Optimization proof for Ax > b. Prove that set is convex

In summary, to prove that the feasible region S is convex, we must show that for any elements x and y in S, the point z = ax + (1-a)y also belongs to S for all 0 < a < 1. This can be proven by using the fact that a convex set is one where the line segment between any two points in the set is also contained in the set. By taking x and y in S, we have Ax < b and Ay < b, which means that ax + (1-a)y is also in S. Then, by using the fact that Az = aAx + (1-a)y, we can show that Az < b, and therefore z is also in S. This proves
  • #1
retspool
36
0

Homework Statement


Consider a feasible region S defined by a set of linear constraints

S = {x:Ax<b}

Prove that S is convex

Homework Equations



All what i know is that, a set is convex if and only if the elements x, and y of S

ax + (1-a)y belongs to S
for all 0 <a < 1

The Attempt at a Solution



I don't even know where to begin!
 
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  • #2
so take x,y in S then Ax < b and Ay <b, now take z = ax + (1-a)y

now see if you can show z is in S, using
Az = aAx + (1-a)Ay
 
  • #3
I still haven't gotten it

I take x,y in S so i have Ax < b and Ay < b

so ax + (1-a)y is in S

now Az = aAx + (1-a)y



what to do now?
 
  • #4
you know a & (1-a) are both >0

now use what you know about Ax and Ay
 
  • #5
Ok, i am seriously lost, i don't know where to bring in z from..

I am attaching the only pages where convex sets have been defined and from the material, and with the information given, it seems difficult to prove the above
 
  • #6
if any point z = ax + (1-a)y is within the set for any x & y, then you have proved it is convex, so all you are trying to show is that
Az < b

Then z is clearly part of the set
 
  • #7
retspool said:
I still haven't gotten it

I take x,y in S so i have Ax < b and Ay < b

so ax + (1-a)y is in S

now Az = aAx + (1-a)y
No, Az= aAx+(1- a)Ay.

Ax< b and Ay< b since they are both in S. So aAx+ (1-a)Ay< ab+ (1- a)b= b.



what to do now?
 

Related to Optimization proof for Ax > b. Prove that set is convex

What is optimization proof for Ax > b?

Optimization proof for Ax > b is a mathematical technique used to prove that a set of linear inequalities, represented by the matrix A and vector b, is convex.

Why is it important to prove that a set is convex?

Proving that a set is convex is important because it guarantees that the set has certain properties, such as having a unique global minimum and being closed under linear combinations. This makes it easier to find the optimal solution for optimization problems.

What are the key steps in an optimization proof for Ax > b?

The key steps in an optimization proof for Ax > b are: 1) defining the set of linear inequalities, 2) showing that the set is convex by using the definition of convexity, and 3) proving that the set has a unique global minimum by showing that any other point in the set will have a higher objective function value.

Can an optimization proof for Ax > b be applied to non-linear problems?

No, an optimization proof for Ax > b is specific to linear problems. For non-linear problems, different techniques such as gradient descent or convex optimization may be used to prove convexity.

What are some real-world applications of optimization proofs for Ax > b?

Optimization proofs for Ax > b are commonly used in various fields such as economics, engineering, and computer science. They can be applied to problems such as resource allocation, production planning, and machine learning algorithms.

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