SOLUTION OF HW6
MINGFENG ZHAO
October 13, 2012
1. [10 Points] Suppose that a < b are extended real numbers (R = R
S
{±∞}) and that f is differen-
tiable on (a, b). If f 0 is bounded on (a, b), prove that f is uniformly continuous on (a, b).
Proof. Since f 0 is bounded on (a, b), then there exists some M > 0 such that
|f 0 (x)| ≤ M,
Now for any > 0, let δ =
∀x ∈ (a, b).
> 0, then for all x, y ∈ (a, b) such that |x − y| < δ. By the mean
M
value theorem, there exists some z ∈ (x, y) such that
f (x) − f (y) = f 0 (z)(x − y).
Hence we get
|f (x) − f (y)| = |f 0 (z)||x − y| ≤ M |x − y| < M δ = .
Hence f is uniformly continuous on (a, b).
2. [10 Points] Let α > 0 and recall (xα )0 = αxα−1 and (ln x)0 =
1
x
for all x > 0. Prove that ln x ≤ xα
for all x large. Prove that there exists some constant Cα such that ln x ≤ Cα xα for all x ∈ [1, ∞) and
Cα → ∞ as α → 0+.
Proof. a. By the L’Hospital’s Rule, then
ln x
x−1
1
=
lim
= lim
= 0.
x→∞ xα
x→∞ αxα−1
x→∞ αxα
lim
1
2
MINGFENG ZHAO
Then there exists some X0 > 0 such that for all x ≥ X0 , we have
ln x
≤ 1.
xα
Hence we get xα ≥ ln x for all x ≥ X0 .
b. Let Cα = max{1, ln X0 }, then for all 1 ≤ x ≤ X0 , we know that
ln x ≤ ln X0 ≤ Cα ≤ Cα xα .
For all x ≥ X0 , then
ln x ≤ xα ≤ Cα xα .
Hence we get
ln x ≤ Cα xα ,
∀x ≥ 1.
c. If ln x ≤ Cα xα for all x ∈ [1, ∞), then
ln x
≤ Cα ,
xα
∀x ≥ 1.
When α & 0, we take 0 < α < 1, then
α1
1
xα =
> 1.
α
Hence we get
1
1
ln xα
1
α ln α
=
= ln → ∞,
−1
xα
α
α
α
as α & 0.
Hence Cα → ∞, as α & 0.
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu