CONVEX ANALYSIS – CONVEX SETS
To prove that open half-spaces and closed half-spaces are convex sets, we can use
the definition of convexity. A set is convex if, for any two points in the set, the line
segment connecting those two points is also contained within the set. In the case of
open and closed half-spaces, we can show this property holds.
Let's start with open half-spaces:
1. Open Half-Space (H+): Consider an open half-space defined as H+ = {x | aᵀx <
b}, where "a" is a nonzero vector, and "b" is a real number. We want to show that
H+ is convex.
2. Take any two points in H+, x₁ and x₂, such that aᵀx₁ < b and aᵀx₂ < b.
3. Now, consider a point x that lies on the line segment between x₁ and x₂. This
point can be expressed as x = λx₁ + (1 - λ)x₂, where λ is a real number between 0
and 1.
4. We need to show that aᵀx < b for this point x:
aᵀx = aᵀ(λx₁ + (1 - λ)x₂)
= λaᵀx₁ + (1 - λ)aᵀx₂
Since both aᵀx₁ < b and aᵀx₂ < b, it follows that:
λaᵀx₁ + (1 - λ)aᵀx₂ < λb + (1 - λ)b
=b
5. Therefore, aᵀx < b, which means x is in the open half-space H+.
Since we have shown that for any two points x₁ and x₂ in H+, the line segment
connecting them is also contained within H+, we have demonstrated that the open
half-space H+ is convex.
Now, let's prove that closed half-spaces are convex:
1. Closed Half-Space (H-): Consider a closed half-space defined as H- = {x | aᵀx ≤
b}, where "a" is a nonzero vector, and "b" is a real number. We want to show that
H- is convex.
2. Take any two points in H-, x₁ and x₂, such that aᵀx₁ ≤ b and aᵀx₂ ≤ b.
3. Similar to the open half-space proof, consider a point x that lies on the line
segment between x₁ and x₂, expressed as x = λx₁ + (1 - λ)x₂, where λ is a real
number between 0 and 1.
4. We need to show that aᵀx ≤ b for this point x:
aᵀx = aᵀ(λx₁ + (1 - λ)x₂)
= λaᵀx₁ + (1 - λ)aᵀx₂
Since both aᵀx₁ ≤ b and aᵀx₂ ≤ b, it follows that:
λaᵀx₁ + (1 - λ)aᵀx₂ ≤ λb + (1 - λ)b
=b
5. Therefore, aᵀx ≤ b, which means x is in the closed half-space H-.
Since we have shown that for any two points x₁ and x₂ in H-, the line segment
connecting them is also contained within H-, we have demonstrated that the closed
half-space H- is convex.
This completes the proofs for both open and closed half-spaces, showing that they
are convex sets.
OTHER QUESTION
The expression you've provided, which is the norm defined as:
‖x‖ = |a_11 * x_1 + a_12 * x_2| + |a_21 * x_1 + a_22 * x_2|
We can start by verifying if this expression satisfies the properties of a valid norm,
namely:
1. Non-negativity: ‖x‖ ≥ 0 for all x ∈ ℝ², and ‖x‖ = 0 if and only if x = 0.
2. Scalar multiplication: ‖αx‖ = |α|‖x‖ for all α ∈ ℝ and x ∈ ℝ².
3. Triangle inequality: ‖x + y‖ ≤ ‖x‖ + ‖y‷ for all x, y ∈ ℝ².
Let's check each property:
1. Non-negativity:
- ‖x‖ = |a_11 * x_1 + a_12 * x_2| + |a_21 * x_1 + a_22 * x_2|
- Each absolute value term is non-negative, so ‖x‖ is non-negative for all x ∈ ℝ².
- ‖x‖ = 0 if and only if both |a_11 * x_1 + a_12 * x_2| = 0 and |a_21 * x_1 + a_22
* x_2| = 0. This is true if and only if x = 0, as it would require all coefficients a_ij
to be 0 for x to be 0.
2. Scalar multiplication:
- ‖αx‖ = |a_11 * (αx_1) + a_12 * (αx_2)| + |a_21 * (αx_1) + a_22 * (αx_2)| =
|α|‖x‖.
- So, the scalar multiplication property holds.
3. Triangle inequality:
- ‖x + y‖ = |a_11 * (x_1 + y_1) + a_12 * (x_2 + y_2)| + |a_21 * (x_1 + y_1) +
a_22 * (x_2 + y_2)|
- Using the triangle inequality for absolute values:
- |a_11 * (x_1 + y_1) + a_12 * (x_2 + y_2)| + |a_21 * (x_1 + y_1) + a_22 * (x_2
+ y_2)| ≤ |a_11 * x_1 + a_12 * x_2| + |a_21 * x_1 + a_22 * x_2| + |a_11 * y_1 +
a_12 * y_2| + |a_21 * y_1 + a_22 * y_2| = ‖x‖ + ‖y‖
- Therefore, the triangle inequality holds.
Since the given expression ‖x‖ satisfies all three properties of a valid norm, it is
indeed a valid norm for ℝ².
The set of coefficients a_ij for which this is a valid norm is the set of all possible
combinations of values that satisfy the non-negativity and triangle inequality
properties. There isn't a unique set of coefficients, as there are many possible
combinations that would make this a valid norm. These coefficients must satisfy
the conditions such that ‖x‖ = 0 if and only if x = 0, and they can vary depending on
the specific choices of values for a_ij.
THE HOMEWORK PART
Let's analyze each of the five expressions to determine if they represent convex
sets:
1. "X={z=(x,y) | x^2 <= y}":
This set is convex because it represents the region below or on the curve y = x^2.
For any two points (x1, y1) and (x2, y2) in this set, the line segment connecting
them lies entirely within the set.
To prove convexity, consider two points (x1, y1) and (x2, y2) in the set. We need to show that
the line segment connecting these two points is entirely within the set. The line segment is
defined by the parameter t in [0, 1]:
(x, y) = (x1, y1) * t + (x2, y2) * (1 - t)
Now, we have: x^2 = (x1t + x2(1 - t))^2 y = y1t + y2(1 - t)
Since x^2 = (x1t + x2(1 - t))^2, and x1^2 <= y1 and x2^2 <= y2 (as both points are in the set),
we have: x1^2t^2 + x2^2(1 - t)^2 <= y1t + y2(1 - t)
Since x1^2t^2 + x2^2(1 - t)^2 is less than or equal to both y1t and y2(1 - t) for any t in [0, 1], this
means the line segment is entirely within the set. Therefore, the set is convex.
2. "X={z=(x,y) | x*y >= 1, x>0}":
This set is not convex. If you take two points in this set, (x1, y1) and (x2, y2),
and draw the line segment connecting them, it's possible for this line segment to
exit the set since the product x*y does not necessarily ensure convexity.
To prove non-convexity, consider the points (1, 1) and (2, 1) in the set. These points satisfy xy >= 1 and x >
0. Now, draw the line segment connecting these two points. It goes outside the set, as (1.5, 1) is not in
the set. Thus, the set is not convex.
3. "X={z=(x,y) | y >= x + 1/x, x>0}":
This set is convex because it represents the region above or on the curve y = x +
1/x. For any two points (x1, y1) and (x2, y2) in this set, the line segment
connecting them lies entirely within the set.
To prove convexity, consider two points (x1, y1) and (x2, y2) in the set. The line segment between them,
(x, y) = (x1, y1) * t + (x2, y2) * (1 - t), will also satisfy y >= x + 1/x because the function y = x + 1/x is
convex for x > 0. Therefore, the set is convex.
4. "X={z=(x,y) | y >= 2x - 1 + 1/(x+1), x>-1}":
This set is convex. It represents the region above or on the curve y = 2x - 1 +
1/(x+1). For any two points (x1, y1) and (x2, y2) in this set, the line segment
connecting them lies entirely within the set.
This set is also convex because the function y = 2x - 1 + 1/(x+1) is convex for x > -1, and the set
represents the region above this curve.
5. "X={z=(x,y) | x-y<=2, x^2+y^2<=4}":
This set is not convex. While the condition x - y ≤ 2 defines a half-plane (a
convex set), the condition x^2 + y^2 ≤ 4 represents a disk in the xy-plane (also
convex). However, the intersection of these two sets is not convex because the
boundary of the disk and the boundary of the half-plane do not form a convex
region when combined. Therefore, the intersection of these two conditions is not a
convex set.
To prove non-convexity, consider the points (0, -2) and (2, 0) in the set. Both points satisfy x - y <= 2 and
x^2 + y^2 <= 4, but if you draw the line segment connecting them, it goes outside the set. Therefore, the
set is not convex.