Acta Math. Hungar., 138 (4) (2013), 329–340DOI: 10.1007/s10474-012-0278-4

First published online November 15, 2012

ON THE FUNCTIONAL EQUATIONS RELATEDTO A PROBLEM OF Z. BOROS AND

Z. DAROCZY

M. BALCEROWSKI

ul. 1 Maja 56/80, 41–200 Sosnowiec, Polande-mail: mbalcerowski@math.us.edu.pl

(Received October 26, 2011; revised July 10, 2012; accepted July 26, 2012)

Abstract. We present some general properties of solutions of an equationwhich appeared in a problem posed by Z. Boros and Z. Daroczy. Using them wesolve this equation in some classes of functions assuming regularity of solutions.We also present some facts concerning regular solutions of two related equations.

Introduction

The problem of solving the equation

(1) f(x + 2f(y)

)+ f

(y + 2f(x)

)= 2f(x) + 2f(y) + x + y

in the class of self-mappings of an abelian group was posed by Z. Borosand Z. Daroczy during the Forty-third International Symposium on Func-tional Equations (see [2], also [1]). The authors of that problem presentedsome forms of solutions of equation (1) (see [2]). Further examples of so-lutions of (1) were found by J. Ratz (oral communication). In this paperwe solve (1) in some classes of functions. In particular, we determine allthe mappings f : R → R satisfying equation (1) such that id + 2f has theDarboux property. We also present some results concerning equations

(2) f(x + 2f(x)

)= x + 2f(x)

and

(3) f(2f(x)

)= f(x) + x.

Key words and phrases: Darboux property, fixed point, functional equation.Mathematics Subject Classification: primary 39B12, 30D05, 39B52, secondary 39B22, 26A18.

0236-5294/$20.00 c© 2012 Akademiai Kiado, Budapest, Hungary

330 M. BALCEROWSKI

In the class of additive self-mappings of an abelian group equations (1), (2)and (3) are equivalent to the equation

(4) f(x + 2f(y)

)= f(x) + y + f(y).

This equation was discussed in [1] and was a motivation for the authors of [1]to consider equation (1).

1. Preliminary results

We start with a few simple facts.

Proposition 1. Let (G,+) be an abelian group with no elements oforder 2 and let f : G → G be a solution of equation (1). Then f satisfiesequation (2). In particular, 3 Fix f ⊂ Fix f . Moreover, f −1

({0}

)⊂ {0}.

Proof. Putting y = x in (1) we see that 2f(x + 2f(x)

)= 2x + 4f(x) =

2(x + 2f(x)

)for every x ∈ G, and thus we obtain the first statement. Now

fix an x ∈ Fix f . We have f(3x) = f(x + 2f(x)

)= x + 2f(x) = 3x, which

shows that 3x ∈ Fix f . Assuming that f(x) = 0 for an x ∈ G we have x =x + 2f(x) = f

(x + 2f(x)

)= f(x) = 0. This proves that f −1

({0}

)⊂ {0}.

�

Proposition 2. Let (G,+) be a set with a binary operation and witha zero. Assume that f : G → G satisfies equation (1) and 2f(0) = 0. If theleft cancellation law holds:

x + y = x + z ⇒ y = z for all x, y, z ∈ G,

then f satisfies equation (3).

Proof. Putting y = 0 in (1) we obtain f(x)+ f(2f(x)

)= 2f(x)+x for

every x ∈ G, whence we deduce that f satisfies equation (3). �

Proposition 3. Let X be a linear space over a field K such that Q ⊂ K

⊂ C, D be a subset of X such that card D � 2 and let f : D → D. Assumethat there exist x0 ∈ X and a ∈ K such that f(x) = ax + x0 for every x ∈ D.Assume that D + D ⊂ D. Then f satisfies equation (1) [equation (2)] if andonly if x0 = 0 and a ∈ {1, −1/2}. Assume that 2D ⊂ D. Then f satisfiesequation (3) if and only if x0 = 0 and a ∈ {1, −1/2}.

Proof. We proceed for solutions of (1) only; for equations (2) and (3)the argument is similar. The function f satisfies equation (1) if and only iffor all x, y ∈ D we have

f(x + 2ay + 2x0) + f(y + 2ax + 2x0) = 2ax + 2x0 + 2ay + 2x0 + x + y.

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ON FUNCTIONAL EQUATIONS 331

This is equivalent to

ax + 2a2y + 2ax0 + x0 + ay + 2a2x + 2ax0 + x0 = 2ax + 2ay + 4x0 + x + y,

i.e.

(2a2 − a − 1)(x + y) + (4a − 2)x0 = 0 for all x, y ∈ D.

Since card D � 2 the last equality means that (1) is satisfied if and only if2a2 − a − 1 = 0 and 2(2a − 1)x0 = 0, and thus the assertion follows. �

Proposition 4. Let (G,+) be a set with a binary operation fulfilling theleft cancellation law and let f : G → G be a solution of equation (3). Thenf is one-to-one. If, in addition, G has a zero then 2f(0) = 0.

Proof. Fix any x, y ∈ G and assume that f(x) = f(y). Then f(2f(x)

)

= f(2f(y)

), whence, using (3) and the cancellation law, we deduce that

x = y. Therefore f is one-to-one. Putting x = 0 in (3) we get f(2f(0)

)

= f(0). Using the first statement we infer that 2f(0) = 0. �

Proposition 5. Let (G,+) be a group and let f : G → G be a solutionof equation (3). Then 2n Fix f ⊂ Fix f for every n ∈ N. If, in addition, Fix f⊂ 2f(G), then Fix f ⊂ 2n Fix f for every n ∈ N.

Proof. Fix an x ∈ Fix f . Using (3) we obtain f(2x) = f(2f(x)

)=

f(x) + x = 2x, whence 2x ∈ Fix f and, by induction, 2nx ∈ Fix f for alln ∈ N. Now assume that Fix f ⊂ 2f(G) and take any y ∈ Fix f . Theny = 2f(x) for some x ∈ G. It follows from (3) that 2f(x) = f

(2f(x)

)

= f(x) + x, which gives f(x) = x. Hence y ∈ 2Fix f . By induction Fix f⊂ 2n Fix f for every n ∈ N. �

The following corollary is an easy consequence of Remark 5.

Corollary 1. Let f : R → R be a solution of equation (3). If thereexists an a ∈ (0, ∞) such that [0, a) ⊂ Fix f

[(−a, 0] ⊂ Fix f

], then [0, ∞)

⊂ Fix f[(− ∞, 0] ⊂ Fix f

].

Proposition 6. Let (G,+) be an abelian group and let f : G → G be asolution of equation (3). Then the set A =

{x ∈ G : 2f(x) = −x

}satisfies

the condition −A ⊂ A.

Proof. Fix any x ∈ A. Applying (3) we obtain

2f(−x) = 2f(2f(x)

)= 2f(x) + 2x = x

which means that −x ∈ A. �

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332 M. BALCEROWSKI

Remark 1. As it has been observed by S. Nabeya (cf. [7]) equation (2)after putting ϕ(x) = x + 2f(x) takes the form ϕ2(x) = 3ϕ(x). This equationwas considered in [4] where the author gave its general solution in the classof continuous functions ϕ : E → E for an arbitrary interval E ⊂ R (cf. The-orem 15.15) and in the class of functions ϕ : E → E where E is an arbitrarysubset of R (cf. Theorem 15.14). Using these facts one can determine the so-lution of equation (2) in the two mentioned classes of functions. Moreover,applying Lemma 15.11 from [4] we deduce that if f : R → R satisfies (2),then the set A = (id + 2f)(R) fulfils the condition 3A ⊂ A. This fact will beused in the proof of Theorem 1 (see below).

The following Theorem R has been formulated and proved by J. Ratz(oral communication).

Theorem R. Let (G,+) be an abelian group and let f : G → G be asolution of equation (1). Then 2f(0) = 0.

From Theorem R, J. Ratz deduced Corollary R below.

Corollary R. Let (G,+) be an abelian group and let f : G → G be asolution of equation (1). Then f satisfies equation (3) and f is one-to-one.

The next three results present some general properties of solutions ofequation (1).

Lemma 1. Let (G,+) be an abelian group and let f : G → G be a solu-tion of equation (1). Then the condition

2x ∈ Fix f and 2f(3x) = 6x =⇒ −x ∈ Fix f

is satisfied for every x ∈ G. In particular, − Fix f ⊂ Fix f .

Proof. Inserting 2f(x) in place of x in (1) we obtain

f(2f(x) + 2f(y)

)= 2f

(2f(x)

)+ 2f(y) + 2f(x) + y − f(y + 2f

(2f(x)

))

for all x, y ∈ G. On account of Corollary R we obtain f(2f(x)

)= f(x) + x

for every x ∈ G. Therefore for all x, y ∈ G we get

f(2f(x) + 2f(y)

)= 4f(x) + 2x + 2f(y) + y − f

(y + 2x + 2f(x)

),

whence, by (1),

f(2f(x) + 2f(y)

)= 4f(x) + 2x + 2f(y) + y + f

(x + 2f(y + 2x)

)

− 2f(x) − 2f(y + 2x) − x − y − 2x

that is

f(2f(x) + 2f(y)

)= 2f(x) + 2f(y) + f

(x + 2f(y + 2x)

)− 2f(y + 2x) − x.

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ON FUNCTIONAL EQUATIONS 333

The left-hand side of that formula is symmetric with respect to x and y, andthus

f(x + 2f(y + 2x)

)− 2f(y + 2x) − x = f

(y + 2f(x + 2y)

)− 2f(x + 2y) − y

for all x, y ∈ G. Putting y = −2x we obtain

f(x) = f(

− 2x + 2f(−3x))

− 2f(−3x) + 3x,

that is

(5) f(2x + 2f(3x)

)= 2f(3x) + f(−x) + 3x for every x ∈ G.

Fix an x ∈ G and assume that 2x ∈ Fix f and 2f(3x) = 6x. Then, by Propo-sition 5, we have 2x + 2f(3x) = 8x ∈ Fix f . Applying (5) we infer thatf(−x) = −x.

Now fix u ∈ Fix f . Using (1) with x = y = u we obtain 2f(3u) = 6u.Moreover Proposition 5 implies 2u ∈ Fix f . On account of the first state-ment −u ∈ Fix f . �

Lemma 2. Let (G,+) be an abelian group and let f : G → G be a solu-tion of equation (1). Then the condition

(6) y, y + x, y + 2x ∈ Fix f ⇒ y − mx ∈ Fix f for all x, y ∈ G

is satisfied for every m ∈ Z.

Proof. Inserting x − 2f(y) in place of x in (1) we obtain

f(x) + f(y + 2f(x − 2f(y)

)) = 2f

(x − 2f(y)

)+ x + y for all x, y ∈ G.

(7)

It follows easily from (7) that if x, y ∈ Fix f and x − 2y ∈ Fix f then 2x − 3y∈ Fix f .

By induction we show (6) for every m ∈ N. Fix arbitrary x, y ∈ G and as-sume that the elements y, y + x and y + 2x belong to Fix f . From Lemma 1we infer that −y ∈ Fix f . Putting x0 = y + 2x and y0 = y + x we obtainx0 − 2y0 = y + 2x − 2y − 2x = −y ∈ Fix f . Hence, in view of the above ob-servation, we deduce that 2x0 − 3y0 ∈ Fix f , that is x − y ∈ Fix f . ApplyingLemma 1 again we get y − x ∈ Fix f . Therefore (6) holds for m = 1. Now as-sume (6) for some m ∈ N and fix x, y ∈ G such that y, y + x and y + 2x arefixed points of the function f . Then y − x ∈ Fix f and putting y′ = y − xwe see that y′, y′ + x and y′ + 2x belong to Fix f . Hence, by inductionhypothesis, y′ − mx ∈ Fix f which yields y − (m + 1)x ∈ Fix f .

Now assume that x, y ∈ G fulfil the left side of implication (6). Definingz = −x and using the first part of the proof we deduce that y + z and y + 2z

Acta Mathematica Hungarica 138, 2013

334 M. BALCEROWSKI

belong to Fix f . Using this again we obtain y − mz ∈ Fix f for every m ∈ N,that is y − (−m)x ∈ Fix f for every m ∈ N. If m = 0, then condition (6) isobvious. �

Corollary 2. Let (G,+) be an abelian group with no elements of or-der two and let f : G → G be a solution of equation (1). If x ∈ Fix f , thenmx ∈ Fix f for every m ∈ Z.

Proof. Fix any x ∈ Fix f . According to Corollary R the function fsatisfies equation (3) and hence, by Proposition 5, the inclusion 2Fix f ⊂Fix f follows. Making use of Theorem R and the assumption about G wededuce that f(0) = 0. Therefore the elements 0, x and 2x are fixed pointsof f . Using Lemma 2 with y = 0 we obtain mx ∈ Fix f for every m ∈ Z. �

2. Solutions of equations (2) and/or (3)

Now we present some theorems concerning equations (2) and (3).

Theorem 1. Let f : R → R be a function such that the set (id + 2f)(R)is an interval. Then f satisfies simultaneously equations (2) and (3) if andonly if either f = id, or f = −id/2.

Proof. Assume that f satisfies equations (2) and (3). Let P =(id + 2f)(R). By the assumption the set P is an interval and, obviously,P ⊂ Fix f . Making use of Proposition 4 we deduce that f(0) = 0 whence0 ∈ P . Moreover, by Remark 1, the inclusion 3P ⊂ P follows. Therefore ei-ther P = {0}, or P is not bounded. If P = {0}, then, of course, f = −id/2.So assume that P is not bounded. We may assume that P is not boundedfrom above. Then [0, ∞) ⊂ P and hence [0, ∞) ⊂ Fix f . By virtue of Re-mark 4 the function f is one-to-one. Therefore f(x) < 0 for every x < 0whence x + 2f(x) < x for every x < 0. So P is not bounded from below.Consequently, P = R which means that f = id. Using Proposition 3 weprove the converse implication. �

Theorem 2. Let f : R → R be a function satisfying the condition Fix f⊂ 2f(R). Assume that there exists a non-degenerated interval I ⊂ R con-taining zero such that (id + 2f)(I) is an interval. If f satisfies simultane-ously equations (2) and (3), then either f(x) = x for every x ∈ (− ∞, 0], orf(x) = x for every x ∈ [0, ∞), or f = −id/2.

Proof. It follows from Proposition 3 that the functions id and −id/2are solutions of equations (2) and (3). So assume that f : R → R satisfiesequations (2) and (3). We consider the case when I is of the form [0, a] forsome a > 0, as in the other cases the proof is analogous.

Let P = (id + 2f)(I). Obviously P ⊂ Fix f . By the assumption P isan interval. From Proposition 4 we infer that f(0) = 0 and thus 0 ∈ P .

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ON FUNCTIONAL EQUATIONS 335

If P = {0}, then f |I = −id/2. Making use of Proposition 6 we obtainf(x) = −x/2 for every x ∈ [−a, a]. Fix any x0 ∈ R and put y0 = x0 +2f(x0).Applying Proposition 5 we get y0/2n ∈ Fix f for every n ∈ N. We can findk ∈ N such that y0/2k ∈ [−a, a]. Then

y0/2k = f(y0/2k) = −y0/2k+1

whence 2y0 = −y0, and therefore y0 = 0. This yields f(x0) = −x0/2, andthus f = −id/2. Now assume that there exists y0 ∈ R\{0} such that y0 ∈ P .If y0 > 0, then [0, y0] ⊂ P and, making use of Corollary 1, we deduce that[0, ∞) ⊂ Fix f . If y0 < 0, then by similar way we show that (− ∞, 0] ⊂ Fix f .�

Proposition 7. Let (G,+) be an abelian group and let f : G → G be asolution of equations (2) and (3). Then 2f is additive on its graph.

Proof. For any x ∈ G, by (2) and (3), we obtain

2f(x + 2f(x)

)= 2x + 4f(x) = 2f(x) + 2

(f(x) + x

)

= 2f(x) + 2f(2f(x)

). �

The following theorem is due to J. Matkowski [6]; for the proof seeW. Jarczyk ([3], Theorem 1.2) and J. Matkowski ([5], Theorem 1).

Theorem M. Let f : (0, ∞) → (0, ∞) be a function such that the map-ping (0, ∞) � x �→ f(x)/x is monotonic. If f is additive on its graph, thenthere exists c ∈ (0, ∞) such that f(x) = cx for every x ∈ (0, ∞).

Making use of Remark 7 and Theorem M we get the following result.

Corollary 3. Let f : (0, ∞) → (0, ∞) be such that the mapping (0, ∞)� x �→ f(x)/x is monotonic. Then f satisfies simultaneously equations (2)and (3) if and only if f(x) = x for every x ∈ (0, ∞).

Proof. Assume that f satisfies simultaneously (2) and (3). On accountof Proposition 7 the function 2f is additive on its graph. By virtue of Theo-rem M there exists c ∈ (0, ∞) such that f(x) = cx for every x > 0. The restis a consequence of Proposition 3. �

It is an easy observation that if f : R → R is a solution of (3), then thefunction g : R → R, given by

(8) g(x) = f(2x),

satisfies the equation

(9) g2(x) = g(x) + 2x.

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336 M. BALCEROWSKI

Using this fact and applying a Tabors’ result concerning (9) (see [9], Theo-rem 3) we obtain the following result.

Theorem 3. Let D ⊂ (− ∞, 0), or D ⊂ (0, ∞). Assume that 2D = D.Let f : D → D be a function fulfilling the condition f(D) = D. Then f sat-isfies equation (3) if and only if f(x) = x for every x ∈ D.

Proof. Assume that f satisfies equation (3). By virtue of Proposition 4the function f is one-to-one. Therefore the function g, given by (8), is abijection mapping D onto D. Applying Theorem 3 from [9] we infer thatg(x) = 2x and hence f(x) = x for every x ∈ D. The converse follows fromProposition 3. �

Theorem 4. Let f : R → R be a function fulfilling the condition Fix f⊂ 2f(R). Assume that f satisfies equations (2) and (3). If I is one of theintervals (− ∞, 0) and (0, ∞), and f(I) is an interval, then either f(x) = xfor every x ∈ I , or f = −id/2.

Proof. We consider only the case I = (0, ∞). Let P ={x + 2f(x) :

x ∈ R}. It follows from (2) that P ⊂ Fix f . By Proposition 4 we know that

f(0) = 0 and f is one-to-one. If P is a singleton, then f = −id/2. So we mayassume that there exists u ∈ R\{0} such that u ∈ P . We show that the setP ∩ (0, ∞) is non-empty. Suppose on the contrary that P ⊂ (− ∞, 0]. Thenu < 0 and f(x) � −x/2 for every x ∈ R. Therefore the interval f

((0, ∞)

)

is contained in (− ∞, 0) and is not bounded from below. Hence we deducethat there exists b < 0 such that (− ∞, b) ⊂ f

((0, ∞)

). We can find n ∈ N

such that 2nu < b. So 2nu = f(y) for some y > 0. As u ∈ Fix f , by virtueof Proposition 5, we obtain 2nu ∈ Fix f . This yields

f(2nu) = 2nu = f(y)

whence 2nu = y, a contradiction. Therefore there exists w > 0 such thatw ∈ P . In particular, f(w) = w which, in view of Proposition 5, implies thatthe set

{2nw : n ∈ Z

}is contained in Fix f . Thus, since Fix f ∩ (0, ∞) ⊂

f((0, ∞)

)and f

((0, ∞)

)is an interval, it follows that (0, ∞) ⊂ f

((0, ∞)

).

Moreover, as 0 ∈ f((0, ∞)

), we get f

((0, ∞)

)⊂ (0, ∞). Hence f

((0, ∞)

)

= (0, ∞). Applying Theorem 3 we infer that f(x) = x for every x ∈ I . �In Theorem 4 it is not sufficient to assume that the set f(R) is an inter-

val.

Example 1. Using Theorem 15.14 [4], or making a direct calculation,we deduce that the function f : R → R, given by

f(x) =

{x for x ∈ Q,

−x/2 for x ∈ R\Q

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ON FUNCTIONAL EQUATIONS 337

is a solution of equation (2). Moreover, f satisfies equation (3), f is contin-uous at 0 and f(R) = R.

Theorem 5. Let I be one of the intervals [0, ∞) and (− ∞, 0] and letf : I → I be a function such that f(I) is an interval. Then f satisfies equa-tion (3) if and only if f(x) = x for every x ∈ I .

Proof. We consider the case I = [0, ∞) only. Assume that f satisfiesequation (3). Since f

(2f(x)

)= f(x) + x � x for every x � 0, the function

f is not bounded from above on I . Using Proposition 4 we deduce that f isone-to-one and f(0) = 0. Hence f

([0, ∞)

)= [0, ∞) and therefore f

((0, ∞)

)

= (0, ∞). Using Theorem 4 we obtain f(x) = x for every x ∈ (0, ∞). Hencewe obtain the required form of f . The converse follows immediately fromProposition 3. �

Using the theorem of S. Nabeya (cf. [7]) J. Ratz solved equation (1) in theclass of all continuous functions f : R → R satisfying condition 2f(0) = 0.In the theorem below we use a similar argument for solving equation (3)itself.

Theorem 6. Let f : R → R be a continuous function. Then f satisfiesequation (3) if and only if either f = id, or f = −id/2.

Proof. Assume that f satisfies equation (3). Then the function g, givenby (8), satisfies equation (9). By virtue of Nabeya’s result [7] either g = 2id,or g = −id. Hence we obtain the required forms of f . The converse is aconsequence of Proposition 3. �

3. Solutions of equation (1)

Before presenting the next results we introduce the following definition.A subset A of a group (G,+) is called absorbing if for every x ∈ G there

exists an n ∈ N such that x ∈ nA.Applying Lemma 2 we can prove the following theorem.

Theorem 7. Let (G,+) be an abelian topological group such that ev-ery neighborhood of zero in G is absorbing. Let f : G → G be a solution ofequation (1). If int Fix f = ∅, then f = id.

Proof. Assume that the set Fix f has a non-empty interior. We shallshow that f = id. Let y be an element of int Fix f . There exists a neighbor-hood U ⊂ G of zero such that y + U ⊂ Fix f . We can find a neighborhoodV ⊂ G of zero such that V + V ⊂ U . For every x ∈ V the elements y + xand y + 2x belong to Fix f . Applying Lemma 2 we obtain y − nx ∈ Fix ffor every x ∈ V and for every n ∈ N. Hence

⋃∞n=1(y − nV ) ⊂ Fix f , that

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338 M. BALCEROWSKI

is y − ⋃∞n=1 nV ⊂ Fix f . But V is absorbing and therefore

⋃∞n=1 nV = G.

Hence Fix f = G which means that f = id. �

Example 2. In Theorem 7 it is not sufficient to assume that f satisfiessimultaneously equations (2) and (3). Let f : R → R be a function given byf(x) = −x/2 for x ∈ (−1, 1) and f(x) = x for x ∈ (− ∞, −1] ∪ [1, ∞). Thenf satisfies equations (2) and (3) and, of course, int Fix f = ∅, but f is notequal to the identity function.

The corollary below is a consequence of Theorem 7.

Corollary 4. Let f : R → R and let A ⊂ R be such that (id + 2f)(A)is an interval. If f satisfies equation (1), then either f = id, or there existsc ∈ R such that f(x) = −x/2 + c for every x ∈ A.

Proof. Assume that f satisfies equation (1). Let P ={x + 2f(x) :

x ∈ A}. It follows from Proposition 1 that P ⊂ Fix f . By assumption the

set P is an interval. If cardP > 1, then intFixf = ∅ and applying Theorem 7we obtain f = id. Otherwise there exists d ∈ R such that x + 2f(x) = d forevery x ∈ A whence f(x) = −x/2 + c for every x ∈ A with c = d/2. �

Corollary 5. Let f : R → R be a function continuous on an intervalI ⊂ R. If f satisfies equation (1), then either f = id, or there exists c ∈ R

such that f(x) = −x/2 + c for every x ∈ I .

Corollary 6. Let f : R → R be a function satisfying the conditionFix f ⊂ 2f(R). Assume that there exists a non-degenerated interval I ⊂ R

containing zero such that (id + 2f)(R) is an interval. Then f satisfies equa-tion (1) if and only if either f = id, or f = −id/2.

Proof. Assume that f satisfies equation (1). From Theorem 2 andLemma 1 we infer that either f = id, or f = −id/2. The converse is a con-sequence of Proposition 3. �

Applying Theorem 4 we deduce

Corollary 7. Let f : R → R be a function such that at least one of thesets f

((− ∞, 0)

)and f

((0, ∞)

)is an interval. Assume that Fix f ⊂ 2f(R).

Then f satisfies equation (1) if and only if either f = id, or f = −id/2.

Proof. Assume that f satisfies equation (1). In view of Proposition 1and Corollary R the function f satisfies also equations (2) and (3). It fol-lows from Theorem 4 that either (− ∞, 0) ⊂ Fix f , or (0, ∞) ⊂ Fix f , orf = −id/2. Hence, applying Theorem 1, we complete the proof. �

Example 3 below shows that in Corollary 7 it is not sufficient to assumethat the set f(R) is an interval.

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ON FUNCTIONAL EQUATIONS 339

Example 3. Let H ⊂ R be a Hamel basis and let ρ : H → {1, −1/2}be a function. Let further f0(h) = ρ(h)h for every h ∈ H . It follows from[1, Corollary 3] that the additive extension f : R → R of the function f0 isa solution of the equation (4). Hence we deduce that f satisfies also equa-tion (1). However, if ρ is non-constant, then f is not continuous althoughf(R) = R.

Finally, we solve equations (1) and (2) in the class of analytic functions.

Theorem 8. Let f : C → C be an analytic function. Then the followingconditions are pairwise equivalent:

(i) f satisfies equation (1);(ii) f satisfies equation (2);(iii) either f = id, or f = −id/2.

Proof. The implication (i) ⇒ (ii) is a consequence of Proposition 1.Applying Proposition 3 we deduce that the implication (iii) ⇒ (i) holds. So itis sufficient to prove the implication (ii) ⇒ (iii). Assume that f satisfies equa-tion (2). Let A =

{z + 2f(z) : z ∈ C

}. From (2) we deduce that A ⊂ Fix f .

If cardA = 1, then there exists a c ∈ C such that f(z) = −z/2 + c for everyz ∈ C. Making use of Remark 3 we obtain c = 0 and therefore f = −id/2. IfcardA > 1 then A is a non-empty and open subset of C. Moreover f(z) = zfor every z ∈ A. Since f is analytic, f must be the identity function. �

Example 4. The assumption that f is analytic is essential for the valid-ity of Theorem 8. Let f : C → C be given by f(z) = aRe z + ibIm z, wherea, b ∈ {1, −1/2} and a = b. A simple calculation shows that f is a solution ofequation (1), of course non-analytic although continuous and different fromid and −id/2. By Proposition 1 it satisfies also equation (2) and, by virtueof Corollary R, f is also a solution of equation (3).

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