Introduction to Unconstrained Optimization - Scilab

C
x 2
x 2
x 1 x 3
x 3
Figure 28: Convex combination of 3 points.
with 0 ≤ θ 3 ≤ 1. This situation is presented in 28. We shall prove that x 3 is a convex combination
of x 1 , x 2 , x 3 . Indeed, we can develop the equation for x 3 so that all the points x 1 , x 2 , x 3 appear:
x 3 = θ 3 θ 2 x 1 + θ 3 (1 − θ 2 )x 2 + (1 − θ 3 )x 3 . (66)
The weights of the convex combination are θ 3 θ 2 ≥ 0, θ 3 (1 − θ 2 ) ≥ 0 and 1 − θ 3 ≥ 0. Their sum is
equal to 1, as proved by the following computation:
θ 3 θ 2 + θ 3 (1 − θ 2 ) + 1 − θ 3 = θ 3 (θ 2 + 1 − θ 2 ) + 1 − θ 3 (67)
= θ 3 + 1 − θ 3 (68)
= 1, (69)
which concludes the proof.
Let us prove that a set is convex if and only if it contains every convex combinations of its
points.
It is direct to prove that a set which contains every convex combinations of its points is convex.
Indeed, such a set contains any convex combination of two points, which implies that the set is
convex.
Let us prove that a convex set is so that every convex combination of its points is convex.
The proof can be done by induction on the number k of points in the set C. The definition of the
convex set C implies that the proposition is true for k = 2. Assume that the hypothesis is true
for k points, and let us prove that every convex combination of k + 1 points is in the set C. Let
{x i } i=1,k+1 ∈ C and let {θ i } i=1,k+1 ∈ R be positive scalars such that θ 1 + . . . + θ k + θ k+1 = 1. Let
us prove that
x k+1 = θ 1 x 1 + . . . + θ k x k + θ k+1 x k+1 ∈ C. (70)
All the weights {θ i } i=1,k+1 cannot be equal to 0, because they are positive and their sum is
1. Without loss of generality, suppose that 0 < θ k+1 < 1 (if not, simply reorder the points
{x i } i=1,k+1 ). Therefore 1 − θ k+1 > 0. Then let us write x k+1 in the following form
(
)
θ 1
θ k
x k+1 = (1 − θ k+1 ) x 1 + . . . + x k + θ k+1 x k+1 (71)
1 − θ k+1 1 − θ k+1
with
= (1 − θ k+1 )x k + θ k+1 x k+1 , (72)
x k =
θ 1
θ k
x 1 + . . . + x k . (73)
1 − θ k+1 1 − θ k+1
The point x k+1 is a convex combination of k points because the weights are all positive and their
sum is equal to one :
θ 1
θ k
+ . . . +
=
1 − θ k+1 1 − θ k+1
1
1 − θ k+1
(θ 1 + . . . + θ k ) (74)
= 1 − θ k+1
1 − θ k+1
(75)
= 1. (76)
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