Chapter 1

Question 1

Metric space

Conditions (check)

Answer 1

Consists of a non-empty set C together with a distance function or metric d:XxX -> R satisfying:

(M1) ∀ x,y ∈X d(x,y) ≥0 & d(x,y) =0 if and only if x=y
(M2) ∀ x,y∈X d(x,y) = d(y,x) SYMMETRY
(M3) ∀ x,y,z ∈X d(x,z) ≤ d(x,y) + d(y,z)

Question 2

Cartesian product

Answer 2

XxX of Ordered pairs (x,y) st x ∈X and y ∈X

Question 3

Lemma 1.3: |x-y|

Answer 3

Function d(x,y) = |x-y| is a metric on R and it measures distance

(M1) by def of modulus
(M2) symmetry of modulus
(M3) triangle inequality

Question 4

Triangle inequality for real numbers

Answer 4

|a+b| ≤ |a| + |b| for all a,b in the reals

Question 5

Modulus

Answer 5

|a| =
{a if a ≥0
{-a if a is less than 0

Also |a| ≤ b
-> -a ≤ b ≤ a

Question 6

Definitions of absolute values examples

Answer 6

• |x-y| ≤ x-y ≤ |x-y|

- |y-z| ≤ y-z ≤|y-z|

Question 7

The Euclidean metric d ₂ on R^k

Answer 7

d ₂ (x,y) = d ₂((a₁, a ₂,…,a_k),(b₁,b₂,…,b_k))

Sqrt(
(a₁−b₁)² +….. + (a_k -b_k)²)

If K equals two then it’s the distance between two points

It’s a metric also on the complex plane
Length
Of the hypotenuse every right angle triangle

Question 8

The taxicab metric. d₁ on R^k

Answer 8

d₁(x,y) = d₁(( a₁, …,a_k),( b₁, …,b_k))

=
|a₁ -b₁| +…+ |a_k -b_k|

When K=2 this metric measures the distance between two points if moving only along the lines of a grid

When K equals two this is the sum of lengths of vertical and horizontal side of a. Right angled triangle

Question 9

The maximum metric or supremum metric

Answer 9

d∞(x,y) = d∞((a₁,a₂,..,a_k), (b₁,b₂,..,b_k))

= max { |a₁- b₁| , …., |a_k-b_k|}

What is the maximum other two lengths of side of the right angled triangle

Question 10

General relationship between the three metrics

Answer 10

In general
For k=2

d ∞((a₁,a₂), (b₁,b₂ ))

≤ d₂(( a₁,a ₂),( b₁, b₂))

≤ d₁(( a₁,a₂),( b₁, b₂))

The equality occurs when k=1

Question 11

Proving the metrics:

Some pointers

Answer 11

•supremum metric on R^k:
(M1) maximum is max of non negative numbers so is non neg and bigger than or equal to 0
(M2) symmetry of modulus |a_i -b_i = |b_i - a_i| for each i.
(M3) for each i the triangle inequality applies, so it does for the max

• Euclidean metric d_2 uses the Cauchy-Schwartz inequality on R^k :
(M1) larger than or equal to 0 and d(a,a) = 0 , d(x,y)= 0 implies x=y.
(M2) symmetry (M3) by Cauchy Schwartz

Question 12

Lemma 1.9 the Cauchy Schwartz inequality

Answer 12

For e ₁,..,e_k, f₁,…,f_k ∈R we have

|e ₁•f₁ + ….. + e_k •f_k|

≤ SQRT(e ² ₁+ …+ e ²_k) • SQRT(f ²₁+…+f ²_k)

Proof:

If e_i values all equal 0 the result follows. For the other case: assume they don’t. Let p(x) = sum of ( (e_i x + f_i) ^2 )from 1 up to k.
This is Ax^2 + 2Bx + C
Where A= sum of e_i ^2
B= sum of e_i•k_i
C= sum of f_i ^2

Since p(x) is bigger than or equal to 0 for all x in reals. We must have 4B^2 -4AC is less than 0 (from det)
Hence |B| is less than or equal to SQRT( AC)... giving the inequality.

Question 13

The three metrics on R^k form a family

Answer 13

If p is in (1, infinity) there is a metric d_p on R^n defined by:

d_p (x,y) = d_p (( a_1,…,a_k),(b_1,..,b_k))

= ( sum from i=1 to k of [ |a_i -b_i|^p ] ) ^1/p

When n=1 these metrics are equal to usual metric on R
P=1 d_p taxicab metric d_1, p=2 Euclidean metric, p tends to infinity then tends to supremum metric

Question 14

The discrete metric.

Answer 14

Let X be a non empty set and define the discrete metric d_0 on X by

d_0(x,y) = {0 of x=y
{1 if x not equal to y

• this metric can be put on any set
• eg give number of positions in which x and y differ

Question 15

Elements in R^n can arise as the limit of a sequence or a solution of a simultaneous equation T or F

Answer 15

TRUE

Question 16

Definition: the space of continuous functions

Answer 16

The space of continuous functions on I :

C(I)

For I= [a,b] a closed bounded interval on R

C(I) = {f: I -> R st f is continuous}

Any continuous function defined on I to R where I is a CLOSED BOUNDED interval .

Question 17

Definition:

Measuring distance between 2 function on a closed interval we can compare points in R^n. And using d_1 for infinitely many points we can approximate to integration

Metric d_1 on C[a,b]

Answer 17

FOR THE METRIC d_1 on CONTINUOUS FUNCTIONS

The metric d_1 on C[a,b] is defined by
d_1(f,g) = integral from a to b of [ |f(x) - g(x)].dx

This exists as f and g are continuous and so f(x) - g(x) and |f(x) -g(x)| are too.

Every continuous function on a closed and bounded interval has a Riemann integral also

Question 18

Supremum Metric d_infinity on C[a,b]

Answer 18

d_infinity defined on C[a,b] (continuous functions)

d_infinity (f,g)= sup{ |f(x) - g(x)|
= max{ |f(x) -g(x) | : x is in I}

Which exists by boundedness theorem and that f-g is a continuous function as f and g are etc

Question 19

Boundedness theorem for continuous function

Answer 19

Every continuous function on a closed and bounded interval is bounded an attains a maximum value.

MUST BE CLOSED AND BOUNDED THEREFORE TO USE THE SUPREMUM METRIC.

Question 20

Example: let I = [0,2] if f(x) = x^2 and g(x) = x+2

d_1 (f,g). And d_infinity (f,g)

Answer 20

d_1(f,g) = integral from 0 to 2 of ( -x^2 +x +2).dx

= 3 and( 1/3)
(Sketch functions over the interval to use modulus)

d_infinity (f,g) = sup {|f(x) -g(x)|: x is in I}

At stationary point this is 2 (1/4)
The end points are less than this so our answer is 2 and a quarter which occurs for x=0.5

Question 21

Finding max of |f(x) -g(x)|

Answer 21

Sketch the functions over the given interval

The maximum occurs either at a stationary point ( find stat points by differentiating)

Or at the end points of the intervals

Check each one then conclude

Question 22

Proposition 1.17: is d_infinity a metric on C(I)

proof

Answer 22

Proof:

We must verify the three axioms for (C(I), d∞) to be a metric space.

Let f, g, h ∈ C(I), so that f, g and h are continuous functions from
I to R. For M1, d_∞(f, g) = sup{|f(x) − g(x)| : x ∈ I} ≥ 0.

Also, d_∞(f, g) = 0 ⇔ |f(x) − g(x)| = 0 for all x ⇔ f(x) = g(x) for all
x ⇔ f = g. This proves M1.

M2 is immediate because |f(x) − g(x)| = |g(x) − f(x)| for all x, so
that
d∞(f, g) = sup{|f(x) − g(x)| : x ∈ I} = sup{|g(x) − f(x)| : x ∈ I}
= d∞(g, f).

For M3, note that, by M3 in R,
|f(x) − h(x)| 6 |f(x) − g(x)| + |g(x) − h(x)| for all x ∈ I. Now we have

d∞(f, h) = sup{|f(x) − h(x)| : x ∈ I}
≤
sup{|f(x) − g(x)| + |g(x) − h(x)| : x ∈ I}
≤
sup{|f(x) − g(x)| : x ∈ I} + sup{|g(x) − h(x)| : x ∈ I}
= d∞(f, g) + d∞(g, h),
as required.

Question 23

prop 1.8: d_infinity is a metric on R^k

Answer 23

Three axioms

We need to check the three axioms for (R^k, d∞) to be a metric space. Let
x = (a1, . . . , ak), y = (b1, . . . , bk), z = (c1, . . . , ck) ∈ R^k

For axiom M1, d∞(x, y) = max{|a1 − b1|, . . . , |ak − bk|} ≥ 0. Also
d∞(x, y) = 0 if and only if |ai −bi
| = 0 for all i if and only if x = y.
Thus M1 holds.

Axiom M2 is clear because |ai− bi| = |bi − ai| for each i, so that
d∞(x, y) = max{|a1 − b1|, . . . , |ak − bk|}
= max{|b1 − a1|, . . . , |bk − ak|}
= d∞(y, x).
For axiom M3, note that, by the proof of Lemma 1.3, |ai − ci| ≤ |ai − bi| + |bi − ci| for each i, so

d∞(x, z) = max{|ai − ci|}
≤  max{|ai − bi| + |bi − ci|}
≤  max{|ai − bi|} + max{|bi − ci|}
= d∞(x, y) + d∞(y, z),
which proves M3.

Question 24

Theorem 1.10: the euclidean metric d_2 is a metric on R^k

Answer 24

Proof:

3 axioms

We must check that d2 satisfies the three axioms for (R^k, d_2) to be a metric space. Let x = (a1, . . . , ak), y = (b1, . . . , bk), z =
(c1, . . . , ck) ∈ R^k

For M1, d_2(x, y) = ( Σ(ai − bi)^2)^0.5 ≥ 0. Also
d2(x, y) = 0 ⇔ (ai−bi)^2 = 0 for each i ⇔ ai = bi for each i ⇔ x = y. This proves axiom M1.

Axiom M2 is clear because (ai− bi)^2 = (bi − ai)^2 for all i, so that
d2(x, y) = [Σ(ai − bi)^2]^0.5
= [Σ(bi − ai)^2]^0.5
= d2(y, x).

Finally let’s prove axiom M3:
(d_2(x, y) + d_2(y, z))^2
=d_2(x, y)^2 + d_2(y, z)^2 + 2d_2(x, y)d_2(y, z)
=
[Σ(ai − bi)^2] +[Σ(bi − ci)^2] + 2 [Σ(ai − bi)^2]^0.5 [Σ(bi − ci)^2]^0.5
≥
[Σ(a_i − b_i)^2] +[Σ(b_i − c_i)^2] + 2 [Σ(a_i − b_i)(b_i − c_i)]
(by Cauchy-Schwarz with e_i = a_i − b_i and f_i = b_i − c_i)
=
Σ[(a_i − b_i) + (b_i +c_i)]^2
=
[Σ(a_i − c_i)^2]
=d_2(x, z)^2
,
so that, taking square roots, d_2(x, z) ≤ d_2(x, y)+d_2(y, z) as required.