Ejercicios y normas en espacios métricos y vectoriales, Ejercicios de Probabilidad

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Exercise Session 6

Arina Rakhaeva

896257@stud.unive.it

November 17, 2023

Exercise 1

Let X = R, explain if and why, the following expressions are metrics on X:

(a) d(x, y) = x^2 + xy + y^2

(b)

d(x, y) =

|x − y| 1 + |x − y|

(c) d(x, y) = (1 + |xy|)|x − y|

(d) d(x, y) = max{ 1 , |x + y|}

(e) d(x, y) = max{ 0 , |x − y|}

Solution

Recall the definition of a Metric Space:

A metric space (X; d) is a (non-empty) space X with a metric d on X satisfying the following properties:

1. Positivity: d(x, y) ≥ 0 for all x, y; with d(x, y) = 0 iff x = y
2. Symmetry: d(x, y) = d(y, x)
3. Triangle inequality: d(x, z) ≤ d(x, y) + d(y, z)

We need to show that either proposed metrics satisfy all of the listed properties and therefore are indeed metric spaces, or that at least one of the properties fails.

(a) • Checking Positivity: Consider f (x, y) = x^2 + xy + y^2 , this function is globally convex, and its minimum value is 0. However, x = y ≠⇒ d(x, y) = 0 Taking x = y:

d(x, y) = x^2 + xy + y^2 = x^2 + x^2 + x^2 = 3x^2 ̸= 0 −→ x = y ̸= 0

therefore the first property fails and the distance is not a metric on X.

Exercise 2

Let X = R, explain if and why, the following function is a metric on X:

d(x, y) = { |x 1 − y 1 |, if x 1 ̸= x 2 |x 2 − y 2 |, if x 1 = x 2

Solution

We can show that positivity fails in this case by taking a case with x = (2, 2) and y = (0, 2). So, with x 1 ̸= x 2 we take: d(x, y) = |x 2 − y 2 | = | 2 − 2 | = 0

The distance is zero even though x ̸= y. So, it is not a metric space.

Exercise 3

Check if the following functions are norms on X = R^3 :

(a) ||x|| = |x 3 | + max{|x 1 |, |x 2 |}

(b) ||x|| = max{|x 1 | + |x 3 |, |x 2 |}

Solution

Recall the definition of a Normed Linear Space:

A normed linear space X is a vector space X with a norm ||.|| (measure of the size of the elements) satisfying the following properties:

1. Positivity: ||x|| ≥ 0; with ||x|| = 0 iff x = 0
2. Homogeneity: ||αx|| = |α|||x|| for all α ∈ R
3. Triangle inequality: ||x + y|| ≤ ||x|| + ||y||

(a) • Positivity is ensured by considering that:

|x 1 | ≥ 0 , |x 2 | ≥ 0 , |x 3 | ≥ 0

we can also show that:

x = 0 =⇒ x 1 = 0, x 2 = 0, x 3 = 0 =⇒ ||x|| = 0

and ||x|| = 0 =⇒ x 3 = 0, max{|x 1 |, |x 2 |} = 0 =⇒ x 1 = 0, x 2 = 0

• Homogeneity can be proven as follows:

||αx|| = |αx 3 |+max{|αx 1 |, |αx 2 |} = |α||x 3 |+max{|α||x 1 |, |α||x 2 |}+|α||x 3 |+|α|max{|x 1 |, |x 2 |} =

= |α|(|x 3 | + max{|x 1 |, |x 2 |}) = |α|∥x∥

• Checking the triangle inequality:

∥x + y∥ = ∥(x 1 + y 1 , x 2 + y 2 , x 3 + y 3 )∥ = |x 3 + y 3 | + max{|x 1 + y 1 |, |x 2 + y 2 |} ≤ |x 3 | + |y 3 | + max{|x 1 | + |y 1 |, |x 2 | + |y 2 |}} ≤ |x 3 | + |y 3 | + max{|x 1 | + |x 2 |}, max{|y 1 | + |y 2 |}} = ∥x∥∥y∥

Therefore the function is indeed a norm.

(b) • The first property is satisfied since:

|x 1 | ≥ 0 , |x 2 | ≥ 0 , |x 3 | ≥ 0

and moreover

∥x∥ =⇒ max{|x 1 | + |x 3 |, |x 2 |} = 0 =⇒ |x 1 | = |x 2 | = |x 3 | = 0 =⇒ x = 0

and x = 0 =⇒ x 1 = x 2 = x 3 = 0 =⇒ ∥x∥ = 0

• Homogeneity follows from:

∥αx∥ = max{|αx 1 | + |αx 3 |, |αx 2 |} = max{|α||x 1 | + |α||x 3 |, |α||x 2 |} =

= max{|α|(|x 1 | + |x 3 |), |α||x 2 |} = |α|max{|x 1 | + |x 3 |, |x 2 |} = |α|∥x∥

• Triangle inequality

||x + y|| = max{|x 1 + y 1 | + |x 3 + y 3 |, |x 2 + y 2 |} ≤ max{|x 1 | + |y 1 | + |x 3 | + |y 3 |, |x 2 | + |y 2 |} ≤

≤ max{|x 1 | + |x 3 |, |x 2 |} + max{|y 1 | + |y 3 |, |y 2 |} = ||x|| + ||y|| Therefore the property is satisfied and the function is indeed a norm.

Exercise 4

Determine whether the following function is a norm on C^2 (X) = {f : X → R : f is differentiable twice } where X = [0, 1]: ||f || = supx∈X {|f ”(x)|} + 2|f (0)|

Solution

Let us evaluate the positivity property. Strictly positivity is ensured by definition of absolute value, however the property fails because ||f || = 0 ≠⇒ f (x) = 0, ∀x ∈ X

we can see this by noticing that ||f || = 0 implies only that f ”(x) = 0 and f (0) = 0. For example, the function f (x) = x has f ′(x) = 1 =⇒ f ”(x) = 0 although f (x) ̸= 0