SOLUTION OF HW9
MINGFENG ZHAO
November 27, 2011
1. [The Problem 27, in Page 139] Prove that there is no real number A such that f (x) = sin
1
→ A,
x
as x → 0.
Proof. If not, then there exists some A ∈ R such that lim sin
x→0
1
= A. Then for = 12 , we know that
x
there exists some δ > 0 such that
sin 1 − A < 1 ,
2
x
Since lim
n→∞
π
2
for all 0 < |x| < δ.
1
= 0, then there exists some large N > 0 such that
+ 2nπ
x1 =
Which implies that x2 =
1
3π
2 +2N π
π
2
1
< δ.
+ 2N π
< δ. So we have
h
i
sin 1 − A = sin π + 2N π − A = |1 − A| = |A − 1| < 1 .
x1
2
2
And
sin 1 − A = sin 3π + 2N π − A = | − 1 − A| = |1 + A| < 1 .
x2
2
2
Since |A − 1| <
1
, then
2
|A + 1| = |A − 1 + 2| ≥ 2 − |A − 1| > 2 −
1
3
1
= > .
2
2
2
Which contradicts with |A + 1| < 21 . Therefore, we know that there is no real number A such that
f (x) = sin
1
→ A, as x → 0.
x
1
2
MINGFENG ZHAO
2. [The Problem 5, in Page 145] Given a real valued function f which is continuous on closed interval
[0, 1]. Assume that 0 ≤ f (x) ≤ 1 for all x ∈ [0, 1]. Prove that there is at least one point c ∈ [0, 1] such
that f (c) = c.
Proof. Consider the function g(x) = f (x) − x for all x ∈ [0, 1]. Since f is continuous on [0, 1], then
g is also continuous on [0, 1]. Also we know that g(0) = f (0) − 0 = f (0), and g(1) = f (1) − 1. Since
0 ≤ f (x) ≤ 1 for all x ∈ [0, 1], in particular, we get f (0) ≥ 0, and f (1) ≤ 1, so we have g(0) ≥ 0, and
g(1) ≤ 0.
If g(0) = 0, then we have f (0) − 0 = 0, that is, f (0) = 0, so we are done. If g(1) = 1, then
f (1) − 1 = 0, that is, f (1) = 1, so we are done. Now assume g(0) > 0, and g(1) < 0, since g is
continuous on [0, 1], then by the Bolzano’s Theorem, we know that there exists some c ∈ [0, 1] such
that g(c) = 0, that is, f (c) − c = 0, so f (c) = c. In a summary, we can conclude that there exists
some c ∈ [0, 1] such that f (c) = c.
3. [The Problem 1, in Page 155] Show the following inequalities:
1
√ ≤
10 2
Z
1
√
0
Proof. Let f (x) = x9 for all [0, 1], and g(x) = √
x9
1
.
dx ≤
10
1+x
1
for all [0, 1]. Then we know that both f and g
1+x
are continuous and nonnegative functions on [0, 1]. Then by the Mean Value Theorem for Integrals,
we know that there exists some c ∈ [0, 1] such that
Z
0
1
√
x9
dx
1+x
Z
=
1
f (x)g(x) dx
0
Z
=
1
g(c) ·
f (x) dx
0
Z
=
g(c) ·
1
x9 dx
0
=
g(c) ·
=
g(c)
.
10
1
1 10 x 10
0
SOLUTION OF HW9
Since g(x) = √
√
3
1
1
1
1
1
for all [0, 1], then g(c) = √
≥ √
= √ , and g(c) = √
≤
1+x
1+c
1+1
1
+c
2
1
= 1. Therefore, we have
1+0
1
√ ≤
10 2
1
Z
√
0
x9
1
dx ≤
.
10
1+x
Z
b
f (x) dx =
4. [The Problem 7, in Page 155] Assume that f is integrable and nonnegative on [a, b]. If
a
0, prove that f (x) = 0 at each point of continuity of f .
Proof. If the above statement is not true, then there exists some c ∈ [a, b] such that f is continuous
at x = c, and f (c) 6= 0. Since f (x) ≥ 0 for all x ∈ [a, b] and f (c) 6= 0, so we know that f (c) > 0. Since
f is continuous at x = c, and f (c) > 0, then for =
f (c)
> 0, there exists some δ > 0 such that for
2
all |x − c| < δ, we have
|f (x) − f (c)| <
f (c)
.
2
That is, we get
3f (c)
f (c)
< f (x) <
,
2
2
for all |x − c| < δ.
If c = a, then we know that
f (c)
3f (c)
< f (x) <
,
2
2
for all |x − c| = |x − a| = x − a < δ.
So we get
Z
0
=
b
f (x) dx
a
Z
a+δ
=
Z
b
f (x) dx +
a
Z
f (x) dx
a+δ
a+δ
≥
f (x) dx
Since f (x) ≥ 0 for all x ∈ [0, 1]
f (c)
dx
2
Since f (x) >
a
Z
≥
a
=
a+δ
f (c)
·δ
2
f (c)
for all x < a + δ
2
4
MINGFENG ZHAO
> 0.
That is, 0 > 0, we get a contradiction. If c = b, then we know
f (c)
3f (c)
< f (x) <
,
2
2
for all |x − c| = |x − b| = b − x < δ.
So we get
b
Z
0
f (x) dx
=
a
b−δ
Z
Z
b
f (x) dx +
=
a
f (x) dx
b−δ
b
Z
≥
f (x) dx
Since f (x) ≥ 0 for all x ∈ [0, 1]
f (c)
dx
2
Since f (x) >
b−δ
b
Z
≥
b−δ
f (c)
for all x > b − δ
2
f (c)
·δ
2
=
> 0.
That is, 0 > 0, we get a contradiction. If a < c < b, then we know that
3f (c)
f (c)
< f (x) <
,
2
2
for all |x − c| < δ.
So we get
Z
0
=
b
f (x) dx
a
Z
c−δ
=
Z
c+δ
f (x) dx +
a
Z
Z
b
f (x) dx +
c−δ
f (x) dx
c+δ
c+δ
≥
f (x) dx
Since f (x) ≥ 0 for all x ∈ [0, 1]
f (c)
dx
2
Since f (x) >
c−δ
Z
c+δ
≥
c−δ
=
f (c)
·δ
2
> 0.
f (c)
for all c − δ < x < c + δ
2
SOLUTION OF HW9
5
That is, 0 > 0, we get a contradiction. In a summary, we can conclude that f (x) ≡ 0 at each point
of continuity of f .
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu