Transcription of The Weierstrass Function
1 The Weierstrass FunctionBrent NelsonWe let sin,cos :R Rbe defined in the usual geometric way, extended to all ofR. We will assume thefollowing facts about these functions:(a) sin and cos are continuous onR;(b)|sin(x)|,|cos(x)| 1 for allx R;(c) sin(x)x 1 for allx R\{0};(d) cos(x) cos(y) = 2 sin(x+y2)sin(x y2)for allx,y R;(e) cos(x+y) = cos(x) cos(y) sin(x) sin(y) for allx,y can be derived by considering, for example, the power series representations:sin(x) = n=0( 1)n(2n+ 1)!x2n+1andcos(x) = n=0( 1)n(2n)! main goal of these notes is to prove the following theorem:Theorem(Karl Weierstrass , 1872).Leta (0,1)and letbbe an odd integer such thatab >1 +3 2. Then theseriesf(x) = n=0ancos(bn x)converges uniformly onRand defines a continuous but nowhere differentiable Function appearing in the above theorem is called the Weierstrass Function . Before we prove the theorem,we require the following lemma:Lemma(The Weierstrass M-test).Let(E,d)be a metric space, and for eachn Nletfn:E Rbe afunction.
2 Suppose that for eachn N, there existsMn>0such that|f(x)| Mn x the series n=1 Mnconverges, then the series n=1fnconverges uniformly >0. The Cauchy criterion for the convergence of a series implies there existsN Nsuch that foralln,m Nwithn < mwe have|Mn+1+Mn+2+ +Mm|=Mn+1+Mn+2+ +Mm< .Consequently, for alln,m Nwithn < mwe have for allx E m i=1fi(x) n i=1fi(x) =|fn+1(x) + +fm(x)| |fn+1(x)|+ +|fm(x)| Mn+1+ +Mm< .That is, the sequence of partial sums ( ni=1fi)n Nsatisfies the Cauchy criterion for functions. So by a propo-sition from lecture we know that these partial sums converge uniformly to the series n= Weierstrass FunctionMath 104 Proof of |ancos(bn x)| anfor allx Rand n=0anconverges, the series converges uni-formly by the Weierstrass M-test. Moreover, since the partial sums are continuous (as finite sums of continuousfunctions), their uniform limitfis also see thatfis nowhere differentiable, we will show for eachx0 Rthatlimx x0f(x) f(x0)x x0does not exist.
3 In particular, we ll show that asxapproachesx0from above and below, the respective differencequotients oscillate wildly between larger and larger positive and negative R. For eachm N, let m Zbe such thatbmx0 m ( 12,12].Definexm:=bmx0 mym:= m 1bmzm:= m+ thatym x0= 1 +xmbm<0<1 xmbm=zm < x0< zm,limm |ym x0|= limm x0 ym= limm 1 +xmbm= 0,andlimm |zm x0|= limm zm x0= limm 1 xmbm= is, (ym)m Nand (zm)m Nare (meticulously constructed) sequences converging tox0, but from above andbelowx0, respectively. We will examine the difference quotients forfproceeding alongx=ym,m N, andx=zm,m N. First,f(ym) f(x0)ym x0= n=0ancos(bn ym) n=0ancos(bn x0)ym x0= n=0ancos(bn ym) cos(bn x0)ym x0=m 1 n=0(ab)ncos(bn ym) cos(bn x0)bn(ym x0)+ n=0an+mcos(bn+m ym) cos(bn+m x0)ym denote the two sums in the last expression byS1andS2, respectively. Roughly speaking, we will show thatS1is small whileS2is big. Using property (d), we haveS1=m 1 n=0(ab)n 2bn(ym x0)sin(bn (ym+x0)2)sin(bn (ym x0)2)=m 1 n=0 (ab)nsin(bn (ym+x0)2)sin(bn (ym x0)2) bn(ym x0) the triangle inequality and properties (b) and (c) we have|S1| m 1 n=0 (ab)n1 1 = (ab)m 1ab 1< (ab)mab , there exists 1 ( 1,1) such thatS1= 1 (ab)mab Brent Nelson 2017 The Weierstrass FunctionMath 104 Next, we handleS2.)
4 First, recall thatym= m 1bm, that mis an integer, and thatbis an odd integer. Thuscos(bn+m ym)= cos (bn ( m 1)) = ( 1)bn( m 1)= ( 1) m 1= ( 1) , recall thatxm=bmx0 mso that using property (e) we havecos(bn+m x0)= cos (bn (xm+ m))= cos(bn xm) cos(bn m) sin(bn xm) sin(bn m)= ( 1)bn mcos(bn xm) 0= ( 1) mcos(bn xm).Using these computations, we haveS2= n=0an+m ( 1) m ( 1) mcos(bn xm)ym x0= n=0an+m( 1)( 1) m1 + cos(bn xm) 1+xmbm= (ab)m( 1) m n=0an1 + cos(bn xm)1 + thatxm ( 12,12]so the terms in the sum in the last expression are non-negative. Consequently, n=0an1 + cos(bn xm)1 +xm 1 + cos( xm)1 +xm 11 +12= there exists 1 1 such thatS2= (ab)m( 1) m our computations forS1andS2together yieldsf(ym) f(x0)ym x0=S1+S2= 1 (ab)mab 1+ (ab)m( 1) m 123= ( 1) m(ab)m 1(23+ ( 1) m 1 1 ab 1)Recall our assumption thatab >1 +3 2, which is equivalent to ab 1<23. Using| 1|<1 and 1, we have23+ ( 1) m 1 1 ab 1>23 ab 1> , the sign off(ym) f(x0)ym x0is completely determined by ( 1) mand f(ym) f(x0)ym x0 >(ab)m(23 ab 1)Thus, not only does the difference quotient alternate signs rapidly, but its magnitude tends to + asm.)
5 Since limm ym=x0, this is enough to show that limx x0f(ym) f(x0)ym x0does not exist. We will show somethingslightly stronger: the same behavior also occurs along (zm)m the same breakdown as before, we can writef(zm) f(x0)zm x0=S 1+S 2,and the same argument yieldsS 1= 2 (ab)mab 1for some 2 ( 1,1). Usingzm x0=1 xmbmwe haveS 2= n=0an+m ( 1) m ( 1) mcos(bn xm)1 xmbm= (ab)m( 1) m n=0an1 + cos(bn xm)1 Brent Nelson 2017 The Weierstrass FunctionMath 104 Sincexm ( 12,12], the terms in the sum in the last expression are non-negative. Consequently, n=0an1 + cos(bn xm)1 xm 1 + cos( xm)1 xm>11 ( 12)= there exists 2 1 such thatS 2= (ab)m( 1) m 223. Thenf(zm) f(x0)zm x0=S 1+S 2= 2 (ab)mab 1 ( 1) m(ab)m 223= ( 1) m(ab)m 2(23 ( 1) m 2 2 ab 1).Just as before we have23 ( 1) m 2 2 ab 1>23 ab 1>0,so that the sign off(zm) f(x0)zm x0has sign completely determined by ( 1) m. Also, f(zm) f(x0)zm x0 >(ab)m(23 ab 1)m + So the same behavior occurs to the right ofx0.)
6 The graph of the Weierstrass Function The rough shape of the graph is determined by then= 0 term in the series: cos( x). The higher-order termscreate the smaller oscillations. Withbcarefully chosen as in the theorem, the graph becomes so jagged thatthere is no reasonable choice for a tangent line at any point; that is, the Function is nowhere Brent Nelson 2017
