SOLUTION OF HW4
MINGFENG ZHAO
September 30, 2012
1. [10 Points] If f : [a, b] → [a, b] is continuous, then the function f has a fixed point, that is, there is
a c ∈ [a, b] such that f (c) = c.
Proof. Define g(x) = f (x) − x for all x ∈ [a, b], since f is continuous on [a, b], then g is also continuous
on [a, b]. On the other hand, since f ([a, b]) ⊂ [a, b], then
g(a) = f (a) − a ≥ a − a = 0,
and g(b) = f (b) − b ≤ b − b = 0.
Case I: g(a) = 0, that is, f (a) = a, we are done.
Case II: g(b) = 0, that is, f (b) = b, we are done.
Case III: g(a) > 0 and g(b) < 0. Since g is continuous on [a, b], then there exists some c ∈ [a, b]
such that g(c) = 0, that is, f (c) = c.
In summary, we know that f (c) = c for some c ∈ [a, b].
2. [10 Points] Decide which of the following statements are true and which are false. Prove the true
ones, and provide counterexamples for the false ones.
a. If f is uniformly continuous on (0, ∞) and g is positive and bounded on (0, ∞), then f g is
uniformly continuous on (0, ∞).
b. The function x ln x1 is uniformly continuous on (0, 1).
c. The function
cos x
is uniformly continuous on (0, 1) for all nonzero m and b ∈ R.
mx + b
1
2
MINGFENG ZHAO
d. If f , g are uniformly continuous on [a, b] and g(x) 6= 0 for all x ∈ [a, b], then
f
g
is uniformly
continuous on [a, b].
Proof. a. False. For example, f (x) ≡ 1 for all x > 0 and
g(x) =




 2,
if x ≥ 2



 1,
if 0 < x < 2.
Then f is uniformly continuous on (0, ∞) and g is positive and bounded on (0, ∞), but f (x)·g(x) ≡
g(x) is not continuous on (0, ∞), in particular, f g is not uniformly continuous on (0, ∞).
b. True. Let f (x) = x ln x1 = −x ln x for all x > 0, then f is continuous on (0, ∞). Since
lim x ln
x&0
ln 1
1
= lim 1 x = 0.
x x&0 x
For any > 0, there exists some δ0 > 0 such that for all 0 < x < 2δ0 , we have
0 < f (x) = x ln
1
< .
x
2
Since f (x) is continuous on (0, ∞), then f is continuous on [δ0 , 1], in particular, f is uniformly
continuous on [δ0 , 1]. Hence for the above > 0, there exists some δ1 > 0 such that for all |x − y| < δ1 ,
we have
|f (x) − f (y)| < .
Let δ = min{δ0 , δ1 } > 0, then for all |x − y| < δ, without loss of generality, we assume 0 < x ≤ y <
x + δ, then we know that
a. Case I: x ≤ δ0 < 2δ0 , then y < x + δ < δ0 + δ0 = 2δ0 . Hence
0 < f (x), f (y) <
.
2
so we get
|f (x) − f (y)| ≤ f (x) + f (y) <
+ = .
2 2
SOLUTION OF HW4
3
b. Case I: x > δ0 , then y ≥ x ≥ δ0 . Hence |x − y| < δ < δ1 , by the definition of δ1 , e know that
|f (x) − f (y)| < .
In summary, for any given > 0, there exists some δ > 0 such that for all |x − y| < δ, we have
|f (x) − f (y)| < .
c. False. For example, let m = 2 and b = −1, then
cos x
cos x
=
,
mx + b
2x − 1
which is not defined at x = 21 , in particular, it is not uniformly continuous on (0, 1).
d. True. Since f and g are uniformly continuous on [a, b], then there exists some M > 0 such that
|f (x)|, |g(x)| ≤ M,
∀x ∈ [a, b].
Since g(x) 6= 0 for all x ∈ [a, b], then there exists some m > 0 such that
0 < m ≤ |g(x)|,
∀x ∈ [a, b].
For any > 0, since f and g are uniformly continuous on [a, b], then there exists some δ > 0 such
that for all |x − y| < δ, we have
|f (x) − f (y)| <
m
m
and |g(x) − g(y)| <
.
2M
2M
Now for any |x − y| < δ, we have
f (x) f (y) g(x) − g(y) =
|f (x)g(y) − f (y)g(x)|
|g(x)g(y)|
=
|[f (x) − f (y)]g(y) − f (y)[g(x) − g(y)]|
|g(x)||g(y)|
≤
|f (x) − f (y)||g(y)| + |f (y)||g(x) − g(y)|
|g(x)||g(y)|
<
m
2M m
· M + 2M
·M
m
4
MINGFENG ZHAO
=
Therefore,
f
g
.
is uniformly continuous on [a, b].
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu