W pracy dowodzi się twierdzenia które orzeka, że jeżeli dane funkcje f i h są monofoniczne
(względnie monotoniczne i wypukłe) to rozwiązanie równania (1) jest monofoniczne (wypukłe)

Matkowski, Jan

English

Repozytorium Uniwersytetu Śląskiego RE-BUŚ

             Title: On some properties of solutions of a functional equation   Author: Janusz Matkowski  Citation style: Matkowski Janusz. (1969). On some properties of solutions of a functional equation. "Prace Naukowe Uniwersytetu Śląskiego w Katowicach. Prace Matematyczne" (Nr 1 (1969), s. 79-82)  P R A C E  N A U K O W E  U N IW E R S Y T E T U  Ś L Ą S K IE G O  W  K A T O W IC A C H  N R  2P R A C E  M A T E M A T Y C Z N E  I ,  1969J a n u s z  M a t k o w s k i  On som e properties o f solutions o f a functional equationIn the present paper we are concerned with the functional equation(1) ę ( x )  = h ( x , ę  [ /(* ) ] ) ,where f ( x )  and h(x ,  y ) are known real-valued functions of real variables and <p(x) is unknown.We assume the following hypotheses:(I) f ( x )  is defined and continuous in an interval <0, a) and(2) 0 < / (x) < x  for x c (0, a) ;(II) h(x ,  y ) is defined in a domain Q such that (0, 0) e £2; moreover, for every x  e <0, a), the set Qx=  { j:(x , y) e 12} is a nonempty open interval and AX<=£2X, where Axd= h (x , Qf M );(III) h(x ,  0) is continuous at x = 0 , h ( 0 , 0 )= 0  and(3) \h(x ,  y 1) - h ( x ,  y2) K » l y i - y 2l. 0 < o < l „holds in a neighbourhood of the point (0 ,0).We shall prove the followingT h e o r e m . Let hypotheses (I)—(III) be fulfilled. Then equation (1) has exactly one solution ę  defined in <0 , a), continuous at x = 0  and such that q>(0) = 0 .1°. If, moreover, f ( x )  is strictly increasing in <0,a )  and h ( x , y )  is increasing with respect to either variable in Q, then ę  is increasing in <0, a) (if, fo r  every f ixed  y , h ( x ,  y ) is a strictly increasing function o f  x, then q> is strictly increasing in <0, a)).2°. If, besides (I)—(III) and hypotheses o f  1°, we assume that f ( x )  is convex in <0, a), Q is a convex domain and h ( x , y )  is a convex function o f  two variables in Q, then (p is convex in <0, a).P ro o f . We choose a c> 0 and a d > 0 such that (3) holds in the set D — {(x, y) : 0 < x < c ,  —By (III) we may assume that c has been chosen in such a manner that(4) \h(x ,  0 ) |3%(1 — u)d for x e < 0 ,c > .Let F  be the set of functions <p defined in <0, c>, continuous at x = 0  and such that(5) <jp(0) =  0 ;  |<p(x)|<d for x e < 0 ,c > .The set F  with the metricQ(<Pi. <Pï ) =  sup |<Pi(x)-ç>2(x)|<o ,c>is a complete metric space. We define the transform(6) lK x) =  h ( x , © [ / (x)]) for <ja e F. It follows from (2) that(7) \<p[ / ( x ) ] |< d  for <peF and x e < 0 ,c > .Now, from (3) and (4) we get(8) |iA (x)| =  I h ( x , ę  [/(x )]) | <  |/i ( x , <p [ /(x )])  -  h ( x , 0)|+  |h (x , 0)| < n |<p [ /(x ) ] | +  \h ( x , 0) | .Hence and from (2) we see that lim >p(x)—0. Since i^(0)= /i(0, ę(0))=h(Q,  0 )= 0 ,*->o+ip(x) is continuous at x = 0 . From  (8) and (7) we obtain| i / f ( x ) K o r f + ( l  —  v ) d  =  d .This proves that (6) transforms F  into itself. Further, we have in virtue of (3) for •Ai(x) =  h ( x ,  (PiUix)]),  ip2(x) = h ( x ,  <p2[ / (x ) ] ) ,  q>i, cp2e F , |i> i(x)-i>2(x)| =  \h(x,<plI f (x )])- h ( x ,  <p2[ /(x )]) |<  u kp i C/(x)3— >^2 [ /(x )]  I ,whencei.e., (6) is a contraction map. On account of B a n a c h ’s  theorem there exists exactly one solution ę  e F  of equation (1). This solution has a unique extension onto the whole interval <0, a) (cf. [1], p. 70, Theorem 3.2).R e m a r k  1. In the book [1] Theorem 3.2 has been proved under the assumption of the continuity of h ( x , y ) in £3 but Theorem 3.2 will remain valid without this assumption (and the proof is the same), except that then the solution need not be continuous.For the proof of 1° we define the space FtF  j =  {<peF :cp is increasing in <0, c>} .Let us take 0 ^ x , < x 2<r/. From the monotonicity of /(x )  and from (2) we get0 < / ( x 1) < / ( x 2) < d .For q  c F x we have0 <  q  [ / (* i) ]  <  ę  U \ x 2)j ^  aNow from (6) and the monotonicity of h ( x , y )  we obtainiA(xi) =  h ( x t , v U ( . x iy])<:h ( x 2 , <p[/(x1) ] )< h (x 2 , q>U(x2y]) = ^ ( x 2) ,i.e., t]/ is increasing in <0, c>. Hence and from the first part of the proof it follows that 1> e F x. Since Ft is a complete metric space, the unique solution q  continuous at x = 0  and such that (0 )= 0  must belong to F l . Thus q  is increasing in <0, c>. Now we shall prove that q  is increasing in <0, a). For this purpose we denote by x0 the supremum of all b such that q  is monotonic in <0, b) and suppose that x 0<a. From (2) we get / ( x 0) < x 0. It follows from the continuity of /(x )  that there exists a number a > x 0 such that for a  e <a-0, a) we have / ( x ) < x 0. Hence for 0 < x x< x 2< a  we obtainq  (Xj) = h ( x l , q  [/(X j)]) <  h (x2 , <p [ / ( x 2)]) =  q  (x2) ,i.e., <p is monotonic in <0, a), a > x 0. This contradiction completes the proof of the monotonicity of q  in <0 , a).If, for every fixed y , h { x , y )  is a strictly increasing function of the variable x, then for 0 < x x< x 2< a  we haveq ( x 1) = h ( x l , q  [ / (x j) ] )  < h ( x 2 , q  [ /(X j)]) <  h (x2 , (p [ / ( x 2)]) =  q  (x2) ,which proves that q  is strictly increasing.R e m a r k  2. It is not sufficient to suppose that h{ x ,y )  is strictly increasing in the variable y. For instance, the unique solution continuous at x = 0  of the equationq  (x) =  \ q  [ /(x )]is q ( x ) = 0 , and h ( x , y ) = \ y  is strictly monotonic with respect to y.For the proof of 2° we define the space F2:F 2 = { q e F 1 : q  is convex in <0 , c>} .From the convexity and monotonicity of / ,  /?, and q  we get for 0 ^ x t ^ c ,  0 <  a*2 s? c and 1/ given by formula (6)J Xl ^ )  =  h X^ , q ^ x-1 +  x2J  ^  < h ^ Xt +  x2 ^|~/(x - |)+ /(x 2)- /; (  X j + x 2 q  [ / ( x t)] +  q  [ /(X 2)]^ < h ( x i , g  [ / ( x j ] )  +  h (x2 , q  [ / ( x 2)]) ^iA(x i) +  iA(x2)Hence (/'(a) is convex in <0, c>. In view of the inclusion F2czF1 and of the preceding part of the proof, (6) transforms F2 into itself. Evidently, F2 is a complete metric space, and thus the unique solution q  continuous at x = 0  and such that q (0 )= 0  must be convex. Similary as in preceding case we can prove that q  is convex in <0, a).R e m a r k  3. The first part of the theorem is implicitly contained in the book [1] (comp, the remark at the end of § 4, Chapter III, p. 54).6 — P ra c e  m a te m a ty c z n e  t, 1869 —  81 —[1] M. K u c z m a : Functional Equations in a  Single Variable, M onografie Matematyczne 46 (PW N, Warszawa 1968).J a n u s z  M a t k o w s k iO PEW NYCH  W ŁASNOŚCIACH RO ZW IĄZAŃ R Ó W N A N IA  FU N K C Y JN E G OS tr e s z c z e n ieW pracy dowodzi się twierdzenia które orzeka, że jeżeli dane funkcje /  i h są monofoniczne (względnie m onotoniczne i wypukłe) to rozwiązanie równania ( 1) jest m onofoniczne (wypukłe).Oddano do Redakcji 1 sierpnia 1969 r.

On some properties of solutions of a functional equation

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On some properties of solutions of a functional equation

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