Real Analysis HW 9 Solutions
Problem 33: Let {fn } be a sequence of functions on [a, b] converging pointwise to f . Then
T V (f ) ≤ lim inf n T V (fn ).
Solution: Let P be any partition of [a, b], then since V (fn , P ) only depends on fn at a finite
number of points,
V (f, P ) = lim V (fn , P ).
n
Furthermore V (fn , P ) ≤ T V (fn ) and therefore
V (f, P ) = lim V (fn , P ) ≤ lim inf T V (fn ).
n
n
Taking the sup over all partitions P gives
T V (f ) ≤ lim inf T V (fn ).
n
Problem 35: For α and β positive numbers, define the function f on [0, 1] by
(
xα sin (1/xβ )
for 0 < x ≤ 1
f (x) =
0
for x = 0.
Show that if α > β, then f is of bounded variation on [0, 1], by showing that f 0 is integrable
over [0, 1]. Then show that if α ≤ β, then f is not of bounded variation on [0, 1].
Solution: If α > β, then
f 0 (x) = αxα−1 sin (1/xβ ) − βxα−β−1 cos (1/xβ ).
Since f is C 1 and bounded on (0, 1) we can use the fundamental theorem of calculus for
Riemann integrals to conclude that for any partition P
Z 1
n
X
V (f, P ) =
|f (xk ) − f (xk−1 )| ≤
|f 0 |
0
k=0
and therefore
Z
T V (f ) ≤
1
0
Z
|f | ≤
0
1
αxα−1 + βxα−β−1 < ∞,
0
1
since α > 0 and α − β > 0.
If α ≤ β, choose a partition Pn = {0, an , an−1 , . . . , a1 }, where
an =
then we see that
V (f, Pn ) =
nπ −1/β
2
−α/β
n X
kπ
k=1
2
.
Note that this series diverges as n → ∞ since α/β ≤ 1. Therefore
lim V (f, Pn ) ≤ T V (f ) = ∞.
n→∞
Problem 37: Let f be a continuous function on [0, 1] that is absolutely continuous on [, 1]
for each 0 < < 1.
(i) Show that f may not be absolutely continuous on [0, 1].
(ii) Show that f is absolutely continuous on [0, 1] if it is increasing.
√
(iii) Show that the function f on [0, 1], defined by f (x) = x for 0 ≤ x ≤ 1, is absolutely
continuous, but not Lipschitz, on [0, 1].
Solution:
(i) Consider
(
x sin (1/x)
f (x) =
0
if 0 < x ≤ 1
if x = 0
Note that on [, 1]
1
1
|f (x)| = sin(1/x) − cos (1/x) ≤ 1 +
x
0
Therefore f is Lipschitz on [, 1] and hence absolutely continuous on [, 1]. However,
we know from Problem 35 that f is not BV on [0, 1] and therefore not abolsolutely
continuous on [0, 1].
(ii) Suppose f is increasing, let η > 0 and choose so that
f () − f (0) < η/2.
Since f is absolutely continuous on [, 1] choose δ > 0 in response to η/2 in the absolute
continuity condition on [,P1]. Now suppose that {(ak , bk )}N
k=1 is a collection of disjoint
N
open intervals such that k=1 |bk − ak | < δ. We may assume that is not contained
2
in any of the intervals since we may always split any such interval (ak0 , bk0 ) into two
consecutive intervals (ak0 , ) ∪ (, bk0 ) such that, by the fact that f is increasing,
|f (bk0 ) − f ()| + |f () − f (ak0 )| = |f (bk0 ) − f (ak0 )|.
It follows that we may divide the set of intervals into n− intervals to the left of ,
+ + n+
− n−
{(a−
k , bk )}k=1 and n+ intervals to the right of , {(ak , bk )}k=1 . We observe that
n+
X
−
|b+
k − ak | < δ
k=1
and so by uniform integrability on [, 1],
n+
X
+
|f (b+
k ) − f (ak )| < η/2.
k=1
Also, since f is increasing,
n−
X
−
|f (b−
k ) − f (ak )| ≤ f () − f (0) < η/2.
k=1
Therefore
n
X
|f (bk ) − f (ak )| < η.
k=1
√
x is not Lipschitz on [0, 1] since its derivative is unbounded as x → 0.
(iii) Clearly √
However, x is increasing and is Lipschitz on [, 1] for any > 0. Therefore by (ii), f
is absolutely continuous on [0, 1].
Problem 39: Use the preceding problem to show that if f is continuous and increasing on
[a, b], then f is absolutely continuous on [a, b] if and only if for each , there is a δ > 0 such
that for a measurable subset E of [a, b],
m∗ (f (E)) < if m(E) < δ.
Solution: Suppose that f is absolutely continuous, and let δ > 0 be chosen in response
to > 0 in the absolute continuity condition as generalized in Problem 38. Supppose that
E ⊆ [a, b] is measurable and m(E) < δ/2. We can find a countable cover of disjoint open
intervals {(ak , bk )}∞
k=1 of E so that
!
∞
[
m
Ik < δ/2 + m(E) < δ
k=1
3
and therefore
∞
X
|f (bk ) − f (ak )| < .
k=1
Since f is increasing and continuous {f (Ik )}∞
k=1 is an open cover of f (E) and m(f (Ik )) =
f (bk ) − f (ak ). By the definition of outer measure we conclude that
∗
m (f (E)) ≤
∞
X
|f (bk ) − f (ak )| < .
k=1
For the converse, let δ > 0 be chosen to satisfy the converse condition with
S > 0 and let
{(ak , bk )}nk=1 be a finite collection of disjoint open interval such that E = nk=1 (ak , bk ) and
m(E) < δ. Since f is increasing we see that {f ((ak , bk ))}nk=1 are disjoint and by continuity
m(f ((ak , bk ))) = f (bk ) − f (ak ). By the convese condition
!
n
n
[
X
(ak , bk ) < .
|f (bk ) − f (ak )| = m
k=1
k=1
Problem 43: Define the functions f and g on [−1, 1] by f (x) = x1/3 for −1 ≤ x ≤ 1 and
(
x2 cos (π/2x)
if x =
6 0, x ∈ [−1, 1]
g(x) =
0
if x = 0.
(i) Show that both f and g are absolutely continuous on [−1, 1]
(ii) For the partition Pn = {−1, 0, 1/2n, 1/[2n − 1], . . . , 1/3, 1/2, 1} of [−1, 1], examine
V (f ◦ g, Pn ).
(iii) Show that f ◦ g fails to be of bounded variation, and hence also fails to be absolutely
continuous, on [−1, 1].
Solution:
(i) To show the absolute continuity of f , we note that it is monotone on [−1, 1], and so it
will be absolutely continuous if
Z 1
f 0 = f (1) − f (−1).
−1
Since f 0 (x) = 13 x−2/3 is Riemann integrable on [−1, −] ∪ [, 1] for any > 0, and
f (1) = 1, and f (−1) = −1, then by monotone convergence
Z
1 1 2/3
x dx = lim (1)1/3 + 1/3 − (−1/3 ) − (−1)1/3 = 2,
→0
3 −1
4
and therefore f must be absolutely continuous.
To show that g is absolutely continuous, we simply show that it is the indefinite integral
of some function over [a, b]. Define g(x) = 2x cos (π/2x) + 21 sin (π/2x), then we claim
that
Z x
g(s)ds = x2 cos (π/2x) = f (x)
−1
Once again using the fact that g(x) is integrable on [−1, −] ∪ [, 1] for any > 0, we
can say that if x < 0, then we already have
Z x
g(s)ds = x2 cos (π/2x) = f (x).
−1
However if x ≥ 0, then
Z x
g(s)ds = lim x2 cos (π/2x) − 2 cos (−π/2) + 2 cos (π/2) − 0
−1
→0
= x2 cos (π/2x) = f (x).
Therefore f (x) is absolutely continuous.
(ii) Let Pn = {−1, 0, 1/2n, 1/[2n−1], . . . , 1/3, 1/2, 1} be a partition of [−1, 1]. We examine
V (f ◦ g, Pn ),
V (f ◦ g, Pn ) = V (x2/3 cos1/3 (π/2x), Pn )
2n−1
X cos1/3 (nπ − kπ/2) cos1/3 (nπ − (k − 1)π/2) =
−
2/3
2/3
(2n
−
k)
(2n
−
k
+
1)
k=1
Using the fact that
(
0
cos (nπ − kπ/2) = cos (nπ) cos (kπ/2) =
(−1)n+k/2
k odd
k even
We can consider the sum over only even indicies in k, each of which is counted twice
in the sum
n−1 X
(−1)(n+k)/3 V (f ◦ g, Pn ) = 2
(2(n − k))2/3 k=0
=2
n−1
X
k=0
1/3
=2
5
1
(2(n − k))2/3
n
X
1
2/3
k
k=1
(iii) The sum in the previous part for diverges as n → ∞ Therefore, f ◦ g has unbounded
variation, and cannot be absolutely continuous.
Problem 45: Let f be absolutely continuous on R and g be absolutely continuous and
strictly monotone on [a, b]. Show that the composition f ◦ g is absolutely continuous on
[a, b].
Solution: Let > 0, {(ak , bk )} be a collection of disjoint intervals of [a, b], Since g is strictly
increasing, {(g(ak ), g(bk )} forms another disjoint colleciton of intervals, and by the absolute
continuity of f , there exists a δ1 > 0 such that whenever
n
X
|f (g(bk )) − f (g(ak ))| < whenever
k=1
n
X
|g(bk ) − g(ak )| < δ1 .
k=1
However, by the absolute continuity of g, there exists a δ2 > 0 such that
n
X
|g(bk ) − g(ak )| < δ1
whenever
k=1
n
X
[bk − ak ] < δ2 .
k=1
6