Aequat. Math. 93 (2019), 137–148c© The Author(s) 20180001-9054/19/010137-12published online September 5, 2018https://doi.org/10.1007/s00010-018-0600-5 Aequationes Mathematicae

On a functional equation related to a problem of G. Derfel

Mariusz Sudzik

Dedicated to Professor Karol Baron on the occasion of his 70th birthday.

Abstract. Two years ago, during the 21st European Conference on Iteration Theory, GregoryDerfel asked: “Does there exist a non-trivial bounded continuous solution of the equation2f(x) = f(x−1)+f(−2x)?” He repeated the question during the 55th International Sympo-sium on Functional Equations. In this paper we present a partial solution of a more generalproblem, connected to the functional equation f(x) = M

(f(x + t1), f(x + t2), . . . , f(x +

tn−1), f(ax)), where n ∈ N, t1, t2, . . . , tn−1 ∈ R\{0}, a ∈ (−∞, 0) and M is a given func-

tion in n variables satisfying some additional properties. In particular, M can be a weightedquasi-arithmetic mean in n variables.

Mathematics Subject Classification. Primary 39B22; Secondary 39B05.

Keywords. Archetypal equation, Quasi-arithmetic mean, Functional equations, Equations

with rescaling.

1. Introduction

In [2,3] Bogachev, Derfel and Molchanov considered the functional equation

f(x) =∫∫

R2f(a(x − b))dμ(a, b), (1.1)

where μ is a given probability measure on R2. The study of Eq. (1.1) was

initiated by Gregory Derfel in 1989 (see [5]). A large number of interestingresults connected to (1.1), its particular cases and equations of a similar formcan be found in [1,6].

In this place it is worth mentioning that one can see Eq. (1.1) in termsof random variables. Indeed, let (Ω,A, P ) be a fixed probability space andassume that (α, β) : Ω → R

2 is a random vector whose distribution is equal toμ. Then equation

f(x) =∫

Ω

f(α(ω)(x − β(ω))

)dP (ω) (1.2)

is equivalent to (1.1).

138 M. Sudzik AEM

Equation (1.1) was called archetypal equation in [2,3] since it is a rich sourceof many famous functional and differential-functional equations. For exampleDerfel showed in [5] that one can obtain many variants of the pantographequation from (1.1). For details and examples the reader is referred to [2,3].Notice that, clearly, every constant function satisfies (1.1); these are trivialsolutions. The authors asked there if Eq. (1.1) has a non-trivial bounded con-tinuous solution. The class of bounded and continuous functions is consideredbecause of an important role played by this class in applications (for examplein astrophysics and risk theory). The interested reader is referred to [3].

The authors answered the problem of the existence of non-constant boundedcontinuous solutions of (1.1) for a large class of probability measures (see[2,3]). The behaviour of solutions depends especially on the value of K :=∫∫

R2 ln |a|dμ(a, b), which was shown by Derfel in [5]. An influence of this num-ber on (1.1) can be exemplified by the following result.

[3, Th. 1.1] Assume that∫∫

R2 ln(max(|b|, 1))dμ(a, b) and K are finite andsuppose that P(α = 0) = 0, P(|α| = 1) < 1 and P(α(c − β) = c)) < 1 for everyc ∈ R.

(i) If K < 0, then any bounded continuous solution of (1.1) is constant.(ii) If K > 0 and α > 0 a.s., then there is a non-constant bounded contin-

uous solution of (1.1).Non-constant solutions which are constructed in (ii) are based on the the-

orems discovered by Grintsevichyus (see [7]). It should be added that theassumption “α > 0 a.s.” is necessary. Under stronger regularity conditions,namely absolute continuity with a Lebesgue integrable derivative, there areno non-constant solutions of the archetypal equation in the case when P (α <0) > 0 (see Theorem 4.5 in [2]). Unfortunately, we still do not have a wideknowledge about the existence of non-trivial bounded continuous solutions of(1.1) for K > 0 and P (α < 0) > 0. It is worth noting that when P (α < 0) > 0we are able to obtain (and, moreover, describe all) non-constant bounded con-tinuous solutions only if P (α = −1) > 0 and |α| = 1 a.s. (see Theorem 2.3 in[3]); each of which is uniformly continuous.

On this account Derfel proposed to consider some special version of Eq. (1.1)with particular coefficients. During the 21st European Conference on IterationTheory, held in Innsbruck, Austria, in 2016 he posed the following

Problem. Does there exist a non-constant bounded continuous solutionf : R → R of the equation

f(x) =12f(x − 1) +

12f(−2x)? (1.3)

The question was repeated by him a year later, during the 55th InternationalSymposium on Functional Equations held in Chengdu, China.

The main result of this paper (Theorem 3.5) asserts in particular thatevery bounded continuous solution of Eq. (1.3), reaching its global extremum,

Vol. 93 (2019) On an equation related to Derfel’s problem 139

is constant. In fact this theorem holds for Eq. (3.1) which is considerably moregeneral than (1.3). In the last part we discuss the case when a solution doesnot attain the global extreme value.

2. Preliminaries

In what follows we fix a postive integer n ≥ 2, a negative a ∈ R, non-zero realnumbers t1, . . . , tn−1 and α, β ∈ R such that α < β. Let M : [α, β]n → [α, β]be an arbitrary function.

We consider the following hypotheses (A1)−(A4):

(A1) The function M(·, . . . , ·, α) is continuous with respect to each variable.

(A2) The function M(·, . . . , ·, α) is strictly increasing with respect to each

variable.

(A3) M(x1, . . . , xn) = α ⇐⇒ x1 = · · · = xn = α for all x1, . . . , xn ∈ [α, β].

(A4) M(x, . . . , x, α) �= x for all x ∈ (α, β].

We also introduce the dual hypotheses:

(B1) The function M(·, . . . , ·, β) is continuous with respect to each variable.

(B2) The function M(·, . . . , ·, β) is strictly increasing with respect to each

variable.

(B3) M(x1, . . . , xn) = β ⇐⇒ x1 = · · · = xn = β for all x1, . . . , xn ∈ [α, β].

(B4) M(x, . . . , x, β) �= x for all x ∈ [α, β).

Remark 2.1. According to the well-known “folk” theorem if M : [α, β]n →[α, β] satisfies hypotheses (A1)−(A2) or (B1)−(B2), then the functionM(·, . . . , ·, α) is continuous.

There exists a very wide and natural class of functions satisfying all theabove hypotheses, namely that of means introduced by J. Matkowski (see [9])and shortly described below, which contains a smaller but better known classof weighted quasi-arithmetic means.

Let I ⊆ R be an interval. Recall that a function M : In → R is calleda mean on I if

min{x1, x2, . . . , xn} ≤ M(x1, x2, . . . , xn) ≤ max{x1, x2, . . . , xn}for all x1, x2, . . . , xn ∈ I. Observe that, in fact, M maps the cube In ontothe interval I. The mean M is called strict if the above inequalities are sharpwhenever xi �= xj for some i, j ∈ {1, 2, . . . , n}.

Hence, we can deduce the following.

140 M. Sudzik AEM

Remark 2.2. If M is a continuous mean, strictly increasing with respect toeach variable, then M satisfies hypotheses (A1)−(A4) and (B1)−(B4).

Let I ⊆ R be an interval and ϕ1, ϕ2, . . . , ϕn : I → R be continuous strictlymonotonic functions of the same type of monotonicity. Then the function M :In → R, defined by

M(x1, . . . , xn) = (ϕ1 + · · · + ϕn)−1(ϕ1(x1) + · · · + ϕn(xn)

),

is a mean on I called the Matkowski mean generated by ϕ1, . . . , ϕn (see [9]).

Remark 2.3. Every Matkowski mean M : [α, β]n → R is strictly increasingwith respect to each variable and continuous, so it serves as an example offunctions satisfying hypotheses (A1)−(A4) and (B1)−(B4).

At the end of this chapter let us note the following simple observation.

Remark 2.4. Let i ∈ {1, 2, 3, 4} be fixed. If M : [α, β]n → [α, β] satisfieshypothesis Bi, then the function N : [−β,−α]n → [−β,−α], defined by

N(x1, x2, . . . , xn) = −M(−x1,−x2, . . . ,−xn), (2.1)

satisfies hypothesis Ai with the parameters −β and − α.

3. The main theorem

This section concerns functions f : R → R satisfying the functional equation

f(x) = M(f(x + t1), f(x + t2), . . . , f(x + tn−1), f(ax)

). (3.1)

Equation (1.3) is a particular case of Eq. (3.1): if we put n = 2, t1 = −1,a = −2 and let M be the arithmetic mean, then (3.1) takes the form (1.3).Moreover, M satisfies hypotheses (A1)−(A4) and (B1)−(B4) with arbitraryα, β ∈ R such that α < β.

In this chapter we consider the functions reaching the global extreme value.We start with a very simple observation.

Remark 3.1. Let f : R → [α, β] be a solution of Eq. (3.1) reaching the globalmaximum. Then the function g := −f reaches the global minimum and satisfiesthe equation

g(x) = N(g(x + t1), g(x + t2), . . . , g(x + tn−1), g(ax)

)

with N given by formula (2.1).

The following two possibilities will be covered by our results:(a) M satisfies (A1)−(A4) and f takes its global minimum,(b) M satisfies (B1)−(B4) and f takes its global maximum.

Vol. 93 (2019) On an equation related to Derfel’s problem 141

Results in this chapter are formulated and proved for the first case only. Thesecond case can be obtained by a similar argumentation or applying Remarks2.3 and 3.1. In particular, the results presented below hold when M is aMatkowski mean.

We are going to prove two useful lemmas.

Lemma 3.2. Assume that M satisfies hypotheses (A1), (A2) and (A4). Thenthe sequence (uk)k∈N0 , defined by u0 = β and

uk = M(uk−1, uk−1, . . . , uk−1, α) for k ∈ N, (3.2)

is strictly decreasing and converges to α.

Proof. Observe that by (A4) we have u1 �= β, and thus u1 < u0. Assume thatfor some k ∈ N we have uk < uk−1. Using (A2) we get

uk+1 = M(uk, uk, . . . , uk, α) < M(uk−1, uk−1, . . . , uk−1, α) = uk.

Therefore the sequence (uk)k∈N0 is strictly decreasing. We also know that(uk)k∈N0 is bounded, and thus convergent. Let

u∞ := limk→∞

uk.

Note that, by (A1)

u∞ = limk→∞

M(uk, uk, . . . , uk, α) = M(u∞, u∞, . . . , u∞, α)

and now hypothesis (A4) implies that u∞ = α. �

Lemma 3.3. Assume that M satisfies hypothesis (A3) and let f : R → [α, β] bea solution of Eq. (3.1). If there exists x0 ∈ R such that f(x0) = α, then

f

(x0 +

n−1∑

i=1

kiti

)= α

for all k1, k2, . . . , kn−1 ∈ N0.

Proof. Fix arbitrary k1, k2, . . . , kn−1 ∈ N0. Applying equality (3.1) to x = x0

we get

α = f(x0) = M(f(x0 + t1), f(x0 + t2), . . . , f(x0 + tn−1), f(ax0)

).

Now (A3) implies that

f(x0 + ti) = α for all i = 1, 2, . . . , n − 1.

Repeating this reasoning ki times for all i = 1, 2, . . . , n − 1 we come to therequired equality. �

142 M. Sudzik AEM

Corollary 3.4. Assume that M satisfies hypothesis (A3) and let f : R → [α, β]be a solution of Eq. (3.1). If there exists x0 ∈ R such that f(x0) = α, then

f

(ax0 +

n−1∑

i=1

(li + aki)ti

)= α

for all k1, k2, . . . , kn−1, l1, l2, . . . , ln−1 ∈ N0.

Proof. Fix k1, . . . , kn−1, l1, . . . , ln−1 ∈ N0. If we apply Lemma 3.3, we obtain

f

(x0 +

n−1∑

i=1

kiti

)= α.

In the next step we can apply equality (3.1) to the point x0 +∑n−1

i=1 kiti.Making use of (A3) we get

f

(ax0 +

n−1∑

i=1

akiti

)= α.

Now, applying Lemma 3.3 to the point ax0 +∑n−1

i=1 akiti, we come to

f

(ax0 +

n−1∑

i=1

(li + aki)ti

)= f

(ax0 +

n−1∑

i=1

liti +n−1∑

i=1

akiti

)= α.

�

Now we are in a position to prove the main theorem.

Theorem 3.5. Assume that a �= −1, M satisfies hypotheses (A1)−(A4) and letf : R → [α, β] be a continuous solution of Eq. (3.1). If f attains the globalminimum, then f is constant.

Proof. Let x0 ∈ R be such that f(x0) = α. The proof depends on the param-eter a, so it is split into three cases.

Case 1: a ∈ Z. We define the sequence (Dj)j∈N0 of sets setting

Dj :=

{ax0 +

∑n−1i=1 siti

aj: s1, . . . , sn−1 ∈ Z

}

and let

D :=∞⋃

j=0

Dj .

Since a < −1, the set D is dense in R. The sequence (Dj)j∈N0 also has thefollowing properties:

Dj + ti ⊂ Dj for all j ∈ N0 and i = 1, 2, . . . , n − 1, (3.3)

aDj+1 ⊂ Dj for all j ∈ N0. (3.4)

Vol. 93 (2019) On an equation related to Derfel’s problem 143

By Corollary 3.4 we have

f

(ax0 +

n−1∑

i=1

(li + aki)ti

)= α for all k1, , . . . , kn−1, l1, . . . , ln−1 ∈ N0.

Since a is a negative integer, the expression li + aki runs through the wholeset Z while i = 1, 2, . . . , n − 1. Therefore

f

(ax0 +

n−1∑

i=1

siti

)= α for every s1, s2, . . . , sn−1 ∈ Z.

This means that f|D0 = α. Assume that f|Dj= α for a j ∈ N0 and let x ∈ Dj+1

be fixed. We show that f(x) = α. Let (uk)k∈N0 be the sequence defined by (3.2)and let y = x+

∑n−1i=1 kiti, where k1, k2, . . . , kn−1 ∈ N0. Then f(y) ≤ β = u0.

Assume that for some p ∈ N0 we have

f

(x +

n−1∑

i=1

kiti

)≤ up for all k1, k2, . . . , kn−1 ∈ N0. (3.5)

Take y = x +∑n−1

i=1 kiti for some k1, k2, . . . , kn−1 ∈ N0. Then, by (3.3) and(3.4), we have ay ∈ Dj . Therefore, using condition (3.5) and hypothesis (A2),we come to

f(y) = M(f(y + t1), f(y + t2), . . . , f(y + tn−1), f(ay)

)

= M(f(y + t1), f(y + t2), . . . , f(y + tn−1), α

)

≤ M(up, up, . . . , up, α

)= up+1.

Thus, using induction, we have

f(x) ≤ up for every p ∈ N0,

hence, by Lemma 3.2,

f(x) ≤ α.

Thus f|Dj+1 = α because the point x was arbitrary. Consequently,

f(x) = α for every x ∈ D.

In this case the proof is finished because of the continuity of f and the densityof the set D.

Case 2: a ∈ Q\Z. Choose p ∈ Z and q ∈ N such that a = p/q and p, q arecoprime. Then obviously p < 0 and q ≥ 2. By Corollary 3.4 we have

f

(ax0 +

n−1∑

i=1

(ki +

pliq

)ti

)= α

144 M. Sudzik AEM

for every k1, . . . , kn−1, l1, . . . , ln−1 ∈ N0. The expression ki+pli/q runs throughthe set Z/q while i = 1, 2, . . . , n − 1. Hence we have

f

(ax0 +

∑n−1i=1 siti

q

)= α for every s1, . . . , sn−1 ∈ Z.

Now we define a sequence (Dj)j∈N of sets putting

Dj :=

{

ajx0 +∑n−1

i=1 sitiqj

: s1, . . . , sn−1 ∈ Z

}

and letting

D :=∞⋃

j=1

Dj .

The set D is dense in R as q > 1. The equality f|D1 = α was shown above.Assume that f|Dj

= α for some j ∈ N. Let x ∈ Dj be fixed. Then there exists1, . . . , sn−1 ∈ Z such that x = ajx0 +

∑n−1i=1 siti/qj . If we apply equality (3.1)

to the point x, we get

α = f(x) = M(f(x + t1), . . . , f(x + tn−1), f(ax)

).

Hence, by hypothesis (A3), we obtain f(ax) = α. This equality means that

f

(aj+1x0 +

∑n−1i=1 psitiqj+1

)= α.

Applying Lemma 3.3 to the point aj+1x0 +∑n−1

i=1 psiti/qj+1, we obtain

f

(aj+1x0 +

n−1∑

i=1

qj+1ki + psiqj+1

ti

)= α for all k1, . . . , kn−1 ∈ N0.

The point x was taken arbitrarily. Therefore we have proved that f takes thevalue α on the set{

aj+1x0 +∑n−1

i=1 (qj+1ki + psi)tiqj+1

: s1, . . . , sn−1 ∈ Z, k1, . . . , kn−1 ∈ N0

}

.

Since qj+1 and p are coprime, the expression qj+1ki +psi runs through the setZ while i = 1, . . . , n − 1. Consequently, the set just above is equal to Dj+1.Hence, by induction, we get f|D = α and the proof is completed as D is adense subset of the real line and f is continuous.

Case 3: a ∈ (−∞, 0)\Q. By Corollary 3.4 we have

f(ax0 + (l + ak)t1

)= α for every k, l ∈ N0.

Kronecker’s density theorem (see Chapter XXIII in [8]) asserts that the setD := {l + ak : l, k ∈ N0} is dense in R. Therefore also the set ax0 + Dt1 isdense in R and thus f is constant. �

Vol. 93 (2019) On an equation related to Derfel’s problem 145

Remark 3.6. Note that hypotheses (A1), (A2) and (A4) are important only inthe first case of the proof. In the case when a /∈ Z hypotheses (A1), (A2) and(A4) can be ommited.

Remark 3.7. Note that if M is a weighted arithmetic mean and |a| < 1, thenevery bounded continuous solution of Eq. (3.1) is constant. This follows as aspecial case of Theorem 3.2 from [2], with no additional assumptions on f .

Remark 3.8. Theorem 3.5 provides a partial answer to the problem of GregoryDerfel:

If f : R → R is a bounded continuous solution of Eq. (1.3) reaching theglobal minimum (maximum), then f is constant.

4. Further results

The situation when a = −1 is completely different. Now Eq. (3.1) takes theform

f(x) = M(f(x + t1), f(x + t2), . . . , f(x + tn−1), f(−x)

). (4.1)

The following obvious fact shows that Eq. (4.1) can have a great number ofnon-constant solutions belonging to the class of bounded continuous functionsand reaching the global minimum and global maximum.

Remark 4.1. If M is reflexive, i.e.

M(x, x, x, . . . , x) = x for every x ∈ [α, β],

and there exists s ∈ R\{0} such that all fractions t1s , t2

s , . . . , tn−1s are integers,

then every continuous even function f : R → R, with period s, is a boundedsolution of Eq. (4.1), attaining both of its global extremes.

If such s cannot be found, we only have the trivial solutions.

Remark 4.2. Assume that M satisfies hypothesis (A3) [(B3)] and there arenumbers p, q ∈ {1, 2, . . . , n − 1} such that tp

tq/∈ Q and let f : R → [α, β] be

a bounded continuous solution of Eq. (4.1). If f attains the global minimum[maximum], then f is constant.

Proof. Let x0 ∈ R be such that f(x0) = α. By Corollary 3.4 we have

f

(− x0 + (kp − lp)tp + (kq − lq)tq

)= α

for every kp, kq, lp, lq ∈ N0, or simply

f

(− x0 + sptp + sqtq

)= α for all sp, sq ∈ Z.

146 M. Sudzik AEM

By Kronecker’s density theorem (see [8]) we know that the set {sq + sptp/tq :sp, sq ∈ Z} is dense in the real line. Consequently, also {sptp + sqtq − x0 :sp, sq ∈ Z} is a dense subset of R, and thus f is constant. �

In the case when M is a weighted arithmetic mean, Remark 4.1, its converseand Remark 4.2 are special cases of Theorem 2.3 in [3], which is an extensionof the Choquet–Deny theorem (see [4]).

If a bounded continuous solution of (3.1) does not attain the global min-imum or maximum, then the problem of its form is much more difficult andis still unsolved. The result below (compare with Theorems 4.2 and 4.3 in [2])describes some properties of possible non-constant solutions of Eq. (3.1).

Proposition 4.3. Let M be a continuous function satisfying hypothesis (A3)[(B3)] and let f : R → R be a continuous solution of Eq. (3.1) such thatinf f(R) = α [sup f(R) = β]. If

α < f(x) < β for every x ∈ R,

then for all c ∈ R

α = inf{f(x) : x > c} = inf{f(x) : x < c}[β = sup{f(x) : x > c} = sup{f(x) : c < x}].

Proof. We prove this theorem under the assumptions (A3) and inf f(R) = αonly. Let c ∈ R and take any sequence (xn)n∈N of real numbers such that

limn→∞ f(xn) = α.

If it was bounded, then one could find a subsequence of (xn)n∈N convergentto some point x0 ∈ R. Hence, by the continuity of f , we would get f(x0) = α,a contradiction. Therefore (xn)n∈N is unbounded. Assume for instance that itis unbounded from above. Then the interval (c,+∞) contains infinitely manyelements of this sequence. Let (xnk

)k∈N be its subsequence such that xnk> c

for every k ∈ N. Since f(xn) → α, we have

limk→∞

f(xnk) = α,

and thus

α = inf{f(x) : x > c}.

It remains to show that α = inf{f(x) : x < c}. Suppose that the sequence(f(axnk

))k∈N

is not convergent to α. The sequences(f(axnk

))k∈N

and(f(xnk

+ti)

)k∈N

are bounded for each i = 1, . . . , n − 1. We can choose a subsequence(xnkl

)l∈N of the sequence (xnk)k∈N such that the sequences

(f(xnkl

+ ti))l∈N

are convergent for every i ∈ {1, 2, . . . , n − 1} and the sequence(f(axnkl

))l∈N

Vol. 93 (2019) On an equation related to Derfel’s problem 147

has a limit different from α. Then, by the continuity of M and hypothesis(A3), we get

liml→∞

M(f(xnkl

+ t1), f(xnkl+ t2), . . . , f(xnkl

+ tn−1), f(axnkl))

> α.

On the other hand f(xnkl) → α. It contradicts the assumption that f is a

solution of Eq. (3.1). Therefore f(axnk) → α and there exists k0 ∈ N such

that axnk< c for every k ≥ k0. Hence inf{f(x) : x < c} = α. �

We immediately get the following

Corollary 4.4. Let M be a continuous function satisfying hypotheses (A3) and(B3) and let f : R → R be a continuous solution of Eq. (3.1) such thatinf f(R) = α and sup f(R) = β. If

α < f(x) < β for every x ∈ R,

then f has no horyzontal asymptotes.

In particular the meaning of these theorems is that every non-constantsolution of Eq. (1.3) should be oscillating in infinity. So the main question isabout the existence or non-existence of such oscillating solutions.

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148 M. Sudzik AEM

Mariusz SudzikFaculty of Mathematics, Computer Science and EconometricsUniversity of Zielona GoraSzafrana 4a65-516 Zielona GoraPolande-mail: M.Sudzik@wmie.uz.zgora.pl

Received: March 27, 2018

Revised: July 12, 2018