A Lipschitz refinement of the Bebutov--Kakutani dynamical embedding theorem

We prove that an $\mathbb{R}$-action on a compact metric space embeds equivariantly in the space of one-Lipschitz functions $\mathbb{R}\to[0,1]$ if its fixed point set can be topologically embedded in the unit interval. This is a refinement of the classical Bebutov--Kakutani theorem (1968).


Introduction
The purpose of this short paper is to refine a classical theorem of Bebutov [Beb40] and Kakutani [Kak68] on dynamical systems. We call (X, T ) a flow if X is a compact metric space and 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 (1.1) 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: Problem 1.2. Is there a compact invariant subset of C(R) satisfying the same universality?
The purpose of this paper is to solve this problem affirmatively. Let L(R) be the set of maps ϕ : R → [0, 1] satisfying the one-Lipschitz condition: 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 Section 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 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].
Acknowledgement. This paper was written when the third named author stayed in the Einstein Institute of Mathematics in the Hebrew University of Jerusalem. He would like to thank the institute for its hospitality. Y.G. was partially supported by the Marie Curie grant PCIG12-GA-2012-334564. Y.G. and L.J. were partially supported by the National Science Center (Poland) grant 2013/08/A/ST1/00275. M.T. was supported by John Mung Program of Kyoto University. Lemma 2.1. Let f ∈ C (X, L[0, a]) and suppose there exists 0 < τ < 1 satisfying

Topological preparations
Then for any δ > 0 there exists g ∈ C (X, L[0, a]) satisfying Proof. We take 0 < b < c < a satisfying b = a − c < δ/4. We take an open covering We take a point p m ∈ U m for each m. We choose a natural number N satisfying We divide the interval [b, c] into (N − 1) intervals of length ∆: b = a 1 < a 2 < · · · < a N = c, a n+1 − a n = ∆ (∀1 ≤ n ≤ N − 1).
Let {h m } M m=1 be a partition of unity on X satisfying supp h m ⊂ U m for all m. For x ∈ X we define a piecewise linear function g(x) : [0, a] → [0, 1] as follows. (We set a 0 = 0 and a N +1 = a.) We apply to each term of the right-hand side the property (1) of This proves g(x) ∈ L[0, a].
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 |g( For every x ∈ X, the function g(x) : [0, a] → [0, 1] is a non-constant function because This proves the statement.
The equation It follows from the property (4) of u m that ε = 0 and h m (x) = h m (y) for all 1 ≤ m ≤ M. Then x, y ∈ U m for some m and hence d(x, y) ≤ diam U m < δ.

Proof of Theorem 1.3
Let (X, T ) be a flow. Set F = Fix(X, T ). We define F L = Fix (L(R)). Namely F L is the space of constant maps ϕ : R → [0, 1], which is homeomorphic to [0, 1]. Suppose there exists a topological embedding h : F → F L . We would like to show that there exists an equivariant embedding f : over all x ∈ X and s, t ∈ R with s = t.
Proof. Consider the map By the Tietze extension theorem, we can extend this function to a continuous map h 0 : ≤ 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, 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 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 |s − t| > a. Notice that if x ∈ F then H(x) = ∅. We denote by Int ([−a, a] · S) the interior of [−a, a] · S. We choose a closed neighborhood A 0 of p in S satisfying A 0 ⊂ Int ([−a, a] · S). 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 We set u(x)(t) = g(x)(t)−f 1 (x)(t) for 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 B 0 ⊂ Int ([−a, a] · S 1 ) and C 0 ⊂ Int ([−a, a] · S 2 ). We take a continuous functionq : X → [0, 1] satisfyingq = 1 on B 0 ∪ C 0 and suppq ⊂ Int ([−a, a] · S 1 ) ∪ Int ([−a, a] · S 2 ). 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 y).
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. 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 [Kak68,Aus88] 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).