Theorem.(Carathéodory’s theorem) Suppose that and its convex hull has finite dimension . Then each point is the convex combination of at most points in .

*Proof.* For each , we can write , where and (distinct points). I claim that . Suppose . Set . Then since , we have that

,

which implies that are linearly dependent in the translated linear space giving . Hence, there exist coefficients , not all zero, such that . Rearranging, we obtain

Observe that , so there exists some , such that . Therefore we can choose such that , for , and for at least one . Hence,

,

since by construction. Since , we obtain that

,

which implies that is a convex combination of points in .

Lemma.If is a convex set in a linear normed space, then both and are convex.

*Proof. *Suppose . Then there exist sequences such that . For , we then have that

To see that is convex, let and choose such that . Since , , so that, for ,

,

since and is convex by hypothesis.

Lemma.If is an open set in a linear normed space , then is open. If is finite dimensional and is compact, then is compact.

*Proof. *Let be the convex combination of points in . For each , there exists such that . Choose such that . For , and therefore

Now suppose that is a -dimensional linear normed space and is compact. Define a function by

is evidently a continuous map, and by Carathéodory’s theorem, the image of is . By repeated application of the Bolzano-Weierstrass theorem, we see that the domain of is compact. Hence, the image of is compact, which shows that is compact.

While the convex hull of an open set is again open, as shown above, the convex hull of a closed subset of a linear normed space need not be closed. Indeed, consider the set in given by . It is evident that is open, hence is closed. I claim that

To see that is closed, we show that is not open. The point , but for any given, we can choose sufficiently large so that and therefore the convex combination

We now show that is convex. Fix and let . Then

which implies that . Hence, . For the reverse inclusion, fix a convex combination . Since , it is immediate that . Analogously,

,

with equality if and only if all the are zero.

It’s worth mentioning that every closed set in has closed convex hull. This result is immediate from an elementary lemma.

Lemma.Let be a nonempty closed, convex subset. Then exactly one of the following holds:

- , where ;
- , where ;
- , where .

*Proof. *Fix a nonempty closed, convex subset . Define and . Suppose first that is bounded, so that and are both finite. Since is closed, and , so that .

Now suppose that is unbounded. If is finite, then and . If is finite, then and .

It is not a coincidence that we had to consider a closed set whose convex hull was unbounded in order to demonstrate that the convex hull of a closed set need not be closed. Indeed, a consequence of Carathéodory’s theorem is that the convex hull of a compact set in a finite-dimensional normed space is also compact. Since the Heine-Borel theorem equates compactness with closed and bounded (in finite-dimensional spaces), a closed set such that is not closed is necessarily unbounded.

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