Solution by Jesse Jenks
Problem 6.2
Problem Statement: A function f satisfies a uniform Lipschitz condition of order ↵ > 0 on [a, b] if
|f (x) f (y)| < M |x y|↵ for every x, y in [a, b]. Show that a) if ↵ > 1, then f is constant, and if ↵ = 1, f
is of bounded variation, b) give an example of a function that satisfies a uniform Lipschitz condition of order
↵ < 1 but is not of bounded variation, and c) give an example of a function that is of bounded variation but
does not satisfy any uniform Lipschitz condition.
a Show that if ↵ > 1, then f is constant, and if ↵ = 1, f is of bounded variation.
Proof. Let f be a function satisfying a uniform Lipschitz condition of order ↵ > 1. Then for x 2 (a, b),
|f (x + ✏) f (x)| < M |x + ✏ x|↵ for some ✏. Since ↵ > 1,
f 0 (x) = lim
✏!0
|f (x + ✏) f (x)|
|✏|
 lim M ✏↵
1
✏!0
=0
which implies that f is constant. If ↵ = 1, then
|f (x + ✏) f (x)|
 lim M,
✏!0
✏!0
|✏|
f 0 (x) = lim
so by theorem 6.6, f is of bounded variation.
b Let xk+1 = xk + 2k
xk = 2 k , and let
1/↵
with x0 = 0 and 0 < ↵ < 1. Then {xk } is Cauchy and converges. Now let
f (x) =
(
|x xk |↵ ,
|xk+1 x|↵ ,
if xk < x  (xk + xk+1 )/2
if (xk + xk+1 )/2 < x  xk+1 ,
Then f satisfies a uniform Lipschitz condition of order ↵ since |b↵ a↵ | < |b a|↵ for 0 < a < b < 1.
Now consider the partition {t2k 1 = xk , t2k = (xk + xk+1 )/2}. That is, t1 = 1, t2 = 2, t3 = 3,
t4 = 3 + 21 1/↵ , and so on. Then
f (t2k ) = |(xk + xk+1 )/2
xk )/2|↵
= |(xk+1
1/↵
= |(2k
= |k
= |k
xk | ↵
)/2|↵
1/↵ ↵
1
|
|
while
f (t2k
This means that | fk | =
1
k,
1)
= 0.
which implies that
Vf (a, b)
1
X
|f (tk )|
k=1
1
X
=2
k=1
1
,
k
which does not converge, and so f is not of bounded variation.
c Let f be defined as
f (x) =
(
x,
x+c
if 1  x < 2
if 2  x  3,
where c > 1. Then f is of bounded variation on [1, 3]. Now let x = 2 + ✏/2 and y = 2 ✏/2 with
0 < ✏ < 1. Then |x y| = ✏, but |f (x) f (y)| > c, so for ↵ > 0, |f (x) f (y)| <
6 |x y|↵ . This shows
that f does not satisfy any uniform Lipschitz condition on [1, 3].