Functions of Bounded Variation and Rectifiable Curves

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1 Fuctios of Bouded Variatio ad Rectifiable Curves Fuctios of bouded variatio 6.1 Determie which of the follwoig fuctios are of bouded variatio o 0, 1. (a) fx x si1/x if x 0, f0 0. (b) fx x si1/x if x 0, f0 0. Proof: (a) Sice f x x si1/x cos1/x for x 0, 1 ad f 0 0, we kow that f x is bouded o 0, 1, i fact, f x 3o0, 1. Hece, f is of bouded variatio o 0, 1. 1 (b) First, we choose 1beaeveitegersothat 1, ad thus cosider a 1 partitio P 0 x 0, x 1 1 we have, x 1,..., x 1 f k, x /k., x 1, the Sice 1/k diverges to, we kow that f is ot of bouded variatio o 0, A fuctio f, defied o a, b, is said to satisfy a uiform Lipschitz coditio of order 0oa, b if there exists a costat M 0 such that fx fy M x y for all x ad y i a, b. (Compare with Exercise 5.1.) (a) If f is such a fuctio, show that 1 implies f is costat o a, b, whereas 1 implies f is of bouded variatio a, b. Proof: As 1, we cosider, for x y, wherex, y a, b, fx fy 0 M x y 1. x y Hece, f x exists o a, b, ad we have f x 0oa, b. So, we kow that f is costat. As 1, cosider ay partitio P a x 0, x 1,..., x b, wehave f k M x x k Mb a. That is, f is of bouded variatio o a, b. (b) Give a example of a fuctio f satisfyig a uiform Lipschitz coditio of order 1oa, b such that f is ot of bouded variatio o a, b. Proof: First, ote that x satisfies uiform Lipschitz coditio of order, where Choosig 1 such that 1 ad let M sice the series k coverges. So, we have M k Defie a fuctio f as follows. We partitio 0, 1 ito ifiitely may subsitervals. Cosider x 0 0, x 1 x 0 M 1 1 1, x x 1 M 1 1,..., x x 1 M 1 1,... Ad i every subiterval x i, x i1,wherei 0,1,..., we defie

2 fx x x i x i1 the f is a cotiuous fuctio ad is ot bouded variatio o 0, 1 sice, 1 M diverges. I order to show that f satisfies uiform Lipschitz coditio of order, we cosider three cases. (1) If x, y x i, x i1,adx, y x i, x ix i1 or x, y x ix i1, x i1, the fx fy x y x y. () If x, y x i, x i1,adx x i, x ix i1 or y x ix i1, x i1, the there is a z x i, x ix i1 such that fy fz. So, fx fy fx fz x z x z x y. (3) If x x i, x i1 ad y x j, x j1,wherei j. If x x i, x ix i1, the there is a z x i, x ix i1 such that fy fz. So, fx fy fx fz x z x z x y. Similarly for x x ix i1, x i1. Remark: Here is aother example. Sice it will use Fourier Theory, we do ot give a proof. We just write it dow as a referece. ft cos3k t 3 k. (c) Give a example of a fuctio f which is of bouded variatio o a, b but which satisfies o uiform Lipschitz coditio o a, b. Proof: Sice a fuctio satisfies uiform Lipschitz coditio of order 0, it must be cotiuous. So, we cosider x if x a, b fx b 1ifx b. Trivially, f is ot cotiuous but icreasig. So, the fuctio is desired. Remark: Here is a good problem, we write it as follows. If f satisfies fx fy K x y 1/ for x 0, 1, wheref0 0. defie fx if x 0, 1 gx 0ifx 0. The g satisfies uiform Lipschitz coditio of order 1/6. Proof: Note that if oe of x, ady is zero, the result is trivial. So, we may cosider 0 y x 1 as follows. Cosider gx gy fx fx fx fy y 1/3 fy fy fy fy fy y 1/3 1 k fy y 1/3. *

3 For the part fx x fy 1 fx fy 1/3 x1/3 K x x 1/3 y 1/ by hypothesis K x y 1/ x y 1/3 sice x x y 0 K x y 1/6. A fy For aother part fy, we cosider two cases. y 1/3 (1) x y x x y y 0, fy fy y 1/3 fy fy y 1/3 xy 1/3 x y 1/3 xy 1/3 sice y 1/3 x y 1/3 for all x, y 0 fy xy 1/3 sice x y 1/3 fy 1 y 1/3 K y 1/ y 1/3 K y 1/6 by hypothesis K x y 1/6 sice y x y. B () x y x y x y 0, fy fy y 1/3 fy fy fy y 1/3 xy 1/3 x y 1/3 xy 1/3 sice y 1/3 x y 1/3 for all x, y 0 x y 1/3 y /3 K y 1/ x y 1/3 y /3 K y 1/6 x y 1/3 sice x y by hypothesis K x y 1/6 x y 1/3 sice y x y K x y 1/6. C So, by (A)-(C), (*) tells that g satisfies uiform Lipschitz coditio of order 1/6. Note: Hereisageeralresult.Let0. Iff satisfies fx fy K x y for x 0, 1, wheref0 0. defie

4 fx x if x 0, 1 gx 0ifx 0. The g satisfies uiform Lipschitz coditio of order. The proof is similar, so we omit it. 6.3 Show that a polyomial f is of bouded variatio o every compact iterval a, b. Describe a method for fidig the total variatio of f o a, b if the zeros of the derivative f are kow. Proof: If f is a costat, the the total variatio of f o a, b is zero. So, we may assume that f is a polyomial of degree 1, ad cosider f x 0 by two cases as follows. (1) If there is o poit such that f x 0, the by Itermediate Value Theorem of Differetiability, we kow that f x 0oa, b, orf x 0oa, b. So, it implies that f is mootoic. Hece, the total variatio of f o a, b is fb fa. () If there are m poits such that f x 0, say a x 0 x 1 x... x m b x m1, where 1 m, the we kow the mootoe property of fuctio f. So, the total variatio of f o a, b is m1 fx i fx i1. i1 Remark: Here is aother proof. Let f be a polyomial o a, b, the we kow that f is bouded o a, b sice f is also polyomial it is cotiuous. Hece, we kow that f is of bouded variatio o a, b. 6.4 A oempty set S of real-valued fuctios defied o a iterval a, b is called a liear space of fuctios if it has the followig two properties: (a) If f S, the cf S for every real umber c. (b) If f S ad g S, the f g S. Theorem 6.9 shows that the set V of all fuctios of bouded variatio o a, b is a liear space. If S is ay liear space which cotais all mootoic fuctios o a, b, prove that V S. This ca be described by sayig that the fuctios of bouded variatio form the samllest liear space cotaiig all mootoic fuctios. Proof: It is directlt from Theorem 6.9 ad some facts i Liear Algebra. We omit the detail. 6.5 Let f be a real-valued fuctio defied o 0, 1 such that f0 0, fx x for all x, adfx fy wheever x y. LetA x : fx x. Prove that sup A A, adthat f1 1. Proof: Note that sice f0 0, A is ot empty. Suppose that sup A : a A, i.e., fa a sice fx x for all x. So, give ay 0, the there is a b A such that a b. * I additio, b fb sice b A. ** So,by(*)ad(**),wehave(let 0 ), a fa fa sice f is mootoic icreasig. which cotradicts to fa a. Hece, we kow that sup A A.

5 Claim that 1 sup A. Suppose NOT, thatis,a 1. The we have a fa f1 1. Sice a sup A, cosider x a, fa, the fx x fa a which cotradicts to a fa. So, we kow that sup A 1. Hece, we have proved that f1 1. Remark: The reader should keep the method i mid if we ask how to show that f1 1 directly. The set A is helpful to do this. Or equivaletly, let f be strictly icreasig o 0, 1 with f0 0. If f1 1, the there exists a poit x 0, 1 such that fx x. 6.6 If f is defied everywhere i R 1, the f is said to be of bouded variatio o, if f is of bouded variatio o every fiite iterval ad if there exists a positive umber M such that V f a, b M for all compact iterval a, b. The total variatio of f o, is the defied to be the sup of all umbers V f a, b, a b, ad deoted by V f,. Similar defiitios apply to half ope ifiite itervals a, ad, b. (a) State ad prove theorems for the iifite iterval, aalogous to the Theorems 6.7, 6.9, 6.10, 6.11, ad 6.1. (Theorem 6.7*) Letf : R R be of bouded variato, the f is bouded o R. Proof: Give ay x R, the x 0, a or x a,0. Ifx 0, a, the f is bouded o 0, a with fx f0 V f 0, a f0 V f,. Similarly for x a,0. (Theorem 6.9*) Assume that f, adg be of bouded variato o R, thesoarethier sum, differece, ad product. Also, we have V fg, V f, V g, ad V fg, AV f, BV g,, where A sup xr gx ad B sup xr fx. Proof: For sum ad differece, give ay compact iterval a, b, wehave V fg a, b V f a, b V g a, b, V f, V g, V fg, V f, V g,. For product, give ay compact iterval a, b, wehave(letaa, b sup xa,b gx, ad Ba, b sup xa,b fx ), V fg a, b Aa, bv f a, b Ba, bv g a, b AV f, BV g,

6 V fg, AV f, BV g,. (Theorem 6.10*) Letf be of bouded variatio o R, ad assume that f is bouded away from zero; that is, suppose that there exists a positive umber m such that 0 m fx for all x R. The g 1/f is also of bouded variatio o R, ad V g, V f, m. Proof: Give ay compacgt iterval a, b, wehave V g a, b V fa, b m V f, m V g, V f,. m (Theorem 6.11*) Letf be of bouded variatio o R, ad assume that c R. The f is of bouded variatio o, c ad o c, ad we have V f, V f, c V f c,. Proof: Give ay a compact iterval a, b such that c a, b. The we have V f a, b V f a, c V f c, b. Sice V f a, b V f, V f a, c V f, ad V f c, b V f,, we kow that the existece of V f, c ad V f c,. Thatis,f is of bouded variatio o, c ad o c,. Sice V f a, c V f c, b V f a, b V f, V f, c V f c, V f,, * ad V f a, b V f a, c V f c, b V f, c V f c, V f, V f, c V f c,, ** we kow that V f, V f, c V f c,. (Theorem 6.1*) Letf be of bouded variatio o R. LetVx be defied o, x as follows: Vx V f, x if x R, adv 0. The (i) V is a icreasig fuctio o, ad (ii) V f is a icreasig fuctio o,. Proof: (i)letx y, the we have Vy Vx V f x, y 0. So, we kow that V is a icreasig fuctio o,. (ii) Let x y, the we have V fy V fx V f x, y fy fx 0. So,

7 we kow that V f is a icreasig fuctio o,. (b) Show that Theorem 6.5 is true for, if mootoic is replaced by bouded ad mootoic. State ad prove a similar modeficatio of Theorem (Theorem 6.5*) Iff is bouded ad mootoic o,, the f is of bouded variatio o,. Proof: Give ay compact iterval a, b, the we have V f a, b exists, ad we have V f a, b fb fa, sicef is mootoic. I additio, sice f is bouded o R, say fx M for all x, we kow that M is a upper bouded of V f a, b for all a, b. Hece, V f, exists. That is, f is of bouded variatio o R. (Theorem 6.13*) Letf be defied o,, the f is of bouded variatio o, if, ad oly if, f ca be expressed as the differece of two icreasig ad bouded fuctios. Proof: Suppose that f is of bouded variatio o,, the by Theorem 6.1*, we kow that f V V f, where V ad V f are icreasig o,. I additio, sice f is of bouded variatio o R, we kow that V ad f is bouded o R V f is bouded o R. So, we have proved that if f is of bouded variatio o, the f ca be expressed as the differece of two icreasig ad bouded fuctios. Suppose that f ca be expressed as the differece of two icreasig ad bouded fuctios, say f f 1 f, The by Theorem 6.9*, adtheorem 6.5*, we kow that f is of bouded variato o R. Remark: The represetatio of a fuctio of bouded variatio as a differece of two icreasig ad bouded fuctios is by o mea uique. It is clear that Theorem 6.13* also holds if icreasig is replaced by strictly icreasig. For example, f f 1 g f g, whereg is ay strictly icreasig ad bouded fuctio o R. Oe of such g is arcta x. 6.7 Assume that f is of bouded variatio o a, b ad let P x 0, x 1,...,x þa, b. As usual, write f k fx k fx, k 1,,...,. Defie AP k : f k 0, BP k : f k 0. The umbers ad p f a, b sup f k : P þa, b kap f a, b sup f k : P þa, b kbp are called respectively, the positive ad egative variatios of f o a, b. For each x i a, b. LetVx V f a, x, px p f a, x, x f a, x, ad let Va pa a 0. Show that we have: (a) Vx px x.

8 Proof: Give a partitio P o a, x, the we have f k f k f k kap kap kbp f k f k, kbp (takig supermum) Vx px x. * Remark: The existece of px ad qx is clear, so we kow that (*) holds by Theorem (b) 0 px Vx ad 0 x Vx. Proof: Cosider a, x, ad sice Vx f k f k, kap we kow that 0 px Vx. Similarly for 0 x Vx. (c) p ad are icreasig o a, b. Proof: Letx, y i a, b with x y, ad cosider py px as follows. Sice py f k f k, kap, a,y kap, a,x we kow that py px. That is, p is icreasig o a, b. Similarly for. (d) fx fa px x. Part (d) gives a alterative proof of Theorem Proof: Cosider a, x, ad sice fx fa f k f k f k kap kbp fx fa f k f k kbp kap fx fa px x. (e) px Vx fx fa, x Vx fx fa. Proof: By (d) ad (a), the statemet is obvious. (f) Every poit of cotiuity of f is also a poit of cotiuity of p ad of. Proof: By (e) ad Theorem 6.14, the statemet is obvious. Curves 6.8 Let f ad g be complex-valued fuctios defied as follows: ft e it if t 0, 1, gt e it if t 0,. (a) Prove that f ad g have the same graph but are ot equivalet accordig to defitio

9 i Sectio 6.1. Proof: Siceft : t 0, 1 gt : t 0, the circle of uit disk, we kow that f ad g have the same graph. If f ad g are equivalet, the there is a 1-1 ad oto fuctio : 0, 0, 1 such that ft gt. That is, e it cost i si t e it cost i si t. I paticular, 1 : c 0, 1. However, fc cosc i si c g1 1 c Z, a cotradictio. (b) Prove that the legth of g is twice that of f. Proof: Sice ad the legth of g 0 g t dt 4 the legth of f 0 1 f t dt, we kow that the legth of g is twice that of f. 6.9 Let f be rectifiable path of legth L defied o a, b, ad assume that f is ot costat o ay subiterval of a, b. Lets deote the arc legth fuctio give by sx f a, x if a x b, sa 0. (a) Prove that s 1 exists ad is cotiuous o 0, L. Proof: ByTheorem 6.19, we kow that sx is cotiuous ad strictly icreasig o 0, L. So, the iverse fuctio s 1 exists sice s is a 1-1 ad oto fuctio, ad by Theorem 4.9, we kow that s 1 is cotiuous o 0, L. (b) Defie gt fs 1 t if t 0, L ad show that g is equivalet to f. Sice ft gst, the fuctio g is said to provide a represetatio of the graph of f with arc legth as parameter. Proof: tisclearbytheorem Let f ad g be two real-valued cotiuous fuctios of bouded variatio defied o a, b, with 0 fx gx for each x i a, b, fa ga, fb gb. Leth be the complex-valued fuctio defied o the iterval a,b a as follows: ht t ift, ifa t b b t igb t, ifb t b a. (a) Show that h describes a rectifiable curve. Proof: It is clear that h is cotiuous o a,b a. Note that t, f ad g are of bouded variatio o a, b, so h a,b a exists. That is, h is rectifiable o a,b a. (b) Explai, by meas of a sketch, the geometric relatioship betwee f, g, adh. Solutio: The reader ca give it a draw ad see the graph lyig o x y plae is a

10 closed regio. (c) Show that the set of poits S x, y : a x b, fx y gx i a regio i R whose boudary is the curve. Proof: It ca be aswered by (b), so we omit it. (d) Let H be the complex-valued fuctio defied o a,b a as follows: Ht t 1 igt ft, ifa t b b t 1 igb t fb t, ifb t b a. Show that H describes a rectifiable curve 0 which is the boudary of the regio S 0 x, y : a x b, fx gx y gx fx. Proof: LetFt 1 gt ft ad Gt 1 gt ft defied o a, b. Itis clear that Ft ad Gt are of bouded variatio ad cotiuous o a, b with 0 Fx Gx for each x a, b, Fb Gb 0, ad Fb Gb 0. I additio, we have Ht t ift, ifa t b b t igb t, ifb t b a. So, by precedig (a)-(c), we have prove it. (e) Show that, S 0 has the x axis as a lie of symmetry. (The regio S 0 is called the symmetrizatio of S with respect to x axis.) Proof: Itisclearsicex, y S 0 x, y S 0 by the fact fx gx y gx fx. (f) Show that the legth of 0 does ot exceed the legth of. Proof: By (e), the symmetrizatio of S with respect to x axis tells that H a, b H b,b a. So, it suffices to show that h a,b a H a, b. Choosig a partitio P 1 x 0 a,...,x b o a, b such that H a, b H P 1 x i x i1 1 f gx i 1 f gx i1 1/ i1 i1 4x i x i1 f gx i f gx i1 1/ ad ote that b a b a b, we use this P 1 to produce a partitio P P 1 x b, x 1 b x x 1,...,x b a o a,b a. The we have *

11 h P hx i hx i1 i1 hx i hx i1 hx i hx i1 i1 i1 i1 i1 x i x i1 fx i fx i1 1/ i1 x i x i1 gx i gx i1 1/ x i x i1 fx i fx i1 1/ xi x i1 gx i gx i1 1/ From (*) ad (**), we kow that H a, b H P 1 h P *** H a,b a H a, b h a,b a. So, we kow that the legth of 0 does ot exceed the legth of. Remark: Defiex i x i1 a i, fx i fx i1 b i,adgx i gx i1 c i, the we have 4a i b i c i 1/ ai b i 1/ a i c i 1/. Hece we have the result (***). Proof: It suffices to square both side. We leave it to the reader. Absolutely cotiuous fuctios A real-valued fuctio f defied o a, b is said to be absolutely cotiuous o a, b if for every 0, there is a 0 such that fb k fa k for every disjoit ope subitervals a k, b k of a, b, 1,,..., the sum of whose legths b k a k is less tha. Absolutely cotiuous fuctios occur i the Lebesgue theory of itegratio ad differetiatio. The followig exercises give some of their elemetary properties Prove that every absolutely cotiuous fuctio o a, b is cotiuous ad of bouded variatio o a, b. Proof: Letf be absolutely cotiuous o a, b. The 0, there is a 0 such that fb k fa k for every disjoit ope subitervals a k, b k of a, b, 1,,..., the sum of whose legths b k a k is less tha. So,as x y, wherex, y a, b, wehave fx fy. That is, f is uiformly cotiuous o a, b. So,f is cotiuous o a, b. I additio, give ay 1, there exists a 0 such that as b k a k, where a k, b k s are disjoit ope itervals i a, b, wehave **

12 fb k fa k 1. For this, ad let K be the smallest positive iteger such that K/ b a. So,we partitio a, b ito K closed subitervals, i.e., P y 0 a, y 1 a /,...,y K1 a K 1/, y K b. So, it is clear that f is of bouded variatio y i, y i1,wherei 0, 1,..., K. It implies that f is of bouded variatio o a, b. Note: There exists fuctios which are cotiuous ad of bouded variatio but ot absolutely cotiuous. Remark: 1. The stadard example is called Cator-Lebesgue fuctio. The reader ca see this i the book, Measure ad Itegral, A Itroductio to Real Aalysis by Richard L. Wheede ad Atoi Zygmud, pp 35 ad pp If we wrtie absolutely cotiuous by ABC, cotiuous by C, ad bouded variatio by B, the it is clear that by precedig result, ABC implies B ad C, adb ad C do NOT imply ABC. 6.1 Prove that f is absolutely cotiuous if it satisfies a uiform Lipschitz coditio of order 1 o a, b. (See Exercise 6.) Proof: Letf satisfy a uiform Lipschitz coditio of order 1 o a, b, i.e., fx fy M x y where x, y a, b. Thegive 0, there is a /M such that as b k a k, wherea k, b k s are disjoit ope subitervals o a, b, k 1,..,, we have fb k fa k M b k a k Mb k a k M. Hece, f is absolutely cotiuous o a, b If f ad g are absolutely cotiuous o a, b, prove that each of the followig is also: f, cf (c costat), f g, f g; alsof/g if g is bouded away from zero. Proof: (1) ( f is absolutely cotiuous o a, b): Give 0, we wat to fid a 0, such that as b k a k, wherea k, b k s are disjoit ope itervals o a, b, wehave fb k fa k. 1* Sice f is absolutely cotiuous o a, b, for this, there is a 0 such that as b k a k, wherea k, b k s are disjoit ope itervals o a, b, wehave fb k fa k (1*) holds by the followig

13 fb k fa k fb k fa k. So, we kow that f is absolutely cotiuous o a, b. () (cf is absolutely cotiuous o a, b): If c 0, it is clear. So, we may assume that c 0. Give 0, we wat to fid a 0, such that as b k a k, where a k, b k s are disjoit ope itervals o a, b, wehave cfb k cfa k. * Sice f is absolutely cotiuous o a, b, for this, there is a 0 such that as b k a k, wherea k, b k s are disjoit ope itervals o a, b, wehave fb k fa k / c (*) holds by the followig cfb k cfa k c fb k fa k. So, we kow that cf is absolutely cotiuous o a, b. (3) (f g is absolutely cotiuous o a, b): Give 0, we wat to fid a 0, such that as b k a k, wherea k, b k s are disjoit ope itervals o a, b, we have f gb k f ga k. 3* Sice f ad g are absolutely cotiuous o a, b, for this, there is a 0 such that as b k a k, wherea k, b k s are disjoit ope itervals o a, b, wehave fb k fa k / ad gb k ga k / (3*) holds by the followig f gb k f ga k fb k fa k gb k ga k fb k fa k gb k ga k. So, we kow that f g is absolutely cotiuous o a, b. (4) (f g is absolutely cotiuous o a, b.): Let M f sup xa,b fx ad M g sup xa,b gx. Give 0, we wat to fid a 0, such that as b k a k, wherea k, b k s are disjoit ope itervals o a, b, wehave

14 f gb k f ga k. 4* Sice f ad g are absolutely cotiuous o a, b, for this, there is a 0 such that as b k a k, wherea k, b k s are disjoit ope itervals o a, b, wehave fb k fa k M g 1 ad gb k ga k (4*) holds by the followig M f. f gb k f ga k fb k gb k ga k ga k fb k fa k gb k ga k M g M f M f 1 M g M g 1 fb k fa k M f 1 Remark: The part shows that f is absolutely cotiuous o a, b, where N, iff is absolutely cotiuous o a, b. (5) (f/g is absolutely cotiuous o a, b): By (4) it suffices to show that 1/g is absolutely cotiuous o a, b. Siceg is bouded away from zero, say 0 m gx for all x a, b. Give 0, we wat to fid a 0, such that as b k a k, where a k, b k s are disjoit ope itervals o a, b, wehave 1/gb k 1/ga k. 5* Sice g is absolutely cotiuous o a, b, for this, there is a 0 such that as b k a k, wherea k, b k s are disjoit ope itervals o a, b, wehave gb k ga k m (4*) holds by the followig 1/gb k 1/ga k gb k ga k gb k ga k 1 gb m k ga k.