Having trouble proving a property of convex sets:

  • Thread starter mcah5
  • Start date
  • #1
38
0
In my econ homework, I was asked to prove that:
A set C is convex iff a C + b C = (a+b) C for all nonnegative scalars a and b.

All that I'm given is that the definition of a convex set is, for x,y elements of a convex set C:
(1-a) x + a y exists in C, for 0<a<1

My thoughts were to first prove it in the forward direction. So suppose C convex. Then a C + b C = {ax+by : x,y exist in C}. Somehow I need to get this to a C + b C = {(a+b) x : x exist in C}. I'm not seeing how to do this using the definition of a convex set. I can see why this is true geometrically by noting the set a C + b C is simply the b C superimposed on a bunch a C's on the edge, so the new "radius" becomes a+b but this isn't rigorous.

I have already proved that if a set C is convex then for every finite subset and nonegative scalars that sum to 1, the linear combination is also in C; that teh sum of two convex sets is convex, and that scalar multiples of convex sets are convex, so I can use those properties but I don't think they help.
 

Answers and Replies

  • #2
HallsofIvy
Science Advisor
Homework Helper
41,847
966
Start by looking at a simple example. Suppose C is an interval in the real numbers (all convex sets in R1 are intervals). Then aC+ bC= {ax1+ bx2} where x1 and x2 are contained in C. You want to prove that there always exist x in C such that ax1+ bx2= (a+ b)x. Well, obviously, x= (ax1+ bx2)/(a+b)= (a/(a+b))x1+ (b/(a+b))x2. Do you see how to prove that x is in C?
 
  • #3
38
0
Ah, I see. Thank you so much for your help.
 
  • #4
In my econ homework, I was asked to prove that:
A set C is convex iff a C + b C = (a+b) C for all nonnegative scalars a and b.

All that I'm given is that the definition of a convex set is, for x,y elements of a convex set C:
(1-a) x + a y exists in C, for 0<a<1

My thoughts were to first prove it in the forward direction. So suppose C convex. Then a C + b C = {ax+by : x,y exist in C}. Somehow I need to get this to a C + b C = {(a+b) x : x exist in C}. I'm not seeing how to do this using the definition of a convex set. I can see why this is true geometrically by noting the set a C + b C is simply the b C superimposed on a bunch a C's on the edge, so the new "radius" becomes a+b but this isn't rigorous.

I have already proved that if a set C is convex then for every finite subset and nonegative scalars that sum to 1, the linear combination is also in C; that teh sum of two convex sets is convex, and that scalar multiples of convex sets are convex, so I can use those properties but I don't think they help.

how?
 

Related Threads on Having trouble proving a property of convex sets:

Replies
0
Views
841
  • Last Post
Replies
9
Views
2K
  • Last Post
Replies
10
Views
4K
Replies
4
Views
902
  • Last Post
Replies
2
Views
5K
Replies
1
Views
256
  • Last Post
Replies
8
Views
3K
Top