Advanced Calculus: MATH 410 Uniform Convergence of Functions Professor David Levermore 11 December 2015


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1 Advnced Clculus: MATH 410 Uniform Convergence of Functions Professor Dvid Levermore 11 December Sequences of Functions We now explore two notions of wht it mens for sequence of functions {f n } n N to converge to function f. The first notion, pointwise convergence, might seem nturl t first, but we will see tht it is not strong enough to do much. The second notion, uniform convergence, is strong enough to do mny things, but might seem less nturl t first. We will explore these notions through exmples tht show the superiority of uniform convergence Pointwise Convergence. We begin with the definition of pointwise convergent. Definition Let S R. Let {f n } n N be sequence of relvlued functions tht re ech defined over S. The sequence {f n } n N is sid to be pointwise convergent or to converge pointwise over S if there exists function f defined over S such tht lim f n(x) = f(x) for every x S. We sy f n converges to f pointwise over S nd cll f the pointwise limit of the sequence {f n } n N over S. We denote this s f n f pointwise over S. Becuse every Cuchy sequence of rel numbers hs unique limit, we hve the following. Proposition Let S R. Let {f n } n N be sequence of relvlued functions tht re ech defined over S such tht for every x S the rel sequence {f n (x)} n N is Cuchy. Then there exists unique relvlued function f defined over S such tht (12.1) f n f pointwise S. Proof. For ech x S define f(x) to be the unique limit of the Cuchy sequence of rel numbers {f n (x)} n N. The following nturl questions rise. If f n f pointwise over [, b] nd ech f n is continuous over [, b] then is f continuous over [, b]? If f n f pointwise over [, b] nd ech f n is Riemnn integrble over [, b] then is f Riemnn integrble over [, b] nd does b f n If f n f pointwise over [, b] nd ech f n is differentible over [, b] then is f differentible over [, b] nd does b f? f n f pointwise over [, b]? 1
2 2 For pointwise convergence the nswer to ech of these three questions is NO, NOT ALWAYS! To understnd why, consider the following exmples. Exmple. Consider f n (x) = x n over [0, 1]. It cn be shown tht { 0 for x [0, 1), f n (x) f(x) = 1 for x = 1. However f is not continuous over [0, 1]. Exercise. Prove the clim in the exmple bove. Exmple. Consider f n (x) = 2nx(1 x 2 ) n 1 over [0, 1]. It cn be shown tht However, for every n Z + we hve f n (x) f(x) = 0 pointwise over [0, 1]. so tht 1 f n = 2n x(1 x 2 ) n 1 dx = (1 x 2 ) n 1 = 1, 1 f n = f = 0. 0 Exercise. Prove the clim in the exmple bove. Exmple. Consider f n (x) = 1 sin(nx) over [ π, π]. It is cler tht n f n (x) f(x) = 0 pointwise over [ π, π]. Becuse sin(nx) is odd nd [ π, π] is symmetric, we hve Therefore However, it cn be shown tht π π π f n = 1 n π π π f n = 0 sin(nx) dx = f = 0. lim f n(x) = lim cos(nx) diverges for every nonzero x [ π, π]. Exercise. Prove the clim in the exmple bove. Exercise. Find sequence {f n } n N of continuously differentible functions over [0, 1] such tht f n 0 pointwise over [0, 1], nd f n(x) diverges for x [0, 1] Q.
3 12.2. Uniform Convergence. In the previous section we sw tht the pointwise limit of sequence of functions my not hve mny properties we might expect. In this section we introduce the uniform limit of sequence of functions, which will behve better. We motivte this notion from the following chrcteriztion of pointwise convergence. Proposition Let S R. Let {f n } n N be sequence of relvlued functions tht re ech defined over S. Let f be relvlued functon tht is defined over S. Then f n f pointwise over S if nd only if for every x S nd every ɛ > 0 there exists n n x,ɛ N such tht Proof. Exercise. n n x,ɛ = f n (x) f(x) < ɛ. We now define uniform convergence, which is stronger notion of convergence. Definition Let S R. Let {f n } n N be sequence of relvlued functions tht re ech defined over S. The sequence {f n } n N is sid to be uniformly convergent or to converge uniformly over S if there exists function f defined over S such tht for every ɛ there exists n n ɛ N such tht for every x S n n ɛ = f n (x) f(x) < ɛ. We sy f n converges to f uniformly over S nd cll f the uniform limit of the sequence {f n } n N over S. We denote this s f n f uniformly over S. It should be cler from this definition nd from Proposition 12.2 tht uniform convergence implies pointwise convergence. Proposition Let S R. Let {f n } n N be sequence of relvlued functions tht re ech defined over S. Let f be relvlued functon tht is defined over S. If f n f uniformly over S then f n f pointwise over S. Proof. Exercise. Remrk. This is why we sy uniform convergence is stronger notion of convergence thn pointwise convergence. The first pyoff of this stronger notion is the following. Proposition Let S R. Let {f n } n N be sequence of relvlued functions tht re ech continuous over S. Let f be relvlued functon tht is defined over S. If f n f uniformly over S then f is continuous over S. Proof. Let x S be rbitrry. We wnt to show tht f is continuous t x. Let ɛ > 0. Becuse f n f uniformly over S, there exists n N such tht 3 f n (z) f(z) < ɛ 3 for every z S. Becuse f n is continuous over S there exists δ > 0 such tht for every y S y x < δ = f n (y) f n (x) < ɛ 3.
4 4 Then y x < δ = f(y) f(x) f(y) f n (y) + f n (y) f n (x) + f n (x) f(x) < ɛ 3 + ɛ 3 + ɛ 3 = ɛ. Therefore f is continuous t x S. Becuse x S ws rbitrry, we conclude tht f is continuous over S. We cn replce continuous by uniformly continuous in the foregoing proposition. Proposition Let S R. Let {f n } n N be sequence of relvlued functions tht re ech uniformly continuous over S. Let f be relvlued functon tht is defined over S. If f n f uniformly over S then f is uniformly continuous over S. Proof. Exercise. Uniform convergence behves s we might hope for integrls. Proposition Let < b. Let {f n } n N be sequence of relvlued functions tht re ech Riemnn integrble over [, b]. Let f be relvlued functon tht is defined over [, b]. If f n f uniformly over [, b] then f is Riemnn integrble over [, b] nd lim b f n = Proof. Let ɛ > 0. Becuse f n f uniformly over [, b] there exists n n N such tht ɛ f n (z) f(z) < for every z [, b]. 3(b ) Becuse f n is Riemnn integrble over [, b] there exists prtition P of [, b] such tht b f. 0 U(f n, P ) L(f n, P ) < ɛ 3. Becuse f n is bounded over [, b] nd becuse for every z [, b] ɛ f n (z) 3(b ) < f(z) < f ɛ n(z) + 3(b ), we conclude tht f is bounded over [, b] nd tht Then L(f n, P ) ɛ 3 < L(f, P ) U(f, P ) < U(f n, P ) + ɛ 3. 0 < U(f, P ) L(f, P ) < U(f n, P ) + ɛ 3 L(f n, P ) + ɛ 3 < ɛ 3 + ɛ 3 + ɛ 3 = ɛ. Therefore f is Riemnn integrble over [, b]. The story for the convergence of derivtives is more complicted. For exmple, consider the sequence of functions over R given by f n (x) = 1 3 n sin(9n x). It should be cler tht f n 0 uniformly over R. Ech f n is differentible over R with f n(x) = 3 n cos(9 n x).
5 5 Notice tht if x = (kπ)/9 m for some k Z nd m N then for every n > m we hve f n(x) = 3 n cos ( 9 n m kπ ) = ( 1) k 3 n. Therefore the rel sequence {f n(x)} diverges to + when k is even nd diverges to when k is odd. But the sets { } } kπ : k is even nd m N 9m, { kπ : k is odd nd m N 9m re ech dense in R. It is cler tht the sequence of functions {f n} is not well behved s n. The foregoing discussion shows tht knowing sequence of continuously differentible functions {f n } converges uniformly to continuously differentible function f does not imply tht the sequence of their derivtives {f n} will converge pointwise to f, or even tht it will converge pointwise t ll! However, if we ssume tht the sequence derivtives {f n} converges uniformly then we cn use the Fundmentl Theorems of Clculus nd Propositions 12.4 nd 12.6 to obtin the following useful theorem. Proposition Let < b. Let {f n } n N be sequence of relvlued continuously differentible functions over [, b]. Let f nd g be relvlued functions over [, b] such tht f n f pointwise over [, b]; f n g uniformly over [, b]. Then g is continuous over [, b] nd f is continuously differentible over [, b] with f = g. Proof. Becuse ech f n is continuously differentible over [, b], the First Fundmentl Theorem of Clculus implies tht f n (x) = f n () + x Becuse f n f pointwise over [, b] we hve f n for every x [, b]. lim f n(x) = f(x) for every x [, b]. Becuse f n g uniformly over [, b] nd ech f n is continuous, Proposition 12.4 implies tht g is continuous over [, b]. Moreover, Propostion 12.6 implies tht Therefore lim x f n = f(x) = f() + x x g for every x [, b]. g for every x [, b]. Becuse g is continuous, the Second Fundmentl Theorem of Clculus implies tht the bove righthnd side is continuously differentible over [, b] nd tht its derivtive is g. Therefore f is continuously differentible over [, b] nd f = g.
6 Uniform Norms. Uniform convergence is best studied with tool clled the uniform norm. Let S R. Let B(S) denote the set of ll bounded functions f : S R. Then for every f B(S) we define its uniform norm f B(S) by (12.2) f B(S) = sup { f(x) : x S }. Clerly, f B(S) if nd only if f B(S) <. Exercise. Let S R. Show tht f B(S) if nd only if f B(S) <. Remrk. Uniform norms were introduced by Krl Weierstrss in the mid 1800s, but he neither cll them tht nor denoted them s we do here. He used the nottion M(f), which he clled the mjorizer of f. The modern nme nd nottion used here where introduced in the 1900s, when it ws relized tht mjorizers re specil cse of the concept of norm. Before we show how the set B(S) nd its uniform norm re connected to uniform convergence, we collect some of their bsic properties. Proposition Let S R. Then for every α R nd every f, g B(S) we hve (12.3) αf B(S), f + g B(S), nd fg B(S). Moreover, for every α R nd every f, g B(S) the uniform norm B(S) stisfies: f B(S) 0, nonegtivity; f B(S) = 0 if nd only if f = 0, definitness; αf B(S) = α f B(S), homogeneity; f + g B(S) f B(S) + g B(S), tringle inequlity; fg B(S) f B(S) g B(S), product inequlity. Proof. Exercise. Exercise. Prove Proposition Remrk. The first two properties in (12.3) stte tht the set B(S) is liner spce over the rels. All three properties in (12.3) stte tht the set B(S) is n lgebr over the rels. The first four properties of B(S) listed in Proposition 12.8 re shred by ll norms. You will meet these bstrctions in lter courses. Remrk. Proposition 12.8 shows tht if f nd g re functions over S such tht f g B(S) then we cn think of f g B(S) s mesure of distnce between f nd g. The connection of the set B(S) nd its uniform norm with uniform convergence is provided by the following chrcteriztion, due to Weierstrss. Proposition Let S R. Let {f n } n N be sequence of relvlued functions tht re ech defined over S. Let f be relvlued functon tht is defined over S. Then f n f uniformly over S s n if nd only if f n f B(S) eventully nd f n f B(S) 0 s n. Proof. Exercise. Exercise. Prove Proposition 12.9.
7 12.4. Uniformly Cuchy. In 1841 Weierstrss gve beutiful extension to sequences of functions of the Cuchy Criterion for convergence of sequences of rel numbers. We begin by defining wht it mens for sequence of functions to be uniformly Cuchy. Definition Let S R. Let {f n } n N be sequence of relvlued functions tht re ech defined over S. We sy tht the sequence {f n } n N is uniformly Cuchy over S if for every ɛ > 0 there exists N ɛ N such tht (12.4) m, n N ɛ = f m f n B(S) < ɛ. Remrk. Notice tht this definition is nlogous to the definition of wht it mens for sequence of rels to be Cuchy, with the uniform norm plying the role of the bsolute vlue. The key fct bout uniformly Cuchy sequences is the following. Proposition Let S R. Let {f n } n N be sequence of relvlued functions tht is uniformly Cuchy over S. Then for every ɛ > 0 there exists N ɛ N such tht for every x S we hve (12.5) m, n N ɛ = f m (x) f n (x) < ɛ. In prticulr, for every x S the rel sequence {f n (x)} n N is Cuchy. Proof. The result follows directly from the definition of uniformly Cuchy (12.4) nd the fct tht for every x S nd every m, n N we hve The detils re left s n exercise. Exercise. Prove Proposition f m (x) f n (x) f m f n B(S). Weierstrss gve the following chrcteriztion of uniformly convergent sequences in terms of sequences being uniformly Cuchy. Proposition (Weierstss Criterion) Let S R. Let {f n } n N be sequence of relvlued functions tht re ech defined over S. Then the sequence {f n } n N is uniformly convergent over S if nd only if it is uniformly Cuchy over S. Proof. The proof of the direction (= ) is similr to tht for the Cuchy Criterion for rel sequences, so we do it first. (= ) Let {f n } n N be uniformly convergent over S. This mens there exists function f : S R such tht f n f uniformly over S. We must show tht {f n } n N is uniformly Cuchy over S. Let ɛ > 0. Becuse f n f uniformly over S, by Proposition 12.9 there exists n N ɛ N such tht Then for every m, n N ɛ we hve n N ɛ = f n f B(S) < ɛ 2. f n f m B(S) = (f n f) (f m f) B(S) Therefore {f n } n N is uniformly Cuchy over S. f n f B(S) + f m f B(S) < ɛ 2 + ɛ 2 = ɛ. 7
8 8 The proof of the direction ( =) uses Proposition nd the Cuchy Criterion for rel sequences to construct the limiting function f. ( =) Let {f n } n N be uniformly Cuchy over S. We wnt to show tht there exists function f : S R such tht f n f uniformly over S. Proposition implies tht for every x S the sequence of rel numbers {f n (x)} n N is Cuchy, nd thereby is convergent by the Cuchy Criterion. Therefore we cn define f : S R by f(x) = lim f n (x) for every x S. This shows tht f n f pointwise over S. We still must show tht f n f uniformly over S. Let ɛ > 0. Let η ɛ (0, ɛ). Becuse {f n } n N is uniformly Cuchy over S there exists N ɛ N such tht m, n N ɛ = f n f m B(S) < η ɛ. Therefore for every x S we hve By letting m in the bove we see tht m, n N ɛ = f n (x) f m (x) < η ɛ. n N ɛ = f n (x) f(x) η ɛ < ɛ, which implies tht n N ɛ = f n f B(S) η ɛ < ɛ. Therefore f n f uniformly over S.
9 13. Series of Functions In the 1800s much mthemticl ttention ws focused on series of simple functions tht defined other functions. The two most importnt exmples of such series re power series nd trigonometric series. Power series hve the form (13.1) c k x k. These define functions s sums of monomils. They re ntrul extensions of polynomils tht were studied long before the time of Newton nd Liebniz. Newton used power series to express solutions of differentil equtions. This technique becme wide spred in the 1800s. Trigonometric series hve the form (13.2) ( k cos(kx) + b k sin(kx) ). k=1 These define 2πperiodic functions s sums of bsic trigonometric functions. They were used in the 1700s by D. Bernoulli nd L. Euler to express certin continuous periodic functions. In 1807 J. Fourier climed tht periodic functions with jump discontinuities could lso be expressed s such sum. At tht time this clim ws controversil becuse then mny (mybe most) mthemticins did not believe tht summing nice functions like cos(kx) nd sin(kx) could ever produce functions with jump discontinuities! This controversy drove much of the work in the 1800s to understnd the sense in which such series converge. For exmple, Riemnn developed his integrl theory s prt of this effort. Eventully Fourier ws proven correct. Trigonometric series re lso clled Fourier series to recognize the importnce of his work Uniform Convergence for Series of Functions. Now consider sequence {h k } of relvlued functions defined over common domin D R. For ech x D consider the infinite series h k (x). Let S = {x D : the bove series converges}. Define function f : S R by (13.3) f(x) = h k (x) for every x S. This sttes tht the ssocited sequence of prtil sums converges to f pointwise over S. The set S is clled the domin of convergence for the series. The following notions of uniform convergence for series of functions re nturl. Definition The series of functions (13.3) is sid to converge uniformly over set S if the ssocited sequence of prtil sums converges to f uniformly over S. It is sid to converge bsolutely uniformly over set S if the sequence of prtil sums ssocited with h k converges uniformly over S. With this definition of uniform convergence for series the following propositions re immedite corollries of Propositions 12.4, 12.5, 12.6, nd 12.7.
10 10 Proposition Let S R. Let {h k } be sequence of functions in B(S) such tht h k converges uniformly over S. Let f : S R be defined by f(x) = h k (x) for every x S. Then we hve the following. If ech h k is continuous over S then f is continuous over S. If ech h k is uniformly continuous over S then f is uniformly continuous over S. Proof. Exercise. The next proposition gives conditions tht permit termbyterm integrtion of infinite series. Proposition Let [, b] R. Let {h k } be sequence of functions in B([, b]) such tht h k converges uniformly over [, b]. Let f : [, b] R be defined by f(x) = h k (x) for every x [, b]. If ech h k is Riemnn integrble over [, b] then f is Riemnn integrble over [, b] with b b f = h k. Proof. Exercise. The next proposition gives conditions tht permit termbyterm differentition of infinite series. Proposition Let [, b] R. Let {h k } be sequence of continuously differentible functions over [, b] such tht nd h k converge uniformly over [, b]. h k Let f : [, b] R be defined by f(x) = h k (x) for every x [, b]. Then f is continuously differentible over [, b] with f (x) = h k(x) for every x [, b].
11 Proof. Exercise. Exercise. Prove Proposition Exercise. Prove Proposition Exercise. Prove Proposition In order to pply Propositions 13.1, 13.2, nd 13.3, we need useful criterion tht tells us when series of functions converges uniformly over set S. Weierstrss gve such criterion for series bsed upon the Weierstrss Criterion for sequences of functions, Proposition Proposition (Weierstrss MTest) Let S R. Let {h k } be sequence of functions in B(S) tht stisfies (13.4) Then (13.5) h k B(S) <. h k converges bsolutely uniformly over S. Remrk. This is clled the Weierstrss MTest becuse he used the nottion M k = h k B(S) in (13.4), becuse he clled M k the mjorizer of h k. Proof. Let f n be the n th prtil sum of (13.4), which for ech n N is the function over S defined by n f n (x) = h k (x) for every x S. We will show tht the sequence {f n } is uniformly Cuchy, nd thereby is uniformly convergent by the Weierstrss Criterion, Proposition Let ɛ > 0. By condition (13.4) there exists n N ɛ N such tht k=n ɛ h k B(S) < ɛ. Let m, n N ɛ. Without loss of generlity we cn ssume tht m < n. Then n n f n f m B(S) = h k h k B(S) h k B(S) < ɛ. k=m+1 B(S) k=m+1 k=n ɛ Therefore the sequence {f n } is uniformly Cuchy, nd thereby is uniformly convergent by the Weierstrss Criterion, Proposition Finlly, becuse the sequence {h k } stisfies condition (13.4), the sequence { h k } lso stisfies condition (13.4). Therefore, by wht we hve lredy proved, the series ssocited with { h k } is uniformly convergent over S. Therefore the series in (13.5) converges bsolutely uniformly over S. 11
12 Power Series. We now pply the propositions from the previous section to power series. In order to pply the Root Test to the power series (13.1), we compute ρ = lim sup ck x k 1 k = x lim sup c k 1 k. k k The Root Test sttes the series converges bsolutely when ρ < 1 nd diverges when ρ > 1. Therefore the series (13.1) converges bsolutely when x < R nd diverges when x > R where (13.6) 1 R = lim sup c k 1 k, k with R = 0 when the lim sup is, nd R = when the lim sup is 0. This result is often clled the CuchyHdmrd Theorem. It ws published by Cuchy in 1821 nd by Hdmrd in The number R is clled the rdius of convergence for the power series (13.1). When R < the intervl of convergence of the series will be either ( R, R), [ R, R), ( R, R], or [ R, R], depending on the convergence or divergence of the series c k ( R) k nd c k R k. When R = the intervl of convergence is R. The power series (13.1) converges to function f(x) over this intervl of convergence. We now consider those power series (13.1) with rdius of convergence R > 0 nd intervl of convergence I. We hve the following. Proposition Let the power series (13.1) hve rdius of convergence R > 0 nd intervl of convergence I. For every x I let f(x) be the sum of this series. Then the series converges bsolutely to f uniformly over [, b] for every [, b] ( R, R). The function f is uniformly continuous over every [, b] ( R, R) nd (13.7) b f = b c k x k = c k b k+1 k+1 k + 1 The function f is infinitely differentible over ( R, R) with f (n) for ech n N given by (13.8) f (n) (k + n)! (x) = c k+n x k for every x ( R, R). n! Remrk. This result sttes tht functions defined by power series with rdius of convergence R re infinitely differentible over ( R, R) with derivtives tht re lso given by power series with the sme rdius of convergence. Moreover, formul (13.8) shows tht the power series for the derivtives re obtined vi termbyterm differenttion of the power series for the function. Proof. Let [, b] ( R, R). Let r (0, R) such tht [, b] [ r, r]. Let s (r, R). Becuse It follows tht lim sup c k 1 1 k = k R < 1 s. c k < 1 s k eventully s k..
13 This implies tht c k x k B([,b]) < rk eventully s k. s k By the Direct Comprison Test, we conclude tht c k x k B([,b]) <. Then by the Weierstrss MTest, the series converges bsolutely to f uniformly over [, b], where [, b] ( R, R) ws rbitrry. Becuse c k x k is uniformly continuous over every [, b] ( R, R), Proposition 13.1 implies tht the function f is uniformly continuous over every [, b] ( R, R). Proposition 13.2 then implies tht (13.7) holds. In order to show tht f is infinitely differentible with f (n) given by (13.8) for every n N, we first show tht for every [, b] ( R, R) we hve (13.9) (k + n)! n! c k+n x k B([,b]) <. Let r (0, R) such tht [, b] [ r, r]. Let s (r, R). By n rgument given erlier we hve c k+n x k B([,b]) < rk eventully s k. s k+n Becuse r < s, the Direct Comprison Test implies tht (13.9) holds becuse for every n N (k + n)! n! r k s k+n = n! s (s r) n+1 <. Therefore the Weierstrss MTest, Proposition 13.4, implies tht (k + n)! c k+n x k converges uniformly over [, b]. n! Hence, if f is ntimes continuously differentible nd f (n) is given by (13.8) for some n N then Proposition 13.3 implies tht f is (n + 1)times continuously differentible nd tht f (n+1) is given by (13.8). Therefore by induction on n we conclude tht f is infinitely differentible nd tht f (n) is given by (13.8) for every n N. 13
14 Everywhere Continuous, Nowhere Differentible Functions In 1872 Krl Weierstrss strtled the mthemticl world by publishing exmples of functions tht where continuous everywhere but differentible nowhere. Specificlly, he considered functions of the form (14.1) f(x) = k cos ( b k x ), where 0 < < 1 < b with b 1. It is cler by the Weierstrss Mtest tht this series converges uniformly over R, nd thereby defines function f tht is continuous over R. It is eqully cler tht if b < 1 then this function would be continuously differentible over R with f (x) = k b k sin ( b k x ). Therefore the condition b 1 is necessry for the function f defined by (14.1) to be nondifferentble somewhere in R. Weierstrss showed f is not differentible nywhere in R when b is n odd integer tht stisfies b > π. These conditions on b were reduced to the necessry 2 condition b 1 in 1916 when Godfrey Hrold Hrdy showed tht every function f defined by (14.1) is not differentible nywhere in R! We will not prove the theorem of Hrdy. Rther, we will give different construction of n everywhere continuous, nowhere differentible function built up from soclled zigzg functions Zigzg Functions. The bsic building block of our construction will be the unit zigzg function z, which we define for every x R by { 2n x when x [2n 1, 2n) for some n Z, (14.2) z(x) = x 2n when x [2n, 2n + 1) for some n Z. Clerly z is continuous, periodic function with (14.3) z(x + 2) = z(x), 0 z(x) 1, for every x R. It is lso piecewise differentible over R with { (14.4) z 1 when x (2n 1, 2n) for some n Z, (x) = 1 when x (2n, 2n + 1) for some n Z. It is not differentible for x Z. Let { z k } k N be the sequence of zigzg functions defined by (14.5) z k (x) = 4 k z ( 4 k x ). Clerly z k is continuous, periodic function with (14.6) z k (x k ) = z k (x), 0 z k (x) 4 k, for every x R. It is piecewise differentible over R with { (14.7) z 1 k(x) = when x ( 2n 1 4 k, 2n 1 when x ( 2n 4 k, 2n+1 It is not differentible t points x R such tht 4 k x Z. ) for some n Z, 4 k ) for some n Z. 4 k
15 14.2. An Everywhere Continuous Function. Becuse the unit zigzg function z stisfies 0 z(x) z(1) = 1 for every x R, we see tht it is in B(R) nd stisfies z B(R) = 1. Becuse z k (x) = 4 k z(4 k x) for every x R nd k N, we see tht z k B(R) for every k N, nd tht (14.8) z k B(R) = 4 k for every k N. The geometric series summtion formul with r = 4 1 then shows tht z k B(R) = 4 k 1 = 1 4 = <. The Weierstrss MTest then implies tht the series (14.9) z k converges bsolutely uniformly over R. Therefore we cn define function h : R R by (14.10) h(x) = z k (x) for every x R. By Proposition 13.1 the function h is continuous over R. In the next section we will show tht h is not differentible t ny point of R. Exercise. Prove tht h defined by (14.10) is periodic with period 2. Exercise. Prove tht h defined by (14.10) is unifromly continuous over R Tht is Nowhere Differentible. We wnt to show tht the function h defined by (14.10) is nowhere differentible. This mens tht h is not differentible t ny point of R. More specificlly, we wnt to show for every x R tht (14.11) lim y x h(y) h(x) y x diverges. It suffices to show for every x R tht there exists sequence {x n } n N R {x} such tht (14.12) lim x n = x, but lim h(x n ) h(x) x n x diverges. The construction of the sequence {x n } n N is bsed upon the observtion tht for every x R the unit zigzg function z either is monotonic on [x, x + 1] or is monotonic on [x 1, x]. The 2 2 grph of z should mke this cler. It follows tht for every x R nd every n N the zigzg function z n either is monotonic on [x, x n ] or is monotonic on [x n, x]. Now let x R. By the prgrph bove, we cn define the sequence {x n } n N by { x + 1 (14.13) x n = 2 4 n if z n is monotonic on [x, x n ], x n otherwise. Becuse x n x = n, it is cler tht lim x n = x. 15
16 16 In order to estblish (14.12) it remins to prove tht (14.14) lim h(x n ) h(x) x n x diverges. The proof of (14.14) is bsed upon two more observtions. First, becuse z is 2periodic, if k > n then z k (x n ) = 4 k z ( 4 k ( x ± n)) = 4 k z ( 4 k x ± 1 2 4k n) = 4 k z ( 4 k x ± 2 4 k n 1) = 4 k z ( 4 k x ) = z k (x). Second, if k n then: if z n is monotonic on [x, x n ] then z k is monotonic on [x, x n ]; if z n is monotonic on [x n, x] then z k is monotonic on [x n, x]. This mens tht if k n then z k (x n ) z k (x) = ±1. x n x Upon combining definition (14.10) of h with these observtions we see tht (14.15) h(x n ) h(x) x n x = z k (x n ) z k (x) x n x = n z k (x n ) z k (x) x n x = n ±1. It follows tht h(x n+1 ) h(x) h(x n) h(x) x n+1 x x n x = 1, whereby (14.14) is stisfied for this sequence. Therefore h is nowhere differentible over R. Remrk. It follows from (14.15) tht { h(x n ) h(x) odd when n is even, = x n x even when n is odd. Remrk. It turns out tht most continuous functions re nowhere differentible. This sense in which this is true is beyond wht we cn do here.
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