Weierstrass’s non-differentiable function Jeff Calder December 12, 2014 In the nineteenth century, many mathematicians held the belief that a continuous function must be differentiable at a large set of points. In 1872, Karl Weierstrass shocked the mathe- matical world by giving the first published example of a continuous function that is nowhere differentiable. His function is given by 1 X W (x) = an cos(bnπx): n=0 In particular, Weierstrass proved the following theorem: Theorem 1 (Weierstrass 1872). Let a 2 (0; 1), let b > 1 be an odd integer, and assume that 3 ab > 1 + π: (1) 2 Then the function W is continuous and nowhere differentiable on R. To see that W is continuous on R, note that jan cos(bnπx)j = anj cos(bnπx)j ≤ an: Since the geometric series P an converges for a 2 (0; 1), the Weierstrass M-test shows that n n the series defining W converges uniformly to W on R. Since each function a cos(b πx) is continuous, each partial sum is continuous, and therefore W is continuous, being the uniform limit of a sequence of continuous functions. To give some motivation for the condition (1), consider the partial sums n X k k Wn(x) = a cos(b πx): k=0 These partial sums are differentiable functions and n 0 X k k Wn(x) = − π(ab) sin(b πx): k=0 0 If ab < 1, then we can again use the Weierstrass M-test to show that (Wn) converges uniformly to a continuous function on R. In this case we can actually prove that W is differentiable 0 0 and Wn ! W uniformly. Therefore, at the very least we need ab ≥ 1 for W to be non- differentiable. In 1916, Godfrey Hardy showed that ab ≥ 1 is sufficient for the nowhere 1 2 2 1.5 1.8 1 1.6 0.5 0 1.4 −0.5 1.2 −1 1 −1.5 −2 0.8 −1 −0.5 0 0.5 1 −0.1 −0.05 0 0.05 0.1 (a) (b) Figure 1: The Weierstrass function W (x) for a = 0:5 and b = 3. Notice that W appears the same on the two different scales shown in (a) and (b). 3 differentiability of W . The more restrictive condition ab > 1 + 2 π present in Weierstrass’s proof is an artifact of the techniques he used. Hardy also relaxed the integral assumption on b, and allowed b to be any real number greater than 1. Figure1 shows a plot of the Weierstrass function for a = 0:5 and b = 3 on two different scales. Notice the similar repeating patterns on each scale. If we were to continue zooming in on W , we would continue seeing the same patterns. The Weierstrass function is an early example of a fractal, which has repeating patterns at every scale. Before giving the proof, we recall a few facts that will be useful in the proof. Let x; y 2 R, and suppose x > y. Then by the fundamental theorem of calculus Z x Z x cos(x) − cos(y) = − sin(t) dt ≤ 1 dt = x − y; y y and Z x cos(x) − cos(y) ≥ −1 dt = −(x − y): y Therefore j cos(x) − cos(y)j ≤ jx − yj: The argument is similar when y ≥ x, so we deduce j cos(x) − cos(y)j ≤ jx − yj for all x; y 2 R: (2) Consider cos(nπ + x) for an integer n and x 2 R. If n is even, then since cosine is 2π-periodic, cos(nπ + x) = cos(x). If n is odd, then n + 1 is even and cos(nπ + x) = cos((n + 1)π + x − π) = cos(x − π) = − cos(x): Draw a graph of cos(x) if the last equality is unclear. Therefore we obtain n cos(nπ + x) = (−1) cos(x) for all x 2 R: (3) We now give the proof of Theorem1. 2 m Proof. Let x0 2 R and let m 2 N. Let us round b x0 to the nearest integer, and call this integer km. Therefore 1 1 bmx − ≤ k ≤ bmx + : (4) 0 2 m 0 2 Let us also set k + 1 x = m : (5) m bm By (4) we see that m 1 m b x0 − 2 + 1 b x0 xm ≥ m > = x0; b bm and m 1 b x0 + + 1 3 x ≤ 2 = x + : m bm 0 2bm Combining these equations we have 3 x < x ≤ x + : (6) 0 m 0 2bm By the squeeze lemma, limm!1 xm = x0. Let us consider the difference 1 1 X n n X n n W (xm) − W (x0) = a cos(b πxm) − a cos(b πx0) n=0 n=0 1 X n n n = a (cos(b πxm) − cos(b πx0)) n=0 = A + B; (7) where m−1 X n n n A = a (cos(b πxm) − cos(b πx0)) ; (8) n=0 and 1 X n n n B = a (cos(b πxm) − cos(b πx0)) : (9) n=m The proof is now split into three steps. 1. The first step is to find an upper bound for jAj. Using the triangle inequality and the identity (2) m−1 m−1 m−1 X n n n X n n X n jAj ≤ a jcos(b πxm) − cos(b πx0)j ≤ a b π(xm − x0) = π(xm − x0) (ab) : n=0 n=0 n=0 Noticing the geometric series, we deduce 1 − (ab)m (ab)m − 1 π(ab)m jAj ≤ π(x − x ) = π(x − x ) ≤ (x − x ): (10) m 0 1 − ab m 0 ab − 1 ab − 1 m 0 3 In the last step we used the hypothesis that ab > 1 + 2 π > 1. 3 2. The second step is to find a lower bound for jBj. By the definition of xm (5) k + 1 cos(bnπx ) = cos bnπ m = cos(bn−m(k + 1)π): m bm m n−m For n ≥ m, b (km + 1) is an integer, and hence n−m n−m km+1 n b (km+1) b km+1 km cos(b πxm) = (−1) = (−1) = (−1) = −(−1) ; (11) where we used the fact that bn−m is odd so that (−1)bn−m = −1. On the other hand, we also have k + bmx − k cos(bnπx ) = cos bnπ m 0 m = cos(bn−mk π + bn−mz π); 0 bm m m m n−m where zm = b x0 − km. Since n ≥ m, b km is an integer and we can use (3) to find that n−m n b km n−m km n−m cos(b πx0) = (−1) cos(b zmπ) = (−1) cos(b zmπ); (12) where, as before, we used the fact that bn−m is odd. We now insert (11) and (12) into (9) to obtain 1 X n km km n−m B = a −(−1) − (−1) cos(b zmπ) n=m 1 km X n n−m = −(−1) a 1 + cos(b zmπ) : n=m n n−m Notice that a > 0 and 1 + cos(b zmπ) ≥ 0. It follows that all the terms in the sum above are non-negative, and therefore 1 X n n−m m jBj = a 1 + cos(b zmπ) ≥ a (1 + cos(zmπ)) : n=m m 1 1 π π Recall that zm = b x0 − km. By (4), zm 2 [− 2 ; 2 ], and therefore πzm 2 [− 2 ; 2 ]. It follows that cos(zmπ) ≥ 0 and jBj ≥ am: 3 By (6), xm − x0 ≤ 2bm , and thus 2bm (x − x ) ≤ 1: 3 m 0 We can combine this with jBj ≥ am to find that 2(ab)m jBj ≥ am · 1 ≥ (x − x ): (13) 3 m 0 This is the desired lower bound on jBj, and completes part 2 of the proof. 3. We now combine the bounds (10) and (13) to complete the proof. Notice by (10), (13) and the reverse triangle inequality that 2(ab)m π(ab)m 2 π jA + Bj ≥ jBj − jAj ≥ (x − x ) − (x − x ) = (ab)m − (x − x ): 3 m 0 ab − 1 m 0 3 ab − 1 m 0 4 By (7) we see that 2 π jW (x ) − W (x )j = jA + Bj ≥ (ab)m − (x − x ): m 0 3 ab − 1 m 0 Since xm − x0 > 0, so that jxm − x0j = xm − x0, we have W (xm) − W (x0) m 2 π ≥ (ab) − : (14) xm − x0 3 ab − 1 We would like this difference quotient to tend to 1 in absolute value as m ! 1. For this we need ab > 1 and 2 π − > 0: 3 ab − 1 Rearranging this for ab we see that we need 3 ab > π + 1; 2 which is exactly the hypothesis (1). Therefore W (xm) − W (x0) lim = +1; m!1 xm − x0 and xm ! x0 as m ! 1. This shows that W is not differentiable at x0. With slight modifications to the proof, we can also show that W (x) − W (x ) lim 0 x!x0 x − x0 does not exist as a real number or ±∞. This rules out the possibility of the Weierstrass function having a vertical tangent line, or an “infinite derivative” anywhere. 5.