Functions of Bounded Variation

1 Functions of Bounded VariationOur main theorem concerning the existence of Riemann Stietjes integrals assures usthat the integral baf(x)d (x) exists whenfis continuous and is monotonic. Our lin-earity theorem then guarantees that the integral baf(x)d (x) exists whenfis continuousand is the difference of two monotonic Functions . In these notes,we prove that is thedifference of two monotonic Functions if and only if it is of Bounded Variation , whereDefinition 1(a) The function : [a, b] IR is said to be of Bounded Variation on [a, b] if and only ifthere is a constantM >0 such thatn i=1 (xi) (xi 1) Mfor all partitions IP ={x0, x1, , xn}of [a, b].(b) If : [a, b] IR is of Bounded Variation on [a, b], then the total Variation of on [a, b]is defined to beV (a, b) = sup{n i=1 (xi) (xi 1) IP ={x0, x1, , xn}is a partition of [a, b]}Example 2If : [a, b] IR is monotonically increasing, then, for any partition IP ={x0, x1, , xn}of [a, b]n i=1 (xi) (xi 1) =n i=1{ (xi) (xi 1)}= (xn) (x0) = (b) (a)Thus is of Bounded Variation andVf(a, b) = (b) (a).

2 Example 3If : [a, b] IR is continuous on [a, b] and differentiable on (a, b) withsupa<x<b| (x)| M, then, for any partition IP ={x0, x1, , xn}of [a, b], we have, bythe Mean Value Theorem,n i=1 (xi) (xi 1) =n i=1 (ti)[xi xi 1] n i=1M[xi xi 1] =M(b a)Thus is of Bounded Variation andVf(a, b) M(b a).c Joel Feldman. 2017. All rights 30, 2017 Functions of Bounded Variation1 Example 4 Define the function : [0,1] IR by (x) ={0ifx= 0xcos xifx6= 0 This function is continuous, but is not of Bounded variationbecause it wobbles too muchnearx= 0. To see this, consider, for eachm IN, the partitionIPm={x0= 0, x1=12m, x2=12m 1, x3=12m 2, , x2m 2=13, x2m 1=12, x2m= 1}The values of at the points of this partition are (IPm) ={0,12m, 12m 1,12m 2, , 13,12, 1}xyy= (x)x2m= 1x2m 1=1213 For this partition,2m i=1 (xi) (xi 1) = 12m 0 + 12m 1 12m + 1(2m 2)+12m 1 + + 13 14 + 12+13 + 1 12 =12m+ 0 +12m 1+12m+1(2m 2)+12m 1+ +13+14+12+13+ 1 +12= 2(12m+12m 1+ +13+12)+ 1 The harmonic series k=21kdiverges.}

3 So given anyM, there is anm IN for which thepartition IPmobeysn i=1 (xi) (xi 1) > MTheorem 5(a) If , : [a, b] IRare of Bounded Variation andc, d IR, thenc +d is of boundedvariation andVc +d (a, b) |c|V (a, b) +|d|V (a, b)c Joel Feldman. 2017. All rights 30, 2017 Functions of Bounded Variation2(b) If : [a, b] IRis of Bounded Variation on[a, b]and[c, d] [a, b], then is of boundedvariation on[c, d]andV (c, d) V (a, b)(c) If : [a, b] IRis of Bounded Variation andc (a, b), thenV (a, b) =V (a, c) +V (c, b)(d) If : [a, b] IRis of Bounded Variation then the functionsV(x) =V (a, x)andV(x) (x)are both increasing on[a, b].(e) The function : [a, b] IRis of Bounded Variation if and only if it is the difference oftwo increasing :We shall use the shorthand notationIP i forn i=1 (xi) (xi 1) where the partition IP ={x0, x1, , xn}.

4 (a) follows from the observation that, for any IP partition of [a, b],IP i(c +d ) |c|IP i +|d|IP i |c|V (a, b) +|d|V (a, b)(b) follows from the observation that, for any partition IP of [c, d],IP i IP {a,b} i V (a, b)(c) If IP ={x0, x1, , xn}is any partition of [a, b] andxi 1 c xi, then (xi) (xi 1) (xi) (c) + (c) (xi 1) so thatIP i IP {c} i =(IP {c}) [a,c] i +(IP {c}) [c,b] i V (a, c) +V (c, b)c Joel Feldman. 2017. All rights 30, 2017 Functions of Bounded Variation3which implies thatV (a, b) V (a, c) +V (c, b). To prove the other inequality, we let >0 and select a partition IP1of [a, c] for whichIP1 i V (a, c) and a partitionIP2of [c, b] for whichIP2 i V (c, b) . ThenIP1 IP2 i =IP1 i +IP2 i V (a, c) +V (c, b) 2 This assures thatV (a, b) V (a, c) +V (c, b) 2 for all >0 and hence thatV (a, b) V (a, c) +V (c, b).

5 (d)Proof thatV(x)is increasing:Leta x1 x2 b. Then, by part (c),V(x2) V(x1) =V (a, x2) V (a, x1) =V (x1, x2) 0(d)Proof thatV(x) (x)is increasing:Leta x1 x2 b. By part (c),{V(x2) (x2)} {V(x1) (x1)}=V (x1, x2) { (x2) (x1)} V (x1, x2) (x2) (x1) =V (x1, x2) {x1,x2} i 0(e) If is of Bounded Variation then (x) =V (a, x) [V (a, x) (x)]expresses asthe difference of two increasing Functions . On the other handif is the difference of two increasing Functions , then and are of Bounded Variation by Example 2 and isof Bounded Variation by part (a).Example 6We know that iffis continuous and is of Bounded Variation on [a, b], thenf R( ) on [a, b]. Iffis of Bounded Variation and is continuous on [a, b], then we havef R( ) on [a, b] with baf d =f(b) (b) f(a) (a) ba dfby our integration by parts theorem.

6 It is possible to havef R( ) on [a, b] even if neitherfnor are of Bounded Variation on [a, b]. For example, we have seen, in Example 4, that (x) ={0ifx= 0xcos xifx6= 0c Joel Feldman. 2017. All rights 30, 2017 Functions of Bounded Variation4is continuous but not of Bounded Variation on [0,1], because of excessive oscillation nearx= 0. Sof(x) = (1 x) (still with the of Example 4) is continuous but not of boundedvariation on [0,1], because of excessive oscillation nearx= 1. Butf R( ) on [0,12],by integration by parts, becausefis of Bounded Variation on [0,12]. Andf R( ) on[12,1], because is of Bounded Variation on [12,1]. Sof R( ) on [0,1], by our Joel Feldman. 2017. All rights 30, 2017 Functions of Bounded Variation5}