A Lipschitz Refinement of the Bebutov–Kakutani Dynamical Embedding Theorem

We prove that an R\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathbb {R}$$\end{document}-action on a compact metric space embeds equivariantly in the space of one-Lipschitz functions R→[0,1]\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathbb {R}\rightarrow [0,1]$$\end{document} if its fixed point set can be topologically embedded in the unit interval. This is a refinement of the classical Bebutov–Kakutani theorem (1968).

is a continuous action of R. We define Fix(X, T ) (sometimes abbreviated to Fix(X )) as the set of x ∈ X satisfying T t x = x for all t ∈ R. We define C(R) as the space of continuous maps ϕ : R → [0, 1]. It is endowed with the topology of uniform convergence over compact subsets of R, namely the topology given by the distance ∞ n=1 2 −n max |t|≤n |ϕ(t) − ψ(t)|, (ϕ,ψ ∈ C(R)). (1.1) The group R continuously acts on it by the translation: A continuous map f : X → C(R) is called an embedding of a flow (X, T ) if f is an R-equivariant topological embedding. Bebutov [2] and Kakutani [4] found that the R-action on C(R) has the following remarkable "universality": The "only if" part is trivial because the set of fixed points of C(R) is homeomorphic to [0, 1]. So the main statement is the "if" part.
Although the Bebutov-Kakutani theorem is clearly a nice theorem, it has one drawback: The space C(R) is not compact (nor locally compact). So it is not a "flow" in the above definition. This poses the following problem: As in the case of the Bebutov-Kakutani theorem, the "only if" part is trivial because the fixed point set Fix(L(R)) is homeomorphic to [0, 1]. Since L(R) is compact, it is a more reasonable choice of such a "universal flow".
The proof of Theorem 1.3 is based on the techniques originally used in the proof of the Bebutov-Kakutani theorem (in particular, the idea of local section). A main new ingredient is the topological argument given in Sect. 2, which has some combinatorial flavor. Remark 1.4 Problem 1.2 asks us to find a universal flow smaller than C(R). If we look for a universal flow larger than C(R), then it is much easier to find an example. Let L ∞ (R) be the set of L ∞ -functions ϕ : R → [0, 1]. (We identify two functions which are equal to each other almost everywhere.) We consider the weak * topology on it. Namely a sequence {ϕ n } in L ∞ (R) converges to ϕ ∈ L ∞ (R) if for every L 1 -function ψ : Then L ∞ (R) is compact and metrizable by Banach-Alaoglu's theorem and the separability of the space of L 1 -functions, respectively. The group R acts continuously on it by translation. So it becomes a flow. Note that Fix (L ∞ (R)) is homeomorphic to [0, 1] and that the natural inclusion map C(R) ⊂ L ∞ (R) is an equivariant continuous injection. Then the Bebutov-Kakutani theorem implies the universality of L ∞ (R): A flow (X, T ) can be equivariantly embedded in L ∞ (R) if and only if Fix(X, T ) can be topologically embedded in [0, 1].

Topological Preparations
Let a be a positive number. We define L[0, a] as the space of maps ϕ : [0, a] → [0, 1] satisfying Let (X, d) be a compact metric space. We define C (X, L[0, a]) as the space of continuous maps f : X → L[0, a], which is endowed with the distance (2.1) Then for any δ > 0 there exists g ∈ C (X, L[0, a]) satisfying We take an open covering We take a point p m ∈ U m for each m. We choose a natural number N satisfying (1) | f ( p m )(a n ) − u m (a n )| < min (δ/4, (1 − τ )b/2) for all 1 ≤ m ≤ M and 1 ≤ n ≤ N .
We apply to each term of the right-hand side the property (1)

1) respectively. Then this is bounded by
Next we show |g(x)(a n ) − f (x)(a n )| < δ/2 for all 0 ≤ n ≤ N + 1. For n = 0, N + 1, this is trivial. For 1 ≤ n ≤ N , we can bound |g(x)(a n ) − f (x)(a n )| from above by Finally, let a n < t < a n+1 . We can bound For every x ∈ X , the function g(x) This proves the statement.
The equation over all x ∈ X and s, t ∈ R with s = t.

Lemma 3.1 The space C T,h (X, L(R)) is not empty. Moreover for any
Proof Consider the map By the Tietze extension theorem, we can extend this function to a continuous map ≤ min (1, δ) .
Lemma 3.2 Let p ∈ X \F. There exist a > 0 and a closed set S ⊂ X containing p such that the map

is a continuous injection whose image contains an open neighborhood of p in X . We call (a, S) a local section around p and denote the image of (3.1) by [−a, a] · S.
Proof We explain the proof for the convenience of readers. We can find c < 0 and a continuous function h : X → [0, 1] satisfying T c p / ∈ supp h and h = 1 on a neighborhood of p. We define f : X → R by We choose 0 < a < |c| and a closed neighborhood A of p satisfying It follows that f (T t x) = f (x)+t for any x ∈ A and |t| ≤ a.
Then (a, S) becomes a local section. Indeed if x, y ∈ S and s, t ∈ [−a, a] satisfy T s x = T t y, then s + f ( p) = f (T s x) = f (T t y) = t + f ( p) and hence s = t and x = y. Thus the map (3.1) is injective. We take 0 < b < a and an open neighborhood U of p satisfying The set (3.2) is an open neighborhood of p.

Lemma 3.3 For any point p ∈ X \F there exists a closed neighborhood A of p in X such that the set
is open and dense in the space C T,h (X, L(R)).
Proof Take a local section (a, S) around p. For x ∈ X we define H (x) ⊂ R (the set of hitting times) as the set of t ∈ R satisfying T t x ∈ S. Any two distinct s, t ∈ H (x) satisfy We define a closed neighborhood A of p in X by We choose a continuous function q : S → [0, 1] satisfying q = 1 on A 0 and supp q ⊂ Int ([−a, a] · S). The set G(A) defined in (3.3) is obviously open. So it is enough to prove that it is dense. Take f ∈ C T,h (X, L(R)) and 0 < δ < 1. By Lemma 3.1 we can find f 0 ∈ C T,h (X, L(R)) satisfying Lip( f 0 ) ≤ 1/2. We define f 1 ∈ C T,h (X, L(R)) by It follows Lip( f 1 ) ≤ 1 − δ/2 < 1. We apply Lemma 2.1 to the map Then we find g ∈ C (X, L[0, a]) satisfying x ∈ X and 0 ≤ t ≤ a. We define g 1 ∈ C T,h (X, L(R)) as follows: Let x ∈ X .
• For each s ∈ H (x), we set This satisfies for all x ∈ X and t ∈ R. If x ∈ A then there exists s ∈ [−a, a] with T s x ∈ A 0 and hence It follows from the property (3) of g that the function g 1 (x) is not constant. Thus g 1 ∈ G(A). Since f and δ are arbitrary, this proves that G(A) is dense in C T,h (X, L(R)).
Lemma 3.4 For any two distinct points p and q in X \F there exist closed neighborhoods B and C of p and q in X respectively such that the set is open and dense in C T,h (X, L(R)).
Proof Take local sections (a, S 1 ) and (a, S 2 ) around p and q respectively. We can assume that [−a, a] · S 1 and [−a, a] · S 2 are disjoint with each other. For x ∈ X we define H (x) as the set of t ∈ R satisfying T t x ∈ S 1 ∪ S 2 . We choose closed neighborhoods B 0 of p in S 1 and C 0 of q in S 2 respectively satisfying We take a continuous functionq : . We define closed neighborhoods B and C of p and q respectively by The set G(B, C) defined in (3.4) is open. We show that it is dense. Take f ∈ C T,h (X, L(R)) and 0 < δ < 1. We can assume that (3.5) We define f 1 ∈ C T,h (X, L(R)) exactly in the same way as in the proof of Lemma 3.3. It satisfies Lip( for all x ∈ X and t ∈ R. We apply Lemma 2.5 to the map Then we find g ∈ C (X, L[0, a]) satisfying (3) If x, y ∈ X and 0 ≤ ε ≤ a/2 satisfy then d(x, y) < δ.
It follows from the property (3) of g that d(T s 1 x, T s 2 y) < δ. Since δ < d(B 0 , C 0 ) ≤ d(T s 1 x, T s 2 y), this is a contradiction. Therefore g 1 (B) ∩ g 1 (C) = ∅. This proves the lemma. Now we can prove Theorem 1.3. Note that X and X × X are hereditarily Lindelöf (that means that every open cover of a subspace has a countable subcover). Using these facts and applying is dense and G δ in C T,h (X, L(R)). In particular it is not empty. Any element f in this set gives an embedding of the flow (X, T ) in L(R).
Remark 3.5 The proof of the Bebutov-Kakutani theorem in [1,4] used the idea of "constructing large derivative". It is possible to prove Theorem 1.3 by adapting this idea to the setting of one-Lipschitz functions. But this approach seems a bit tricky and less flexible than the proof given above. The above proof possibly has a wider applicability to different situations (e.g. other function spaces).