SOLUTION OF HW6
MINGFENG ZHAO
October 13, 2012
1. [4 Points, Exercises 27, on Page 72] Let f be continuous on [a, b), and define
F (x) = max f (t),
a≤t≤x
a ≤ x < b.
Show that F is continuous on [a, b).
Proof. For any a ≤ x < y < b, by the definition of F , then
F (x) = max f (t) ≤ F (y) = max f (t)
a≤t≤x
a≤t≤y
That is, F is an increasing function on [a, b).
Claim I: F is right continuous at a.
Since f is continuous on [a, b), then for any > 0, there exists some δ > 0 such that for all
a ≤ x < a + 2δ, we have
|f (x) − f (a)| < .
Now for all a ≤ x < a + δ, by the definition of F (x) and f is continuous on [a, a + δ], then there
exists some yx ∈ [a, a + δ] such that F (x) = f (yx ). But y0 ∈ [a, a + δ] ⊂ [a, a + 2δ), then
|F (x) − F (a)| = |f (yx ) − f (a)| < .
Hence F is right continuous at a.
Claim II: F is continuous on (a, b).
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MINGFENG ZHAO
For any x ∈ (a, b) and fix, then there exists some 0 > 0 such that [x − 0 , x + 0 ] ⊂ (a, b). Since
f is continuous on [a, b), then F is an increasing and bounded function on [a, x + 0 ], which implies
that F (x±) exist and
F (x−) ≤ F (x) ≤ F (x+).
Take a sequence xn such that xn increases and converges to x, as n → ∞. Since F is increasing,
then F (xn ) is increasing. By the definition of F and f is continuous on [a, b), then there exists some
yn ∈ [a, xn ] ⊂ [a, x + 0 ] such that F (xn ) = f (yn ) for all n ≥ 1. By taking a subsequence of yn ,
without loss of generality, assume yn itself, such that yn → y for some y ∈ [a, x], as n → ∞. Since f
is continuous, then F (xn ) = f (yn ) → f (y) as n → ∞, that is, F (x−) = f (y). Since y ∈ [a, x], then
f (y) ≤ F (x). On the other hand, for all z ∈ [a, x), since xn → x as n → ∞, then there exists some
xN such that z ∈ [a, xN ]. Hence f (z) ≤ f (xN ) ≤ F (xN ) = f (yn ) ≤ f (y), that is, F (x) ≤ f (y). Hence
F (x) = f (y), and F (xn ) → F (x), as n → ∞, which implies that F (x−) = F (x).
Take a sequence xn such that xn decreases and converges to x, as n → ∞, also we assume xn ∈
[x, x + 0 ] for all n ≥ 1. Since F is increasing, then F (xn ) is decreasing. By the definition of F and f
is continuous on [a, b), then there exists some yn ∈ [a, xn ] ⊂ [a, x + 0 ] such that F (xn ) = f (yn ) for all
n ≥ 1. By taking a subsequence of yn , without loss of generality, assume yn itself, such that yn → y
for some y ∈ [a, x], as n → ∞. Since f is continuous, then F (xn ) = f (yn ) → f (y) as n → ∞, that
is, F (x+) = f (y). Since y ∈ [a, x], then F (x+) = f (y) ≤ F (x). On the other hand, we know that
F (x) ≤ F (x+), hence F (x+) = F (x).
In summary, we get F (x−) = F (x) = F (x+), which implies that F is continuous at x. Since x is
arbitrary, then F is continuous on (a, b).
Therefore, we can conclude that F is continuous on [a, b).
SOLUTION OF HW6
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2. [3 Points, Exercises 32, on Page 72] Prove: If f is continuous on [a, ∞) and f (∞) exists, then f is
uniformly continuous on [a, ∞).
Proof. Since f (∞) exists, then for any > 0, there exists some X0 > 0 such that for all x, y ≥ X0 − 1,
we have
|f (x) − f (y)| < (1)
Since f is continuous on [a, ∞), then f is uniformly continuous on [a, X0 + 1]. Hence for the above
> 0, there exists some
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> δ > 0 such that for all x, y ∈ [a, X0 + 1] with |x − y| < δ, we have
|f (x) − f (y)| < (2)
Now for any x, y ∈ [a, ∞) with |x − y| < δ, without loss of generality, assume 0 ≤ x − y < δ.
If x ≤ X0 + 1, then y ≤ x ≤ X0 − 1. By (2), then |f (x) − f (y)| < . If x ≥ X0 + 1, then
y > x − δ ≥ X0 + 1 −
1
2
≥ X0 − 1. By (1), then |f (x) − f (y)| < . In summary, we know that for all
x, y ∈ [a, ∞) with |x − y| < , we have
|f (x) − f (y)| < .
Hence f is uniformly continuous on [a, ∞).
3. [2 Points, Exercises 3, on Page 96] Find the indicated limit:
lim
x→0
1 − cos x
ln(1 + x2 )
Proof. By the L’Hospital’s Rule, then
lim
x→0
1 − cos x
ln(1 + x2 )
=
=
lim
sin x
x→0
2x
1+x2
lim
sin x
· (1 + x2 )
2x
x→0
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MINGFENG ZHAO
=
=
=
=
lim
sin x
· lim (1 + x2 )
2x x→0
lim
sin x
2x
lim
cos x
2
x→0
x→0
x→0
1
.
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Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu