On complemented copies of the space c0 in spaces Cp(X,E)$C_p(X,E)$

We study the question for which Tychonoff spaces X and locally convex spaces E the space Cp(X,E)$C_p(X,E)$ of continuous E‐valued functions on X contains a complemented copy of the space (c0)p={x∈Rω:x(n)→0}$(c_0)_p=\lbrace x\in \mathbb {R}^\omega : x(n)\rightarrow 0\rbrace$ , both endowed with the pointwise topology. We provide a positive answer for a vast class of spaces, extending classical theorems of Cembranos, Freniche, and Domański and Drewnowski, proved for the case of Banach and Fréchet spaces Ck(X,E)$C_k(X,E)$ . Also, for given infinite Tychonoff spaces X and Y, we show that Cp(X,Cp(Y))$C_p(X,C_p(Y))$ contains a complemented copy of (c0)p$(c_0)_p$ if and only if any of the spaces Cp(X)$C_p(X)$ and Cp(Y)$C_p(Y)$ contains such a subspace.

Both of the theorems have important consequences.For example, since for each Tychonoff space  and locally compact space , the spaces   ( × ) and   (,   ()) are linearly homeomorphic (in short, isomorphic), see [28,Corollary 2.5.7],Theorem 1.1 yields that for infinite compact spaces  and  the Banach space ( × ) contains a complemented copy of  0 .Similarly, if  is a Tychonoff space containing an infinite compact subset and  is a locally compact and -compact Tychonoff space, then   () is an infinite-dimensional Fréchet space, which is not a Montel space [16,Theorem 11.8.7], and hence Theorem 1.2 implies that   ( × ) ≈   (,   ()) contains a complemented copy of  0 .
We first answer question ( * ) in the case of spaces of the form   (,   ())-in Theorem 5.2, we show that   (,   ()) contains a complemented copy of ( 0 )  if and only if any of the spaces   () and   () contains such a copy.For characterizations and examples of classes of Tychonoff spaces  for which   () contains a complemented copy of ( 0 )  , see, for example, [4,[19][20][21].
Regarding a more general case, we show in Theorem 6.2 that for every Tychonoff space  containing an infinite compact subset and barrelled locally convex space  with the Josefson-Nissenzweig property (JNP; see Section 4 for the definition) the space   (, ) contains a complemented copy of ( 0 )  .This not only generalizes Theorem 1.1 (see Corollary 6.5) and provides a   -variant of Theorem 1.2 (Corollary 6.4), but also has several applications.For example, we get that if  is an infinite Tychonoff space and  is an infinite compact space, then the space   (, ()) contains a complemented copy of ℝ  or a complemented copy of ( 0 )  (Corollary 6.6; see also Corollary 6.7 for a   (,   ())-version).We immediately get from this that the space   (, ()) is isomorphic to   (, ()) × ℝ (Corollary 6.9), which extends the number of known cases of locally convex spaces  for which  ≈  × ℝ, see the remark following Corollary 6.9 and concerning a problem of Arkhangel'ski.
For the case of spaces of the form   (,   ), where   means a locally convex spaces  with its weak topology, we show in Theorem 7.2 that, for every Tychonoff space  containing an infinite compact subset and locally convex space  containing a copy of the Banach space  1 , the space   (,   ) contains a complemented copy of ( 0 )  .Hence, we get that if  is a Tychonoff space, which contains an infinite compact subset, and  is an infinite Tychonoff space containing a nonscattered compact subset, then also   (,   ()  ) contains a complemented copy of ( 0 )  , see Corollary 7.3.This applies to show that, for example, there is no continuous linear surjection from   (,   ()) onto   (, ()  ), as well as that for every infinite compact space  the spaces   () and ()  are not isomorphic (which is a special case of [25, Corollary 3.2], cf.Remark 3.6).

PRELIMINARIES
The cardinality of a set  is denoted by ||.By  we denote the cardinality of the space of natural numbers ℕ and, as usual, we identify  with ℕ.  denotes the Čech-Stone compactification of , and  * =  ⧵ .
We assume that all topological spaces we consider are Tychonoff, that is, completely regular and Hausdorff.If (, ) is a topological space (with the topology ) and  ⊆ , then by  ↾  we mean the restriction of  on , that is,  ↾  = { ∩  ∶  ∈ }.
For a locally convex space ,  ′ denotes its topological dual and   denotes the pair (, (,  ′ ))), that is, the space  endowed with its weak topology.
If  is a Tychonoff space, then   () = (  ())  holds.We write   () =   () ′ .Recall that each  ∈   () may be uniquely represented as a finite linear combination of point-measures on , that is,  = If  and  are both locally convex, we write  ≈  if  and  are isomorphic (i.e., linearly homeomorphic).We also say that  contains a complemented copy of  (or that  is complemented in ) if there are closed linear subspaces  1 and  2 of  such that  is a direct algebraic sum of  1 and  2 (i.e.,  =  1 ⊕  2 as a vector space and  1 ∩  2 = {0}),  1 is isomorphic to , and the natural projection from  onto  1 is continuous.

3
( × ) VERSUS   (,   ()) In this section, we will briefly recall standard and folklore facts concerning spaces of separately continuous functions on products  ×  of two Tychonoff spaces and their relations to the spaces   (,   ()), which will be useful in the next sections.Fix Tychonoff spaces  and .Recall that a function  ∶  ×  → ℝ is separately continuous if the functions  ∋  ↦ ( 0 , ) and  ∋  ↦ (,  0 ) are continuous for every  0 ∈  and  0 ∈ .Of course, every continuous function is separately continuous, but the converse may not hold (e.g., for every infinite compact spaces  and , there is a separately continuous noncontinuous  ∶  ×  → ℝ).By ( × ), we denote the space of all real-valued separately continuous functions on  × , and by   ( × ), we mean ( × ) endowed with the pointwise topology.Recall that ( × ) is a linear subspace of ( × ) and that   ( × ) is dense in   ( × ).
By the last listed result and Theorem 3.1, we immediately get the following theorem, which allows us to treat spaces of the form   (,   ()) the same way as spaces of the form   ( × ).Theorem 3.2.Let  and  be Tychonoff spaces.Then, the space   ( × , ) is isomorphic to the space   (,   ()).
The following folklore theorem follows immediately from the main result of Uspensky [32].
Theorem 3.3.Let  and  be infinite pseudocompact spaces such that the product  ×  is pseudocompact.Then, the space   ( × ) cannot be mapped onto the space   (,   ()) by a continuous linear map.In particular,   ( × ) is not isomorphic to   (,   ()).
Recall that the product of a compact space and a pseudocompact space is pseudocompact.
Corollary 3.4.If  is an infinite compact space and  an infinite pseudocompact space, then the space   ( × ) cannot be mapped onto the space   (,   ()) by a continuous linear map.
Note that one cannot drop the compactness assumption in Corollary 3.5.Indeed, if  and  are infinite and  is discrete, then the canonical mapping  ∶   ( × ) →   (,   ()) defined above is easily seen to be an isomorphism.Also, [14,Corollary 6.15] states that if  is a (locally) separable Tychonoff space and  is a P-space, then ( × ) = ( × ) and hence  coincides with the product topology on  × -it follows by Theorem 3.2 that the spaces   ( × ) and   (,   ()) are isomorphic.In particular, if  is a metric compact space or  =  and  a P-space, then   ( × ) ≈   (,   ()).
Remark 3.6.Krupski and Marciszewski [25,Corollary 3.2] proved that for all infinite compact spaces  and  the locally convex spaces   () and ()  are never isomorphic.It follows that   ( × ) and ( × )  are not isomorphic.Since ( × )  is isomorphic to (, ())  , we get that   ( × ) is not isomorphic to (, ())  .Thus, Corollary 3.5 may be thought of as a   -counterpart of the result of Krupski and Marciszewski.This observation leads to the following relevant question, some issues of which we will study in Section 7.
Problem 3.7.For which infinite compact spaces  and  are the spaces   ( × ) and   (, ()  ) isomorphic?Note that if  and  are both infinite countable compact spaces, then   ( × ) is metrizable, whereas   (, ()  ) is not, so these function spaces are not isomorphic.
Let us now provide an example and a question motivated by it.
In Corollary 5.5, we describe a class of infinite compact spaces  such that   ( × ) is not complemented in   (,   ()).Note that if  is an infinite compact space such that   ( × ) is complemented in   (,   ()), then the latter may be linearly mapped onto the former-a phenomenon which by Corollary 3.4 cannot occur in the reverse direction.
We finish the section with the following observation.
Proposition 3.10.Let  be a normed space and  a Tychonoff space.Then, the original pointwise topology of   (, ) coincides with the weak topology of   (, ) if and only if  is finite-dimensional.
Proof.If  is finite-dimensional, say  is isomorphic to ℝ  for some  ∈ , then we have:   (, ) ≈   (, ℝ  ) ≈   ()  .Hence, the pointwise topology and weak topology coincide in   (, ), as they do on   ().Conversely, if both of the topologies coincide, then  is finite-dimensional.Indeed, by [31,Theorem I.4.4], is isomorphic to a complemented subspace of   (, ) and the weak topology on this complemented subspace agrees with the subspace topology inherited from the weak topology of   (, ).Since  is a normed space on which the norm topology and the weak topology agree, it is finite-dimensional.□ Corollary 3.11.If  is a Tychonoff space and  is an infinite compact space, then the spaces   ( × ) and   (, ()) are not isomorphic.

THE JNP
Following Banakh and Gabriyelyan [3], we say that a locally convex space  has the JNP, if the identity map ( ′ , ( ′ , )) → ( ′ ,  * ( ′ , )) is not sequentially continuous, that is, if the dual space  ′ contains a weak * null sequence, which does not converge in the topology  * ( ′ , ).The topology  * ( ′ , ) can be described as the topology of uniform convergence on so-called barrel-bounded sets, that is, on sets that have the property that they are absorbed by all barrels-see [3, p. 3] for more details.Note that the JNP is preserved by isomorphisms, since the transpose of an isomorphism is also an isomorphism for the above topologies-see, for example, Corollary 8.6.6 in [16, p. 161].By the famous Josefson-Nissenzweig Theorem, see, for example, [7,17,30], every infinite-dimensional Banach space has the JNP.On the other hand, no Montel space enjoys the JNP.Indeed: First, since Montel spaces are barrelled, the topologies  * ( ′ , ) and ( ′ , ), where the strong topology ( ′ , ) is the topology of uniform convergence on bounded subsets of , coincide.Second, the strong dual of a Montel space is again Montel and hence on bounded sets the topologies ( ′ , ) and ( ′ , ) coincide.In [5], Bonet, Lindström, and Valdivia showed that for Fréchet spaces this is already a characterization of the JNP, that is, a Fréchet space has the JNP if and only if it is not Montel; see also [26] for a similar characterization of non-Schwarz Fréchet spaces.
In the case of   ()-spaces, Banakh and Gabriyelyan proved the following characterization of the JNP.
If a sequence (  ) ∈ of functionals on a given space   () has the properties as in the above theorem, then it will be called a JN-sequence on .The existence of JN-sequences on arbitrary spaces was intensively studied in [4,19,20], and [21], where several classes of Tychonoff spaces were recognized to (not) admit such sequences (e.g., it appears that for every infinite compact spaces  and  their product admits a JN-sequence, see [20,Theorem 1.2]).Note that if  is pseudocompact and   () has the JNP, then every JN-sequence on  satisfies already the conclusion of the Josefson-Nissenzweig Theorem for the Banach space () (by virtue of the Riesz representation of continuous functionals on ()).
The following theorem provides a simple criterion for a space   ( × , ) to have the JNP.
Theorem 4.2.Let  and  be infinite Tychonoff spaces.Then, the space   ( × , ) has the JNP if and only if the space   () has the JNP or the space   () has the JNP.
If  is contained in finitely many horizontal sections of the product  × , then we proceed similarly and prove that   () has the JNP.
We prove similarly that if   () has the JNP, then   ( × , ) has the JNP.□ The first part of the proof suggests the following result.We will now study the JNP of the spaces   (, ) for  Tychonoff and  locally convex.Note that by [11,  .By Theorem 3.1, the operator  ∶   ( × , ) →   (,   ()) is an isomorphism, so the sequence (  ) ∈ given for every  ∈  by the formula: where  ′ is the transpose of , is weakly * convergent to 0 (see [24,Theorem 21.5]).Obviously, ‖  ‖ = 1 for every  ∈ .
Let us now assume that, conversely, the space   (,   ()) admits a weak * null sequence (  ) ∈ in   () ⊗   () such that ‖  ‖ = 1 for every  ∈ .By a very similar argument as above (this time applied to the transpose of  −1 ), one may obtain a JN-sequence on the space ( × , ).This way we obtain the following: Proposition 4.4.For every Tychonoff spaces  and  the space   ( × , ) has the JNP if and only if there exists a sequence (  ) ∈ in the dual   () ⊗   (), which is weakly * convergent to 0 and such that ‖  ‖ = 1 for every  ∈ .
Since the JNP for locally convex spaces is preserved under topological isomorphisms, Proposition 4.4 and Theorem 4.1 yield the following characterization of the property for   (,   ())-spaces.Proposition 4.5.For every Tychonoff space  and , the space   (,   ()) has the JNP if and only if there exists a sequence (  ) ∈ in   () ⊗   () such that ‖  ‖ = 1 for every  ∈  and   () → 0 for every  ∈   (,   ()).
The following corollary is an immediate consequence of Theorems 4.2 and 3.2.Corollary 4.6.For every infinite Tychonoff spaces  and , the space   (,   ()) has the JNP if and only if   () has the JNP or   () has the JNP.
For general locally convex spaces , we still have one of the implications observed in the above corollary.Proposition 4.7.Let  be a Tychonoff space and let  be a locally convex space.If   () has the JNP or  has the JNP, then the same is true for the space   (, ).
We now exhibit a barrel-bounded subset of   (, ) on which (  ) ∈ does not converge uniformly.The tensor product   () ⊗  embedds (as a vector space) into the space   (, ) by the formula Thus, the subspace of   () ⊗  given by the formula  ⊗  = { ⊗  ∶  ∈ } may be thought of as a subspace of   (, ).Moreover, by [ Having in mind Theorem 4.1 and Proposition 4.5, we also ask the following.
Problem 4.9.Can the JNP of the spaces   (, ) for  Tychonoff and  an arbitrary locally convex space be characterised in terms of some special "JN-sequences" in   () ⊗  ′ , like it is done in the case of  =   () for  Tychonoff?
Example 5.4.The space   (,   ()) does not have any complemented copy of ( 0 )  , even though the space   ( × ) does have such a copy.
As a corollary to Theorem 5.2, we obtain also the following result providing a large class of counterexamples to Problem 3.9.
Corollary 5.5.If  is such an infinite compact space that   () does not have the JNP, then   (,   ()) does not contain any complemented copy of   ( × ).
If  is a Tychonoff space and  a cardinal number, then we assume that   ()  is equipped with the standard product topology, that is, the topology given by basic open sets of the form ∏ <   , where each   is open in   () and for all but finitely many  < , we have   =   ().Using Theorem 5.1, it is easy to see that for every infinite Tychonoff space  the space   () 2 ≈   ( ⊔ ), where  ⊔  denotes the topological disjoint union of two copies of , contains a complemented copy of ( 0 )  if and only if   () contains such a copy.The following proposition generalizes this result to any infinite power of   ().Proposition 5.7.Let  be an infinite Tychonoff space and  an infinite cardinal number.Then, the space   ()  contains a complemented copy of ( 0 )  if and only if   () contains a complemented copy of ( 0 )  .
Conversely, assume that   () does not contain any complemented copies of ( 0 )  .Note that for any infinite discrete space , the space   () does not have the JNP, so if we equip the set  = { ∶  < } with the discrete topology, then the space   () does not have the JNP.By Theorem 5.1, it follows that   () does not contain complemented copies of ( 0 )  .Theorem 5.2 yields that the space   (,   ()) does not contain such a copy as well.Since   () ≈ ℝ  , we get the following equalities:   ()  =   (, ℝ)  ≈   (, ℝ  ) ≈   (,   ()), which implies that   ()  does not contain any complemented copies of the space ( 0 )  .□

COMPLEMENTED COPIES OF (𝒄 𝟎 ) 𝒑 IN 𝑪 𝒑 (𝑿, 𝑬)
We start with the following simple observation.Proposition 6.1.Let  be a locally convex space and  a Tychonoff space containing an infinite compact subset.Then,   (, ) contains a closed copy of the space ( 0 )  .
Proof.It is known that for such a space , the space   () contains a closed copy of the space ( 0 )  (see the proof of [22,Theorem 5] for  = ℝ).The thesis follows now from the fact that   () is complemented in   (, ) (see [31,Theorem I.4.4]).□ Let us now prove the main result of this section.The proof is inspired by the arguments presented in the proof of Theorem 1 in Domański and Drewnowski [10].Theorem 6.2.Let  be a barrelled locally convex space with the JNP and let  be a Tychonoff space containing an infinite compact set.Then,   (, ) contains a complemented copy of ( 0 )  .
Proof.Since  has the JNP, there is a weak * null sequence ( *  ) ∈ in  ′ , which is not convergent in the  * ( ′ , )topology.Taking into account that  is barrelled, this means that the sequence is not strongly convergent and hence we may choose a bounded sequence (  ) ∈ in  with ⟨ *  ,   ⟩ = 1 for every  ∈ .Let  ⊂  be an infinite compact set.Using the Tychonoff property, we may choose a sequence of continuous functions   ∶  → [0, 1] with disjoint supports and   (  ) = 1 for some   ∈ .
We We now check that these mappings are continuous.Let  be a continuous seminorm on  and  > 0 such that (  ) ≤  for all  (such  exists, since (  ) ∈ is a bounded sequence).For each  ∈ , since the functions   have disjoint supports, there is at most one   with    () ≠ 0. We pick this   if it exists and set   = 1 otherwise (of course, any other   works as well in this case), and obtain that for every (  ) ∈ ∈ ( 0 )  and  ∈  we have: we have that  is an isomorphic embedding and  is a surjection.Since  = , the mapping  is a projection onto the image of .□ Remark 6.3.Note the following comments concerning Theorem 6.2.
(1) If  is infinite discrete and  is a Banach space, then   (, ) ≈   , which does not contain any closed copies of ( 0 )  , as the latter is not complete.(2) The existence of an infinite compact subspace in  is not necessary for Theorem 6.2.Indeed, if  is a pseudocompact space with all compact subsets being finite and  is an infinite compact space, then   (, ()) ≈   (,   ()) ≈   ( × ), which, by [19, Proposition 2.10], contains a complemented copy of ( 0 )  .(3) If   () has the JNP for an infinite Tychonoff space , then   () is not barrelled (by the Buchwalter-Schmets theorem [8] and [19, Proposition 4.1]), so it does not satisfy the assumptions of Theorem 6.2, even though the space   (,   ()) still has the JNP for every space  (by Theorem 5.2).
As a corollary we immediately obtain the following   -variant of Domański-Drewnowski's Theorem 1.2.
Corollary 6.4.Let  be a Tychonoff space containing an infinite compact set and  a Fréchet space, which is not a Montel space.Then,   (, ) contains a complemented copy of ( 0 )  .
Proof.Since by [3, Theorem 2.3] and [5] a Fréchet space has the JNP if and only if it is not a Montel space, the claim is a special case of Theorem 6.2.□ Using the Closed Graph Theorem, one can easily derive a special case of Theorem 1.2 for Fréchet spaces   (, ).Thus, if  is compact and  infinite-dimensional Banach, we obtain Cembranos' and Freniche's Theorem 1.1.Corollary 6.5.Let  be a Tychonoff space containing an infinite compact subspace and let  be a Fréchet space, which is not Montel.If   (, ) is an infinite-dimensional Fréchet space, then   (, ) contains a complemented copy of the Banach space  0 .
Corollary 6.6.Let  and  be infinite Tychonoff spaces.Assume that  is compact.Then,   (, ()) contains a complemented copy of ℝ  or a complemented copy of ( 0 )  .
Proof.We need to consider the following three cases: Case 1.  is not pseudocompact.Then,   (, ()) contains a complemented copy of ℝ  (since   () contains such a copy and is complemented in   (, ())).
Case 2.  is pseudocompact and  contains an infinite compact subset.We apply Theorem 6.2 to get a complemented copy of ( 0 )  .
Case 3.  is pseudocompact and every compact set in  is finite.Since  ×  is pseudocompact, we apply [19, Proposition 2.10] to deduce that   (, ()) ≈   ( × ) contains a complemented copy of ( 0 )  .□ Recall that a Tychonoff space  is -compact if  is covered by an increasing sequence of compact subsets.We provide the following extension of [19,Proposition 2.10].Corollary 6.7.Let  and  be infinite locally compact -compact spaces.Then,   ( × ) contains a complemented copy of ℝ  or a complemented copy of the Banach space  0 .
Corollary 7.5.Let  be an infinite compact space.Then, the spaces   () and ()  are not isomorphic.
By Corollary 7.4, the space   () contains a complemented copy of ( 0 )  .Since  is an isomorphism, also the space ()  contains a complemented copy  of ( 0 )  .As both the weak topology   of () and the norm topology  ∞ of () have the same bounded sets, we have   ↾  =  ∞ ↾ , a contradiction (since ()  does not contain an infinitedimensional normed subspace).

A C K N O W L E D G M E N T S
The first and the third authors are supported by the Austrian Science Fund (FWF): I 4570-N.The second author is supported by the GAČR project 20-22230L RVO: 67985840.
[15,mapping  is well-defined, since the sequence (    ) ∈ is bounded and the functions   have disjoint supportsthis implies that the series converges uniformly on , and hence it defines a continuous function from  into .Note that the property of the supports of   s implies that this function only takes values of the form     ()  .Let us see why  is well-defined, too.Since  is continuous, the set () ⊂  is compact.Since on equicontinuous subsets of  ′ the weak * topology coincides with the topology of uniform convergence on precompact subsets of  (see, e.g.,[15, Proposition 3.9.8])and the sequence ( *  ) ∈ , being a weak * null sequence, is equicontinuous, we may conclude that ⟨ *  , ⟩ → 0 uniformly for  ∈ ().