Sandwich results for periodicity and conjugacy

Let P be a nonconstant selfmap of a set M\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathcal {M}$$\end{document}. A sandwich-type theorem for generalized sub-P-periodic functions defined on M\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\mathcal {M}$$\end{document} with values in a reflexive Banach space is proved. In particular, given functions f,g:M→R\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$f,g:\mathcal {M}\rightarrow \mathbb {R}$$\end{document}, we obtain necessary and sufficient conditions for the existence of a generalized P-periodic function F:M→R\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$F:\mathcal {M}\rightarrow \mathbb {R}$$\end{document} such that f≤F≤g\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$f\le F\le g$$\end{document}. The formula for F is given and its Lipschitz constant is discussed. Moreover the solvability of the functional equation f∘p=r∘f\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$f\circ p= r\circ f$$\end{document} with the help of a new sandwich method, is considered.


Introduction
In a recent paper Izumi et al. [5] proved the following sandwich (separation) type result: functions f, g : R → R satisfy the inequalities for some p ∈ R\{0} if, and only if, there is a p-periodic function F : R → R separating f and g, that is such that F (x + p) = F (x) for all x ∈ R and f ≤ F ≤ g. In Sect. 2 of this paper, applying Gajda's theorem on the existence of a translative invariant mean in the Banach space of bounded vector valued functions [4], we prove far reaching generalizations of this result. Let M be a nonempty set, P : M → M a nonconstant mapping, and X a reflexive Banach space. Assuming that a function f : M → X and a set-valued function Γ : M → 2 X with bounded closed and convex values are such that (ii) for every ϕ ∈ B(S, X) and closed convex bounded subset V of X, if ϕ(S) ⊆ V , then M (ϕ) ∈ V .
The linear operator M is called an invariant mean.
In the sequel M is a nonempty set, P : M → M is a nonconstant function, X is a reflexive Banach space, and Γ : M → 2 X a set-valued function such that, for every x, the value Γ(x) X is nonempty and convex.
Putting P k (x) in place of x in the above equality we get F (P k (x)) ∈ conv{f (P n (P k (x))) : n ∈ N 0 }. On the other hand, we also have F (P k (x)) = F (x), hence F (x) ∈ conv{f (P n+k (x))) : n ∈ N 0 }. Since k ∈ N 0 is arbitrarily chosen, we conclude that F (x) ∈ m∈N conv{f (P n (x))) : n > m}. Theorem 3. Let M be a nonempty set and assume that P : M → M is not a constant mapping. Suppose that h : M → R is a P -super-periodic function, i.e. h(P (·)) ≥ h(·), and g : M → R is a P -sub-periodic, i.e. g(P (·)) ≤ g(·). Then the functions h, g : M → R satisfy the condition if and only if there is a P -periodic function F : . By the assumptions we have h(P n (x)) ≥ h(x) and g(P n (x)) ≤ g(x). It follows that . Now applying Theorem 2 we arrive at the desired assertion. The converse implication has a trivial verification.
The above result can be strengthened (in some circumstances) as follows.
if and only if there is a P -periodic function F : . Applying Theorem 3, we complete the proof. The converse result is immediate.
The next result shows that the Lipschitz property of a P -Γ-periodic function can be inherited by some P -periodic function.  Proof. Given y ∈ M we define a function ϕ y : Now Theorem 1 implies that there is an invariant mean M : Hence, by (9) and Finally, we show that f is a Lipschitz function. Recall that the function M is a continuous operator with M = 1. From the definition of F , we have The Proof of Theorem 5 is complete.

Sandwich with functional equation f • p = r • f
In the previous section we saw that the problem of determining periodic (resp. p-super-periodic) functions leads to the functional equation f (p(x)) = f (x) (resp. the functional inequality f (p(x)) ≥ f (x)). So, it becomes quite natural to consider the functional equation f (p(x)) = r(f (x)) (resp. the functional inequality f (p(x)) ≥ r(f (x))).
The results presented in this section are of the type as in the previous section, but focused on the functional equation f (p(x)) = r(f (x)) instead of periodic functions. First we prove a sandwich type result for the functional equation f (p(x)) = g(f (x)).

Theorem 6.
Let I ⊆ R be a compact interval. Assume that functions p : I → I and r : I → I satisfy the following conditions: p is invertible and If there exist continuous functions f, g : I → I such that and for some h 0 : for all x, y ∈ I, then there is a continuous function h : and satisfying the equality Proof. Let C (I) be the Banach space of all continuous functions ϕ : I → R with the supremum norm ϕ ∞ := sup {|ϕ (x)| : x ∈ I}, and denote by F the family of all continuous functions ϕ : and By the assumption, F is nonempty. Define the mapping T : F → C (I) by Then, for every ϕ and x ∈ I, applying in turn: inequality (14) together with the invertibility of p; the monotonicity of r and the inequality in (17); again the monotonicity of r and the second inequality in (17); the inequality in (15), we obtain which shows that f ≤ T (ϕ) ≤ g.
Moreover, using in turn: the definition of T ; inequality (11); inequality (10); and (12), we get for arbitrary ϕ ∈ F and for all x, y ∈ I, which shows that, T (ϕ) satisfies inequality (18) for every ϕ ∈ F. This proves that T maps F into itself. Clearly, F is a closed subset of C (I). It follows from the continuity of γ at 0 and γ (0) = 0, and the Arzela Theorem that F is a compact subset of C (I). Moreover, one should be able to verify with little effort that F is convex. Now we show that T is continuous. Take ϕ n ∈ F, n ∈ N; ϕ ∈ F such lim n→∞ ϕ n = ϕ. In turns making use the definition of the norm · ∞ ; the bijectivity of p; inequality (11); the monotonicity of β and the continuity of the involved functions on the compact interval, we have Next we would like to present an application of the above theorem to Hölder functions. For a number q ∈ (0, 1], by H q (I) denote the space of Hölder continuous functions ϕ : I → R, that is such that for some C ≥ 0. As an immediate consequence of the above result we obtain: Corollary 7. Let I ⊆ R, p, r, f, g, h 0 satisfy the conditions of Theorem 6 with α (t) = α (1) t, β (t) = β (1) t and γ (t) = γ (1) t q , for some q ∈ (0, 1]. If then there exists a function h ∈ H q (I) such that and satisfying the equality We conclude this work by proving two results concerning the functional equation αf (p(x)) = g(f (x)) and the functional inequality αf (p n (x)) ≥ g n (f (x)).

Theorem 8.
Let M ⊆ (R n , · ) be a compact subset with intM = ∅, and suppose that X, W, Z are nontrivial Banach spaces such that X = W + Z and W ∩ Z = {0}, i.e. X is the direct sum of W and Z. Assume that p : M → M AEM is continuous. Suppose that g : X → X is a continuous surjection such that the restriction g| W : W → X is a surjective isometry (not necessarily linear), and Z ⊆ {x ∈ X : g(x) = 0}. Then for every α ∈ (0, 1) there exists a continuous function f : M → X such that and f is not a constant function. Moreover, if dim Z ≥ n, then the solution can be chosen in such a way that f is injective.
Proof. Let N : R n → Z be a continuous linear mapping with N ≤ 1. The operator N can be chosen in such a way that N | M is not constant. Since N (M) is compact and N (M) ⊆ Z ⊆ X, without loss of generality, it may be assumed that the set M is contained in the closed unit ball B X := {y ∈ X : y ≤ 1}. By C(M, X) we denote the space of continuous X-valued functions with the supremum norm · ∞ , and set B C := {f ∈ C(M, X) : f ∞ ≤ 1}.
Put g 1 := (g| W ) −1 : X → W and define a mapping ϕ : It is easy to verify that ϕ is a contraction. By the Banach Fixed-Point Theorem, there is a function Since g| W : W → X is a surjective isometry, obviously, so is g 1 : X → W. The Mazur-Ulam Theorem implies that g 1 is an affine mapping. Since g • g 1 = Id X , we have whence g • f α = αf α • p. Now, we prove that f α is not constant. Assume, contrary to our claim, that f α (·) = c for some c ∈ X. It follows that (g 1 • f α • p)(·) = g 1 (c), whence (1 − α)N | M (·) = c − αg 1 (c). Thus the function N | M : M → Z would be constant, which is a contradiction. Now assume that dim Z ≥ n. Then there is a continuous injective linear operator N : R n → Z. Repeating the above reasoning we obtain the equality Note that X = W ⊕ Z, g 1 (X) ⊆ W and N (x), N(y) ∈ Z. So the uniqueness of the decomposition implies that N (x) = N (y). Since N is injective, x = y.
Bearing the Proof of Theorem 8 in mind, we can formulate and prove another result. More precisely, now we want to study the existence of solutions f of the functional inequality f (p(x)) ≥ g(f (x)) .
The mapping ϕ is well-defined. Indeed, since the range of of f •p is bounded from below by b, the monotonicity of g 1 implies that ϕ(f )(t) = (1 − α)μ(t) + αg 1 (f (p(t))) ≥ αg 1 (b) ≥ b, so ϕ (f ) (t) ∈ [b, a] for every t ∈ M. By the Banach Fixed-Point Theorem we get for some f ∈ W . Since g is convex and g • μ = 0, it follows from (19) that g • f ≤ αf • p. Combining this inequality and again (19), we get Since g is increasing in [b, c] and convex, we obtain g • g • f ≤ α • f • p 2 , that g 2 • f ≤ α • f • p 2 . Now, taking into account that g n is increasing for n ≥ 2, we can prove, by induction, that and consider a similar mapping ϕ : W → W . The rest run similarly. Finally, in this case it is easy to notice that the fixed point f is a nondecreasing function.