A Proof on the Non-differentiability of Weierstrass in An Uncountable Dense Set

Boyu Chen

Adviser Zixiang Zhou

Shanghai Shibei High School 2010.8.3

E17 ------1 Abstract

The Weierstrass function is the first function in history which is continuous everywhere, but differentiable nowhere. In this paper, we prove that the Weierstrass function with a condition weaker than Weierstrass’s original one cannot be differentiable in an uncountable dense set in the framework of .

E17 ------2 Preface ∞ X Weierstrass function f(x) = an cos(bnπx) is one of the most famous “pathological n=0 function” in mathematical history. Weierstrass proved that if 0 < a < 1, b is an odd integer 3π and ab > 1+ , then f(x) is non-differentiable everywhere for x ∈ [1]. This broke people’s 2 R original guess that the continuous functions are always nearly differentiable, and makes the gradual development of from analysis. In 1916, G.H.Hardy proved in the framework of real analysis that Weierstrass function f(x) is non-differentiable everywhere for x ∈ R when 0 < a < 1 and ab > 1[3]. But from the fundamental mathematics analysis, the best conclusion as far I know is that f(x) is non-differentiable everywhere for x ∈ R when 0 < a < 1, b is an odd integer and ab > (1 − a)π 1 + [2]. 2 In this paper, we will use a different method to prove that the function f(x) is non- differentiable at least in an uncountable dense set when 0 < a < 1, b is an integer greater than 5 and ab > 1. (The condition is a bit weaker than Weierstrass’s original one). The main theorem is as follow.

Theorem. Assume that the real number a and the natural number b satisfy the condi- ∞ X n n tion that 0 < a < 1, b > 6 and ab > 1, then the function f(x) = a cos(b πx) is n=0 non-differentiable in an uncountable dense set.

Proof

First of all, let us prove three lemmas. 0 0 bn − x Lemma 1. Suppose {an}, {bn} satisfy an > bn > x , = k is a constant for any an − bn df positive integer n, and a → x0, b → x0. Moreover, suppose (x0) = A exists. Then n n dx f(a ) − f(b ) lim n n = A . (1) n→∞ an − bn

Proof. f(x) − f(x0) df lim = (x0) = A. (2) x→x0 x − x0 dx

Let x = an and x = bn respectively, we have

0 0 f(an) − f(x ) f(bn) − f(x ) lim 0 = A, lim 0 = A, (3) n→∞ an − x n→∞ bn − x

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E17 ------3 0 0 since an → x , bn → x . From the definition of limit, for any positive number ε, there must be an N such that for any n > N,

0 0 f(an) − f(x ) f(bn) − f(x ) A − ε < 0 < A + ε, A − ε < 0 < A + ε. (4) an − x bn − x Hence 0 0 f(an) − f(x ) 0 f(bn) − f(x ) 0 0 (an − x ) − 0 (bn − x ) f(an) − f(bn) an − x bn − x = 0 0 an − bn (an − x ) − (bn − x ) 0 0 (A + ε)(an − x ) − (A − ε)(bn − x ) < 0 0 (an − x ) − (bn − x ) 0 0 (an − x ) + (bn − x ) = A + ε 0 0 = A + ε + 2kε. (5) (an − x ) − (bn − x ) f(a ) − f(b ) Similarly, n n > A − ε − 2kε. From the definition of limit, an − bn f(a ) − f(b ) lim n n = A, (6) n→∞ an − bn Lemma 1 is proved.

Now expand any real number x0 in base b, that is α α α α x0 = α + 1 + 2 + 3 + ··· + n + ··· , (7) 0 b b2 b3 bn

where α0, α1, α2, ··· , αn, ··· are all integers such that 0 6 αi 6 b − 1 for i > 1 and αi is 1 not always equal to b − 1 when i is large enough. Similarly, we can also expand in base b, 3 which is 1 β β β β = 1 + 2 + 3 + ··· + n + ··· (8) 3 b b2 b3 bn b with 0 β b − 1 for i 1. Since b 6, we always have β 2 and β . We call a real 6 i 6 > > 1 > 1 6 3 number x0 a decidable number if there are infinitely many

αn1 , αn2 , αn3 , ··· , αni , ··· (9)

in α0, α1, α2, ··· , αn, ··· satisfying αni < β1 − 2 or αni > b − β1 − 3.

Lemma 2. For any decidable number x0, there are infinitely many positive integers k such that k−1 0 2π  cos b πx + > 0.5 . (10) b

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E17 ------4 Proof. We only need to prove that for any decidable number x0, there are infinitely many k such that 1 2 1 2 bk−1x0 ∈ m − − , m + − , (11) 3 b 3 b where m is an integer decided by k. Using (8), we have

1 2 β − 2 β β β m + − = m + 1 + 2 + 3 + ··· + n + ··· , (12) 3 b b b2 b3 bn and 1 2 b − β − 3 b − 1 − β b − 1 − β b − 1 − β m − − = m − 1 + 1 + 2 + 3 + ··· + n + ··· . (13) 3 b b b2 b3 bn Since x0 is a decidable number, we can find infinitely many numbers

αn1 , αn2 , αn3 , ··· , αni , ··· (14)

which satisfy αni < β1 − 2 or αni > b − β1 − 3. Taking k = ni, we have

αn αn +1 bk−1x0 = (bni−1α + bni−2α + bni−3α + ··· + α ) + i + i + ··· . (15) 0 1 2 ni−1 b b2

ni−1 ni−2 ni−3 If αni < β1 − 2, then let m = b α0 + b α1 + b α2 + ··· + αni−1 and we have 1 2 β − 2 1 2 m − − < m bk−1x0 < m + 1 m + − . (16) 3 b 6 b 6 3 b

ni−1 ni−2 ni−3 Otherwise αni > b − β1 − 3, then let m − 1 = b α0 + b α1 + b α2 + ··· + αni−1 and we have 1 2 1 2 m − − < bk−1x0 < m m + − . (17) 3 b 6 3 b

Lemma2 is true because there are infinitely many such ni.

Lemma 3. For a given integer b ≥ 6, the set of its decidable numbers is dense in R and has the same cardinality as R. Proof. The set of decidable numbers belongs to R. Let C = {α | 0 6 α < β1 −2 or b−β1 − 3 < α 6 b − 1}. Let c be the number of elements in C. Let ϕ be a one to one correspondence between C and {0, 1, 2, ··· , c − 1}which maps b − 1 to c − 1. Let S be the set of all x0 whose α1, α2, ··· satisfy that either αk < β1 − 2 or αk > b − β1 − 3 for any k > 1. Obviously any element of S is a decidable number. For x0 ∈ S, let

ϕ(α ) ϕ(α ) ϕ(α ) ϕ(α ) ψ(x0) = α + 1 + 2 + 3 + ··· + n + ··· . (18) 0 c c2 c3 cn

It is a one to one correspondence between S and R. Hence S and R have the same cardinality.

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E17 ------5 ∞ X aj Meanwhile, for any real number x = and any positive number ε, let n be large bj j=0 −n enough such that b < ε. Assume the integer c satisfies c < β1 − 2 or c > b − β1 − 3, then

n ∞ X aj X c x0 = + (19) bj bj j=0 j=n+1 is a decidable number, and

∞ ∞ X |aj − c| X b − 1 |x0 − x| ε. (20) 6 bj 6 bj 6 j=n+1 j=n+1

So the set of decidable number is dense in R. Lemma 3 is proved. Proof of the theorem. Let x0 be a decidable number. Assume that f(x) had a finite at x0, then f(x) − f(x0) lim = A. (21) x→x0 x − x0 0 0 2 0 f(an) − f(x ) Let ak = x + k . Obviously ak → x , so lim 0 = A. On the other hand, b n→∞ an − x

0 k ∞ f(ak) − f(x ) b X 2 = an cos bnπx0 +  − cos(bnπx0) a − x0 2 bk k n=0 k−1 bk X 2 = an cos bnπx0 +  − cos(bnπx0) 2 bk n=0 ∞ bk X 2 + an cos bnπx0 +  − cos(bnπx0). (22) 2 bk n=k 2 Since cos bnπx0 +  − cos(bnπx0) = 0 when n k, we have bk >

0 k k−1 f(ak) − f(x ) b X 2 = an cos bnπx0 +  − cos(bnπx0) a − x0 2 bk k n=0 k−1 X = − anbk sin(bnπx0 + bn−kπ) sin(bn−kπ). (23) n=0 Hence

k−1 X lim anbk sin(bnπx0 + bn−kπ) sin(bn−kπ) = −A. (24) k→∞ n=0

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E17 ------6 4 2 Similarly, by setting a = x0 + , b = x0 + , Lemma 1 leads to k bk k bk

k−1 X lim anbk sin(bnπx0 + 3bn−kπ) sin(bn−kπ) = −A. (25) k→∞ n=0 Therefore,

k−1 X lim anbk cos(bnπx0 + 2bn−kπ) sin2(bn−kπ) k→∞ n=0 k−1 1 X = lim anbk sin(bnπx0 + 3bn−kπ) sin(bn−kπ) 2 k→∞ n=0 k−1 1 X − lim anbk sin(bnπx0 + bn−kπ) sin(bn−kπ) 2 k→∞ n=0 1 = (−A + A) = 0. (26) 2 For each decidable number x0, Lemma 2 implies that there are either infinitely many k 2π 2π such that cos bk−1πx0 +  > 0.5, or infinitely many k such that cos bk−1πx0 +  < −0.5. b b 2π Suppose there are infinitely many k such that cos bk−1πx0 +  > 0.5, then for any b k ∈ N+,

k−1 X anbk cos(bnπx0 + 2bn−kπ) sin2(bn−kπ) n=0 k−2 2π π X ak−1bk cos bk−1πx0 +  sin2  − anbk sin2(bn−kπ) > b b n=0 k−2 2π π X > ak−1bk cos bk−1πx0 +  sin2  − anb2n−kπ2 b b n=0 π 2π [(ab2)k−1 − 1]π2 = b(ab)k−1 sin2  cos bk−1πx0 +  − b b (ab2 − 1)bk h π 2π π2 i > (ab)k−1 b sin2  cos bk−1πx0 +  − . (27) b b b(ab2 − 1)

2 ! 2 π π 2 b − 1 ab − 1 Noticing that π < < < , we have b sin  2 2 2 b 1 π π2 b sin2  > . (28) 2 b b(ab2 − 1)

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E17 ------7 Hence there are infinitely many k such that π 2π π2 b sin2  cos bk−1πx0 +  − > λ, (29) b b b(ab2 − 1)

1 π π2 where λ = b sin2  − > 0 is a constant. However, λ(ab)k−1 diverges as k → ∞, 2 b b(ab2 − 1) which contradicts (26). 2π Similarly, if there are infinitely many k such that cos bk−1πx0 +  < −0.5, then for b each k ∈ N+,

k−1 X anbk cos(bnπx0 + 2bn−kπ) sin2(bn−kπ) n=0 h π 2π π2 i < (ab)k−1 b sin2  cos bk−1πx0 +  + . (30) b b b(ab2 − 1) Similar to (32), there are infinitely many k such that

π 2π π2 b sin2  cos bk−1πx0 +  + < −λ. (31) b b b(ab2 − 1)

This also contradicts (26) since −λ(ab)k−1 diverges as k → ∞. ∞ X Therefore, the function f(x) = an cos(bnπx) cannot be differentiable at any decidable n=0 number. Lemma 3 implies that f(x) is non-differentiable in an uncountable dense set.

Acknowledgement:

I am grateful to Professor Zixiang Zhou for his suggestions to let me avoid some infeasible ideas, as well as his help on translation. I also thank the other three students and three instructors, who accompanied me in the seminars. Thanks the RSI and Fudan University, who gave us such a great opportunity to experience university life. Furthermore, thanks my class sponsor Rongsheng Jin who gave some good advice and help in the process of preparation of the defense and Keliang Pan who help me to translate.

References

[1] Γ.M.Fihtengol~c: The Differential and Integral Calculus (Volume II)(Version 8), Higher Education Press, 2006, p.403.

[2] Liu Wen: A proof of the differentiability of Weierstrass function, Studies in College Mathematics, 2002, Vol.5, No.2.

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E17 ------8 [3] G. H. Hardy: Weistrass’s nondifferentiable function, Transactions of the American Math- ematical Society, Vol. 17, No. 3, (Jul., 1916), pp. 301-325

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