Assignment #7
Morgan Schreffler
1 of 4
Page 83
Exercise 12 Suppose (S, d) is a nonempty, complete metric space, and suppose f : S → S
is a strict contraction, i.e., there is an r, 0 < r < 1, such that d(f (p), f (q)) ≤ r · d(p, q) for
all p, q ∈ S. Prove that f has a unique fixed point.
Proof. First, let > 0 be given, and let δ = r . Then d(a, b) < δ implies
d(f (a), f (b)) ≤ r · d(a, b) < r · δ = r = ,
r
so f is uniformly continuous. Next, define f n as follows: Let f 1 = f , and for n ≥ 2, let
f n = f ◦ f n−1 . Now, let p0 ∈ S be arbitrary, pj = f j (p0 ), and consider the sequence {pj }∞
j=1 .
If i ≥ 1, then
d(pi , pi+1 ) = d(f (pi−1 ), f (pi )) ≤ r · d(pi−1 , pi ).
Similarly, if i ≥ 2, d(pi , pi+1 ) ≤ r2 · d(pi−2 , pi−1 ). By induction, for i ≥ 1, d(pi , pi−1 ) ≤
ri · d(p0 , p1 ). Let ζ > 0 be given. Since 0 < r < 1, it follows that there exists an N ∈ N such
that
rN
d(p0 , p1 )
1−r
< ζ. Now, let m ≥ n ≥ N . It follows that
d(pn , pm ) ≤ d(pn , pn+1 ) + d(pn+1 , pn+2 ) + . . . + d(pm−1 , pm )
≤ rn · d(p0 , p1 ) + rn+1 · d(p0 , p1 ) + . . . + rm−1 · d(p0 , p1 )
= (rn + rn+1 + . . . + rm−1 )d(p0 , p1 )
∞
X
<(
rk )d(p0 , p1 )
k=n
n
r
d(p0 , p1 )
1−r
rN
≤
d(p0 , p1 )
1−r
= ζ,
=
making pj a Cauchy sequence in the complete metric space S. As a result, pj → p for some
p ∈ S. Since f is uniformly continuous, it is continuous, so f (pj ) → f (p). But, for each
∞
j+1
j, f (pj ) = f (f j (p0 )) = f j+1 (p0 ), so f (pj )∞
(p0 )j=1 = pj ∞
j=1 = f
j=2 , which converges to p.
Hence, f (pj ) → f (p) and f (pj ) → p, so f (p) = p. Now, suppose p and q are fixed points of f .
Then f (p) = p and f (q) = q, so d(f (p), f (q)) = d(p, q). But, d(f (p), f (q)) ≤ r · d(p, q), which
is less than d(p, q) when p 6= q. As a result, d(p, q) = r · d(p, q), which implies d(p, q) = 0, or
p = q.
1
Assignment #7
Morgan Schreffler
2 of 4
Page 89
Exercise 1 Prove that if f : (S, dS ) → (T, dT ) is continuous at a point p ∈ S and pn → p,
then f (pn ) → f (p).
Proof. Let > 0 be given. Since f is continuous at p, there exists a δ > 0 such that for
some q ∈ S, if dS (p, q) < δ, then dT (f (p), f (q)) < . Further, since pn → p, there is an
N ∈ N such that
n ≥ N ⇒ dS (pn , p) < δ ⇒ dT (f (pn ), f (p)) < ,
so f (pn ) → f (p).
Exercise 3 Let f : (S, dS ) → (T, dT ) be continuous at a point p ∈ S, and let g : (T, dT ) →
(U, dU ) be continuous at f (p). Prove that g ◦ f is continuous at p.
Proof. Let > 0 be given. Since g is continuous at f (p), there is a δ > 0 such that for
q ∈ T , if dT (f (p), q) < δ, then dU ((g ◦ f )(p), g(q)) < . Also, since f is continuous at p, there
is a γ > 0 such that for r ∈ S, if dS (p, r) < γ, then dT (f (p), f (r)) < δ. Thus, for p0 ∈ S,
dS (p, p0 ) < γ ⇒ dT (f (p), f (p0 )) < δ ⇒ dU ((g ◦ f )(p), (g ◦ f )(p0 )) < ,
so g ◦ f is continuous at p.
Exercise 4 Suppose a ∈ S is fixed in the metric space (S, d). Define g : S → R by g(p) =
d(a, p). Prove that g is uniformly continuous.
Proof. By the Triangle Inequality, we obtain the following:
d(a, p) ≤ d(a, q) + d(p, q) =⇒ d(a, p) − d(a, q) ≤ d(p, q),
and
d(a, q) ≤ d(a, p) + d(p, q) =⇒ d(a, q) − d(a, p) ≤ d(p, q).
Together, these inequalities imply |d(a, p) − d(a, q)| = |g(p) − g(q)| ≤ d(p, q). Now, let > 0
be given, and let δ = . Then for any p, q ∈ S, d(p, q) < δ implies
|g(p) − g(q)| ≤ d(p, q) < δ = ,
so g is uniformly continuous.
2
Assignment #7
Morgan Schreffler
3 of 4
Exercise 6 Suppose (S, d) is a complete metric space and B ⊆ S is not closed. Show that
there exists a bounded, uniformly continuous, real-valued function on B which does not attain
a minimum.
Before we begin the proof, we first require two lemmas.
Lemma 1 Let R+ = (0, ∞). The function Arctan : R+ → R+ is uniformly continuous.
Proof. Let 0 < α < β. Since Arctan is strictly increasing and bounded above by
0 < Arctan(α) < Arctan(β) <
π
,
2
so 0 < Arctan(β) − Arctan(α) <
π
.
2
π
,
2
Knowing this, we
can set x = Arctan(β) − Arctan(α), and we know tan(x) = tan(Arctan(β) − Arctan(α)) is
well-defined. Using trigonometric identities, we have the following:
β−α
tan(Arctan(β)) − tan(Arctan(α))
=
.
1 + tan(Arctan(β)) tan(Arctan(α))
1 + βα
β−α
Since β − α > 0, we can conclude that x = Arctan 1+βα
. Furthermore, since 1 + βα > 1,
tan(x) =
we obtain the following useful inequality:
Arctan(β) − Arctan(α) = x = Arctan
β−α
1 + βα
< Arctan(β − α).
Finally, to prove uniform continuity, let > 0 be given. If ≥
π
,
2
will produce the desired result, since |Arctan(β) − Arctan(α)| <
π
2
then any choice of δ
≤ for any choice of
α, β ∈ R+ . If < π2 , then let δ = tan(). Without loss of generality, if we assume α ≤ β,
then β − α < δ implies
Arctan(β) − Arctan(α) < Arctan(β − α)
< Arctan(δ)
= Arctan(tan())
= ,
proving that Arctan is uniformly continuous.
Lemma 2 If f : S → T and g : T → U are uniformly continuous, then so is g ◦ f .
Proof. The proof of this lemma is almost identical to the proof of Exercise 3 on Page 89.
3
Assignment #7
Morgan Schreffler
4 of 4
We are now prepared to prove the proposition.
Proof. Let B ⊆ S not be closed. Then there is an a ∈ S which is a limit point of B but
which is not itself in B. Define g : S → R+ ∪ {0} as in Exercise 4, using a as the fixed point
in S. Notice that g(p) = 0 iff p = a. Also, note that Arctan(x) = 0 iff x = 0. Now, consider
the function h = Arctan ◦ g. For each δ > 0, since a is a limit point of B and a ∈
/ B, there is
a point b ∈ B such that b ∈ Nδ (a) or, equivalently, d(a, b) < δ. As we showed in Exercise 4,
this tells us that |g(a) − g(b)| = g(b) < δ. Further, h(b) = Arctan(g(b)) < Arctan(δ). Since
δ was arbitrary, this tells us that 0 is the minimum of h on B, but g(b) > 0 for all b ∈ B,
so h(b) = Arctan(g(b)) > 0 as well, so h does not attain its minimum. Finally, h is bounded
above by
π
2
and below by 0, and h is uniformly continuous by our two previous lemmas, so h
is a bounded, uniformly continuous, real-valued function which does not attain its minimum.
4