# Having trouble proving a property of convex sets:

1. Mar 29, 2006

### mcah5

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.

2. Mar 29, 2006

### HallsofIvy

Staff Emeritus
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. Mar 29, 2006

### mcah5

Ah, I see. Thank you so much for your help.

4. Jan 6, 2010

how?