Math 341 Homework # 8
P104. 13, 14, 17. P105. 19, 20, 21, 22.
13. Let f : D → R be continuous at x0 ∈ D. Prove that there is M > 0 and
a neighborhood Q of x0 such that |f (x)| ≤ M for all x ∈ Q ∩ D.
Proof: Since f is continuous at x0 , ∀ > 0, ∃ δ > 0, for x ∈ D with
|x − x0 | < δ, we have
|f (x) − f (x0 )| < .
Let = 1, M = |f (x0 )| + 1 and Q = (x0 − δ, x0 + δ). Then for x ∈ Q ∩ D,
|f (x)| = |f (x) − f (x0 ) + f (x0 )|
≤ |f (x) − f (x0 )| + |f (x0 )|
< 1 + |f (x0 )| = M.
14. If f : D → R is continuous at x0 ∈ D, prove that |f | : D → R such that
|f |(x) = |f (x)| is continuous at x0 .
Proof: Since f is continuous at x0 , ∀ > 0, ∃ δ > 0, for x ∈ D with
|x − x0 | < δ, we have
|f (x) − f (x0 )| < .
Hence
||f |(x) − |f |(x0 )| ≤ |f (x) − f (x0 )| < .
Thus, |f | is continuous at x0 .
17. Suppose f : D →√R with f (x) ≥ 0 for all x ∈ D. Show that, if f is
continuous at x0 , then f is continuous at x0 .
Proof: Since f is continuous at x0 , for any sequence {xn } q
⊂ D converging to
x0 , we have {f (xn )} converges to f (x0 ). By a hw earlier, { f (xn )} converges
q
√
to f (x0 ). Hence, f is continuous at x0 .
(Remark: You can also prove it directly using -δ argument.)
19. Let f, g : D → R be uniformly continuous. Prove that f + g : D → R is
uniformly continuous. What can be said about f g? Justify.
Proof: Since f is uniformly continuous, ∀ > 0, ∃ δ1 > 0, for |x − y| < δ1 ,
we have
|f (x) − f (y)| < .
2
1
Similarly, ∃ δ2 > 0, for |x − y| < δ2 , we have
|g(x) − g(y)| < .
2
Let δ = min(δ1 , δ2 ). For |x − y| < δ, we have
|(f + g)(x) − (f + g)(y)| = |f (x) − f (y) + g(x) − g(y)|
≤ |f (x) − f (y)| + |g(x) − g(y)|
+ = .
<
2 2
Thus, f + g is uniformly continuous.
f g is not necessarily uniformly continuous. For example, f (x) = g(x) =
x. Both f and g are uniformly continuous, but f g(x) = x2 is not uniformly
continuous.
20. Let f : A → B and g : B → C be uniformly continuous. What can be
said about g ◦ f : A → C? Justify.
Solution: g ◦ f is uniformly continuous.
Since g is uniformly continuous, ∀ > 0, ∃ δ1 > 0, for |z − w| < δ1 , we
have
|g(z) − g(w)| < .
Since f is uniformly continuous, for the δ1 above, ∃ δ > 0, for |x − y| < δ,
we have
|f (x) − f (y)| < δ1 .
Thus
|(g ◦ f )(x) − (g◦)(y)| = |g(f (x)) − g(f (y))| < .
2
21. Define f : [3.4, 5] → R by f (x) = x−3
. Show that f is uniformly
continuous on [3.4, 5] without using Theorem 3.8.
Proof: ∀ > 0, ∃ δ = .08 > 0, for x, y ∈ [3.4, 5] with |x − y| < δ, we have
2
x − 3
2 −
|f (x) − f (y)| =
y − 3
2|x − y|
=
(x − 3)(y − 3)
2δ
<
= .
.42
2
Thus, f is uniformly continuous.
22. Define f : (2, 7) → R by f (x) = x3 − x + 1. Show that f is uniformly
continuous on (2, 7) without using Theorem 3.8.
> 0, for x, y ∈ (2, 7) with |x − y| < δ, we have
Proof: ∀ > 0, ∃ δ = 146
|f (x) − f (y)| = x3 − x + 1 − (y 3 − y + 1)
= |x − y||x2 + xy + y 2 − 1|
< δ(3 × 72 − 1) = .
Thus, f is uniformly continuous.
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