Homework 3
Math 501
Due September 19, 2014
Exercise 1
Let I be an interval. A function f : I → R is uniformly continuous if for
each > 0 there exists δ > 0 (which depends only on ) such that |f (x)−f (t)| < for all t ∈ I such that |t − x| < δ. [Recall that the definition of continuity allows
for δ to depend on x, but here we require that a single δ work for all x.]
The first part of the problem below asks you to show that all uniformly
continuous functions are continuous. In class we will show a partial converse to
this fact. We will show that all continuous functions defined on closed intervals
are uniformly continuous. The rest of the parts show that when we consider
domains which are not closed intervals, things become less clear.
(a) Prove that if f is uniformly continuous then it is continuous.
(b) Consider the function f : R → R given by f (x) = 2x. Show that f is
uniformly continuous.
(c) Consider the function g : R → R given by g(x) = x2 . In a previous assignment (Exercise 1 of Homework 2), we showed that g is continuous. Show
here that g is not uniformly continuous.
(d) Consider the function h : (0, 1) → R given by h(x) = sin(1/x). Show that
h is not uniformly continuous.
Exercise 2
Let (an ), (bn ) ⊂ R be sequences. Prove that
lim sup an + bn ≤ lim sup an + lim sup bn
n→∞
n→∞
n→∞
as long as the right hand side makes sense, ie. it’s not −∞ + ∞ nor ∞ − ∞.
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[Hint: Recall that lim supn→∞ an = limn→∞ xn where xn = sup{ak : k ≥
n}. First prove that the inequality
sup{ak + bk : k ≥ n} ≤ sup{ak : k ≥ n} + sup{bk : k ≥ n}
holds for all n ≥ 1. For this, you might find the following elementary fact useful:
if S is a set then
{(n, m) ∈ S × S : n = m} ⊂ S × S.]
Exercise 3
Recall that an inner product on a vector space V is a map h·, ·i : V × V → R
that satisfies the following three properties:
1. hu, vi = hv, ui for all u, v ∈ V ,
2. hu + tv, wi = hu, wi + thv, wi for all u, v, w ∈ V and t ∈ R,
3. hu, ui ≥ 0 for all u ∈ V and equality holds if and only if u = 0.
A norm on a vector space V is a map k·k : V → R that satisfies the following
three properties:
1. ktvk = |t| kvk for all v ∈ V and t ∈ R
2. ku + vk ≤ kuk + kvk for all u, v ∈ V
3. If kvk = 0, then v = 0.
It is not difficult to check that if one is given an
p inner product space (V, h·, ·i),
then if one defines a map k·k : V → R by kvk = hv, vi, then this map is indeed
a norm– it satisfies the three properties above. Thus an inner product always
yields a norm on a vector space. A natural question then is whether every norm
arises from an inner product. The answer, which we’ll show below, is no.
(a) Given p
an inner product space (V, h·, ·i), define a norm k·k : V → R by
kvk = hv, vi. Show that this norm satisfies the parallelogram law:
2
2
2
2
ku + vk + ku − vk = 2(kuk + kvk )
for all u, v ∈ V . [If you’re wondering about the name “parallelogram law”,
it’s drawn from a fact from elementary geometry: the sum of the squares
of the side lengths of a parallelogram must equal the sum of the squares of
the lengths of its two diagonals.]
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(b) The above part shows that if a norm arises from an inner product, then
it satisfies the parallelogram law. Now let’s prove the converse. Given a
normed vector space (V, k·k) whose norm satisfies the parallelogram, prove
that the map h·, ·i : V × V → R given by
1
2
2
hu, vi =
ku − vk − ku + vk
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is an inner product.
(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) Explain why kf k∞ < ∞ for all f ∈ C[0, 1].
(ii) Prove that this is a norm. [We call this the supremum- or sup-norm.]
(d) Show that the sup-norm does not satisfy the parallelogram law. That is,
find f, g ∈ C[0, 1] such that
2
2
2
2
kf + gk∞ + kf − gk∞ 6= 2 kf k∞ + kgk∞ .
Exercise 4
Recall that a function f is called sequentially continuous if whenever (xn )
is a sequence that converges to say x, then f (xn ) converges to f (x). The
following problem shows (by contrapositive) that sequential continuity in R
implies continuity.
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 we have that
|f (p) − f (x)| ≥ for some x such that |x − p| < δ.
Construct a sequence (xn ) which has the following two properties:
1. xn converges to p,
2. f (xn ) does not converge to f (p).
Exercise 5
Let M be a set. A metric on M is a map d : M × M → R such that
1. d(x, y) = d(y, x) for all x, y ∈ M .
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2. d(x, y) ≥ 0 for all x, y ∈ M with equality if and only if x = y.
3. d(x, y) ≤ d(x, z) + d(z, y) for all x, y, z ∈ M .
A metric space (M, d) is a set M equipped with a metric d.
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) Explain why d(f, g) ∈ R for all f, g ∈ M .
(b) Prove that d is a metric.
(c) Was there anything special about C[0, 1] here? Would this procedure work
for any normed vector space?
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