ON A FUNCTIONAL EQUATION BASED UPON A RESULT OF GASPARD MONGE

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1 ON A FUNCTIONAL EQUATION BASED UPON A RESULT OF GASPARD MONGE C. Alsina, M. Sablik, J. Sikorska Abstract. We present and solve completely a functional equation motivated by a classical result of Gaspard Monge. 1. INTRODUCTION Gaspard Monge ( ) was a great geometer. His ideas and results on Descriptive Geometry are, still today, a rigourous way to deal with graphical constructions. Our aim in this paper is to focus our attention on a property discovered by Monge concerning some triangles associated to any straight line of positive slope (in the rst quadrant). Precisely, let 0 denote the origin of a cartesian reference in the plane and let A = (A 1 ;A ) and B = (B 1 ;B ) with midpoint M (see gure 1). Then Monge noted that the areas S;S 1 and S of the triangles 0AB, A 1 MB 1 and A MB, respectively, satisfy the relationjs 1 S j = S. Figures 1 and 1

2 Based upon this equality we want to examine for which strictly increasing functions we can assure a similar result. Figure 1 suggests a possible generalization shown in Figure. Of course whenf is an a±ne function then gure reduces to gure 1, i.e., to Monge's case. Following the notations of gure, Monge's result may be stated as follows, for any x y, x +y 1 (y x)f 1 + y Z y (f(y) f(x))x = f(t)dt+ 1 x xf(x) 1 yf(y) (1) Our chief concern in this paper is to solve (1), showing that only a±ne functions may present this behaviour. Note that the formulation of (1) forces that the expression on the right hand side is non-negative, i.e., that we are in the situation where f(y) y x f(x), i.e., f(t)=t is non-increasing. But as we will see in our arguments this fact will be derived from (1).. MAIN RESULT We begin by considering a lemma which will be of interest in the next theorem. LEMMA 1. Let f be a function from R + into R + such that f(t)=t is nonincreasing on (0;1). Then, for all x y we have x + y x + y (f(y) f(x)) (y x)f : (3) Proof. Obviously (3) holds for x = y. So let us consider the case 0 < x < y which yields f(x) f(y), i.e., yf(x) xf(y). x y Therefore, since x < x+y < y, x +y x + y (f(y) f(x)) = 1 x + y 1 (yf(y) yf(x)) (yf(y) xf(y)) y y = x + y 1 x +y (y x)f(y) = (y x)f(y) y y x + y x+y x +y (y x)f = (y x)f : x+y

3 Now we can show the following result concerning (1): THEOREM 1. A strictly increasing function f from R + into R + satis es (1) if and only if f(x) = ax + b, for some arbitrary constants a and b such that a > 0 and b 0. Proof. Let us assume that f satis es (1), i.e., for x y x + y (y x)f x + y Z y (f(y) f(x)) = f(t)dt +xf(x) yf(y): x (1) Since f is strictly increasing when y tends to x from the right we have x + y lim f = lim f(y) = f(x+) y!x+ y!x+ so taking limits in (1) when y! x+ we obtain 0 xjf(x+) f(x)j = x(f(x) f(x+)); i.e., f(x+) f(x) and since f(x) f(x+) we deduce f(x) = f(x+). Analogously, taking limits when x!y from the left we deduce f(y) = f(y ). Therefore f is continuous, and being strictly increasingf(t) is di erentiable almost everywhere and the same applies to f(t) t. If F (t) denotes a primitive function of f then F is almost everywhere di erentiable (F 0 = f) non-decreasing and (absolutely) continuous on R + and we can write Z y Presenting (1) in the form: x +y f x + y f(y) f(x) y x x f(t)dt = F(y) F(x): when x tends to y from the left we deduce = F (y) F(x) y x yf(y) xf(x) ; y x 0 jf(y) yf 0 (y)j = f(y) (yf 0 (y) + f(y)) = f(y) yf 0 (y) 3

4 so d dy f(y) y = yf0 (y) f(y) y 0; and f(y)=y is non-increasing on (0;1). By virtue of Lemma 1, (3) holds, so we can omit the absolute value in (1) and present our equation in the form: x +y (y x)f x + y (f(y) f(x)) = (F(y) F(x)) +xf(x) yf(y); or equivalently F(y) F(x) = (y x) f µ x + y + f(x) + f(y) : (4) Since F andf are di erentiable, taking partial derivatives in (4) we obtain x +y f(x) +f(y) 1 x + y f(x) = f + +(y x) f0 + 1 f0 (x) and f(y) = f µ x + y + f(x) + f(y) 1 + (y x) f0 x + y + 1 f0 (y) so subtraction of the rst equality from the second one yields x +y f(y) + f(x) = f + f(x) + f(y) + (y x) 1 (f0 (y) f 0 (x)); i.e., x +y f = f(x) +f(y) + 1 (y x)(f0 (x) f 0 (y)): (5) Fixing x = x 0 > 0 into the above equation (5) we immediately see that for any y in [x 0 ;1) f 0 (y) = f(x 0) + f(y) + 1 (y x 0)f 0 (x 0 ) f((x 0 + y)=) ; (y x 0 )= whence f 0 is also di erentiable in [x 0 ;1). Since x 0 is arbitrary, f 00 exists in R + so we can take again partial derivatives in (5) to obtain: x + y f 0 = f 0 (x) 1 (f0 (x) f 0 (y)) + 1 (y x)f00 (x); 4

5 and x + y f 0 = f 0 (y) + 1 (f0 (x) f 0 (y)) 1 (y x)f00 (y); i.e., f 00 (x) = f 00 (y) so xingy = x 0 we get thatf 00 is constant, i.e., f 00 (x) = k and k = k yields k = 0. That f(x) = ax + b with a > 0;b 0. The theorem is proved. Remark 1. Let us note that one can obtain the same solutions of (4) with no regularity assumptions on f or F by means of results to be found in (Sablik, 000), (Sablik, 004) and (Pawlikowska, 00). However these general results are stated in a very general framework of group theory and therefore here we decided to give a short and direct proof. Similarly one can consider the functional equation x +y 1 (y x)f 1 + y (f(y) f(x))x = 1 Z y yf(y) 1 xf(x) f(t)dt; x (6) which is based on the case where the graph of f is below the line g(t) = (f(y)=y) t. Then one obtains the following THEOREM. Fixed [a;b] with a > 0, a strictly increasing function f from [a;b] into R + such that f(a) > 0, satis es (6) if and only if f(x) = mx + n for some arbitrary constants m;n in R, (m > 0). In amore general situation if f is a continuous strictly increasing function from [a;b] into R + with a > 0 and f(a) > 0 and f can oscillate above and below the line y(t) = (f(b)=b), then one can consider the sets ½ A(b) := x [a;b] : f(x) f(b) ¾ b x > 0 ; ½ B(b) := x [a;b] : f(x) f(b) ¾ b x < 0 ; ½ C(b) := x [a;b] : f(x) f(b) ¾ b x : A(b) and B(b) are open sets so they can be represented as union of open intervals. For each of such intervals easier results can be applied and for C(b) we have the result of Monge. So each consideration leads to a linear 5

6 function f(x) = mx + n on an interval which together with continuity of f gives f(x) = f(b) f(a) b a (x a) + f(a). ACKNOWLEDGMENT The author thanks Prof. J. Garcia-Roig his remarks and to Prof. A. Monreal for making the gures. REFERENCES [1] ACZ EL, J., 1966, Lectures on Functional Equations and Their Applications. Academic Press, New York. [] MONGE, G., 1897, Application de l'analyse a la G eom etrie. Imprimerie de H. Perronneau, Paris [3] MONGE, G., 1996, Geometr ³a descriptiva. Edici on facsimil. Colegio Ingenieros de Caminos, Canales y Puertos, Madrid. [4] I. Pawlikowska, 00, Charakteryzacja odwzorowa n poprzez twierdzenia o warto sci sredniej (in Polish), Doctoral thesis, Uniwersytet Sl»aski. [5] M. Sablik, 000, Taylor's theorem and functional equations. Aequationes Math. 60, no. 3, 58{67. [6] M. Sablik, 004, Remark to a result of C. Alsina. 4nd ISFE. Aequationes Math. C. Alsina M. Sablik and T. Sikorska Sec. Matemµatiques. ETSAB-UPC, mssablik@us.edu.pl Diagonal 649, 0808 Barcelona, Spain. claudio.alsina@upc.es 6