Functions of Bounded Variation

Our main theorem concerning the existence of Riemann–Stietjes integrals assures us

Rb
that the integral a f (x) dα(x) exists when f is continuous and α is monotonic. Our lin-
Rb
earity theorem then guarantees that the integral a f (x) dα(x) exists when f is continuous
and α is the difference of two monotonic functions. In these notes, we prove that α is the
difference of two monotonic functions if and only if it is of bounded variation, where

Definition 1

(a) The function α : [a, b] → IR is said to be of bounded variation on [a, b] if and only if
there is a constant M > 0 such that
n
X  
α(xi ) − α(xi−1 ) ≤ M
i=1

for all partitions IP = {x0 , x1 , · · · , xn } of [a, b].

(b) If α : [a, b] → IR is of bounded variation on [a, b], then the total variation of α on [a, b]
is defined to be
n P n    o
Vα (a, b) = sup α(xi ) − α(xi−1 )  IP = {x0 , x1 , · · · , xn } is a partition of [a, b]

i=1

Example 2 If α : [a, b] → IR is monotonically increasing, then, for any partition IP =

{x0 , x1 , · · · , xn } of [a, b]
n
X n
  X 
α(xi ) − α(xi−1 ) = α(xi ) − α(xi−1 ) = α(xn ) − α(x0 ) = α(b) − α(a)
i=1 i=1

Thus α is of bounded variation and Vf (a, b) = α(b) − α(a).

Example 3 If α : [a, b] → IR is continuous on [a, b] and differentiable on (a, b) with

supa<x<b |α′ (x)| ≤ M , then, for any partition IP = {x0 , x1 , · · · , xn } of [a, b], we have, by
the Mean Value Theorem,
n
X n n
  X  ′  X
α(xi ) − α(xi−1 ) = α (ti )[xi − xi−1 ] ≤ M [xi − xi−1 ] = M (b − a)
i=1 i=1 i=1

Thus α is of bounded variation and Vf (a, b) ≤ M (b − a).

c Joel Feldman. 2017. All rights reserved. January 30, 2017 Functions of Bounded Variation 1
Example 4 Define the function α : [0, 1] → IR by

0 if x = 0
α(x) =
x cos πx if x =
6 0

This function is continuous, but is not of bounded variation because it wobbles too much
near x = 0. To see this, consider, for each m ∈ IN, the partition

1 1 1 1 1

IPm = x0 = 0 , x1 = 2m
, x2 = 2m−1
, x3 = 2m−2
, · · · , x2m−2 = 3
, x2m−1 = 2
, x2m = 1

The values of α at the points of this partition are

1 1 1
α(IPm ) = {0, 2m , − 2m−1 , 2m−2 , · · · , − 13 , 1
2, −1}

1
3 x2m = 1
x2m−1 = 1 x
2

y = α(x)

For this partition,

2m
X  
α(xi ) − α(xi−1 )
i=1
 1 1
  1
1  1 
+ · · · +  − 13 − 14  +  21 + 13  +  −1− 12 
        
=  2m −0 +  − 2m−1 − 2m +  (2m−2) + 2m−1
1 1 1 1 1 1 1 1 1 1
= 2m +0 + 2m−1 + 2m + (2m−2) + 2m−1 + · · · + 3 + 4 + 2 + 3 + 1+ 2
1 1
· · · + 13 + 12 + 1


=2 2m + 2m−1 +

∞
1
P
The harmonic series k diverges. So given any M , there is an m ∈ IN for which the
k=2
partition IPm obeys
n
X  
α(xi ) − α(xi−1 ) > M
i=1

Theorem 5

(a) If α, β : [a, b] → IR are of bounded variation and c, d ∈ IR, then cα + dβ is of bounded

variation and
Vcα+dβ (a, b) ≤ |c|Vα (a, b) + |d|Vβ (a, b)

c Joel Feldman. 2017. All rights reserved. January 30, 2017 Functions of Bounded Variation 2
(b) If α : [a, b] → IR is of bounded variation on [a, b] and [c, d] ⊂ [a, b], then α is of bounded
variation on [c, d] and
Vα (c, d) ≤ Vα (a, b)

(c) If α : [a, b] → IR is of bounded variation and c ∈ (a, b), then

Vα (a, b) = Vα (a, c) + Vα (c, b)

(d) If α : [a, b] → IR is of bounded variation then the functions V (x) = Vα (a, x) and
V (x) − α(x) are both increasing on [a, b].

(e) The function α : [a, b] → IR is of bounded variation if and only if it is the difference of
two increasing functions.

Proof: We shall use the shorthand notation

IP
X n
X
   
 ∆i α  for α(xi ) − α(xi−1 )
i=1

where the partition IP = {x0 , x1 , · · · , xn }.

(a) follows from the observation that, for any IP partition of [a, b],

IP
X IP
X IP
X
     
∆i (cα + dβ) ≤ |c| ∆i α + |d| ∆i β  ≤ |c|Vα (a, b) + |d|Vβ (a, b)

(b) follows from the observation that, for any partition IP of [c, d],

IP
X   IP∪{a,b}
X  
 ∆i α  ≤ ∆i α ≤ Vα (a, b)


(c) If IP = x0 , x1 , · · · , xn is any partition of [a, b] and xi−1 ≤ c ≤ xi , then
     
α(xi ) − α(xi−1 ) ≤ α(xi ) − α(c) + α(c) − α(xi−1 )

so that

IP
X   IP∪{c}
X   (IP∪{c})∩[a,c]
X   (IP∪{c})∩[c,b]
X  
 ∆i α  ≤  ∆i α  = ∆i α + ∆i α ≤ Vα (a, c) + Vα (c, b)

c Joel Feldman. 2017. All rights reserved. January 30, 2017 Functions of Bounded Variation 3
which implies that Vα (a, b) ≤ Vα (a, c) + Vα (c, b). To prove the other inequality, we let
IP
P1  
ε > 0 and select a partition IP1 of [a, c] for which ∆i α ≥ Vα (a, c) − ε and a partition
IP
P2  
IP2 of [c, b] for which ∆i α ≥ Vα (c, b) − ε. Then

IPX
1 ∪IP2 IP1 IP2
  X   X  
∆i α = ∆i α + ∆i α ≥ Vα (a, c) + Vα (c, b) − 2ε

This assures that Vα (a, b) ≥ Vα (a, c) + Vα (c, b) − 2ε for all ε > 0 and hence that Vα (a, b) ≥
Vα (a, c) + Vα (c, b).

(d) Proof that V (x) is increasing: Let a ≤ x1 ≤ x2 ≤ b. Then, by part (c),

V (x2 ) − V (x1 ) = Vα (a, x2 ) − Vα (a, x1 ) = Vα (x1 , x2 ) ≥ 0

(d) Proof that V (x) − α(x) is increasing: Let a ≤ x1 ≤ x2 ≤ b. By part (c),

{V (x2 ) − α(x2 )} − {V (x1 ) − α(x1 )} = Vα (x1 , x2 ) − {α(x2 ) − α(x1 )}

 
≥ Vα (x1 , x2 ) − α(x2 ) − α(x1 )
{x1 ,x2 }
X  
= Vα (x1 , x2 ) − ∆i α

≥0

 
(e) If α is of bounded variation then α(x) = Vα (a, x) − Vα (a, x) − α(x) expresses α as
the difference of two increasing functions. On the other hand if α is the difference β − γ
of two increasing functions, then β and γ are of bounded variation by Example 2 and α is
of bounded variation by part (a).

Example 6 We know that if f is continuous and α is of bounded variation on [a, b], then
f ∈ R(α) on [a, b]. If f is of bounded variation and α is continuous on [a, b], then we have
f ∈ R(α) on [a, b] with
Z b Z b
f dα = f (b)α(b) − f (a)α(a) − α df
a a

by our integration by parts theorem. It is possible to have f ∈ R(α) on [a, b] even if neither
f nor α are of bounded variation on [a, b]. For example, we have seen, in Example 4, that

0 if x = 0
α(x) =
x cos πx if x =
6 0

c Joel Feldman. 2017. All rights reserved. January 30, 2017 Functions of Bounded Variation 4
is continuous but not of bounded variation on [0, 1], because of excessive oscillation near
x = 0. So f (x) = α(1 − x) (still with the α of Example 4) is continuous but not of bounded
variation on [0, 1], because of excessive oscillation near x = 1. But f ∈ R(α) on [0, 12 ],
by integration by parts, because f is of bounded variation on [0, 12 ]. And f ∈ R(α) on
[ 21 , 1], because α is of bounded variation on [ 21 , 1]. So f ∈ R(α) on [0, 1], by our linearity
theorem.

c Joel Feldman. 2017. All rights reserved. January 30, 2017 Functions of Bounded Variation 5