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On a functional equation involving iterates and powers

Abstract

We present a complete list of all continuous solutions f:(0,+)(0,+) of the equation f 2 (x)=γ [ f ( x ) ] α x β , where α, β and γ>0 are given real numbers.

MSC:39B22, 39B12, 26A18.

1 Introduction

In this note we give a complete list of all continuous solutions f:(0,+)(0,+) of the equation

f 2 (x)=γ [ f ( x ) ] α x β ,
(1.1)

where α, β and γ>0 are given real numbers; here and throughout, f 2 denotes the second iterate of f. The motivation for writing this note was two problems concerning continuous solutions f:(0,+)(0,+) 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 [58]). 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 β0 equation (1.1) can be rewritten in the form

f(x) [ f 1 ( x ) ] β =γ x α .

Remark 1.1 Assume β0. Then every continuous solution f:(0,+)(0,+) of equation (1.1) is strictly monotone and maps (0,+) onto (0,+).

Proof Fix x,y(0,+) and assume that f(x)=f(y). Then by (1.1) we get

γ [ f ( x ) ] α x β = f 2 (x)= f 2 (y)=γ [ f ( y ) ] α y β =γ [ f ( x ) ] α y β ,

and since β0, we obtain x=y. Thus f is injective. This jointly with continuity implies strict monotonicity.

Now suppose that, contrary to our claim, lim x 0 f(x)(0,+). Then, by the continuity of f and (1.1), we obtain

f ( lim x 0 f ( x ) ) = lim x 0 f 2 (x)= lim x 0 γ [ f ( x ) ] α x β {0,+},

a contradiction. Thus lim x 0 f(x){0,+}. In the same manner we can prove that lim x + f(x){0,+}. □

2 Main results

To give a complete list of all continuous solutions f:(0,+)(0,+) of equation (1.1), we will split our consideration into the following three cases: β=0, α=0β and α0β. 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 β=0

If β=0, then equation (1.1) reduces to the equation

f 2 (x)=γ [ f ( x ) ] α .
(2.1)

Equation (2.1) was examined in [911]; cf. also [12, 13] and the references therein.

We begin with a (rather obvious and simple) characterization of general solutions f:(0,+)(0,+) of equation (2.1).

Proposition 2.1 A function f:(0,+)(0,+) satisfies (2.1) if and only if

f(x)=γ x α
(2.2)

for all xf((0,+)).

Proof () If a function f:(0,+)(0,+) satisfies (1.1), then for every x=f(z)f((0,+)) we have f(x)= f 2 (z)=γ [ f ( z ) ] α =γ x α .

() Fix a function f:(0,+)(0,+) satisfying (2.2) for all xf((0,+)). Then, for every x(0,+), we have f(x)f((0,+)). Now, putting f(x) in place of x in (2.2), we obtain (2.1). □

From Proposition 2.1 we obtain the following description of all continuous solutions f:(0,+)(0,+) of equation (2.1).

Corollary 2.2 Let f:(0,+)(0,+) be a continuous solution of equation (2.1). Then either f has form (2.2) for all x(0,+) 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 (0,+) satisfying

γ x α Ifor all xI
(2.3)

such that f has form (2.2) for all xI,

lim x y f(x)=γ y α for all y{infI,supI}{0,+}
(2.4)

and

f ( ( 0 , + ) I ) I.
(2.5)

Moreover:

  1. (i)

    If α<1, then I={ γ 1 1 α };

  2. (ii)

    If α=1, then clI=[A, γ A ] with arbitrary A(0, γ ];

  3. (iii)

    If α(1,0), then clI=[A,B] with arbitrary A(0, γ 1 1 α ] and B[ γ 1 1 α ,+);

  4. (iv)

    If α[0,1), then either clI=[0,B] or clI=[A,B] or clI=[A,+) with arbitrary A(0, γ 1 1 α ] and B[ γ 1 1 α ,+);

  5. (v)

    If α=1<γ, then clI=[A,+) with arbitrary A(0,+);

  6. (vi)

    If α=1=γ, then no restriction on I;

  7. (vii)

    If α=1>γ, then clI=(0,B] with arbitrary B(0,+);

  8. (viii)

    If α(1,+), then either I={ γ 1 1 α } or clI=[0,B] or clI=[A,+) with arbitrary A[ γ 1 1 α ,+) and B(0, γ 1 1 α ].

Proof Put I=f((0,+)).

If I=(0,+), then (2.2) holds for all x(0,+) by Proposition 2.1. Therefore, to the end of the proof, we assume that I(0,+).

Since f is continuous, it follows that I is an interval. By Proposition 2.1 we see that (2.2) holds for all xI. 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 A=infI and B=supI. Since I(0,+), it follows that 0<A or B<+.

Assume first that α<0.

We will show that 0<A and B<+.

If B=+, then (2.3) implies 0= lim x B γ x α clI, which contradicts I(0,+). Similarly, if A=0, then (2.3) implies += lim x A + γ x α clI, which contradicts I(0,+).

Applying condition (2.3), we get 0<Aγ B α and γ A α B<+. Hence γ A α B γ 1 α A 1 α .

  1. (i)

    If α<1, then γ 1 1 α AB γ 1 1 α .

  2. (ii)

    If α=1, then γ A B γ A .

  3. (iii)

    If α(1,0), then A γ 1 1 α B.

Assume now that α0.

Then (2.3) yields Aγ A α provided that A>0 and γ B α B provided that B<+.

  1. (iv)

    If α[0,1), then A γ 1 1 α if A>0 and γ 1 1 α B if B<+.

  2. (v)

    If α=1<γ, then B=+ and no restriction on A.

  3. (vi)

    If α=1=γ, then no restriction on A and B.

  4. (vii)

    If α=1>γ, then A=0 and no restriction on B.

  5. (viii)

    If α>1, then A γ 1 1 α if A>0 and B γ 1 1 α if B<+.

 □

As a consequence of Corollary 2.2, we have the following direct construction of all continuous solutions f:(0,+)(0,+) 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 αR, and let f 0 :II be a function given by f 0 (x)=γ x α . Then every extension of f 0 to a continuous function f:(0,+)(0,+) satisfying (2.5) is a solution of equation (2.1).

2.2 The case α=0β

If α=0β, then equation (1.1) reduces to the equation

f 2 (x)=γ x β .
(2.6)

Charles Babbage was probably the first who looked for solutions of equation (2.6) in the case where β=γ=1 (see [14]). For the case where γ=β=1, see [1517]; cf. also [18]. Equation (2.6) is a particular case of the equation of iterative roots (see [12, 13, 1921]). According to known results on the equation of iterative roots, we can formulate a theorem on continuous solutions f:(0,+)(0,+) of equation (2.6).

Denote by I(x,y) the closed interval [min{x,y},max{x,y}] with x,y(0,+).

Theorem 2.4

  1. (i)

    Assume β<0. Then equation (2.6) has no continuous solution f:(0,+)(0,+).

  2. (ii)

    Assume β=γ=1.

(ii1) Then the formula f(x)=x defines the unique continuous solution f:(0,+)(0,+) of equation (2.6).

(ii2) If f:(0,+)(0,+) is a continuous and decreasing solution of equation (2.6), then there exists x 0 (0,+) such that f( x 0 )= x 0 and f maps (0, x 0 ] bijectively onto [ x 0 ,+). Conversely, if x 0 (0,+), then every decreasing bijection f 0 :(0, x 0 ][ x 0 ,+) such that f( x 0 )= x 0 can be uniquely extended to a continuous and decreasing solution f:(0,+)(0,+) of equation (2.6).

  1. (iii)

    Assume β=1γ.

(iii1) Let x 0 (0,+). If f:(0,+)(0,+) is a continuous and increasing solution of equation (2.6), then f maps I( x 0 ,f( x 0 )) bijectively onto I(f( x 0 ),γ x 0 ). Conversely, if x 1 IntI( x 0 ,γ x 0 ), then every increasing bijection f 0 :I( x 0 , x 1 )I( x 1 ,γ x 0 ) can be uniquely extended to a continuous and increasing solution f:(0,+)(0,+) of equation (2.6).

(iii2) Equation (2.6) has no continuous and decreasing solution from f:(0,+)(0,+).

  1. (iv)

    Assume 0<β1.

(iv1) Let x 0 (0, γ 1 1 β ) and let y 0 ( γ 1 1 β ,+). If f:(0,+)(0,+) is a continuous and increasing solution of equation (2.6), then f maps I( x 0 ,f( x 0 ))I( y 0 ,f( y 0 )) bijectively onto I(f( x 0 ),γ x 0 β )I(f( y 0 ),γ y 0 β ). Conversely, if x 1 IntI( x 0 ,γ x 0 β ) and y 1 IntI( y 0 ,γ y 0 β ), then every increasing bijection f 0 :I( x 0 , x 1 )I( y 0 , y 1 )I( x 1 ,γ x 0 β )I( y 1 ,γ y 0 β ) can be uniquely extended to a continuous and increasing solution f:(0,+)(0,+) of equation (2.6).

(iv2) If f:(0,+)(0,+) is a continuous and decreasing solution of equation (2.6), then f maps (0, γ 1 1 β ] bijectively onto [ γ 1 1 β ,+) and γ [ f ( x ) ] β =f(γ x β ) for all x(0,+). Conversely, every decreasing bijection f 0 :(0, γ 1 1 β ][ γ 1 1 β ,+) such that γ [ f 0 ( x ) ] β = f 0 (γ x β ) for all x(0, γ 1 1 β ] can be uniquely extended to a continuous and decreasing solution f:(0,+)(0,+) 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.

  1. (i)

    Suppose that, contrary to our claim, equation (2.6) has a continuous solution f:(0,+)(0,+). Then the second iterate f 2 of f is strictly increasing. Now by (2.6) we conclude that β>0, a contradiction.

(ii1) It is clear that the identity function on (0,+) satisfies (2.6). Suppose the contrary, and let f:(0,+)(0,+) be another increasing solution of equation (2.6). Then there exists x(0,+) such that f(x)x. By the monotonicity of f, we conclude that f 2 (x)x, which contradicts (2.6).

(ii2) The first assertion is clear. To prove the second one, fix a decreasing bijection f 0 :(0, x 0 ][ x 0 ,+) such that f( x 0 )= x 0 and extend it to a function f:(0,+)(0,+) putting f(x)= f 0 1 (x) for all x( x 0 ,+). 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 f:(0,+)(0,+) is a continuous and decreasing solution of equation (2.6). Then there exists x(0,+) such that f(x)=x. Hence f 2 (x)=xγx, 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 x 0 (0,+) such that f( x 0 )= x 0 . Hence by (2.6) we get x 0 = f 2 ( x 0 )=γ x 0 β , and thus x 0 = γ 1 1 β . Consequently, f maps (0, γ 1 1 β ] bijectively onto [ γ 1 1 β ,+). Moreover, (2.6) yields f(γ x β )= f 3 (x)=γ [ f ( x ) ] β for all x(0,+). To prove the second part of the assertion, fix a decreasing bijection f 0 :(0, γ 1 1 β ][ γ 1 1 β ,+) such that γ [ f 0 ( x ) ] β = f 0 (γ x β ) for all x(0, γ 1 1 β ] and extend it to a function f:(0,+)(0,+) putting f(x)= f 0 1 (γ x β ) for all x( γ 1 1 β ,+). It is easy to calculate that f is a decreasing bijection satisfying (2.6). □

2.3 The case α0β

In this case an explicit description of all continuous solutions f:(0,+)(0,+) 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 f:(0,+)(0,+) is a solution of equation (1.1), then the function g:RR given by g(x)=logf( e x ) satisfies

g 2 (x)=logγ+αg(x)+βx.
(2.7)

Conversely, if g:RR is a solution of equation (2.7), then the function f:(0,+)(0,+) given by f(x)= e g ( log x ) satisfies (1.1).

Equation (2.7) is a special case of the polynomial-like iterative inhomogeneous equation

n = 0 N a n f n (x)=b,
(2.8)

where all a n ’s and b are real numbers and f is an unknown self-mapping; here f n denotes the n th iterate of f. For the theory of equation (2.8) and its generalizations, we refer the readers to [2230]. The problem of finding all continuous solutions of equation (2.8) seems to be very difficult. It is completely solved in [31] for N=2, but it is still open even in the case where N=3 (see [32]). It turns out that the nature of continuous solutions of equation (2.8) depends on the behavior of complex roots r 1 ,, r N of its characteristic equation n = 0 N α n r n =0. This characteristic equation is obtained by putting f(x)=rx into (2.8) with b=0; 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 f:(0,+)(0,+) of equation (1.1), denote by r 1 and r 2 the complex roots of the equation

r 2 αrβ=0.

By our assumption, we have r 1 + r 2 =α0 and r 1 r 2 =β0.

Combining Lemma 2.5 with the results from [31], we get the following theorem.

Theorem 2.6

  1. (i)

    Assume r 1 , r 2 CR. Then equation (1.1) has no continuous solution f:(0,+)(0,+).

  2. (ii)

    Assume r 1 , r 2 R{1}. Let ξ= γ 1 ( r 1 1 ) ( r 2 1 ) and f 1 , f 2 :(0,+)(0,+) be defined by

    f 1 (x)= γ 1 r 2 1 x r 1 , f 2 (x)= γ 1 r 1 1 x r 2 .

(ii1) Assume either 1< r 1 < r 2 or 0< r 1 < r 2 <1. Let 0< x 0 <ξ< y 0 . If f:(0,+)(0,+) is a continuous solution of equation (1.1), then f(ξ)=ξ, f 0 =f | I ( x 0 , f ( x 0 ) ) I ( y 0 , f ( y 0 ) ) is continuous,

f 0 ( x 0 )I ( f 1 ( x 0 ) , f 2 ( x 0 ) ) , f 0 2 ( x 0 )=γ [ f 0 ( x 0 ) ] α x 0 β ,
(2.9)
f 0 ( y 0 )I ( f 1 ( y 0 ) , f 2 ( y 0 ) ) , f 0 2 ( y 0 )=γ [ f 0 ( y 0 ) ] α y 0 β
(2.10)

and

( y x ) r 1 f 0 ( y ) f 0 ( x ) ( y x ) r 2
(2.11)

for all x,yI( x 0 ,f( x 0 ))I( y 0 ,f( y 0 )). Conversely, every continuous function f 0 :I( x 0 , f 0 ( x 0 ))I( y 0 , f 0 ( y 0 ))R such that (2.9) and (2.10) are satisfied and (2.11) holds for all x,yI( x 0 , f 0 ( x 0 ))I( y 0 , f 0 ( y 0 )) can be uniquely extended to a continuous solution f:(0,+)(0,+) of equation (1.1).

(ii2) Assume 0< r 1 <1< r 2 . Let x 0 >0. If f:(0,+)(0,+) is a continuous solution of equation (1.1), then either f | ( 0 , ξ ] { f 1 | ( 0 , ξ ] , f 2 | ( 0 , ξ ] } and f | ( ξ , + ) { f 1 | ( ξ , + ) , f 2 | ( ξ , + ) } or f 0 =f | I ( x 0 , f ( x 0 ) ) is continuous,

f 0 ( x 0 )I ( f 1 ( x 0 ) , f 2 ( x 0 ) ) , f 0 2 ( x 0 )=γ [ f 0 ( x 0 ) ] α x 0 β
(2.12)

and (2.11) holds for all x,yI( x 0 ,f( x 0 )). Conversely, every continuous function f 0 :I( x 0 , f 0 ( x 0 ))R such that (2.12) is satisfied and (2.11) holds for all x,yI( x 0 , f 0 ( x 0 )) can be uniquely extended to a continuous solution f:(0,+)(0,+) of equation (1.1).

(ii3) Assume either r 1 < r 2 <1 or 1< r 1 < r 2 <0. Let x 0 >ξ. If f:(0,+)(0,+) is a continuous solution of equation (1.1), then f(ξ)=ξ and f 0 =f | I ( x 0 , f 2 ( x 0 ) ) is continuous,

f 0 ( x 0 )I ( f 1 ( x 0 ) , f 2 ( x 0 ) ) , f 0 3 ( x 0 )=γ [ f 0 2 ( x 0 ) ] α x 0 β ,
(2.13)

and (2.11) holds for all x,yI( x 0 , f 2 ( x 0 )). Conversely, every continuous function f 0 :I( x 0 , f 0 2 ( x 0 ))R such that (2.13) is satisfied and (2.11) holds for all x,yI( x 0 , f 0 2 ( x 0 )) can be uniquely extended to a continuous solution f:(0,+)(0,+) of equation (1.1).

(ii4) Assume either r 1 < r 2 =1 or 1= r 2 < r 1 <0. Then every continuous solution f:(0,+)(0,+) of equation (1.1) is of the form

f(x)= { a r 1 + 1 ξ 1 r 1 x r 1 for  x ( 0 , a 1 ξ ] , ξ 2 x 1 for  x ( a 1 ξ , a ξ ) , a r 1 1 ξ 1 r 1 x r 1 for  x [ a ξ , + )

with some a[1,+].

(ii5) Assume either r 1 <0< r 2 or r 1 <1< r 2 <0. Then f 1 and f 2 are the only continuous solutions from (0,+) to (0,+) of equation (1.1).

  1. (iii)

    Assume r 2 =1 r 1 . Let f 3 :(0,+)(0,+) be defined by

    f 3 (x)= γ 1 r 1 1 x.

(iii1) Assume r 1 >1γ. If f:(0,+)(0,+) is a continuous solution of equation (1.1), then f 0 =f | I ( x 0 , f ( x 0 ) ) is continuous,

f 0 ( x 0 ) f 3 ( x 0 ), f 0 2 ( x 0 )=γ [ f 0 ( x 0 ) ] α x 0 β
(2.14)

and

y x f 0 ( y ) f 0 ( x ) ( y x ) r 1
(2.15)

for all x,yI( x 0 ,f( x 0 )). Conversely, every continuous function f 0 :I( x 0 , f 0 ( x 0 ))R such that (2.14) is satisfied and (2.15) holds for all x,yI( x 0 , f 0 ( x 0 )) can be uniquely extended to a continuous solution f:(0,+)(0,+) of equation (1.1).

(iii2) Assume 0< r 1 <1γ. If f:(0,+)(0,+) is a continuous solution of equation (1.1), then f 0 =f | I ( x 0 , f ( x 0 ) ) is continuous,

f 0 ( x 0 ) f 3 ( x 0 ), f 0 2 ( x 0 )=γ [ f 0 ( x 0 ) ] α x 0 β
(2.16)

and

( y x ) r 1 f 0 ( y ) f 0 ( x ) y x
(2.17)

for all x,yI( x 0 ,f( x 0 )). Conversely, every continuous function f 0 :I( x 0 , f 0 ( x 0 ))R such that (2.16) is satisfied and (2.17) holds for all x,yI( x 0 , f 0 ( x 0 )) can be uniquely extended to a continuous solution f:(0,+)(0,+) of equation (1.1).

(iii3) Assume r 1 >0 and γ=1. Then every continuous solution f:(0,+)(0,+) of equation (1.1) is of the form

f(x)= { a 1 r 1 x r 1 for  x ( 0 , a ] , x for  x ( a , b ) , b 1 r 1 x r 1 for  x [ b , + )

with some 0ab+.

(iii4) Assume r 1 <0 and γ1. Then f 3 is the unique continuous solution from (0,+) to (0,+) of equation (1.1).

(iii5) Assume r 1 <0 and γ=1. Then f 3 is a continuous solution of equation (1.1) and every other continuous solution f:(0,+)(0,+) of equation (1.1) is of the form

f(x)=a x r 1

with some a(0,+).

  1. (iv)

    Assume r 1 = r 2 .

(iv1) Assume r 1 1. Then the formula

f(x)= γ 1 r 1 1 x r 1

defines the unique continuous solution f:(0,+)(0,+) of equation (1.1).

(iv2) Assume r 1 =1γ. Then equation (1.1) has no continuous solution f:(0,+)(0,+).

(iv3) Assume r 1 =γ=1. Then every continuous solution f:(0,+)(0,+) of equation (1.1) is of the form

f(x)=ax

with some a(0,+).

Author’s contributions

The author wrote the whole article, read and approved the final manuscript.

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Acknowledgements

This research was supported by Silesian University Mathematics Department (Iterative Functional Equations and Real Analysis program).

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Correspondence to Janusz Morawiec.

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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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Keywords

  • iterative functional equation
  • continuous solution
  • bijection