iterative roots of piecewise monotonic functions of nonmonotonicity height not less than 2

Nonlinear Analysis 75 (2012) 286–303

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Nonlinear Analysis

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Iterative roots of piecewise monotonic functions of nonmonotonicityheight not less than 2Liu Liu a, Witold Jarczyk b,∗, Lin Li c, Weinian Zhang a

a Yangtze Center of Mathematics and Department of Mathematics, Sichuan University, Chengdu, Sichuan 610064, PR Chinab Faculty of Mathematics, Computer Science and Econometrics, University of Zielona Góra, Szafrana 4a, 65-516 Zielona Góra, Polandc Department of Mathematics, Physics and Information Engineering, Jiaxing University, Jiaxing, Zhejiang 314001, PR China

a r t i c l e i n f o

Article history:Received 9 May 2011Accepted 11 August 2011Communicated by Ravi Agarwal

MSC:39B1237E0526A18

Keywords:Continuous iterative rootPiecewise monotonic functionNonmonotonicity height

a b s t r a c t

It is known that any piecewisemonotonic function of nonmonotonicity height not less than2 has no continuous iterative roots of order greater than the number of forts of the function.So the following problem arises naturally: does such a function have an iterative root oforder n not greater than the number of forts?We consider the case that the number of fortsis equal to n, in which there appear possibly only two types T1 and T2 of iterative roots, i.e.,the roots strictly increasing on the interval stretched on all forts of the given function andthe roots strictly decreasing on such an interval, respectively. We characterize all type T1roots of order n and give a necessary condition for a root of order n to be of type T2. A fulldescription of type T2 is still an open question.

© 2011 Elsevier Ltd. All rights reserved.

1. Introduction

Given a set X , a self-mapping F of X and a number n ∈ N, i.e. a positive integer n, a function f : X → X is said to be aniterative root (or: fractional iterate) of order n of F if

f n(x) = F(x), x ∈ X, (1.1)

where f n denotes the nth iterate of f (cf. [1–4]). The problem of finding iterative roots is still alive in the literature, startingfrom the paper [5] by Ch. Babbage published almost two hundred years ago. Being a weak version of the problem ofembedding a function into a flow [6] or into a semiflow, it attracts great interest of specialists in the fields of functionalequations [1,3] iteration theory [7,2,8], dynamical systems [9–11], and information processing [12,13]. There have beenplenty of results (see [1,3,14–16], and references therein) on iterative roots; among them there are some classical results byBödewadt [17], Łojasiewicz [18], and Kuczma [19].

A big part of the researchwas devoted to roots ofmonotonic functions. In 1961, in the paper [19], Kuczma gave a completedescription of iterative roots of continuous strictly monotonic self-mappings of a given interval (see also [1,3]). In particular,each such function has infinitely many roots of a given order. Strictly increasing roots of continuous strictly increasingfunctions were determined by Bödewadt [17] in 1944 and a description of strictly decreasing roots of continuous strictlyincreasing functions is due to Haidukov [20].

Unexpectedly the situation changes completely if we resign the assumption that the monotonicity of F is strict. Forinstance consider the setFk of all continuous increasing self-mappings of a given compact interval, having a continuous root

∗ Corresponding author.E-mail address:[email protected] (W. Jarczyk).

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L. Liu et al. / Nonlinear Analysis 75 (2012) 286–303 287

of order k. As follows from the paper [21] by Humke and Laczkovich, for every integer k ≥ 2 the set Fk is an analytic andnon-Borel subset of the spaceC of all continuous self-mappings of the interval, endowedwith sup-norm. The papers [22,23]by Simon and Blokh, respectively, show that this set is small in C both from the category and measure-theoretical points ofview. Nonexistence of roots is typical also for some regular functions which was observed in the paper [24].

It seems that Jingzhong Zhang and Lu Yang in 1983 started to study roots of non-monotonic functions, more precisely:piecewise monotonic functions. Their method, based on the notion of characteristic interval for piecewise monotonicfunctions, was described in Chinese in [25] and then developed in [26], and also [27]. In the present paper we continue thatresearch. Another important article on iterative roots of piecewise monotonic functions was published in 1991 by Blokhet al. (cf. [28]).

2. Preliminaries

Let a, b ∈ R, a < b, and let F : [a, b] → R be a continuous function. A point c ∈ (a, b) is called a fort of F if F is strictlymonotonic in no neighborhood of c. The set of forts of F is denoted by S(F). We accept also the convention that the set offorts of a function defined on a singleton is empty. The function F is said to be piecewise monotonic if it has only finitelymanyforts. By PM(a, b) we denote the set of all piecewise monotonic self-mappings of [a, b].

Lemma 2.1. Let F : [a, b] → R be a continuous function. Then F is strictly monotonic in each connected component of the set[a, b] \ S(F) and every point of (a, b) where F takes a local extremum is a fort of F . In particular, if S(F) = ∅ then F is strictlymonotonic. Moreover, F takes a proper local extremum at every isolated fort.

Proof. Take any connected component I of [a, b] \ S(F). Then every point of I has a neighborhood in which F is strictlymonotonic. Thus, as I is connected, F is strictly monotonic in I . The second part of the first assertion is obvious.

Now let c ∈ (a, b) be an isolated fort of F . Choose a positive ε such that (c − ε, c), (c, c + ε) ⊂ [a, b] \ S(F). By the firstassertion F is strictly monotonic in each of the intervals (c − ε, c), (c, c + ε). Since F is continuous at c and c ∈ S(F), thekind of monotonicity of F in these intervals changes. Consequently, F takes a proper local extremum at c. �

Example 2.2. Clearly 0 is a fort of the function F : [−1/π, 1/π ] → R given by

F(x) =

x sin1x, if x ∈ [−1/π, 1/π ] \ {0},

0, if x = 0,

but F does not take any extremum at 0.

The next lemma describes the set of forts of the composition of two functions.

Lemma 2.3. Let p, q ∈ R be such that p < q and F([a, b]) ⊂ [p, q] and let F : [a, b] → R and G : [p, q] → R be continuousfunctions. Then

S(G ◦ F) = S(F) ∪ {c ∈ (a, b) : F(c) ∈ S(G)}.

In particular, the function G ◦ F is piecewise monotonic if and only if so are the functions F and G|F([a,b]).

Proof. Take any c ∈ S(G ◦ F) and assume that c ∈ S(F). Then F is strictly monotonic in a neighborhood W ⊂ [a, b] of c. Inparticular, F(c) ∈ (p, q). Let V ⊂ [p, q] be an arbitrary neighborhood of F(c) and choose a neighborhood U ⊂ W of c suchthat F(U) ⊂ V . Since c ∈ S(G◦ F) the function G◦ F is not strictly monotonic in U , that is G is not strictly monotonic in F(U),and thus in V as F(U) is a subset of V and is not a singleton. Consequently, F(c) ∈ S(G).

Conversely: if c ∈ (a, b) \ S(G ◦ F) then G ◦ F is strictly monotonic in an open interval I ⊂ [a, b] containing c , andthus F is one-to-one in I , that is strictly monotonic there. This proves that S(F) ⊂ S(G ◦ F). Now let c ∈ (a, b) \ S(F) besuch that F(c) ∈ S(G) and take any neighborhood U ⊂ [a, b] of c . Since c ∈ S(F) there is a neighborhood V ⊂ U of csuch that F is strictly monotonic in V . Then F(V ) is a neighborhood of F(c), and thus G is not strictly monotonic in F(V )as F(c) ∈ S(G). In other words, G ◦ F is not strictly monotonic in V and so is in U , and thus c ∈ S(G ◦ F). Consequently,{c ∈ (a, b) : F(c) ∈ S(G)} \ S(F) ⊂ S(G ◦ F). �

By Lemma 2.3, every continuous iterative root of a piecewise monotonic (strictly monotonic) self-mapping of [a, b] isalso piecewise monotonic (strictly monotonic).

Corollary 2.4. Let p, q ∈ R be such that p < q and F([a, b]) ⊂ [p, q], and let F : [a, b] → R and G : [p, q] → R be continuousfunctions. Then

S(F) ⊂ S(G ◦ F).

If G|F([a,b]) is strictly monotonic, then

S(G ◦ F) = S(F). (2.1)

If F is piecewise monotonic and (2.1) holds, then G|F([a,b]) is strictly monotonic.

288 L. Liu et al. / Nonlinear Analysis 75 (2012) 286–303

Proof. If G|F([a,b]) is strictly monotonic, then S(G|F([a,b])) = ∅ and (2.1) follows by Lemma 2.3.Now assume that (2.1) holds and, in addition, F is piecewise monotonic. Suppose that S(G|F([a,b])) = ∅ and let c ∈

S(G|F([a,b])), then, on account of Lemma 2.3 and (2.1), we have F−1({c}) ∩ (a, b) ⊂ S(F). Therefore, according to Lemma 2.1,the function F takes only extreme values at the points of F−1({c}). Moreover, as S(F) is finite, the set F−1({c}) is also finite.Consequently, the local extrema at consecutive points of F−1({c}) are of the same type: all are minimas or all are maximas,and thus

either F(x) ≤ c, x ∈ [a, b], or F(x) ≥ c, x ∈ [a, b].

However, this is impossible as c ∈ intF([a, b]). Consequently, S(G|F([a,b])) = ∅, i.e. G|F([a,b]) is strictly monotonic. �

Corollary 2.5. Let F : [a, b] → [a, b] be a continuous function. Then

S(F k) ⊂ S(F k+1)

for every k ∈ N ∪ {0}. If, in addition, F ∈ PM(a, b) and there is a k ∈ N ∪ {0} such that

S(F k) = S(F k+1) (2.2)

then

S(F k) = S(F k+i)

for every i ∈ N.

Proof. Let k ∈ N ∪ {0}. The first assertion follows immediately from Corollary 2.4. Now assume (2.2). Then, again byCorollary 2.4, the function F is strictly monotonic in F k([a, b]) and so is also in F k+i([a, b]) for every k ∈ N. To completethe proof it is enough to use Corollary 2.4 once more. �

Given a function F ∈ PM(a, b)we define the nonmonotonicity height (or simply height)H(F) of F as the least k ∈ N∪{0}satisfying (2.2) if such a k exists and ∞ otherwise.

Corollary 2.6. Let F ∈ PM(a, b) and k ∈ N∪{0}. Then H(F) ≤ k if and only if F is strictly monotonic in the interval F k([a, b]).In particular, H(F) = 0 if and only if F is strictly monotonic.

Proof. The claim follows directly from the second assertion of Corollary 2.4. �

Remark 2.7. It follows from Corollary 2.5 that

S(f ) ⊂ S(f n), n ∈ N,

for each continuous function f : [a, b] → [a, b]. Therefore every continuous iterative root of a piecewisemonotonic (strictlymonotonic) self-mapping of [a, b] is also piecewise monotonic (strictly monotonic).

Corollary 2.8. If F ∈ PM(a, b) then

S(F k) = S(FH(F))

for every integer k ≥ H(F).

Example 2.9. For the classical hat function F : [0, 1] → [0, 1], given by

F(x) = min{2x, 2 − 2x},

we have

S(F k) =

12k

, . . . ,2k

− 12k

, k ∈ N,

whence H(F) = ∞.

The main difficulties in finding iterative roots of piecewise monotonic functions come from the sharply increasingnumber of forts under iteration. Fortunately that process stabilizes ifH(F) is finite. AsH(F) = 0 if and only if F is monotonic,we have a full description of iterative roots of F (see [19], also [1,3]). When H(F) = 1 the problem can be reduced to amonotone case by the method of the characteristic interval (see [26,25], also [27]).

In the case H(F) ≥ 2 it follows from Theorem 1 in [25] (or Theorem 1 in [26]) that F has no continuous iterative rootsof order n > #S(F), so a natural problem, as raised in [25], also [26], is: Does F ∈ PM(a, b) with H(F) ≥ 2 have aniterative root of order n ≤ #S(F)? In the case #S(F) = 2 this problem has been solved in [29,30] although it is difficult tobe observed when reading these papers, as the desired result is formulated in a little bit sophisticated, not quite immediate,


Fig. 1. f ∈ T1 with S(f ) = {c1}.

Fig. 2. f ∈ T1 with S(f ) = {cn}.

form. Moreover, both papers are written in Chinese, and thus are not available for a wide audience. When #S(F) ≥ 3 theproblem was discussed also in [31] for a special class of piecewise monotonic functions.

In this paper we generally consider the case H(F) ≥ 2 and look for continuous iterative roots of order n = #S(F) ≥ 3,although most of the results, in particular Theorem 4.1, fully describing the type T1 of iterative roots, i.e., the roots strictlyincreasing on the interval stretched on all forts of the given function F , admit also the case n = 2.

3. Properties of roots

Let F ∈ PM(a, b). By N(F) we denote the number of forts of F :

N(F) = #S(F).

It follows from Corollaries 2.5 and 2.8 that

0 = NF 0

≤ NF k

≤ NF k+1 , k ∈ N,

and NF k

= N

FH(F)

for every integer k ≥ H(F).

We are interested in continuous iterative roots of F of order nwhich is equal to N(F), so assume that n ≥ 2 and write

S(F) = {c1, . . . , cn},

where a < c1 < · · · < cn < b.The following lemma yields a necessary condition for a continuous self-mapping of [a, b] to be a root of order n of

F ∈ PM(a, b).

Lemma 3.1. Let F ∈ PM(a, b) with H(F) ≥ 2 and assume that N(F) ≥ 2. If f ∈ PM(a, b) is an iterative root of ordern = N(F) of F , then N(f ) = 1 and there is a permutation (l1, . . . , ln) of {1, . . . , n} such that

f (clk) = clk−1 , k = 2, . . . , n, (3.1)

and, S(f ) = {cl1}. Moreover, one of the following cases holds:T1. Either S(f ) = {c1}, the function f reaches the minimum value at c1 and f (c1) < c1 (see Fig. 1), or S(f ) = {cn}, the functionf reaches the maximum value at cn and f (cn) > cn (see Fig. 2);T2. Either S(f ) = {c1}, the function f reaches the maximum value at c1 and f (c1) > c1 (see Fig. 3), or S(f ) = {cn}, the functionf reaches the minimum value at cn and f (cn) < cn (see Fig. 4).


Fig. 3. f ∈ T2 with S(f ) = {c1}.

Fig. 4. f ∈ T2 with S(f ) = {cn}.

From Lemma 3.1 we see that iterative roots of type T1 are increasing on the interval stretched on all forts of the givenfunction, i.e., the interval [c1, cn], and the roots of type T2 are decreasing on that interval.

Proof of Lemma 3.1. It follows from Corollary 2.5 that N(f ) ≤ N(f 2) ≤ · · · ≤ N(f n) = N(F) = n. If N(f k) = N(f k+1) for ak ∈ {1, . . . , n − 1}, then, using Corollary 2.5 again, we would have N(f k) = · · · = N(f n) = N(F) = · · · = N(f 2n) = N(F 2),i.e. N(F 2) = N(F), whence H(F) ≤ 1, contrary to the assumption. Thus N(f ) < N(f 2) < · · · < N(f n) = n, that is

N(f k) = k, k = 1, . . . , n. (3.2)

In particular, N(f ) = 1.Put

S1 = S(f ), Sk = {x ∈ (a, b) : f (x) ∈ S(f k−1)} \ S(f k−1), k = 2, . . . , n. (3.3)

It follows from Lemma 2.3 that for every k ∈ {2, . . . , n} we have

S(f k) = S(f ) ∪ {x ∈ (a, b) : f (x) ∈ S(f k−1)},

whence, as S(f ) ⊂ S(f k−1) by Corollary 2.5,

S(f k) \ S(f k−1) = S(f ) ∪ {x ∈ (a, b) : f (x) ∈ S(f k−1)} \ S(f k−1)

= {x ∈ (a, b) : f (x) ∈ S(f k−1)} \ S(f k−1) = Sk.

Therefore,

S(f k) = S(f k−1) ∪ Sk, k = 1, . . . , n, (3.4)

and thus, according to (3.2) and Corollary 2.5, the sets Sk are pairwise disjoint singletons summing up to S(f n) = S(F). Thismeans that there is a permutation (k1, . . . , kn) of {1, . . . , n} such that

{ci} = Ski , i = 1, . . . , n.

Let (l1, . . . , ln) be its inverse permutation. Then

Sk = {clk}, k = 1, . . . , n.


We prove (3.1). Take any k ∈ {2, . . . , n}. Then clk ∈ Sk, whence, by (3.3), f (clk) ∈ S(f k−1) and clk ∈ S(f k−1). By virtue ofCorollary 2.5 we have clk ∈ S(f k−2), so if f (clk) ∈ S(f k−2) then (3.3) would yield clk ∈ Sk−1, contrary to the disjointness ofSk−1 and Sk. Therefore f (clk) ∈ S(f k−1) \ S(f k−2), that is, according to (3.4) and Corollary 2.5, we come to f (clk) ∈ Sk−1, i.e.f (clk) = clk−1 .

Now we claim that

either S(f ) = {c1}, or S(f ) = {cn}. (3.5)

Suppose, on the contrary, that S(f ) = {ci} with some i ∈ {2, . . . , n − 1}. Then ki = 1, and thus k1 ≥ 2. Depending onwhether f takes the maximum or minimum value at ci, only the following cases are possible: either

f (cn) ≤ f (c1) < f (ci) or f (ci) < f (c1) ≤ f (cn),

or

f (c1) < f (cn) < f (ci) or f (ci) < f (cn) < f (c1).

Without loss of generality we may assume the first possibility. Then, by the continuity of f , there is a ξ ∈ (ci, cn] such thatf (ξ) = f (c1). Since c1 ∈ Sk1 , we have f (c1) ∈ S(f k1−1), and thus also f (ξ) ∈ S(f k1−1). Hence either ξ ∈ Sk1 , or ξ ∈ S(f k1−1).The first possibility cannot hold as Sk1 = {c1} and ξ = c1. Therefore ξ ∈ S(f k1−1) = S1 ∪ · · · ∪ Sk1−1, that is ξ ∈ Sj for somej ∈ {1, . . . , k1 − 1}. Since c1 ∈ Sk1 , we have c1 ∈ S(f k1−1), whence Corollary 2.5 yields c1 ∈ S(f j−1). On the other hand thecondition ξ ∈ Sj implies f (c1) = f (ξ) ∈ S(f j−1), and thus c1 ∈ Sj which is impossible as c1 ∈ Sk1 and j = k1. This proves(3.5).

Consider the case where f takes the maximum value at cl, where l ∈ {1, n}. We will prove that f (cl) > cl. Suppose thatf (cl) ≤ cl. Then f ([a, b]) ⊂ [a, f (cl)] ⊂ [a, cl]. Hence, as f strictly increases in [a, cl], it strictly increases also in f ([a, b]).Therefore H(f ) ≤ 1 by Corollary 2.6, and thus N(f ) = N(f 2) = · · · = N(f n) = N(F) = · · · = N(F 2) implying that H(F) ≤ 1contrary to the assumption. Similarly one can prove that f (cl) < cl when f takes the minimum value at cl, where l ∈ {1, n}.This completes the proof. �

Remark 3.2. Given a mapping F : [a, b] → [a, b] define F∗ : [a, b] → [a, b] by

F∗(x) = a + b − F(a + b − x)

and observe that (F∗)∗ = F ; moreover, if F ∈ PM(a, b) and S(F) = {c1, . . . , cn}, where a < c1 < · · · < cn < b, thenS(F∗) = {d1, . . . , dn} with dk = a + b − cn+1−k for every k ∈ {1, . . . , n}, also

N (F∗) = N(F) and H (F∗) = H(F).

Since i : [a, b] → [a, b], given by i(x) = a+ b− x, is an involution, we have i−1= i, and thus F∗ = i−1

◦ F ◦ i. Consequently,F k∗

= i−1◦ F k

◦ i, i.e.

F k∗(x) = a + b − F k(a + b − x), x ∈ [a, b],

for every k ∈ N.Now let f ∈ PM(a, b) be an iterative root of order n of F . Then f∗ ∈ PM(a, b) is an iterative root of order n of F∗.

Moreover,

(i) S(f ) = {cn}, the function f reaches the maximum value at cn and f (cn) > cn if and only if S(f∗) = {d1}, the function f∗reaches the minimum value at d1 and f∗(d1) < d1;

(ii) S(f ) = {cn}, the function f reaches the minimum value at cn and f (cn) < cn if and only if S(f∗) = {d1}, the function f∗reaches the maximum value at d1 and f∗(d1) > d1.

This observation allows us to confine ourselves to the case S(f ) = {c1} when studying roots of types T1 and T2.The next two results describe the behavior of a root on forts of the function F depending on which types of T1 and T2

holds.

Lemma 3.3. Let F ∈ PM(a, b) with H(F) ≥ 2, assume that N(F) ≥ 2 and let f ∈ PM(a, b) be an iterative root of ordern = N(F) of F . Assume that f is of type T1. If S(f ) = {c1} then f (ci) = ci−1 for every i ∈ {2, . . . , n}. If S(f ) = {cn} thenf (ci) = ci+1 for every i ∈ {1, . . . , n − 1}.

Proof. Due to Remark 3.2wemay assume that S(f ) = {c1}. Then f takes theminimumvalue at c1 and f (c1) < c1. Lemma 3.1provides a permutation (l1, . . . , ln) of {1, . . . , n} such that condition (3.1) holds and S(f ) = {cl1}.

We claim that

cl1 < cl2 < · · · < cln . (3.6)


Since l1 = 1, l2 = l1 and c1 is the least fort of F , we have cl1 = c1 < cl2 . In the case when n ≥ 3 take any k ∈ {3, . . . , n}and assume that cl1 < · · · < clk−1 . Observe that {c1, . . . , cn} ⊂ [c1, b] and f strictly increases in [c1, b]. Thus, as (3.1) givesf (clk) = clk−1 > clk−2 = f (clk−1), we have clk > clk−1 . Consequently, we come to (3.6) by induction.

Since c1 < · · · < cn it follows from (3.6) that lk = k, k ∈ {1, . . . , n}. Therefore (3.1) means that f (ck) = ck−1 for everyk ∈ {2, . . . , n} which completes the proof. �

Lemma 3.4. Let F ∈ PM(a, b) with H(F) ≥ 2, assume that N(F) ≥ 2 and let f ∈ PM(a, b) be an iterative root of ordern = N(F) of F . Assume that f is of type T2. If S(f ) = {c1} then f (ci) = cn+2−i for every i ∈

2, . . . ,

n+12

and f (ci) = cn+1−i

for every i ∈ n+1

2

+ 1, . . . , n

. If S(f ) = {cn} then f (cn−i) = ci for every i ∈

1, . . . ,

n−12

and f (cn−i) = ci+1 for every

i ∈ n−1

2

+ 1, . . . , n − 1

.

Proof. On account of Remark 3.2 we may confine ourselves to the case S(f ) = {c1}. Then, as f is of type T2, it strictlydecreases in [c1, b]. By Lemma 3.1 there is a permutation (l1, . . . , ln) of {1, . . . , n} such that condition (3.1) holds and l1 = 1.

At first assume that n is even, that is n = 2m for an m ∈ N. Since l1 = 1, l2 = l1 and c1 is the least fort of F , wehave cl1 < cl2 . Assume that m ≥ 2. Then, for a similar reason, cl1 < cl3 . Further, assuming that cl2k−1 < cl2k+1 for ak ∈ {1, . . . ,m− 1}, we get f (cl2k) < f (cl2k+2) by (3.1), whence cl2k+2 < cl2k since f strictly decreases in [c1, b]. If, in addition,k ≤ m − 2 then an analogous argument yields cl2k+1 < cl2k+3 . In such a way, making use of induction, we come to

cl1 < cl3 < · · · < cl2m−1 and cl2m < cl2m−2 < · · · < cl2 .

If cl2m−1 ≥ cl2m then, using (3.1) and having in mind that f strictly decreases in [c1, b], we would get cl1 ≥ cl2 , which isimpossible. Therefore cl2m−1 < cl2m , and thus

cl1 < cl3 < · · · < cln−1 < cln < cln−2 < · · · < cl2 .

Now, taking into account that c1 < c2 < · · · < cn and making use of (3.1), we obtain

f (ci) = c2m+2−i, i = 2, . . . ,m,

and

f (ci) = c2m+1−i, i = m + 1, . . . , 2m,

which is what we need.In the case when n is odd, say n = 2m+ 1 with somem ∈ N, we proceed analogously. Making use of induction we prove

that

cl1 < cl3 < · · · < cl2m+1 and cl2m < cl2m−2 < · · · < cl2and then we show the inequality cl2m+1 < cl2m arguing as previously. Finally we conclude that

f (ci) = c2m+3−i, i = 2, . . . ,m + 1,

and

f (ci) = c2m+2−i, i = m + 2, . . . , 2m + 1,

and complete the proof. �

4. Existence of roots

We have a full characterization of those functions F ∈ PM(a, b) that have a root of order n = N(F) of type T1. Toformulate the main result we need one more notion.

Let I be compact interval. A strictly increasing function φ mapping I into itself is said to be a reversing correspondence ifthere are a ξ ∈ Fixφ, i.e. a fixed point of φ, and a strictly decreasing function ω mapping Fixφ onto itself such that ω(ξ) = ξand the expression φ(x) − x has opposite signs in the intervals (ξ1, ξ2) and (ω (ξ2) , ω (ξ1)) for every ξ1, ξ2 ∈ Fixφ withξ1 < ξ2 and (ξ1, ξ2) ∩ Fixφ = ∅.

The main result of this section reads as follows.

Theorem 4.1. Let F ∈ PM(a, b) with H(F) ≥ 2 and assume that N(F) ≥ 2. Then F has a continuous iterative root of ordern = N(F) of type T1 if and only if one of the following conditions is fulfilled:

(i) n is even, F |[a,c1] is a reversing correspondence,

F(a) ≥ F(c2) ≥ · · · ≥ F (cn−2) ≥ F(cn) ≥ a, (4.1)F(c1) ≤ F(c3) ≤ · · · ≤ F (cn−1) ≤ c1

and either all inequalities of (4.1) are equalities (see Fig. 5), or at most one of them, namely F(a) ≥ F(c2) or F(cn) ≥ a, isan equality (see Fig. 6);


Fig. 5. All are equalities in Case (i).

Fig. 6. One is an equality in Case (i).

Fig. 7. All are equalities in Case (ii).

(ii) n is odd, F |[a,c1] is decreasing,

F(a) ≤ F(c2) ≤ · · · ≤ F (cn−1) ≤ c1, (4.2)F(c1) ≥ F(c3) ≥ · · · ≥ F (cn) ≥ a

and either all inequalities of (4.2) are equalities (see Fig. 7), or at the most one of them, namely F(a) ≤ F(c2) or F (cn) ≥ a,is an equality (see Fig. 8);

(iii) n is even, F |[cn,b] is a reversing correspondence,

F(b) ≤ F (cn−1) ≤ · · · ≤ F (c1) ≤ b,F(cn) ≥ F (cn−2) ≥ · · · ≥ F (c2) ≥ cn

and either all of the above inequalities are equalities, or at the most one of them, namely F(b) ≤ F(cn−1) or F (c1) ≤ b, is anequality;


Fig. 8. One is an equality in Case (ii).

(iv) n is odd, F |[cn,b] is decreasing,

F(b) ≥ F (cn−1) ≥ · · · ≥ F(c2) ≥ cn,F(cn) ≤ F (cn−2) ≤ · · · ≤ F(c1) ≤ b

and either all of the above inequalities are equalities, or at the most one of them, namely F(b) ≥ F(cn−1) or F (c1) ≤ b, is anequality.

To prove Theorem 4.1 we need the following two results. The first one provides a necessary condition for a piecewisemonotonic function to have a continuous iteration root. As it deals with both the types T1 and T2 it is of interestindependently of Theorem 4.1.

Proposition 4.2. Let F ∈ PM(a, b) with H(F) ≥ 2 and assume that n = N(F) ≥ 2.

(i) If F has a continuous iterative root f of order n of type T1, then F([a, c1]) ⊂ [a, c1] in the case S(f ) = {c1} andF([cn, b]) ⊂ [cn, b] in the case S(f ) = {cn}.

(ii) If F has a continuous iterative root of order n of type T2 and n ≥ 3, then F has a fixed point in (c1, cn); moreover,F([a, c1]) ⊂ [a, c1] and F([cn, b]) ⊂ [cn, b] whenever n is even and F([a, c1]) ⊂ [cn, b] and F([cn, b]) ⊂ [a, c1] whenn is odd.

Sometimes, using Proposition 4.2, one can easily prove the nonexistence of iterative roots.

Proof of Proposition 4.2. Let f : [a, b] → [a, b] be a continuous root of F of order n.

(i) Assume that f is of type T1. On account of Remark 2.7, or Lemma 2.3 (also Corollary 2.3 in [26]) we know thatf ∈ PM(a, b), and thus, by Lemma 3.1 and Remark 3.2, we may assume that S(f ) = {c1} and f (c1) < c1. It suffices toprove that f (a) ≤ c1, since f |[a,c1] is decreasing.Observe that if f (a) > c1 then we would have c1 = f (c) for a c ∈ (a, c1), whence f (c) ∈ S(f ) and by Lemma 2.3, wherewe put p = a, q = b,G = F = f , and Corollary 2.5, we would get c ∈ S

f 2

⊂ S (f n) = S(F) which is impossible as c1

is the least fort of F . Thus f (a) ≤ c1. Since f decreases in [a, c1] it follows that f maps [a, c1] into itself and so is f k forevery k ∈ N and, in particular, F .

(ii) Now assume that f is of type T2 and n ≥ 3. We may again assume that S(f ) = {c1}. Then, according to Lemma 3.1, thefunction f reaches the maximum value at c1 and f (c1) > c1, whereas, by Lemma 3.4, we have f (cn) = c1 and f (c2) = cn.Since f decreases in [cn, b] the first equality implies that

f ([cn, b]) ⊂ [a, c1]. (4.3)

On the other hand, as f (c1) is the strict maximum of f , the second equality yields f (c1) > cn. If f (a) < cn then wewould have cn = f (c) for a c ∈ (a, c1), whence f (c) ∈ S(f ) and, by Lemma 2.3 and Corollary 2.5, we would getc ∈ S(f 2) ⊂ S(f n) = S(F) contrary to the fact that c1 is the least fort of F . Thus f (a) ≥ cn and since f increases in[a, c1] it follows that f ([a, c1]) ⊂ [cn, b] which together with (4.3) implies the second assertion of (ii). Moreover, sincef (cn) ≤ c1 < cn ≤ f (c1) it follows that (c1, cn) contains a fixed point of f which, clearly, is also a fixed point of F . �

Lemma 4.3. Let p, q ∈ R, p < q, let H : [p, q] → [p, q] be a continuous function and x0, . . . , xn−1 ∈ [p, q]. Assume that H isstrictly increasing and

p = x1 < x3 < · · · < xn−1 < H(p) < H(q) < xn−2 < · · · < x2 < x0 = q (4.4)

if n is even, and H is strictly decreasing and

p = x1 < x3 < · · · < xn−2 < H(q) < H(p) < xn−1 < · · · < x2 < x0 = q


if n is odd. Then H has a continuous strictly decreasing root h : [p, q] → [p, q] of order n such that

h(xi) = xi+1, i = 0, . . . , n − 2, and h(xn−1) = H(q).

Proof. Consider the case of even n. Denote by ξ− and ξ+ the least and the greatest fixed point of H , respectively. Of course,in view of (4.4), we have

p < H(p) < ξ− ≤ ξ+ < H(q) < q. (4.5)

Define a sequence (xi)i≥n recursively by

xi = H(xi−n), i ∈ N, i ≥ n. (4.6)

Since H is strictly increasing, it follows from (4.4)–(4.6) that

x1 < x3 < · · · < xn−1 < xn+1 = H(x1) < · · · < ξ−

≤ ξ+ < · · · < H(x0) = xn < xn−2 < · · · < x2 < x0,

i.e.,

x2i−1 < x2i+1 < ξ− ≤ ξ+ < x2i < x2i−2, i ∈ N. (4.7)

Moreover,

limi→∞

x2i−1 = ξ− and limi→∞

x2i = ξ+. (4.8)

Denote by Ii, i ∈ N0, the closed interval ended in xi and xi+2, and observe that, by (4.7), (4.8) and the equalities x1 = p andx0 = q, we get

∞i=0

Ii = [p, ξ−) ∪ (ξ+, q]. (4.9)

For each i ∈ {0, 1, . . . , n − 2} choose any decreasing homeomorphism hi mapping Ii onto Ii+1. Now the formula

hi = H|Ii−n+1 ◦ h−1i−n+1 ◦ · · · ◦ h−1

i−1 (4.10)

recursively defines a sequence (hi)i∈N0 such that hi, i ∈ N0, is a strictly decreasing function mapping continuously Ii ontoIi+1. Clearly we have

hi (xi) = xi+1 and hi (xi+2) = xi+3, i ∈ N0. (4.11)

Moreover, as H(ξ−) = ξ− and H(ξ+) = ξ+, by virtue of [19, Th. III] or [1, Th. 15.9], the function H|[ξ−,ξ+] has a continuousstrictly decreasing root h∞ : [ξ−, ξ+] → [ξ−, ξ+] of order n; clearly

h∞ (ξ−) = ξ+ and h∞ (ξ+) = ξ−. (4.12)

It follows from (4.9) and (4.11) that the formula

h(x) =

hi(x), if x ∈ Ii, i ∈ N0,h∞(x), if x ∈ [ξ−, ξ+],

defines a function h : [p, q] → [p, q]. Observe that, according to (4.8), (4.11) and (4.12), we get

h (ξ−−) = limi→∞

h2i−1(x2i−1) = limi→∞

x2i = ξ+ = h∞(ξ−) = h(ξ−)

and

h (ξ++) = limi→∞

h2i(x2i) = limi→∞

x2i+1 = ξ− = h∞(ξ+) = h(ξ+),

so h is continuous and, consequently, also strictly decreasing.Take any x ∈ [p, q]. If x ∈ [ξ−, ξ+] then

hn(x) = hn∞

(x) = H(x).

Otherwise, by (4.9), we have x ∈ Ii for some i ∈ N0 and, on account of (4.10),

hn(x) = (hi+n−1 ◦ hi+n−2 ◦ · · · ◦ hi) (x)

=H|Ii ◦ h−1

i ◦ · · · ◦ h−1i+n−2 ◦ hi+n−2 ◦ · · · ◦ hi

(x)

= H|Ii(x) = H(x).


Therefore h is a root of H of order n. The rest of the assertion in that case follows immediately from the definition of h andproperty (4.11), as xn = H(x0) = H(q) by virtue of (4.6).

In the case of odd n the proof is quite similar. The only difference is that now H has a unique fixed point ξ , and thusξ− = ξ+ = ξ . Consequently, h∞ is defined on the singleton {ξ} and h∞(ξ) = ξ , so there is no need to use either [19, Th. III]nor [1, Th. 15.9]. �

Remark 4.4. The proof of Theorem 4.1 describes how to construct desired roots. The reader is asked to follow the procedurestep by step carefully, to ensure that it allows us to obtain, in fact, all the roots of type T1. The construction shows that rootsdepend on an arbitrary function, i.e. any continuous strictly monotonic function given on an interval can be extended to aroot.

Proof of Theorem 4.1. We first prove the necessity. Let F have a continuous root f : [a, b] → [a, b] of order n = N(F) of typeT1. Then (cf. Remark 2.7, or Lemma 2.3, also Corollary 2.3 in [26]) f ∈ PM(a, b), and thus, by Lemma 3.1 and Remark 3.2,we may assume that S(f ) = {c1}, the function f reaches the minimum value at c1 and f (c1) < c1. Therefore, according toLemma 3.3, we have also f (ci) = ci−1, i ∈ {2, . . . , n}, whence

F(ci) = f n−i+1(c1), i = 1, . . . , n. (4.13)

Moreover, Proposition 4.2(i) gives F([a, c1]) ⊂ [a, c1].Assume that n is even. Then [19, the first paragraph on p. 171] (see also [1, Th. 15.10] or [3, Remark 11.2.2]) implies that

F |[a,c1] is a reversing correspondence. Note that f (a) ≤ c1, otherwise, the continuity of f implies the existence of a fort of Fbetween a and c1, which contradicts the fact that c1 is the least fort of F . This implies that f is a decreasing self-mapping on[a, c1]. Hence, f n−1 decreases in [a, c1] when n is even. It follows from (4.13) that

F(a) = f n−1(f (a)) ≥ f n−1(c1) = F(c2). (4.14)

Moreover, as f 2([a, c1]) ⊂ [a, c1] we have f 2(c1) ≤ c1, and thus, if i ∈ {1, . . . , n− 2} is even then f n−i−1 decreases in [a, c1]and, by (4.13),

F(ci) = f n−i+1(c1) = f n−i−1(f 2(c1)) ≥ f n−i−1(c1) = F(ci+2), (4.15)

and similarly, if i ∈ {1, . . . , n − 2} is odd then f n−i−1 increases in [a, c1], whence

F(ci) = f n−i+1(c1) = f n−i−1(f 2(c1)) ≤ f n−i−1(c1) = F(ci+2). (4.16)

Clearly we have also F(cn) ≥ a and F(cn−1) = f 2(c1) ≤ c1 which completes the proof of the inequalities stated in (i). Nowassume additionally that F(a) = F(c2) and F(cn) = a. Since f |[a,c1] is one-to-one, the first equality and (4.14) give f (a) = c1,whereas the second equality and (4.13) yield f (c1) = a. Then f 2(c1) = c1 and, by (4.15) and (4.16), we obtain F(ci+2) = F(ci)for every i ∈ {1, . . . , n − 2}. Moreover, (4.13) implies that F(cn−1) = f 2(c1) = c1. On the other hand, if at least one of theinequalities F(a) ≥ F(c2) and F(cn) ≥ a is sharp, then f 2(c1) < c1, whence, as f |[a,c1] is one-to-one, so are all the inequalitiesin conditions (4.15) and (4.16); also F(cn−1) = f 2(c1) < c1 then.

Now assume that n is odd. Then F |[a,c1] is strictly decreasing. Using similar arguments as in the preceding part of the proofone can get

F(a) = f n−1(f (a)) ≤ f n−1(c1) = F(c2),F(ci) ≤ F(ci+2)

for every even i ∈ {1, . . . , n − 2} and

F(ci) ≥ F(ci+2)

whenever i ∈ {1, . . . , n − 2} is odd; moreover, F(cn) ≥ a and F(cn−1) = f 2(c1) ≤ c1. The remaining conclusion can beobtained analogously as in case (i).

In what follows we prove the sufficiency. According to Remark 3.2 we may confine ourselves to the cases (i) and (ii). Atfirst we shall find a continuous strictly decreasing root f0 of F |[a,c1] of order n. Then we shall extend it continuously to a rootdefined on the whole interval [a, b].

To ensure the existence of such an extension we need to choose a strictly decreasing iterative root f0 : [a, c1] → [a, c1]of F |[a,c1] of order n, fulfilling the condition

f0(F(ci)) = F(ci−1), i = 2, . . . , n, and f0(c1) = F(cn). (4.17)

We give a detailed argument in case (i). Then n is even and F |[a,c1] is a reversing correspondence. If all inequalities (4.1)are equalities, then F(a) = a and F(c1) = c1, so, according to [19, Th. III] or [1, Th.15.9], the function F |[a,c1] has acontinuous strictly decreasing root f0 : [a, c1] → [a, c1] of order n; clearly f0(a) = c1 and f0(c1) = a, and thus (4.17)follows. Thus we may assume that at most one of inequalities (4.1), namely F(a) ≥ F(c2) or F(cn) ≥ a, is an equality. Putp = F(cn), q = x0 = c1 and xi = F(cn+1−i) for each i ∈ {1, . . . , n − 1}. Then p < q as the function F |[a,c1] strictly increases


and F(cn) ≤ F(a) by virtue of (4.1), whence p = F(cn) ≤ F(a) < F(c1) ≤ c1 = q. Further x0, . . . , xn−1 ∈ [p, q] and it followsfrom (4.1) that

xn−1 = F(c2) ≤ F(a) ≤ F(F(cn)) = F(p).

Moreover, since the monotonicity of F |[a,c1] is strict and either F(c2) < F(a), or a < F(cn), we have xn−1 < F(p).Consequently, (4.1) yields

p = x1 < x3 < · · · < xn−1 < F(p) < F(q) < xn−2 < · · · < x2 < x0 = q,

whence, making use of Lemma 4.3, we find a continuous strictly decreasing root f0 : [F(cn), c1] → [F(cn), c1] of F |[F(cn),c1]of order n satisfying

f0(c1) = F(cn) and f0(F(cn+1−i)) = F(cn−i), i = 1, . . . , n − 1,

that is condition (4.17). A simple argument applies in case (ii). The main difference is that nowwemake use of Th. II insteadof Th. III from [19] or Th. 15.8 instead of Th. 15.9 from [1].

If F(cn) = a, we have obtained the desired root f0 : [a, c1] → [a, c1] of F |[a,c1]. Now, if F(cn) > a, we must extend f0 tothe interval [a, c1]. Observe first that, by (4.17) and a simple induction, for every i ∈ {1, . . . , n− 1} the image f i0([F(cn), c1])is the closed interval conv{F(cn−i), F(cn+1−i)} ended in F(cn−i) and F(cn+1−i); in particular,

f n−10 ([F(cn), c1]) = conv{F(c1), F(c2)}.

Therefore, using (4.1) when F increases in [a, c1] and (4.2) when it decreases there, we come to

F([a, F(cn))) = conv{F(a), F(F(cn))}⊂ conv{F(c1), F(c2)} = f n−1

0 ([F(cn), c1]).

Thus the formula

f0(x) = f −(n−1)0 (F(x)), x ∈ [a, F(cn)), (4.18)

allows us to extend f0 to the interval [a, c1]. Then, because of the condition f n0 (x) = F(x), x ∈ [F(cn), c1], we have

f0(F(cn)−) = limx→F(cn)

f −(n−1)0 (F(x)) = f −(n−1)

0 (F(F(cn)))

= f −(n−1)0 (f n0 (F(cn))) = f0(F(cn)).

This proves that the extended f0 is continuous, and it is easy to verify that f0 is strictly decreasing on [a, F(cn)] and,consequently, on the whole interval [a, c1]. Moreover, taking into account also condition (4.18), we infer that f0 is a rootof F |[a,c1] of order n.

In what follows our attention is paid to the extension of f0 on [a, b]. We construct a root on the interval [c1, b] in twocases depending on whether F has a fixed point in [cn, b] or not. Firstly we make a partition of [a, b] into some subintervals,define the root on each of them, and then prove its continuity.

Put ξ = b if F has no fixed points in [cn, b] and let

ξ = inf{x ∈ [cn, b] : F(x) = x}

otherwise. Then, since F(cn) < c1 < cn, we have ξ ∈ (cn, b] and

F(x) < x, x ∈ [cn, ξ).

Let y0 = a and y1 = c1, . . . , yn = cn. Of course y0 < · · · < yn < ξ . If y1 ≤ F(ξ) then

F(yn) = F(cn) < c1 = y1 ≤ F(ξ),

whence there is a yn+1 ∈ (yn, ξ ] such that F(yn+1) = y1. Take any integer k ≥ n + 1 and assume that we have definedy0, . . . , yk ∈ [a, ξ ] such that y0 < · · · < yk and F(yi) = yi−n for each i ∈ {n + 1, . . . , k}. Assuming that yk+1−n ≤ F(ξ) weget

F(yk) = yk−n < yk+1−n ≤ F(ξ),

and thus yk+1−n = F(yk+1) for some yk+1 ∈ (yk, ξ ]. Otherwise we stop the recurrence. In such a way we construct a strictlyincreasing sequence (yi)i∈I of points of the interval [a, ξ ] such that y0 = a, y1 = c1, . . . , yn = cn, and

F(yi) = yi−n, i ∈ I, i ≥ n + 1, (4.19)

where either I = N0 if yi−n ≤ F(ξ) for all i ≥ n+1, or I = {0, . . . , l} if l is the greatest positive integer such that yi−n ≤ F(ξ)for all n + 1 ≤ i ≤ l. Observe that if I is finite then F(ξ) < yi for some i ∈ I , whence F(ξ) < ξ , and thus F has no fixed


points in [cn, b] and ξ = b. On the other hand, if I = N0 then (yi)i∈N0 increases to a limit which, by (4.19), is a fixed point ofF , viz. ξ .

Now we define the partition of [a, b] into subintervals, announced earlier. Denote by Ji the interval [yi, yi+1] wheneveri ∈ I = N0, or if i ∈ I \ {l} when I = {0, . . . , l}; moreover, put Jl = [yl, b] in the second case. We claim that

F(Ji) ⊂ J0, i = 0, . . . , n − 1, and F(Jn) = [F(cn), c1], (4.20)F(Ji) = Ji−n, i ∈ N, i ≥ n + 1, (4.21)

if I = N0, and

F(Ji) = Ji−n, i = n + 1, . . . , l − 1, and F(Jl) ⊂ Jl−n (4.22)

in the case when I = {0, . . . , l}, and

f0(F(Ji)) ⊂ F(Ji−1), i = 1, . . . , n. (4.23)

In fact, for every i ∈ {0, . . . , n − 1} we have F(Ji) = conv{F(ci), F(ci+1)}, and thus F(Ji) ⊂ J0. Condition (4.17) implies alsothat f0(F(Ji)) = conv{F(ci−1), F(ci)}, that is F(Ji−1), whenever i ∈ {2, . . . , n − 1}. Since f0 is a root of F |J0 it follows thatf0 ◦ F |J0 = F |J0 ◦ f0, and thus, by (4.17), we have also

f0(F(J1)) = conv{f0(F(c1)), f0(F(c2))} = conv{F(f0(c1)), F(c1)}= conv{F(F(cn)), F(c1)} ⊂ conv{F(a), F(c1)} = F(J0).

Moreover, since F increases in [cn, b], condition (4.19) gives

F(Jn) = [F(yn), F(yn+1)] = [F(yn), y1] = [F(cn), c1] ⊂ J0,

whence

f0(F(Jn)) = f0([F(cn), c1]) = [f0(c1), f0(F(cn))]= [F(cn), F(cn−1)] = F(Jn−1).

Next, (4.19) yields also

F(Ji) = F([yi, yi+1]) = [F(yi), F(yi+1)] = [yi−n, yi+1−n] = Ji−n

for each i ∈ N, i ≥ n + 1, if I = N0, and for each i ∈ {n + 1, . . . , l − 1} in the case when I = {0, . . . , l}. Finally, in the lastcase we see that F(b) < yl+1−n, and thus

F(Jl) = F([yl, b]) = [F(yl), F(b)] ⊂ [yl−n, yl+1−n] = Jl−n.

Therefore the claimed (4.20)–(4.23) are proved.We are in a position to define roots fi : Ji → Ji−1, i ∈ I . In view of (4.23) the formula

fi =F |Ji−1

−1◦ f0 ◦ F |Ji , i = 1, . . . , n,

defines functions f1 : J1 → J0, . . . , fn : Jn → Jn−1, which are continuous and strictly increasing. By virtue of (4.17) we obtain

fi(yi) =F |Ji−1

−1(f0(F(yi))) = yi−1, i = 2, . . . , n,

fi(yi+1) =F |Ji−1

−1(f0(F(yi+1))) = yi, i = 1, . . . , n − 1,

and, using also (4.19),

fn(yn+1) =F |Jn−1

−1(f0(F(yn+1))) =

F |Jn−1

−1(f0(c1)) = cn = yn.

Moreover, as f0 ◦ F |J0 = F |J0 ◦ f0, we get

f1(y1) =F |J0

−1(f0(F(y1))) =

F |J0

−1(F(f0(y1))) = f0(y1).

Consequently

fi(Ji) = Ji−1 (4.24)

for every i ∈ {2, . . . , n} and fi−1(yi) = fi(yi) for each i ∈ {1, . . . , n}. Observe also that, according to (4.21), (4.22) and (4.17)and the second condition of (4.20),

f1(F(Jn+1)) = f1(J1) = [f1(y1), f1(y2)] = [f0(y1), y1]

= [F(cn), c1] = F(Jn). (4.25)


Fig. 9. F1 with N(F1) = 2, f ∈ T1 .

Now, proceeding recurrently, using the inclusion fi−n(F(Ji)) ⊂ F(Ji−1) following from (4.21)–(4.24) and putting

fi =F |Ji−1

−1◦ fi−n ◦ F |Ji , i > n,

we define a sequence (fi)i∈I such that fi : Ji → Ji−1, i ∈ I , is a continuous strictly increasing function, satisfying (4.24) foreach i ∈ I, i ≥ 2, excluding the case when I = {0, . . . , l} and i = l. It can be deduced that

fi−1(yi) = fi(yi) = yi−1, i ∈ I \ {0}. (4.26)

Finally, we come to the desired root f defined on [a, b]. The formula

f (x) = fi(x), x ∈ Ji, i ∈ I,

defines a self-mapping f of

i∈I Ji which, according to (4.26), is continuous. We know thatf |J0

n= f n0 = F |J0 . Moreover, if

i ∈ {1, . . . , n − 1} thenf |Ji

n= f n−i

0 ◦ f1 ◦ · · · ◦ fi

= f n−i0 ◦

F |J0

−1◦ f0 ◦ F |J1 ◦

F |J1

−1◦ f0 ◦ F |J2 ◦ · · · ◦

F |Ji−1

−1◦ f0 ◦ F |Ji

= f n−i0 ◦

f n0

−1◦ f i0 ◦ F |Ji = F |Ji

and, iff |Ji−1

n= F |Ji−1 for an integer i ≥ n, then

f |Jin

= fi−n+1 ◦ · · · ◦ fi = f −1i−n ◦ (fi−n ◦ · · · ◦ fi−1) ◦ fi

= f −1i−n ◦

f |Ji−1

n◦

F |Ji−1

−1◦ fi−n ◦ F |Ji

= f −1i−n ◦ F |Ji−1 ◦

F |Ji−1

−1◦ fi−n ◦ F |Ji = F |Ji .

This shows that f is a root of F |

i∈I Ji of order n. If ξ = b then, putting additionally f (b) = f (b−), we get a continuous rootf : [a, b] → [a, b] of F of order n. So assume that ξ < b. Then F(ξ) = ξ and I = N0, whence, according to (4.26), we getf (ξ−) = limn→∞ fn(yn) = limn→∞ yn−1 = ξ . Using [19, Th. I] or [1, Th. 15.7], we can find a continuous strictly increasingroot f∞ : [ξ, b] → [ξ, b] of F[ξ,b] of order n. Since F(ξ) = ξ we have f∞(ξ) = ξ . Extending f from [a, ξ ] to [a, b] with theuse of f∞ we obtain again a continuous root f : [a, b] → [a, b] of F of order n. Observe that in both cases f |[a,c1] = f0 isstrictly decreasing, whereas f |[c1,b] is strictly increasing. Moreover, f (c1) = f0(c1) < c1 and, consequently, f is of type T1.This completes the proof of the theorem. �

5. Examples

Example 5.1. Consider F1 : [0, 1] → [0, 1] (see Fig. 9), given by

F1(x) =

x, if x ∈

[0,

14

,

−12x +

38, if x ∈

[14,34

,

3x −94, if x ∈

[34, 1

].


It follows from Corollary 2.6 that H(F1) ≥ 2. The assumption (i) in Theorem 4.1 is satisfied with n = 2, a = 0, b = 1, c1 =

1/4 and c2 = 3/4. The sequence (yi)i∈I defined in the proof of Theorem 4.1 is (0, 1/4, 3/4, 5/6, 1). One can verify that themapping f : [0, 1] → [0, 1] (see Fig. 9), defined by

f (x) =

−x +14, if x ∈

[0,

14

,

12x −

18, if x ∈

[14,34

,

6x −174

, if x ∈

[34,56

,

12x +

13, if x ∈

[56, 1

],

satisfies f 2 = F1. Clearly it is of type T1.

Example 5.2. Let F2 : [0, 1] → [0, 1] (see Fig. 10) be given by

F2(x) :=

8x, if x ∈

[0,

18

,

−2x +54, if x ∈

[18,14

,

12x +

58, if x ∈

[14,12

,

−18x +

1516

, if x ∈

[12, 1

].

Corollary 2.6 implies that H(F2) ≥ 2. Now the assumption (iv) in Theorem 4.1 is satisfied with n = 3, a = 0, b = 1, c1 =

1/8, c2 = 1/4 and c3 = 1/2. One can check that the mapping f : [0, 1] → [0, 1] (see Fig. 10), defined by

f (x) =

2x, if x ∈

[0,

12

,

−12x +

54, if x ∈

[12, 1

],

(5.1)

satisfies f 3 = F2. The sequence (yi)i∈I appearing in the construction of f (see the proof of Theorem 4.1 and Remark 3.2) isnow (1/2i)i∈N. Clearly it is of the type T1.

Example 5.3. Consider the function F3 : [0, 1] → [0, 1] (see Fig. 11), defined by

F3(x) =

−32x +

34, if x ∈

[0,

14

,

x +18, if x ∈

[14,34

,

−32x + 2, if x ∈

[34, 1

].

According to Corollary 2.6 we have H(F3) ≥ 2. It follows from Proposition 4.2(i) that F3 has no continuous root of type T1 oforder 2. Suppose that it has a continuous square root f of type T2. As n = 2 we cannot apply Proposition 4.2(ii). However,we can proceed as follows. We know that either f reaches its maximum value at 1/4, f (1/4) > 1/4 and f strictly decreasesin [1/4, 1], or f takes the minimum value at 3/4, f (3/4) < 3/4 and f strictly decreases in [0, 3/4]. Thus f has a fixed pointξ ∈ (0, 1). Clearly F3(ξ) = f (f (ξ)) = ξ , and thus ξ = 4/5. Since f , being of type T2, is strictly monotonic in [3/4, 1],it follows that F3 = f ◦ f strictly increases in a neighborhood of 4/5 which contradicts the definition of F3 in [3/4, 1].Consequently, F3 has no continuous square root.


Fig. 10. F2 with N(F2) = 3, f ∈ T1 .

Fig. 11. F3 with N(F3) = 2.

Fig. 12. F4 with N(F4) = 3.

Example 5.4. It follows from Corollary 2.6 that the function F4 : [0, 1] → [0, 1] (see Fig. 12), given by

F4(x) =

−12x +

12, if x ∈

[0,

14

,

2x −18, if x ∈

[14,12

,

−32x +

138

, if x ∈

[12,34

,

12x +

18, if x ∈

[34, 1

],

has the height at least 2. By virtue of Proposition 4.2 it has no continuous root of order 3.


Fig. 13. F with N(F) = 3.

Fig. 14. F with N(F) = 5.

Example 5.5. Similarly as in Example 5.4, using Proposition 4.2 only, one can check that the functions presented below onFigs. 13 and 14 have no continuous roots of order 3 and 5, respectively.

6. Conclusion and future directions

In the main Theorem 4.1 we completely answered the question: When does a piecewise monotonic function F withH(F) ≥ 2 have a continuous iterative root of order N(F) of type T1? It is an open question to describe the case of such roots oftype T2. When H(F) = 2 an answer has been given by Taixiang Sun and Hongjian Xi [29] and by the first of them in [30]. Thenecessary conditions proved in Lemma 3.4 and (when H(F) ≥ 3) in Proposition 4.2(ii) are practically the only informationthat we have about the problem. It is expected that a unified description of the existence of roots of type T2 should coveralso the case H(F) = 2 presented in [29,30].

References

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iterative roots of piecewise monotonic functions of nonmonotonicity height not less than 2

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