# Free Subspaces of Free Locally Convex Spaces

Authors: Saak S. Gabriyelyan; Sidney A. Morris
Journal: Journal of Function Spaces
 (2018)

---

## Abstract

IfX and Y are Tychonoff spaces, let L(X) and L(Y) be the free locally convex space over X and Y, respectively. For general X and Y, the question of whether L(X) can be embedded as a topological vector subspace of L(Y) is difficult. The best results in the literature are that if L(X) can be embedded as a topological vector subspace of L(I), where I=[0,1], then X is a countable-dimensional compact metrizable space. Further, if X is a finite-dimensional compact metrizable space, then L(X) can be embedded as a topological vector subspace of L(I). In this paper, it is proved that L(X) can be embedded in L(R) as a topological vector subspace if X is a disjoint union of a countable number of finite-dimensional locally compact separable metrizable spaces. This is the case if X=Rn,n∈N. It is also shown that if G and Q denote the Cantor space and the Hilbert cube IN, respectively, then (i) L(X) is embedded in L(G) if and only if X is a zero-dimensional metrizable compact space; (ii) L(X) is embedded in L(Q) if and only if Y is a metrizable compact space.


---

## Body

## 1. Introduction

For a Tychonoff spaceX, we denote by L(X), V(X), F(X), and A(X) the free locally convex space, the free topological vector space, the free topological group, and the free abelian topological group over X, respectively. These spaces and groups always exist and are essentially unique; see [1–5].LetX and Y be Tychonoff spaces. The questions of whether L(X) and V(X) can be embedded as a topological vector space in L(Y) and V(Y), respectively, and whether F(X) and A(X) can be embedded as a topological group in F(Y) and A(Y), respectively, have been studied for many years; see for example [6–8]. The case Y=I≔[0,1] was studied in [7] where the following result is obtained.Theorem 1 (see [7]).
IfK is a finite-dimensional metric compact space, then L(K) is embedded in L(I) as a topological vector space.The proof of this theorem in [7] is nontrivial and uses the deep and powerful Kolmogorov Superposition Theorem which answered Hilbert’s 13th Problem. In the last section of this paper, we give a simpler application of the Kolmogorov Superposition Theorem to obtain Theorem 1.In this paper we answer similar questions for whenI is replaced with the Cantor space G or the Hilbert cube Q=IN. We determine what Tychonoff spaces Y are such that L(Y) can be embedded as a topological vector subspace in L(G) and L(Q), respectively.It is well-known thatG is universal for all zero-dimensional compact metrizable spaces and Q is universal for all compact metrizable spaces. So it is straightforward to prove that if Z is a compact metrizable space, then L(Z) can be embedded as a topological vector subspace in L(Q), and if Z is a zero-dimensional compact metrizable space, then L(Z) can be embedded as a topological vector subspace of L(G). Of course both of these embeddings are the natural ones. But are there other Tychonoff spaces Z such that L(Z) can be embedded as a topological vector subspace of L(Q) or L(G) in a less natural way? A partial answer to this question was given in [8] where this was answered in the negative for compact spaces.Theorem 2 (see [8]).
IfZ is a compact Hausdorff space, then (i)
L(Z) is embedded as a topological vector subspace of L(Q) if and only if Z is metrizable;(ii)
L(Z) is embedded as a topological vector subspace of L(G) if and only if Z is metrizable and zero-dimensional.In Theorem11 of this paper we prove that the condition “compact” is unnecessary, more precisely, if L(Z) is embedded in L(Q) or L(G), then Z must to be compact.Related questions are as follows: letY be a compact metrizable space and X a subspace of Y. What are the conditions on X under which each of the following is true: (i) L(X) can be embedded as a topological vector subspace of L(Y); (ii) V(X) can be embedded as a topological vector subspace of V(Y); F(X) can be embedded as a topological subgroup of F(Y); A(X) can be embedded as a topological subgroup of A(Y). For the special case Y=I complete answers to this question are given in [7, 9].Theorem 3.
For a subspaceX of I the following are equivalent: (i)
([7]) A(X) is embedded into A(I) as a topological subgroup;(ii)
([9]) V(X) is embedded into V(I) as a topological vector subspace;(iii)
X is locally compact.
In particular,A(0,1) can be embedded in A(I) as a topological subgroup and V(0,1) can be embedded in V(I) as a topological vector subspace.For the case of free locally convex spaces on compact metrizable spacesY, our Theorem 8 gives a complete description of those subspaces X of Y with the property that L(X) can be embedded as a topological vector subspace of L(Y). Our proof is based on significant generalizations of some results in [7].


## 2. The Free Locally Convex Spaces on the Cantor Space and the Hilbert Cube

Let us recall (see [1]) that thefree locally convex spaceL(X) on a Tychonoff space X is a pair consisting of a locally convex space L(X) and a continuous map i:X→L(X) such that every continuous map f from X to any locally convex space E gives rise to a unique continuous linear operator f¯:L(X)→E with f=f¯∘i. The free locally convex space L(X) always exists and is essentially unique. The set X forms a Hamel basis for L(X), and the map i is a topological embedding [3, 10].It is well-known (see [3]) that the dual space of L(X) is canonically isomorphic to the space C(X) of all continuous real-valued functions on X. The space C(X) endowed with the pointwise topology or with the compact-open topology is denoted by Cp(X) and Ck(X), respectively. Denote by Lp(X) the space L(X) endowed with the weak topology. Then the spaces Lp(X) and Cp(X) are in duality.Firstly we note the following useful necessary condition for the existence of an embedding for free locally convex spaces which easily follows from general facts of locally convex space theory.Proposition 4.
LetX and Y be Tychonoff spaces. (i)
If the spaceL(X) can be embedded into L(Y) as a locally convex subspace, then also the space Lp(X) can be embedded into Lp(Y) as a locally convex subspace.(ii)
The spaceLp(X) can be embedded into Lp(Y) as a locally convex subspace if and only if Cp(X) is an image of Cp(Y) under a linear continuous surjection.
Consequently, ifL(X) can be embedded into L(Y) as a locally convex subspace, then there is a linear continuous surjection T from Cp(Y) onto Cp(X).Proof.
(i) follows from Theorem8.12.2 of [11], and (ii) follows from Theorems 8.10.5 and 8.13.2 of [11] applying to X≔Cp(Y) and Y≔Cp(X).Note that the converse assertion in (i) of Proposition4 holds; for example, if L(X) is barrelled, see Corollary 11.3.7 of [11]. However, the condition on L(X) to be a barrelled space is very restrictive: the space L(X) is barrelled if and only if X is discrete; see Theorem 6.4 of [4].LetX and Y be Tychonoff spaces. To obtain sufficient conditions on X and Y for which L(X) is embedded into L(Y), by Proposition 4, it is sufficient to find conditions on X and Y under which the converse assertion in (i) of Proposition 4 holds. The case when Y is compact (or, more generally, Y is a kω-space) is the most natural and important.Recall that a Tychonoff spaceX is calledDieudonné complete if its topology is induced by a complete uniformity. Every Lindelöf space (in particular, every kω-space) and every metrizable space is Dieudonné complete. We shall use the following result of Arhangelskii.Theorem 5 (see [12]).
LetX and Y be Dieudonné complete spaces. If a linear map h:Cp(X)→Cp(Y) is continuous, then h is continuous as a map Ck(X)→Ck(Y).Recall that a mapf:X→Y is calledcompact-covering if for every compact subset C of Y there is a compact subset K of X such that f(K)=C.Proposition 6.
LetX be a Dieudonné complete space and let Y be a kω-space. If T:Cp(Y)→Cp(X) is a linear continuous surjection, then X is also a kω-space and T:Ck(Y)→Ck(X) is a quotient compact-covering map. If Y is additionally a compact space, then X is a compact space. If Y is compact metrizable, then so is X.Proof.
Since anykω-space is Dieudonné complete, the map T is also continuous with respect to the compact-open topologies on both spaces by Theorem 5. As Ck(Y) is a Fréchet space, the Open Mapping Theorem [11, Theorem 14.4.6] implies that T:Ck(Y)→Ck(X) is open, and hence it is a quotient map. Therefore Ck(X) is also a Fréchet space, so X is a kω-space. Since Ck(Y) and Ck(X) are completely metrizable and T is open, the map T is compact-covering by Theorem 1.2 of [13]. If, in addition, Y is a (metrizable) compact space, then Ck(Y) is a (separable) Banach space, so is Ck(X). Thus X is also a (metrizable) compact space.The equivalence (i)⇔(ii) in the next theorem generalizes Lemma 3.2 of [7] and has an essentially simpler proof.Proposition 7.
LetX be a Dieudonné complete space and let Y be a kω-space (a compact space or a metrizable compact space). Then the following assertions are equivalent: (i)
L(X) can be embedded into L(Y) as a locally convex subspace.(ii)
Lp(X) can be embedded into Lp(Y) as a locally convex subspace.(iii)
There is a linear continuous surjection ofCp(Y) onto Cp(X).
If (i)–(iii) hold, the spaceX is a kω-space (a compact space or a metrizable compact space, resp.).Proof.
(i)⇒(ii) and (ii)⇔(iii) follow from Proposition 4. Let us prove (ii)⇒(i).
Leth:Lp(X)→Lp(Y) be an embedding of locally convex spaces. Then the dual linear map h∗:Cp(Y)→Cp(X) is a continuous surjection by Theorems 8.10.5 and 8.11.3 of [11]. By Proposition 6, X is a kω-space (a compact space or a metrizable compact space, resp.) and h∗:Ck(Y)→Ck(X) is a quotient compact-covering map. Hence the dual continuous map h∗∗ of h∗ from the space Ck(Ck(X)) to Ck(Ck(Y)) is an embedding of Ck(Ck(X)) into Ck(Ck(Y)) as a locally convex subspace. Observe that since X and Y are kω-spaces, L(X) and L(Y) are locally convex subspaces of Ck(Ck(X)) and Ck(Ck(Y)), respectively, by [10, 14] (for a more general assertion, see Theorem 1.2 of [15]). This observation and the fact that h coincides with the restriction of h∗∗ to L(X) imply that the map h is an embedding of L(X) into L(Y).We can now easily deduce the following main result from Proposition7 and Lemma 3.3 of [16].Theorem 8.
LetX be a subspace of a compact metrizable space K. Then L(X) can be embedded into L(K) as a locally convex subspace if and only if X is closed.Proof.
Assume thatL(X) embeds into L(K). Then X is compact by Proposition 7, and so X is closed. The converse assertion follows from Lemma 3.3 of [16].It follows immediately from Theorem8 that L(0,1) is not embedded in L(I). This fact contrasts with the facts, mentioned in Theorem 3, that A(0,1) can be embedded as a topological subgroup of A(I) and V(0,1) can be embedded as a topological vector subspace of V(I).It turns out that the existence of a linear continuous surjectionT from Cp(Y) onto Cp(X) in Propositions 4 and 7 is also a sufficiently strong condition as the following easy corollary of Uspenskiĭ’s theorems [17] shows.Theorem 9 (see [7]).
LetT:Cp(Y)→Cp(X) be a linear continuous surjection. If Y is a metrizable compact space, then so is X.Proposition4 and Theorem 9 immediately imply the following result.Corollary 10.
LetX be a Tychonoff space and let K be a compact metrizable space. If L(X) can be embedded into L(K) as a locally convex subspace, then X is compact and metrizable.Corollary10 and Theorem 2 imply the following surprising complete descriptions of Tychonoff spaces X for which L(X) is embedded in L(Q) and L(X) is embedded in L(G).Theorem 11.
For a Tychonoff spaceX the following assertions are equivalent: (i)
L(X) can be embedded into L(Q) as a locally convex subspace.(ii)
There is a linear continuous surjection ofCp(Q) onto Cp(X).(iii)
X is a compact metrizable space.Proof.
(i)⇒(ii) follows from Proposition 4, (ii)⇒(iii) follows from Theorem 9, and (iii)⇒(i) follows from Theorem 2.Theorem 12.
For a Tychonoff spaceX the following assertions are equivalent: (i)
L(X) can be embedded into L(G) as a locally convex subspace.(ii)
There is a linear continuous surjection ofCp(G) onto Cp(X).(iii)
X is a compact zero-dimensional metrizable space.Proof.
(i)⇒(ii) follows from Proposition 4. (ii)⇒(i): by Theorem 9, X is compact and metrizable and Proposition 7 applies.
(i)⇒(iii): If L(X) can be embedded into L(G), then X is compact and metrizable by Corollary 10. Therefore X is zero-dimensional by Theorem 2. (iii)⇒(i) follows from Theorem 2.Corollary 13.
LetK be a zero-dimensional metrizable compact space and X a Tychonoff space. If L(X) can be embedded into L(K) as a locally convex subspace, then X is a compact zero-dimensional metrizable space.Proof.
By Theorem12, L(K) is embedded into L(G) and Theorem 12 applies once again.Note that the locally compact pseudocompact spaceω1(=[0,ω1)) is not Dieudonné complete and its Dieudonné completion is the compact space ω1+1. By [18, Example 3.1.27]; the restriction continuous map R:Cp(ω1+1)→Cp(ω1) is bijective. However R is not open because the set(1)Rf∈Cω1+1:fω1≤1is not a neighborhood of zero in Cp(ω1).Question 14.
Do there exist a compact (nonmetrizable) spaceK and a non-Dieudonné complete subspace X of K such that L(X) cannot be embedded into L(K)? What can be said about the specific case: K=ω1+1 and X=ω1?


## 3. Embedding intoL(X) and the Kolmogorov Superposition Theorem

Now we consider an important case when a linear continuous surjection ofCp(Y) onto Cp(X) can be easily constructed using the following deep result of Kolmogorov [19] in the version of Lorentz [20, Theorem 11.1].Theorem 15 (Kolmogorov superposition theorem).
For everyn∈N there exist strictly increasing functions φ1,…,φ2n+1∈C(I) with values in I and constants λ1,…,λn∈(0,1] such that for every function f(x1,…,xn)∈C(In) there is a function g∈C[0,n] such that(2)f=∑q=12n+1g∑i=1nλiφqxi.Corollary 16.
LetK be a finite-dimensional compact metrizable space. Then there is a linear continuous surjective map T from Cp(I) onto Cp(K).Proof.
By the Embedding Theorem [21, Theorem 1.11.4], every finite-dimensional compact metrizable space K is embedded into Rn for some n∈N. Now the Tietze–Urysohn Theorem shows that to prove the corollary it is sufficient to construct a linear continuous surjective map T from Cp(I) onto Cp(In).
Define the homeomorphismh:[0,n]→I by h(x)≔x/n. Now, using (2) we define the operator T:Cp(I)→Cp(In) by(3)Tgx1,…,xn≔∑q=12n+1g∘h∑i=1nλiφqxi,g∈CI.It is clear that T is linear, and T is surjective by the Kolmogorov Theorem. To check that T is continuous, fix zj=(x1j,…,xnj),j=1,…,m, and ε>0. Denote by F the finite family of all points in I of the following form:(4)h∑i=1nλiφqxij,j=1,…,m,q=1,…,2n+1.Then, if g belongs to the standard open pointwise neighborhood {f∈C(I):|f(x)|<ε/(2n+1)∀x∈F} of zero function in Cp(I), we obtain(5)Tgzi≤∑q=12n+1g∘h∑i=1nλiφqxij<ε,j=1,…,m.Thus T is continuous.Remark 17.
Levin informed the authors that a linear continuous surjective mapT from Cp(I) onto Cp(K) can even be chosen to be open; see [22].We now observe the following proof of Theorem1.Proof.
Theorem1 follows immediately from Proposition 7 and Corollary 16.Now we consider anoncompact case, namely, when X=R. First we recall the following definition motivated by the Kolmogorov Superposition Theorem 15.Definition 18 (see [23]).
LetX be a Tychonoff space. A family Φ⊆C(X) is said to be basic for X if each f∈C(X) can be written as(6)f=∑q=1ngq∘Φq,for  someΦ1,…,Φn∈Φ,g1,…,gn∈CR,n∈N.We shall use the following noncompact version of the Kolmogorov Superposition Theorem.Theorem 19 (see [24]).
IfX is a finite-dimensional locally compact separable metrizable space, then X has a finite basic family.Analogously to Corollary16 we easily obtain the following result.Corollary 20.
LetX be a finite-dimensional locally compact separable metrizable space. Then there is an n∈N such that Cp(X) is the image of (Cp(R))n under a continuous linear surjective map T.Proof.
By Theorem19 there exists a basic family {Φn,…,Φn} for X. Then the map(7)T:CpRn⟶CpX,Tg1,…,gn≔∑i=1ngi∘Φi,is linear and surjective. Let us check that T is continuous.
Fix a standard neighborhood{f∈C(X):|f(x)|<ε∀x∈F} of zero in Cp(X), where F⊆X is finite and ε>0. Denote by F~ the finite set of all points in R of the form Φi(z), where z∈F and 1≤i≤n. If g1,…,gn∈{h∈C(R):|h(z)|<ε/n∀z∈F~}, we obtain(8)Tg1,…,gnz=∑i=1ngi∘Φiz<ε,∀z∈F.Therefore T is continuous.Proposition 21.
LetX=⨆n∈NXn be a disjoint union of finite-dimensional locally compact separable metrizable spaces. Then there is a continuous linear surjective map T from Cp(R) onto Cp(X).Proof.
By Corollary20, for every n∈N, we can choose an sn∈N and a continuous linear surjective map Tn:(Cp(R))sn→Cp(Xn). Then the map(9)T:CpRN=∏n=1∞CpRsn⟶∏n=1∞CpXn=CpX,Tfn≔Tnfn,is continuous, linear, and surjective. It is known (see [25, Section 177]) that there is a linear homeomorphism S from Cp(R) onto (Cp(R))N. Thus the map T∘S is as desired.Now Propositions7 and 21 immediately imply the main result of this section.Theorem 22.
IfX=⨆n∈NXn is a disjoint union of finite-dimensional locally compact separable metrizable spaces, then L(X) can be embedded into L(R) as a locally convex subspace. In particular, for every n∈N, L(Rn) can be embedded into L(R) as a locally convex subspace.We conclude by noting that Theorem22 implies that if X is the countable-dimensional locally compact separable metrizable space ⨆n∈NRn, then L(X) can be embedded as a locally convex subspace of L(R).



---

*Source: 2924863-2018-01-24.xml*