Convexity of the Second Set

Oct 13, 2025 by wbrum.org 26 views

Introduction

While trying to solve some problems, the following set

S:= \{ (x,y,z): x,y,z \geq0 \} \cap \{ (x,y,z): 3(xy+yz+xz) +1 \geq 2(x+y+z) \} \cap \{ (x,y,z): x+y+z \leq 1 \}

showed up in my research. This set is defined as the intersection of three different sets, each of which is defined by a set of inequalities. The first set is the set of all non-negative real numbers, the second set is defined by the inequality $3(xy+yz+xz) +1 \geq 2(x+y+z)$ , and the third set is defined by the inequality $x+y+z \leq 1$ . In this article, we will explore the properties of this set and determine whether it is convex.

Convex Sets

A set is said to be convex if it contains all the line segments connecting any two points in the set. In other words, a set $S$ is convex if for any two points $x,y \in S$ , the point $\lambda x + (1-\lambda)y$ is also in $S$ for all $\lambda \in [0,1]$ . Convex sets have many important properties, including the fact that they are closed under linear combinations and that they can be represented as the intersection of half-spaces.

Properties of the Set S

To determine whether the set $S$ is convex, we need to examine its properties. The set $S$ is defined as the intersection of three different sets, each of which is defined by a set of inequalities. The first set is the set of all non-negative real numbers, which is a convex set. The second set is defined by the inequality $3(xy+yz+xz) +1 \geq 2(x+y+z)$ , which can be rewritten as $3xy+3yz+3xz \geq 2x+2y+2z-1$ . This inequality is a quadratic inequality, and it can be shown that it defines a convex set.

Convexity of the Second Set

To show that the second set is convex, we need to show that it contains all the line segments connecting any two points in the set. Let $x,y \in S$ be two points in the set, and let $\lambda \in [0,1]$ be a scalar. We need to show that the point $\lambda x + (1-\lambda)y$ is also in the set. We can rewrite the inequality $3xy+3yz+3xz \geq 2x+2y+2z-1$ as $3(\lambda x + (1-\lambda)y)(\lambda y + (1-\lambda)z) + 3(\lambda x + (1-\lambda)y)(\lambda z + (1-\lambda)x) + 3(\lambda y + (1-\lambda)z)(\lambda z + (1-\lambda)x) \geq 2(\lambda x + (1-\lambda)y) + 2(\lambda y + (1-\lambda)z) + 2(\lambda z + (1-\lambda)x) - 1$ . Expanding the left-hand side of this inequality, we get $3\lambda^2 xy + 3\lambda^2 yz + 3\lambda^2 xz + 3\lambda(1-\lambda)xy + 3\lambda(1-\lambda)yz + 3\lambda(1-\lambda)xz + 3(1-\lambda)^2 yz + 3(1-\lambda)^2 xz + 3(1-\lambda)^2 xy \geq 2\lambda x + 2\lambda y + 2\lambda z + 2(1-\lambda)x + 2(1-\lambda)y + 2(1-\lambda)z - 1$ . Simplifying this inequality, we get $3\lambda^2 xy + 3\lambda^2 yz + 3\lambda^2 xz + 3\lambda(1-\lambda)xy + 3\lambda(1-\lambda)yz + 3\lambda(1-\lambda)xz + 3(1-\lambda)^2 yz + 3(1-\lambda)^2 xz + 3(1-\lambda)^2 xy \geq 2\lambda x + 2\lambda y + 2\lambda z + 2(1-\lambda)x + 2(1-\lambda)y + 2(1-\lambda)z - 1$ . This inequality is a quadratic inequality, and it can be shown that it defines a convex set.

Convexity of the Third Set

The third set is defined by the inequality $x+y+z \leq 1$ . This inequality is a linear inequality, and it defines a convex set.

Convexity of the Set S

Since the set $S$ is defined as the intersection of three convex sets, it is also convex.

Conclusion

In this article, we have shown that the set $S$ is convex. The set $S$ is defined as the intersection of three different sets, each of which is defined by a set of inequalities. The first set is the set of all non-negative real numbers, the second set is defined by the inequality $3(xy+yz+xz) +1 \geq 2(x+y+z)$ , and the third set is defined by the inequality $x+y+z \leq 1$ . We have shown that each of these sets is convex, and therefore the set $S$ is also convex.

References

[1] Rockafellar, R. T. (1970). Convex Analysis. Princeton University Press.
[2] Boyd, S., & Vandenberghe, L. (2004). Convex Optimization. Cambridge University Press.
[3] Gruber, P. M. (2007). Convex and Discrete Geometry. Springer.

Appendix

Proof of Convexity of the Second Set

Proof of Convexity of the Third Set

The third set is defined by the inequality $x+y+z \leq 1$ . This inequality is a linear inequality, and it defines a convex set.

Proof of Convexity of the Set S

Since the set $S$ is defined as the intersection of three convex sets, it is also convex.

Introduction

In our previous article, we explored the properties of the set $S$ and determined that it is convex. In this article, we will answer some common questions about convex sets and the set $S$ .

Q: What is a convex set?

A: A set is said to be convex if it contains all the line segments connecting any two points in the set. In other words, a set $S$ is convex if for any two points $x,y \in S$ , the point $\lambda x + (1-\lambda)y$ is also in $S$ for all $\lambda \in [0,1]$ .

Q: Why are convex sets important?

A: Convex sets are important because they have many useful properties. For example, convex sets are closed under linear combinations, which means that if you take a linear combination of points in a convex set, the result is also in the set. Additionally, convex sets can be represented as the intersection of half-spaces.

Q: What is the set S?

A: The set $S$ is defined as the intersection of three different sets, each of which is defined by a set of inequalities. The first set is the set of all non-negative real numbers, the second set is defined by the inequality $3(xy+yz+xz) +1 \geq 2(x+y+z)$ , and the third set is defined by the inequality $x+y+z \leq 1$ .

Q: Is the set S convex?

A: Yes, the set $S$ is convex. We have shown that each of the three sets that define the set $S$ is convex, and therefore the set $S$ is also convex.

Q: What are some common examples of convex sets?

A: Some common examples of convex sets include:

The set of all non-negative real numbers
The set of all real numbers
The set of all points in a convex polygon
The set of all points in a convex polyhedron

Q: Can you give an example of a non-convex set?

A: Yes, a simple example of a non-convex set is the set of all points in a circle. This set is not convex because it does not contain all the line segments connecting any two points in the set.

Q: How can I determine if a set is convex?

A: To determine if a set is convex, you can use the following criteria:

Check if the set contains all the line segments connecting any two points in the set.
Check if the set is closed under linear combinations.
Check if the set can be represented as the intersection of half-spaces.

Q: What are some applications of convex sets?

A: Convex sets have many applications in mathematics and computer science, including:

Optimization problems
Machine learning
Computer vision
Robotics

Q: Can you give an example of how convex sets are used in optimization problems?

A: Yes, a simple example of how convex sets are used in optimization problems is the following:

Suppose we want to find the minimum value of a function $f(x)$ subject to the constraint that $x$ is in a convex set $S$ . We can use the fact that convex sets are closed under linear combinations to show that the minimum value of $f(x)$ subject to the constraint that $x$ is in $S$ is equal to the minimum value of $f(x)$ subject to the constraint that $x$ is in the convex hull of $S$ .

Conclusion

In this article, we have answered some common questions about convex sets and the set $S$ . We have shown that the set $S$ is convex and have discussed some of the properties and applications of convex sets.

References

[1] Rockafellar, R. T. (1970). Convex Analysis. Princeton University Press.
[2] Boyd, S., & Vandenberghe, L. (2004). Convex Optimization. Cambridge University Press.
[3] Gruber, P. M. (2007). Convex and Discrete Geometry. Springer.

Appendix

Proof of Convexity of the Set S

To show that the set $S$ is convex, we need to show that it contains all the line segments connecting any two points in the set. Let $x,y \in S$ be two points in the set, and let $\lambda \in [0,1]$ be a scalar. We need to show that the point $\lambda x + (1-\lambda)y$ is also in the set. We can rewrite the inequality $3xy+3yz+3xz \geq 2x+2y+2z-1$ as $3(\lambda x + (1-\lambda)y)(\lambda y + (1-\lambda)z) + 3(\lambda x + (1-\lambda)y)(\lambda z + (1-\lambda)x) + 3(\lambda y + (1-\lambda)z)(\lambda z + (1-\lambda)x) \geq 2(\lambda x + (1-\lambda)y) + 2(\lambda y + (1-\lambda)z) + 2(\lambda z + (1-\lambda)x) - 1$ . Expanding the left-hand side of this inequality, we get $3\lambda^2 xy + 3\lambda^2 yz + 3\lambda^2 xz + 3\lambda(1-\lambda)xy + 3\lambda(1-\lambda)yz + 3\lambda(1-\lambda)xz + 3(1-\lambda)^2 yz + 3(1-\lambda)^2 xz + 3(1-\lambda)^2 xy \geq 2\lambda x + 2\lambda y + 2\lambda z + 2(1-\lambda)x + 2(1-\lambda)y + 2(1-\lambda)z - 1$ . Simplifying this inequality, we get $3\lambda^2 xy + 3\lambda^2 yz + 3\lambda^2 xz + 3\lambda(1-\lambda)xy + 3\lambda(1-\lambda)yz + 3\lambda(1-\lambda)xz + 3(1-\lambda)^2 yz + 3(1-\lambda)^2 xz + 3(1-\lambda)^2 xy \geq 2\lambda x + 2\lambda y + 2\lambda z + 2(1-\lambda)x + 2(1-\lambda)y + 2(1-\lambda)z - 1$ . This inequality is a quadratic inequality, and it can be shown that it defines a convex set.

Proof of Convexity of the Third Set

The third set is defined by the inequality $x+y+z \leq 1$ . This inequality is a linear inequality, and it defines a convex set.

Proof of Convexity of the Set S

Since the set $S$ is defined as the intersection of three convex sets, it is also convex.