Let A,B be any two convex subsets of R^n. That is, for any element
x,y of A (resp. of B), and any real scalar t in [0,1], the "convex
tx + (1-t)y
again belongs to A (resp. to B).
First we will show that the set A+B (formed by taking all terms a+b
where a in A, b in B) is again a convex set. For suppose x+y and
x'+y' are any two points in A+B, naturally with x,x' in A and y,y' in
B. Given real t in [0,1], we know:
tx + (1-t)x' belongs to A
ty + (1-t)y' belongs to B
t(x+y) + (1-t)(x'+y') = tx + (1-t)x' + ty + (1-t)y'
will also belong to A+B. Therefore A+B is shown to be convex.
Finally we can show the difference A-B of convex sets is convex by a
similar argument, or by noting that A-B can be considered a sum of two
convex sets, A and -B, and applying what has already been shown.
If B is convex, then so is any scalar multiple cB, e.g. in particular
(-1)B = -B. For x belongs to B if and only if -x belongs to -B. From
this it is easy to see that given two typical elements -x, -x' of -B
and scalar t in [0,1]:
t(-x) + (1-t)(-x') = -( tx + (1-t)x' )
again belongs to -B (because tx + (1-t)x' clearly belongs to convex
Now any element a-b of A-B, where a in A and b in B, is just a + (-b),
the sum of an element from A and one from -B, and conversely. Thus
A-B = A + (-B), so the difference of two convex sets may be considered
a "case" of the sum of two convex sets.