Math 501
Introduction to Real Analysis
Instructor: Alex Roitershtein
Iowa State University
Department of Mathematics
Summer 2015
Homework #6
Solutions
1. Suppose that f : R → R is continuous on [a, b] and P is a partition of [a, b]. Show that
Rb
there exists a Riemann sum of f over P that equals a f (x)dx.
Solution: Let P = {a = x0 , x1 , . . . , xn = b} be a partition of [a, b]. Denote ∆i = xi+1 − xi ,
mi =
inf
f (x),
and
x∈[xi ,xi+1 ]
Mi =
sup
f (x).
x∈[xi ,xi+1 ]
By the intermediate value theorem for integrals,
Z xi+1
mi ∆i ≤
f (x)dx ≤ Mi ∆i ,
i = 0, . . . , n − 1.
xi
Since f is continuous, there exists si , ti ∈ [xi , xi+1 ] such that
mi = f (si )
and
Mi = f (ti ),
i = 0, . . . , n − 1.
Since
1
f (si ) ≤
∆i
xi+1
Z
f (x)dx ≤ f (ti )
xi
and f is continuous, there exist ri ∈ [xi , xi+1 ] such that f (ri ) =
the additivity property of integrals,
n−1
X
i=0
f (ri )∆i =
n−1 Z
X
i=0
xi+1
Z
f (x)dx =
xi
1
∆i
R xi+1
xi
f (x)dx. Then, by
b
f (x)dx,
a
as desired.
2. Compute supP L(P, fi ) and inf P U (P, fi ) on [0, 1] for i = 1, 2, and the following two
functions defined on [0, 1] :
x
if x ∈ Q
1 if x ∈ Q
f1 (x) =
and
f2 (x) =
−x if x 6∈ Q
x if x 6∈ Q.
Solution: Both Q and its complement are dense in any interval of R. Hence L(P, f1 ) = 0
while U (P, f1 ) = 1 for any partition P of [0, 1]. Similarly, U (P, f2 ) = 1 for any partition P.
1
Observe next that for an arbitrary partition P = {x0 = 0, x1 , . . . , xn−1 , xn = 1},
n−1
X
L(P, f2 ) =
xi (xi+1 − xi )
i=0
is a Riemann sum of
R1
0
xdx = 1/2. Thus supP L(P, f2 ) = supP L(P, x) =
R1
0
xdx = 1/2.
3 [20 Points].
(a) Prove that if f : R → R is absolutely integrable on [0, +∞) then
Z ∞
f (xn )dx = 0.
lim
n→∞
1
(b) Suppose that f : R → R is continuously differentiable and one-to-one on [a, b]. Prove
that
Z b
Z f (b)
f (x)dx +
f −1 (x)dx = bf (b) − af (a).
a
f (a)
where f −1 is the inverse function of f.
Solution:
(a) Using change of variables xn = t and the fact that
dt
= nxn−1 = nt(n−1)/n ,
dx
we obtain
Z
∞
1
Z
f (x )dx =
n
f (xn )dx
1
1
=
n
Hence limn→∞
∞
R∞
1
∞
Z
1
1
f (t)t− n−1
n dt ≤
n
Z
∞
f (t)dt < +∞.
1
f (xn )dx = 0, by the squeeze theorem.
(b) Using change of variables x = f (t) and integrating by part, we obtain
Z f (b)
Z b
Z b
0
−1
−1
f (x)dx =
f
f (t)) f (t)dt =
tf 0 (t)dt
f (a)
a
a
Z b
b Z b
= tf (t) −
f (t)dt = bf (b) − af (a) −
f (t)dt,
a
a
a
from which the claim follows.
Remark: The identity that we proved admits a simple geometric interpretation in terms
of areas under the curves y = f (x) and y = f −1 (x) .... To figure it out, use the fact
that the graph of f is the reflection of the graph of f −1 across the line y = x ...
2
4. Suppose that f : R → R is bounded and f 2 is integrable on [a, b]. Does it follow that f
is integrable on [a, b]?
Solution: Let
f (x) =
1
if x ∈ Q
−1 if x 6∈ Q
Both the set of rational numbers and the set of irrational numbers are dense in any interval.
Therefore, U (P, f ) = 1 whereas L(P, f ) = −1 for any partition P of [a, b]. Thus f is not
integrable on 0, 1] even though f 2 = 1 is.
5. Suppose that f : R → R is continuous and non-negative on [0, 1]. Show that if
then f (x) = 0 for every x ∈ [0, 1].
R1
0
f (x) = 0,
Rb
Rd
Solution: Since f ≥ 0, we have a f (x)dx ≥ c f (x)dx for any a ≤ c ≤ d ≤ b. However,
if f (x0 ) > 0 for some x0 ∈ [a, b], then, since f is continuous, there exists [c, d] ⊂ [a, b] such
that x0 ∈ [c, d] and f (x) ≥ f (x0 )/2. Then
b
Z
Z
f (x)dx ≥
a
d
f (x)dx ≥
c
f (x0 ) · (d − c)
> 0,
2
which yields a contradiction.
6. Solve Exercise 37 in Chapter 2 of the textbook.
Solution: Consider, for instance, f = 0 and
0
if x ∈ [a, b − 1/n]
fn (x) =
n3/2 (x − b + 1/n) if x ∈ [b − 1/n, b]
Then
b
Z
3/2
Z
1/n
|fn (x) − f (x)| dx = n
a
0
1
x dx = √ ,
2 n
and hence limn→∞ fn = f in Cint . On the other hand,
max |fn (x) − f (x)| = |fn (b) − f (b)| =
x∈[a,b]
and hence fn − f is a divergent sequence in Cmax .
3
√
n,