Alienation of the logarithmic and exponential functional equations

Transcription

1 Aequat. Math. 90 (2016), The Author(s) This artile is published with open aess at Springerlink.om /16/ published online July 11, 2015 DOI /s Aequationes Mathematiae Alienation of the logarithmi and exponential funtional equations Zygfryd Kominek and Justyna Sikorska Dediated to Professor Roman Ger on the oasion of his 70-th birthday Abstrat. We study the logarithmi exponential funtional equation and hek whether the alienation phenomenon takes plae. Mathematis Subjet Classifiation. 39B22. Keywords. logarithmi equation, exponential equation, alienation. 1. Introdution Let E 1 =0andE 2 = 0 be arbitrary funtional the equations. Consider equation of the form E 1 = E 2. We ask whether or not this equation is equivalent to the system of funtional equations E 1 =0andE 2 = 0. If the answer is yes then we say that equations E 1 =0andE 2 = 0 are alien. The phenomenon of alienation was first disussed by Dhombres [2]. Later on, several papers and letures have appeared on this subjet (see [3 12]). The aim of the present paper is to solve the funtional equation f(xy) f(x) f(y) =g(x + y) g(x)g(y), whih is stritly onneted with the problem of alienation of logarithmi and exponential Cauhy funtional equations for real funtions of a real variable. We solve this equation both in the ase where we onsider all real variables, and taking into aount the nature of a logarithmi funtion for non-zero variables. In the latter ase the problem turns out to be muh more ompliated. The ruial point in this ase is a very tehnial Lemma 2. Generally, we do not assume any regularity onditions on f and g. Butifg(1) = 1 and f(1) = 0, unfortunately, the method of the proof whih we use fores us to assume the ontinuity of g at the origin.

2 108 Z. Kominek and J. Sikorska AEM 2. Funtions defined on the whole real line This setion we devote to solving our equation onsidered for funtions of all real variables, i.e., we solve f(xy) f(x) f(y) =g(x + y) g(x)g(y), x,y R. (1) We start the setion with presenting a result whih is an easy onsequene of Dhombres [2, Theorem 11] or Ger and Reih [8, Proposition 3] in the ase of real funtions (see also [1, Proposition 5.6]). Proposition 1. Given three nonzero real onstants a, b,, ifamapf : R R is a solution of the equation af(xy)+bf(x)f(y)+ ( f(x + y) f(x) f(y) ) =0, x,y R then (i) f(x) 0, or (ii) f(x) ( a)b 1,or (iii) f(x) = ab 1 x for all x R. Now we are able to proeed with our main result of this setion. Theorem 1. Let f,g: R R be funtions satisfying (1). Then f(x) 0 and g is an arbitrary exponential funtion, or there exists a nonzero real onstant α suh that f(x) α(α +1), g(x) α +1, or f(x) = αx 2 + α(α +1), g(x) = αx + α +1, x R. Conversely, eah of the above pairs of funtions is a solution of (1) with any α R. Proof. Put y =0in(1). Then f(x) = ( g(0) 1 ) g(x), x R. Let α := g(0) 1. If α = 0, i.e., g(0) = 1, then f(x) = 0 for all x R, andg is a nonzero exponential funtion. Assume that α 0. We may write (1) intheform αg(xy) αg(x) αg(y) =g(x + y) g(x)g(y), x,y R. Setting G(x) :=g(x) g(0), x R, we obtain G(0) = 0 and αg(xy)+g(x)g(y) G(x + y)+g(x)+g(y) =0, x,y R.

3 Vol. 90 (2016) Alienation of the logarithmi and exponential By Proposition 1 it follows that G(x) 0, or G(x) 1 α, org(x) = αx for all x R, whih yields g(x) α +1,org(x) 0org(x) = αx + α +1 for all x R and f(x) α(α + 1), or f(x) 0, or f(x) = α 2 x + α 2 + α for all x R, respetively. The onverse we obtain by a diret heking. 3. Equation on a restrited domain The situation beomes more ompliated if we assume that f is not defined on the whole real line but (whih is natural for a logarithmi funtion) on R\{0}. In what follows let f : R\{0} R and g : R R be funtions satisfying f(xy) f(x) f(y) =g(x + y) g(x)g(y), x,y R \{0}. (2) For the rest of the paper we denote := g(1), d := f(1). We begin our onsiderations from a simple ase where =0. Lemma 1. Let f : R\{0} R, g : R R satisfy (2) and let =0.Then g =0 and f(xy) =f(x)+f(y), x,y 0. Proof. Put y = 1 into (2). Then, by the assumption that = 0, we have g(x +1)=d for all x 0, whene g(x) =d, x 1. (3) Substituting first x = y = 2, and then x = y = 2 in(2), and omparing the so obtained results, by (3) we onlude that f(2) = f( 2). (4) Put now in turn, x =2,y= 1 andx = 2, y= 1 in(2). Comparing the results, by (3) and (4) weget d 2 = d d 2, whene d = 0. This finishes the proof. From now on we will assume that 0 and we start with a tehnial lemma. Lemma 2. Let f : R\{0} R, g : R R satisfy (2) and assume that 0. Then the following ases are the only possible ones (i) 0arbitrary, d =0, g(0) = 1, (ii) =1, d arbitrary, g(0) = 1 d, (iii) = 1, d arbitrary, g(0) = 1 + d, (iv) 0arbitrary, d = 2, g(0) =.

4 110 Z. Kominek and J. Sikorska AEM Proof. Substituting y = 1 into (2), we get g(x +1)=g(x)+d, Hene, n 1 g(n +1)= n+1 + k d, k=0 g( n) = g(0) n x R\{0}. n k=1 d, n N. (5) k Put now x = y = 2 andx = y = 2 into (2) and subtrat the so obtained results side by side to obtain 2f(2) 2f( 2) = g( 4) g( 2) 2 g(4) + g(2) 2. (6) Set now x = 1 andy =2in(2): f( 2) f( 1) f(2) = g(1) g( 1)g(2). (7) Multiplying (7) by 2 and adding it to (6) we obtain 2f( 1) = g( 4) g( 2) 2 g(4) + g(2) 2 +2g(1) 2g( 1)g(2). (8) Substitute x = y = 1 in(2): f(1) 2f( 1) = g( 2) g( 1) 2. (9) By (8) and (9) it follows that f(1) + g( 4) g( 2) 2 g(4) + g(2) 2 +2g(1) 2g( 1)g(2) = g( 2) g( 1) 2. (10) Set x = 1 andy = 2 in(2): f(2) f( 1) f( 2) = g( 3) g( 1)g( 2). (11) Adding (7) and (11) side by side, we get 2f( 1) = g(1) g( 1)g(2) + g( 3) g( 1)g( 2), whih put into (9), gives f(1) + g(1) g( 1)g(2) + g( 3) g( 1)g( 2) = g( 2) g( 1) 2. (12) Finally, put x = 1, y=3andx = 1, y= 3 in(2), f( 3) f( 1) f(3) = g(2) g( 1)g(3), f(3) f( 1) f( 3) = g( 4) g( 1)g( 3). Adding the above equalities side by side yields 2f( 1) = g(2) g( 1)g(3) + g( 4) g( 1)g( 3), whene, on aount of (8) wehave g( 2) 2 g(4)+g(2) 2 +2g(1) 2g( 1)g(2) = g(2) g( 1)g(3) g( 1)g( 3). (13)

5 Vol. 90 (2016) Alienation of the logarithmi and exponential Taking into aount our notations and formula (5), by (10), (12) and (13) we obtain the system of equations with unknown variables, d and g(0): d+d+2d g(0) 4 + d 4 + d 3 ( 2 + d ) ( 2 g(0) + 2 d 2 d ) 2 + g(0) 2 ( g(0) d ) 2 ( g(0) +2 d ) ( 2 + d ) 2 =0 ( g(0) g(0) 2 d ) 2 ( g(0) + d ) ( 2 + d ) g(0) 3 + d ( 3 g(0) + d )( g(0) 2 d 2 d ) + d = d+d+2d ( 2 + d ) ( 2 g(0) + 2 d 2 d ) 2 ( g(0) d ) ( 2 +d ) ( g(0) d ) ( 3 +d+d ) ( g(0) d )( g(0) 3 d 3 d 2 d ) 2=0, whene, 6 d+ 5 d 2 5 g(0) d d+2 3 d dg(0)+ 2 g(0) 2 2 g(0) 2 2 dg(0) + 2dg(0) 2d 2 d d 2 +2dg(0) d g(0) 2 + g(0) = d 4 g(0) 3 d + 2 d 2 2 dg(0) + g(0) 2 dg(0) g(0) d 2 +2dg(0) d g(0) 2 + g(0) = g(0) d 2 4 g(0) 2 3 d + 3 dg(0) + 2 d 2 2 dg(0) dg(0) + dg(0) d 2 =0. (14) From the last equation we have ( d 2 d d + d ) g(0)= d +2 3 d 2 d 2 +d 2. (15) Denote w := d 2 d d + d and assume first that w =0,thatis, d ( ) =2 4 5, d ( 1) 2 ( +1)= 4 (2 ). For = 1or = 1 we would get = 0or = 2, whih is impossible. Consequently, in this ase, it must be ±1 and then 4 (2 ) d = ( 1) 2 ( +1). Substituting this into the right-hand side of (15), whih equals zero, we get (after simplifying) 4 ( 2)(2 1) ( 1) 3 =0, ( +1)

6 112 Z. Kominek and J. Sikorska AEM whene, = 1 2 or = 2. Then, d = 1 4 or d = 0, respetively. Substituting the pair (, d) = ( 1 2, 4) 1 into the first two equations of (14) we obtain { 12g(0) 2 20g(0) + 7 = 0 4g(0) 2 6g(0) + 2 = 0, whih yields g(0) = 1 2. Putting (, d) =(2, 0) into the first two equations in (14) gives { 3g(0) 2 67g(0) + 64 = 0 g(0) 2 17g(0) + 16 = 0, and g(0) = 1. Assume now that w 0. Then d (1 ) 2 (1 + ) 4 (2 ) and g(0) = d +2 3 d 2 d 2 + d d 2. d d + d Substituting this into the first two equations in (14), we obtain (after simplifiation) d 3 ( 1)( +1) ( d + 2)( d d d d d d d 2 +2d 2 ) =0 d 3 ( 1)( +1) ( d + d 2 )( d + 2) ( d 2 d +1 ) =0. (16) Solving (16), we get the following ases d =0and 2 arbitrary; then g(0) = 1. This, together with the earlier obtained triple (, d, g(0)) = (2, 0, 1), gives assertion (i). =1andd arbitrary; then g(0) = 1 d and we get (ii). = 1 andd arbitrary; then g(0) = 1 + d, whih gives (iii) d + d 2=0,sod( 1) = (1 + 2 )(2 ). Hene, 1and d = (1 + 2 )(2 ). 1 Then for 2wehavew 0 and substituting the above d into the first equation of (16), we obtain 2 3 ( + 1)( 2) 2 ( 2 +1) =0, 1 whene, = 1. But then d = 3 andg(0) = 2, and suh a triple is already inluded in (iii).

7 Vol. 90 (2016) Alienation of the logarithmi and exponential d = 2 with an arbitrary suh that w 0, whih means 1 2. Then g(0) =. This ase with the earlier obtained triple (, d, g(0)) = ( 1 2, 1 4, ) 1 2 gives (iv) d 2 d + 1 = 0, i.e., d(1 2 )= , whene, ±1 and d = Then for 2 we have w 0 and substituting suh d into the first equation of (16), we obtain 2 3 ( )(2 1) 2 2 =0, 1 whene, = 1 2 (then d = 1 4 and g(0) = 1 2 )or = 0. In the latter ase we have d = 0, whih even with an arbitrary value of gives g(0) = 1. This finishes the proof. Corollary 1. Let f : R\{0} R, g : R R satisfy (2) and 0.Then g(x +1)=g(x)+d, x R. (17) Lemma 3. Let f : R\{0} R, g : R R satisfy (2). If 0and d =0, then f(rx) =f(r)+f(x), r Q\{0}, x 0 (18) and g(r + x) =g(r)g(x), r Q\{0}, x 0. (19) Proof. By Lemma 2 and (17), we have g(0) = 1 and g( 1) = 1. Hene, g(x + p) = p g(x) for all p Z and x R. Therefore, g(x + p) =g(x)g(p), p Z, x R. On aount of (2), f(px) f(p) f(x) =0, p Z\{0}, x 0. Substituting x = 1 p, p Z\{0} in the above equality gives ( ) 1 f(p)+f =0, p Z\{0}. p Hene, for p, q Z\{0} and x 0, ( ) ( ) ( ) ( ) p 1 1 p f q x = f(p)+f q x + f(q)+f = f + f(x), q q so, we have (18), and onsequently, by (2), we obtain (19). Corollary 2. Let f : R\{0} R, g : R R and d =0.If<0 then (2) does not have solutions.

8 114 Z. Kominek and J. Sikorska AEM Proof. Putting x = r = 1 2 whih is a ontradition. in (19) weget = g(1) = g ( 1 2) 2 0, Under the assumptions of the following lemma we get the full alienation of logarithmi and exponential Cauhy funtional equations. Lemma 4. Let f : R\{0} R, g : R R satisfy (2). If >0, 1and d =0, then g(x + y) =g(x)g(y), x,y R and f(xy) =f(x)+f(y), x,y 0. Proof. On aount of (2), f(2xy) f(2x) f(y) =g(2x + y) g(2x)g(y), x,y 0, (20) and by (18), f(2xy) =f(2) + f(xy), x,y 0, and f(2x) =f(2) + f(x), x 0. Therefore, from (20), f(xy) f(x) f(y) =g(2x + y) g(2x)g(y), x,y 0. By (2) and Lemma 2 we have g(0) = 1 and g(x + y) g(x)g(y) =g(2x + y) g(2x)g(y), x,y R. (21) Putting x+1 in the plae of x and realling that g(x+1) = g(x) for all x R, we get ( g(x + y) g(x)g(y) ) = 2( g(2x + y) g(2x)g(y) ), x,y R. This together with (21) and the assumption that 1 yields g(x + y) =g(x)g(y), x,y R and, as a onsequene, f(xy) =f(x)+f(y), x,y 0, whih was to be proved. In the next lemma we obtain a partiular ase of alienation of logarithmi and exponential Cauhy funtional equations (with g(x) 1), but we assume the ontinuity of g at the origin. Without this assumption we are not able to get suh type of results. But on the other side, we also do not have an example showing that in suh a ase the phenomenon of alienation does not our.

9 Vol. 90 (2016) Alienation of the logarithmi and exponential Lemma 5. Let f : R\{0} R, g : R R satisfy (2). If =1, d =0,andg is ontinuous at the origin, then Proof. By (18), and we get By (17) and onsequently, g(x) 1 and f(xy) =f(x)+f(y), x,y R. On aount of (19), ( g n p ) = g q whene thus f( x) =f(x)+f( 1), 0=f(1) = 2f( 1), f(x) =f( x), x 0. g(x +1)=g(x), x R, ( 1=g q p ) = g q g(p) =1, p Z. x R\{0} ( ) n p, n,q N, p Z, q ( ) q p, p Z, q N, q g(r) =1, r Q. Therefore g(x + r) =g(x) for all x R and r Q. We shall show that g(x) 1. Suppose to the ontrary that g(x 0 ) 1for some x 0 R. Putε := g(x 0). By the ontinuity of g at the origin it follows that there exists δ>0 suh that if x <δthen g(x) 1 <ε. Observe that we an take x = x 0 + w with a suitably hosen rational number w. Hene, 2ε = g(x 0 ) 1 = g(x 0 + w) 1 <ε. This ontradition shows that g(x) 1, whih was our assertion. In the remaining ases with d 0 the phenomenon of alienation does not our, although we are not very far from it. We may observe that funtions f and g are logarithmi and exponential, respetively, up to linear funtions.

10 116 Z. Kominek and J. Sikorska AEM Lemma 6. Let f : R\{0} R, g : R R satisfy (2). If = 1 and d 0, then g(x) 1 and f(x) =F (x)+2 for all x R\{0}, wheref : R\{0} R satisfies F (xy) =F (x)+f (y), x,y 0. Proof. By Lemma 2, g(0) = d + 1 and on aount of (17), whene, g( 1) = 1 and Substituting y = 1 into(2) we obtain g(x +1)= g(x)+d, x R, (22) g(x +2)=g(x), x R. (23) f( x) f(x) f( 1) = g(x 1) g(x)g( 1), x,y 0, whene, by use of (22) weget f( x) f(x) =d + f( 1), x 0. Interhanging the roles of x and x we see that f( 1) = d, and, onsequently, f( x) =f(x), x 0. (24) By (2) and (23), f(2x) f(x) f(2) = g(x +2) g(x)g(2) = dg(x), x 0, whene, on aount of (24), Therefore, for all x 0,and g(x) =g( x), x R. g(2x) g(x) 2 = f(x 2 ) 2f(x) =f( x 2 ) f(x) f( x) = g(0) g(x)g( x) =1+d g(x) 2 g(x) =1+d, x R. Sine g(1) = 1, we have g(x) = 1 for all x R, and f(xy) f(x) f(y) = 2, x,y 0. Finally, if we define F (x) :=f(x) 2 for all x 0, then F (xy) =F (x)+f (y), x,y 0, and the proof is finished. In the proof of Lemmas 8 and 9 we will use the following result.

11 Vol. 90 (2016) Alienation of the logarithmi and exponential Lemma 7. Let f : R\{0} R, g : R R satisfy (2). If {0, 1} and d 0, then 2g(x+y) = K2 + d(1 + ) [ ] g(x)+g(y) +K K 2 2 (1 + )d( d d + 3 ) 4, for all x, y 0,whereK = + 3 d d 2. Proof. By (17), we have (25) g(0) = d, 2 g( 1) = d d, g(x 1) = g(x) d, x R. (26) On aount of (2) and (26), 2[ f( x) f(x) f( 1) ] = d(1 + )g(x) d, x 0. (27) Substituting x in the plae of x in the above equation and adding these two equations we obtain 2 2 f( 1) = d(1 + ) [ g(x)+g( x) ] 2d, x 0. (28) Set x =1in(28) in order to obtain f( 1) = d [ (1 ) (1+)(d + d ) ]. (29) Putting it bak to (28) weget (1 + ) [ g(x)+g( x)] = 1+ [ ( ) d(1 + ) ], x R, whih, by the assumption 1, gives g(x)+g( x) = + 3 d d 2 =: K, x 0. (30) On the other side, substituting (29) into(27), yields d(1 + ) [ f( x) f(x) = g(x) K], x 0. (31) Using again (2), on aount of (30), we obtain f(xy) f( x) f( y) =K g(x + y) (K g(x))(k g(y)) = K K 2 g(x + y)+kg(x)+kg(y) g(x)g(y), x,y 0. Subtrating (2) from the above equality, we get f(x)+f(y) f( x) f( y) = 2g(x + y)+k K 2 + Kg(x)+Kg(y), x,y 0.

12 118 Z. Kominek and J. Sikorska AEM By (31) we may write d(1 + ) d(1 + ) 2 g(x) 2 g(y)+ (1 + )d( d d + 3 ) 4 = 2g(x + y)+k K 2 + Kg(x)+Kg(y), x,y 0, whih is equivalent to our assertion (25). Lemma 8. Let f : R\{0} R, g : R R satisfy (2). If =1and d 0, then g(x) =dx +(1 d), x R and there exists a funtion F : R\{0} R satisfying suh that F (xy) =F (x)+f (y), x,y 0, f(x) =F (x) d 2 x d(1 d), x 0. Proof. We will use Lemma 7. Setting =1wegetK =2 2d and (25) takes the form g(x + y) =g(x)+g(y)+d 1, x,y 0. Sine by Lemma 2, g(0) = 1 d, wehaveinfat, g(x + y) =g(x)+g(y)+d 1, x,y R. Therefore, the funtion G: R R defined by G(x) :=g(x)+d 1 for all x R is additive. By the additivity of G it follows that G(2x) =2G(x) for all x R, whene, by the definition of G we get g(x + y) g(x)g(y) =G(x + y)+1 d (G(x)+(1 d))(g(y)+(1 d)) Consequently, (2) takes the form = G(x + y)+1 d G(x)G(y) (1 d)(g(x) + G(y)) (1 d) 2 = dg(x) + dg(y) G(x)G(y)+(1 d)d. f(xy) f(x) f(y) =dg(x)+dg(y) G(x)G(y)+d(1 d), x,y 0. (32) Substituting y = 2 in the above equality, we get f(2x) f(x) f(2) = dg(x)+d 2 + d, x 0.

13 Vol. 90 (2016) Alienation of the logarithmi and exponential Therefore, aording to (32), for all x, y 0 we have dg(xy) =d 2 + d [f(2xy) f(xy) f(2)] = d 2 + d [f(2xy) f(2x) f(y)+f(2x) f(2) f(x) + f(x)+f(y) f(xy)] = d 2 + d [dg(2x)+dg(y) G(2x)G(y)+d(1 d) + dg(x) + dg(2) G(x)G(2) + d(1 d) dg(x) dg(y) +G(x)G(y) d(1 d)] = G(x)G(y). The funtion 1 d G, being additive, is of the form 1 dg(x) =x for all x R (see [1, Proposition 5.6]), whene, Consequently, we may rewrite (32) as Defining G(x) =dx, x R. f(xy) f(x) f(y) =d 2 x + d 2 y d 2 xy + d(1 d), x,y 0. we observe that Hene, F (x) :=f(x)+d 2 x + d(1 d), x 0, F (xy) =F (x)+f (y), x,y 0. g(x) =dx +(1 d), x R and f(x) =F (x) d 2 x d(1 d), x 0, and the proof is finished. Lemma 9. Let f : R\{0} R, g : R R satisfy (2). If {0, 1, 1} and d 0, then g(x) and there exists F : R\{0} R satisfying F (xy) =F (x)+f (y), x,y 0, so that f(x) =F (x)+ 2 for all x 0. Proof. We will apply Lemma 7. Puty =1in(25). By (17), for all x 0we have 2(g(x)+d) = K2 + d(1 + ) 2 g(x)+ K2 + d(1 + ) + K K 2 (1 + )d( d d + 3 ) 4,

14 120 Z. Kominek and J. Sikorska AEM whene, ( ) 2 K2 + d(1 + ) 2 g(x) = 2d + K2 + d(1 + ) + K K 2 (1 + )d( d d + 3 ) 4, for all x 0. This together with the definition of K yields ( 2 1)g(x) =A, x 0, where A := 2d 2 + K 3 + d(1 + )+K 2 K 2 2 (1 + )d( d d + 3 ) 2. Therefore, A g(x) = ( 2 1), x 0. Hene, g(x) =g(1) = for all x 0. By Lemma 2, g(0) =. Consequently, g(x) = for all x R. By(2), the funtion F (x) :=f(x) + 2, x 0, satisfies F (xy) =F (x)+f(y), x,y 0, so, f(x) =F (x) + 2, x 0, with {0, 1, 1}. Consequently, onsidering all the above results, we may formulate the final theorem. Theorem 2. Assume that f : R\{0} R and g : R R satisfy (2). If g(1) 1or f(1) 0, then g(x + y) =g(x)g(y), x,y R and f(xy) =f(x)+f(y), x,y R\{0}, (33) or there exist α R\{0} andafuntionf : R\{0} R satisfying F (xy) = F (x)+f(y) for all x, y R\{0} suh that g(x) =αx+(1 α), x R and f(x) =F (x) α 2 x α(1 α), x R\{0}, (34) or there exist β R\{1} andafuntionf : R\{0} R satisfying F (xy) = F (x)+f(y) for all x, y R\{0} suh that g(x) =β, x R and f(x) =F (x)+β 2 β, x R\{0}. (35) If g(1) = 1, f(1) = 0 and g is ontinuous at the origin, then g(x) =1, x R and f(xy) =f(x)+f(y), x,y R\{0}.

15 Vol. 90 (2016) Alienation of the logarithmi and exponential Conversely, eah pair of funtions desribed by (33), (34) and (35) is a solution of (2) with any real α and β. We terminate our paper with the following open problem. Problem 1. Assume g(1) = 1 and f(1) = 0. Find all solutions of (2) without assuming the ontinuity of g at the origin (f., Lemma 5). Open Aess. This artile is distributed under the terms of the Creative Commons Attribution 4.0 International Liense ( whih permits unrestrited use, distribution, and reprodution in any medium, provided you give appropriate redit to the original author(s) and the soure, provide a link to the Creative Commons liense, and indiate if hanges were made. Referenes [1] Azél, J., Dhombres, J.: Funtional Equations in Several Variables. Enylopedia Math. Appl. vol. 31. Cambridge University Press, Cambridge (1989) [2] Dhombres, J.: Relations de dépendane entre les équations fontionnelles de Cauhy. Aequationes Math. 35, (1988) [3] Ger, R.: On an equation of ring homomorphisms. Publ. Math. Debreen 52, (1998) [4] Ger, R.: Ring homomorphisms equation revisited. Roznik-Naukowo-Dydaktyzny Akademii Pedagogiznej w Krakowie, Prae Matematyzne 17, (2000) [5] Ger, R.: Additivity and exponentiality are alien to eah other. Aequationes Math. 80, (2010) [6] Ger, R.: Alienation of additive and logarithmi equations. Ann. Univ. Si. Budapest Set. Comp. 40, (2013) [7] Ger, R.: The alienation phenomenon and assoiative rational operations. Ann. Math. Sil. 27, (2013) [8] Ger, R., Reih, L.: A generalized ring homomorphisms equation. Monatsh. Mat. 159, (2010) [9] Maksa, Gy.: On multipliative Cauhy differenes that are Hosszú differenes. Report of Meeting. The 14th Debreen-Katowie Winter Seminar, Hajdúszoboszló (Hungary), January 29 February 1, Ann. Math. Sil. 28, p. 105 (2014) [10] Maksa, Gy.: On exponential Cauhy differenes that are Hosszú differenes. Talk on the 52nd ISFE, Innsbruk, Austria (2014) [11] Sablik, M.: Additivity of insurane premium. Talk on the 51st ISFE, Rzeszów, Poland, 2013 and on Zaanen Centennial Conferene, Leiden, Netherlands (2013) [12] Sablik, M.: On alien funtional equations. Talk on the 52nd ISFE, Innsbruk, Austria (2014) Zygfryd Kominek and Justyna Sikorska Institute of Mathematis University of Silesia Bankowa 14, Katowie Poland zkominek@math.us.edupl; justyna.sikorska@us.edu.pl Reeived: Marh 29, 2015 Revised: May 29, 2015