Math 501
Introduction to Real Analysis
Instructor: Alex Roitershtein
Iowa State University
Department of Mathematics
Summer 2015
Homework #2
Solutions
EXERCISES FROM CHAPTER 1
39.
(a) It is straightforward that if f is uniformly continuous then it is continuous. Furthermore, we will learn later in the course that any continuous function on a closed and
bounded interval [a, b] is uniformly continuous. However, a continuous function defined on an open (a, b) interval might not be uniformly continuous. For instance, as
the textbook suggests, f (x) = sin(1/x) is not uniformly continuous on (0, 1). Indeed,
by the mean value theorem, for any y ∈ (0, 1) and x ∈ (0, y) there exists ξx,y ∈ (x, y)
such that
(y − x)
(y − x)
≥ cos y ·
.
|f (x) − f (y)| = f 0 (ξx,y )(x − y) = cos ξx,y ·
2
ξx,y
y2
Thus, if f is uniformly continuous than for any ε > 0 and x, y as above there exists
δε > 0 such that |x − y| < δε implies
ε > |f (x) − f (y)| ≥ cos y ·
(y − x)
.
y2
In particular, if y < δε and x = y/2,
ε > |f (x) − f (y)| ≥ cos y ·
y
cos y
=
.
2
2y
2y
The above inequality however cannot hold for any y ∈ (0, δε ) because
cos y
lim
= +∞.
y→0 2y
The contradiction show that f (x) = sin(1/x) is not uniformly continuous on (0, 1).
Remark. We give here a short alternative solution which is more tailored to the specific
form of the function f than the above given standard argument. Since sin x is a periodic
function with period 2π, function f (x) = sin 1/x will admit all values within the interval
[−1, 1] when x varies within any interval x01+π , x10 , x0 > 0. However
1
1
2π
−
=
→x0 →∞ 0.
x0 x0 + 2π
x0 (x0 + 2π)
Thus, given
any δ > 0, one can find x0 > 0 such that the length of the interval
1
1
,
< δ. Since the range of f within this small interval is [−1, 1], f cannot be
x0 +π x0
uniformly continuous.
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(b) Yes. In fact, one can put δ = ε/2.
Remark. Functions (for instance, linear functions f (x) = cx) that for some constant
c > 0 satisfy the condition
|f (x) − f (y)| ≤ c|x − y|,
∀ x, y,
(1)
are called Lipschitz-continuous. Lipschitz-continuity is, in general, a stronger condition
than the uniform continuity. In fact, if (1) holds, one can choose δ = ε/c to verify the
uniform continuity of f.
(c) No. Indeed, if f (x) = x2 then for any δ > 0,
f (x + δ) − f (x) = δ 2 + 2xδ.
Thus limx→∞ [f (x + δ) − f (x)] = +∞ for any δ > 0. However, if f (x) would be uniform
continuous on R, the difference gδ (x) := f (x + δ) − f (x) would be a bounded function
of x for all small enough values of the parameter δ > 0.
EXERCISES FROM CHAPTER 2
18.
(a) Let p ∈ N and a constant ε > 0 be given. Put δ = ε. Then, since dN (f p, f q) = dM (p, q),
dM (p, q) < δ
implies
dN (f p, f q) < ε,
showing that f is continuous at p.
(b) Since f is a bijection, the inverse f −1 is well-defined and is also a bijection. Consider now arbitrary x, y ∈ N and denote p = f −1 x, q = f −1 y. The isometry identity
dN (f p, f q) = dM (p, q) implies for x = f p and y = f q,
dN (x, y) = dM (f −1 x, f −1 y).
Thus f −1 is also an isometry. By the result in (a) both f and f −1 are continuous, and
hence f is a homeomorphism by definition.
(c) We have seen that if f : [0, 1] → [0, 2] is an isometry then f −1 is also an isometry. This
implies, in particular,
|f −1 (0) − f −1 (2)| = 2,
which is impossible since both f −1 (0) and f −1 (2) are points within the interval [0, 1].
84.
2
(a) Function ρ is bounded by 0 from below and by 1 from above. Clearly, ρ(x, y) = ρ(y, x).
Furthermore, ρ(x, y) = 0 if and only if d(x, y) = 0 if and only if x = y. It remains to
verify the triangle inequality. We have:
d(x, y)
d(y, z)
+
1 + d(x, y) 1 + d(y, z)
d(x, y) + d(y, z) + 2d(x, y)d(y, z)
=
1 + d(x, y) + d(y, z) + d(x, y)d(y, z)
d(x, y) + d(y, z) + d(x, y)d(y, z)
≥
1 + d(x, y) + d(y, z) + d(x, y)d(y, z)
d(x, y) + d(y, z)
≥
,
1 + d(x, y) + d(y, z)
ρ(x, y) + ρ(y, z) =
where in the last step we used the fact that function f (t) =
increasing for t ≥ 0. Using the triangle inequality
t
1+t
1
= 1 − 1+t
is monotone
d(x, y) + d(y, z) ≥ d(x, z)
and again the monotonicity of f (t) =
ρ(x, y) + ρ(y, z) ≥
t
,
1+t
we obtain
d(x, y) + d(y, z)
d(x, z)
≥
= ρ(x, z),
1 + d(x, y) + d(y, z)
1 + d(x, z)
as required.
(b) The identity map is clearly a bijection. Assume that limn→∞ xn = x in the metric
space (M, d). Then limn→∞ d(xn , x) = 0, and hence limn→∞ ρ(xn , x) = 0. This shows
that the identity map is a continuous function from (M, d) to (M, ρ). Conversely, if
limn→∞ ρ(xn , x) = 0, then
0
ρ(xn , x)
=
= 0.
n→∞ 1 − ρ(xn , x)
1−0
lim d(xn , x) = lim
n→∞
This shows that the identity map is a continuous function from (M, ρ) to (M, d).
(c) The result in (b) implies that boundedness of M is not necessarily preserved under
homeomorphisms.
(d) The pair (R, ρ) and (R, d) can serve as an example when d is the usual Euclidean
metric.
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