continuous but nowhere differentiable

MASTER’S THESIS2003:320 CIV Continuous Nowhere Differentiable Functions MASTER OF SCIENCE PROGRAMME Department of Mathematics 2003:320 CIV • ISSN: 1402 - 1617 • ISRN: LTU - EX - - 03/320 - - SEJOHAN THIM Continuous Nowhere Differentiable Functions Johan Thim December 2003 Master Thesis Supervisor: Lech Maligranda Department of Mathematics Abstract In the early nineteenth century, most mathematicians believed that a contin- uous function has derivative at a significant set of points. A. M. Amp` ere even tried to give a theoretical justification for this (within the limitations of the definitions of his time) in his paper from 1806. In a presentation before the Berlin Academy on July 18, 1872 Karl Weierstrass shocked the mathematical community by proving this conjecture to be false. He presented a function which was continuous everywhere but differentiable nowhere. The function in question was defined by W(x) =∞/summationdisplay k=0akcos(bkπx), whereais a real number with 0 <a< 1,bis an odd integer and ab> 1+3π/2. This example was first published by du Bois-Reymond in 1875. Weierstrass also mentioned Riemann, who apparently had used a similar construction (which was unpublished) in his own lectures as early as 1861. However, neither Weierstrass’ nor Riemann’s function was the first such construction. The earliest known example is due to Czech mathematician Bernard Bolzano, who in the years around 1830 (published in 1922 after being discovered a few years earlier) exhibited a continuous function which was nowhere differen- tiable. Around 1860, the Swiss mathematician Charles Cell´ erier also discov- ered (independently) an example which unfortunately wasn’t published until 1890 (posthumously). After the publication of the Weierstrass function, many other mathemati- cians made their own contributions. We take a closer look at many of these functions by giving a short historical perspective and proving some of their properties. We also consider the set of all continuous nowhere differentiable functions seen as a subset of the space of all real-valued continuous functions. Surprisingly enough, this set is even “large” (of the second category in the sense of Baire). Acknowledgement I would like to thank my supervisor Lech Maligranda for his guidance, help and support during the creation of this document. His input was invaluable and truly appreciated. Also the people I have had contact with (during all of my education) at the Department of Mathematics here in Lule˚ a deserves a heartfelt thank you. On another note, I would like to extend my gratitude to Dissection, Chris Poland and Spawn of Possession for having provided some quality music that made the long nights of work less grating. Thanks to Jan Lindblom for helping me with some French texts as well. Contents 1 Introduction 4 2 Series and Convergence 7 3 Functions Through the Ages 11 3.1 Bolzano function ( ≈1830) . . . . . . . . . . . . . . . . . . . . 11 3.2 Cell´ erier function ( ≈1860) . . . . . . . . . . . . . . . . . . . . 17 3.3 Riemann function ( ≈1861) . . . . . . . . . . . . . . . . . . . . 18 3.4 Weierstrass function (1872) . . . . . . . . . . . . . . . . . . . 20 3.5 Darboux function (1873) . . . . . . . . . . . . . . . . . . . . . 28 3.6 Peano function (1890) . . . . . . . . . . . . . . . . . . . . . . 32 3.7 Takagi (1903) and van der Waerden (1930) functions . . . . . 36 3.8 Koch “snowflake” curve (1904) . . . . . . . . . . . . . . . . . . 39 3.9 Faber functions (1907, 1908) . . . . . . . . . . . . . . . . . . . 41 3.10 Sierpi´ nski curve (1912) . . . . . . . . . . . . . . . . . . . . . . 44 3.11 Knopp function (1918) . . . . . . . . . . . . . . . . . . . . . . 45 3.12 Petr function (1920) . . . . . . . . . . . . . . . . . . . . . . . 47 3.13 Schoenberg function (1938) . . . . . . . . . . . . . . . . . . . 48 3.14 Orlicz functions (1947) . . . . . . . . . . . . . . . . . . . . . . 52 3.15 McCarthy function (1953) . . . . . . . . . . . . . . . . . . . . 55 3.16 Katsuura function (1991) . . . . . . . . . . . . . . . . . . . . . 57 3.17 Lynch function (1992) . . . . . . . . . . . . . . . . . . . . . . 62 3.18 Wen function (2002) . . . . . . . . . . . . . . . . . . . . . . . 64 4 How “Large” is the Set ND[a,b] 71 4.1 Metric spaces and category . . . . . . . . . . . . . . . . . . . . 71 4.2 Banach-Mazurkiewicz theorem . . . . . . . . . . . . . . . . . . 74 4.3 Prevalence of ND[0,1] . . . . . . . . . . . . . . . . . . . . . . 78 Bibliography 85 Index 92 Index of Names . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 92 Index of Subjects . . . . . . . . . . . . . . . . . . . . . . . . . . . . 93 2 List of Figures 3.1 The three first elements in the “Bolzano” sequence {Bk(x)} with [a,b] = [0,20] and [A,B] = [4,16]. . . . . . . . . . . . . . 13 3.2 Cellerier’s function C(x) witha= 2 on [0,π]. . . . . . . . . . . 18 3.3 Riemann’s function Ron [−1,5]. . . . . . . . . . . . . . . . . 20 3.4 Weierstrass’ function Wwitha=1 2andb= 5 on [0,3]. . . . . 22 3.5 Darboux’s function D(x) on [0,3]. . . . . . . . . . . . . . . . . 29 3.6 First four steps in the geometric generation of Peano’s curve. . 33 3.7 The component φpof Peano’s curve. . . . . . . . . . . . . . . . 34 3.8 Takagi’s and van der Waerden’s functions on [0 ,1]. . . . . . . 36 3.9 First four steps in the construction of Koch’s “snowflake”. . . 40 3.10 The functions f1(dashed) and f2(whole). . . . . . . . . . . . 42 3.11 Faber’s functions Fi(x) on [0,1]. . . . . . . . . . . . . . . . . . 43 3.12 Polygonal approximations (of order n) to Sierpi´ nski’s curve. . 45 3.13 The “saw-tooth” function φ(x) on [−3,3]. . . . . . . . . . . . 46 3.14 Petr’s function in a 4-adic system. . . . . . . . . . . . . . . . . 48 3.15 First four approximation polygons in the construction of Scho- enberg’s curve (sampled at tk=m/3n,m= 0,1,..., 3n). . . . 49 3.16 Schoenberg’s function φsand the auxiliary function p. . . . . . 49 3.17 McCarthy’s function Mand the auxiliary function g(x). . . . 56 3.18 The graphs of the first four “iterations” of the Katsuura func- tion and the corresponding mappings of X(the rectangles). . . 58 3.19 Line segment with band neighborhoods for Lynch’s function. . 63 3.20 Wen’s function WLwithan= 2−nandpn= 6nforx∈[0,2]. . 65 3.21 Two of Liu Wen’s functions with 0 ≤x≤1. . . . . . . . . . . 67 3 Chapter 1 Introduction I turn away with fear and horror from the lamentable plague of continuous functions which do not have derivatives... – Hermite, letter to Stieltjes dated 20 May 18931. Judging by the quote above, some mathematicians didn’t like the possibility of continuous functions which are nowhere differentiable. Why was these functions so poorly received? Observing the situation today, many students still find it strange that there exists a continuous function which is nowhere differentiable. When I first heard of it myself I was a bit perplexed, at least by the sheer magnitude of the number of such functions that actually exist. Usually beginning students of mathematics get the impression that continuous functions normally are differentiable, except maybe at a few especially “nasty” points. The standard example of f(x) =|x|, which only lacks derivative at x= 0, is one such function. This was also the situation for most mathematicians in the late 18th and early 19th century. They were not interested in the existence of the derivative of some hypothetical function but rather just calculating the derivative as some explicit expression. This was usually successful, except at a few points in the domain where the differentiation failed. These actions led to the belief that continuous functions have derivatives everywhere, except at some particular points. Amp` ere even tried to give a theoretical justification for this statement in 1806 (cf. Amp` ere [1]), although it is not exactly clear 1Quote borrowed from Pinkus [57]. 4 if he attempted to prove this for all continuous functions or for some smaller subset (for further discussion see Medvedev [48], pages 214-219). Therefore, with all this in mind, the reaction of a 19th century mathematician to the news of these functions doesn’t seem that strange anymore. These functions caused a reluctant reconsideration of the concept of a continuous function and motivated increased rigor in mathematical analysis. Nowadays the existence of these functions is fundamental for “new” areas of research and applications like, for example, fractals, chaos and wavelets. In this report we present a chronological review of some of the continu- ous nowhere differentiable functions constructed during the last 170 years. Properties of these functions are discussed as well as traits of more general collections of nowhere differentiable functions. The contents of the thesis is as follows. We start in Chapter 2 with sequences and series of functions defined on some interval I⊂Rand convergence of those. This is important for the further development of the subject since many constructions are based on infinite series. In Chapter 3 we take a stroll through the last couple of centuries and present some of the functions constructed. We do this in a concise manner, starting with a short historical background before giving the construction of the function and showing that it has the desired properties. Some proofs has been left out for various reasons, but in those cases a clear reference to a proof is given instead. Chapter 4 continues with an examination of the set of all continuous nowhere differentiable functions. It turns out that the “average” continuous function normally is nowhere differentiable and not the other way around. We do this both by a topological argument based on category and also by a measure theoretic result using prevalence (considered by Hunt, Sauer and York). Table 1.1 gives a short timeline for development in the field of continuous nowhere differentiable functions. 5 Discoverer Year Page What B. Bolzano ≈1830 11 First known example M. Ch. Cell´ erier ≈1830 17 Early example B. Riemann ≈1861 18 “Nondifferentiable” function K. Weierstrass 1872 20 First published example H. Hankel 1870 29 “Condensation of singularities” H. A. Schwarz 1873 28 Not differentiable on a dense subset M. G. Darboux 1873-5 28 Example (’73) and generalization (’75) U. Dini 1877 25 Large class including Weierstrass K. Hertz 1879 27 Generalization of Weierstrass function G. Peano 1890 32 Space-filling curve (nowhere differentiable) D. Hilbert 1891 33 Space-filling curve (nowhere differentiable) T. Takagi 1903 36 Easier (than Weierstrass) example H. von Koch 1904 39 Continuous curve with tangent nowhere G. Faber 1907-8 41 “Investigation of continuous functions” W. Sierpi´ nski 1912 44 Space-filling curve (nowhere differentiable) G. H. Hardy 1916 27 Generalization of Weierstrass conditions K. Knopp 1918 45 Generalization of Takagi-type functions M. B. Porter 1919 27 Generalization of Weierstrass function K. Petr 1922 47 Algebraic/arithmetic example A. S. Besicovitch 1924 78 No finite orinfinite one-sided derivative B. van der Waerden 1930 36 Takagi-like construction S. Mazurkiewicz 1931 74 ND[0,1] is of the second category S. Banach 1931 74 ND[0,1] is of the second category S. Saks 1932 78 The set of Besicovitch-functions is Ist category I. J. Schoenberg 1938 48 Space-filling curve (nowhere differentiable) W. Orlicz 1947 52 Intermediate result J. McCarthy 1953 55 Example with very simple proof G. de Rham 1957 36 Takagi generalization H. Katsuura 1991 57 Example based on metric-spaces M. Lynch 1992 62 Example based on topology B. R. Hunt 1994 78 ND[0,1] is a prevalent set L. Wen 2002 64 Example based on infinite products Table 1.1: Timelime (partial) of the development in the field of continuous nowhere differentiable functions. 6 Chapter 2 Series and Convergence Many constructions of nowhere differentiable continuous functions are based on infinite series of functions. Therefore a few general theorems about series and sequences of functions will be of great aid when we continue investigating the subject at hand. First we need a clear definition of convergence in this context. Definition 2.1. A sequence Snof functions on the interval Iis said to converge pointwise to a function SonIif for every x∈I lim n→∞Sn(x) =S(x), that is ∀x∈I∀/epsilon1>0∃N∈N∀n≥N|Sn(x)−S(x)|</epsilon1. The convergence is said to be uniform on I if lim n→∞sup x∈I|Sn(x)−S(x)|= 0, that is ∀/epsilon1>0∃N∈N∀n≥Nsup x∈I|Sn(x)−S(x)|</epsilon1. Uniform convergence plays an important role to whether properties of the elements in a sequence are transfered onto the limit of the sequence. The fol- lowing two theorems can be of assistance when establishing if the convergence of a sequence of functions is uniform. 7 Theorem 2.1. The sequence Snconverges uniformly on Iif and only if it is a uniformly Cauchy sequence on I, that is lim m,n→∞sup x∈I|Sn(x)−Sm(x)|= 0 or ∀/epsilon1>0∃N∈N∀m,n≥Nsup x∈I|Sn(x)−Sm(x)|</epsilon1. Proof. First, assume that Snconverges uniformly to SonI, that is ∀/epsilon1>0∃N∈N∀n≥Nsup x∈I|Sn(x)−S(x)|</epsilon1 2. For such/epsilon1>0 and form,n∈Nwithm,n≥Nwe have sup x∈I|Sn(x)−Sm(x)| ≤sup x∈I(|Sn(x)−S(x)|+|S(x)−Sm(x)|) ≤sup x∈I|Sn(x)−S(x)|+ sup x∈I|S(x)−Sm(x)|<2/epsilon1 2=/epsilon1. Conversely, assume that {Sn}is a uniformly Cauchy sequence, i.e. ∀/epsilon1>0∃N∈N∀m,n≥Nsup x∈I|Sn(x)−Sm(x)|</epsilon1 2. For any fixed x∈I, the sequence {Sn(x)}is clearly a Cauchy sequence of real numbers. Hence the sequence converges to a real number, say S(x). From the assumption and the pointwise convergence just established we have ∀/epsilon1>0∃N∈N∀m,n≥Nsup x∈I|Sn(x)−Sm(x)|</epsilon1 2 and ∀/epsilon1>0∀x∈I∃mx>N|Smx(x)−S(x)|</epsilon1 2. If/epsilon1>0 is arbitrary and n>N , then sup x∈I|Sn(x)−S(x)| ≤sup x∈I(|Sn(x)−Smx(x)|+|Smx(x)−S(x)|)</epsilon1 2+/epsilon1 2=/epsilon1. Hence the convergence of SntoSis uniform on I. 8 Theorem 2.2 (Weierstrass M-test). Letfk:I→Rbe a sequence of functions such that supx∈I|fk(x)| ≤Mkfor everyk∈N. If/summationtext∞ k=1Mk<∞, then the series/summationtext∞ k=1fk(x)is uniformly convergent on I. Proof. Letm,n∈Nwithn>m . Then sup x∈I|Sn(x)−Sm(x)|= sup x∈I/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsinglen/summationdisplay k=1fk(x)−m/summationdisplay k=1fk(x)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle = sup x∈I/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsinglen/summationdisplay k=m+1fk(x)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle≤n/summationdisplay k=m+1sup x∈I|fk(x)| ≤n/summationdisplay k=m+1Mk=n/summationdisplay k=1Mk−m/summationdisplay k=1Mk. SinceM=/summationtext∞ k=1Mk<∞it follows that n/summationdisplay k=1Mk−m/summationdisplay k=1Mk→M−M= 0 asm,n→ ∞ which gives that {Sn}is a uniformly Cauchy sequence on I. Using Theo- rem 2.1 we obtain that the series/summationtext∞ k=1fk(x) is uniformly convergent on I. We are often interested in establishing the continuity of a limit of a sequence of continuous functions. To accomplish this, the following theorem and its corollary can be helpful. Theorem 2.3. If{Sn}is a sequence of continuous functions on IandSn converges uniformly to SonI, thenSis a continuous function on I. Proof. Letx0∈Ibe arbitrary. By assumption we have ∀/epsilon1>0∃N∈N∀n≥Nsup x∈I|Sn(x)−S(x)|</epsilon1 3 and ∀/epsilon1>0∃δ>0 such that |x−x0|<δ⇒ |Sn(x)−Sn(x0)|</epsilon1 3. 9 Let/epsilon1>0 be given, x∈I,n∈Nwithn>N and|x−x0|<δ. Then |S(x)−S(x0)| ≤ |S(x)−Sn(x)|+|Sn(x)−Sn(x0)|+|Sn(x0)−S(x0)|<3/epsilon1 3=/epsilon1 and therefore Sis continuous at x0. Sincex0∈Iwas arbitrary, Sis contin- uous onI. Corollary 2.4. Iffk:I→Ris a continuous function for every k∈N and/summationtext∞ k=1fk(x)converges uniformly to S(x)on I, then Sis a continuous function on I. 10 Chapter 3 Functions Through the Ages 3.1 Bolzano function ( ≈1830; published in 1922) Probably the first example of a continuous nowhere differentiable function on an interval is due to Czech mathematician Bernard Bolzano. The his- tory behind this example is filled with unfortunate circumstances. Due to these circumstances, Bolzano’s manuscript with the name “Functionenlehre” , which was written around 1830 and contained the function, wasn’t published until a century later in 1930. The publication came to since in 1920, after the first World War, another Czech mathematician Martin Jaˇ sek discov- ered a manuscript in the National Library of Vienna belonging to Bernard Bolzano (a photocopy is also in the archives of the Czech Academy of Sci- ences). It was named “Functionenlehre” and it was dated 1830. Originally it was supposed to be a part of Bolzano’s more extensive work “Gr¨ ossenlehre” . The manuscript “Functionenlehre” was published in Prague in 1930 (in the “Schriften I” ), having 183 pages and containing an introduction and two parts. Bolzano proved in it that the set of points where the function is non- differentiable is dense in the interval where it is defined. The continuity was also deduced, however not completely correct. The full story on “Functio- nenlehre” can be found in Hyksˇ ov´ a [33] who also has written the following: The first lecture of M. Jaˇ sek reporting on Functionenlehre was given on December 3, 1921. Already on February 3, 1922 Karel Rychl´ ık presented to K ˇCSN [Royal Czech Science Soci- ety] his treatise [61] where the correct proof of the continuity of 11 Bolzano’s function was given as well as the proof of the assertion that this function does not have a derivative at any point of the interval (a,b) (finite or infinite). The same assertion was proved by Vojtˇ ech Jarn´ ık (1897 - 1970) at the same time but in a differ- ent way in his paper [34]. Both Jarn´ ık and Rychl´ ık knew about the work of the other. Giving reference to Rychl´ ık’s paper, Jarn´ ık did not prove the continuity of Bolzano’s function; on the other hand, Rychl´ ık cited the work of Jarn´ ık (an idea of another way to the same partial result). Unlike many other constructions of nowhere differentiable functions, Bolz- ano’s function is based on a geometrical construction instead of a series ap- proach. The Bolzano function, B, is constructed as the limit of a sequence {Bk}of continuous functions. We can choose the domain of B1(which will be the domain of Bas well) and the range of B1. Let the interval [ a,b] be the desired domain and [ A,B] the desired range. Each piecewise linear and continuous function in the sequence is defined as follows. (i)B1(x) =A+B−A b−a(x−a); (ii)B2(x) is defined on the intervals I1=/bracketleftbigg a,a+3 8(b−a)/bracketrightbigg ,I2=/bracketleftbigg a+3 8(b−a),1 2(a+b)/bracketrightbigg , I3=/bracketleftbigg1 2(a+b),a+7 8(b−a)/bracketrightbigg ,I4=/bracketleftbigg a+7 8(b−a),b/bracketrightbigg as the piecewise linear function having the values B2(a) =A,B2/parenleftbigg a+3 8(b−a)/parenrightbigg =A+5 8(B−A), B2/parenleftbigg1 2(a+b)/parenrightbigg =A+1 2(B−A), B2/parenleftbigg a+7 8(b−a)/parenrightbigg =B+1 8(B−A),B2(b) =B at the endpoints; 12 (iii)B3(x) is constructed by the same procedure as in (ii) on each of the four subintervals Ii(with the corresponding values for a,b,AandB). This continues for k= 4,5,6,...and the limit of Bk(x) ask→ ∞ is the Bolzano function B(x). Bk(x) x1020 10 20 (a)B1andB2.Bk(x) x1020 10 20 (b)B1(dotted),B2(dashed) and B3(whole). Figure 3.1: The three first elements in the “Bolzano” sequence {Bk(x)}with [a,b] = [0,20] and [A,B] = [4,16]. A fitting closing remark, before the proof of continuity and nowhere differ- entiability, can be found in Hyksˇ ov´ a [33]: “Already the fact that it occurred to Bolzano at all that such a function might exist, deserves our respect. The fact that he actu- ally succeeded in its construction, is even more admirable”. Theorem 3.1. The Bolzano function Bis continuous and nowhere differ- entiable on the interval [a,b]. Proof. First we want to show that the function Bis continuous. For fixed k∈Nconsider the function Bk. Let us find the slopes Mk={Mk,m}of each of the linear functions on the subintervals. Not to have too many indices we will just write Mkinstead ofMk,m. Fork= 1 it is immediate from the definition that M1=B−A b−afor all of [a,b]. Letk≥2. For each linear part [ak,bk] ofBkwe have the following 13 1. ForI= [t1,t2] =/bracketleftbig ak,ak+3 8(bk−ak)/bracketrightbig , M(1) k+1=Bk(t2)−Bk(t1) t2−t1=5 8(Bk−Ak) 3 8(bk−ak)=5 3Bk−Ak bk−ak=5 3Mk; 2. forI= [t2,t3] =/bracketleftbig ak+3 8(bk−ak),1 2(ak+bk)/bracketrightbig , M(2) k+1=Bk(t3)−Bk(t2) t3−t2=/parenleftbig1 2−5 8/parenrightbig (Bk−Ak)/parenleftbig1 2−3 8/parenrightbig (bk−ak)=−1 8 1 8Bk−Ak bk−ak=−Mk; 3. forI= [t3,t4] =/bracketleftbig1 2(ak+bk),ak+7 8(bk−ak)/bracketrightbig , M(3) k+1=Bk(t4)−Bk(t3) t4−t3=/parenleftbig 1+1 8−1 2/parenrightbig (Bk−Ak)/parenleftbig7 8−1 2/parenrightbig (bk−ak)=5 8 3 8Bk−Ak bk−ak=5 3Mk; 4. forI= [t4,t5] =/bracketleftbig ak+7 8(bk−ak),bk/bracketrightbig , M(4) k+1=Bk(t5)−Bk(t4) t5−t4=−1 8(Bk−Ak)/parenleftbig 1−7 8/parenrightbig (bk−ak)=−1 8 1 8Bk−Ak bk−ak=−Mk. Let{In,k}={[In(sk),In(tk)]}be the collection of subintervals of [ a,b] where Bnis linear and define Ln= sup I∈{In+1,k}(I(tk)−I(sk)) andMn= sup I∈{In+1,k} i=1,2,3,4|M(i) n(I)|. That is,Lnis the maximal length of an interval where Bn+1is linear and Mn is the maximum slope (to the absolute value) of Bn+1. Clearly Ln≤/parenleftbigg3 8/parenrightbiggn+1 |b−a|andMn≤/parenleftbigg5 3/parenrightbiggn+1/vextendsingle/vextendsingle/vextendsingle/vextendsingleB−A b−a/vextendsingle/vextendsingle/vextendsingle/vextendsingle which gives that the maximum increase/decrease of the function from step nton+ 1 is bounded by MnLn≤/parenleftbig5 8/parenrightbign+1|B−A|. Hence, for k∈N, sup x∈[a,b]|Bk+1(x)−Bk(x)| ≤/parenleftbigg5 8/parenrightbiggk+1 |B−A|. 14 Letm,n∈Nwithm>n . We have sup x∈[a,b]|Bm(x)−Bn(x)| ≤sup x∈[a,b]/parenleftBiggm/summationdisplay k=n+1|Bk(x)−Bk−1(x)|/parenrightBigg ≤m/summationdisplay k=n+1sup x∈[a,b]|Bk(x)−Bk−1(x)| ≤m/summationdisplay k=n+1/parenleftbigg5 8/parenrightbiggk |B−A| =|B−A|/parenleftBiggm/summationdisplay k=1/parenleftbigg5 8/parenrightbiggk −n/summationdisplay k=1/parenleftbigg5 8/parenrightbiggk/parenrightBigg → |B−A|/parenleftbigg5 3−5 3/parenrightbigg = 0 asm,n→ ∞ . Thus {Bk}is a uniformly Cauchy sequence on the interval [ a,b] and since eachBkis continuous it follows from Theorems 2.1 and 2.3 that Bolzano’s function is continuous on [ a,b]. Secondly, we show that Bis not differentiable at any x∈[a,b]. Again, let {In,k}={[In(sk),In(tk)]}be the collection of subintervals of [ a,b] whereBn is linear and define Mas the set of all endpoints in {In,k}, i.e. M={s,t|[s,t]∈ {In,k}}. We show that Mis dense in [ a,b]. That is, for any x0∈[a,b],∃xn∈Msuch thatxn→x0. Letx0∈[a,b] be arbitrary but fixed. If x0=bwe are done sinceb∈M. Assume that x0/negationslash=b, we proceed as follows. (i) Step 1: let L=b−aand define J(0) 0=/bracketleftbigg a,a+3 8L/parenrightbigg , J(1) 0=/bracketleftbigg a+3 8L,a+1 2L/parenrightbigg , J(2) 0=/bracketleftbigg a+1 2L,a+7 8L/parenrightbigg andJ(3) 0=/bracketleftbigg a+7 8L,b/parenrightbigg . Clearly there exists i0∈ {0,1,2,3}such thatx0∈J(i0) 0. We take J0=J(i0) 0. 15 (ii) Stepn: we havex0∈In−1= [an,bn]. LetLn=bn−anand define J(0) n=/bracketleftbigg an,an+3 8Ln/parenrightbigg , J(1) n=/bracketleftbigg an+3 8Ln,an+1 2Ln/parenrightbigg , J(2) n=/bracketleftbigg an+1 2Ln,an+7 8Ln/parenrightbigg andJ(3) n=/bracketleftbigg an+7 8Ln,bn/parenrightbigg . As before, there exists in∈ {0,1,2,3}such thatx0∈J(in) n. We take Jn=J(in) n. HenceMis dense in [ a,b] since |x0−an+1| ≤/parenleftbigg3 8/parenrightbiggn+1 |b−a| →0 asn→ ∞1. Now we show that Bis non-differentiable for every x0∈M. Letx0∈Mbe arbitrary but fixed, we consider two cases that exhaust all possibilities. Forx0=a: Letxn=a+/parenleftbig3 8/parenrightbign|b−a|. Thenxn→aasn→ ∞ and xn∈Mfor everyn∈N. By the construction of the function Bit is clear thatB(xn) =Bn+1(xn) for every n∈N. Also,B(a) =AandBn+1(xn) = A+/parenleftbig5 3/parenrightbign/parenleftbig3 8/parenrightbign|b−a|. Hence B(xn)−B(a) xn−a=A+/parenleftbig5 3/parenrightbign/parenleftbig3 8/parenrightbign|b−a| −A/parenleftbig3 8/parenrightbign|b−a|=/parenleftbigg5 3/parenrightbiggn → ∞ asn→ ∞ and therefore B/prime(x0) does not exist. Forx0∈M\ {a}: letxn=x0−/parenleftbig1 8/parenrightbign+q|b−a|,q∈N. Sincex0∈M, there exists r∈Nsuch thatB(x0) =Bp(x0) for allp≥r. We can choose q > r so thatxn∈(a,b] for every n∈N. From the construction of Bwe see thatB(xn) =Bn+1(xn) =Bn(x0) + (−1)nK/parenleftbig1 8/parenrightbign+qwhereK∈Rwith K≥ |b−a|/|B−A| /negationslash= 0. Moreover, since q >r ,B(x0) =Bn(x0) for every n∈N. This implies that (B(x0)−Bn(x0)) 8n+q= (Bn(x0)−Bn(x0)) 8n+q= 0. So forn∈N, B(x0)−B(xn) x0−xn= 8n+q/parenleftBigg B(x0)−Bn(x0)−(−1)nK/parenleftbigg1 8/parenrightbiggn+q/parenrightBigg = (B(x0)−Bn(x0)) 8n+q−(−1)nK= (−1)n+1K. 1And also |x0−bn+1| →0 asn→ ∞ . 16 But (−1)n+1Kdoes not converge as n→ ∞ and thusB/prime(x0) does not exist. In no way is it clear from this that Bis nowhere differentiable, only that it is non-differentiable on a dense subset of [ a,b] (which theoretically means that it might still be possible that Bis differentiable almost everywhere). We will not complete the proof here but merely give a reference: the complete proof can be found in Jarn´ ık [34]. 3.2 Cell´ erier function ( ≈1860; published in 1890) Charles Cell´ erier had proposed the function Cdefined as C(x) =∞/summationdisplay k=11 aksin(akx),a>1000 earlier than 1860 but the function wasn’t published until 1890 (posthu- mously) in Cell´ erier [10]. When the manuscripts were opened after his death they were found to be containing sensational material. In an undated folder (according to the historians it is from around 1860) with heading “Very important and I think new. Correct. Can be published as it is written.” there was a proof of the fact that the function Cis continuous and nowhere differentiable if ais a sufficiently large even number. The publication of Cell´ eriers example in 1890 came as only a curiosity since it was already generally known from Weierstrass (see Section 3.4). Cell´ erier’s function is strikingly similar to Weierstrass’ function and its nowhere differentiability follows from Hardy’s generalization of that function (see the remark to The- orem 3.4). In Cell´ erier’s paper (which, roughly translated, has the title “Notes on the fundamental principles of analysis”) there is a section called “Example de fonctions faisant exception aux r` egles usuelles” – “Example of functions mak- ing departures from the usual rules”. In this section Cell´ erier proposed the functionCdefined above and states that this function will provide an ex- ample of a function that is continuous, differentiable nowhere and never has any periods of growth or decay. 17 Cell´ erier’s original condition on awasa >1000 where ais an even integer (for nowhere differentiability) or a>1000 where ais an odd integer (for no periods of growth or decay). According to Hardy [27], for the case of nowhere differentiability, the condition can be weakened to a>1 (not necessarily an integer). C(x) x0.5 −0.52.0 Figure 3.2: Cellerier’s function C(x) witha= 2 on [0,π]. Theorem 3.2. The Cell´ erier function C(x) =∞/summationdisplay k=11 aksin(akx),a>1 is continuous and nowhere differentiable on R. Proof. The continuity of Cfollows exactly like in the proof for Weierstrass’ function (Theorem 3.4). That Cis nowhere differentiable follows from Hardy [27] (see the remark to Theorem 3.4) since a·a−1≥1 and that if g(x) is nowhere differentiable than so is g(x/π). 3.3 Riemann function ( ≈1861) In a thesis from 1854 ( Habilitationsschrift ), Riemann [59] attempted to find necessary and sufficient conditions for representation of a function by Fourier series. In this paper he also generalized the definite integral and gave an example of a function that between any two points is discontinuous infinitely 18 often but still is integrable (with respect to the Riemann-integral). The function he defined was f(x) =∞/summationdisplay k=1(nx) n2, where ( x) =/braceleftBigg 0, if x=p 2,p∈Z x−[x], elsewhere, and [x] is the integer part of x. This function is interesting in this context for another reason. Consider, for x∈[a,b], the function F: [a,b]→Rdefined by the indefinite integral of f, F(x) =/integraldisplayx af(τ)dτ. It can quite easily be seen that this function is continuous and it is also clear that it is not differentiable on a dense subset of [ a,b]. This, however, is not the function we will be concerned with here. What we will refer to as Riemann’s function in this framework is the function R:R→Rdefined by R(x) =∞/summationdisplay k=11 k2sin(k2x). Interesting to note is that there seems to be no other known sources for the claim that this was Riemann’s construction than those that can be traced back to Weierstrass (cf. Butzer and Stark [7], Ullrich [74] and Section 3.4). Riemann’s function isn’t actually a nowhere differentiable function. It has been shown that Rpossess a finite derivative ( R/prime(x0) =−1 2) at points of the form x0=π2p+ 1 2q+ 1,p,q∈Z. These points however, are the only points where Rhas a finite derivative (cf. Gerver [24], [25] and Hardy [27] or for a more concise proof based on number theory see Smith [71]). According to Weierstrass, Riemann used this function as an example of a “nondifferentiable” function in his lectures as early as 1861. It is unclear whether he meant that the function was nowhere differentiable or something else. Riemann claimed to have a proof, obtained from the theory of elliptic functions, but it was never presented nor was it found anywhere in his notes after his death (cf. Neuenschwander [49] and Segal [68]). 19 R(x) x1.0 −1.02.0 4 .0 Figure 3.3: Riemann’s function Ron [−1,5]. Theorem 3.3. The Riemann function R(x) =∞/summationdisplay k=11 k2sin(k2x) is continuous on all of Rand only has a derivative at points of the form x0=π2p+ 1 2q+ 1,p,q∈Z. Proof. We start with showing that the function Ris continuous. Since/summationtext∞ k=11 k2=π2 6<∞and supx∈R|1 k2sin(k2x)|=1 k2, the Weierstrass M-test (Theorem 2.2) proves that the convergence is uniform and the Corollary 2.4 gives the continuity of RonR. Secondly, the only points where Rhas a finite derivative (cf. Gerver [24], [25] and Hardy [27] or Smith [71]) is points of the form x0=π2p+ 1 2q+ 1,p,q∈Z. 3.4 Weierstrass function (1872; published in 1875 by du Bois-Reymond) On July 18, 1872 Karl Weierstrass presented in a lecture at the Royal Aca- demy of Science in Berlin an example of a continuous nowhere differentiable 20 function, W(x) =∞/summationdisplay k=0akcos(bkπx), for 0< a < 1,ab > 1 + 3π/2 andb >1 an odd integer. On the lecture Weierstrass said As I know from some pupils of Riemann, he as the first one (around 1861 or earlier) suggested as a counterexample to Am- p` ere’s Theorem [which perhaps could be interpreted2as: every continuous function is differentiable except at a few isolated poi- nts]; for example, the function Rdoes not satisfy this theorem. Unfortunately, Riemann’s proof was unpublished and, as I think, it is neither in his notes nor in oral transfers. In my opinion Riemann considered continuous functions without derivatives at any point, the proof of this fact seems to be difficult... Weierstrass’ function was the first continuous nowhere differentiable function to be published, which happened in 1875 by Paul du Bois-Reymond [19]. At this time, du Bois-Reymond was a professor at Heidelberg University in Germany and in 1873 he sent a paper to Borchardt’s Journal [“Journal f¨ ur die reine und angewandte Mathematik”]. This paper dealt with the function Weierstrass had discussed earlier (among several other topics). Borchardt gave the paper to Weierstrass to read through. Weierstrass wrote in a letter to du Bois-Reymond (dated 23 of November, 1873; cf. Weierstrass [77]) that he had made no new progress, except for some remarks about Riemann’s function. In the letter, du Bois-Reymond had Weierstrass’ function presented in the form f(x) =∞/summationdisplay k=0sin(anx) bn,a b>1, which apparently was changed before the paper was published. Du Bois- Reymond accepted Weierstrass’ remarks and put them in his paper together with some more historical notes about the subject and in 1875 the paper was published in Borchardt’s Journal. Since this was the first published continuous nowhere differentiable function it has been regarded by many as the first such function exhibited. This regardless of the fact that Weierstrass’ function was not the earliest such 2See Medvedev [48], pages 214-219. 21 construction. Several others3had done it earlier, although non of those are believed to have been published before the publication of the Weierstrass function. W(x) x1.0 −1.01.0 2.0 Figure 3.4: Weierstrass’ function Wwitha=1 2andb= 5 on [0,3]. In 1916, Hardy [27] proved that the function Wdefined above is continuous and nowhere differentiable if 0 <a< 1,ab≥1 andb>1 (not necessarily an odd integer). Theorem 3.4. The Weierstrass function, W(x) =∞/summationdisplay k=0akcos(bkπx), for0<a< 1,ab≥1andb>1, is continuous and nowhere differentiable on R. Proof. Starting with establishing the continuity, observe that 0 < a < 1 implies/summationtext∞ k=0ak=1 1−a<∞. This together with supx∈R|ancos(bnπx)| ≤an gives, using the Weierstrass M-test (Theorem 2.2), that/summationtext∞ k=0ancos(bnπx) converges uniformly to W(x) onR. The continuity of Wnow follows from the uniform convergence of the series just established and from the Corollary 2.4. During the rest of this proof we assume that Weierstrass original assumptions hold, i.e.ab > 1 +3 2πandb >1 an odd integer. For a general proof with ab≥1 andb > 1 we refer to Hardy [27]. The rest of the proof follows, 3For example, Cell´ erier’s and Bolzano’s functions both described in earlier sections were constructed much earlier than Weierstrass’ function. 22 quite closely, from the original proof of Weierstrass (as it is presented in du Bois-Reymond [19]). Letx0∈Rbe arbitrary but fixed and let m∈Nbe arbitrary. Choose αm∈Z such thatbmx0−αm∈/parenleftbig −1 2,1 2/bracketrightbig and definexm+1=bmx0−αm. Put ym=αm−1 bmandzm=αm+ 1 bm. This gives the inequality ym−x0=−1 +xm+1 bm<0<1−xm+1 bm=zm−x0 and therefore ym<x 0<zm. Asm→ ∞ ,ym→x0from the left and zm→x0 from the right. First consider the left-hand difference quotient, W(ym)−W(x0) ym−x0=∞/summationdisplay n=0/parenleftbigg ancos(bnπym)−cos(bnπx0) ym−x0/parenrightbigg =m−1/summationdisplay n=0/parenleftbigg (ab)ncos(bnπym)−cos(bnπx0) bn(ym−x0)/parenrightbigg +∞/summationdisplay n=0/parenleftbigg am+ncos(bm+nπym)−cos(bm+nπx0) ym−x0/parenrightbigg =S1+S2. We treat these sums separately, starting with S1. Since/vextendsingle/vextendsingle/vextendsinglesin(x) x/vextendsingle/vextendsingle/vextendsingle≤1 we can, using a trigonometric identity, bound the sum by |S1|=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsinglem−1/summationdisplay n=0(ab)n(−π) sin/parenleftbiggbnπ(ym+x0) 2/parenrightbiggsin/parenleftBig bnπ(ym−x0) 2/parenrightBig bnπym−x0 2/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle ≤m−1/summationdisplay n=0π(ab)n=π((ab)m−1) ab−1≤π(ab)m ab−1.(3.1) Considering the sum S2we can use (since b>1 is an odd integer and αm∈Z) cos(bm+nπym) = cos/parenleftbigg bm+nπαm−1 bm/parenrightbigg = cos(bnπ(αm−1)) =/bracketleftbig (−1)bn/bracketrightbigαm−1=−(−1)αm 23 and cos(bm+nπx0) = cos/parenleftbigg bm+nπαm+xm+1 bm/parenrightbigg = cos(bnπαm) cos(bnπxm+1)−sin(bnπαm) sin(bnπxm+1) =/bracketleftbig (−1)bn/bracketrightbigαmcos(bnπxm+1)−0 = (−1)αmcos(bnπxm+1) to express the sum as S2=∞/summationdisplay n=0am+n−(−1)αm−(−1)αmcos(bnπxm+1) −1+xm+1 bm = (ab)m(−1)αm∞/summationdisplay n=0an1 + cos(bnπxm+1) 1 +xm+1. Each term in the series above is non-negative and xm+1∈/parenleftbig −1 2,1 2/bracketrightbig so we can find a lower bound by ∞/summationdisplay n=0an1 + cos(bnπxm+1) 1 +xm+1≥1 + cos(πxm+1) 1 +xm+1≥1 1 +1 2=2 3. (3.2) The inequalities (3.1) and (3.2) ensures the existence of an /epsilon11∈[−1,1] and anη1>1 such that W(ym)−W(x0) ym−x0= (−1)αm(ab)mη1/parenleftbigg2 3+/epsilon11π ab−1/parenrightbigg . As with the left-hand difference quotient, for the right-hand quotient we do pretty much the same, starting by expressing the said fraction as W(zm)−W(x0) zm−x0=S/prime 1+S/prime 2. As before, it can be deduced that |S/prime 1| ≤π(ab)m ab−1. (3.3) The cosine-term containing zmcan be simplified as (again since bis odd and αm∈Z) cos(bm+nπzm) = cos/parenleftbigg bm+nπαm+ 1 bm/parenrightbigg = cos(bnπ(αm+ 1)) =/bracketleftbig (−1)bn/bracketrightbigαm+1=−(−1)αm, 24 which gives S/prime 2=∞/summationdisplay n=0am+n−(−1)αm−(−1)αmcos(bnπxm+1) 1−xm+1 bm =−(ab)m(−1)αm∞/summationdisplay n=0an1 + cos(bnπxm+1) 1−xm+1. As before, we can find a lower bound for the series by ∞/summationdisplay n=0an1 + cos(bnπxm+1) 1−xm+1≥1 + cos(πxm+1) 1−xm+1≥1 1−/parenleftbig −1 2/parenrightbig=2 3. (3.4) By the same argument as for the left-hand difference quotient (but by using the inequalities (3.3) and (3.4) instead), there exists an /epsilon12∈[−1,1] and an η2>1 such that W(zm)−W(x0) zm−x0=−(−1)αm(ab)mη2/parenleftbigg2 3+/epsilon12π ab−1/parenrightbigg . By the assumption ab> 1+3 2π, which is equivalent toπ ab−1<2 3, the left- and right-hand difference quotients have different signs. Since also ( ab)m→ ∞ asm→ ∞ it is clear that Whas no derivative at x0. The choice of x0∈R was arbitrary so it follows that W(x) is nowhere differentiable on R. Remark 1 (Dini). In a series of publications (cf. Dini [15], [16], [17] and [18]) in the years 1877-78, Italian mathematician Ulisse Dini proposed a more general class of continuous nowhere differentiable functions (under which Weierstrass function happen to fall). Our presentation here is largely based on Knopp’s summary (cf. Knopp [38], pp. 23-26). Let {fn}be a sequence of differentiable functions fn: [0,1]→Rthat have bounded derivative on [0 ,1] and such that WD(x) =∞/summationdisplay n=1fn(x) converges uniformly on [0 ,1]. We also require that (i) each function fnhas a finite number of extrema and if δnis the maxi- mum distance between two successive extrema then δn→0 asn→ ∞ ; 25 (ii) ifγnis the (to the absolute value) greatest difference between two successive extreme values then lim n→∞δn γn= 0; (iii) ifhn,xdenotes the two increments (one which is positive and one which is negative) for which x+hn,xgives the first right (respectively left) extremum for which |fn(x+hn,x)−fn(x)| ≥1 2γn, then we can define a sequence {rn}of positive numbers such that sup x∈[0,1]|Rn(x+hn,x)−Rn(x)| ≤2rn whereRn(x) is the remainder of the series defining the function WD; (iv) if {cn}is a sequence of positive numbers such that supx∈[0,1]|f/prime n(x)| ≤cn then from some index on 4δn γnn/summationdisplay k=1ck+4rn γn≤θ,θ∈[0,1); (v) the sign of fn(x+hn,x)−fn(x) is independent of hn,xfrom some n0 onward for all x∈[0,1]. Then the function WDis continuous and nowhere differentiable on [0 ,1]. As two concrete examples of functions in Dini’s classification, consider for |a|>1 + 3π/2 WD1(x) =∞/summationdisplay k=1an 1·3·5···(2n−1)cos(1·3·5···(2n−1)πx) and fora>1 + 3π/2 WD2(x) =∞/summationdisplay k=1an 1·5·9···(4n+ 1)sin(1·5·9···(4n+ 1)πx). 26 Remark 2 (Hertz). Polish mathematician Karol Hertz gave in his paper [28] from 1879 a generalization of Weierstrass function, namely WH(x) =∞/summationdisplay k=1akcosp(bkπx), wherea>1,p∈Nis odd,ban odd integer and ab> 1 +2 3pπ. Remark 3 (Hardy). Hardy proved (in Hardy [27]) that if 0 < a < 1, b>1 andab≥1 then both W1(x) =∞/summationdisplay k=0aksin(bkπx) and W2(x) =∞/summationdisplay k=0akcos(bkπx) are continuous and nowhere differentiable on all of R. Remark 4 (Porter). M. B. Porter generalized Weierstrass function in an article (Porter [58]) published in 1919. He proposed two classes of functions Wi: [a,b]→Rdefined by W1(x) =∞/summationdisplay k=0uk(x) sin(bnπx) and W2(x) =∞/summationdisplay k=0uk(x) cos(bnπx) where {bn}is a sequence of integers and {uk}is a sequence of differentiable functions. We have the following requirements: (i)Wiconverges uniformly on [ a,b] fori= 1,2; (ii)bndividesbn+1and for an unlimited number of n’s,bn+1/bnmust be divisible by four or increase to infinity with n; (iii)/summationtext∞ k=0u/prime n(x) converge uniformly on [ a,b] by the Weierstrass M-test; (iv) (3π/2)/summationtextN−1 k=0|bnun(x)|<|bNuN(x)|for allx∈[a,b]. If this holds then both W1andW2are continuous and nowhere differentiable. The following concrete functions are examples that falls under Porter’s gen- 27 eralization. (a)∞/summationdisplay k=0an n!sin(n!πx) and∞/summationdisplay k=0an n!cos(n!πx), where |a|>1 +3 2π; (b)∞/summationdisplay k=01 ansin(n!anπx) and∞/summationdisplay k=01 ancos(n!anπx), where |a| ∈N\ {1}; (c)∞/summationdisplay k=1ak 10ksin(103kπx) and∞/summationdisplay k=1ak 10kcos(103kπx), whereakis chosen such that/summationtext∞ k=1ak 10kis a non-terminating decimal. Both Dini functions in the remark above falls under this generalization as well. 3.5 Darboux function (1873; published in 1875) Darboux’s function, discovered independently of Weierstrass, was presented on 19 March 1873 (two years earlier than the first publication of Weierstrass’ function) and was published two years later in Darboux [11]. In this publica- tion (whose title translates to “paper on the discontinuous functions”) Dar- boux spends much of the discussion on the subject of Riemann-integration of discontinuous functions but he also investigated when a continuous function possess a finite derivative. Contained in this document is his description of a continuous function which is nowhere differentiable and this function is defined as the infinite series D(x) =∞/summationdisplay k=11 k!sin ((k+ 1)!x) . Darboux constructed this function after having analyzed and generalized results from Schwarz and Hankel, who in the years before had studied and made suggestions about the subject. One of Schwarz ideas, proposed in 1873 in Schwarz [67], was a function S: (0,∞)→Rdefined by S(x) =∞/summationdisplay k=0ϕ(2kx) 4k, where ϕ(x) = [x] +/radicalbig x−[x] 28 and [x] means the integer part of x. The function Sis continuous and monotonically increasing, but there is no derivative at infinitely many points in any interval so Sis not differentiable on a dense subset of (0 ,M) (which we will prove). Interesting to note is that Schwarz (and many others) seem to have considered these types of functions “without derivative”, but today, with measure theoretic background, we call many of them differentiable almost everywhere. Hankel had introduced the concept of “Condensation of singularities” some years before (cf. Hankel [26]). This is a process where by letting each term in an absolutely convergent series have a singularity, a function with singu- larities at all rational points4is created. An example of this procedure could be the function gdefined by g(x) =∞/summationdisplay n=1ψ(sin(nπx)) ns, where ψ(x) =/braceleftBigg xsin/parenleftbig1 x/parenrightbig ,x/negationslash= 0 0, x= 0 ands > 1. Hankel’s treatment of the subject, however, wasn’t entirely accurate as other mathematicians pointed out after the publication. Darboux writes in his paper that he thought it was a shame that Hankel had died before he had a chance to correct some of his ideas himself. D(x) x1.0 −1.01.0 2 .0 Figure 3.5: Darboux’s function D(x) on [0,3]. In a subsequent paper (Darboux [12]), Darboux generalized his example. He 4Ifxis a rational number, say x=p/q, then sin(nπx) = sin(nπp/q ) = sin( ±pπ) = 0 forn=q. Hencexis a singular point of ψ(sin(nπx)) (for a special n) and this behavior can be shown to transfer onto the sum of the series as well. 29 considered the series ϕ(x) =∞/summationdisplay k=1f(anbnx) an whereanandbnare sequences of real numbers and f:R→Ris a bounded continuous function with a bounded second derivative. By adding some re- strictions to the two sequences {an}and{bn}, lim n→∞an+1 an= 0 and for some fixed k∈N lim n→∞/summationtextn−k m=1amb2 m an= 0, Darboux states that ϕis a continuous function. Moreover, it is possible to make some additional restrictions on the parameters to ensure that ϕis nowhere differentiable as well as continuous. For example, with bn= 1 and k= 1 it is enough to have lim n→∞/summationtextn−1 m=1am an= 0 forϕto be nowhere differentiable for an infinite number of functions f. For example with an=n! andf(x) = cos(x). Another example would be bn=n+ 1,an=n!,k= 3 andf(x) = sin(x) which is the function D introduced by Darboux in his earlier paper (Darboux [11]) and which was defined at the beginning of this section. Theorem 3.5. The Darboux function D(x) =∞/summationdisplay k=11 k!sin ((k+ 1)!x) is continuous and nowhere differentiable on R. Proof. Since/summationtext∞ k=01 k!=eit is clear that/summationtext∞ k=11 k!<∞. This and the fact that supx∈R/vextendsingle/vextendsingle1 k!sin((k+ 1)!x)/vextendsingle/vextendsingle≤1 k!implies, by the Weierstrass M-test (Theorem 2.2), that the convergence is uniform. The Corollary 2.4 gives the continuity ofD. A proof of the fact that Dis nowhere differentiable can be found in Darboux [12]. 30 Theorem 3.6. The Schwarz function S: (0,M)→Rdefined by S(x) =∞/summationdisplay k=0ϕ(2kx) 4k, where ϕ(x) = [x] +/radicalbig x−[x], is continuous and non-differentiable on a dense subset of (0,M). HereM > 0 is any real number. Proof. We start by proving that Sis continuous. The only possible discon- tinuities of the function ϕis forx∈N. Letp∈N, we show that ϕis both left and right continuous at p. From the right we have lim x→p+ϕ(x) = lim x→p+/parenleftBig [x] +/radicalbig x−[x]/parenrightBig =p+√p−p=p and from the left lim x→p−ϕ(x) = lim x→p−/parenleftBig [x] +/radicalbig x−[x]/parenrightBig =p−1 +/radicalbig p−(p−1) =p. Henceϕis continuous on (0 ,M) (andϕ(p) =pforp∈N). Now we show that the series converge uniformly so that also Sis continuous on (0 ,M). Leth∈(0,1) andp∈N∪ {0}. Then ϕ(p+h) = [p+h] +/radicalbig p+h−[p+h] =p+√ h. Defineq(h) =ϕ(p+h)−(p+h), thenq(h)≤p+h+ 1/4 since q/prime(h) =1 2√ h−1 = 0 ⇒h=1 4 andq/prime/prime(1/4)<0 so the maximum is attained at h= 1/4 (q(0) =q(1) = 0). From this we get the inequality ϕ(x)≤x+1 4. Now it follows that sup x∈(0,M)/vextendsingle/vextendsingle/vextendsingle/vextendsingle1 4nϕ(2nx)/vextendsingle/vextendsingle/vextendsingle/vextendsingle≤sup x∈(0,M)/vextendsingle/vextendsingle/vextendsingle/vextendsingle2nx+ 1/4 4n/vextendsingle/vextendsingle/vextendsingle/vextendsingle≤M 2n+1 4n+1 31 and since∞/summationdisplay n=0/parenleftbiggM 2n+1 4n+1/parenrightbigg <∞, the Weierstrass’ M-test (Theorem 2.2) and the Corollary 2.4 gives that Sis continuous on (0 ,M). We turn to the non-differentiable part. Let x0,x1∈(0,M) withx0<x 1be arbitrary. We show that between any two such points there exists a point whereSis without derivative (which implies that Sis non-differentiable on a dense subset of (0 ,M)). Letxbe a dyadic rational such that x0< x < x 1. Thenx=i2−mfor somei,m∈N. Let 0< h < 2−m, then, since each term in the series is non-negative, S(x+h)−S(x) h=∞/summationdisplay k=0ϕ(2n(x+h))−ϕ(2nx) 4nh≥ϕ(2m(x+h))−ϕ(2mx) 4mh. Since 2mh<1 and 2mx=i∈Nwe see that ϕ(2m(x+h))−ϕ(2mx) = [2mx+ 2mh] +/radicalbig 2mx+ 2mh−[2mx+ 2mh] −[2mx]−/radicalbig 2mx−[2mx] =i+√ i+ 2mh−i−i−√ i−i=√ 2mh. Hence S(x+h)−S(x) h≥√ 2mh 4mh=1 2m√ 2m·1√ h→ ∞ ash→0 and therefore S/prime(x) does not exist. 3.6 Peano function (1890) Lett= (t1t2t3···)3be a ternary representation of t∈[0,1] (that is, t=/summationtext∞ k=1tk3−kwithtk∈ {0,1,2}). Then Peano’s function Pis expressed as   P: [0,1]→[0,1]×[0,1], (t1t2t3···)3/mapsto→/parenleftbigg(t1(kt2t3)(kt2+t4t5)(kt2+t4+t6t7)···)3 ((kt1t2)(kt1+t3t4)(kt1+t3+t5t6)···)3/parenrightbigg , 32 where the operator kis defined as ktj= 2−tj,tj= 0,1,2 andkltjis thel’th element in the sequence {ktj,k(ktj),k(k(ktj)),...}(and we adhere to the convention that k0tj=tj). It can be shown (cf. Sagan [64], pp. 32-33) that Pis independent of which5ternary representation of tis chosen and that Pis surjective (i.e. a space-filling curve, that is a “1-dimensional” curve that fills two-dimensional space6). (a)n= 1. (b)n= 1. (c)n= 2. (d)n= 3. Figure 3.6: First four steps in the geometric generation of Peano’s curve. Peano’s curve was the first space-filling curve discovered and it was published in 1890 (in Peano [54]). After his publication several other mathematicians proposed new examples and among those were Hilbert’s function (published in 1891, see Sagan [64]) and Schoenberg’s curve (proposed in 1938). Both of those happen to be nowhere differentiable (and in Section 3.13 we take a closer look on Schoenberg’s curve). It is not, however, the case that all space-filling curves are nowhere differentiable (although Peano’s turns out to be). For example, Lebesgue’s space filling curve7is differentiable almost everywhere (it is differentiable everywhere except on the Cantor set, which incidentally has Lebesgue measure zero). 5The representation is not unique, e.g. (1) 3= 1/3 and also (022 ···)3= 1/3. 6Or more generally, a curve that passes through every point of some subset of n- dimensional Euclidean space (or even more general as is stated in the Hahn-Mazurkiewicz theorem). 7Henri Lebesgue constructed his curve in 1904 as a continuous extension of a known mapping. The original mapping had the Cantor set as domain and mapped it onto [0 ,1]× [0,1]. The extension is done by linear interpolation, see Sagan [64]. 33 Letφpandψpbe the component functions of P. Peano stated in his pre- sentation that both components were continuous and nowhere differentiable but left the proof of nowhere differentiability out of his paper. The proof presented here is due to Sagan [64], pp. 33-34. φp(x) x1.0 0.5 1.0 0 .5 Figure 3.7: The component φpof Peano’s curve. Theorem 3.7. The components φpandψpof the Peano function Pare continuous and nowhere differentiable on the interval [0,1]. Proof. First we establish the continuity of φp. We do this in two steps, first we show that φpis continuous from the right. Fort0∈[0,1), lett0= (t1t2t3···t2nt2n+1···)3be the ternary representation oft0that doesn’t end in infinitely many 2’s. Choose δ= 3−2n−(00···t2n+1t2n+2···)3. Clearlyδ→0 asn→ ∞ . The definition of δgives t0+δ= (t1t2t3···t2nt2n+1···)3+ 3−2n−(00···t2n+1t2n+2···)3 = (t1t2t3···t2n00···)3+ 3−2n= (t1t2t3···t2n22···)3. So for any t∈[t0,t0+δ), the first 2 ndigits in the ternary expansion are equal, i.e.t= (t1t2t3···t2nτ2n+1τ2n+2···)3. Let/epsilon1n=/summationtextn i=1t2i. We have |φ(t)−φ(t0)|=|(t1(kt2t3)···(k/epsilon1nτ2n+1)···)3−(t1(kt2t3)···(k/epsilon1nt2n+1)···)3| ≤∞/summationdisplay i=n1 3i+1|k/epsilon1iτ2i+1−k/epsilon1it2i+1| ≤∞/summationdisplay i=n2 3i+1 =2 3n+1∞/summationdisplay i=01 3i=1 3n→0 asn→ ∞ . 34 Henceφpis continuous from the right. Now we show that φpis also continuous from the left. The argument follows similarly as above. Fort0∈(0,1], lett0= (t1t2t3···t2nt2n+1···)3be the ternary representation with infinitely many non-zero terms. Pick δ= (00 ···0t2n+1t2n+1···)3. Then t0−δ= (t1t2t3···t2n00···)3. Hence, for t∈(t0−δ,t0],t’s ternary representation has the same first 2 n digits ast0. Thus |φ(t)−φ(t0)|=|(t1(kt2t3)···(k/epsilon1nτ2n+1)···)3 −(t1(kt2t3)···(k/epsilon1nt2n+1)00···)3| ≤2 3n+1∞/summationdisplay i=01 3i=1 3n→0 asn→ ∞ . Soφpis continuous from the left on (0 ,1]. Since we established that φpalso is continuous from the right on [0 ,1) it is clear that φpis continuous on [0 ,1]. Next we show that φpis nowhere differentiable on [0 ,1]. For arbitrary t∈ [0,1], lett= (t1t2t3···t2nt2n+1···)3be a ternary representation of t. Define the sequence {tn}bytn= (t1t2t3···t2nτ2n+1t2n+2···)3, whereτ2n+1is chosen asτ2n+1= (t2n+1+ 1) mod 2. This implies that |t−tn|=1 32n+1. From the definition of Pandtn,φp(t) andφp(tn) only differs at position n+1 in the ternary representation. Therefore we have |φp(t)−φp(tn)|=1 3n+1|k/epsilon1nt2n+1−k/epsilon1nτ2n+1|=1 3n+1. Analyzing the differential quotient we see that /vextendsingle/vextendsingle/vextendsingle/vextendsingleφp(t)−φp(tn) t−tn/vextendsingle/vextendsingle/vextendsingle/vextendsingle=1 3n+132n+1 1= 3n→ ∞ asn→ ∞ . Henceφpis not differentiable at t. Sincet∈[0,1] was arbitrary it follows thatφpis nowhere differentiable on [0 ,1]. Moreover, since ψp(t) = 3φp(t/3), the fact that ψpis continuous and nowhere differentiable on [0 ,1] follows from what we just established for φp. 35 3.7 Takagi (1903) and van der Waerden (1930) functions Takagi’s and van der Waerden’s functions are very similar in their construc- tion. Takagi presented his example in 1903 (cf. Takagi [73]) as an example of a “simpler” continuous nowhere differentiable function than Weierstrass. Van der Waerden published his function in 1930 (van der Waerden [75]), apparently unaware of Takagi’s very similar idea. The definition of Takagi’s function is expressed as the infinite series T(x) =∞/summationdisplay k=01 2kdist/parenleftbig 2kx,Z/parenrightbig =∞/summationdisplay k=01 2kinf m∈Z/vextendsingle/vextendsingle2kx−m/vextendsingle/vextendsingle and Van der Waerden’s function is defined as V(x) =∞/summationdisplay k=01 10kdist/parenleftbig 10kx,Z/parenrightbig =∞/summationdisplay k=01 10kinf m∈Z/vextendsingle/vextendsingle10kx−m/vextendsingle/vextendsingle. T(x) x0.5 0.5 (a) Takagi’s function.V(x) x0.5 0.5 (b) Van der Waerden’s function. Figure 3.8: Takagi’s and van der Waerden’s functions on [0 ,1]. The function φ:R→Rdefined byφ(x) = dist(x,Z) = inf m∈Z|x−m|, which both series above are superpositions of, can be seen graphically in figure 3.13. More variations have been developed and in 1918 Knopp [38] did a general- ization, which we consider in Section 3.11. Also de Rham treated these kinds of functions in his article de Rham [14]. He gives a proof that the function referred to as Takagi’s function here is continuous and nowhere differentiable. 36 De Rham also considers the function f(x) =∞/summationdisplay k=0a−kφ(akx) whereais an even positive integer. He claims that his proof can be adapted to show that fis continuous and nowhere differentiable as well. Moreover, he points out that the function fis a solution to the functional equation f(x)−1 af(ax) =φ(x) and that it is the only solution that is bounded. He proceeds to generalize this equation to F(x)−bF(ax) =g(x) wheregis a given function and aandbare constants. De Rham claims that the only bounded solution for b∈(0,1) is F(x) =∞/summationdisplay k=0bkg(akx). Interesting to note is that for g(x) = cos(x) andaan odd integer with ab> 1 + 3π/2 we have the Weierstrass function (see Section 3.4). We will use the following lemma when proving that Takagi’s function (and several others) is nowhere differentiable. Lemma 3.8. Leta < a n< x < b n< b for alln∈Nand letan→xand bn→x. Iff: [a,b]→Ris a continuous function and f/prime(x)exists then lim n→∞f(bn)−f(an) bn−an=f/prime(x). Proof. Since /vextendsingle/vextendsingle/vextendsingle/vextendsinglebn−x bn−an/vextendsingle/vextendsingle/vextendsingle/vextendsingle≤bn−an bn−an= 1 and/vextendsingle/vextendsingle/vextendsingle/vextendsinglex−an bn−an/vextendsingle/vextendsingle/vextendsingle/vextendsingle≤bn−an bn−an= 1 37 we can estimate by /vextendsingle/vextendsingle/vextendsingle/vextendsinglef(bn)−f(an) bn−an−f/prime(x)/vextendsingle/vextendsingle/vextendsingle/vextendsingle=/vextendsingle/vextendsingle/vextendsingle/vextendsinglebn−x bn−an/parenleftbiggf(bn)−f(x) bn−x−f/prime(x)/parenrightbigg +x−an bn−an/parenleftbiggf(an)−f(x) an−x−f/prime(x)/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle ≤/vextendsingle/vextendsingle/vextendsingle/vextendsinglef(bn)−f(x) bn−x−f/prime(x)/vextendsingle/vextendsingle/vextendsingle/vextendsingle+/vextendsingle/vextendsingle/vextendsingle/vextendsinglef(an)−f(x) an−x−f/prime(x)/vextendsingle/vextendsingle/vextendsingle/vextendsingle →0 asn→ ∞ . Hence lim n→∞f(bn)−f(an) bn−an=f/prime(x). Theorem 3.9. Both the Takagi function and the van der Waerden function are continuous and nowhere differentiable on R. Proof. That both TandVare continuous follows from the proof of the continuity of the Knopp function in Section 3.11. We show that Tis nowhere differentiable. The proof is based on an argument by Billingsley [5] and in a similar way it can be shown that also V(x) is nowhere differentiable (cf. van der Waerden [75]). Letx∈Rbe arbitrary and assume that T/prime(x) exists. By Lemma 3.8, if un≤x≤vn(withun<vn) andvn−un→0, then T(vn)−T(un) vn−un→T/prime(x). We will define two sequences that contradicts this. Let φ(x) = inf m∈Z|x−m|. Then T(x) =∞/summationdisplay k=01 2kinf m∈Z/vextendsingle/vextendsingle2kx−m/vextendsingle/vextendsingle=∞/summationdisplay k=01 2kφ/parenleftbig 2kx/parenrightbig . LetD={i2−n|i,n∈Z}be the dyadic rationals. If u∈Dis of ordernthen, for every integer k≥n, 2ku∈Z. Hence, since φ(p) = 0 forp∈Z, we have T(u) =n−1/summationdisplay k=01 2kφ/parenleftbig 2ku/parenrightbig . 38 Letun,vn∈Dbe successive numbers of order nfor whichun≤x < v n. Thenvn−un=i2−n−(i−1)2−n= 2−nand T(vn)−T(un) vn−un=n−1/summationdisplay k=01 2kφ(2kvn)−φ(2kun) vn−un. Obviouslyφ(x) is linear for x∈/bracketleftbig 2kun,2kvn/bracketrightbig since/bracketleftbig 2kun,2kvn/bracketrightbig =/bracketleftbigi−1 2l,i 2l/bracketrightbig wherel=n−k∈N. Hence, for 0 ≤k<n , 1 2kφ(2kvn)−φ(2kun) vn−un=±2−l 2−l=±1 which gives T(vn)−T(un) vn−un=n−1/summationdisplay k=0±1. Asn→ ∞ , the series on the right does not converge. This contradicts the assumption that T/prime(x) exists. Since x∈Rwas arbitrary, the function Tis nowhere differentiable. Note that Cater [8] has shown that Thas no one-sided derivative at any point (whereas what we just proved above is that Thas no two-sided derivative at any point). A further property of Takagi’s function is deduced in Shidfar and Sabet- fakhri [69], it turns out that it’s also continuous in the H¨ older sense for 0<α< 1 (or Lipschitz class of order α). That is, for every α∈(0,1) there existsMα>0 such that for every x,y∈R |T(x)−T(y)| ≤Mα|x−y|α. 3.8 Koch “snowflake” curve (1904) In 1904, Swedish mathematician Helge von Koch published (in Koch [39]) an article about a curve of infinite length with tangent nowhere. It was re- published two years later with some added pages in Koch [40]. Koch writes8 about the previous misconception that all continuous curves has a well de- termined tangent except at some isolated points: 8Translation from Edgar [20]. 39 “Even though the example of Weierstrass [Section 3.4] has cor- rected this misconception once and for all, it seems to me that his example is not satisfactory from the geometrical point of view since the function is defined by an analytic expression that hides the geometrical nature of the corresponding curve and so from this point of view one does not see why the curve has no tangent.” Koch’s “snowflake” curve (named after its shape) is constructed as follows: Take an equilateral triangle and split each line in three equal parts. Re- place the middle segments by two sides of a new equilateral triangle that is constructed with the removed segment as its base. Repeat this procedure on each of the four new lines (for each of the original three sides). Repeat indefinitely. The limit of the process gives rise to a curve that is continuous and has a tangent nowhere (which is shown in Koch’s paper). In figure 3.9 the first few iterations are shown graphically. Koch also shows that there exists a parameterization /braceleftBigg x=f(t) y=g(t) of the curve for t∈[0,1] and that both functions fandgare continuous and nowhere differentiable (on [0 ,1]). (a)n= 0. (b)n= 1. (c)n= 2. (d)n= 3. Figure 3.9: First four steps in the construction of Koch’s “snowflake”. 40 3.9 Faber functions (1907, 1908) In 1907, German mathematician Georg Faber [21] presented an example of a continuous nowhere differentiable function defined by F1(x) =∞/summationdisplay k=11 10kinf m∈Z/vextendsingle/vextendsingle2k!x−m/vextendsingle/vextendsingle=∞/summationdisplay k=11 10kdist/parenleftbig 2k!x,Z/parenrightbig . In 1908, Faber proceeded to publish a second article [22] named “ ¨Uber stetige Funktionen” and two years later a longer article [23] about a similar subject. In the article from 1908, Faber presented another continuous nowhere differ- entiable function: F2(x) =∞/summationdisplay k=11 k!inf m∈Z/vextendsingle/vextendsingle2k!x−m/vextendsingle/vextendsingle=∞/summationdisplay k=11 k!dist/parenleftbig 2k!x,Z/parenrightbig . Faber set out to do an investigation of what he referred to as the “deep gap” between the differentiable and the merely continuous functions. By doing this, Faber intended to give a better insight on the infinitesimal structure of continuous functions in general. His investigation makes it possible to construct, by superposition of piecewise linear functions, examples of continuous functions with special properties like, for example, nowhere differentiability. The construction is to a large degree geometrical and the fact that Faber’s function has the desired properties is proven as it is constructed. This is in contrast with most other proofs of this kind, which are done on a fixed analytical expression that is given. Unfortunately we won’t go through all of Faber’s construction but merely give a brief description leading to a fixed expression and thereby neglecting some of the beauty in his architecture. To make the understanding easier, Faber wished to characterize his functions by a countable (and dense) subset of the interval [0 ,1], namely the set M=/braceleftbiggk 2n/vextendsingle/vextendsingle/vextendsingle/vextendsinglek,n∈N, k≤2n/bracerightbigg . He then proceeded to define the real numbers δk 2nsuch that δ1 2=F2/parenleftbigg1 2/parenrightbigg −F2(0) +F2(1) 2 41 and form,n∈Nwith 2m+ 1<2n δ2m+1 2n=F2/parenleftbigg2m+ 1 2n/parenrightbigg −F2/parenleftbigm 2n−1/parenrightbig +F2/parenleftbigm+1 2n−1/parenrightbig 2 whereF2is the function we will show is continuous and nowhere differen- tiable. We take F2(0) =F2(1) = 0. From the relations above it is clear that we can recursively express F2(xi) in terms of “ δ’s” for any xi∈M. To give a more geometrical view, we consider a sequence of piecewise linear continuous functions {fk}, where, for any k∈N,fk: [0,1]→Ris the function that binds together the points (0,0),/parenleftbigg1 2n,δ 1 2n/parenrightbigg ,/parenleftbigg2 2n,0/parenrightbigg ,/parenleftbigg3 2n,δ 3 2n/parenrightbigg , ...,/parenleftbigg2n−1 2n,δ2n−1 2n/parenrightbigg ,(1,0) in a continuous and piecewise linear manner. Figure 3.11(a) shows two func- tions graphically. δ1 2 δ1 4 δ3 4 x 1 41 23 41y Figure 3.10: The functions f1(dashed) and f2(whole). Now, for any xi∈Mwe can deduce that F2(xi) =∞/summationdisplay k=1fk(xi). We choose a subsequence {nk}and for each nkwe choose all “ δ’s” equal to 1 k!(for simplicity; Faber made the same choice in his article). Naturally, for 42 arbitraryx∈[0,1] we define F2(x) =∞/summationdisplay k=1fnk(x). This gives rise to the function which we refer to as Faber’s function: F2(x) =∞/summationdisplay k=11 k!inf m∈Z/vextendsingle/vextendsingle2k!x−m/vextendsingle/vextendsingle=∞/summationdisplay k=11 k!dist/parenleftbig 2k!x,Z/parenrightbig . F(x) x0.05 1.0 0 .5 (a) Faber’s first function F1.F(x) x0.5 0.5 1 .0 (b) Faber’s second function F2(x). Figure 3.11: Faber’s functions Fi(x) on [0,1]. Theorem 3.10. The Faber functions F1(x) =∞/summationdisplay k=11 10kinf m∈Z/vextendsingle/vextendsingle2k!x−m/vextendsingle/vextendsingleandF2(x) =∞/summationdisplay k=11 k!inf m∈Z/vextendsingle/vextendsingle2k!x−m/vextendsingle/vextendsingle are continuous and nowhere differentiable on R. We will not repeat Faber’s proof here but instead opt for a direct proof based on the analytic expression given above. The proof is based on the same argument that was used in the proof of Theorem 3.9 (for Takagi’s function). Proof. We only prove this for F2. The proof for F1follows similarly. First, that F2is continuous follows from the Weierstrass M-test (Theorem 2.2) and the Corollary 2.4 since sup x∈[0,1]1 k!inf m∈Z/vextendsingle/vextendsingle2k!x−m/vextendsingle/vextendsingle≤1 2k! 43 and/summationtext∞ k=11/(2k!)<∞. We show that F2also is nowhere differentiable. Let x∈Rbe arbitrary. As before we construct sequences unandvnof successive dyadic rationals (of the same order) such that un≤x≤vn(withun< v n) andvn−un= 2−n. Then we show that F2(vn)−F2(un) vn−un does not converge as n→ ∞ which implies that F/prime 2(x) does not exist (by Lemma 3.8). Letφ(x) = inf m∈Z|x−m|, then F2(x) =∞/summationdisplay k=11 k!φ/parenleftbig 2k!x/parenrightbig . Ifu∈Dis a dyadic rational of order n, then F2(u) =/summationdisplay k!<n1 k!φ/parenleftbig 2k!u/parenrightbig . Now, F2(vn)−F2(un) vn−un=/summationdisplay k!<n1 k!φ(2k!vn)−φ(2k!un) vn−un. As before,φ(x) is linear for x∈/bracketleftbig 2k!un,2k!vn/bracketrightbig . Hence, for 0 ≤k!<n, 1 k!φ(2k!vn)−φ(2k!un) vn−un=±2k! k!→ ±∞ asn→ ∞ which gives F2(vn)−F2(un) vn−un=/summationdisplay k!<n±2k! k!. This series does not converge as n→ ∞ soF/prime 2(x) does not exist. Since x∈R was arbitrary, F2is nowhere differentiable. 3.10 Sierpi´ nski curve (1912) Wac/suppress law Sierpi´ nski published another example of a space-filling curve in 1912 in his paper Sierpi´ nski [70]. He found a bounded, continuous and even func- 44 tionSWsuch that, for t∈[0,1], the mapping /braceleftBigg x=SW(t) y=SW(t−1/4) is surjective onto [ −1,1]. Sierpi´ nski deduced the following expression for SW: SW(t) =Θ(t) 2/parenleftBigg 1 +∞/summationdisplay k=1(−1)k/producttextk l=1Θ(τl(t)) 2k/parenrightBigg where both Θ and τare periodic functions with period 1 defined by Θ(t) =/braceleftBigg −1 ift∈[1/4,3/4), 1 ift∈[0,1/4)∪[3/4,1) and τl(t) =/braceleftBigg 1/8 + 4tift∈[0,1/4)∪[1/2,3/4), 1/8−4tift∈[1/4,1/2)∪[3/4,1), τl+1(t) =τl(τ1(t)), for every l∈N. Moreover, he demonstrated that SWis the limit of a sequence of polygonal curves, of which the first four can be seen graphically in figure 3.12. (a)n= 1. (b)n= 2. (c)n= 3. (d)n= 4. Figure 3.12: Polygonal approximations (of order n) to Sierpi´ nski’s curve. 3.11 Knopp function (1918) Define the function K:R→Ras K(x) =∞/summationdisplay k=0akφ(bkx), 45 where φ(x) = inf m∈Z|x−m|= dist(x,Z) anda∈(0,1),ab> 4 andb>1 an even integer. φis a “saw-tooth” function and can be seen graphically in figure 3.13. φ(x) x0.5 1.0 2 .0 −1.0 −2.0 Figure 3.13: The “saw-tooth” function φ(x) on [−3,3]. Kis the Knopp function which was introduced by Konrad Knopp in 1918 (cf. Knopp [38]). Both Takagi’s and van der Waerden’s functions are special cases of this function. Originally Knopp had the restrictions 0<a< 1,ab> 4 andb>1 an even integer on the parameters but in an article published in 1994 Baouche and Dubuc [3] weakened the restrictions to 0<a< 1,ab> 1 wherebis not necessarily an integer. Further investigations were done on the case when ab= 1 and in another article published in 1994 by Cater [9], F.S. Cater proved9that ∞/summationdisplay k=0b−nφ(bnx) is nowhere differentiable if b≥10. Theorem 3.11. The Knopp function K(x) =∞/summationdisplay k=0akφ(bkx) =∞/summationdisplay k=0akdist/parenleftbig bkx,Z/parenrightbig 9Actually, both in Baouche and Dubuc [3] and in Cater [9] the results are proved in a more general case when an additional phase sequence {cn}is added to the argument, i.e. /hatwideK(x) =/summationtext∞ k=0anφ(bnx+cn). Moreover, Cater actually proves that for non-zero sequences {an}and{bn}withbn>0 and |anbn|= 1,/hatwideKhas no unilateral derivative if bn+1≥10bn. 46 is continuous and nowhere differentiable on Rfora∈(0,1)andab> 1. Proof. We can establish the continuity of Ksimilarly as for Weierstrass’ function. In fact, if 0 <a< 1 then/summationtext∞ k=0ak<∞and sinceφis a bounded function, supx∈R|φ(x)| ≤1 2, it follows that supx∈R|akφ(bkx)| ≤1 2ak. The Weierstrass M-test (Theorem 2.2) shows that Kconverges uniformly on R. As before, the continuity of Know follows from the Corollary 2.4. The proof of nowhere differentiability can be found in Baouche and Dubuc [3] forab> 1 or in Knopp [38] for the original constraints ab> 4 andb>1 an even integer. 3.12 Petr function (1920) The Czech mathematician Karel Petr published, in 1920, a simple example of a continuous nowhere differentiable function. The Petr function Pk: [0,1]→ Rin question is defined as follows. For any x∈[0,1], let x=∞/summationdisplay k=1ak 10k, where ak∈ {0,1,..., 9}, be a decimal expansion of xand define PK(x) =∞/summationdisplay k=1ckbk 2k wherebk=akmod 2,c1= 1 and for k≥2 ck=/braceleftBigg −ck−1ifak−1∈ {1,3,5,7}, ck−1 else. In the same year as Petr’s function was published, another Czech mathe- matician, Karel Rychl´ ık, gave a generalization (cf. Rychl´ ık [60],[62]) where he carried over the definition from Rto the ring10Qpofp-adic numbers. For x∈Qp, that is, x=∞/summationdisplay k=rakpk, where ak∈ {0,1,...,p −1}, 10Or field when certain properties hold, like, for example, if pis a prime number. 47 we define the function fby f(x) =∞/summationdisplay k=0ar+2kpr+2k. Rychl´ ık proves in his papers that fis continuous and nowhere differentiable inQp. Pk(x) x1.0 0.5 1.0 0 .5 Figure 3.14: Petr’s function in a 4-adic system. Theorem 3.12. The Petr function PKis continuous and nowhere differen- tiable on (0,1). Proof. Petr proved this result himself in Petr [56]. 3.13 Schoenberg function (1938) Schoenberg’s two functions φsandψsare defined as   φs(x) =∞/summationdisplay k=01 2kp(32kx), ψs(x) =∞/summationdisplay k=01 2kp(32k+1x),wherep(x) =  0x∈[0,1/3], 3x−1x∈[1/3,2/3], 1x∈[2/3,4/3], 5−3x x∈[4/3,5/3], 0x∈[5/3,2] andp(x+ 2) =p(x) for every x∈R. Figure 3.16(a) gives a more intuitive description of the function p. Schoenberg’s curve is actually another example 48 of a space-filling curve (like Peano’s curve, which was discussed earlier). The “space-filling” is accomplished by the parameterization (for t∈[0,1]) /braceleftBigg x=φs(t), y=ψs(t). (a)n= 1. (b)n= 2. (c)n= 3. (d)n= 3. Figure 3.15: First four approximation polygons in the construction of Scho- enberg’s curve (sampled at tk=m/3n,m= 0,1,..., 3n). Schoenberg constructed the curve in 1938 as an extension of the same map that Henri Lebesgue had used in the construction of his space-filling function decades earlier. Schoenberg’s curve resulted in a much easier proof of the continuity (that is, easier than for Lebesgue’s case) and the curve also turned out to be nowhere differentiable (a fact proven later). For more discussion, see Sagan [64], pp, 119-130. p(x) x1 −3−2−1 3 2 1 (a)pfor−3≤x≤3.φs(x) x1.0 0.5 1.0 0 .5 (b) Schoenberg’s function φs. Figure 3.16: Schoenberg’s function φsand the auxiliary function p. 49 Theorem 3.13. Both the Schoenberg functions φsandψsare continuous and nowhere differentiable on the interval (0,1). The proof of this theorem is based on Sagan’s proof from [63]. Proof. The continuity of both φsandψsfollows immediately from the Weier- strass M-test (Theorem 2.2) and Corollary 2.4 since sup |(1/2k)p(x)| ≤1/2k. We turn to the nowhere differentiable part. Let t∈(0,1) be arbitrary and assume that φ/prime s(x) exists. By Lemma 3.8, if 0 <a n<t<b n<1,an→tand bn→tthen φs(bn)−φs(an) bn−an→φ/prime s(x) asn→ ∞ . We construct two sequences {an}and{bn}that contradicts this. Take ˆkn= [9nt], where [x] denotes the integer part of x, and put ˆan=ˆkn9−n andˆbn=ˆkn9−n+ 9−n. Thenan→tandbn→twith 0< a n< t < b n<1 fornlarge enough. Now, infinitely many ˆknare even, odd or both. We consider two cases. (i) If there are infinitely many even ˆknthen takeknas the corresponding subsequence of ˆkn(and the same subsequences anandbnof ˆanandˆbnrespec- tively). Then we have φs(bn)−φs(an) =1 2∞/summationdisplay k=01 2kp(9k−nkn+ 9k−n)−1 2∞/summationdisplay k=01 2kp(9k−nkn) =1 2n−1/summationdisplay k=01 2k/parenleftbig p(9k−nkn+ 9k−n)−p(9k−nkn)/parenrightbig +1 2∞/summationdisplay k=n1 2k/parenleftbig p(9k−nkn+ 9k−n)−p(9k−nkn)/parenrightbig =M1+M2. For 0 ≤k < n , 9k−n≤1 9and from the definition of p(x) we can obtain the lower bound p/parenleftbig 9k−nkn+ 9k−n/parenrightbig −p/parenleftbig 9k−nkn/parenrightbig ≥ −3·9k−n. This gives a lower bound for M1by M1≥ −3 2n−1/summationdisplay k=01 2k9k−n=−3 2·9nn−1/summationdisplay k=0/parenleftbigg9 2/parenrightbiggk =−3 7·9n/parenleftbigg/parenleftbigg9 2/parenrightbiggn −1/parenrightbigg . 50 Fork≥n, 9k−n≥1 is odd which implies that uk= 9k−nknis even and vk= 9k−nkn+ 9k−nis odd. Hence M2=1 2∞/summationdisplay k=n1 2k(p(vk)−p(uk)) =1 2∞/summationdisplay k=n1 2k(1−0) =1 2n. Now, φs(bn)−φs(an) bn−an= 9n(M1+M2)≥9n/parenleftbigg1 2n−3 7·9n/parenleftbigg/parenleftbigg9 2/parenrightbiggn −1/parenrightbigg/parenrightbigg =4 7/parenleftbigg9 2/parenrightbiggn +3 7→ ∞ asn→ ∞ . (ii) If there are infinitely many odd ˆkninstead, we can define the corre- sponding subsequences kn,anandbnsimilarly as before but with the odd subsequence. Instead of a lower bound for M1we estimate by M1≤3 2n−1/summationdisplay k=01 2k9k−n=3 7·9n/parenleftbigg/parenleftbigg9 2/parenrightbiggn −1/parenrightbigg and fork≥nwe haveuk= 9k−nknodd andvk= 9k−nkn+ 9k−neven, which gives M2=1 2∞/summationdisplay k=n1 2k(p(vk)−p(uk)) =1 2∞/summationdisplay k=n1 2k(0−1) =−1 2n. From this we get φs(bn)−φs(an) bn−an≤9n/parenleftbigg3 7·9n/parenleftbigg/parenleftbigg9 2/parenrightbiggn −1/parenrightbigg −1 2n/parenrightbigg =−4 7/parenleftbigg9 2/parenrightbiggn −3 7→ −∞ asn→ ∞ . Henceφ/prime s(t) does not exist. Since t∈(0,1) was arbitrary, φsis nowhere differentiable on (0 ,1). Moreover, since ψs(t) =φs(3t) it is clear that ψsis nowhere differentiable on (0 ,1) as well. Remark. It can quite easily be shown, with similar technique, that both φs andψslacks derivative at both t= 0 andt= 1 as well (cf. Sagan [63]). 51 3.14 Orlicz functions (1947) In 1947 Polish mathematician W/suppress ladys/suppress law Orlicz [50] put forth a slightly dif- ferent approach to continuous functions without derivative. Instead of going for a pure existence proof or a direct construction he presents a form of an intermediate result in terms of both a fairly general construction and Baire category. Orlicz did more research in this area and in two of his subsequent papers (cf. Orlicz [51],[52]) he dealt with a more general form of Lipschitz conditions from which several new constructions of continuous nowhere dif- ferentiable functions sprang to life. Let{fn}be a sequence of functions fn: [a,b]→Rfor which/summationtext∞ n=1|fn(x)| converges uniformly on [ a,b] and let Nbe the metric space of all sequences η={ηn},ηn∈ {0,1}for everyn∈N, with the metric ddefined by d(x,y) =∞/summationdisplay k=11 2k|xk−yk|. It can be seen that ( N,d) is complete, and by Baire’s category theorem (Theorem 4.2) it is therefore of the second category in itself. The metric space terminology used here can be reviewed in Section 4.1. We define the first Orlicz function as: O1(x) =∞/summationdisplay n=1ηnfn(x). Letϕ:R→Rbe a continuous periodic function with period l=b−aand let{αn}and{βn}be sequences of positive numbers such that/summationtext∞ n=1αn<∞ andβ1<β 2<···<β n→ ∞ . We define the second Orlicz function as: O2(x) =∞/summationdisplay n=1ηnαnϕ(βnx). Letψ:R→Rbe a continuous, periodic function satisfying a Lipschitz condition (on R) and lets >1 be a real number. Let also αandβbe real numbers such that α∈(0,1) andαβ > 1. We define the third and fourth Orlicz functions as: O3(x) =∞/summationdisplay n=11 2n2ψ(2sn2x) and O4(x) =∞/summationdisplay n=1αnψ(βnx). 52 Theorem 3.14. (i) If (a)f/prime+ n(x)exists for every x∈[a,b)and is continuous except for possibly on a finite subset of [a,b); (b) there exists λ > 0and a sequence {δn}of positive numbers with δn→0such that /vextendsingle/vextendsingle/vextendsingle/vextendsinglefn(x+h)−fn(x) h−f/prime+ n(x)/vextendsingle/vextendsingle/vextendsingle/vextendsingle>λ for everyx∈[a/prime,b/prime]⊂[a,b) for someh(possibly dependent on xandn) with 0<h<δ nand h<b−x, thenO1has no right-hand derivative at any x∈[a/prime,b/prime]for anyηin a residual subset of N. (ii) If (a)fnsatisfies a Lipschitz condition on [a,b]; (b) There exists a real sequence {kn}withkn→ ∞ such that /vextendsingle/vextendsingle/vextendsingle/vextendsinglefn(x+h)−fn(x) h/vextendsingle/vextendsingle/vextendsingle/vextendsingle≥kn is satisfied for some h(possibly dependent on x) with 0<h<b −x, then for any ηin a residual subset of Nwe have lim sup h→0+/vextendsingle/vextendsingle/vextendsingle/vextendsingleO1(x+h)−O1(x) h/vextendsingle/vextendsingle/vextendsingle/vextendsingle=∞ which implies that there exist no right-hand derivative. (iii) Ifϕis a non-constant function with a continuous derivative everywhere andαnβn>c> 0for everyn∈N, then for each ηin a residual subset ofN,O2has no right-hand derivative. (iv) The third Orlicz function O3is continuous and nowhere differentiable. 53 (v) IfKis the Lipschitz constant for ψand 0<α<1 1 +4σmax|ϕ(x)| crτ,αβ > 1 +2Kσ (1−c)r, wherec∈(0,1)is arbitrary and σ,τandrare suitable11real numbers, then the fourth Orlicz function O4is continuous and nowhere differen- tiable. Proof. The proof of (i) can be found in the proof of Theorem 7 in Orlicz [50], (ii) in the proof of Theorem 8 and (iii) in the proof of Theorem 9. (iv) and (v) are proven in Section 4 of Orlicz [51]. Remark 1. Withψ(x) = cos(x),r= 2,K=π,σ= 1,c= 1/2 and τ= 1/2 in (v) above we get Weierstrass function Wbut with slightly different conditions on αandβ. Remark 2. Orlicz also considered series with sequences /epsilon1from a metric space Econsisting of all sequences /epsilon1={/epsilon1n}, where/epsilon1n∈ {− 1,1}, with the same metric as for ( N,d). Remark 3. In Orlicz [50], Orlicz also gave measure theoretic results on the differentiability of especially O1andO2. The term “almost every” when applied to the metric spaces EandNhas the following meaning: Let /epsilon1n(t) = sgn[sin(2nπt)] (the Rademacher system) and ηn(t) =1 2(1−/epsilon1n(t)). Both {/epsilon1n(t)}and{ηn(t)}are orthonormal sequences in [0 ,1]. By neglecting a countable set in E(orN) and a countable set in the interval [0 ,1] there is a bijective mapping between these two sets. One says “for almost every” sequence in E(orN) if the set of numbers from [0 ,1] for which the sequence {/epsilon1n(t)}(or{ηn(t)}) does not have this property is of measure zero. With this terminology, Orlicz stated and proved the following: (i) Iff/prime+ n(x) exists for almost every x∈[a,b] and lim sup h→0+∞/summationdisplay n=1/parenleftbiggfn(x+h)−fn(x) h−f/prime+ n(x)/parenrightbigg2 >0 for almost every x∈[a,b], then both O1,/epsilon1(t,x) =∞/summationdisplay n=1/epsilon1n(t)fn(x) andO1,η(t,x) =∞/summationdisplay n=1ηn(t)fn(x) 11See Orlicz [51], the existence of suitable constants are given by an existence proof. 54 have no right-hand derivative almost everywhere for almost every t∈ [0,1]. (ii) Ifϕis also absolutely continuous, 0</integraldisplay [a,b][ϕ/prime(x)]2dx<∞and∞/summationdisplay n=1α2 nβ2 n=∞, then both O2,/epsilon1(t,x) =∞/summationdisplay n=1/epsilon1n(t)αnϕ(βnx) andO2,η(t,x) =∞/summationdisplay n=1ηn(t)αnϕ(βnx) have no derivative almost everywhere for almost every t∈[0,1]. 3.15 McCarthy function (1953) McCarthy’s function Mis defined as the infinite series M(x) =∞/summationdisplay k=11 2kg/parenleftBig 22kx/parenrightBig , where g(x) =/braceleftBigg 1 +x,x∈[−2,0] 1−x,x∈[0,2] andg(x+ 4) =g(x) for anyx∈R. John McCarthy [47] writes that this function has the easiest proof of conti- nuity and nowhere differentiabillity of any such function he has seen and I’m inclined to agree that the proof indeed is one of the shorter I have seen. Theorem 3.15. The McCarthy function M(x) =∞/summationdisplay k=11 2kg/parenleftBig 22kx/parenrightBig , where g(x) =/braceleftBigg 1 +x,x∈[−2,0] 1−x,x∈[0,2] andg(x+ 4) =g(x)for anyx∈R, is continuous and nowhere differentiable onR. 55 g(x) x1 −1−4 −2 2 4 (a)g(x) for−5≤x≤5.M(x) x1.0 0.5 −0.51.00.5 (b) McCarthy’s function M. Figure 3.17: McCarthy’s function Mand the auxiliary function g(x). Proof. First we show that Mis continuous on R. Obviously gis continuous and since supx∈R|2−kg(22kx)|= 2−kwith/summationtext∞ k=12−k<∞it follows from the Weierstrass M-test (Theorem 2.2) and the Corollary 2.4 that Mis continuous. Secondly, we show that Mis nowhere differentiable on R. Letx∈Rbe arbitrary but fixed and let n∈Nbe arbitrary. Choose hn=±2−2nwhere the sign is chosen such that xandx+hnare on the same linear segment of g(22nx). Letk∈N, fork>n we have g/parenleftBig 22k(x+hn)/parenrightBig −g/parenleftBig 22kx/parenrightBig =g/parenleftBig 22kx/parenrightBig −g/parenleftBig 22kx/parenrightBig = 0 sinceghas period 4 and 22khn= 4qfor someq∈Z. Fork=nwe obtain /vextendsingle/vextendsingleg/parenleftbig 22n(x+hn)/parenrightbig −g/parenleftbig 22nx/parenrightbig/vextendsingle/vextendsingle=/vextendsingle/vextendsingleg/parenleftbig 1 + 22nx/parenrightbig −g/parenleftbig 22nx/parenrightbig/vextendsingle/vextendsingle= 1. Fork<n we can estimate by sup k=1,...,n−1/vextendsingle/vextendsingle/vextendsingleg/parenleftBig 22k(x+hn)/parenrightBig −g/parenleftBig 22kx/parenrightBig/vextendsingle/vextendsingle/vextendsingle≤22n−12−2n= 2−2n−1 and therefore /vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsinglen−1/summationdisplay k=12−k/parenleftBig g/parenleftBig 22k(x+hn)/parenrightBig −g/parenleftBig 22kx/parenrightBig/parenrightBig/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle≤(n−1)2−2n−1<2n2−2n−1≤1. 56 Hence, /vextendsingle/vextendsingle/vextendsingle/vextendsingleM(x+hn)−M(x) hn/vextendsingle/vextendsingle/vextendsingle/vextendsingle= 22n/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle∞/summationdisplay k=12−k/parenleftBig g/parenleftBig 22k(x+hn)/parenrightBig −g/parenleftBig 22kx/parenrightBig/parenrightBig/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle = 22n/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsinglen/summationdisplay k=12−k/parenleftBig g/parenleftBig 22k(x+hn)/parenrightBig −g/parenleftBig 22kx/parenrightBig/parenrightBig/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle ≥22n/parenleftBigg 1−/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsinglen−1/summationdisplay k=12−k/parenleftBig g/parenleftBig 22k(x+hn)/parenrightBig −g/parenleftBig 22kx/parenrightBig/parenrightBig/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/parenrightBigg ≥22n/parenleftBig 1−2n2−2n−1/parenrightBig = 22n−1/parenleftBig 22n−1−2n/parenrightBig → ∞ asn→ ∞ . It is now clear that the function Mcannot be differentiable at xand since x∈Rwas arbitrary it follows that Mis nowhere differentiable. 3.16 Katsuura function (1991) Hidefumi Katsuura claims in his paper Katsuura [37], published in 1991, that he got the idea for this function when attending a master’s thesis defense about attractors of contraction mappings. We construct the function as follows. Let X= [0,1]×[0,1] be the closed unit square and let F(X) by the collection of all non-empty closed subsets of X. Fori= 1,2,3, define the mappingsTi:X→Xby T1(x,y) =/parenleftbiggx 3,2y 3/parenrightbigg , T2(x,y) =/parenleftbigg2−x 3,1 +y 3/parenrightbigg and T3(x,y) =/parenleftbigg2 +x 3,1 + 2y 3/parenrightbigg . We define the mapping T:F(X)→F(X) byT(A) =T1(A)∪T2(A)∪T3(A). LetD0={(x,x)∈X}(i.e. the diagonal) and for n∈NdefineDn= T(Dn−1). EachDnis the graph of a function Kn: [0,1]→[0,1] and Hidefumi Katsuura’s function KH: [0,1]→[0,1] is the function whose graph Dis the limit of this process. Figure 3.18 shows a few steps of graphically. 57 y x1.0 1.0 (a)D0andX.y x1.0 1.0 (b)D1andT(X). y x1.0 1.0 (c)D2andT2(X).y x1.0 1.0 (d)D3andT3(X). Figure 3.18: The graphs of the first four “iterations” of the Katsuura function and the corresponding mappings of X(the rectangles). 58 A few things about this construction are interesting to consider. It can be shown that the metric space ( F(X),dH) is complete (where dHis the Hausdorff12metric induced by the Euclidean metric) and also that Tis a contraction13mapping on this space (again with respect to the Hausdorff metric). Since these properties are fulfilled, Banach’s fixed point theorem implies that Thas a unique fixed point in F(X) and no matter what set A∈F(X) we start with, the sequence {Tn(A)}always converge to Din the Hausdorff metric. Theorem 3.16. The Katsuura function KHis continuous and nowhere dif- ferentiable on the interval (0,1). The proof is based on Hidefumi’s proof in Katsuura [37]. Proof. First we show that KHis continuous. For m≤n,Dn⊂Tm(X) and Tm(X) is the union of 3mrectangles of height bounded by (2 /3)m. Hence sup x∈[0,1]|Km(x)−Kn(x)| ≤/parenleftbigg2 3/parenrightbiggm →0 asm,n→ ∞ . Thus the sequence {Kn}of functions is uniformly Cauchy and therefore the convergence is uniform by Theorem 2.1. Since each function Knobviously is continuous it follows by Theorem 2.3 that the limit function KHalso is continuous We show that the function KHis nowhere differentiable on (0 ,1) in two steps. (i) Forx∈(0,1) whenxis a ternary rational which has a finite ternary representation. Then, for some n∈N, x=n/summationdisplay k=1xk 3k, where xk∈ {0,1,2}andxn/negationslash= 0. Let the sequence {yk}be defined by yk=x+3−(n+k)fork∈N. Thenyk→x andyk−x= 3−(n+k). 12The Hausdorff metric dHcan be defined by dH(A,B) = max/braceleftbigg sup x∈A/parenleftbigg inf y∈Bd(x,y)/parenrightbigg , sup y∈B/parenleftbigg inf x∈Ad(x,y)/parenrightbigg/bracerightbigg , wheredis the Euclidean metric (in our case). 13Meaning that ∃α∈(0,1) such that ∀x,y∈F(X)dH(Tx,Ty )≤αdH(x,y). 59 We claim that /vextendsingle/vextendsingle/vextendsingle/vextendsingleKH(yk)−KH(x) yk−x/vextendsingle/vextendsingle/vextendsingle/vextendsingle≥2k−1for allk∈N (3.5) which will imply that KHhas no derivative at x. The proof of the claim is by induction on k. Fork= 1 we have, by the geometry of the construction, /vextendsingle/vextendsingle/vextendsingle/vextendsingleKH(y1)−KH(x) y1−x/vextendsingle/vextendsingle/vextendsingle/vextendsingle= 3n+1|KH(y1)−KH(x)| ≥3n+11 3n+1= 1. Assume that equation (3.5) holds for k=q. The geometry of the construction implies that   |KH(yq+1)−KH(x)|=2 3|KH(yq)−KH(x)|, |yq+1−x| =1 3|yq−x|,(3.6) which gives /vextendsingle/vextendsingle/vextendsingle/vextendsingleKH(yq+1)−KH(x) yq+1−x/vextendsingle/vextendsingle/vextendsingle/vextendsingle=(2/3)|KH(yq)−KH(x)| (1/3)|yq−x|≥2·2q−1= 2(q+1)−1 and this completes the proof of the claim. (ii) Now, if x∈(0,1) isn’t a ternary rational with a finite ternary represen- tation, then x=∞/summationdisplay k=1xk 3kwhere infinitely many xkare non-zero. We choose two sequences {yn}and{zn}of ternary rationals with finite ternary representation such that yn< x < z nandzn−yn= 3−(n+q)for someq∈N. We takeqas the smallest element in Nfor which there exists natural numbers r1≤qandr2≤qfor whichxr1/negationslash= 2 andxr2/negationslash= 0 (these ele- ments exists since 0 <t< 1 and are necessary to ensure that our sequences will satisfy yn>0 andzn<1). Define the sequences by yn=n+q/summationdisplay k=1xk 3kandzn=yn+1 3n+q. 60 Then both ynandznare ternary rationals with n+qdigits and clearly 0< y n< t < z n<1. Moreover, that zn−yn= 3−(n+q)is immediate from the definition of the sequences. We claim that /vextendsingle/vextendsingle/vextendsingle/vextendsingleKH(zn)−KH(yn) zn−yn/vextendsingle/vextendsingle/vextendsingle/vextendsingle≥1 for all n∈N (3.7) and prove this by induction. For n= 1 we have /vextendsingle/vextendsingle/vextendsingle/vextendsingleKH(z1)−KH(y1) z1−y1/vextendsingle/vextendsingle/vextendsingle/vextendsingle= 3q+1|KH(z1)−KH(y1)| ≥3q+11 3q+1= 1 by the construction of KH. Assume that equation (3.7) holds for n=k. We consider two cases. First, if zk=zk+1or ifyk=yk+1it follows, similarly as for equation (3.6), that /vextendsingle/vextendsingle/vextendsingle/vextendsingleKH(zk+1)−KH(yk+1) zk+1−yk+1/vextendsingle/vextendsingle/vextendsingle/vextendsingle=(2/3)|KH(zk)−KH(yk)| (1/3)|zk−yk|≥1. where the last inequality is the induction assumption. Secondly, if zk/negationslash=zk+1 andyk/negationslash=yk+1, then the geometry of the construction implies that   KH(zk+1)−KH(yk+1) =−1 3(KH(zk)−KH(yk)), zk+1−yk+1 =1 3(zk−yk)(3.8) and thus /vextendsingle/vextendsingle/vextendsingle/vextendsingleKH(zk+1)−KH(yk+1) zk+1−yk+1/vextendsingle/vextendsingle/vextendsingle/vextendsingle=(1/3)|KH(zk)−KH(yk)| (1/3)|zk−yk|≥1 by the induction assumption. Sincexis not a ternary rational we must have zk/negationslash=zk+1andyk/negationslash=yk+1for infinitely many k. Taking this subsequence of {yk}and{zk}(with the same index to avoid sub-subscripts) it follows from equation (3.8) that KH(zk+1)−KH(yk+1) zk+1−yk+1=−KH(zk)−KH(yk) zk−yk. 61 Hence the only possible limit would be zero, but by equation (3.7) this is not possible. Thus the limit lim n→∞KH(zn)−KH(yn) zn−yn does not exist and therefore KHhas no derivative at xsince this would contradict Lemma 3.8. 3.17 Lynch function (1992) In an article from 1992 (Lynch [43]), Mark Lynch presented a function which is continuous and nowhere differentiable by using a topological argument. As a result, no theorems involving infinite series and uniform convergence were needed. We define the mapping T:R2→Rby (x,y)/mapsto→x(i.e. the projection on the first coordinate). For any x∈Rand anyA⊂R2letA[x] ={y|(x,y)∈A}. We will define a sequence {Cn}of compact sets with Cn+1⊂Cn⊂R2for all n∈Nsuch that (i)T(Cn) = [0,1] for alln∈N; (ii) diam(Cn[x])<1/nfor eachx∈[0,1] andn∈N;14 (iii) for each x∈[0,1] there exists y∈[0,1] with 0<|x−y|<1/nsuch thatp∈Cn[x] andq∈Cn[y] implies that /vextendsingle/vextendsingle/vextendsingle/vextendsinglep−q x−y/vextendsingle/vextendsingle/vextendsingle/vextendsingle>n. We choose the elements in the sequence {Cn}as the closures of band neigh- borhoods of the graph of polygonal arcs defined on [0 ,1]. It is quite easy to see that (i) and (ii) holds (the first is trivial and the second one can be obtained by choosing the thickness of the bands appropriately to compensate 14The diameter diam( A) of a setAis defined by diam(A) = sup x,y∈Ad(x,y). 62 for the steepnes of each segment). To show that (iii) will hold, we start by considering a linear segment of a polygonal arc. Letf(x) =mx+bwith|m|> n and let both δ > 0 andx∈[0,1] be arbitrary. If m>n (the other case, when m<−n, is handled similarly) we choosey=x+δand take a band neighborhood N/epsilon1(f) of the graph of f. For p∈N/epsilon1(f)[x] andq∈N/epsilon1(f)[y] it is obvious from figure 3.19 that |p−q|/|x−y| f(x) =mx+b p q N/epsilon1(f)[x]N/epsilon1(f)[y] x y Figure 3.19: Line segment with band neighborhoods for Lynch’s function. is the absolute value of the slope of the line between the points ( x,p) and (y,q). The minimum of this slope is attained when we choose p=mx+b+/epsilon1 andq=m(x+δ) +b−/epsilon1and for this case we can choose /epsilon1>0 small enough so that /vextendsingle/vextendsingle/vextendsingle/vextendsinglep−q x−y/vextendsingle/vextendsingle/vextendsingle/vextendsingle=/vextendsingle/vextendsingle/vextendsingle/vextendsinglem−2/epsilon1 δ/vextendsingle/vextendsingle/vextendsingle/vextendsingle>n sincem>n by assumption. So, assuming that Cn−1is constructed, we construct Cnin the following man- ner. First, take a polygonal arc Pin the interior of Cn−1where each segment Pnhas a slope whose absolute value exceeds n. For each i∈ {0,1,...k} chooseδisuch that 0<δi<min/braceleftbigg|T(Pi)| 2,1 n/bracerightbigg where |T(Pi)|is the length of the interval T(Pi). From our result above for linear segments we get an /epsilon1i-neighborhood for each Pi(and since we have δi<|T(Pi)|/2 we can always choose y∈T(pi)). Let/epsilon1= min {/epsilon11,...,/epsilon1 k}, then N/epsilon1(P) is a closed neighborhood of Pthat clearly satisfies (iii). Furthermore, if we should happen to be unlucky enough so that N/epsilon1(P)/negationslash⊂Cn−1we can 63 always choose a smaller /epsilon1 >0 so that both N/epsilon1(P)⊂Cn−1and (i)-(iii) are satisfied. We take Cn=N/epsilon1(P). In other words, we construct a sequence of bands that get steeper and nar- rower for each element in the sequence. This results in a “zig-zag” pattern which is transferred onto our function. Theorem 3.17. The sequence {Cn}defines a continuous function L: [0,1]→Rthat is nowhere differentiable on the interval [0,1]. Proof. LetC=/intersectiontext nCn. Since diam( Cn[x])<1/nfor anyx∈[0,1] it follows that diam(C[x]) = 0 for any x∈[0,1] as well. Hence Cis the graph of a well-defined function L: [0,1]→R. Since each Cnis compact (and non- empty) and Cis a nested intersection of {Cn}, it is clear that also Cmust be compact (and non-empty). Thus the graph of Lis compact and therefore Lis continuous. We prove that Lis also nowhere differentiable. Let both x∈[0,1] andδ>0 be arbitrary. We can choose n∈Nso that 1/n < δ . By (iii) there exists y∈[0,1] with 0<|x−y|<1/nsuch thatp∈Cn[x] andq∈Cn[y] implies that /vextendsingle/vextendsingle/vextendsingle/vextendsinglep−q x−y/vextendsingle/vextendsingle/vextendsingle/vextendsingle>n. SinceL(x)∈Cn[x] andL(y)∈Cn[y] the difference quotient /vextendsingle/vextendsingle/vextendsingle/vextendsingleL(x)−L(y) x−y/vextendsingle/vextendsingle/vextendsingle/vextendsingle is unbounded (as we let δ→0). HenceLis not differentiable at x. 3.18 Wen function (2002) The Chinese mathematician Liu Wen has during the last few years proposed several continuous nowhere differentiable functions. One of these is an in- teresting function that is based on an infinite product instead of a series (Wen [80]). Let WL:R→Rbe the function defined by WL(x) =∞/productdisplay n=1(1 +ansin(bnπx)), 64 where the parameters anandbnare chosen such that 0 <a n<1 for alln, ∞/summationdisplay k=1ak<∞ andbn=n/productdisplay k=1pk, andpkis an even integer for all k∈N. Moreover, we require that lim n→∞2n anpn= 0. We show later that WLis both continuous and nowhere differentiable. WL(x) x2.04.0 1.0 2 .0 Figure 3.20: Wen’s function WLwithan= 2−nandpn= 6nforx∈[0,2]. In two other articles, from 2000 (Wen [78]) and 2001 (Wen [79]), Liu Wen presented two other continuous nowhere differentiable functions. Both are based on an expansion of the real numbers in [0 ,1] in a different base than the usual base-10 representation. In the first article the base- brepresentation is used (where b∈N\{1}). The construction of this function is done as follows. Let b≥2 be an integer and forx∈[0,1] let (x1x2···)bbe the base- bexpansion of x, i.e. x=∞/summationdisplay k=1xk bkwherexk∈ {0,1,...,b −1}. Letλ>1 be a real number and define the sequence {un}byu1= 1 and for n>1 let un=/braceleftBigg un−1 ifxn=xn−1, φ(un−1) ifxn/negationslash=xn−1 65 whereφis a function that is chosen so that f1is continuous and nowhere differentiable when f1: [0,1]→Ris defined by f1(x) =∞/summationdisplay k=1uk λk. In this article, Liu choose φ(u) = (1 −λ)(u−c) wherecis any real constant. This article also presents a proof that f1is right-continuous but lacks a finite right-hand derivative (which can be extended similarly to the left-hand side). In figure 3.21(a) an example (with fixed parameters) of f1is shown graphically. The second article uses the Cantor series representation to construct a func- tion. Letqn≥2 be an integer for all nand letx∈[0,1]. Then the Can- tor series expansion of xis defined as x=∞/summationdisplay n=1xn q1q2···qnwherexn∈ {0,1,...,q n−1}. The function f2: [0,1]→Ris expressed as f2(x) =∞/summationdisplay n=1un n(n+ 1), where the sequence {un}is defined as u1= 1 and for n∈N un+1=  −un n,if (xn+1= 0 andxn/negationslash= 0) or if (xn+1=qn+1−1 andxn/negationslash=qn−1), un, else. In Wen [79] it is shown that f2is well-defined and is right continuous but lacks right-hand derivative on [0 ,1) (the argument can be done similarly for the left-hand side). We turn to prove that the function that was based on an infinite product, WL, is continuous and nowhere differentiable on R. Theorem 3.18. The Wen function WLis continuous and nowhere differen- tiable on R. The proof follows Liu Wen’s proof in Wen [80] but is a bit more explicit. 66 f1(x) x0.5 0.5 1 .0 (a)f1withb= 3,λ= 3 andc= 1/10.f2(x) x1.0 0.5 1 .0 (b)f2. Figure 3.21: Two of Liu Wen’s functions with 0 ≤x≤1. Proof. We start by establishing the continuity of WL. The following well- known inequality is needed, x 1 +x≤ln(1 +x)≤xifx>−1. (3.9) Leta= max n≥1an. By the restrictions on anit is clear that 0 < a < 1 so from the inequality (3.9) we get |ln(1 +ansin(bnπx))| ≤an|sin(bnπx)|max/braceleftbigg1 |1 +ansin(bnπx)|,1/bracerightbigg ≤anmax/braceleftbigg1 1−a,1/bracerightbigg ≤an 1−a. Since/summationtext∞ k=1ak<∞it follows from the Weierstrass’ M-test (Theorem 2.2) and the Corollary 2.4 that ∞/summationdisplay k=1ln(1 +ansin(bnπx)) converges to a continuous function and therefore WL(x) =∞/productdisplay n=1(1 +ansin(bnπx)) = exp/parenleftBigg∞/summationdisplay k=1ln(1 +ansin(bnπx))/parenrightBigg is also continuous. 67 We turn to prove that WLis nowhere differentiable. For every x∈Rthere exists a sequence {Nn}withNn∈Zsuch that x∈/bracketleftbiggNn bn,Nn+ 1 bn/parenrightbigg for alln∈N. Define the sequences {yn}and{zn}by yn=Nn+ 1 bnandzn=Nn+ 3/2 bn. Clearlyx<y n<znand 0<zn−x<3/(2bn). Moreover, zn−yn= 1/(2bn) and also zn−x=Nn+ 3/2 bn−x≤Nn+ 3/2 bn−Nn bn=3 2bn= 3(zn−yn). From the relations above we have the inequalities zn−yn≥1 3(zn−x)>1 3(yn−x). (3.10) We definea,bandLn:R→Ras a=∞/productdisplay k=1(1−ak),b=∞/productdisplay k=1(1 +ak) andLn(x) =n/productdisplay k=1(1 +aksin(bkπx)). We will consider the expression ∆n=WL(zn)−WL(yn) =∞/productdisplay k=1(1 +aksin(bkπzn))−∞/productdisplay k=1(1 +aksin(bkπyn)). First, fork >n it is obvious that bk/bnis an even integer. Thus, for k >n and someqk∈Z, we have sin(bkπyn) = sin/parenleftbiggbk bnπ(Nn+ 1)/parenrightbigg = sin(2qk(Nn+ 1)π) = 0 and sin(bkπzn) = sin/parenleftbiggbk bnπ(Nn+ 3/2)/parenrightbigg = sin(3qkπ) = 0. 68 Moreover, for k=nwe obtain sin(bnπyn) = sin(π(Nn+ 1)) = 0 and sin(bnπzn) = sin(π(Nn+ 3/2)) = −(−1)Nn. With these equalities in mind we can rewrite ∆ nas ∆n=Ln−1(zn)(1 +ansin(bnπzn))−Ln−1(yn)(1 +ansin(bnπyn)) =Ln−1(zn)−Ln−1(yn)−(−1)NnanLn−1(zn). Now, fork<n we have |aksin(bkπzn)−aksin(bkπyn)|= 2ak/vextendsingle/vextendsingle/vextendsingle/vextendsinglesin/parenleftbigg bkπzn−yn 2/parenrightbigg cos/parenleftbigg bkπzn+yn 2/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle ≤ak|bkπ(zn−yn)|=akbkπ 2bn<π 2pn so there exists σk∈Rwith|σk|<π/ (2pn)<1 such that aksin(bkπzn) =aksin(bkπyn) +σk. Now we can estimate ∆ n, but first we need the following bound |Ln−1(zn)−Ln−1(yn)|=/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsinglen−1/productdisplay k=1[(1 +aksin(bkπzn)) +σk] −n−1/productdisplay k=1[1 +aksin(bkπyn)]/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle =/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle2(n−1)−1/summationdisplay i=1σli/parenleftBigg/productdisplay j∈Iiσj/parenrightBigg/parenleftBigg/productdisplay j∈Ji(1 +ajsin(bjπyn))/parenrightBigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle ≤2(n−1)−1/summationdisplay i=1|σli|/parenleftBigg/productdisplay j∈Ii|σj|/parenrightBigg/parenleftBigg/productdisplay j∈Ji|1 +ajsin(bjπyn)|/parenrightBigg ≤π 2pn2(n−1)−1/summationdisplay i=1/parenleftBigg/productdisplay j∈Ji|1 +aj|/parenrightBigg ≤bπ 2pn(2n−1−1)≤bπ pn2n−2 69 where, for 0 < i < n ,Ii⊂NandJi⊂Nare some index sets and lisome index (we also adhere to the convention that/producttext j∈∅xj= 1). Thus we can find a lower bound for |∆n|by |∆n|=|Ln−1(zn)−Ln−1(yn)−(−1)NnanLn−1(zn)| ≥anLn−1(zn)− |Ln−1(zn)−Ln−1(yn)| ≥ana−bπ pn2n−2=an/parenleftbigg a−2n−2 anpnbπ/parenrightbigg where the last inequality follows from the bound above and the fact that a<L n(x)<b. Now, since lim n→∞2n/(anpn) = 0 by assumption (which also implies that anbn→ ∞ sinceanbn≥anpn), we have lim n→∞/vextendsingle/vextendsingle/vextendsingle/vextendsingleWL(zn)−WL(yn) zn−yn/vextendsingle/vextendsingle/vextendsingle/vextendsingle= lim n→∞|2bn∆n| ≥lim n→∞/vextendsingle/vextendsingle/vextendsingle/vextendsingle2anbn/parenleftbigg a−2n−2 anpnbπ/parenrightbigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle=∞ ·a=∞. By the triangle inequality and inequality (3.10) we can estimate /vextendsingle/vextendsingle/vextendsingle/vextendsingleWL(zn)−WL(yn) zn−yn/vextendsingle/vextendsingle/vextendsingle/vextendsingle≤|WL(zn)−WL(x)| zn−yn+|WL(yn)−WL(x)| zn−yn ≤3|WL(zn)−WL(x)| zn−x+3|WL(yn)−WL(x)| yn−x. If we letn→ ∞ it is clear that WLis not differentiable at x. Sincex∈R was arbitrary it follows that WLis nowhere differentiable. 70 Chapter 4 How “Large” is the Set ND[a,b] From the previous chapter it is clear that there exists continuous nowhere differentiable functions but how many are there? A simple answer would be “infinitely many” but could we perhaps say something else about the size of the set of continuous nowhere differentiable functions? As a matter of fact we can. One such way is based on topology in metric spaces and for this we will need some definitions and theorems. But first, where does all continuous nowhere differentiable functions live? We refer to this place as ND[a,b], or more exactly as in the following definition. Definition 4.1. LetND[a,b](a < b ) be the set of all continuous nowhere differentiable functions f: [a,b]→R. 4.1 Metric spaces and category We collect a few ideas from the theory of metric spaces, starting with defining exactly what we mean by a metric space. Definition 4.2. A metric space is a pair (X,d)of a setXand a metric d defined on X. A metric d:X×X→[0,∞)is a mapping that, for any x,y,z ∈X, satisfies (i)d(x,y)≥0is a real number; (ii)d(x,y) = 0 if and only if x=y; (iii)d(x,y) =d(y,x); 71 (iv)d(x,y)≤d(x,z) +d(z,y). A metric space (X,d)is said to be complete if every Cauchy sequence in X converges. That is, if {xn}is a Cauchy sequence in (X,d), i.e. ∀/epsilon1>0∃N∈N∀m,n≥Nd(xm,xn)</epsilon1, then there exists x∈Xsuch that lim n→∞d(xn,x) = 0 . Remark. We often write Xinstead of ( X,d) when the metric is implicit. A metric space is a very general construction, a bit too general for our ap- plication, so we will need to make some more restrictions. We introduce the concept of a normed vector space. Definition 4.3. A normed space is a pair (X,/bardbl · /bardbl)of a vector space Xand a norm /bardbl · /bardbl defined onX. A norm /bardbl · /bardbl:X→[0,∞)is a mapping that for anyx,y∈Xand anyα∈R(orCwhenXis a complex space) satisfies (i)/bardblx/bardbl ≥0is a real number; (ii)/bardblx/bardbl= 0 if and only if x= 0; (iii)/bardblαx/bardbl=|α|/bardblx/bardbl; (iv)/bardblx+y/bardbl ≤ /bardblx/bardbl+/bardbly/bardbl. A Banach space (X,/bardbl·/bardbl)is a normed space which is complete seen as a metric space (X,d)with the metric dinduced by the norm, that is for x,y∈X d(x,y) =/bardblx−y/bardbl. Remark. As with metric spaces, we often write Xinstead of ( X,/bardbl·/bardbl) when the norm is implicit. Our results in the next section will be presented in a specific normed space, namely the vector space of all continuous functions with supremum norm. Definition 4.4. LetC[a,b](a<b ) be the normed (real) vector space of all continuous functions f: [a,b]→Rwith the supremum norm, i.e. /bardblf/bardbl= sup x∈[a,b]|f(x)|. 72 Remark. Clearly ND[a,b]⊂C[a,b] properly. As it turns out, this normed space is actually a Banach space. Theorem 4.1. The spaceC[a,b]of all real-valued (or complex-valued) con- tinuous functions on [a,b]with the supremum norm, /bardblf/bardbl= sup x∈[a,b]|f(x)|, is a Banach space. Proof. See Kreyszig [42], pages 36-37. To get some understanding of the “size” of subsets of a metric space we start by giving a definition of some topological properties. We will later establish that the set ND[a,b] is of the second category (actually, what we will show is that it is residual). Definition 4.5. We say that a set Min a metric space Xis (i)nowhere dense if the closure Mcontains no non-empty open sets, (ii) of the first category if M=∞/uniondisplay k=1Mk, where each Mkis nowhere dense (in X), (iii) of the second category ifMis not of the first category. A set which is a complement (in X) of a set of the first category is called residual and a property that holds on a residual set is called a (topologically) generic property. We will rely heavily on the following theorem when proving that the set ND[a,b] is of the second category. This will be possible since we know that C[a,b] is complete. Theorem 4.2 (Baire’s Category Theorem). If a metric space X/negationslash=∅is complete it is of the second category in itself. Proof. See Kreyszig [42], pages 247-248. 73 Remark. The formulation of Theorem 4.2 is equivalent with the following: IfX/negationslash=∅is a complete metric space and X=∞/uniondisplay k=1Mk where each Mkis closed, then at least one Mk0contains a nonempty open subset. The equivalence is obvious since (i), if no Mk(=MksinceMkis closed) contains a non-empty open subset then Xwould be of the first category in itself and (ii), if Xis of the second category in itself we cannot write Xas a countable union of nowhere dense sets (hence there is a non-empty open subset in some Mk). 4.2 Banach-Mazurkiewicz theorem We can get some topological results on the size of ND[0,1] from a theorem that was originally done in 1931 by Banach (cf. Banach [2]) and Mazurkiewicz (cf. Mazurkiewicz [45]). In 1929, H. Steinhaus posed the question “of what category is the set of all continuous nowhere differentiable functions in the space of all continuous functions” in his paper Steinhaus [72], pp. 81. This as a reaction to his statement (in the same paper, pp. 63) that the set of all 2 π- periodic continuous nowhere differentiable functions is of the second category seen as a subset of all 2 π-periodic continuous functions (with supremum norm). The papers of Banach and Mazurkiewicz gives an answer to Steinhaus question. The Banach-Mazurkiewicz theorem is based on Baire’s category theorem (Theorem 4.2) which states that a complete metric space is of the second category in itself. The proof presented here is largely due to Oxtoby [53] and we need the following lemma. Lemma 4.3. The set P[a,b]of all piecewise linear continuous functions defined on the interval [a,b]is dense in C[a,b]. Proof. Letg∈C[a,b] be arbitrary but fixed. Put hnas the piecewise linear function on the partition Pn:a=t0<t 1<···<tn=bdefined by hn(x) =g(ti)ti+1−x ti+1−ti+g(ti+1)x−ti ti+1−ti,x∈[ti,ti+1]. 74 Clearlyhn∈ P[a,b] for every partition Pn. Let/epsilon1>0 be given, we show that /bardblg−hn/bardbl</epsilon1for some partition Pn. The function gis continuous on [ a,b], i.e. ∀/epsilon1>0∃δ>0 such that |x−x0|<δ⇒ |Sn(x)−Sn(x0)|</epsilon1 4. Choose the partition Pnso that max i=0...n−1|ti+1−ti|<δ. Forx∈[ti,ti+1] we have |ti+1−ti|<δand |g(x)−hn(x)|=/vextendsingle/vextendsingle/vextendsingle/vextendsingleg(x)−1 ti+1−ti[ti+1g(ti)−tig(ti+1) +x(g(ti+1)−g(ti))]/vextendsingle/vextendsingle/vextendsingle/vextendsingle =/vextendsingle/vextendsingle/vextendsingle/vextendsingleg(x)−ti+1g(ti)−tig(ti+1) ti+1−ti−xg(ti+1)−g(ti) ti+1−ti/vextendsingle/vextendsingle/vextendsingle/vextendsingle =/vextendsingle/vextendsingle/vextendsingle/vextendsingleg(x)−g(ti+1)−ti+1−x ti+1−ti(g(ti)−g(ti+1))/vextendsingle/vextendsingle/vextendsingle/vextendsingle ≤ |g(x)−g(ti+1)|+/vextendsingle/vextendsingle/vextendsingle/vextendsingleti+1−x ti+1−ti/vextendsingle/vextendsingle/vextendsingle/vextendsingle|g(ti+1)−g(ti)| ≤/epsilon1 4+ 1·/epsilon1 4=/epsilon1 2. Now, /bardblg−hn/bardbl ≤ max i=0,...,n−1/parenleftBigg sup x∈[ti,ti+1]|g(x)−hn(x)|/parenrightBigg ≤/epsilon1 2</epsilon1 and we are done. Remark From the lemma above and the construction in Section 3.17 it is clear that ND[a,b] is dense in C[a,b]. Consider the following (for the interval [0,1]): Letg∈C[0,1] and/epsilon1>0 be arbitrary but fixed. Let Pbe a polygonal arc (a piecewise linear function) within /epsilon1/2 ofg. This is no problem because of what we just established in Lemma 4.3. We can, as in Section 3.17, construct Cn⊂N/epsilon1/2(P) that satisfies conditions (i)-(iii) in said section. In the same manner as in Theorem 3.17,/intersectiontext nCndefines a well-defined, continuous and nowhere differentiable function on [0 ,1] that clearly is within /epsilon1/2 ofPand henceforth within /epsilon1ofg. Thus ND[0,1] is dense in C[0,1]. 75 The main theorem of this section states that in the normed (real) vector space of all real-valued continuous functions on [ a,b] with the supremum norm, nowhere differentiability is a topologically generic property. This implies that the set ND[a,b] is of the second category in C[a,b]. Theorem 4.4 (Banach-Mazurkiewicz Theorem). The set ND[a,b]of all nowhere differentiable continuous functions on [a,b]is of the second cat- egory inC[a,b]. Proof. It is enough to prove the theorem for [ a,b] = [0,1]. Let En=/braceleftbigg f/vextendsingle/vextendsingle/vextendsingle/vextendsingle∃x∈/bracketleftbigg 0,1−1 n/bracketrightbigg s.t.∀h∈(0,1−x)|f(x+h)−f(x)| ≤nh/bracerightbigg wheref∈C[0,1]. We show that the sets Enare closed for all n∈N. Takef∈En, then ∃fk∈Ensuch thatfk→funiformly on [0 ,1]. Since fk∈En,∃xk∈[0,1−1 n] for every k∈Nby the definition of En. The sequence {xk}is clearly bounded so by the Bolzano-Weierstrass theorem it has a convergent subsequence, say {xkl}, that converges to some element x∈[0,1−1 n]. Let {fkl}be the corresponding subsequence of {fk}. By the construction, |fkl(xkl+h)−fkl(xkl)| ≤nhfor all 0< h < 1−xkl. Since xkl→xand 0<h< 1−xwe can always choose some l0∈Nlarge enough so that 0<h< 1−xklforl>l 0. Then (for llarge enough) |f(x+h)−f(x)| ≤ |f(x+h)−f(xkl+h)|+|f(xkl+h)−fkl(xkl+h)| +|fkl(xkl+h)−fkl(xkl)|+|fkl(xkl)−f(xkl)| +|f(xkl)−f(x)| ≤ |f(x+h)−f(xkl+h)|+/bardblf−fkl/bardbl+nh+/bardblfkl−f/bardbl +|f(xkl)−f(x)|. If we letl→ ∞ then the continuity of fatxandx+hand the convergence offkl(in the norm) gives the inequality |f(x+h)−f(x)| ≤nhfor every 0<h< 1−xand thusf∈En. HenceEnis closed. Now consider the set P[0,1] of all piecewise linear continuous functions on the interval [0 ,1]. This set is dense in C[0,1] by Lemma 4.3. The setsEnare nowhere dense if we show that for any g∈ P[0,1] and any /epsilon1>0 there exists h∈C[0,1]\Ensuch that /bardblg−h/bardbl</epsilon1. 76 Let/epsilon1 >0 be given and let Mbe the maximum slope of any “piece” of g. Choosem∈Nsuch thatm/epsilon1>n +M. Letφ(x) = inf k∈Z|x−k|(“saw-tooth” function, see figure 3.13 as well) and take h(x) =g(x) +/epsilon1φ(mx). Clearly h∈C[0,1]. Then, for all x∈[0,1),h(x) has a right-hand side derivative, h/prime+(x), such that |h/prime+(x)|=|g/prime+(x) +/epsilon1mφ/prime+(mx)|>n since we have chosen m/epsilon1>n +M. Henceh∈C[0,1]\En. We also have /bardblg−h/bardbl= sup x∈[0,1]|g(x)−(g(x) +/epsilon1φ(mx))|=/epsilon1sup x∈[0,1]|φ(mx)|=/epsilon1 2</epsilon1 and thusEnis clearly nowhere dense in C[0,1]. SinceEnis nowhere dense, we see that E=/uniontext∞ k=1Ekis of the first category inC[0,1]. This is the set of all elements in C[0,1] with bounded right hand difference quotients at some point x∈[0,1] (i.e. the complement to E in C[0,1] does not possess a finite right-hand derivative anywhere in [0 ,1]). SinceC[0,1] is complete and thus by Baire’s theorem (Theorem 4.2) of the second category it is clear that the set of functions in C[0,1] which are nowhere differentiable constitutes a set of the second category. Remark 1. Banach and Mazurkiewicz did not prove exactly the same thing in their respective articles, however their results coincide when formulated as in the theorem above. Mazurkiewicz shows that the set of continuous functions which have a bounded one-sided derivative at some point is of the first category while Banach proved that the set of functions which have a bounded Dini-derivative1at some point is of the first category. This makes the theorem of Banach stronger than Mazurkiewicz’s similar result. Remark 2. What was shown in the proof of the theorem above is that the set of continuous functions that have a finite right-hand derivative at some pointx∈[0,1] is of the first category. It can similarly be shown that the subset ofC[0,1] having a finite left-hand derivative at some point x∈[0,1] is of the first category. Thus the subset of C[0,1] consisting of functions with a finite one-sided derivative at some point is also of the first category. 1The four Dini-derivatives are defined as one-sided derivatives with limes superior and limes inferior instead of only limes. 77 From the second remark above, the following question arise: what about the set of continuous functions without finite orinfinite one-sided derivative ev- erywhere? Saks solved the question in a paper published in 1932 (Saks [65]). He proved that the set of continuous functions which have a finite or infinite right-hand derivative at some point is of the second category. This is the complement of the set above so that set is of the first category. The first example of such a function wasn’t constructed until 1922 when Besicovitch managed the feat (and published the function in 1924, cf. Besicovitch [4]2). These types of functions are usually referred to as functions of the Besicovitch type. 4.3 Prevalence of ND[0,1] Prevalence3is a concept that can be used when one is interested in a measure theoretic result of how “large” a set in an infinite dimensional vector space is. It enables us to use terms such as “almost every” and “measure zero” on these spaces (without a specific measure like, for example, the Wiener measure). Its development was partially motivated by wanting to keep some of the properties that the Lebesgue measure on finite dimensional spaces possess, one of which is the translation invariance. Prevalence is a more useful property than topological properties like category and denseness when a probabilistic result on the likelihood of a given property is desired. This in part due to the fact that it actually turns out that a property that is topologically generic in Rncan have very low probability4(and also that a first category set can contain almost every [Lebesgue] point in the space). In Hunt, Sauer and Yorke [31] there are a few examples of this phenomenon as well as detailed development of the concept of prevalence. We borrow a few definitions from this paper and show that ND[0,1] constitutes a prevalent set inC[0,1]. The main part of this section is gathered from Hunt [30]. Definition 4.6. LetXbe a complete metric vector space. A measure µis said to be transverse to a Borel set S⊂Xif the following conditions hold. 2In 1928, E. Pepper published an article (Pepper [55]) about functions of the Besicovitch type where he produced the same function as Besicovitch but with simpler reasoning. 3After the publication of Hunt, Sauer and Yorke it became clear that this was closely related to another concept, more specifically, that so called shy sets are very closely related to the notion of a Haar zero set for Abelian polish groups (cf. Hunt, Sauer and Yorke [32]) 4Actually, in some cases, probability equal to zero. 78 (i) There exists a compact set K⊂Xfor which 0<µ(K)<∞. (ii)µ({x+s|s∈S}) = 0 for everyx∈X. Definition 4.7. A Borel set S⊂Xis called shy if there exists a measure transverse to S. If a setWis contained in a shy Borel set then Wis also said to be shy. The complement of a shy set is called a prevalent set. Definition 4.8. We call a finite-dimensional subspace P⊂C[0,1]aprobe for a setS⊂C[0,1]if Lebesgue measure supported on Pis transverse to a Borel set which contains Sc=C[0,1]\S. The following inequality is central for our main result. Lemma 4.5. Letg(x) =/summationtext∞ k=11 k2cos(2kπx)andh(x) =/summationtext∞ k=11 k2sin(2kπx). Then there exists c>0such that for every α,β∈Rand any closed interval I⊂[0,1]with length /epsilon1≤1 2, sup x∈I(αg(x) +βh(x))−inf x∈I(αg(x) +βh(x))≥c/radicalbig α2+β2 (log/epsilon1)2. Proof. Letf=αg+βh. Then, for some θ∈[0,2π], f(x) =∞/summationdisplay k=11 k2/parenleftbig αcos(2kπx) +βsin(2kπx)/parenrightbig =/radicalbig α2+β2∞/summationdisplay k=11 k2cos(2kπx+θ). We may assume that α2+β2= 1 without loss of generality. Let Ibe some closed interval in [0 ,1] with length 2−m, wherem∈N. We claim that for any continuous function f, sup x∈If(x)−inf x∈If(x)≥sup j∈N2mπ/integraldisplay If(x) cos(2m+jπx+θ)dx. (4.1) We may assume that supx∈If(x) =−infx∈If(x) =Kfor someK≥0 since adding a constant to both sides of (4.1) does not change the inequality (since/integraltext Icos(2m+jπx)dx= 0). Then |f| ≤1 onIand hence 2mπ/integraldisplay If(x) cos(2m+jπx+θ)dx≤2mπ/integraldisplay IK|cos(2m+jπx+θ)|dx = 2mπK2 π2−m= 2K, 79 which is equivalent to (4.1). We then have, for fdefined as above and for anyj∈N, that sup x∈If(x)−inf x∈If(x)≥2mπ/integraldisplay I∞/summationdisplay k=11 k2cos(2kπx+θ) cos(2m+jπx+θ)dx =∞/summationdisplay k=12mπ k2/integraldisplay Icos((2m+j−2k)πx) + cos((2m+j+ 2k)πx+ 2θ) 2dx. (4.2) SinceIhas length 2−m,/integraltext Icos((2m+j±2k)πx+ϕ)dx= 0 whenever k >m , except when k=m+j(with the “-” sign). For k≤m, letω=±2kand let ybe the left endpoint of I. Then /integraldisplay Icos((2m+j+ω)πx+ϕ)dx =sin((2m+j+ω)π(y+ 2−m) +ϕ)−sin((2m+j+ω)πy+ϕ) (2m+j+ω)π =sin((2m+j+ω)πy+ϕ+ 2−mπω)−sin((2m+j+ω)πy+ϕ) (2m+j+ω)π ≥ −|2−mπω| (2m+j+ω)π=−|ω| 2m(2m+j+ω). It then follows from (4.2) that sup x∈If(x)−inf x∈If(x)≥π 2(m+j)2−m/summationdisplay k=1π 2k2/parenleftbigg2k 2m+j−2k+2k 2m+j+ 2k/parenrightbigg ≥π 2(m+j)2−π 2m(2j−1)m/summationdisplay k=12k k2. (4.3) Now we claim that m/summationdisplay k=12k k2≤52m m2(4.4) for allm∈N. Form= 1,2,3,4 it can quite easily be seen to hold and for m≥4 we prove by induction. Assume that equation (4.4) holds for m=n. 80 Letm=n+ 1, n+1/summationdisplay k=12k k2≤52n n2+2n+1 (n+ 1)2=/parenleftbigg5(n+ 1)2 2n2+ 1/parenrightbigg2n+1 (n+ 1)2 ≤/parenleftbigg125 32+ 1/parenrightbigg2n+1 (n+ 1)2≤52n+1 (n+ 1)2 and the claim follows by induction on m. This gives a new estimate for (4.3), sup x∈If(x)−inf x∈If(x)≥π 2(m+j)2−5π (2j−1)m2. We letj= 10 and assume that m≥2. Then sup x∈If(x)−inf x∈If(x)≥π 2(m+j)2−5π (2j−1)m2 ≥π 2(6m)2−π 200m2=2π 225m2. Finally, ifI⊂[0,1] has arbitrary length /epsilon1≤1/2, choose an m≥2 such that 21−m≥/epsilon1 >2−m. Then, for any closed subinterval J⊂Iwith length 2−m, we have sup x∈If(x)−inf x∈If(x)≥sup x∈Jf(x)−inf x∈Jf(x) ≥2π 225m2≥π 450(m−1)2≥(log 2)2π 450(log/epsilon1)2, which proves the lemma. It turns out that we can’t work directly with the set ND[0,1] (since it’s not a Borel set, which was proved by Mazurkiewicz [46] in 1936, see Mauldin [44]) so we will instead consider the set of nowhere Lipschitz functions. As we shall see, this set is actually a subset of the set of continuous nowhere differentiable functions and the results we prove in this section will therefore hold for a class of functions that is actually smaller than ND[0,1]. Definition 4.9. A function f∈C[a,b]is said to be M-Lipschitz at x∈[a,b] if ∃M > 0such that ∀y∈[a,b]|f(x)−f(y)| ≤M|x−y|. 81 We define NL M[a,b]as the set of nowhere M-Lipschitz functions on [a,b], i.e. NL M[a,b] ={f∈C[a,b]|∀x∈[a,b]∀y∈[a,b]|f(x)−f(y)|>M|x−y|}. We collect some properties of these sets of nowhere Lipschitz functions. Lemma 4.6. Let NL[a,b] =/intersectiondisplay M∈NNL M[a,b], i.e. the set of all nowhere Lipschitz functions. Then the following properties hold. (i)NL M[a,b]is an open set for every M∈N. (ii)NL[a,b]is a Borel set. (iii)NL[a,b]⊂ ND [a,b]. Proof. It is enough to prove the theorem for [ a,b] = [0,1]. (i) LetM∈Nbe arbitrary. Take f∈C[0,1]\ NL [0,1]M, then ∃fn∈ C[0,1]\ NL M[0,1] such that fn→funiformly on [0 ,1]. For every n∈N, there exists xn∈[0,1] such that fnisM-Lipschitz at xn. That is, ∀y∈[0,1]|fn(xn)−fn(y)| ≤M|xn−y|. The sequence {xn}is bounded so by the Bolzano-Weierstrass theorem there exists a convergent subsequence {xnk}, sayxnk→x∈[0,1]. Lety∈[0,1] be arbitrary, |f(x)−f(y)| ≤ |f(x)−f(xnk)|+|f(xnk)−fnk(xnk)| +|fnk(xnk)−fnk(y)|+|fnk(y)−f(y)| ≤ |f(x)−f(xnk)|+/bardblf−fnk/bardbl+M|xnk−y|+/bardblfnk−f/bardbl →M|x−y|ask→ ∞ . Hencefis M-Lipschitz at xand henceforth f∈C[0,1]\ NL M[0,1]. Thus C[0,1]\ NL M[0,1] is closed and therefore NL M[0,1] is open. (ii) This is obvious from the definition of a Borel set and (i). (iii) Takef∈ NL [0,1]. Then for every M∈Nand for every x,y∈[0,1] |f(x)−f(y)|>M|x−y| ⇒|f(x)−f(y)| |x−y|>M 82 and so the difference quotient is unbounded for all x,y∈[0,1]. Hencefhas no derivative and thus f∈ ND [0,1]. Now follows the two main theorems of this section. The first ensures the existence of a probe and the second proves the desired prevalence of ND[0,1]. Theorem 4.7. There exists g,h∈C[0,1]such that for all f∈C[0,1] m/parenleftBigg R2\/braceleftBigg (λ,ν)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle(λ,ν)∈R2,(f+λg+νh)∈/intersectiondisplay m∈NNL m[0,1]/bracerightBigg/parenrightBigg = 0, wheremis the Lebesgue measure on R2. Proof. Takegandhas in Lemma 4.5. That both functions are continuous is shown similarly as for Weierstrass’ function in Section 3.4. Let f∈C[0,1] be arbitrary and put S=/braceleftbig (α,β)∈R2|f+αg+βhis Lipschitz at some x∈[0,1]/bracerightbig . We want to show that Shas Lebesgue measure zero. Let SM=/braceleftbig (α,β)∈R2|f+αg+βhis M-Lipschitz at some x∈[0,1]/bracerightbig . From this definition it is clear that S=/uniontext M∈NSM. So ifmis the Lebesgue measure on R2and we show that m(SM) = 0 for each M∈Nit is clear that m(S) = 0 (by the countable sub-additivity of m). LetN∈N\ {1}and cover [0 ,1] byNclosed intervals of length /epsilon1=1 N. Let Ibe anyone of those intervals and put JI={(α,β)∈SM|f+αg+βhis M-Lipschitz at some x∈I}. Let (α1,β1),(α2,β2)∈JIbe arbitrary. Let fi=f+αig+βihandxi∈Ibe an M-Lipschitz point for fiwherei= 1,2. Then sup x∈I|fi(x)−fi(xi)| ≤sup x∈IM|x−xi| ≤M/epsilon1 which gives sup x∈I|f1(x)−f2(x)−[f1(x1)−f2(x2)]| ≤sup x∈I|f1(x)−f1(x1)| + sup x∈I|f2(x)−f2(x2)| ≤2M/epsilon1. 83 This gives sup x∈I(f1(x)−f2(x))−inf x∈I(f1(x)−f2(x))≤4M/epsilon1 and sincef1−f2= (α1−α2)g+ (β1−β2)h, Lemma 4.5 gives the bound /radicalbig (α1−α2)2+ (β1−β2)2≤4M/epsilon1log(/epsilon1)2 cI. Letc= min I{cI}. Since the points were arbitrary, it follows that JIis enclosed by a disk of radius4M c/epsilon1log(/epsilon1)2(for all the intervals I). It now follows that SMcan be covered by N=1 /epsilon1such discs, giving that the total area of the covering is bounded by A=π/parenleftbigg4M c/epsilon1(log/epsilon1)2/parenrightbigg21 /epsilon1=π16M2 c2/epsilon1(log/epsilon1)4. As/epsilon1→0,A→0. HenceSMhas measure zero and thus Shas measure zero. Theorem 4.8. Almost every function in C[0,1]is nowhere differentiable; that is, ND[0,1]is a prevalent subset of C[0,1]. Proof. LetNL[0,1] be the set of all nowhere Lipschitz functions. By The- orem 4.7 there exists a probe P(spanned by gandh) forNL[0,1]. This is clear since, by the conclusion of said theorem, Lebesgue measure supported onPis transverse to NL[0,1]c=C[0,1]\ NL [0,1] (which is a Borel set by Lemma 4.6(ii) since NL[0,1] is a Borel set and obviously there exists a compact set K⊂C[0,1] such that 0 <m(K)<∞(wheremis the Lebesgue measure)). Hence NL[0,1]cis shy and therefore NL[0,1] is a prevalent set. 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Monthly 109(2002), 378–380. 91 Index of Names Andr´ e Marie Amp` ere ( 1775–1836 ), 4 Stefan Banach ( 1892–1945 ), 74 Abram Besivocitch ( 1891–1970 ), 78 Bernhard Bolzano ( 1781–1848 ),11 Carl Borchardt ( 1817–1880 ), 21 Charles Cell´ erier (1818–1890) ,17 Gaston Darboux ( 1842–1917 ),28 Georges de Rham ( 1903–1990 ), 36 Ulisse Dini ( 1845–1918 ), 25 Paul du Bois-Reymond ( 1831–1889 ), 21 Georg Faber ( 1877–1966 ),41 Hermann Hankel ( 1839–1873 ), 29 Godfrey Harold Hardy ( 1877–1947 ), 17, 19, 22, 27 Karol Hertz ( 1843–1904 ), 27 David Hilbert ( 1862–1943 ), 33 Vojtˇ ech Jarn´ ık ( 1897–1970 ), 12 Martin Jaˇ sek ( 1879–1945 ), 11 Hidefumi Katsuura, 57 Konrad Knopp ( 1882–1857 ),45 Helge von Koch ( 1870–1924 ),39 Henri Lebesgue ( 1875–1941 ), 33, 49 Mark Lynch, 62 Stefan Mazurkiewicz ( 1888–1945 ), 74 John McCarthy, 55 W/suppress ladys/suppress law Orlicz ( 1903–1990 ),52Giuseppe Peano ( 1858–1932 ),32 Karel Petr ( 1868–1950 ),47 M. B. Porter, 27 Bernhard Riemann ( 1826–1866 ),18, 21 Karel Rychl´ ık ( 1885–1968 ), 11, 47 Stanis/suppress law Saks ( 1897–1942 ), 78 Isaac Schoenberg ( 1903–1990 ), 33, 48 Hermann Schwarz ( 1843–1921 ), 28 Wac/suppress law Sierpi´ nski ( 1882–1969 ),44 Hugo Steinhaus ( 1887–1972 ), 74 Teiji Takagi ( 1875–1960 ),36 Bartel van der Waerden ( 1903–1996 ), 36 Karl Weierstrass ( 1815–1897 ), 17, 19,20, 28, 40, 54 Liu Wen, 64 92 Index of Subjects Baire’s Category Theorem, 52, 73 Banach space, 72 Banach-Mazurkiewicz Theorem, 76 base-brepresentation, 65 Besicovitch type, 78 Bolzano function, 13 Borchardt’s Journal, 21 Borel set, 81, 82 C[a,b], 72 Cantor series, 66 Cantor set, 33 category, 73 Cauchy sequence, 72 uniformly, 7 Cell´ erier function, 18 complete, 72, 73 continuity H¨ older sense, 39 of limit, 9 contraction mapping, 57, 59 convergence of sequence, 7 uniform, 7 Darboux function, 30 decimal representation, 47 dense ND[a,b], 75 P[a,b], 74 Dini-derivatives, 77 dyadic rational, 32, 38, 44 Faber functions, 43 functional equation, 37generic (topologically), 73 Hausdorff metric, 59 Hilbert curve, 33 Katsuura function, 57 Knopp function, 46 Koch “snowflake” curve, 39 Lebesgue curve, 33 Lipschitz class, 39 Lipschitz condition, 52, 81 Lynch function, 62 McCarthy function, 55 metric, 52, 71 metric space, 52, 54, 59, 71 ND[a,b], 71 NL[a,b], 82 NL M[a,b], 82 norm, 72 normed space, 72 nowhereM-Lipschitz, 82 nowhere dense, 73 nowhere Lipschitz, 82 Orlicz functions, 52 p-adic numbers, 47 Peano function, 32 Petr function, 48 prevalence, 78, 79 probe, 79 residual, 53, 73 Riemann function, 20 93 Schoenberg function, 48 Schwarz function, 31 shy, 79 Sierpi´ nski curve, 44 space-filling curve, 33, 49 supremum norm, 72 Takagi function, 36 ternary representation, 32 transverse, 78 van der Waerden function, 36 Weierstrass function, 22 M-test, 9 Wen functions, 64 94