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A Proof on the Non-Differentiability of Weierstrass Function in An

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A Proof on the Non-Differentiability of Weierstrass Function in An A Proof on the Non-differentiability of Weierstrass Function 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 mathematical analysis. E17 ------ 2 Preface 1 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 2 [1]. This broke people's 2 R original guess that the continuous functions are always nearly differentiable, and makes the gradual development of real analysis from mathematics analysis. In 1916, G.H.Hardy proved in the framework of real analysis that Weierstrass function f(x) is non-differentiable everywhere for x 2 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 2 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- 1 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 fang, fbng 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!1 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!1 an − x n!1 bn − x 1 E17 ------ 3 0 0 since an ! x , bn ! x . From the definition of limit, for any positive number ", there must be an N such that for any n > N, 0 0 f(an) − f(x ) f(bn) − f(x ) A − " < 0 < A + "; A − " < 0 < A + ": (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 + ")(an − x ) − (A − ")(bn − x ) < 0 0 (an − x ) − (bn − x ) 0 0 (an − x ) + (bn − x ) = A + " 0 0 = A + " + 2k": (5) (an − x ) − (bn − x ) f(a ) − f(b ) Similarly, n n > A − " − 2k". From the definition of limit, an − bn f(a ) − f(b ) lim n n = A; (6) n!1 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 2 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 2 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 = fα j 0 6 α < β1 −2 or b−β1 − 3 < α 6 b − 1g. Let c be the number of elements in C. Let ' be a one to one correspondence between C and f0; 1; 2; ··· ; c − 1gwhich 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 2 S, let '(α ) '(α ) '(α ) '(α ) (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. 3 E17 ------ 5 1 X aj Meanwhile, for any real number x = and any positive number ", let n be large bj j=0 −n enough such that b < ". Assume the integer c satisfies c < β1 − 2 or c > b − β1 − 3, then n 1 X aj X c x0 = + (19) bj bj j=0 j=n+1 is a decidable number, and 1 1 X jaj − cj X b − 1 jx0 − xj ": (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 derivative 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!1 an − x 0 k 1 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 1 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!1 n=0 4 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!1 n=0 Therefore, k−1 X lim anbk cos(bnπx0 + 2bn−kπ) sin2(bn−kπ) k!1 n=0 k−1 1 X = lim anbk sin(bnπx0 + 3bn−kπ) sin(bn−kπ) 2 k!1 n=0 k−1 1 X − lim anbk sin(bnπx0 + bn−kπ) sin(bn−kπ) 2 k!1 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 2 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) 5 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.
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