Abstract
We present a complete list of all continuous solutions of the equation , where α, β and are given real numbers.
MSC:39B22, 39B12, 26A18.
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1 Introduction
In this note we give a complete list of all continuous solutions of the equation
where α, β and are given real numbers; here and throughout, denotes the second iterate of f. The motivation for writing this note was two problems concerning continuous solutions of some special cases of equation (1.1) (see [[1], Problem 11, p.312] and [[2], Problem 5, p.22]) as well as conference reports and papers on both problems (see [3, 4] and [5–8]). Let us mention that the problem from booklet [2] is wrongly solved in this booklet. To see that the problem from [1] concerns really equation (1.1) observe that from Remark 1.1 below it follows that in the case where equation (1.1) can be rewritten in the form
Remark 1.1 Assume . Then every continuous solution of equation (1.1) is strictly monotone and maps onto .
Proof Fix and assume that . Then by (1.1) we get
and since , we obtain . Thus f is injective. This jointly with continuity implies strict monotonicity.
Now suppose that, contrary to our claim, . Then, by the continuity of f and (1.1), we obtain
a contradiction. Thus . In the same manner we can prove that . □
2 Main results
To give a complete list of all continuous solutions of equation (1.1), we will split our consideration into the following three cases: , and . It turns out that the description of all continuous solutions of equation (1.1) in the first case is quite easy; whereas in the third case it is much more complicated than in the second one.
2.1 The case
If , then equation (1.1) reduces to the equation
Equation (2.1) was examined in [9–11]; cf. also [12, 13] and the references therein.
We begin with a (rather obvious and simple) characterization of general solutions of equation (2.1).
Proposition 2.1 A function satisfies (2.1) if and only if
for all .
Proof (⇒) If a function satisfies (1.1), then for every we have .
(⇐) Fix a function satisfying (2.2) for all . Then, for every , we have . Now, putting in place of x in (2.2), we obtain (2.1). □
From Proposition 2.1 we obtain the following description of all continuous solutions of equation (2.1).
Corollary 2.2 Let be a continuous solution of equation (2.1). Then either f has form (2.2) for all or there exists a proper subinterval I (open or closed or closed on one side; possible infinite or degenerated to a single point) of the half-line satisfying
such that f has form (2.2) for all ,
and
Moreover:
-
(i)
If , then ;
-
(ii)
If , then with arbitrary ;
-
(iii)
If , then with arbitrary and ;
-
(iv)
If , then either or or with arbitrary and ;
-
(v)
If , then with arbitrary ;
-
(vi)
If , then no restriction on I;
-
(vii)
If , then with arbitrary ;
-
(viii)
If , then either or or with arbitrary and .
Proof Put .
If , then (2.2) holds for all by Proposition 2.1. Therefore, to the end of the proof, we assume that .
Since f is continuous, it follows that I is an interval. By Proposition 2.1 we see that (2.2) holds for all . Thus (2.3) holds. Condition (2.4) follows from the continuity of f and condition (2.5) is a consequence of the definition of I. This completes the proof of the main part of the result.
To prove the moreover part put and . Since , it follows that or .
Assume first that .
We will show that and .
If , then (2.3) implies , which contradicts . Similarly, if , then (2.3) implies , which contradicts .
Applying condition (2.3), we get and . Hence .
-
(i)
If , then .
-
(ii)
If , then .
-
(iii)
If , then .
Assume now that .
Then (2.3) yields provided that and provided that .
-
(iv)
If , then if and if .
-
(v)
If , then and no restriction on A.
-
(vi)
If , then no restriction on A and B.
-
(vii)
If , then and no restriction on B.
-
(viii)
If , then if and if .
□
As a consequence of Corollary 2.2, we have the following direct construction of all continuous solutions of equation (2.1) (cf. [[13], Theorem 15.15]).
Corollary 2.3 Let I be a suitably chosen interval from Corollary 2.2 for a given , and let be a function given by . Then every extension of to a continuous function satisfying (2.5) is a solution of equation (2.1).
2.2 The case
If , then equation (1.1) reduces to the equation
Charles Babbage was probably the first who looked for solutions of equation (2.6) in the case where (see [14]). For the case where , see [15–17]; cf. also [18]. Equation (2.6) is a particular case of the equation of iterative roots (see [12, 13, 19–21]). According to known results on the equation of iterative roots, we can formulate a theorem on continuous solutions of equation (2.6).
Denote by the closed interval with .
Theorem 2.4
-
(i)
Assume . Then equation (2.6) has no continuous solution .
-
(ii)
Assume .
(ii1) Then the formula defines the unique continuous solution of equation (2.6).
(ii2) If is a continuous and decreasing solution of equation (2.6), then there exists such that and f maps bijectively onto . Conversely, if , then every decreasing bijection such that can be uniquely extended to a continuous and decreasing solution of equation (2.6).
-
(iii)
Assume .
(iii1) Let . If is a continuous and increasing solution of equation (2.6), then f maps bijectively onto . Conversely, if , then every increasing bijection can be uniquely extended to a continuous and increasing solution of equation (2.6).
(iii2) Equation (2.6) has no continuous and decreasing solution from .
-
(iv)
Assume .
(iv1) Let and let . If is a continuous and increasing solution of equation (2.6), then f maps bijectively onto . Conversely, if and , then every increasing bijection can be uniquely extended to a continuous and increasing solution of equation (2.6).
(iv2) If is a continuous and decreasing solution of equation (2.6), then f maps bijectively onto and for all . Conversely, every decreasing bijection such that for all can be uniquely extended to a continuous and decreasing solution of equation (2.6).
Proof All the assertions can be derived from [[13], Chapter XV]) as it has been noticed earlier. However, most of the assertions have evident proofs, so we present them for the convenience of the reader.
-
(i)
Suppose that, contrary to our claim, equation (2.6) has a continuous solution . Then the second iterate of f is strictly increasing. Now by (2.6) we conclude that , a contradiction.
(ii1) It is clear that the identity function on satisfies (2.6). Suppose the contrary, and let be another increasing solution of equation (2.6). Then there exists such that . By the monotonicity of f, we conclude that , which contradicts (2.6).
(ii2) The first assertion is clear. To prove the second one, fix a decreasing bijection such that and extend it to a function putting for all . It is easy to see that f is a decreasing bijection satisfying (2.6).
(iii1) The first assertion is evident. The second one can be deduced from [[13], Lemma 15.6].
(iii2) Suppose, to derive a contradiction, that is a continuous and decreasing solution of equation (2.6). Then there exists such that . Hence , a contradiction.
(iv1) The first assertion is easy to verify. The second one can be inferred from [[13], Theorem 15.7].
(iv2) Let f be a continuous and decreasing solution of equation (2.6). Then there exists such that . Hence by (2.6) we get , and thus . Consequently, f maps bijectively onto . Moreover, (2.6) yields for all . To prove the second part of the assertion, fix a decreasing bijection such that for all and extend it to a function putting for all . It is easy to calculate that f is a decreasing bijection satisfying (2.6). □
2.3 The case
In this case an explicit description of all continuous solutions of equation (1.1) is much more involved than in both the previous cases. We begin with an observation which allows us to rewrite equation (1.1) in an equivalent form.
Lemma 2.5 If is a solution of equation (1.1), then the function given by satisfies
Conversely, if is a solution of equation (2.7), then the function given by satisfies (1.1).
Equation (2.7) is a special case of the polynomial-like iterative inhomogeneous equation
where all ’s and b are real numbers and f is an unknown self-mapping; here denotes the n th iterate of f. For the theory of equation (2.8) and its generalizations, we refer the readers to [22–30]. The problem of finding all continuous solutions of equation (2.8) seems to be very difficult. It is completely solved in [31] for , but it is still open even in the case where (see [32]). It turns out that the nature of continuous solutions of equation (2.8) depends on the behavior of complex roots of its characteristic equation . This characteristic equation is obtained by putting into (2.8) with ; in this way we can determine all linear solutions of the homogeneous counterpart of equation (2.8), and then all affine solutions of equation (2.8). Therefore, to formulate our result on continuous solutions of equation (1.1), denote by and the complex roots of the equation
By our assumption, we have and .
Combining Lemma 2.5 with the results from [31], we get the following theorem.
Theorem 2.6
-
(i)
Assume . Then equation (1.1) has no continuous solution .
-
(ii)
Assume . Let and be defined by
(ii1) Assume either or . Let . If is a continuous solution of equation (1.1), then , is continuous,
and
for all . Conversely, every continuous function such that (2.9) and (2.10) are satisfied and (2.11) holds for all can be uniquely extended to a continuous solution of equation (1.1).
(ii2) Assume . Let . If is a continuous solution of equation (1.1), then either and or is continuous,
and (2.11) holds for all . Conversely, every continuous function such that (2.12) is satisfied and (2.11) holds for all can be uniquely extended to a continuous solution of equation (1.1).
(ii3) Assume either or . Let . If is a continuous solution of equation (1.1), then and is continuous,
and (2.11) holds for all . Conversely, every continuous function such that (2.13) is satisfied and (2.11) holds for all can be uniquely extended to a continuous solution of equation (1.1).
(ii4) Assume either or . Then every continuous solution of equation (1.1) is of the form
with some .
(ii5) Assume either or . Then and are the only continuous solutions from to of equation (1.1).
-
(iii)
Assume . Let be defined by
(iii1) Assume . If is a continuous solution of equation (1.1), then is continuous,
and
for all . Conversely, every continuous function such that (2.14) is satisfied and (2.15) holds for all can be uniquely extended to a continuous solution of equation (1.1).
(iii2) Assume . If is a continuous solution of equation (1.1), then is continuous,
and
for all . Conversely, every continuous function such that (2.16) is satisfied and (2.17) holds for all can be uniquely extended to a continuous solution of equation (1.1).
(iii3) Assume and . Then every continuous solution of equation (1.1) is of the form
with some .
(iii4) Assume and . Then is the unique continuous solution from to of equation (1.1).
(iii5) Assume and . Then is a continuous solution of equation (1.1) and every other continuous solution of equation (1.1) is of the form
with some .
-
(iv)
Assume .
(iv1) Assume . Then the formula
defines the unique continuous solution of equation (1.1).
(iv2) Assume . Then equation (1.1) has no continuous solution .
(iv3) Assume . Then every continuous solution of equation (1.1) is of the form
with some .
Author’s contributions
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This research was supported by Silesian University Mathematics Department (Iterative Functional Equations and Real Analysis program).
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Morawiec, J. On a functional equation involving iterates and powers. Adv Differ Equ 2014, 271 (2014). https://doi.org/10.1186/1687-1847-2014-271
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DOI: https://doi.org/10.1186/1687-1847-2014-271