Homework 3 Solutions
Math 501
Due September 19, 2014
Exercise 1
(a) This is obvious.
(b) Let f (x) = 2x. Let > 0, let x ∈ R, and let δ = /2. Observe that for any
t ∈ R such that |x − t| < δ, we have
|f (x) − f (t)| = |2x − 2t| = 2|x − t| < 2δ = .
Since δ is independent of x, it follows that f is uniformly continuous.
(c) Let g(x) = x2 . We must show that for any δ > 0 there exist > 0 and
x, t ∈ R with |x − t| < δ such that |g(x) − g(t)| ≥ .
Let δ > 0 and let = 1. Let x = δ −1 and let t = x + δ/2. Then |x − t| =
δ/2 < δ. However, observe that
|f (x) − f (t)| = |x2 − t2 | = |xδ + δ 2 /4| = 1 + δ 2 /4 ≥ 1 = .
(d) Let h(x) = sin(1/x). Let δ > 0 and let = 1. For each n ≥ 1, let
xn =
π
2
1
,
+ 2πn
tn =
1
.
− π2 + 2πn
It’s easy to see that there exist N1 , N2 ∈ N such that |xn | < δ/2 for n ≥ N1
and |tn | < δ/2 for n ≥ N2 . Let N = max{N1 , N2 } and observe that
|xN − tN | ≤ |xN | + |tN | < δ.
Moreover, observe that
|h(xN ) − h(tN )| = | sin(π/2 + 2πN ) − sin(−π/2 + 2πN )| = 2 > .
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Exercise 2
Let
xn = sup{ak : k ≥ n},
yn = sup{bk : k ≥ n},
zn = sup{ak + bk : k ≥ n},
and let
x = lim xn ,
n→∞
y = lim yn ,
n→∞
z = lim zn .
n→∞
[Note that it was shown in a previous assignment that these limits do in fact
exist and the method by which this was proved showed that the limits were the
infimums shown above.] We claim that the inequality
zn ≤ xn + yn
holds for all n ≥ 1. We shall prove this fact below, but in the meantime note
that this implies z ≤ x + y, the desired inequality.
To see that the claimed inequality holds, let n ≥ 1, let k ≥ n, and observe
that
ak + bk ≤ sup{ak : k ≥ n} + sup{bk : k ≥ n}.
It follows that sup{ak : k ≥ n} + sup{bk : k ≥ n} is an upper bound for the set
{ak + bk : k ≥ n}. Since sup{ak + bk : k ≥ n} is the least upper bound for this
set the desired inequality follows.
Exercise 3
(a) Let (V, h·, ·i) and define a norm k·k : V → R by kvk =
that for u, v ∈ V
2
p
hv, vi. Observe
2
ku + vk + ku − vk = hu + v, u + vi + hu − v, u − vi
= hu, ui + 2hu, vi + hv, vi + hu, ui − 2hu, vi + hv, vi
= 2hu, ui + 2hv, vi
2
2
= 2 kuk + kvk .
(b) Skip.
(c) Let V = C[0, 1] be the space of continuous functions defined on the unit
interval [0, 1]. Define a map k·k∞ : C[0, 1] → R by
kf k∞ = sup |f (x)| = sup{|f (x)| : x ∈ [0, 1]}.
x∈[0,1]
(i) The boundedness theorem tells us that functions in C[0, 1] are bounded,
and hence have finite sup-norm.
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(ii) Before proving that k·k∞ is a norm, we prove the following facts about
supremum.
Lemma 1. Let S ⊂ R be a nonempty set, let c > 0, and let f, g : S →
R be functions. The following hold:
i. supx∈S cx = c sup S.
ii. supx∈S |f (x) + g(x)| ≤ supx∈S |f (x)| + supx∈S |g(x)|
Proof of Lemma. i. Let x ∈ S, let α = sup S, and let β = supx∈S cx.
Then x ≤ α and hence cx ≤ cα. Since x was arbitrary it follows
that cα is an upper bound of the set {cx : x ∈ S} which implies
β ≤ cα.
On the other hand, note that β ≥ cx for all x ∈ S, which implies
c−1 β ≥ x for all x ∈ S. Therefore, c−1 β is an upper bound for S,
which implies c−1 β ≥ α. Hence β ≥ cα. It follows that β = cα.
ii. Let x ∈ S and let α = supx∈S |f (x)| and β = supx∈S |g(x)|.
|f (x) + g(x)| ≤ |f (x)| + |g(x)| ≤ α + β
which implies α + β is an upper bound for {|f (x) + g(x)| : x ∈ S}.
Therefore,
sup |f (x) + g(x)| ≤ α + β.
x∈S
Now, we proceed to show that k·k∞ is a norm. Let f, g ∈ C[0, 1] and
let c ∈ R. By the first part of the lemma, we have that
kcf k∞ = sup |cf (x)| = |c| sup |f (x)| = |c| kf k∞ .
x∈[0,1]
x∈[0,1]
Next observe that
kf + gk∞ ≤ kf k∞ + kgk∞
directly by the second part of the lemma.
Finally, it is clear that |f (x)| ≥ 0 for all x ∈ [0, 1], and hence kf k∞ ≥ 0.
As for equality, if f ≡ 0 then |f (x)| = 0 for all x ∈ [0, 1] and hence
kf k∞ = 0. Conversely, if kf k∞ = 0, then 0 ≤ |f (x)| ≤ kf k∞ = 0 for
all x ∈ [0, 1], which implies f ≡ 0.
(d) We claim that the sup-norm does not satisfy the parallelogram law. Consider f (x) = 1 and g(x) = x − 1. Then observe that (f + g)(x) = x and
(f − g)(x) = 2 − x and
kf k∞ = 1,
Hence
kgk∞ = 1,
kf + gk∞ = 1,
kf − gk∞ = 2
2
2
2
2
kf + gk∞ + kf − gk∞ 6= 2 kf k∞ + kgk∞ .
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Exercise 4
Let p ∈ R and suppose that f : R → R is not continuous at p. That is,
suppose that there exists > 0 such that for every δ > 0 there exists x such
that |x − p| < δ and |f (x) − f (p)| ≥ .
This implies that for every n ≥, there exists xn such that |xn − p| < 1/n
and |f (xn ) − f (p)| ≥ . It’s clear that the sequence xn converges to p. Indeed,
for any α > 0, if we let N = dα−1 e then |xn − p| < α for all n ≥ N . Moreover,
since |f (xn ) − f (p)| ≥ for all n ≥ 1, it cannot be that f (xn ) converges to f (p).
Exercise 5
Consider the space M = C[0, 1] of continuous, real-valued functions defined
on the unit interval [0, 1]. Define a map d on M × M by
d(f, g) = sup |f (x) − g(x)| = sup {|f (x) − g(x)| : x ∈ [0, 1]} .
x∈[0,1]
(a) That d(f, g) ∈ R for all f, g ∈ C[0, 1] is again a consequence of the boundedness (or extreme value) theorem.
(b) Note that d(f, g) = kf − gk∞ , where k·k∞ denotes the sup-norm on C[0, 1].
Since kf − gk∞ = kg − f k∞ is easy to see, it clearly follows that
d(f, g) = d(g, f ).
Moreover, note that because k·k∞ is a norm, it follows that khk∞ ≥ 0 for
all h ∈ C[0, 1] with equality holding if and only if h ≡ 0. This holds in
particular for h = f − g. This implies,
d(f, g) = kf − gk∞ ≥ 0
with equality if and only if f − g ≡ 0; that is, if and only if f = g. Finally
note that
d(f, g) = kf − gk∞
= kf − h + h − gk∞
≤ kf − hk∞ + kh − gk∞
= d(f, h) + d(h, g).
It follows that d is a metric.
(c) The above arguments show that any norm on a vector space naturally induce
a metric structure on the vector space.
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