Smooth Interpolation, Hölder Continuity, and the Takagi van der Waerden Function

Transcription

1 Theorem. Let r.ifp, p,...,p r are the first r primes of the form p i = 4k i +, then the interval (p r, r p i) contains at least log (4(r )) primes congruent to modulo 4. Proof. For r = or, the result can be checked by hand, so we assume that r 4. Since p 4 = 9 > 4 4, we have 4r < p r.if j log (4r) +, it follows that j+ < 4p r. Hence r p i j+ > 7p r 4p r > p r. If r is odd, set n k = r p i k+.thenn k (4) (since (4)). But no p i divides n k for i r, so the integer n k has some prime factor q k (4) with q k > p r.if j = k,say j > k, the assumption that q k also divides n j leads to the same contradiction as earlier: since n k n j = j+ k+ = k+ ( j k ), wehaveq k j k and hence q k < p r. Thus, there are at least log (4r) + distinct primes of the form 4l + in(p r, r p i).ifr is even, the same argument applied to r shows that there are at least log (4(r )) +distinct primes of the form 4l + in(p r, r p i ). Since the first of these is p r, there are at least log (4(r )) distinct primes of the form 4l + in(p r, r p i ), a subinterval of (p r, r p i). Universidad de La Rioja, 6004 Logroño, La Rioja, Spain. aldaz@dmc.unirioja.es Departament of Mathematics, University of Michigan, Ann Arbor MI anabz@math.lsa.umich.edu Smooth Interpolation, Hölder Continuity, and the Takagi van der Waerden Function Jack B. Brown and George Kozlowski. INTRODUCTION. Consider the classes of functions f :[0, ] R indicated in the following diagram (where 0 <β<α<): C Lip Lip γ Lip α Lip β C, 0<γ < in which C denotes the class of continuous functions, C the class of continuously differentiable functions, and Lip α ={f C : K > 0 f (x) f (y) < K x y α for x, y [0, ]} the class of Lipschitz (or Hölder continuous) functions of order α. The symbol λ signifies Lebesgue measure. Marcinkiewicz [7] showed that there is a strong interpolation 4 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 0

2 link between the classes C and Lip when he proved: if f belongs to Lip, then for every ɛ>0 there exists g in C such that λ({x : f (x) = g(x)}) <ɛ. Marcinkiewicz actually required only that f be pointwise Lip (i.e., that f (x + t) = f (x) + O(t) at each x) and obtained similar results for higher order smoothness. Federer [5, p. 44] obtained the analogous Lip -C result in higher dimensions, and Whitney [] extended Federer s results to higher order smoothness (see also [6, p.7]).whitney described an example of a function φ of one variable that was designed to show that one could not weaken the requirement of f being in Lip in the Marcinkiewicz theorem. He stated the following [, p. 44]: Let φ 0 (t) be the distance from t to the nearest integer. Using any sufficiently large integer a, set φ i (t) = i φ 0 (a i t)/a i, φ(t) = φ i (t). Then φ satisfies a Lipschitz condition of order α, foranyα>0; but [the conclusion of Marcinkiewicz s theorem] is not true for it. The function φ 0 is the familiar rooftop function, with roofs having slopes ±. For n > 0 the roofs of φ n have slopes ± n. The graphs of φ and φ on [0,], with a =, are displayed in Figure. Whitney s statement would seem to suggest that for sufficiently large a in N the resulting function φ would belong to 0<γ < Lip γ. Such, however, is not the case. f 9 f 0 0 Figure. Theorem. Let a in N be greater than, let φ be the function described by Whitney, and let t 0 = ln()/ln(a). Thenφ has the following properties:. φ is not a member of Lip α for any α satisfying t 0 <α ; in fact, for τ N = /(a N ) the quotients φ(τ N ) φ(0) τ N α are unbounded as N.. φ belongs to Lip β if 0 <β t 0 ; in fact, with L = 4(a )/(a ), itistrue that φ(s) φ(t) L s t β for all such β and for all s and t in [0, ]. Proof. Suppose that t 0 <α. If i N, φ i (τ N ) = i φ 0 (/(a N i )/a i = i τ N.If i > N, there are two possibilities: if a is even, φ i (τ N ) = 0; if a is odd, φ i (τ N ) = i /a i = ( i /a i N )τ N. For our purposes, it will be enough to know that φ i (τ N ) 0 February 00] NOTES 4

3 for i > N.Then φ(τ N ) φ(0) τ N α = φ i(τ N ) τ α N N τ α N ( ) N. α a α Since a α >, these quotients tend to as N. Suppose next that 0 <β t 0. For given s and t in [0, ], letn be the nonnegative integer such that Note that φ(s) φ(t) < s t a N+ a. N N φ i (s) φ i (t) + i=n+ φ i (s) φ i (t) and that each of the functions φ i is polygonal, with graph consisting of line segments of slope ± i and with values in the interval [0,(/a) i ]. Majorizing terms of the first sum by i s t and terms of the second by (/a) i shows that Thus, φ(s) φ(t) N i s t +(/a) N+ (/a) i = ( N+ ) s t +(/a) N+ a a ( s t N+ + a ) = s t N+ a a a. ( ) φ(s) φ(t) N s t β N L L L, s t β a β because the hypothesis on β implies that a β.. THE TAKAGI VAN DER WAERDEN FUNCTION. Since lim ln() a ln(a) = and since it can be shown (the proof is similar to the proof of Theorem ) that λ({x : φ(x) = g(x)}) = 0 for each g in C, it would follow that one could not replace the hypothesis that f belongs to Lip in Marcinkiewicz s theorem with the weaker requirement that f be in Lip α for any specific choice of α in (0, ). However, it does not show that one could not replace that requirement with the assumption that f belongs to 0<γ < Lip γ. It is conceivable that the leading i in Whitney s description of the function φ i is actually a typographical error and that Whitney was really referring to the so-called van der Waerden function v(t) = φ 0 (a i t) a i 44 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 0

4 with parameter a in N, a >. B. L. van der Waerden [] described this function (with a = 0) in 90, providing a simple example of a continuous, nowhere differentiable function. The function v was described independently by other authors, including T. Takagi [] in 90 (as was pointed out in []). We refer to it as the Takagi van der Waerden function. It has been shown [], [4] thatv has more pathological nondifferentiability properties than were pointed out in []. It turns out that v does indeed satisfy the conditions described by Whitney. Theorem. The Takagi van der Waerden function v belongs to 0<γ < Lip γ,butv agrees with no function g from C on any set of positive measure. In fact, if M is a subset of [0,] with λ(m) >0, then the set { } v(y) v(x) D(v, M) = : x M, y M, y = x y x is unbounded. Proof. The fact that v is a member of 0<γ < Lip γ was established by Shidfar and Sabetfakhri for a = in[9], and the same fact for all integers a greater than follows from the theorem in [0]. For each i in N,let f i (t) = φ 0 (a i t)/a i,andlet v n (t) = n f i (t) for 0 t. Then f i is a polygonal function with each segment in its graph having slope ±, and v n is likewise a polygonal function, each segment in its graph having slope some integer in the interval [ n, n]. In order to prove the second claim in the theorem, let M be a subset of [0, ] with λ(m) >0 and suppose that b > 0 (assume without loss of generality that b is an integer). Let z in M be a Lebesgue density point of M (i.e., for every ɛ>0 there exists δ>0 such that for every subinterval [c, d] of [0,] that contains z and has length less than δ, itistruethatλ([c, d]\m) <ɛ (d c) [8, pp. ], [, pp. 5 6]). Choose δ>0 corresponding to ɛ = a b /8 in this definition, and consider an arbitrary subinterval [c, d] of [0, ] that contains z and has d c <δ,sothat λ([c, d]\ M) < ɛ (d c). Note that this ensures that if a subinterval (u,v) of [c, d] has length v u ɛ (d c),then(u,v)must contain a point x of M.Fixn in N such that /a n <δ. We now specify [c, d] to be the base (of length /a n ) of one of the roofs of f n that contains z. Write m = n + b and e = c + a n,sothat[c, e] is the base of one of the roofs (illustrated in Figure ) of f n+. The polygonal function v n has constant slope M n over the interval [c, e]. There will be two cases to consider. Assume first that M n 0, and consider h = c + (/a m ). Note that h is the abscissa of the top of a roof of f m. The graph of v m has constant slope b + M n on [c, h]. Set = 4 (h c) = 4 a = m 8 a = n+b 8 a = ɛ (d c). b an The interval (u,v) with left endpoint c and length = ɛ (d c) must contain a point x of M. There must also exist an element y of M in the open interval of length with right endpoint h (see Figure ). We then have February 00] NOTES 45

5 f n f n + f n + f m f m c h e d Figure. v(y) v(x) = v m (y) v m (x) + i=m+ f i (y) i=m+ f i (x), so we can estimate v(y) v(x) (b + M n ) (y x) (b + M n ) h c i=m+ i=m+ a i f i (y) f i (x) = (b + M n ) h c a a m+ a ( b + Mn = ) (h c). a Accordingly, for each integer b > 0 there exist points x and y of M for which v(y) v(x) y x v(y) v(x) h c b + M n a b a. Since b can be arbitrarily large, D(v, M) is unbounded when M n 0. The case where M n < 0 is similar. Take h = d /(a m ), pick an element x of M belonging to the open interval of length with left endpoint h, and select a point y of M belonging the open interval of length with right endpoint e. The polygonal function v m has constant slope b + M n on [h, e]. Proceeding through inequalities similar to those given in the preceding argument, one arrives at 46 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 0

6 v(y) v(x) y x v(y) v(x) d h b + M n a b a in this case as well. Again, the unboundedness of D(v, M) follows. REFERENCES. Y. Baba, On maxima of Takagi van der Waerden functions, Proc. Amer. Math. Soc. 9 (984) P. Billingsley, Van der Waerden s continuous nowhere differentiable function, this MONTHLY 89 (98) 69.. A. M. Bruckner, J. B. Bruckner, and B. S. Thomson, Real Analysis, Prentice Hall, Upper Saddle River, NJ, F. S. Cater, On van der Waerden s nowhere differentiable function, this MONTHLY 9 (984) H. Federer, Surface area II, Trans. Amer. Math. Soc. 55 (944) , Geometric Measure Theory, Die Grundlehren der mathematischen Wissenschaften, vol. 5, Springer-Verlag, New York, J. Marcinkiewicz, Sur les séries de Fourier, Fund. Math. 7 (96) F. Riesz and B. Sz.-Nagy, Functional Analysis, Fredrick Unger, New York, A. Shidfar and K. Sabetfakhri, On the continuity of van der Waerden s function in the Hölder sense, this MONTHLY 9 (986) , On the Hölder continuity of certain functions, Exposition. Math. 8 (990) T. Takagi, A simple example of the continuous function without derivative, Proc. Phys.-Math. Soc. Tokyo Ser. II (90) B. W. van der Waerden, Ein einfaches Beispiel einer nicht-differenzierbaren stetigen Funktion, Math. Z. (90) H. Whitney, On totally differentiable and smooth functions, Pacific J. Math. (95) Department of Mathematics, Auburn University, Auburn, AL 6849 brownj4@auburn.edu kozloga@auburn.edu When Does the Position Vector of a Space Curve Always Lie in Its Rectifying Plane? Bang-Yen Chen. INTRODUCTION. Let E denote Euclidean three-space, with its inner product,. Consider a unit-speed space curve x : I E,whereI = (α, β) is a real interval, that has at least four continuous derivatives. Let t denote x. It is possible, in general, that t (s) = 0forsomes; however, we assume that this never happens. Then we can introduce a unique vector field n and positive function κ so that t = κn. We call t the curvature vector field, n the principal normal vector field, andκ the curvature of the given curve. Since t is a constant length vector field, n is orthogonal to t. The binormal vector field is defined by b = t n. It is a unit vector field orthogonal to both t and n. One defines the torsion τ by the equation b = τn. The famous Frenet-Serret equations are given by (see, for instance, [4] or[6]): February 00] NOTES 47