Real Analysis Test 2 Fall 2004 SOLUTIONS 1. (20 points) Suppose ...

Real Analysis
Test 2
Fall 2004
SOLUTIONS
1. (20 points) Suppose that f and g are increasing on an interval I and that f(x) > g(x) for
all x ∈ I. Denote the inverses of f and g by F and G and their domains by J 1 and J 2 ,
respectively. Prove that F (x) < G(x) for each x ∈ J 1 ∩ J 2 .
Proof. Let x 0 ∈ J 1 ∩ J 2 . Then there exist a 1 , a 2 ∈ I such that f(a 1 ) = x 0 = g(a 2 ).
Equivalently, F (x 0 ) = a 1 , G(x 0 ) = a 2 . Claim that a 1 = F (x 0 ) < G(x 0 ) = a 2 . If not, then
a 1 ≥ a 2 . Now if a 1 = a 2 , then x 0 = f(a 1 ) > g(a 1 ) = g(a 2 ) = x 0 , which is absurd. On the
other hand, if a 1 > a 2 , then x 0 = f(a 1 ) > f(a 2 ) > g(a 2 ) = x 0 , yet another contradiction.
This proves our claim that for x 0 ∈ J 1 ∩ J 2 , F (x 0 ) < G(x 0 ).
2. (a) (10 points) lim sec x − tan x
x→
π
2
lim
x→ π 2
1 − sin x
cos x
− cos x
= lim
x→
π
2
− sin x = 0
(b) (10 points) lim
x→0 + xx
Let
y = x x
ln y = x ln x
lim ln y = ln lim y
x→0 + x→0 + ln x
= lim
x→0 + 1
x
(since ln is continuous).
Hence
ln lim
x→0 + y = lim
1
x
x→0 + − 1
x 2
1
= lim
x→0 + x · −x2
1
lim y =
x→0 e0 = 1.
= 0

3. (20 points) Suppose that f and g are uniformly continuous on a subset S of R (not a
closed interval). Prove that the function h = f − g is uniformly continuous.
Proof. Since f, g are uniformly continuous on S ⊂ R, given ε > 0, there exists δ 1 , δ 2 > 0
such that for x 1 , x 2 ∈ S with
|x 1 − x 2 | < δ 1 ⇒ |f(x 1 ) − f(x 2 )| < ε 2
(1)
|x 1 − x 2 | < δ 2 ⇒ |g(x 1 ) − g(x 2 )| < ε 2 . (2)
Let δ = min(δ 1 , δ 2 ). Then (1) and (2) hold if δ 1 , δ 2 are replaced by S. Now for |x 1 −x 2 | < δ,
we have
Thus h is uniformly continuous on S.
4. Give examples of
|h(x 1 ) − h(x 2 )| = |f(x 1 ) − g(x 1 ) − f(x 2 ) + g(x 2 )|
= |f(x 1 ) − f(x 2 )| + |g(x 1 ) − g(x 2 )|
< ε 2 + ε 2 = ε.
(i) (5 points) A bounded sequence that is not Cauchy.
Consider the sequence {(−1) n }, n = 1, 2, 3, . . . . It is bounded but not Cauchy.
(ii) (5 points) A nested sequence {I n } of intervals such that ∞ ∩
n=1
= I n = ∅
Let I n = ( 0, 1 n)
, n = 1, 2, 3, . . . . Then In ⊃ I n+1 . But ∞ ∩
n=1
I n = ∅.
(iii) (5 points) A continuous but not uniformly continuous function.
f(x) = 1 x
on (0, ∞) is continuous but not uniformly continuous.
(iv) (5 points) A Cauchy sequence {x n } and a function f such that {f(x n )} is not Cauchy.
{
1
x
Define f(x) =
, x > 0
2, x = 0 on [0, ∞). Let x n = 1 . Then x n n → 0 as n → ∞,
but f(x n ) = n → ∞ ≠ f(0) = 2.

5. Prove directly from the definition that the function f(x) = x 4 is uniformly continuous on
the closed interval [0, 1]. Is f uniformly continuous on R?
Proof. (a) We will show that f is continuous at each point x ∈ [0, 1]. Let x 0 ∈ [0, 1] and
ε > 0 be given. Put δ = ε. Then for |x − x 4 0| < δ, we have
|f(x) − f(x 0 )| = |x 4 − x 4 0 |
= |x − x 0 ||x + x 0 ||x 2 + x 2 0 |
< δ · 2 · 2 = ε.
Hence f is continuous on x 0 . By Theorem 3.13, f is uniformly continuous on [0, 1].
(b) Let x n = n, y n = n + 1 . Then |x n n − y n | = 1 → 0 as n → ∞. However,
n |f(x n ) − f(y n )| =
∣ n4 + 4 n3
n + 6n2 n + 4 n 2 n + 1 ∣ ∣∣∣
3 n − 4 n4 =
∣ 4n2 + 6 + 4 n + 1 ∣ ∣∣∣
→ ∞ as n → ∞.
2 n 4
Hence f is not uniformly continuous on R.