Lecture 18: Weierstrass’s Example of a Continuous and Nowhere Differentiable Function

18.100A: Complete Lecture Notes Lecture 18: Weierstrass’s Example of a Continuous and Nowhere Differentiable Function Theorem 1 If f : I → R is differentiable at c ∈ I, then f is continuous at c. Proof : Since every point of I is a cluster point of I, f is continuous at c ∈ I ⇐⇒ limx→c f (x) = f (c). Now, lim f (x) = lim (f (x) − f (c) + f (c)) x→c   f (x) − f (c) = lim (x − c) + f (c) x→c x−c x→c = 0 · f 0 (c) + f (c) = f (c). Question 2. Is the converse true? Does f being continuous imply that f is differentiable? The answer, is no. Example 3 Let f (x) = |x|. Then, f is not differentiable at 0. Proof : We find a sequence xn → 0 such that lim n→∞ Let xn = (−1)n n . f (xn ) − f (0) does not exist. xn − 0 Then, limn→∞ xn = 0. However, f (xn ) − f (0) |(−1)n /n| = = (−1)n , xn − 0 (−1)n /n and limn→∞ (−1)n does not exist.  Question 4. If f : R → R is continuous, then does there exist a c ∈ R such that f is differentiable at c? The answer is again no! This was shown by Weierstrass, aka the Godfather. The basic idea is to build a continuous function that is a sum of highly oscillating functions. Remark 5. Note that we number the upcoming theorems so we may reference them a bit later in this lecture. 1 Theorem 6 (Theorem I) We will show the following 1. ∀x, y ∈ R, | cos x − cos y| ≤ |x − y|. 2. Let c ∈ R. Then, for all K ∈ N, ∃y ∈ (c + π/K, c + 3π/K) such that | cos(Kc) − cos(Ky)| ≥ 1. Proof : 1. In the proof of continuity of sin x, we showed that ∀x, y ∈ R, | sin x − sin y| ≤ |x − y|. Thus, | cos x − cos y| = | sin(x + π/2) − sin(y + π/2)| ≤ |x − y|. 2. The function f (x) = cos(Kx) is a 3π/K). 2π K -periodic function. In particular, ([−1, 1] \ cos(Kc)) ⊂ f (c + π/K, c + If cos Kc ≥ 0, then we choose y such that cos(Ky) = −1. If cos(Kc) < 0, then we choose y such that cos(Ky) = 1. This completes the proof Theorem 7 (Theorem II) For all a, b, c ∈ R, |a + b + c| ≥ |a| − |b| − |c|. Proof : We apply the Triangle Inequality twice: |a| = |a + b + b + (−b) + (−c)| ≤ |a + b + b| + |b + c| ≤ |a + b + c| + |b| + |c|. Theorem 8 (Theorem III) We will show the following: k P∞ x) 1. ∀x ∈ R, k=0 cos(160 is absolutely convergent. 4k k P∞ x) 2. The function f : R → R, f (x) = k=0 cos(160 is bounded and continuous. 4k Proof : 1. ∀k, cos(160k x) 4k ≤ 4−k . Hence, by the Comparison Test, ∞ X cos(160k x) 4k converges. k=0 2. For all x ∈ R, |f (x)| ≤ P∞ k=0 | cos(160k x)| 4k ≤ P 4−k = 43 . Therefore, f is bounded. We now show that f is continuous over R. Suppose c ∈ R and xn → c. Note that {|f (xn ) − f (c)|}n is bounded, and thus lim |f (xn ) − f (c)| = 0 ⇐⇒ lim sup |f (xn ) − f (c)| = 0. n→∞ n→∞ 2 We claim that for all  > 0, lim supn→∞ |f (xn )−f (c)| ≤ . Let  > 0. Choose M0 such that Then, lim sup |f (xn ) − f (c)| = lim sup n→∞ n M0 X cos(160k xn ) ≤ lim sup n M0 X cos(160k c) + 4k ∞ X n ∞ X 4−k | cos(160k xn ) − cos(160k c)| + M0 X k=M0 +1 4−k < 2 . cos(160k xn ) cos(160k c) − 4k 4k k=M0 +1 k=0 ≤ lim sup Theorem 9 (Weierstrass) P∞ The function f (x) = k=0 4k k=0 − P∞ 4−k (| cos(160k xn )| + | cos(160k c)|) k=M0 +1 ! 40k |xn − c| +  = . k=0 cos(160k x) 4k is nowhere differentiable. Proof : Let c ∈ R. We will construct a sequence xn → c such that π 160n ∀n ∈ N there exists an xn such that a) < xn − c < 3π By a), xn 6= 0∀n and |xn − c| ≤ 160n → 0. Let fk (x) f (c) − f (xn ) = fn (c) − fn (xn ) + n o f (xn )−f (c) xn −c n n 3π 160n and b) k x) = cos(160 4k is unbounded. By Theorem I 2), | cos(160 c) − cos(160n xn )| ≥ 1. P so f (x) = fk (x). Let n ∈ N. Thus, denote n−1 X ∞ X k=0 k=n (fk (c) − fk (xn )) + (fk (c) − fk (xn )) := an + bn + cn . Therefore, by Theorem II, |f (c) − f (xn )| ≥ |an | − |bn | − |cn |. By b), |an | = 4−n | cos(160k xn ) − cos(160k c)| ≥ 4−n . Furthermore, we have |bn | ≤ n−1 X 4−k | cos(160k c) − cos(160k xn )| ≤ k=0 n−1 X 4−k · 160k |xn − c| ≤ k=0 n−1 3π X k 3π 40n − 1 4−n+1 40 = · ≤ . 160n 160n 39 13 k=0 Finally, we have |cn | ≤ ∞ X 4−k (| cos(160k c)| + | cos(160k xn )|) ≤ 2 k=n+1 ∞ X 4−k = 2 · 4−n−1 · k=n+1 4 2 = 4−n . 3 3 Therefore, by the above inequalities, we have |f (c) − f (xn )| ≥ 4 −n   4 2 1 1− − = 4−n · . 13 3 39 Therefore, |f (c) − f (xn )| 160n −n 1 40n ≥ ·4 · = . |c − xn | 3π 39 117π Thus, n f (xn )−f (c) xn −c o is unbounded. n Remark 10. In other words, this proof by Weierstrass shows that there exists a continuous function that is nowhere differentiable! 3