# Functions of bounded variation

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1 Division for Mthemtics Mrtin Lind Functions of bounded vrition Mthemtics C-level thesis Dte: Supervisor: Viktor Kold Exminer: Thoms Mrtinsson Krlstds universitet Krlstd Tfn Fx

2 Introduction In this pper we investigte the functions of bounded vritions. We stud bsic properties of these functions nd solve some problems. I m ver grteful to m supervisor Viktor Kold for his guidnce. 1

3 Contents Introduction 1 1 Monotone functions 3 2 Functions of bounded vrition Generl properties Positive nd negtive vrition Conditions for bounded vrition The function v(x) Two limits Jump functions 29 2

4 1 Monotone functions The properties of monotone functions will be useful to us becuse lter we shll see tht some of them cn be extended directl to the functions of bounded vrition. First definition: Definition Let f : [, b] R be function. Then f is sid to be incresing on [, b] if for ever x, [, b] x < f(x) f() decresing on [, b] if for ever x, [, b] x < f(x) f() monotone if f is either incresing or decresing on [, b] If the intervl [, b] cn be divided into finite number of intervls such tht f is monotone on ech of them then f is sid to be piecewise monotone on [, b]. The following theorem is given in [2], we refer to it for proof. Theorem [2, p. 95]. Let f : [, b] R be incresing on [, b] nd suppose tht c (, b). Then f(c + 0) nd f(c 0) 1 exists nd sup{f(x) : x < c} = f(c 0) f(c) f(c + 0) = inf{f(x) : x > c} If f is decresing the theorem bove still holds, with opposite inequlities of course. Thus we cn stte tht if f is monotone nd c (, b) then both f(c + 0) nd f(c 0) exists. Therefore the following definition mkes sense: Definition Let f : [, b] R be monotone on [, b] nd let c [, b]. (i) If c [, b) we define the right-hnd jump of f t c to be σ + c = f(c + 0) f(c) (ii) If c (, b] we define the left-hnd jump of f t c to be σ c = f(c) f(c 0) 1 We denote the right-hnd nd left-hnd limits lim f(c + h) = f(c + 0) lim h 0 f(c h) = f(c 0) h 0 where h tends to 0 from the positive side. 3

5 (iii) If c [, b] we define the jump of f t c to be σ c + + σc if c (, b) σ c = σ c + if c = if c = b σ c We now mke the following simple observtion: if f : [, b] R is monotone on [, b], then f is continuous t c [, b] if nd onl if σ c = 0. Here, the necessit is obvious nd the sufficienc follows t once from theorem Of course, monotone function needn t be continuous. However we cn now prove tht if f : [, b] R is monotone then it cn t be too discontinuous. Theorem [2, p. 96]. Let f : [, b] R be monotone on [, b] nd let D be the set of ll points of discontinuit of f. Then D is t most countble. Proof: Suppose tht f is incresing on [, b]. A point c [, b] is point of discontinuit of f if nd onl if σ c 0. Since f is incresing on [, b] clerl σ c 0 for ever c [, b] nd hence D = {x [, b] : σ x > 0} For n given n N tke points x 1, x 2,..., x n stisfing x 1 < x 2 <... < x n b. For 0 j n tke points t j such tht = t 0 x 1 < t 1 < x 2 < t 2 <... < t n 1 < x n t n = b Then, since f is incresing we hve σ xj f(t j ) f(t j 1 ) for 1 j n nd it follows tht σ x1 + σ x σ xn n (f(t j ) f(t j 1 )) j=1 = f(b) f() So if D k = {x : σ x (f(b) f())/k} then D k cn hve t most k elements. Furthermore, D = k=1 D k nd since ever D k is finite D is t most countble. 4

6 2 Functions of bounded vrition 2.1 Generl properties Let f : [, b] R be function nd Π = {x 0, x 1,..., x n } prtition of [, b]. We denote V Π (f) = n 1 f(x k+1) f(x k ) nd set (f) = sup V Π (f) V b where the supremum is tken over ll prtitions of [, b]. We clerl hve 0 V b (f). The quntit V b (f) is clled the totl vrition of f over [, b]. Definition A function f : [, b] R is sid to be of bounded vrition on [, b] if V b (f) is finite. If f is of bounded vrition on [, b] we write f V [, b]. Occsionll we shll s tht function is of bounded vrition, leving out the specifiction of intervl, when the intervl in question is cler. We shll stte nd prove some importnt properties of functions of bounded vrition nd their totl vrition but first we need theorem concerning refinements of prtitions. Theorem Let f : [, b] R be function nd Π n prtition of [, b]. If Π is n refinement of Π then V Π (f) V Π (f). Proof: Since n refinement of Π cn be obtined b dding points to Π one t time it s enough to prove the theorem in the cse when we dd just one point. Tke Π = {x 0, x 1,..., x n } nd dd the point c to Π nd denote the result Π. Assume tht x j < c < x j+1 for some 0 j n 1, then the tringle inequlit gives tht nd hence V Π (f) = f(x j+1 ) f(x j ) = f(x j+1 ) f(c) + f(c) f(x j ) f(x j+1 ) f(c) + f(c) f(x j ) n 1 f(x k+1 ) f(x k ) = f(x j+1 ) f(x j ) + n 1 Π,k j f(x j+1 ) f(c) + f(c) f(x j ) + = V Π (f) 5 f(x k+1 ) f(x k ) n 1,k j f(x k+1 ) f(x k )

7 The theorem bove ssures us tht dding points to prtition Π will onl mke the sum V Π (f) lrger or perhps leve it unchnged, useful fct tht we shll use in the proof of the following theorem, given in [1]. Theorem [1, p. 120]. Let f : [, b] R nd c n rbitrr point in (, b). Then f V [, b] if nd onl if f V [, c] nd f V [c, b]. Furthermore, if f V [, b] then V b (f) = V c (f) + V b c (f) Proof: Assume tht f V [, b]. We will show tht f V [, c], the proof is similr to prove tht f V [c, b]. Tke n rbitrr prtition Π of [, c] nd dd the point b to Π nd denote the result Π, which is prtition of [, b]. We then hve V Π (f) = V Π (f) + f(b) f(c) V b (f) V Π (f) V b (f) f(b) f(c) Since V b (f) is finite, the sums V Π (f) re bounded bove nd thus sup Π V Π (f) is finite, tht is f V [, c]. Now ssume tht f V [, c] nd f V [c, b]. Let Π be n prtition of [, b]. Add the point c to Π nd denote the result Π 1. Then Π 1 = Π Π where Π is prtition of [, c] nd Π is prtition of [c, b]. Then we hve V Π (f) V Π1 (f) = V Π (f) + V Π (f) V c (f) + V b c (f) nd since both V c (f) nd Vc b (f) re finite the sums V Π (f) re bounded bove nd thus f V [, b] nd V b (f) V c (f) + Vc b (f). Now we tke n two prtitions Π nd Π of [, c] nd [c, b] respectivel nd let Π be the union of Π nd Π, then Π is prtition of [, b]. We hve V Π (f) + V Π (f) = V Π (f) V b (f) nd thus V Π (f) V b (f) V Π (f). For n fixed prtition Π of [c, b] the number V b (f) V Π (f) is n upper bound for the sums V Π (f) nd therefore V c (f) V b (f) V Π (f). This is equivlent to V Π (f) V b (f) V c (f) nd thus V b (f) V c (f) is n upper bound for the sums V Π (f) nd therefore Vc b (f) V b (f) V c (f) whence V c (f) + Vc b (f) V b (f). But then we must hve V b (f) = V c (f) + Vc b (f). 6

8 The following two theorems re given in [1] nd re strightforwrd to prove nd therefore their proofs re omitted. Theorem [1, p. 120]. Let f, g : [, b] R be of bounded vrition on [, b]. Then (f + g) V [, b] nd V b (f + g) V b (f) + V b (g). Theorem [1, p. 120]. Let f : [, b] R be of bounded vrition on [, b]. Then cf V [, b] for n c R nd V b (cf) = c V b (f) Theorem [1, p. 119]. If f : [, b] R is monotone on [, b] then f V [, b] nd V b (f) = f(b) f() Proof: We will give the proof in the cse when f is incresing, it is similr when f is decresing. Let f be incresing on [, b], then f(b) f() = f(b) f(). Tke n rbitrr prtition Π = {x 0, x 1,..., x n } of [, b]. Since f is incresing we hve f(x k+1 ) f(x k ) = f(x k+1 ) f(x k ) nd hence V Π (f) = n 1 f(x k+1 ) f(x k ) = n 1 (f(x k+1 ) f(x k )) = f(b) f() Since the sum V Π (f) is independent of the prtition Π we conclude tht V b (f) = f(b) f() Combining the theorem bove with theorem we see tht n piecewise monotone function defined on compct intervl is of bounded vrition. However, the converse is certinl not true. Indeed, there exists functions of bounded vrition tht ren t monotone on n subintervl. Even so, functions of bounded vrition cn be chrcterized in terms of monotone functions, s the following theorem due to Jordn shows. Theorem (Jordn s theorem) [1, p. 121]. Let f : [, b] R, f is of bounded vrition if nd onl if f is the difference of two incresing functions. Proof: Assume tht f V [, b] nd let v(x) = V x (f), x (, b] nd v() = 0. Then clerl f(x) = v(x) [v(x) f(x)]. We will show v(x) nd v(x) f(x) re incresing. For n x 1 < x 2 we hve v(x 2 ) v(x 1 ) = V x 2 x 1 (f) 0 v(x 2 ) v(x 1 ) so v(x) is incresing. Furthermore, 7

9 f(x 2 ) f(x 1 ) f(x 2 ) f(x 1 ) V x 2 x 1 (f) = v(x 2 ) v(x 1 ) v(x 1 ) f(x 1 ) v(x 2 ) f(x 2 ) nd thus v(x) f(x) is incresing. Conversel, suppose tht f(x) = g(x) h(x) with g nd h incresing. Since h is incresing h is decresing nd thus f(x) = g(x)+( h(x)) is the sum of two monotone functions so theorem together with theorem gives tht f V [, b] Since n function of bounded vrition cn be written s the sum of two monotone functions mn of the properties of monotone functions re inherited b functions of bounded vrition. If f V [, b], then The limits f(c + 0) nd f(c 0) exists for n c (, b). The set of points where f is discontinuous is t most countble. We shll return to these fcts lter on. The following two problems demonstrte how one cn use the theorems given bove in computtions. Problem Represent f(x) = cos 2 x, 0 x 2π s difference of two incresing functions. Solution: The proof of Jordns theorem shows tht the functions v(x) f(x) nd v(x) will do, so the problem is to determine v(x). Divide [0, 2π] into four subintervls I 1 = [0, π], I 2 2 = [ π, π], I 2 3 = [π, 3π] nd I 2 4 = [ 3π, 2π]. 2 The function f(x) decreses from 0 to 1 on I 1 nd I 3 nd increses from 0 to 1 on I 2 nd I 4, so the totl vrition of f over n of these subintervls is 1. To determine V0 x (f) we need to stud seprte cses depending on which intervl x lies in. To demonstrte the principle, ssume tht x I 3, then V0 x (f) = V π 2 0 (f) + V π π (f) + f(x) f(π) 2 = cos 2 x 1 = 3 cos 2 x Similr clcultions for the other subintervls gives tht 1 cos 2 x 0 x π cos v(x) = 2 x π x π 2 3 cos 2 x π x 3π cos 2 x 3π x 2π 2 8

10 Problem Represent the function x 2 0 x < 1 f(x) = 0 x = < x 2 s difference of two incresing functions. Solution: As bove, we determine V0 x (f). On [0, 1) the function f(x) is decresing so if x [0, 1) then V x 0 (f) = x 2 0 = x 2 To determine V 1 0 (f), let Π = {x 0,..., x n } be n prtition of [0, 1] nd consider V Π (f). We hve V Π (f) = = = n 1 f(x k+1 ) f(x k ) n 2 f(x k+1 ) f(x k ) + f(1) f(x n 1 ) n 2 (x 2 k+1 x 2 k) + x 2 n 1 = 2x 2 n 1 B tking the point x n 1 close enough to 1, V Π (f) cn be mde rbitrr close to but less thn 2, nd thus V 1 0 (f) = 2. Finll, if x (1, 2] we hve V x 0 (f) = V 1 0 (f) + V x 1 (f) = 2 + V x 1 (f) Let Π = {x 0,..., x n } be n prtition of [1, x] nd consider V Π (f) n 1 V Π (f) = f(x k+1 ) f(x k ) n 1 = f(x 1 ) f(x 0 ) + f(x k+1 ) f(x k ) = 1 0 = 1 k=1 9

11 Clerl V Π (f) is independent of the prtition Π, so V1 x (f) = 1 nd thus V0 x (f) = 3, x (1, 2]. Hence V x 0 (f) = x 2 0 x < 1 2 x = < x 2 Then f(x) = V0 x (f) (V0 x (f) f(x)) where V x 0 (f) f(x) = 2x 2 0 x < 1 2 x = < x 2 10

12 2.2 Positive nd negtive vrition For n R set + = mx{, 0} nd = mx{, 0} We begin b noticing the following equlities: + + = (1) + = (2) Indeed, if > 0 then + = = nd = 0 so + + = nd + =. The cse < 0 is treted similrl. Equtions 1 nd 2 gives tht + = ( + )/2. Then we hve (α + β) + = α + β + α + β 2 α + β + α + β 2 = α + + β + tht is (α + β) + α + + β + (3) Let f : [, b] R nd Π = {x 0, x 1,..., x n } n prtition of [, b]. Denote n 1 n 1 P Π (f) = [f(x k+1 ) f(x k )] + nd Q Π (f) = [f(x k+1 ) f(x k )] nd set P b (f) = sup Π P Π (f) nd Q b (f) = sup Q Π (f) Π P(f) b nd Q b (f) will be referred to s the positive respectivel negtive vrition of f on [, b]. There is connection between P(f), b Q b (f) nd V b (f), s the following problem shows: Problem If one of the mgnitudes P b (f), Q b (f) nd V b (f) is finite then so re the two others. Proof: For n prtition Π of [, b] we hve P Π (f) + Q Π (f) = V Π (f) nd P Π (f) Q Π (f) = f(b) f() ccording to equtions 1 nd 2. The equlit P Π (f) Q Π (f) = f(b) f() gives tht (i) (ii) P Π (f) Q b (f) + f(b) f() Q Π (f) P b (f) + f() f(b) 11

13 Hence P b (f) is finite if nd onl if Q b (f) is finite. Suppose tht V b (f) is finite. For n prtition Π of [, b] we hve P Π (f) + Q Π (f) = V Π (f) V b (f) nd since P Π (f) 0 nd Q Π (f) 0 we lso hve P Π (f) V b (f) nd Q Π (f) V b (f) whence it follows tht P(f) b nd Q b (f) is finite. Suppose now tht one of P b (f), Q b (f) is finite, then the other one is finite s well. Then, for n prtition Π so V b (f) must be finite. V Π (f) = P Π (f) + Q Π (f) P b (f) + Q b (f) Let f be function of bounded vrition. Then the dditive propert (theorem 2.1.3) holds lso for the positive nd negtive vrition of f, the problem below shows this for the positive vrition. Problem Let f V [, b] nd < c < b. Then P b (f) = P c (f) + P b c (f) Proof: We will prove n nlogue of the theorem on refinements of prtitions for the positive vrition. Once this is done the proof is exctl s the proof of theorem As in theorem we tke n rbitrr prtition Π = {x 0, x 1,..., x n } of [, b] nd dd one dditionl point c, where x j < c < x j+1 for some j, nd denote the result Π. B ( 3) we hve nd thus P Π (f) = [f(x j+1 ) f(x j )] + [f(x j+1 ) f(c)] + + [f(c) f(x j )] + n 1 [f(x k+1 ) f(x k )] + = [f(x j+1 ) f(x j )] + + n 1,k j [f(x k+1 ) f(x k )] + [f(x j+1 ) f(c)] + + [f(c) f(x j )] + + = P Π (f) 12 n 1,k j [f(x k+1 ) f(x k )] +

14 With the useful dditivit propert estblished we consider the following problem. Problem Find P x (f), Q x (f) nd V x (f) if: ) f(x) = 4x 3 3x 4, 2 x 2 b) f(x) = x + 2[x], 0 x 3 Solution: ) The derivtive f (x) = 12x 2 12x 3 = 12x 2 (1 x) gives tht f(x) is incresing on [ 2, 1] nd decresing on [1, 2]. If x [ 2, 1] then V x 2(f) = f(x) f( 2) = 4x 3 3x since f(x) is incresing. If x (1, 2] then V 2(f) x = V 2(f) 1 + V1 x (f) = f(1) f( 2) + f(x) f(1) = x 3 3x 4 1 = x 3 + 3x 4 Thus V x 2(f) = { 4x 3 3x x [ 2, 1] 82 4x 3 + 3x 4 x (1, 2] Since f is incresing on [ 2, 1] then [f(x) f()] + = f(x) f() nd [f(x) f()] = 0 for n x, [ 2, 1]. It follows tht P 2(f) x = V 2(f) x nd Q x 2(f) = 0 when x [ 2, 1]. Since f is decresing on [1, 2] then [f(x) f()] + = 0 nd [f(x) f()] = f(x) f() for n x, [1, 2]. It follows tht P1 x (f) = 0 nd Q x 1(f) = V1 x (f) nd then P 2(f) x = P 2(f) 1 + P1 x (f) = 81 nd Q x 2(f) = Q 1 2(f) + Q x 1(f) = V1 x (f) = 1 4x 3 + 3x 4 for x [1, 2]. Thus we hve { 4x P 2(f) x = 3 3x x [ 2, 1] 81 x (1, 2] { 0 x [ 2, 1] Q x 2(f) = 1 4x 3 + 3x 4 x (1, 2] b) The function f(x) is incresing on [0,3] so V x 0 (f) = P x 0 (f) nd Q x 0(f) = 0 for n x [0, 3]. Furthermore V x 0 (f) = f(x) f(0) = x + 2[x] Thus, V x 0 (f) = P x 0 (f) = x + 2[x] nd Q x 0(f) = 0. 13

15 We shll finish this section with nother chrcteriztion of functions of bounded vrition. Problem Let f be defined on [, b]. Then f V [, b] if nd onl if there exists n incresing function ϕ on [, b] such tht for n x < x b f(x ) f(x ) ϕ(x ) ϕ(x ) Proof: Suppose tht f V [, b]. We tke ϕ(x) = V x (f), this function is incresing nd for n x < x b we hve ϕ(x ) ϕ(x ) = V x x (f) f(x ) f(x ) f(x ) f(x ) Conversel, suppose tht there exist n incresing function ϕ on [, b] such tht f(x ) f(x ) ϕ(x ) ϕ(x ) for n x < x b. Since ϕ is incresing on [, b] we hve ϕ V [, b] nd V b (ϕ) = P(ϕ). b Furthermore, since [f(x ) f(x )] + equls to either f(x ) f(x ) or 0 we hve [f(x ) f(x )] + ϕ(x ) ϕ(x ) for n x < x b. Therefore P Π (f) V b (ϕ) for n prtition Π of [, b] whence P(f) b V b (ϕ) Since V b (ϕ) is finite P b (f) must be finite s well. But then V b (f) is finite ccording to problem 2.2.1, tht is f V [, b]. 14

16 2.3 Conditions for bounded vrition We know tht piecewise monotone functions nd n function tht cn be expressed s the difference of two incresing functions re of bounded vrition. In this section we shll give some dditionl conditions which will gurntee tht function is of bounded vrition. Definition Let f : [, b] R, f is sid to stisf Lipschitz condition if there exists constnt M > 0 such tht for ever x, [, b] we hve f(x) f() M x Theorem [1, p. 119]. If f : [, b] R stisfies Lipschitz condition on [, b] with constnt K, then f V [, b] nd V b (f) K(b ). Proof: Suppose tht f(x) f() K x for ever x, [, b]. Tke n rbitrr prtition Π = {x 0, x 1,..., x n } of [, b]. Then V Π (f) = n 1 f(x k+1 ) f(x k ) n 1 K x k+1 x k = K(b ) Since Π ws rbitrr the inequlit bove is vlid for n prtition, which mens tht the sums V Π (f) re bounded bove b K(b ) whence it follows tht f V [, b] nd V b (f) K(b ). Theorem [1, p. 119]. If f : [, b] R is differentible on [, b] nd if there exists M > 0 such tht f (x) M on [, b] then f V [, b] nd V b (f) M(b ) Proof: For n x, [, b] we hve f(x) f() = f (c)(x ) for some c between x nd ccording to the Men vlue theorem. Hence f(x) f() = f (c) x M x for n x, [, b], tht is f stisfies Lipschitz condition on [, b] with constnt M nd thus Theorem gives the result. Problem Prove tht f V [0, 1] if f(x) = { x 3/2 cos(π/ x) 0 < x 1 0 x = 0 15

17 Solution: For n x (0, 1] the function f(x) hs the derivtive nd in the point 0 the derivtive f (x) = 3 2 x1/2 cos(π/ x) + π 2 sin(π/ x) f (0) = lim x 0 f(x) f(0) x 0 Now, for n x [0, 1] we hve = lim x 0 x 1/2 cos(π/ x) = 0 f (x) 3 2 x1/2 cos(π/ x) + π 2 sin(π/ x) π 2 Since f hs bounded derivtive on [0, 1] we hve tht f V [0, 1]. With the dditionl ssumption tht the derivtive f is continuous on [, b] we cn give formul for the totl vrition: Theorem If f is continuousl differentible on [, b] then f V [, b] nd the totl vrition is given b V b (f) = b f (x) dx Proof: Since f is continuous then f is bounded so f V [, b]. Furthermore, the continuit of f implies tht f is Riemnn-integrble on [, b]. Set I = b f (x) dx nd let ɛ > 0 be rbitrr. Then there exists δ > 0 such tht for n prtition P with d(p) < δ the Riemnn-sum S( f, P), with rbitrr intermedite points, stisf the following inequlities: I ɛ < S( f, P) < I + ɛ So let Π be prtition of [, b] with d(π) < δ. According to the Men Vlue theorem f(x k+1 ) f(x k ) = f (t k ) (x k+1 x k ) for some t k (x k, x k+1 ) Hence n 1 n 1 V Π (f) = f(x k+1 ) f(x k ) = f (t k ) (x k+1 x k ) 16

18 The right hnd side is Riemnn sum S( f, Π) nd since d(π) < δ we hve I ɛ < V Π (f) < I + ɛ I ɛ < V b (f) whence I V b (f) since ɛ is rbitrr. Further, for n prtition Π = {x 0, x 1,..., x n } we hve n 1 n 1 xk+1 V Π (f) = f(x k+1 ) f(x k ) = f (x)dx x k n 1 xk+1 x k f (x) dx = b f (x) dx Since Π is n rbitrr prtition it follows tht V b (f) I. Then we must hve I = V b (f). In the following problem we will estblish fct tht will be ver useful together with theorem Problem Let f be defined on [, b]. If f V [, c] for n < c < b nd if there exists number M such tht V c (f) M for n < c < b then f V [, b]. Proof: Let Π = {x 0, x 1,..., x n } be n rbitrr prtition of [, b]. Set Π = {x 0, x 1,..., x n 1 }. Then Π is prtition of [, x n 1 ] nd since V x n 1 (f) M we hve V Π (f) = V Π + f(b) f(x n 1 ) V x n 1 (f) + f(b) f(x n 1 ) M + f(b) f(x n 1 ) = M + f(b) + (f() f()) f(x n 1 ) M + f(b) f() + f() f(x n 1 ) M + f(b) f() + V x n 1 (f) 2M + f(b) f() Thus V Π (f) is bounded bove b 2M + f(b) f() nd therefore f V [, b]. The problem below demonstrtes how one cn use theorem nd problem 2.3.6: Problem Prove tht f V [0, 1] if { x f(x) = 3/2 cos(π/x) 0 < x 1 0 x = 0 17

19 Solution: For n x (0, 1] the function f(x) hs the derivtive f (x) = 3 2 x cos(π/x) + π x sin(π/x) For n 0 < < 1 the function f (x) is continuous on [, 1] nd thus f V [, 1] nd the totl vrition of f over [, 1] is given b Furthermore V 1 (f) = 1 f (x) dx so it follows tht f (x) = 3 π x cos(π/x) + x sin(π/x) 3 π x + x 2 2 V 1 (f) = 1 f (x) dx 1 ( 3 2 x + π x )dx < 1 + 2π tht is V 1 (f) 1 + 2π for n > 0. problem Then f V [0, 1] ccording to 18

20 2.4 The function v(x) In the proof of Jordns theorem we introduced the function of totl vrition, v(x) = V x (f). In this section we shll stud some properties of this function, more specificll continuit nd differentibilit. The following theorem is given in [1]: Theorem [1, p. 125]. Let f V [, b], then v(x) is continuous t point c if nd onl if f(x) is continuous t c. Insted of proving we shll give result tht is bit stronger: Problem Let f V [, b], then for n x (, b) we hve v(x + 0) v(x) = f(x + 0) f(x) nd v(x) v(x 0) = f(x) f(x 0) Proof: We shll prove the first equlit, the proof of the second one is similr. Tke fixed but rbitrr x 0 (, b) nd set L = f(x 0 + 0) f(x 0 ) Given ɛ > 0, there exists δ > 0 such tht x 0 < x < x 0 + δ L ɛ 2 < f(x) f(x 0) < L + ɛ 2 Let Π = {x 0, x 1,..., x n } be prtition of [x 0, b] such tht (i) (ii) Vx b 0 (f) ɛ < V 2 Π(f) x 1 < x 0 + δ The point x 1 cn be treted s n rbitrr point stisfing x 0 < x 1 < x 0 +δ. Now set Π = Π \ {x 0 }, clerl Π is prtition of [x 1, b] nd we hve Furthermore, V Π (f) V Π (f) = f(x 1 ) f(x 0 ) V Π (f) V Π (f) (Vx b 0 (f) ɛ ) V 2 Π (f) Vx b 0 (f) Vx b 1 (f) ɛ 2 = V x 1 x 0 (f) ɛ 2 = V x 1 (f) V x 0 (f) ɛ 2 = v(x 1 ) v(x 0 ) ɛ 2 Thus v(x 1 ) v(x 0 ) ɛ 2 f(x 1 ) f(x 0 ) < L + ɛ 2 f(x 1) f(x 0 ) nd since x 1 < x 0 + δ we hve whence v(x 1 ) v(x 0 ) < L + ɛ 19

21 But we lso hve tht f(x 1 ) f(x 0 ) V x 1 x 0 (f) = v(x 1 ) v(x 0 ) L ɛ 2 < v(x 1) v(x 0 ) L ɛ < v(x 1 ) v(x 0 ) Since we regrd the point x 1 s n rbitrr point stisfing x 0 < x 1 < x 0 +δ we hve x 0 < x < x 0 + δ L ɛ < v(x) v(x 0 ) < L + ɛ nd hence v(x 0 + 0) v(x 0 ) = L We shll proceed to stud how the differentibilit of the function v(x) is relted to tht of f(x). Problem If f hs continuous derivtive on [, b], then the function v(x) is differentible nd hs continuous derivtive on [, b]. Proof: First of ll, since f is continuous on [, b] then f is bounded on [, b] nd therefore f V [, b]. Further, b theorem v(x) = x f (t) dt nd ccording to the fundmentl theorem of clculus v (x) = f (x) which is continuous on [, b]. If f V [, b] then the existence of derivtive f on [, b] is not sufficient to gurntee tht v is differentible on [, b], s we shll demonstrte in problem. But first we need lemm: Lemm Set ϕ(x) = x sin t dt, then 0 Proof: For n n N ϕ(x) lim x x = 2 π ϕ(nπ) = nπ 0 sin t dt = n 1 (k+1)π kπ sin t dt To evlute (k+1)π sin t dt for given k, we mke the chnge of vrible kπ t = kπ + u, then 0 u π nd since the function sin is periodic with period π then (k+1)π sin t dt = π kπ 0 20 sin udu = 2

22 nd thus ϕ(nπ) = 2n. Assume tht nπ x < (n + 1)π for some n N, then 2n (n + 1)π < 2n x = ϕ(nπ) x ϕ(x) x since ϕ is n incresing function. On the other hnd, ϕ(x) x ϕ(x) nπ ϕ((n + 1)π)) nπ nd therefore 2 π n n + 1 < ϕ(x) x 2 π n + 1 n Letting x gives the result. = 2(n + 1) nπ Problem Let f(x) = x 2 cos( x 3/2 ) for 0 < x 1 nd f(0) = 0. Then f V [ 1, 1] nd f(x) is differentible on [ 1, 1] but v(x) is not differentible t the point x = 0. Proof: If x [ 1, 1]\{0} then the stndrd rules of differentition pplies, for exmple if x (0, 1] then f (x) = 2x cos(x 3/2 ) x 1/2 sin(x 3/2 ). Using the definition of the derivtive we cn esil show tht f (0) = 0. B ppling similr rgument s in problem one shows tht f V [ 1, 0] nd f V [0, 1] nd thus f V [ 1, 1]. To show tht v(x) isn t differentible in x = 0 it s sufficient to show tht the limit lim x +0 (v(x) v(0))/x doesn t exist. For n such tht 0 < < x we hve v(x) v() = x f (t) dt nd since v(x) is continuous t the point x = 0 it follows tht v(x) v(0) = x f (t) dt where the integrl is 0 improper. Thus we need to show tht the limit 1 lim x +0 x x doesn t exist. It s es to see tht x 0 2t cos(t 3/2 ) t 1/2 sin(t 3/2 ) dt = nd therefore we consider 1 x 0 2t cos(t 3/2 ) t 1/2 sin(t 3/2 ) dt x x 0 t 1/2 sin(t 3/2 ) dt t 1/2 sin(t 3/2 ) dt + O(x 2 ) 21

23 Mke the chnge of vrible u = t 3/2, then t = u 2/3 nd dt = 2 3 u 5/3 du, the integrl bove becomes 1 x x 3/2 2 3 sin u u 4/3 du Set = x 3/2, then s x +0 nd therefore we shll show tht 2/3 sin u u 4/3 du s We shll integrte sin u u 4/3 du b using integrtion b prts. Set ϕ(u) = u sin t dt, then ϕ is primitive of sin nd integrtion b prts 0 gives nd therefore 2/3 sin u u 4/3 du = ϕ(u)u 4/ = 4 3 ϕ(u)u 7/3 du ϕ(u)u 7/3 du ϕ() 4/3 sin u u 4/3 du = 1/3 ( 4 3 1/3 ϕ(u)u 7/3 du ϕ() ). ϕ() According to lemm lim = 2 nd lim π 1/3 ϕ(u)u 7/3 du = which is shown b using L Hospitl s rule: 6 π ( ϕ(u)u 7/3 du) ϕ() 7/3 3ϕ() lim = lim ( 1/3 ) 1 = lim 3 4/3 = 6 π b lemm Thus lim ( 4 3 1/3 nd therefore 1/3 ( 4 3 1/3 ϕ(u)u 7/3 du ϕ() ) exists ϕ(u)u 7/3 du ϕ() ) s. According to problem 2.4.3, if f is continuous on [, b] then v is differentible on [, b]. The bove problem demonstrtes tht onl the existence of derivtive f t point is not sufficient to gurntee tht v is differentible. We shll show tht if f is continuous t point, then v will be differentible t the point. We cnnot use the sme pproch s in problem since we onl ssume tht f is continuous t one point. Indeed, f might not even be Riemnn-integrble. Problem Let f V [, b] be differentible on [, b]. If f is continuous t point x 0 [, b] then v(x) is differentible t x 0. 22

24 Proof: Continuit of f in x 0 implies tht f is continuous in x 0. Let ɛ > 0 be given, there exists δ 1 > 0 such tht if x x 0 < δ 1 then f (x 0 ) ɛ < f (x) < f (x 0 ) + ɛ Tke δ = δ 1 2, for n x [x 0 δ, x 0 + δ] we hve f (x) < f (x 0 ) + ɛ nd since f hs bounded derivtive on [x 0 δ, x 0 +δ] then f stisfies Lipschitz condition on [x 0 δ, x 0 + δ] with constnt f (x 0 ) + ɛ. If x 0 < x < x 0 + δ then v(x) v(x 0 ) = Vx x 0 (f) ( f (x 0 ) + ɛ)(x x 0 ) whence v(x) v(x 0 ) x x 0 < f (x 0 ) + ɛ (4) In the sme w one shows tht the inequlit bove lso holds for x such tht x 0 δ < x < x 0. On the other hnd, if x > x 0 we hve v(x) v(x 0 ) f(x) f(x 0) x x 0 x x 0 = f (x 1 ) for some x 1 between x nd x 0 ccording to the Men vlue theorem. Since x 1 x 0 < δ we hve f (x 1 ) > f (x 0 ) ɛ nd therefore f (x 0 ) ɛ < v(x) v(x 0) x x 0 (5) A similr inequlit holds for x < x 0. Combining inequlit 4 with inequlit 5 the following impliction follows: 0 < x x 0 < δ v(x) v(x 0) x x 0 f (x 0 ) < ɛ nd hence v is differentible in x 0, with the derivtive f (x 0 ). We know now tht continuit of f in point is sufficient for v to be differentible in the point. However, it s not necessr condition. Problem Let f(x) = x 2 cos(1/x) for 0 < x 1 nd f(0) = 0. Then f(x) nd v(x) re differentible everwhere on [ 1, 1] but both f nd v re discontinuous t the point x = 0. Proof: Clerl f is differentible with derivtive f (x) = 2x cos(1/x) + sin(1/x) for x [ 1, 1] \ {0}. We hve (f(x) f(0))/x = x cos(1/x) 0 s x 0 nd thus f (0) = 0. However, since lim x 0 f (x) doesn t exist f is 23

25 not continuous t the point x = 0. It follows from problem tht v is differentible nd hs the derivtive v (x) = f (x) for ever x [ 1, 1]\{0} since f (x) is continuous there. Now we shll show tht lim x 0 (v(x) v(0))/x exists. Let x > 0, s in problem we hve v(x) v(0) = x f (t) dt nd therefore the derivtive 0 in x = 0 is given b 1 x lim 2t cos 1 x +0 x t + sin 1 t dt Further, x 0 2t cos 1 0 t + sin 1 x t dt = sin 1 0 t dt + O(x2 ) 1 x so it ll be sufficient to show the existence of lim x +0 sin 1 dt. x 0 t Mke the chnge of vrible u = 1, then dt = 1 du nd we get t u 2 1 lim x +0 x 1/x sin u 1 u 2 du Set = 1, then + s x +0 which ields x lim sin u 1 u du 2 Let ϕ(u) be s in problem nd use integrtion b prts: nd thus Write sin u u 2 du = ϕ(u)u = 2 sin u u 2 du = 2 ϕ(u)u 3 du = ϕ(u)u 3 du ϕ(u)u 3 du ϕ() 2 ϕ(u)u 3 du ϕ() 1 ϕ(u)u 3 du/ 1 nd ppl L Hospitl s rule: ( ϕ(u)u 3 du) ϕ() 3 ϕ() lim = lim = lim ( 1 ) 2 b lemm nd therefore L Hospitl s rule. Thus lim sin u u 2 du = lim (2 = 2 π lim ϕ(u)u 3 du = 2 π ccording to = 4 π 2 π = 2 π ϕ(u)u 3 du ϕ() ) Therefore v (0) = 2 π. 24

26 2.5 Two limits In this section we shll consider continuous functions of bounded vrition. If f is continuous then the totl vrition is given s limit of the sums V Π (f). We shll introduce some nottions: Let Π = {x 0, x 1,..., x n } be n prtition of [, b]. Then we set d(π) = mx (x k+1 x k ) 0 k n 1 Now let f : [, b] R be continuous nd Π = {x 0, x 1,..., x n } n prtition of [, b]. Let M k = mx xk x x k+1 f(x) nd m k = min xk x x k+1 f(x) nd set n 1 Ω Π (f) = (M k m k ) Problem If f C[, b] then both sums V Π (f) nd Ω Π (f) tend to V b (f) s d(π) 0. Proof: We del with V Π (f) first. We re going to show tht for ever ɛ > 0 there exists δ > 0 so tht if Π is n prtition of [, b] with d(π) < δ then V b (f) ɛ < V Π (f). Let ɛ > 0 be given, then there exists prtition Π 0 = { 0, 1,..., m+1 } (the reson for this nottion will be cler lter) such tht V b (f) ɛ 2 < V Π 0 (f) Since f is continuous on compct intervl [, b] f is uniforml continuous on [, b]. Therefore there exists δ 1 > 0 such tht x < δ 1 f(x) f() < ɛ 4m Now let Π = {x 0, x 1,..., x n } be n prtition with d(π) < δ where δ < δ 1 but lso smll enough to ensure tht there is onl one point j Π 0 in ever intervl [x k, x k+1 ]. Tke the prtition Π consisting of the points in Π together with the points in Π 0. Then V Π (f) V Π0 (f) since Π is refinement of Π 0. We will now compre the sums V Π (f) nd V Π (f): V Π (f) consists of terms f(x k+1 ) f(x k ) nd the coincides with the terms of V Π (f) except for those terms tht corresponds to those intervls [x k, x k+1 ] which contins point j with 1 j m. Here the term f(x k+1 ) f(x k ) is exchnged for f( j ) f(x k ) + f(x k+1 ) f( j ). Thus V Π (f) V Π (f) consists of m terms on the form 25

27 nd therefore f( j ) f(x k ) + f(x k+1 ) f( j ) f(x k+1 ) f(x k ) 0 f( j ) f(x k ) + f(x k+1 ) f( j ) f(x k+1 ) f(x k ) f( j ) f(x k ) + f(x k+1 ) f( j ) < ɛ/4m + ɛ/4m = ɛ/2m where the lst inequlit is true becuse x k+1 j, j x k < δ 1. Therefore V Π (f) V Π (f) < m ɛ/2m = ɛ/2 nd thus V b (f) ɛ/2 < V Π (f) < V Π (f) + ɛ/2 V b (f) ɛ < V Π (f) We conclude tht V Π (f) V b (f) when d(π) 0 For the second prt of the problem, let ɛ > 0 be given, there exists δ > 0 such tht d(π) < δ V b (f) ɛ < V Π (f). Tke n prtition Π with d(π) < δ. We notice tht for n prtition we hve V Π (f) Ω Π (f) nd thus d(π) < δ V b (f) ɛ < Ω Π (f) On the other hnd, for n prtition Π = {x 0, x 1,..., x n } let t k, s k be points in which f ttins it s mximum respectivel minimum on [x k, x k+1 ], tht is M k = f(t k ) nd m k = f(s k ). Now dd t k nd s k, 0 k n 1, to the prtition Π nd denote the resulting prtition Π. The sum V Π (f) will then contin terms f(t k ) f(s k ) = M k m k so we hve n 1 V b (f) V Π (f) = (M k m k ) + dditionl positive terms V b (f) Ω Π (f) for n prtition Π. Hence, for ever ɛ > 0 there exists δ > 0 such tht if Π is n prtition with d(π) < δ then V b (f) ɛ < Ω Π (f) V b (f), tht is Ω Π (f) V b (f) s d(π) 0. The totl vrition of continuous function of bounded vrition is lso given b nother limit, s the following problem shows: Problem Let f V [, b] nd let f be continuous on [, b]. Then V b 1 (f) = lim h +0 h b h 26 f(x) f(x + h) dx

28 Proof: 1 h b h Since f(x) f(x + h) v(x + h) v(x) it follows tht f(x) f(x + h) dx 1 h = 1 h = 1 h b h b h b [v(x + h) v(x)]dx v(x + h)dx 1 h +h v(x)dx 1 h b = 1 v(x)dx 1 h b h h v(b) = V b (f) The inequlit bove is vlid for ever h nd thus lim sup h +0 1 h b h b h +h f(x) f(x + h) dx V b (f) b h v(x)dx v(x)dx v(x)dx Now let Π = {x 0, x 1,..., x n } be n prtition of [, b]. For n h smll enough to ensure tht x n 1 < b h nd for j N such tht 0 j n 2 the following is true: 1 h xj+1 x j f(x) f(x + h) dx 1 h = 1 h = 1 h xj+1 x j xj+1 x j xj +h x j f(x) f(x + h)dx f(x)dx 1 h f(x)dx 1 h xj+1 +h x j +h xj+1 +h x j+1 f(x)dx f(x)dx The right-hnd side tends to f(x j ) f(x j+1 ) s h +0 ccording to the fundmentl theorem of clculus. So, for 0 j n 2 lim inf h +0 1 h xj+1 In the sme w one shows tht lim inf h +0 1 h x j f(x) f(x + h) dx f(x j+1 ) f(x j ) b h Summing these inequlities gives lim inf h +0 1 h b h x n 1 f(x) f(x + h) dx f(b) f(x n 1 ) f(x) f(x + h) dx 27 n 1 f(x k+1 ) f(x k ) = V Π (f)

29 for n prtition Π. It follows tht lim inf h +0 1 h b h f(x) f(x + h) dx V b (f) Thus we ve obtined the following inequlities: nd therefore V b 1 (f) lim inf h +0 h 1 lim h +0 h lim sup h +0 V b (f) b h 1 h b h b h f(x) f(x + h) dx f(x) f(x + h) dx f(x) f(x + h) dx = V b (f) 28

30 3 Jump functions In this section we shll introduce the so clled jump function of function of bounded vrition. Before we give definition we shll estblish some fcts. One of the consequences of Jordn s theorem is tht if f is of bounded vrition, then the set of points t which f is discontinuous is t most countble. Let {d n } n 1 be the sequence of points of discontinuit of f nd let σ n + be the right-hnd jump of f t d n nd σn the left-hnd jump of f t d n, tht is σ + n = f(d n + 0) f(d n ), nd σ n = f(d n ) f(d n 0) Problem If f V [, b] then ( σ n + + σn ) V b (f) Proof: Set n=1 S n = n ( σ + k + σ k ) k=1 It is sufficient to show tht the sequence {S n } n N is bounded bove b V b (f), tht is S n V b (f) for ever n. Let {d k } k N be the sequence of ll points of discontinuit of f in [, b]. For the ske of simplicit ssume tht neither nor b is point of discontinuit. The problem tht rises when or b is point of discontinuit is purel nottionl, the resoning is the sme. Let ɛ > 0 be given nd tke n N. Consider the points d k, 1 k n, we m reorder these points so tht d 1 < d 2 <... < d n. For 1 k n let r k, l k be points such tht (i) r k 1 < l k < d k < r k < l k+1, 2 k n 1 (ii) f(d k ) f(l k ) > σ k ɛ 2n (iii) f(r k ) f(d k ) > σ + k ɛ 2n Now let Π be the prtition of [, b] consisting of,b nd the points l k, d k, r k for 1 k n. Then we hve n V b (f) V Π (f) ( f(d k ) f(l k ) + f(r k ) f(d k ) ) > n k=1 k=1 ( σ k ɛ 2n + σ+ k ɛ 2n ) = S n ɛ Since ɛ is rbitrr S n V b (f) for ever n. 29

31 Definition Set s() = 0 nd for < x b s(x) = σ + k + d k <x <d k x The function s(x) is clled the jump function of f. It follows from problem tht the series k=1 σ+ k nd k=1 σ k re bsolutel convergent nd thus the subseries d k <x σ+ k nd <d k x σ k re bsolutel convergent. This is crucil in order for the jump function to be well-defined. In the following problem some properties of the jump function is proved. Problem If s is the jump function of f V [, b] then: (1) the function s is continuous t ever point x d k nd t ever point d k hs the right-hnd jump σ + k nd the left-hnd jump σ k. (2) s V [, b] nd for x b we hve V x (s) = d k <x σ + k + <d k x σ k σ k (3) the difference f s is continuous function of bounded vrition on [, b]. (4) if f is incresing then f s lso is incresing. Proof: (1) Let x 0 d k, since f is continuous in x 0 the function of totl vrition v(x) is continuous in x 0. Then, given ɛ > 0 there exists δ > 0 such tht x x 0 < δ v(x) v(x 0 ) < ɛ. Tke x such tht x 0 < x < x 0 + δ. Then we hve: s(x) s(x 0 ) = σ + k + σ k x 0 d k <x x 0 d k <x x 0 d k <x σ + k + v(x) v(x 0 ) x 0 <d k x x 0 <d k x σ + k + x 0 <d k x σ k σ k where the lst inequlit is true becuse of problem Furthermore, v(x) v(x 0 ) < ɛ nd thus x 0 < x < x 0 + δ s(x) s(x 0 ) < ɛ. The sme inequlit holds when x 0 δ < x < x 0 nd thus s is continuous in x 0. 30

32 For the second prt, tke fixed point of discontinuit d j. We shll show tht lim x +dj s(x) s(d j ) = σ + j, one shows tht lim x d j s(d j ) s(x) = σ j in the sme w. For x > d j we set ω (x) = d j d k <x σ+ k nd ω (x) = d j <d k x σ k. Then s(x) s(d j ) = ω (x) + ω (x) We shll show tht ω (x) σ + j nd ω (x) 0 s x +d j. B problem both series k=1 σ+ k nd k=1 σ k re bsolutel convergent nd therefore we hve tht for given ɛ > 0 there exists nturl numbers N 1, N 2 such tht k N 1 σ + k < ɛ nd k N 2 σ k < ɛ. For N = mx{n 1, N 2 } there exists δ > 0 such tht if k N then the point d k does not belong to (d j, d j + δ). Hence, if d j < x < d j + δ then nd ω (x) σ + j = ω (x) = d j <d k <x d j <d k x σ + k σ k d j <d k <x d j <d k x σ + k k N σ k k N σ + k < ɛ σ k < ɛ nd we conclude tht lim x +dj ω (x) = σ + j nd lim x +dj ω (x) = 0. (2) Let x (, b], ccording to (1) the function s(x) hs the sme discontinuities s f(x) nd lso the sme left nd right-hnd jumps s f(x) t these points. Then problem pplied to the function s(x) gives tht d k <x σ + k + <d k x We shll show the opposite inequlit. Set t(x) = σ + j + d j <x σ k V x (s) <d j x σ j nd let Π = {x 0, x 1,..., x n } be n prtition of [, x]. For 0 j n 1 we hve s(x k+1 ) s(x k ) = σ + j + σ j x k d j <x k+1 x k <d j x k+1 σ + j + σ j x k d j <x k+1 x k <d j x k+1 = t(x k+1 ) t(x k ) 31

33 nd therefore V Π (s) = n 1 s(x k+1 ) s(x k ) n 1 (t(x k+1 ) t(x k )) = t(x) It follows tht V x (s) t(x) nd therefore V x (s) = σ + j + d j <x <d j x σ j (3) If x [, b] is point of continuit for f then s is continuous in x s well, b (1), nd thus (f s) is continuous in x. If x is point of discontinuit for f, tht is x = d k for some k, then (f s)(d k + 0) = f(d k + 0) s(d k + 0) = f(d k + 0) s(d k + 0) + s(d k ) s(d k ) + f(d k ) f(d k ) = f(d k + 0) f(d k ) [s(d k + 0) s(d k )] + f(d k ) s(d k ) = σ + k σ+ k + f(d k) s(d k ) = f(d k ) s(d k ) One shows tht (f s)(d k 0) = (f s)(d k ) in the sme mnner, so (f s) is continuous in d k. (4) Let x, [, b] nd < x. In the proof of (1) bove we showed tht s(x) s() s(x) s() V x (f) Since f is incresing V x (f) = f(x) f() nd thus we hve s(x) s() f(x) f() which is equivlent with f() s() f(x) s(x) if < x, hence f s is incresing. We cn give n lterntive definition of the jump function tht will be useful: Problem For those n such tht < d n < b set 0 x < d n σ n (x) = σn x = d n σn + σ n + d n < x b 32

34 If d n = for some n set σ n (x) = { σ + n < x b 0 x = Finll, if d n = b for some n set { 0 x < b σ n (x) = x = b Then s(x) = σ n σ n (x) n=1 Proof: Set s 1 (x) = n=1 σ n(x), we shll prove tht s 1 = s. Since σ n () = 0 for ever n clerl s 1 () = s(). Tke x (, b] fixed but rbitrr, for those n N such tht < d n < x we hve σ n (x) = σ n + + σn nd for those n N such tht x < d n we hve σ n (x) = 0. We get different cses depending on wether or not the points x nd re points of discontinuit. We ll do the cse when is point of discontinuit, tht is = d i for some i, nd x isn t point of discontinuit: s 1 (x) = σ i (x) + σ n (x) = σ n + + = s(x) <d n<x d n<x <d n<x σ n We shll conclude this section with two problems on how the totl vrition of the function f s reltes to the totl vrition of f. Problem Let f V [, b] nd let s be the jump function of f. Then Proof: We hve V b (f) = V b (f s) + ( σ n + + σn ) n=1 V b (f) = V b (f s+s) V b (f s)+v b (s) = V b (f s)+ For the opposite inequlit, set for ever n N s n (x) = n σ k (x) k=1 33 ( σ + k + σ k ) k=1

35 where σ k (x) is s defined in problem For nottionl simplicit we ssume tht nd b re not points of discontinuit of f. We shll show tht (i) (ii) V b (f s n ) = V b (f) n k=1 ( σ+ k + σ k ) V b (s s n ) = V b (s) n k=1 ( σ+ k + σ k ) In order to show (i), first consider the cse when n = 1. We shll show tht V b (f s 1 ) = V b (f) ( σ σ1 ). Let h be n rbitrr number stisfing 0 < h < min{c, b c}. The function s 1 is constnt on [, c h] nd [c+h, b] nd thus V c h (f s 1 ) = V c h (f) nd Vc+h b (f s 1) = Vc+h b (f) nd the following equlit follows: V b (f) V b (f s 1 ) = V c+h c h (f) V c+h c h (f s 1) Let h +0, with v(x) = V x (f) problem gives tht v(c + 0) v(c) = σ + nd v(c) v(c 0) = σ so therefore v(c+0) v(c 0) = σ + + σ. Further, since f s 1 is continuous in the point c then V c+h c h (f s 1) 0 nd thus we obtin V b (f) V b (f s 1 ) = σ σ1. For the generl cse of (i), order the points of discontinuit d 1,.., d n such tht d 1 < d 2 <... < d n nd divide [, b] into n compct, non-overlpping subintervls I 1,..., I n such tht d k lies in the interior of I k = [x k 1, x k ] for 1 k n. Then ppling the cse n = 1 to ech intervl [x k 1, x k ] ields V x k x k 1 (f s n ) = V x k x k 1 (f) ( σ + k + σ k ) for 1 k n Summing these equlities gives (i). Since the function s hs the sme points of discontinuit s f nd the sme left-hnd nd right-hnd jumps t these points (ii) follows directl b ppling (i) to the function s. Now, we hve V b (f s) = V b (f s n + s n s) V b (f s n ) + V b (s s n ) = V b (f) = V b (f) n k=1 ( σ + k + σ k ) + V b (s) n ( σ + k + σ k ) + k=1 k=n+1 n ( σ + k + σ k ) k=1 ( σ + k + σ k ) Given n ɛ there exists nturl numbers N 1 nd N 2 such tht if n N 1 then ( σ + k + σ k ) < ɛ 2 k>n 34

36 nd if n N 2 then n ( σ + k + σ k ) ( σ + k + σ k ) ɛ 2 k=1 Thus if n mx{n 1, N 2 } then V b (f s) V b (f) < V b (f) [ = V b (f) k=1 n ( σ + k + σ k ) + k=1 k=n+1 ( σ + k + σ k ) ɛ 2 ] + ɛ 2 k=1 ( σ + k + σ k ) + ɛ k=1 ( σ + k + σ k ) whence it follows tht V b (f s) + k=1 ( σ+ k + σ k ) V b (f) nd thus we must hve V b (f) = V b (f s) + ( σ + k + σ k ) Problem If f V [, b] nd if function ϕ is such tht f ϕ is continuous on [, b] nd k=1 then ϕ s = constnt V b (ϕ) = ( σ n + + σ n ) n=1 Proof: If we let h = f ϕ then ϕ = f h nd since h is continuous on [, b] we see tht if f is continuous in point then so is ϕ. On the other hnd f = h + ϕ so if ϕ is continuous in point then so is f. Thus, for n x 0 [, b], f(x) is continuous in x 0 if nd onl if ϕ(x) is continuous in x 0. It follows tht f nd ϕ hve exctl the sme sequence of points of discontinuit. Furthermore, if d k is n point of discontinuit of f we hve, since f ϕ is continuous, tht 0 = (f ϕ)(d k + 0) (f ϕ)(d k ) = f(d k + 0) f(d k ) [ϕ(d k + 0) ϕ(d k )] = σ + k [ϕ(d k + 0) ϕ(d k )] 35

37 Tht is, the function ϕ hs the sme right-hnd jump s f t d k. In the sme w we cn show tht ϕ lso hs the sme left-hnd jump s f t d k. Then problem gives tht V b (ϕ) = V b (ϕ s ϕ ) + ( σ + k + σ k ) where s ϕ is the jump function of ϕ. However, f nd ϕ hve the sme points of discontinuit nd the sme right-hnd nd left-hnd jumps t these points so then s f = s ϕ. Thus V b (ϕ) = V b (ϕ s f ) + k=1 ( σ + k + σ k ) Finll we re given tht V b (ϕ) = k=1 ( σ+ k + σ k ) nd it follows tht V b (ϕ s f ) = 0 ϕ s f = constnt k=1 36

38 References [1] R. Knnn nd C.K. Kreuger, Advnced nlsis on the rel line, Springer-Verlg, 1996 [2] W. Rudin, Principles of mthemticl nlsis, Third edition, McGrw- Hill,

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