`convexhull(1, 2, 3) = [1, 3]`

. Note that `2`

was used to generate the
convex hull, but is not a vertex since it is the convex combination of `1, 3`

.
We wish to prove that such a situation does not happen in the case of the
permutahedron.
I have not seen an elementary proof that each point that is a permutation of
the original is a vertex, so I'm recording that proof out of interest.
`i`

of the "largest value" `N`

in the point `P`

.
Since it's the largest point, if it's a convex combination, all other points in
the convex combination must have the same "largest value" at that index `i`

. So
we can now ignore the index `i`

and get permutations of `[1..N-1]`

Repeat this process to prove that `P`

. But this is absurd, because all vertices are distinct.
Hence, the only way for this to work out is if we have only one point that we
are convex-summing over. That is, `P`

`[1, 2, ..., N]`

. Pick point `P`

. Assume
`P`

is in the convex sum of some `{ Xi }`

.
Let `index(N, P)`

be the index of number `N`

in `P`

. For example, if
`P = [2, 3, 1]`

, then
```
index(2, P) = 0 (assuming 0-based indexing)
index(3, P) = 1
index(1, P) = 2
```

If `P`

is the convex sum of vertices `{ Xi }`

, all of them ```
Xi[index(N, P)] = N forall Xi in the convex sum
```

since `N`

is the largest value that `X_I`

can have at any index. The only way
to get the maximum value by a convex sum is for all values to be that maximum
value.
So, we can now ignore dimension `index(N, P)`

, since `X_i`

involved in the convex combination and `P`

have `index(N, P) = N`

. If we
project out this dimension, we are left with permutation of `(1..N-1)`

.
Repeat the same process. We will need to have all `X_i`

to have their value at
`index(N-1, P)`

to be `N-1`

.
Keep doing this till we get that the dimensions of all points in `Xi`

and `P`

are equal. But all the points in `{ Xi }`

are distinct since they are
permutation of `[1..n]`

. Hence, `{ Xi }`

can contain only a single point, and
this point is `P`

.
Hence, `P`

cannot be written as the convex combination of other points. Hence,
`P`

is a vertex.
`{v1, v2, ... vn}`

where without loss of generality,
`v1 <= v2 <= ... <= vn`

. We can always relabel otherwise.
Now we want to write `vn`

as the convex sum of `{v1, v2, ... vn}`

. We can draw this on
the number line as:
```
---[v1--v2--v3--...--vn]---
```

- A convex sum must be inside this line segment between
`[v1--...--vn]`

(`vn`

is to the rightmost since it's known to be the largest value in`{v1...vn}`

). - So, if we try to write
`vn`

as the convex sum, we*must*have the coefficient of`vn=1`

and the coefficient of all other points`=0`

, since any other combination will "pull the convex sum away from the right (where`vn`

sits), towards the left".