For example, consider a set P = {(0,0), (0,1), (1,0), (1,1)} which is a subset of R2. The convex hull of this set is a square. Consider now a point x = (1/4, 1/4), which is in the convex hull of P. We can then construct a set {(0,0),(0,1),(1,0)} = P ′, the convex hull of which is a triangle and encloses p, and thus the theorem works for this instance, since |P′| = 3. It may help to visualise Carathéodory's theorem in 2 dimensions, as saying that we can construct a triangle consisting of points from P that encloses any point in P.
Let x be a point in the convex hull of P. Then, x is a convex combination of a finite number of points in P :
where every xj is in P, every λj is positive, and .
Suppose k > d + 1 (otherwise, there is nothing to prove). Then, the points x2 − x1, ..., xk − x1 are linearly dependent, so there are real scalars μ2, ..., μk, not all zero, such that
If μ1 is defined as
then
and not all of the μj are equal to zero. Therefore, at least one μj>0. Then,
for any real α. In particular, the equality will hold if α is defined as
Note that α>0, and for every j between 1 and k,
In particular, λi − αμi = 0 by definition of α. Therefore,
where every is nonnegative, their sum is one , and furthermore, . In other words, x is represented as a convex combination of at most k-1 points of P. This process can be repeated until x is represented as a convex combination of at most d + 1 points in P.