Chapter 1 Flashcards
Metric space
Conditions (check)
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)
Cartesian product
XxX of Ordered pairs (x,y) st x ∈X and y ∈X
Lemma 1.3: |x-y|
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
Triangle inequality for real numbers
|a+b| ≤ |a| + |b| for all a,b in the reals
Modulus
|a| =
{a if a ≥0
{-a if a is less than 0
Also |a| ≤ b
-> -a ≤ b ≤ a
Definitions of absolute values examples
- |x-y| ≤ x-y ≤ |x-y|
- |y-z| ≤ y-z ≤|y-z|
The Euclidean metric d ₂ on R^k
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
The taxicab metric. d₁ on R^k
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
The maximum metric or supremum metric
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
General relationship between the three metrics
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
Proving the metrics:
Some pointers
•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
Lemma 1.9 the Cauchy Schwartz inequality
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.
The three metrics on R^k form a family
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
The discrete metric.
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
Elements in R^n can arise as the limit of a sequence or a solution of a simultaneous equation T or F
TRUE
Definition: the space of continuous functions
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 .
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]
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
Supremum Metric d_infinity on C[a,b]
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
Boundedness theorem for continuous function
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.
Example: let I = [0,2] if f(x) = x^2 and g(x) = x+2
d_1 (f,g). And d_infinity (f,g)
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
Finding max of |f(x) -g(x)|
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
Proposition 1.17: is d_infinity a metric on C(I)
proof
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.
prop 1.8: d_infinity is a metric on R^k
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.
Theorem 1.10: the euclidean metric d_2 is a metric on R^k
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.
