On Flett's mean value theorem

Transcription

1 P. J. ŠAFÁRIK UNIVERSITY FACULTY OF SCIENCE INSTITUTE OF MATHEMATICS Jesenná 5, Košice, Slovki O. Hutník nd J. Molnárová On Flett's men vlue theorem IM Preprint, series A, No. 4/2012 September 2012

2 On Flett s men vlue theorem Ondrej HUTNÍK nd Jn MOLNÁROVÁ Institute of Mthemtics, Fculty of Science, Pvol Jozef Šfárik University in Košice, Jesenná 5, SK Košice, Slovki, E-mil ddress: ondrej.hutnik@upjs.sk, jn.molnrov88@gmil.com Abstrct. This pper reviews the current stte of the rt of the men vlue theorem due to Thoms M. Flett. We present the results with detiled proofs nd provide mny new proofs of known results. Moreover, some new observtions nd yet unpublished results re included. Key words nd phrses. Flett s men vlue theorem, rel-vlued function, differentil function, Tylor s polynomil, Pwlikowsk s theorem Mthemtics Subject Clssifiction (2010) 26A24, 26D15 1 Introduction nd preliminries Motivtions nd bsic im Men vlue theorems of differentil nd integrl clculus provide reltively simple, but very powerful tool of mthemticl nlysis suitble for solving mny diverse problems. Every student of mthemtics knows the Lgrnge s men vlue theorem which hs ppered in Lgrnge s book Theorie des functions nlytiques in 1797 s n extension of Rolle s result from More precisely, Lgrnge s theorem sys tht for continuous (relvlued) function f on compct set, b which is differentible on (, b) there exists point η (, b) such tht f (η) = f(b) f(). b Geometriclly Lgrnge s theorem sttes tht given line l joining two points on the grph of differentible function f, nmely [, f()] nd [b, f(b)], then there exists point η (, b) such tht the tngent t [η, f(η)] is prllel to the given line l, see Fig. 1. Clerly, Lgrnge s theorem reduces to Rolle s theorem corresponding uthor 1

3 2 IM Preprint series A, No. 4/2012 y y f(x) f(x) η b x η 1 η 2 b x Figure 1: Geometricl interprettion of Rolle s nd Lgrnge s theorem if f() = f(b). In connection with these well-known fcts the following questions my rise: Are there chnges if in Rolle s theorem the hypothesis f() = f(b) refers to higher-order derivtives? Then, is there ny nlogy with the Lgrnge s theorem? Which geometricl consequences do such results hve? These (nd mny other) questions will be investigted in this pper in which we provide survey of known results s well s of our observtions nd obtined new results. Nottion Throughout this pper we will use the following unified nottion: C (M), resp. D n (M), will denote the spces of continuous, resp. n-times differentible rel-vlued functions on set M R. Usully we will work with compct set of the rel line, i.e., M =, b with < b. Therefore, we recll tht under continuity of function on, b we understnd its continuity on (, b) nd one-sided continuity t the end points of the intervl. Similrly we will understnd the notion of differentibility on closed intervl. For functions f, g on n intervl, b (for which the following expression hs its sense) the expressions of the form f (n) (b) f (n) () g (n) (b) g (n) (), n N {0}, will be denoted by the symbol b K ( f (n), g (n)). If the denomintor is equl to b, we will write only b K ( f (n)). So, Lgrnge s theorem in the introduced nottion hs the following form: if f C, b D(, b), then there exists η (, b) such tht f (η) = b K (f), where we use the usul convention f (0) := f. Structure of this pper The orgniztion of this pper is s follows: in Section 2 we present the originl result of Flett s well s its generliztion due to Riedel nd Shoo removing the boundry condition. Further sufficient conditions of Trhn nd Tong for vlidity of ssertion of Flett s theorem re described in Section 3 together with proving two new extensions nd the detiled comprison of ll the presented conditions. Section 4 dels with integrl version

4 O. Hutník, J. Molnárová: On Flett s men vlue theorem 3 of Flett s theorem nd relted results. In the lst Section 5 we give new proof of higher-order generliztion of Flett s men vlue theorem due to Pwlikowsk. 2 Flett s men vlue theorem Let us begin with the following esy observtion from [5]: if g C, b, then from the integrl men vlue theorem there exists η (, b) such tht g(η) = 1 b b g(t) dt. Moreover, if we consider the function g C, b with the properties nd define the function g() = 0, b g(t) dt = 0, (1) ϕ(x) = { 1 x x g(t) dt, x (, b, 0, x =, then ϕ C, b D(, b) nd ϕ() = 0 = ϕ(b). Thus, by Rolle s theorem there exists η (, b) such tht ϕ (η) = 0, i.e., g(η) η 1 (η ) 2 η g(t) dt = 0 g(η) = 1 η η g(t) dt. The ltter formul resembles the one from integrl men vlue theorem replcing formlly b by η. It is well-known tht the men vlue ϕ (known s the integrl men) of function g on the intervl, x is in generl less irregulr in its behviour thn g itself. When defining the function g we my sk whether the second condition in (1) my be replced by simpler condition, e.g., by the condition g(b) = 0. Lter we will show tht it is possible nd the result in this more generl form is consequence of Drboux s intermedite vlue theorem, see the proof of Flett s theorem. If we define the function f(x) = x g(t) dt, x, b, from our considertions we get n equivlent form of result to which this pper is devoted. This result is n observtion of Thoms Muirhed Flett ( ) from 1958 published in his pper [5]. Indeed, it is vrition on the theme of Rolle s theorem where the condition f() = f(b) is replced by f () = f (b), or, we my sy tht it is Lgrnge s type men vlue theorem with Rolle s type condition.

5 4 IM Preprint series A, No. 4/2012 Theorem 2.1 (Flett, 1958) If f D, b nd f () = f (b), then there exists η (, b) such tht f (η) = η K (f). (2) For the ske of completeness (nd becuse it is not wide known) we give the originl proof of Flett s theorem dpted from [5] nd rewritten in the sense of introduced nottion. Proof of Flett s theorem Without loss of generlity ssume tht f () = f (b) = 0. If it is not the cse we tke the function h(x) = f(x) xf () for x, b. Put { x K (f), x (, b g(x) = (3) f (), x =. Obviously, g C, b D(, b) nd g x (x) = K (f) f (x) = x x K (g) + x K (f ), x (, b. It is enough to show tht there exists η (, b) such tht g (η) = 0. From the definition of g we hve tht g() = 0. If g(b) = 0, then Rolle s theorem gurntees the existence of point η (, b) such tht g (η) = 0. Let g(b) 0 nd suppose tht g(b) > 0 (similr rguments pply if g(b) < 0). Then g (b) = b g(b) K (g) = b < 0. Since g C, b nd g (b) < 0, i.e., g is strictly decresing in b, then there exists x 1 (, b) such tht g(x 1 ) > g(b). From continuity of g on, x 1 nd from reltions 0 = g() < g(b) < g(x 1 ) we deduce from Drboux s intermedite vlue theorem tht there exists x 2 (, x 1 ) such tht g(x 2 ) = g(b). Since g C x 2, b D(x 2, b), from Rolle s theorem we hve g (η) = 0 for some η (x 2, b) (, b). A different proof of Flett s theorem using Fermt s theorem (necessry condition for the existence of locl extremum) my be found in [18, p.225]. Geometricl mening of Flett s theorem If curve y = f(x) hs tngent t ech point of, b nd tngents t the end points [, f()] nd [b, f(b)] re prllel, then Flett s theorem gurntees the existence of such point η (, b) tht the tngent constructed to the grph of f t tht point psses through the point [, f()], see Fig. 2. Remrk 2.2 Clerly, the ssertion of Flett s theorem my be vlid lso in the cses when its ssumption is not fulfilled. For instnce, function f(x) = x on

6 O. Hutník, J. Molnárová: On Flett s men vlue theorem 5 y η b x y = f() + f (η)(x ) Figure 2: Geometricl interprettion of Flett s theorem the intervl, b, with < 0 < b, is not differentible, but there exist infinite mny points η (, 0) for which the tngent constructed in the point η psses through the point [, ] (since the tngent coincides with the grph of function f(x) for x (, 0)). Another exmple is the function g(x) = sgn x nd h(x) = [x] (sign function nd floor function) on the intervl 1, 1 which re not differentible on 1, 1. Finlly, the function k(x) = rcsin x on 1, 1 is not differentible t the end points, but ssertion of Flett s theorem still holds (we will consider other sufficient conditions for vlidity of (2) in Section 3, nmely k fulfills Tong s condition). We cn observe tht the functions g nd k hve improper derivtives t the points in which re not differentible, i.e., g + (0) = g (0) = k + ( 1) = k (1) = +. Therefore, we stte the conjecture tht Flett s theorem still holds in tht cse. Conjecture 2.3 If f hs proper or improper derivtive t ech point of the intervl, b nd the tngents t the end points re prllel, then there exists η (, b) such tht (2) holds. Remrk 2.4 Assertion of Flett s theorem my be written in the following equivlent forms: f f(η) f() f (η) 1 0 (η) = f() = T 1 (f, η)() η f() 1 f(η) η 1 = 0. In the second expression T 1 (f, x 0 )(x) is the first Tylor s polynomil (or, in other words tngent) of function f t the point x 0 s function of x. The lst expression resembles n equivlent formultion of the ssertion of Lgrnge s theorem in the form of determinnt, i.e., f (η) 1 0 f() 1 f(b) b 1 = 0.

7 6 IM Preprint series A, No. 4/2012 This motivtes us to stte the following question: Question 2.5 Is it possible to find similr proof (s derivtive of function given in the form of determinnt) of Flett s theorem? In connection with pplicbility of Flett s theorem there exists mny interesting problems proposed nd solved by vrious uthors, see e.g. the problems nd solutions section of journls s Americn Mthemticl Monthly, Electronic Journl of Differentil Equtions, etc. A nice ppliction of Flett s theorem for investigting some integrl men vlue theorems is given in [9] nd similr pproch is used in [6]. We give here only one representtive exmple of this kind. The problem ( in Electronic Journl of Differentil Equtions) ws proposed by Duong Viet Thong, Vietnm. The solution to this problem is our own. Problem 2.6 Let f C 0, 1 nd 1 0 f(x) dx = Prove tht there exists η (0, 1) such tht η 2 f(η) = η xf(x) dx. xf(x) dx. Solution. Consider the differentible function G(t) = t 0 xf(x) dx, t 0, 1. Clerly, G (t) = tf(t) for ech t 0, 1. By [9, Lemm 2.8] there exists ζ (0, 1) such tht G(ζ) = ζ xf(x) dx = 0. Since G(0) = 0, then by Rolle s theorem there 0 exists θ (0, ζ) such tht G (θ) = 0. From G (0) = 0 = G (θ) by Flett s theorem there exists η (0, θ) such tht G (η) = η 0K (G) ηf(η) = G(η) η η 2 f(η) = η 0 xf(x) dx. Nturlly, we my sk whether the Lgrnge s ide to remove the equlity f() = f(b) from Rolle s theorem is pplicble for Flett s theorem, i.e., whether the ssumption f () = f (b) my be removed for the purpose to obtin more generl result. First result of tht kind hs ppered in the book [22]. Theorem 2.7 (Riedel-Shoo, 1998) If f D, b, then there exists η (, b) such tht η K (f) = f (η) b K (f ) η 2.

8 O. Hutník, J. Molnárová: On Flett s men vlue theorem 7 In their originl proof [22] Riedel nd Shoo consider the uxiliry function ψ given by ψ(x) = f(x) b K (f (x )2 ), x, b, (4) 2 nd pply Flett s theorem to it. Indeed, function ψ is constructed s difference of f nd its qudrtic pproximtion A+B(x )+C(x ) 2 t neighbourhood of. From ψ () = ψ (b) we get C = 1 b K (f ), nd becuse A nd B my be 2 rbitrry, they put A = B = 0. Of course, the function ψ is not the only function which does this job. For instnce, the function ( ) x Ψ(x) = f(x) b K (f 2 ) 2 x, x, b, does the sme, becuse Ψ (x) = ψ (x) for ech x, b. In wht follows we provide different proof of Riedel-Shoo s theorem with n uxiliry function of different form. New proof of Riedel-Shoo s theorem. Let us consider the function F defined by f(x) x 2 x 1 F(x) = f() 2 1 f () 2 1 0, x, b. f (b) 2b 1 0 Clerly, F D, b nd F (x) = f (x) 2x 1 0 f() 2 1 f () f (b) 2b 1 0, x, b. Thus, F () = F (b) = 0, nd by Flett s theorem there exists η (, b) such tht F (η) = η K (F), which is equivlent to the ssertion of Riedel-Shoo s theorem. Remrk 2.8 As in the cse of Flett s theorem it is esy to observe tht the ssertion of Riedel-Shoo s theorem my be equivlently written s follows f (η) 1 0 f() 1 f(η) η 1 = b K (f (η ) ). 2 The geometricl fct behind Flett s theorem is source of interesting study in [4] we would like to mention here in connection with Riedel-Shoo s theorem.

9 8 IM Preprint series A, No. 4/2012 Following [4] we will sy tht the grph of f C, b intersects its chord in the extended sense if either there is number η (, b) such tht η K (f) = b x K (f), or lim K (f) = b x + K (f). Now, for f C, b denote by M the set of ll points x, b in which f is non-differentible nd put m = M. Define the function F(x) := 1 x (f (x) x K (f)), x (, b \ M. Then the ssertion of Flett s theorem is equivlent to F(η) = 0. Clerly, if m = 0, then by Riedel-Shoo s theorem there exists η (, b) such tht So, wht if m > 0? F(η) = 1 2 bk (f ). Theorem 2.9 (Powers-Riedel-Shoo, 2001) Let f C, b. (i) If m n for some non-negtive integer nd / M, then there exist n + 1 points η 1,...,η n+1 (, b) nd n + 1 positive numbers α 1,..., α n+1 with n+1 i=1 α i = 1 such tht n+1 α i F(η i ) = 1 ( b K (f) f () ). b i=1 (ii) If m is infinite nd the grph of f intersects its chord in the extended sense, then there exist η (, b) nd two positive numbers δ 1, δ 2 such tht either F 1 (η, h) 0 F 2 (η, k), or F 2 (η, k) 0 F 1 (η, h), holds for h (0, δ 1 nd k (0, δ 2, where F 1 (η, h) := (η ) ( η η h K (f) η K (f)), F 2 (η, k) := (η ) ( η+k η K (f) η K (f)). In item (i) we note tht if f () = b K (f), i.e., the second condition for the grph of f intersecting its chord in the extended sense holds, then the convex combintion of vlues of F t points η i, i = 1,...,n + 1, is simply zero. If, in item (ii), f is differentible t η, then lim h 0 + F 1 (η, h) (η ) 2 = lim k 0 + F 2 (η, k) (η ) 2 = F(η).

10 O. Hutník, J. Molnárová: On Flett s men vlue theorem 9 The proof of item (i) cn be found in [17] nd the proof of (ii) is given in [4]. Note tht in the pper [17] uthors extended the results of Theorem 2.9 in the context of topologicl vector spces X, Y for clss of Gteux differentible functions f : X Y. Flett s nd Riedel-Shoo s theorem give n opportunity to study the behviour of intermedite points from different points of view. Recll tht points η (depending on the intervl, b ) from Flett s, or Riedel-Shoo s theorem re clled the Flett s, or the Riedel-Shoo s points of function f on the intervl, b, respectively. The questions of stbility of Flett s points ws firstly investigted in [3], but the min result therein ws shown to be incorrect. In pper [8] the correction ws mde nd the following results on Hyers-Ulm s stbility of Riedel-Shoo s nd Flett s points were proved. Theorem 2.10 (Lee-Xu-Ye, 2009) Let f D, b nd η be Riedel-Shoo s point of f on, b. If f is twice differentible t η nd f (η)(η ) 2f (η) + 2 η K (f) 0, then to ny ε > 0 nd ny neighborhood N (, b) of η, there exists δ > 0 such tht for every g D, b stisfying g(x) g() (f(x) f()) < δ for x N nd g (b) g () = f (b) f (), there exists point ξ N such tht ξ is Riedel-Shoo s point of g nd ξ η < ε. As corollry we get the Hyers-Ulm s stbility of Flett s points. Theorem 2.11 (Lee-Xu-Ye, 2009) Let f D, b with f () = f (b) nd η be Flett s point of f on, b. If f is twice differentible t η nd f (η)(η ) 2f (η) + 2 η K (f) 0, then to ny ε > 0 nd ny neighborhood N (, b) of η, there exists δ > 0 such tht for every g D, b stisfying g() = f() nd g(x) f(x) < δ for x N, there exists point ξ N such tht ξ is Flett s point of g nd ξ η < ε. Another interesting question is the limit behviour of Riedel-Shoo s points (Flett s points re not interesting becuse of the condition f () = f (b)). We demonstrte the min ide on the following esy exmple: let f(t) = t 3 for t 0, x with x > 0. By Riedel-Shoo s theorem for ech x > 0 there exists point η x (0, x) such tht 3ηx 2 = η3 x + 3x2 η x x ηx 2 4η 2 x = 3xη x η x = 3 4 x.

11 10 IM Preprint series A, No. 4/2012 Thus, we hve obtined dependence of Riedel-Shoo s points on x. If we shorten the considered intervl, we get η x 0 lim x 0 + x 0 = lim x x x = 3 4. So, how do Flett s points behve for the widest clss of function? In pper [16] uthors proved the following result. Theorem 2.12 (Powers-Riedel-Shoo, 1998) Let f D, + x be such tht f(t) = p(t) + (t ) α g(t), α (1, 2) (2, + ), where p is polynomil t most of second order, g is bounded on the intervl (, + x nd g() = lim g( + x) 0. Then x 0 + η x lim x 0 + x ( = α 2(α 1) ) 1 α 2, where η x re the corresponding Riedel-Shoo s points of f on, + x. Problem 2.13 Enlrge the Power-Riedel-Shoo s fmily of functions for which it is possible to stte the exct formul for limit properties of corresponding intermedite points. 3 Further sufficient conditions for vlidity of (2) 3.1 Trhn s inequlities Probbly the first study bout Flett s result nd its generliztion is dted to the yer 1966 by Donld H. Trhn [24]. He provides different condition for the ssertion of Flett s theorem under some inequlity using comprison of slopes of secnt line pssing through the end points nd tngents t the end points. Theorem 3.1 (Trhn, 1966) Let f D, b nd ( f (b) b K (f)) (f () b K (f)) 0. (5) Then there exists η (, b such tht (2) holds. Donld Trhn in his proof gin considers the function g given by (3). Then g C, b D(, b nd g (x) = 1 x (f (x) x K (f)), x (, b.

12 O. Hutník, J. Molnárová: On Flett s men vlue theorem 11 Since [g(b) g()] g (b) = 1 ( f (b) b b K (f)) (f () b K (f)), then by (5) we get [g(b) g()] g (b) 0. Now Trhn concludes tht g (η) = 0 for some η (, b, which is equivlent to (2). The only step here is to prove Trhn s lemm, i.e., the ssertion of Rolle s theorem under the conditions g C, b D(, b nd [g(b) g()] g (b) 0. Esily, if g() = g(b), then Rolle s theorem gives the desired result. If g (b) = 0, putting η = b we hve g (η) = 0. So, let us ssume tht [g(b) g()] g (b) < 0. This mens tht either g (b) < 0 nd g(b) > g(), or g (b) > 0 nd g(b) < g(). In the first cse, since g C, b, g(b) > g() nd g is strictly decresing in b, then g hs its mximum t η (, b) nd by Fermt s theorem we get g (η) = 0. Similrly, in the second cse g hs minimum t the sme point η (, b), thus g (η) = 0. Remrk 3.2 Obviously, the clss of Trhn s functions, i.e., differentible functions on, b stisfying Trhn s condition (5), is wider thn the clss of Flett s functions f D, b stisfying Flett s condition f () = f (b). Indeed, for f () = f (b) Trhn s condition (5) is trivilly fulfilled. On the other hnd the function y = x 3 for x 1, 1 does not stisfy Flett s condition, nd it is esy 2 to verify tht it stisfies Trhn s one. Geometricl mening of Trhn s condition Clerly, Trhn s inequlity (5) holds if nd only if [ f (b) b K (f) f () b K (f)] [ f (b) b K (f) f () b K (f)]. Since b K (f) gives the slope of the secnt line between [, f()] nd [b, f(b)], Trhn s condition requires either both slopes of tngents t the end points re greter or equl, or both re smller or equl to b K (f). We consider two cses: (i) if f (b) = b K (f), then the tngent t b is prllel to the secnt, nd the tngent t my be rbitrry (prllel to the secnt, lying bove or under the grph of secnt on (, b)), nlogously for f () = b K (f); (ii) if f (b) b K (f) nd f () b K (f), then one of the tngents t the end points hs to lie bove nd the second one under the grph of secnt line on (, b), or vice vers, see Fig. 3. More precisely, let tngent t intersect the line x = b t the point Q = [b, y Q ] nd tngent t b intersect the line x = t the point P = [, y P ]. Then either y Q > f(b) nd y P < f(), or y Q < f(b) nd y P > f(). For prllel tngents t the end points, i.e., for f () = f (b), this geometricl interprettion provides new insight which leds to the lredy mentioned pper [4].

13 12 IM Preprint series A, No. 4/2012 y Q y Q f(x) f() f(b) b x y P P Figure 3: Geometricl interprettion of Trhn s condition Moreover, Trhn in his pper [24] provides other generliztion of Flett s theorem. Nmely, he proves certin,,cuchy form of his result for two functions which will be source of our results lter in Section 3.3. Theorem 3.3 (Trhn, 1966) Let f, g D, b, g (x) 0 for ech x, b nd ( ) f () ( ) g () b b K (f, g) K (g) f (b) b K (f) g (b) 0. Then there exists η (, b such tht f (η) g (η) = η K (f, g). Its proof is bsed on ppliction of Trhn s lemm [24, Lemm 1] for function { x K (f, g), x (, b h(x) = f (), x =. g () 3.2 Tong s discrete nd integrl mens Another sufficient condition for vlidity of (2) ws provided by JingCheong Tong in the beginning of 21st century in his pper [23]. Tong does not require differentibility of function f t the end points of the intervl, b, but he uses certin mens of tht function. Indeed, for function f : M R nd two distinct points, b M denote by A f (, b) = f() + f(b) 2 nd I f (, b) = 1 b b f(t) dt the rithmetic (discrete) nd integrl (continuous) men of f on the intervl, b, respectively.

14 O. Hutník, J. Molnárová: On Flett s men vlue theorem 13 Theorem 3.4 (Tong, 2004) Let f C, b D(, b). If A f (, b) = I f (, b), then there exists η (, b) such tht (2) holds. In his proof Tong defines the uxiliry function { (x )[A f (, x) I f (, x)], x (, b h(x) = 0, x =. Esily, h C, b D(, b) nd h() = 0 = h(b). Then Rolle s theorem for h on, b finishes the proof. Geometricl mening of Tong s condition The condition A f (, b) = I f (, b) is not so evident geometriclly in comprison with the Flett s condition f () = f (b). In some sense we cn demonstrte it s the re under the grph of f on, b is exctly the volume of rectngle with sides b nd f()+f(b). 2 Let us nlyze Tong s condition A f (, b) = I f (, b) for f C, b D(, b) in detil. It is importnt to note tht this equlity does not hold in generl for ech f C, b D(, b). Indeed, for f(x) = x 2 on 0, 1 we hve A f (0, 1) = = 1 2, I f(0, 1) = x 2 dx = 1 3. A nturl question is how lrge is the clss of such functions? For f C (M) D(M) denote by F primitive function to f on n intervl M nd let, b be interior points of M. Then the condition A f (x, b) = I f (x, b), x M, is equivlent to the condition f(x) + f(b) = x K (F), x. (6) 2 Since f D(M), then F D 2 (M) nd f (t) = F (t) for ech t M. Differentiting the equlity (6) with respect to x we get f (x) 2 which is equivlent to the eqution = F (x) x K (F), x F (x)(x ) 2 = 2(F (x)(x ) + F() F(x)). Solving this differentil eqution on intervls (, b) M nd (b, + ) M, nd using the second differentibility of F t b we hve F(x) = α 2 x2 + βx + γ, x M, α, β, γ R, α 0, nd therefore f(x) = αx + β, x M.

15 14 IM Preprint series A, No. 4/2012 So, the clss of functions fulfilling Tong s condition A f (, b) = I f (, b) for ech intervl, b is quite smll (ffine functions, in fct). Of course, if we do not require the condition on ech intervl, b, then we my use, e.g., the function y = rcsin x on the intervl 1, 1 which does not stisfy neither Flett s nor Trhn s condition (becuse it is not differentible t the end points). Remrk 3.5 The reltions mong the clsses of Flett s, Trhn s nd Tong s functions re visulized in Fig. 4, where ech clss is displyed s rectngle with the corresponding nme below the left corner. Moreover, i, i = 1,...,6, re the clsses of functions of possible reltionships of Flett s, Trhn s nd Tong s clsses of functions. For instnce, 6 denotes clss of (not necessrily differentible or continuous) functions on, b for which none of the three conditions is fulfilled, but the ssertion of Flett s theorem still holds. Immeditely, 1 is non-empty, becuse it contins ll ffine functions on, b. Thus, (i) Flett s nd Trhn s conditions were compred in Remrk 3.2 yielding tht Trhn s clss of functions is wider thn Flett s one, i.e., ; (ii) Tong s condition nd Flett s condition re independent ech other, becuse for π f(x) = sin x, x 2, 5 2 π, we hve f ( π) = f ( 5π) = 0, but 1 = A 2 2 f( π, 5π) I 2 2 f( π, 5 π) = 0; on the 2 2 other hnd f(x) = rcsin x for x 1, 1 fulfills Tong s condition, but does not stisfy Trhn s one; lso for f(x) = x 3, x 1, 1, we hve A f ( 1, 1) = I f ( 1, 1) nd f ( 1) = f (1) which yields tht the clsses of functions 1, 2 nd 3 re non-empty; (iii) similrly, Trhn s condition nd Tong s condition re independent, e.g., the function f(x) = x 3 on the intervl 1, 1 stisfies Trhn s condition, 2 but A f ( 1, 1) I 2 f( 1, 1), so 2 4 is non-empty, nd for f(x) = rcsin x, x 1, 1 we hve f 5 ; (iv) for the function sgn on 2, 1 none of the three conditions is fulfilled, but the ssertion (2) still holds, i.e., 6 is non-empty. Question 3.6 Is ech clss i, i {3, 5, 6}, non-empty when considering the stronger condition f D, b in Tong s ssumption? Removing the condition A f (, b) = I f (, b) Tong obtined the following result which no more corresponds to the result of Riedel-Shoo s theorem. Theorem 3.7 (Tong, 2004) Let f C, b D(, b). Then there exists η (, b) such tht f (η) = η K (f) + 6[A f(, b) I f (, b)] (b ) 2 (η ).

16 O. Hutník, J. Molnárová: On Flett s men vlue theorem Flett 3 Trhn 5 6 Tong Figure 4: The reltions mong Flett s, Trhn s nd Tong s fmilies of functions Tong s proof uses the uxiliry function H(x) = f(x) 6[A f(, b) I f (, b)] (b ) 2 (x )(x b), x, b. It is esy to verify tht H C, b D(, b), H() = f() nd H(b) = f(b). Thus, A H (, b) = A f (, b) nd I H (, b) = A H (, b). Then by Theorem 3.4 there exists η (, b) such tht H (η) = η K (H) which is equivlent to the ssertion of theorem. Question 3.8 Anlogously to Riedel-Shoo s Theorem 2.7 we my sk the following: Wht is the limit behviour of Tong s points η of function f on the intervl, b? 3.3 Two new extensions: inequlities in ction In this section we present other sufficient conditions for vlidity of (2) nd its extension. As fr s we know they re not included in ny literture we were ble to find. The bsic ide is mixture of Trhn s results with (lthough not explicitly stted) Diz-Výborný s concept of intersecting the grphs of two functions [4]. We will lso present nice geometricl interprettions of these results. A prticulr cse of our second result is discussed in the end of this section. Lemm 3.9 If f, g D, b, g(b) g() nd [ f () b K (f, g) g () ] [f (b) b K (f, g) g (b) ] 0, (7) then there exists ξ (, b) such tht f(ξ) f() = b K (f, g)(g(ξ) g()). (8)

17 16 IM Preprint series A, No. 4/2012 Proof. Let us consider the function ϕ(x) = f(x) f() b K (f, g) (g(x) g()), x, b. Then ϕ () = f () b K (f, g) g () nd ϕ (b) = f (b) b K (f, g) g (b). If ϕ () 0, then ccording to ssumption we get ϕ (b) 0. So, there exist points α, β (, b) such tht ϕ(α) > 0 nd ϕ(β) < 0. Thus, ϕ(α)ϕ(β) < 0 nd by Bolzno s theorem there exists point ξ (α, β) such tht ϕ(ξ) = 0. The cse ϕ () 0 nd ϕ (b) 0 is nlogous. Theorem 3.10 If f, g D, b, g(b) g() nd the inequlity (7) holds, then there exists η (, b) such tht f (η) η K (f) = b K (f, g) [g (η) η K (g)]. (9) Proof. Let us tke the uxiliry function F(x) = { x K (f) b K (f, g) xk (g), x (, b, f () b K (f, g) g (), x =. Observe tht F(x) = x K (ϕ) for x (, b, where ϕ is the uxiliry function from the proof of Lemm 3.9. Thus, by Lemm 3.9 there exists point ξ (, b) such tht F(ξ) = 0 = F(b). Since F C ξ, b D(ξ, b), then by Rolle s theorem there exists η (ξ, b) (, b) such tht F (η) = 0 which is equivlent to the desired result. In wht follows we denote by T n (f, x 0 )(x) := f(x 0 ) + f (x 0 ) 1! (x x 0 ) + + f(n) (x 0 ) (x x 0 ) n n! the n-th Tylor s polynomil of function f t point x 0. Rewriting the ssertion of Theorem 3.10 in terms of Tylor s polynomil yields f() T 1 (f, η)() = b K (f, g) (g() T 1 (g, η)()). Geometricl mening of Theorem 3.10 Relize tht T 1 (f, x 0 )(x) is the tngent to the grph of f t the point x 0, i.e., T 1 (f, x 0 )(x) = f(x 0 ) + f (x 0 )(x x 0 ). Then the eqution (9) my be equivlently rewritten s follows ( η (ξ, b) (, b)) f() T 1 (f, η)() = b K (f, g) (g() T 1 (g, η)()). (10)

18 O. Hutník, J. Molnárová: On Flett s men vlue theorem 17 y g(x) f(x) ξ η b x P Figure 5: Geometricl interprettion of Theorem 3.10 Since f(b) f() = g(b) g() f(b) g(b) = f() g(), then the eqution (8) hs the form ( ξ (, b)) f(ξ) g(ξ) = f() g() = f(b) g(b), (11) nd (10) my be rewritten into ( η (ξ, b)) T 1 (f, η)() T 1 (g, η)() = f() g() = f(b) g(b). (12) Thus, considering function g such tht g() = f() nd g(b) = f(b) the eqution (11) yields ( ξ (, b)) f(ξ) = g(ξ) nd the eqution (12) hs the form ( η (ξ, b)) T 1 (f, η)() = T 1 (g, η)(). Geometriclly it mens tht tngents t the points [η, f(η)] nd [η, g(η)] pss though the common point P on the line x =, see Fig. 5. Remrk 3.11 Observe tht in the specil cse of secnt joining the end points, i.e., the function g(x) = f() + b K (f)(x ), where f is function fulfilling ssumptions of Theorem 3.10, we get the originl Trhn s result of Theorem 3.1 (in fct, generliztion of Flett s theorem) with the explicit geometricl interprettion on Fig. 6.

19 18 IM Preprint series A, No. 4/2012 y f(x) ξ η b x Figure 6: Geometricl interprettion of Theorem 3.10 for secnt Lemm 3.12 Let f, g D, b nd f, g be twice differentible t the point. If g() g(b) nd [ f () b K (f, g) g () ] [ f () b K (f, g) g () ] > 0, (13) then there exists ξ (, b) such tht f () ξ K (f) = b K (f, g) [ g () ξ K (g)]. Proof. Consider the function 0, x =, F(x) = f () x K (f) + b K (f, g) [x K (g) g ()], x (, b), f () b K (f, g) g (), x = b. Then F C, b D, b) with F () = lim x + f ()(x ) (f(x) f()) + b K (f,g)[g(x) g() g ()(x )] (x ) 2 f () f (x) + b = lim K (f,g) [g (x) g ()] x + 2(x ) = 1 ( f 2 lim (x) f () b x + x K (f,g) g (x) g ) () x = 1 [ ] f () b 2 K (f,g) g (), where L Hospitl rule hs been used. Suppose tht F(b) = [ f () b K (f, g) g () ] > 0, nlogous rguments pply if F(b) < 0. Then [ f () b K (f, g) g () ] > 0

20 O. Hutník, J. Molnárová: On Flett s men vlue theorem 19 by ssumption of theorem which implies F () < 0. Since F() = 0, then there exists α (, b) such tht F(α) < 0. According to Bolzno s theorem there exists point ξ (α, b) such tht F(ξ) = 0, which completes the proof. Theorem 3.13 Let f, g D, b nd f, g be twice differentible t the point. If g() g(b) nd the inequlity (13) holds, then there exists η (, b) such tht (9) holds. Proof. Consider the function F s in the proof of Lemm Then by Lemm 3.12 there exists ξ (, b) such tht F(ξ) = 0 = F() nd by Rolle s theorem there exists η (, ξ) such tht F (η) = 0. Geometricl mening of Theorem 3.13 Using the Tylor s polynomil we cn rewrite the ssertion of Lemm 3.12 nd Theorem 3.13 s follows nd ( ξ (, b)) f(ξ) T 1 (f, )(ξ) = b K (f, g) (g(ξ) T 1(g, )(ξ)) (14) ( η (, ξ)) f() T 1 (f, )(η) = b K (f, g) (g() T 1 (g, )(η)), (15) respectively. Since f(b) f() = g(b) g() f(b) g(b) = f() g(), then (14) my be rewritten s ( ξ (, b)) f(ξ) g(ξ) = T 1 (f, )(ξ) T 1 (g, )(ξ). Similrly (15) my be rewritten s follows ( η (, ξ)) f() g() = T 1 (f, η)() T 1 (g, η)()). If f nd g hve the sme vlues t the end points, the lst eqution reduces to ( η (, b)) T 1 (f, η)() = T 1 (g, η)(). Geometriclly it mens tht tngents t points [η, f(η)] nd [η, g(η)] pss through the common point P on the line x =, see Fig. 7. Remrk 3.14 Agin, Theorem 3.13 for the secnt g(x) = f() + b K (f)(x ) gurntees the existence of point η (, b) such tht T 1 (f, η)() = f(), i.e., tngent t [η, f(η)] psses through the point A = [, f()] which is exctly the

21 20 IM Preprint series A, No. 4/2012 y g(x) P f(x) x = η b x Figure 7: Geometricl interprettion of Theorem 3.13 geometricl interprettion of Flett s theorem in Fig. 2. The ssumption (13) reduces in the secnt cse to the inequlity [ f () b K (f)] f () > 0, (16) i.e., [ f () > b K (f) f () > 0 ] [ f () < b K (f) f () < 0 ]. Considering the first cse yields f () > b K (f) f () > 0 f(b) < f() + f ()(b ) f () > 0 f(b) < T 1 (f, )(b) f () > 0. This mens tht there exists point X = [ξ, f(ξ)] such tht the line AX is tngent to the grph of f t A = [, f()]. Then from the ssertion of Theorem 3.13 we hve the existence of point E = [η, f(η)], where η (, ξ), such tht the tngent to the grph of f t E psses through the point A = [, f()], see Fig. 8. Similrly for the second cse. Remrk 3.15 Observe tht if f () = b K (f), then the condition (16) is not fulfilled, but the ssertion (2) of Flett s theorem still holds by Trhn s condition. On the other hnd, if f () b K (f) nd f () = 0, then the ssertion (2) does not need to hold, e.g., for f(x) = sin x on the intervl 0, π we hve (f (0) π 0 K (f)) f (0) = 1 0 = 0, but there is no such point η (0, π) which is solution of the eqution η cos η = sin η.

22 O. Hutník, J. Molnárová: On Flett s men vlue theorem 21 y E f(x) X A η ξ b x Figure 8: Geometricl interprettion of Theorem 3.13 for secnt We hve to point out tht the inequlity (16) ws observed s sufficient condition for vlidity of (2) in [10], but strting from different point, therefore our generl result of Theorem 3.13 seems to be new. Indeed, Mleševi`c in [10] considers some itertions of Flett s uxiliry function in terms of n infinitesiml function, i.e., for f D, b which is differentible rbitrry number of times in right neighbourhood of the point he defines the following functions { { x K (f) f (), x (,b x K (α k ) α k (), x (,b α 1 (x) =,... α k+1 (x) = 0, x = 0, x =, for k = 1, 2,... Then he proves the following result. Theorem 3.16 (Mleševi`c, 1999) Let f D, b nd f be (n + 1)-times differentible in right neighbourhood of the point. If either α n(b)α n (b) < 0, or α n()α n (b) < 0, then there exists η (, b) such tht α n(η) = 0. For n = 1 Mleševi`c s condition α 1 (b)α 1(b) < 0 is equivlent to Trhn s condition (5) where the second differentibility of f in right neighbourhood of is superfluous constrint. The second Mleševi`c s condition α 1()α 1 (b) < 0 is equivlent to our condition (16), becuse α 1() f(x) f() f ()(x ) f (x) f () = lim = lim x + (x ) 2 x + 2(x ) nd then the inequlity 0 > α 1 ()α 1(b) = 1 2 f () ( b K (f) f () ) = 1 2 f ()

23 22 IM Preprint series A, No. 4/2012 Λ 3 Λ 2 Λ 1 Λ 5 Λ4 Flett Λ 8 Λ 7 Λ 6 Trhn Λ 10 Λ 9 Λ 11 Tong Mlešević Λ 12 Figure 9: The digrm of four fmilies of functions holds if nd only if (16) holds. However, we require only the existence of f () in (16). Note tht for n > 1 Mleševi`c s result does not correspond to Pwlikowsk s theorem ( generliztion of Flett s theorem for higher-order derivtives), see Section 5, but it goes different wy. Fig. 9 shows ll the possible cses of reltions of clsses of functions stisfying ssumptions of Flett, Trhn, Tong nd Mlešević, respectively. Some exmples of functions belonging to sets Λ 1,...,Λ 12 were lredy mentioned (e.g. y = x 3, x 1, 1, belongs to Λ 1 ; y = sin x, x π 2, 5 2 π, belongs to Λ 2; y = x 3, x 2 3, 1, belongs to Λ 3; y = rcsin x, x 1, 1, belongs to Λ 9, nd y = sgn x, x 2, 1, belongs to Λ 12 ), other (nd more sophisticted) exmples is not so difficult to find. Remrk 3.17 Agin, if we strengthen our ssumption nd consider only the functions f D, b which re twice differentible t, we my sk the legitimte question: Is ech of the sets Λ i, i = 1,...,12, in Fig. 9 non-empty? In the positive cse, it would be interesting to provide complete chrcteriztion of ll the clsses of functions. Problem 3.18 All the presented conditions re only sufficient for the ssertion of Flett s theorem to hold. Provide necessry condition(s) for the vlidity of (2). 4 Integrl Flett s men vlue theorem Nturlly s in the cse of Lgrnge s theorem we my sk whether Flett s theorem hs its nlogicl form in integrl clculus. Consider therefore function f C, b. Putting F(x) = x f(t) dt, x, b,

24 O. Hutník, J. Molnárová: On Flett s men vlue theorem 23 the fundmentl theorem of integrl clculus yields tht F D, b with F () = f() nd F (b) = f(b). If f() = f(b), then the function F on the intervl, b fulfils the ssumptions of Flett s theorem nd we get the following result. It ws proved by Stnley G. Wyment in 1970 nd it is nothing but the integrl version of Flett s theorem. Although our presented reflection is trivil proof of this result, we dd here the originl Wyment s proof dopted from [25] which does not use the originl Flett s theorem. Theorem 4.1 (Wyment, 1970) If f C, b with f() = f(b), then there exists η (, b) such tht f(η) = I f (, η). (17) Proof. Consider the function { (t )[f(t) I f (, t)], t (, b, F(t) = 0, t =. If f is constnt on, b, then F 0 nd the ssertion of theorem holds trivilly. Thus, suppose tht f is non-constnt. Since f C, b, then by Weierstrss theorem on the existence of extrem there exist points t 1, t 2, b such tht ( t, b ) f(t 1 ) f(t) f(t 2 ). From f() = f(b) we deduce tht f cnnot chieve both extrem t the end points nd b. If t 2, then F(b) < 0 < F(t 2 ) nd by Bolzno s theorem there exists η (t 2, b) such tht F(η) = 0. If t 1, then F(t 1 ) < 0 < F(b) nd nlogously s bove we conclude tht there exists η (t 1, b) such tht F(η) = 0. Finlly, consider the cse when none of t 1 nd t 2 is equl to. Then < min{t 1, t 2 } < mx{t 1, t 2 } < b, nd so F(t 1 ) 0 F(t 2 ). From Bolzno s theorem pplied to function F on the intervl t 1, t 2 we hve tht there exists point η (, b) such tht F(η) = 0. Geometricl mening of Wyment s theorem Geometriclly Wyment s theorem sys tht the re under the curve f on the intervl, η is equl to (η )f(η), i.e., volume of rectngle with sides η nd f(η), see Fig. 10. Removing the condition f() = f(b) yields the following integrl version of Riedel-Shoo s theorem. Its proof is bsed on using Riedel-Shoo s theorem for function F(x) = x f(t) dt, x, b.

25 24 IM Preprint series A, No. 4/2012 y y f() f(η) f(η) y = (η )f(η) η b x η b x Figure 10: Geometricl interprettion of Wyment s theorem Theorem 4.2 If f C, b, then there exists η (, b) such tht f(η) = I f (, η) + η b K (f). 2 In wht follows we present some results from Section 3 in their integrl form to show some sufficient conditions for vlidity of (17) with short ide of their proofs. The first one is Trhn s result. Proposition 4.3 If f C, b nd [f() I f (, b)] [f(b) I f (, b)] 0, then there exists η (, b such tht (17) holds. For the proof it is enough to consider the function { I f (, x), x (, b g(x) = f(), x =, nd pply Trhn s lemm [24, Lemm 1]. To show n nlogy with Tong s result consider the following mens B f (, b) = b 2 I f(, b), J f (, b) = bi f (, b) 1 b b tf(t) dt. Proposition 4.4 Let f C, b. If B f (, b) = J f (, b), then there exists η (, b) such tht f(η) = I f (, η). In the proof we consider the uxiliry function { (x )[B f (, x) J f (, x)], x (, b, h(x) = 0, x =, nd the further steps coincide with the originl Tong s proof of Theorem 3.4. Removing the condition B f (, b) = J f (, b) we obtin the following result.

26 O. Hutník, J. Molnárová: On Flett s men vlue theorem 25 Proposition 4.5 Let f C, b. Then there exists η (, b) such tht f(η) = I f (, η) + 6[B f(, b) J f (, b)] (b ) 2 (η ). Proof, in which we use the following uxiliry function H(x) = f(x) 6[B f(, b) J f (, b)] (b ) 2 (2x b), x, b, is gin nlogous to the proof of Tong s Theorem 3.7. In the end of this chpter we formlly present integrl nlogies of new sufficient conditions of vlidity of Flett s theorem. Lemm 4.6 Let f, g C, b nd [f()i g (, b) g()i f (, b)] [f(b)i g (, b) g(b)i f (, b)] 0. (18) Then there exists ξ (, b) such tht I f (, ξ) I g (, b) = I g (, ξ) I f (, b). Proof. Considering the function { I f (, x)i g (, b) I g (, x)i f (, b), x (, b, ϕ(x) = 0, x =, we hve ϕ () = f()i g (, b) g()i f (, b), ϕ (b) = f(b)i g (, b) g(b)i f (, b). Further steps re nlogous to the proof of Lemm 3.9. Theorem 4.7 Let f, g C, b nd the inequlity (18) holds. Then there exists η (, b) such tht I g (, b) (f(η) I f (, η)) = I f (, b) (g(η) I g (, η)). (19) Proof. Tke the uxiliry function { I g (, b) I f (, x) I f (, b) I g (, x), x (, b, F(x) = f()i g (, b) g()i f (, b), x =. By Lemm 4.6 there exists ξ (, b) such tht F(ξ) = 0 = F(b). Then by Rolle s theorem for F on the intervl ξ, b there exists η (ξ, b) such tht F (η) = 0. Similrly we my prove the following integrl versions of Lemm 3.12 nd Theorem 3.13.

27 26 IM Preprint series A, No. 4/2012 Lemm 4.8 Let f, g C, b nd f, g be differentible t. If [f()i g (, b) g()i f (, b)] [f ()I g (, b) g ()I f (, b)] > 0, (20) then there exists ξ (, b) such tht I g (, b) (f() I f (, ξ)) = I f (, b) (g() I g (, ξ)). Theorem 4.9 Let f, g C, b nd f, g re differentible t. If (20) holds, then there exists η (, b) such tht (19) holds. 5 Flett s theorem for higher-order derivtives The previous sections delt with the question of replcing the condition f() = f(b) in Rolle s theorem by f () = f (b). In this section we will consider nturl question of generlizing Flett s theorem for higher-order derivtives. We will provide the originl solution of Pwlikowsk nd present new proof of her result together with some other observtions. The problem of generlizing Flett s theorem to higher-order derivtives ws posed first time by Zsolt Ples in 1997 t the 35th interntionl symposium of functionl equtions in Grz. Solution hs lredy ppered two yers lter by polish mthemticin Iwon Pwlikowsk in her pper [13] nd it hs the following form. Theorem 5.1 (Pwlikowsk, 1999) If f D n, b with f (n) () = f (n) (b), then there exists η (, b) such tht n η K (f) = i=1 ( 1) i+1 (η ) i 1 f (i) (η). (21) i! Pwlikowsk in her pper [13] generlized originl Flett s proof in such wy tht she uses (n 1)-th derivtive of Flett s uxiliry function g given by (3) nd Rolle s theorem. More precisely, the function { g (n 1) (x), x (, b G f (x) = 1 n f(n) (), x =. plys here n importnt role. Indeed, G f C, b D(, b) nd ( ) g (n) (x) = ( 1)n n! n x (x ) n K (f) + ( 1) i (x ) i 1 f (i) (x) i! i=1 = 1 ( f (n) (x) n g (n 1) (x) ) x (22)

28 O. Hutník, J. Molnárová: On Flett s men vlue theorem 27 for x (, b which cn be verified by induction. Moreover, if f (n+1) () exists, then lim g(n) (x) = 1 x + n + 1 f(n+1) (). Further steps of Pwlikowsk s proof is nlogous to the originl proof of Flett s theorem using Rolle s theorem. Similrly we my proceed using Fermt s theorem. We hve found new proof of Pwlikowsk theorem (it ws not published yet, it exists only in the form of preprint [11]) which dels only with Flett s theorem. The bsic ide consists in itertion of Flett s theorem using n pproprite uxiliry function. New proof of Pwlikowsk s theorem. function For k = 1, 2,..., n consider the ϕ k (x) = k ( 1) i+1 (k i)(x ) i f (n k+i) (x) + xf (n k+1) (), x, b. i! i=0 Running through ll indices k = 1, 2,..., n we show tht its derivtive fulfills ssumptions of Flett s men vlue theorem nd it implies the vlidity of Flett s men vlue theorem for l-th derivtive of f, where l = n 1, n 2,...,1. Indeed, for k = 1 we hve ϕ 1 (x) = f (n 1) (x) + xf (n) () nd ϕ 1 (x) = f(n) (x) + f (n) () for ech x, b. Clerly, ϕ 1() = 0 = ϕ 1(b), so pplying Flett s theorem for ϕ 1 on, b there exists u 1 (, b) such tht ϕ 1(u 1 ) = u 1 K (ϕ 1 ) u 1 K ( f (n 1)) = f (n) (u 1 ). (23) Then for ϕ 2 (x) = 2f (n 2) (x) + (x )f (n 1) (x) + xf (n 1) () we get ϕ 2 (x) = f(n 1) (x) + (x )f (n) (x) + f (n 1) () nd ϕ 2 () = 0 = ϕ 2 (u 1) by (23). So, by Flett s theorem for ϕ 2 on, u 1 there exists u 2 (, u 1 ) (, b) such tht ϕ 2(u 2 ) = u 2 K (ϕ 2 ) u 2 K ( f (n 2)) = f (n 1) (u 2 ) 1 2 (u 2 )f (n) (u 2 ). Continuing this wy fter n 1 steps, n 2, there exists u n 1 (, b) such tht n 1 u n 1 K (f ) = i=1 ( 1) i+1 (u n 1 ) i 1 f (i+1) (u n 1 ). (24) i!

29 28 IM Preprint series A, No. 4/2012 Considering the function ϕ n we get ϕ n (x) = f (x) + f () + = f () + Clerly, ϕ n () = 0 nd then n 1 i=0 u n 1 K (ϕ n ) = n 1 u n 1 K (f ) + i=1 n 1 i=1 ( 1) i+1 (x ) i f (i) (x) i! ( 1) i+1 (x ) i f (i+1) (x). i! ( 1) i+1 (u n 1 ) i 1 f (i+1) (u n 1 ) = 0 i! by (24). From it follows tht ϕ n (u n 1) = 0 nd by Flett s theorem for ϕ n on, u n 1 there exists η (, u n 1 ) (, b) such tht ϕ n (η) = η K (ϕ). (25) Since nd ϕ n(η) = f () + n i=1 ( 1) i (i 1)! (η )i 1 f (i) (η) n η K (ϕ) = f () n η K (f) + the equlity (25) yields i=1 i=1 n n η K (f) = ( 1) i (i 1)! (η )i 1 f (i) (η) = n ( 1) i+1 (n i)(η ) i 1 f (i) (η), i! n ( 1) i (η ) i 1 f (i) (η), i! i=1 ( 1 + n i ) i which corresponds to (21). Remrk 5.2 Recll tht the ssertion of Flett s theorem hs n equivlent form f() = T 1 (f, η)(). Now, in the ssertion of Pwlikowsk s theorem we cn observe deeper (nd very nturl) reltion with Tylor s polynomil. Indeed, f() = T n (f, η)() is n equivlent form of (21). Geometriclly it mens tht Tylor s polynomil T n (f, η)(x) intersects the grph of f t the point A = [, f()].

30 O. Hutník, J. Molnárová: On Flett s men vlue theorem 29 Remrk 5.3 Equivlent form of Pwlikowsk s theorem in the form of determinnt is s follows f (n) (η) h (n) n (η) h (n) n 1 (η)... h(n) 0 (η) f (n 1) (η) h n (n 1) (η) h (n) n 1 (η)... h(n 1) 0 (η) = 0, h i (x) = xi f (η) h n(η) h n 1(η)... h i! 0(η). f(η) h n (η) h n 1 (η)... h 0 (η) f() h n () h n 1 ()... h 0 () Verifiction of this fct is not s complicted s it is rther long. It is bsed on n-times ppliction of Lplce s formul ccording to the lst column. Agin we my sk whether it is possible to remove the condition f (n) () = f (n) (b) to obtin Lgrnge s type result. The first proof of this fct ws given in Pwlikowsk s pper [13]. Here we present two other new proofs. Theorem 5.4 (Pwlikowsk, 1999) If f D n, b, then there exists η (, b) such tht f() = T n (f, η)() + ( η)n+1 (n + 1)! b K ( f (n)). Proof I. Consider the uxiliry function ψ k (x) = ϕ k (x) + ( 1)k+1 (x ) k+1 (k + 1)! b K ( f (n)), k = 1, 2..., n, where ϕ k is the function from our proof of Pwlikowsk s theorem. Then nd so ψ 1 (x) = ϕ 1 (x) + (x )2 2 bk ( f (n)), x, b, ψ 1 (x) = ϕ 1 (x) + (x ) b K ( f (n)). Thus, ψ 1 () = 0 = ψ 1 (b) nd by Flett s theorem for function ψ 1 on, b there exists u 1 (, b) such tht u 1 K ( f (n 1)) = f (n) (u 1 ) + u 1 b 2 K ( f (n)). After n 1 steps we conclude tht there exists u n 1 (, b) such tht n 1 u n 1 K (f ) = i=1 ( 1) i+1 (u n 1 ) i 1 f (i+1) (u n 1 ) ( u n 1) n 1 b i! n! K ( f (n)).

31 30 IM Preprint series A, No. 4/2012 By Flett s theorem for ψ n on the intervl, u n 1 we get the desired result (ll the steps re identicl with the steps of previous proof). Proof II. Applying Pwlikowsk s theorem to the function f(x) x n+1 x n 1 F(x) = f() n+1 n 1 f (n) () (n + 1)! n! 0, x, b f (n) (b) (n + 1)! b n! 0 we get the result. Remrk 5.5 By Remrk 5.3 we my rewrite the ssertion of Theorem 5.4 in the form of determinnt s follows f (n) (η) h (n) n (η) h (n) n 1 (η)... h(n) 0 (η) f (n 1) (η) h n (n 1) (η) h (n) n 1 (η)... h(n 1) 0 (η) = b f (η) h n(η) h n 1(η)... h K ( f (n)) (η )n (n + 1)! 0(η), f(η) h n (η) h n 1 (η)... h 0 (η) f() h n () h n 1 ()... h 0 () where h i (x) = xi i! for i = 0, 1,..., n. A reltively esy generliztion of Pwlikowsk s theorem my be obtined for two functions (s kind of Cuchy version of it). Theorem 5.6 Let f, g D n, b nd g (n) () g (n) (b). Then there exists η (, b) such tht f() T n (f, η)() = b K ( f (n), g (n)) [g() T n (g, η)()]. Proof I. Considering the function h(x) = f(x) b K ( f (n), g (n)) g(x), x, b, we hve h D n, b nd h (n) () = h (n) (b). By Pwlikowsk s theorem there exists η (, b) such tht n η K (h) = i=1 which is equivlent to the stted result. ( 1) i+1 (η ) i 1 h (i) (η), i! Nturlly, s in the cse of Flett s theorem we would like to generlize e.g. Trhn s result for higher-order derivtives. Our ide of this generliztion is bsed on ppliction of Trhn s pproch to Pwlikowsk s uxiliry function (22).

32 O. Hutník, J. Molnárová: On Flett s men vlue theorem 31 Theorem 5.7 Let f D n, b nd ( ) ( f (n) ()( b) n f (n) (b)( b) n + T f () n! n! ) + T f () 0, where T f () := T n 1 (f, b)() f(). Then there exists η (, b such tht (21) holds. Proof. Consider the function g given by (3) nd function G f given by (22). Clerly, G f C, b D(, b nd g (n) (x) = ( 1)n n! (x ) n+1 (T n(f, x)() f()), x (, b. To pply Trhn s lemm [24, Lemm 1] we need to know signum of ( [G f (b) G f ()] G f(b) = g (n 1) (b) 1 ) n f(n) () g (n) (b), i.e., ( n!(n 1)! f (n) ()( b) n (b ) 2n+1 n! ) ( ) f (n) (b)( b) n + T f () + T f () 0 n! by ssumption. Then by Trhn s Lemm [24, Lemm 1] there exists η (, b such tht G f (η) = 0, which is equivlent to the ssertion of theorem. Here we present nother proof of Cuchy type Theorem 5.6 which is independent on Flett s theorem, but uses gin Trhn s lemm [24, Lemm 1]. Proof of Theorem 5.6 II. For x (, b put ϕ(x) = x K (f) nd ψ(x) = K (g). Consider the uxiliry function { ϕ (n 1) (x) b F(x) = K ( f (n), g (n)) ψ (n 1) (x), x (, b [ f (n) () b K ( f (n), g (n)) g (n) () ], x =. x 1 n Then F C, b D(, b nd for x (, b we hve F (x) = ϕ (n) (x) b K ( f (n), g (n)) ψ (n) (x) = ( 1)n n! (x ) n+1 ( Tn (f, x)() f() b K ( f (n), g (n)) (T n (g, x)() g()) ). Then it is esy to verify tht [F(b) F()] F (b) = 1 b (F(b) F())2 0, thus by Trhn s lemm [24, Lemm 1] there exists η (, b such tht F (η) = 0, i.e., f() T n (f, η)() = b K ( f (n), g (n)) [g() T n (g, η)()].

33 32 IM Preprint series A, No. 4/2012 Remrk 5.8 Similrly s in the cse of Flett s nd Riedel-Shoo s points it is possible to give the stbility results for the so clled nth order Flett s nd Riedel- Shoo s points. These results were proved by Pwlikowsk in her pper [15], which is recommended to interested reder. Also, some results which concern the connection between polynomils nd the set of (nth order) Flett s points re proven in [14]. 6 Concluding remrks In this pper we provided results relted to Flett s men vlue theorem of differentil nd integrl clculus of rel-vlued function of one rel vrible. Indeed, we showed tht for f D, b the ssertion of Flett s theorem holds in ech of the following cses: (i) f () = f (b) (Flett s condition); (ii) (f () b K (f)) (f (b) b K (f)) 0 (Trhn s condition); (iii) A f (, b) = I f (, b) (Tong s condition); (iv) (f () b K (f)) f () > 0 provided f () exists (Mleševi`c s condition). Then we discussed possible generliztion of Flett s theorem to higher-order derivtives nd provided new proof of Pwlikowsk s theorem nd relted results. Up to few questions nd open problems explicitly formulted in this pper, there re severl problems nd directions for the future reserch. The survey of results relted to Flett s men vlue theorem should be continued in [7], becuse we did not mention here ny known nd/or new generliztions nd extensions of Flett s theorem mde t lest in two directions: to move from the rel line to more generl spces (e.g. vector-vlued functions of vector rgument [22], holomorphic functions [2], etc.), nd/or to consider other types of differentibility of considered functions (e.g. Dini s derivtives [19], symmetric derivtives [21], v-derivtives [12], etc.). Also, chrcteriztion of ll the functions tht ttin their Flett s men vlue t prticulr point between the endpoints of the intervl [20], other functionl equtions nd mens relted to Flett s theorem should be mentioned in the future. Acknowledgement This reserch ws prtilly supported by Grnt VEGA 1/0090/13 nd VVGS- PF