SOLUTION OF HW8
MINGFENG ZHAO
October 29, 2012
1. (10 Points) Suppose a < b and f : [a, b] → R is bounded.
a. Prove that if f is continuous at x0 ∈ [a, b] and f (x0 ) 6= 0, then
Z
b
|f (x)| dx > 0
a
b. Show that if f is continuous on [a, b], then
Z
b
|f (x)| dx = 0
if and only if f (x) = 0, ∀x ∈ [a, b]
a
c. Does part b hold when the absolute value is removed?
Proof. a. Since f (x0 ) > 0 and f is continuous at x0 , then for there exist some δ > 0 such that
|f (x) − f (x0 )| <
|f (x0 )|
,
2
∀|x − x0 | < δ
Hence for all |x − x0 | < δ, we have
|f (x)| = |f (x) − f (x0 ) + f (x0 )| ≥ |f (x0 )| − |f (x) − f (x0 )| > |f (x0 )| −
|f (x0 )|
|f (x0 )|
=
>0
2
2
Hence we know that
Z
b
Z
x+δ
|f (x)| dx ≥
a
Z
x+δ
|f (x)| dx ≥
x−δ
x−δ
|f (x0 )|
dx = δ|f (x0 )| > 0
2
Hence we get
Z
b
|f (x)| dx > 0
a
1
2
MINGFENG ZHAO
Z
b
|f (x)| dx = 0. If f (x) 6≡ 0 in [a, b], that is, there exitsts some x0 ∈ [a, b]
b. (=⇒) Assume that
a
such that f (x0 ) 6= 0. By the result of part a, we have
b
Z
|f (x)| dx > 0
a
Contradiction. Hence we must have f (x) ≡ 0 for all x ∈ [a, b].
(⇐=) Assume that f (x) ≡ 0 for all x ∈ [a, b], then
b
Z
Z
|f (x)| dx =
a
b
0 dx = 0
a
c. When the absolute value is removed, the part b is not true. In fact, we take a = −1, b = 1 and
f (x) = x for all x ∈ [a, b], then f (x) 6≡ 0 for all x ∈ [a, b], but we have
Z
1
1 2 1 1
x = − =0
2 −1
2 2
1
x dx =
−1
2. (10 Points) a. Suppose gn ≥ 0 is a sequence of integrable functions, wehre
b
Z
lim
gn (x) dx = 0
n→∞
a
Show that if f : [a, b] → R is integrable on [a, b], then
b
Z
lim
n→∞
f (x)gn (x) dx = 0
a
b. Prove that if f is integrable on [0, 1], then
Z
lim
n→∞
1
xn f (x) dx = 0
0
Proof. a. Since f (x) and gn (x) are integrable on [a, b], then f (x)gn (x) is also integrable on [a, b], and
f (x) is bounded on [a, b], Hence there exists some −∞ < m < M < ∞ such that
m ≤ f (x) ≤ M,
∀x ∈ [a, b]
SOLUTION OF HW8
3
Since gn (x) ≥ 0 for all x ∈ [a, b], then
mgn (x) ≤ f (x)gn (x) ≤ M gn (x),
∀x ∈ [a, b]
Integrating from a to b, we get
b
Z
f (x)gn (x) dx ≤ M
a
a
b
gn (x) dx
a
b
gn (x) = 0, by taking n → ∞ in the above identity, we get
Since lim
n→∞
Z
gn (x) dx ≤
m
Z
b
Z
a
b
Z
0 ≤ lim
f (x)gn (x) dx ≤ 0.
n→∞
a
Hence we have
b
Z
lim
f (x)gn (x) dx = 0
n→∞
a
b. Let gn (x) = xn for all x ∈ [0, 1], then gn (x) ≥ 0 and
Z
1
1
Z
gn (x) dx =
0
1
1
1
n+1 x dx =
x
=
n+1
n+1
0
n
0
Hence we get
b
Z
lim
gn (x) dx = 0
n→∞
a
Since f (x) is integrable on [0, 1], by the result of part a, we know that
b
Z
lim
n→∞
f (x)gn (x) dx = 0
a
That is, we get
Z
lim
n→∞
1
xn f (x) dx = 0
0
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu