Let φ be a continuous and periodic function on ℝ with period 1 and φ(0)=0. We consider the generalized Takagi function ƒφ defined by ƒφ(x)=Σ[n=0,∞]1/2ⁿφ(2ⁿx) and the set Mᵩ of maximum points of ƒᵩ in the interval [0,1]. When φ₀(x) is the function defined by the distance from x to the nearest integer, ƒᵩ₀ is just the Takagi function. Our aim is to seek a condition on φ in order that Mᵩ⊂Mᵩ₀

Fujita Yasuhiro

Saito Yusuke

University of Toyama Repository

Toyama Math. J.Vol. 39(2017), 87-94On the sets of maximum points forgeneralized Takagi functionsYasuhiro Fujita and Yusuke SaitoAbstract. Let ' be a continuous and periodic function on R withperiod 1 and '(0) = 0. We consider the generalized Takagi functionf' dened by f'(x) =1Xn=012n'(2nx) and the set M' of maximumpoints of f' in the interval [0; 1]. When '0(x) is the function denedby the distance from x to the nearest integer, f'0 is just the Takagifunction. Our aim is to seek a condition on ' in order thatM' M'0 .1. Introduction and the resultThe Takagi function is dened byf'0(x) =1Xn=012n'0(2nx); x 2 R; (1)where '0(x) is the function dened by the distance from x to the nearestinteger (cf. [1, 6]). It is a pathological function in the sense that it iseverywhere continuous but nowhere dierentiable on R. Let M'0 be theset of maximum points in the interval [0; 1] for the Takagi function f'0 . ByKahane [4], we haveM'0 =8<:1Xj=1aj4jj aj 2 f1; 2g9=; : (2)2010Mathematics Subject Classication. Primary 26A03, 26A27; Secondary 39B22.Key words and phrases. The Takagi function, Set of maximum points, Concavity.The rst author is supported in part by JSPS KAKENHI # 15K04949.8788 Yasuhiro Fujita and Yusuke SaitoThus, M'0 is uncountable. It is important to note that minM'0 =13 andmaxM'0 =23 .As a generalization of the Takagi function, we consider the function f'of the formf'(x) =1Xn=012n'(2nx); x 2 R; (3)where ' is a continuous and periodic function on R with period 1. We callthe function of the form (3) a generalized Takagi function. The functionsof the form (3) have been considered since the work of [5]. Behrend [3]studied non-dierentiability of the functions of the form (3).In the following, for the function f' of (3), we use the notationsM' = fx 2 [0; 1] j f'(x) = m'g ; m' = maxy2[0;1]f'(y): (4)Note that f' is also periodic with period 1.Our aim is to compare M' with M'0 . In particular, we are interested ina condition on ' in order that M' M'0 .For a function ' : R! R, we consider the following conditions:(C1) ' is continuous and periodic on R with period 1 and '(0) = 0.Furthermore, ' 6 0.(C2) '(x) = '(1  x); x 2 [0; 1].(C3) ' is concave on [0; 1], that is, for all x; y 2 [0; 1] and  2 [0; 1],'(x) + (1  )'(y)  '(x+ (1  )y):(C4) '(x) < '13; x 2h0;13.There exist many functions ' with (C1), (C2), (C3) and (C4).Example 1.1. Let  2 13 ; 12. Dene the periodic function  : R ![0;1) with period 1 by (x) = minf'0(x); g for x 2 R. Note that 1=2 '0. It is easy to see that, for each  213 ;12,  fullls (C1), (C2), (C3)and (C4).On the sets of maximum points for generalized Takagi functions 89Theorem 1.2. Let ' be a function with (C1), (C2), (C3) and (C4).Then, M' M'0.Now, we consider the conditions (C1), (C2), (C3) and (C4). The condi-tion (C1) is natural. The condition (C2) implies that ' is symmetric withrespect to the line x = 12 . The condition (C3) is necessary in Theorem 1.2,since we have an example of a function '^ with (C1), (C2) and (C4) suchthat '^ does not fulll (C3) and M'^ 6 M'0 (see Example 3.1 below). Thecondition (C4) is necessary in Theorem 1.2, since we have an example of afunction ^ with (C1), (C2) and (C3) such that ^ does not fulll (C4) andM^ 6M'0 (see Example 3.2 below).The contents of the present paper are as follows: In Section 2, we proveTheorem 1.2. In Section 3, we provide two examples which show that theconditions (C3) and (C4) for ' are necessary in order that M' M'0 .2. Proof of Theorem 1.2In this section, we prove Theorem 1.2. We use the notations of Section1.Lemma 2.1. Assume that ' fullls (C1) and (C3). Then, ' > 0 in theopen interval (0; 1).Proof. We derive a contradiction by supposing that there exists an x0 2(0; 1) such that '(x0) = 0. Since ' 6 0 by (C1), we nd an x1 2 (0; 1)nfx0gsuch that '(x1) 6= 0. First, we consider the case that 0 < x1 < x0. By(C3), we have'(x1) = 'x1x0x0 +1  x1x0 0 x1x0'(x0) +1  x1x0'(0) = 0:Thus, '(x1) > 0. On the other hand, let  =1 x01 x1 . Then, 0 <  < 1 andx1 + (1  )  1 = x0. Thus, we have0 = '(x0) = '(x1 + (1  )  1)  '(x1) + (1  )'(1) = '(x1) > 0:This is a contradiction. When x0 < x1 < 1, we can derive a contradictionsimilarly. Therefore, ' > 0 in (0; 1).90 Yasuhiro Fujita and Yusuke SaitoIn the following, we assume that ' fullls (C1), (C2) and (C3), otherwisestated. We dene the function p' : R! R of period 1 byp'(x) = '(x) +12'(2x); x 2 R: (5)Note thatf'(x) =1Xn=014np'(4nx); x 2 R: (6)LetE(n)' = fx 2 [0; 1] j p'(4nx) = 'g ; n 2 N [ f0g; (7)where ' = maxy2[0;1]p'(y). By (6) and (7), it is clear to see that1\n=0E(n)' M': (8)It is also clear that if1\n=0E(n)' 6= ;, the inclusion relation in (8) is reducedto the equality.Since the following two lemmas are clear, we omit their proofs.Lemma 2.2. For each m 2 N, there exist integers qm 2 N and rm 2N [ f0g such that13(42m   1) = 5qm; 13(42m 1   1) = 20rm + 1:Lemma 2.3. Assume that ' fullls (C1), (C2) and (C3). Then,f'(x) = f'(1  x); p'(x) = p'(1  x); x 2 [0; 1]:Lemma 2.4. Assume that ' fullls (C1), (C2) and (C3). Then,p'(x)  p'13= p'23; x 2 [0; 1]:Proof. Let 0  x  12 . By (C2) and (C3), we havep'(x) =3223'(1  x) + 13'(2x) 32'23(1  x) + 13(2x)=32'23= '23+12'2  13= '13+12'2  13= p'13:By Lemma 2.3, we conclude the lemma.On the sets of maximum points for generalized Takagi functions 91Proposition 2.5. Assume that ' fullls (C1), (C2) and (C3). Then,13;232M': (9)Proof. By Lemma 2.4, we have 13 2 E(0)' . Let n 2 N. When n is even,it follows from Lemma 2.2 that there exists an integer qn 2 N such that4n  13   5qn = 13 2 E(0)' . Thus, we have13 2 E(n)' . When n is odd, it fol-lows from Lemma 2.2 that there exists an integer rn 2 N [ f0g such that4n  13   (20rn + 1) = 13 2 E(0)' . Thus, we have13 2 E(n)' . Hence,13 2 E(n)'for all n 2 N. Therefore, 13 2M' by (8). By Lemma 2.3, we see that23 2M'.Proposition 2.6. Assume that ' fullls (C1), (C2) and (C3). Then,M' =1\n=0E(n)' . Furthermore, m' = 2' 13, where m' is the constant of(4).Proof. Note that 13 21\n=0E(n)' by the proof of Proposition 2.5. Since1\n=0E(n)' 6= ;, it is easy to see that M' =1\n=0E(n)' by (8). Furthermore, by(6), (8) and Proposition 2.5, we havem' =1Xn=014np'13= 2'13:Here, we used p' 13= 32 ' 13.Proposition 2.7. Assume that ' fullls (C1), (C2), (C3) and (C4).Then, minM' =13 and maxM' =23 .Proof. Let 0  x < 14 . Then, by (C3) and (C4), we havep'(x) =3223'(x) +13'(2x) 32'23x+13(2x)=32'43x<32'13= '13+12'2  13= p'13:92 Yasuhiro Fujita and Yusuke SaitoThus, [0; 14) \ E(0)' = ;. By Proposition 2.6, [0; 14) \M' = ;. For eachx 2 [0; 14) and n 2 N, we havep'4n  nXi=114i+x4n= p'(x) < p'13:Thus,h nXi=114i;n+1Xi=114i\E(n)' = ;. By Proposition 2.6,h nXi=114i;n+1Xi=114i\M'= ;. Since1Xi=114i= 13 , we conclude that [0;13) \M' = ;. Since 13 2M' byProposition 2.5, we have shown that minM' =13 . By Lemma 2.3, we havemaxM' =23 .Lemma 2.8. Assume that ' and  fulll (C1), (C2) and (C3). If E(0)' E(0) , then M' M .Proof. Let x 2 M'. Then, by Proposition 2.6, x 2 E(n)' for all n 2N [ f0g. Thus, p'(4nx) = p' 13for all n 2 N [ f0g, which implies that4nx  [4nx] 2 E(0)' for all n 2 N[f0g, where [y] denotes the greatest integernot exceeding y 2 R. Since E(0)'  E(0) , we have 4nx   [4nx] 2 E(0) forall n 2 N [ f0g. Then, x 2 E(n) for all n 2 N [ f0g. Hence, x 2 M byProposition 2.6.Now, we prove Theorem 1.2.Proof of Theorem 1.2. Let ' be a function with (C1), (C2), (C3) and(C4). By the proof of Proposition 2.7, we have [0; 14) \M' = ;, so thatE(0)' 14 ;34by Lemma 2.3. Since E(0)'0 =14 ;34, we conclude the theoremby Lemma 2.8.3. ExamplesIn this section, we provide two examples which show that the conditions(C3) and (C4) for ' are indispensable in order that M' M'0 .On the sets of maximum points for generalized Takagi functions 93Example 3.1. Let '^ : R ! R be the periodic function with period 1such that '^(x) = ('0(x))2 in R. In this case, '^ fullls (C1), (C2) and (C4).However, '^ does not fulll (C3). We show that M'^ 6M'0 .Indeed, we see that1Xn=012n '0(2nx)2= x  x2; x 2 [0; 1];since  0(x) = x and  n(x) = 2'0(2n 1x) (n 2 N) in the notations of theexample of [8, p.335]. Thus, f'^(x) = x  x2 in [0; 1]. Hence, M'^ =12	.Example 3.2. Dene the periodic function ^ : R ! [0;1) with period1 by ^(x) = min'0(x);115	for x 2 R. In this case, ^ fullls (C1), (C2)and (C3). However, ^ does not fulll (C4). We show that M^ 6M'0 .Indeed, we have, on0; 12,p^(x) =8>>>>>>>>>><>>>>>>>>>>:2x; 0  x  130;x+130;130 x  115;110;115 x  715; x+ 1730;715 x  12:Thus, E(0)^ =115 ;715 [ [ 815 ; 1415 ].We show that 115 2M^. Note that 115 2 E(0)^ . Fix n 2 N arbitrarily.When n is even, it follows from Lemma 2.2 that there exists a numberqn 2 N such that 4n  115 = qn + 115 . Since 115 2 E(0)^ , we have115 2 E(n)^ .When n is odd, it follows from Lemma 2.2 that there exists a number rn 2N [ f0g such that 4n  115 = 4rn + 415 . Since 415 2 E(0)^ , we have115 2 E(n)^ .Therefore, 115 2 E(n)^ for all n 2 N, and 115 2M^. Since minM'0 = 13 , wesee that 115 62M'0 . Thus, we conclude the assertion.References[1] P. Allaart and K. Kawamura, The Takagi function: a survey,Real Anal. Exchange 37 (2011/12), 1{54.94 Yasuhiro Fujita and Yusuke Saito[2] Y. Baba , On maxima of Takagi-van der Waerden functions , Proc.Amer. Math. Soc. 91 (1984), 373{376.[3] F. A. Behrend , Some remarks on the construction of continuousnon-dierentiable functions , Proc. London Math. Soc. 50, (1949), 463{481.[4] J.-P. Kahane, Sur l'exemple, donne par M. de Rham, d'une fonctioncontinue sans derivee, Enseignement Math., 5 (1959), 53{57.[5] K. Knopp, Ein einfaches Verfahren zur Bildung stetiger nirgends dif-ferenzierbarer Funktionen, Math. Z. 2 (1918), 1{26.[6] J. C. Lagarias, The Takagi function and its properties, RIMSKo^kyu^roku Bessatsu, B34 (2012), 153{189.[7] T. Takagi, A simple example of the continuous function withoutderivative, Phys.-Math. Soc. Japan 1 (1903), 176{177. The CollectedPapers of Teiji Takagi, S. Kuroda, Ed., Iwanami (1973), 5{6.[8] M. Yamaguti and M. Hata, Weierstrass's function and chaos,Hokkaido Math. J. 12 (1983), 333{342.Yasuhiro FujitaDepartment of MathematicsFaculty of ScienceUniversity of ToyamaGofuku, Toyama 930-8555, JAPANe-mail: yfujita@sci.u-toyama.ac.jpYusuke SaitoGraduate School of Science and Engineering for EducationUniversity of ToyamaGofuku, Toyama 930-8555, JAPANe-mail: d1771001@ems.u-toyama.ac.jp(Received July 12, 2017)

On the sets of maximum points for generalized Takagi functions

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On the sets of maximum points for generalized Takagi functions

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