x)| < 1/n for every x ∈ n k=1 supp(λ k )}, n ∈ N.
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 + U n ⊂ Z + U. Clearly we can assume that U = x∈F {f ∈ C p (X) : |f (x)| < δ} for some finite subset F of X and some δ > 0. By the continuity of the operator Φ : c 0 → C p (X), there exists n ∈ N such that for any y = (y k ) ∈ c 0 with max 1≤k≤n |y k | ≤ 1/n we get Φ(y) ∈ 1 2 U. Replacing n by a larger number, we can assume that 1 n < 1 2 δ and and h(x) = 0 for every x ∈ n k=1 supp(λ k ). Put g = h|X. Then g ∈ L, since λ k (g) = λ k (h) → k 0. Put y = S(g) and ξ = Φ(y). Since g(x) = 0 for x ∈ n k=1 suppλ k , we have |λ k (g)| = 0 < 1 n for 1 ≤ k ≤ n, so max 1≤k≤n |y k | < 1 n . Hence ξ = Φ(y) ∈ 1 2 U, so max x∈F |ξ(x)| < 1 2 δ. For ς = g − ξ we obtain Lemma 2. Let X be a Tychonoff space. Each metrizable continuous image of C p (X) is separable.
Proof. It is well-known [10, 2.3.18] that the Tychonoff product R X has countable cellularity, which means that R X contains no uncountable family of pairwise disjoint non-empty open sets. Then the dense subspace C p (X) of 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 [10].

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∈N in S and consider the continuous map By the pseudocompactness of X and the metrizability of R N , the image M := f (X) is a compact metrizable space. The continuous surjective map f : X → M induces an isomorphic embedding So, we can identify the space C p (M) with its image C p f (C p (M)) in C p (X). We claim 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 O ϕ ⊂ 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 The pseudocompactness of X implies that the image h(X) ⊂ M ×R is a compact closed subset of M × R. Let pr M : h(X) → M and pr R : h(X) → R be the coordinate projections. It follows that which implies that pr R = ψ • pr X . The map pr M : h(X) → M between the compact metrizable spaces h(X) and M is closed and hence is quotient. Then the continuity of the map pr R = ψ • pr X implies the continuity of ψ. Now we see that the function ϕ = ψ • 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 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 {f n } 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 c 0 ; (3) C p (X) has a quotient isomorphic to c 0 . The implication (1) ⇒ (2) follows from Lemma 1, applied to the Stone-Čech compactification K = βX of X. The implication (2) ⇒ (3) is trivial.
To prove the implication (3) ⇒ (1), assume that C p (X) has a quotient isomorphic to c 0 .

Then it admits an open continuous linear operator
be the sequence of coordinate functional. By definition of c 0 , e * n (y) → n 0 for every y ∈ c 0 . For every n ∈ N consider the linear continuous functional which can be thought as a finitely supported sign-measure on X. It follows that for every f ∈ C p (X) we have λ n (f ) = e * n (T f ) → n 0. If λ n → n 0, then we can find an infinite subset Ω ⊂ N such that inf n∈Ω λ n > 0. For every n ∈ Ω put µ n := λ n λ n ∈ C p (X) * and observe that the sequence {µ n } n∈N witnesses that the function space C p (X) has the JNP. It remains to consider the case when λ n → n 0. We are going to prove that the assumption λ n → n 0 leads to a contradiction.
First we show that the union S := n∈N supp(λ n ) is bounded in X in the sense that for any continuous function ϕ : X → [0, +∞] the image ϕ(S) is bounded in R. To derive a contradiction, assume that for some function ϕ ∈ C p (X) the image ϕ(S) is unbounded. Then we can find an increasing number sequence (n k ) k∈N such that max ϕ(supp(λ n k )) > 3 + max ϕ(supp(λ n i )) for any i < k.
For every k ∈ N choose a point x k ∈ supp(λ n k ) with ϕ(x k ) = max ϕ(supp(λ n k )).
It follows that ϕ(x k ) > 3 + ϕ(x i ) for every i < k. Since the space X is Tychonoff, for every k ∈ N we can find an open neighborhood U k ⊂ {x ∈ X : |ϕ(x) − ϕ(x k )| < 1} of x k such that U k ∩ supp(λ n k ) = {x k }. Also find a continuous function ψ k : X → [0, 1] such that ψ k (x k ) = 1 and ψ k (X \ U k ) ⊂ {0}.
Inductively, choose a sequence of positive real numbers (r k ) k∈N such that for every k ∈ N Since the family (U k ) k∈N is discrete, the function is well-defined and continuous. It follows that for every k ∈ N and i > k we have U i ∩ supp(λ n k ) = ∅ and hence But this contradicts λ n (ψ) → n 0. This contradiction shows that the set S = k∈N supp(λ k ) is bounded in X and so is its closureS in X.
Consider the space C p (X↾S) = {f ↾S : f ∈ C p (X)} ⊂ RS and observe that the restriction operator R : This norm is well-defined since the setS is bounded in X. The completion C(X↾S) of the normed space C(X↾S) can be identified with a closed subspace of the Banach space C b (S) of bounded continuous functions onS, endowed with the sup-norm. It follows from λ n → n 0 and λ n (f ) → n 0 for all f ∈ C(X↾S) that λ n (f ) → n 0 for all So, Λ : C(X↾S) → c 0 , Λ : f → (λ n (f )) n∈N , is a well-defined continuous operator such that T = Λ • R. It follows that the operator to c 0 endowed with its standard norm x = sup n∈N |e n (x)| has closed graph and hence is continuous and open (being surjective). Then the image Λ(B 1 ) of the unit ball contains some closed ε-ball B ε := {x ∈ c 0 : x ≤ ε} in the Banach space (c 0 , · ). Since λ n → n 0, we can find n ∈ N such that λ n < ε. Next, find an element y ∈ B ε ⊂ c 0 such that y = e * n (y) = ε. Since y ∈ B ε ⊂ Λ(B 1 ), there exists a point x ∈ B 1 such that Λ(x) = y. Then ε = e * n (y) = λ n (x) ≤ λ n · x < ε and this contradiction completes the proof of the implication (3) ⇒ (1). Now assuming that the space X is pseudocompact, we shall prove that the conditions (1)-(3) are equivalent to (4) C p (X) contains a complemented infinite-dimensional metrizable subspace; (5) C p (X) contains a complemented infinite-dimensional separable subspace; (6) C p (X) has an infinite-dimensional Polishable quotient.
(6) ⇒ (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 τ p the quotient topology of C p (X)/Z and by τ 0 ⊃ τ p a stronger separable Fréchet locally convex topology on E. Denote by τ ∞ the topology of the quotient Banach space C(X)/Z. Here C(X) is endowed with the sup-norm f ∞ := sup x∈X |f (x)| (which is well-defined as X is pseudocompact).
The identity maps between (E, τ 0 ) and (E, τ ∞ ) have closed graphs, since τ p ⊂ τ 0 ∩ τ ∞ . Using the Closed Graph Theorem we infer that the topologies τ 0 and τ ∞ are equal. Let G be 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 is dense in C(X). Let (g n ) n∈N be a linearly independent sequence in G such that its linear span G 0 has G 0 ∩ Z = {0} and G 0 + Z = G + Z. Let f 1 = g 1 and ν 1 ∈ C p (X) * with ν 1 |Z = 0 such that ν 1 (f 1 ) = 1.
Continuing on this way we can construct inductively a biorthogonal sequence (f n , ν n ) n∈N in C p (X) × C p (X) * such that lin{f n : n ∈ N} = lin{g n : n ∈ N} and ν n |Z = 0, ν n (f m ) = δ n,m for all n, m ∈ N. Then lin{f n : n ∈ N} + Z is dense in C u (X). Let µ n = ν n / ν n for n ∈ N. Then µ n = 1 and µ n (f m ) = 0 for all n, m ∈ N with n = m.
We prove that µ n (f ) → n 0 for every f ∈ C p (X). Given any f ∈ C(X) and ε > 0, find m ∈ N and g ∈ lin{f 1 , . . . , f m } + Z with d(f, g) < ε; clearly d(f, g) = f − g ∞ . Then µ n (g) = 0 for n > m, so for n > m. Thus µ n (f ) → n 0, which means that the space C p (X) has the JNP.

An example of Plebanek
In this section we describe the following example suggested to the authors by Grzegorz Plebanek [23].
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.
We need some facts to present the construction of the space K. By definition, the asymptotic density of a subset A ⊂ N is the limit 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; then x = {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 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 For an infinite set X ⊂ K, we have two cases: 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 x n ∈ X ∩ (K \ N) such that x n = p for every n. Then for every n we have Z n ∈ x n for some Z n ∈ Z. Take B ∈ Z as in Fact 1. Then B ∈ x n because x n is a nonprincipial ultrafilter on A so A n \ B / ∈ x n . Again, we conclude that {x n : n ∈ N} is βN.
Proof. Observe ν n (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 ν n as measures on K then ν n (V ) converges to δ p (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.

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 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) X is a submetrizable hemicompact k-space; (2) C k (X) is Polish; (3) C p (X) is Polishable.
(1) ⇒ (2): If X is a submetrizable hemicompact k-space, then X = n∈ω X n for some increasing sequence (X n ) n∈ω 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 {(f n ) n∈ω ∈ n∈ω C k (X n ) : ∀n ∈ ω f n+1 ↾X n = f n } of the countable product ∞ n=1 C k (X n ) of separable Banach spaces. (2) ⇒ (1) If the function space C k (X) is Polish, then by Theorem 4.2 in [21], 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 [14, 2.9], the space X is submetrizable.
The implication (2) ⇒ (3) follows from the continuity of the identity map C k (X) → C p (X).
(3) ⇒ (2): Assume that the space C p (X) is Polishable and fix a stronger Polish locally convex topology τ on C p (X). Let C τ (X) denote the separable Fréchet space (C p (X), τ ). By τ 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 τ (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 µ-space. By Theorem 10.1.20 in [22, Theorem 10.1.20] the function space C k (X) is barrelled. The continuity of the identity maps C k (X) → C p (X) and C τ (X) → C p (X) implies that the identity map C k (X) → C τ (X) has closed graph. Since C k (X) is barelled and C τ (X) is Fréchet, we can apply the Closed Graph Theorem 4.1.10 in [22] and conclude that the identity map C k (X) → C τ (X) is continuous.
Next, we show that the identity map C τ (X) → C k (X) is continuous. Given any compact set K ⊂ X and any ε > 0 we have to find a neighborhood U ⊂ C τ (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 C τ (X) → C p (X) imply that the restriction operator R : C τ (X) → C p (K) is continuous and hence has closed graph. The continuity of the identity map C k (K) → C p (K) implies that R seen as an operator R : C τ (X) → C k (K) still has closed graph. Since the spaces C τ (X) and C k (K) are Fréchet, the Closed Graph Theorem 1.2.19 in [22] implies that the restriction operator R : C τ (X) → C k (K) is continuous. So, there exists a neighborhood U ⊂ C τ (X) of zero such that R(U) ⊂ {f ∈ C k (K) : f (K) ⊂ (−ε, ε)}.