This is a question 1.6a from the book Computational Geometry Algorithms and Applications
1.6 In many situatuions we need to compute convex hulls of objects other than points.
a. Let S be a set of n lines segments in the plane. Prove that the convex hull is exactly the same as the convex hull of the 2n end points of the segments.
However I need to do this with induction proof using convex combinations.
I've made the base case where we only have 2 points so that $a_1 + a_2 = 1 => a_1*x_1 + (1-a_1)x_2$ defines the convex combination
so we get $\sum_i^n a_i*x_i$ is the convex combination för n points. n = p and we get $\sum_i^p a_i*x_i$
$\sum_i^{p+1}a_i*x_i = \sum_i^{n}a_i*x_i + a_{i+1}*x_{i+1}$ is the induction step for the convex combination.
However as of my understanding this only proves a convex hull exists for the points and not set of lines necessarily have the same convex hull as the points
Edit:
Ok, so i found a possible solution with the help from @SomeCallMeTim
So $\sum_{i=1}^{n} (a_i*x_i + a_{2i}*x_{2i})$ is the convex hull of S
and $\sum_{i=1}^{2n} (a_i*x_i) $ is the convex hull of S'
Therefore to prove $S' = S \rightarrow \sum_{i=1}^{n} (a_i*x_i + a_{2i}*x_{2i}) = \sum_{i=1}^{2n} (a_i*x_i)$ must be true.
Induction proof:
Base case: $n = 1 \rightarrow a_1*x_1+a_2*x_2 = a_1*x_1+a_2*x_2$
$n = p \rightarrow \sum_{i=1}^{p} (a_i*x_i + a_{2i}*x_{2i}) = \sum_{i=1}^{2p} (a_i*x_i)$
$n = p + 1 \rightarrow \sum_{i=1}^{p+1} (a_i*x_i + a_{2i}*x_{2i}) = \sum_{i=1}^{2p+2} (a_i*x_i) \rightarrow$
$\sum_{i=1}^{p+1} (a_i*x_i) +\sum_{i=p+2}^{2p+2}( a_{i}*x_{i}) =\sum_{i=1}^{2p+2} (a_i*x_i) \rightarrow$
$\sum_{i=p+2}^{2p+2}(a_i*x_i) = \sum_{i=p+2}^{2p+2}(a_i*x_i)$
Is this correct or does anything need improvement