BookRags.com Literature Guides Literature
Guides
Criticism & Essays Criticism &
Essays
Questions & Answers Questions &
Answers
Lesson Plans Lesson
Plans
My Bibliography Periodic Table U.S. Presidents Shakespeare Sonnet Shake-Up
Research Anything:        
History | Encyclopedias | Films | News | Create a Bibliography | More... Login | Register | Help
Not What You Meant?  There are 42 definitions for Hull.

Carathéodory's theorem (convex hull)

Print-Friendly
About 2 pages (530 words)

Bookmark and Share Know this topic well? Help others and get FREE products!
See also Carathéodory's theorem for other meanings

In convex geometry Carathéodory's theorem states that if a point x of Rd lies in the convex hull of a set P, there is a subset P′ of P consisting of d+1 or fewer points such that x lies in the convex hull of P′. Equivalently, x lies in a r-simplex with vertices in P, where <math>r \leq d</math>. The result is named for Constantin Carathéodory, who proved the theorem in 1911 for the case when P is compact. In 1914 Steinitz expanded Carathéodory's theorem for any sets P in Rd. 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.

Proof

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 :

<math>\mathbf{x}=\sum_{j=1}^k \lambda_j \mathbf{x}_j</math>

where every xj is in P, every λj is positive, and <math>\sum_{j=1}^k\lambda_j=1</math>. 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

<math>\sum_{j=2}^k \mu_j (\mathbf{x}_j-\mathbf{x}_1)=\mathbf{0}.</math>

If μ1 is defined as

<math>\mu_1:=-\sum_{j=2}^k \mu_j</math>

then

<math>\sum_{j=1}^k \mu_j \mathbf{x}_j=\mathbf{0}</math>
<math>\sum_{j=1}^k \mu_j=0</math>

and not all of the μj are equal to zero. Therefore, at least one μj>0. Then,

<math>\mathbf{x} = \sum_{j=1}^k \lambda_j \mathbf{x}_j-\alpha\sum_{j=1}^k \mu_j \mathbf{x}_j = \sum_{j=1}^k (\lambda_j-\alpha\mu_j) \mathbf{x}_j</math>

for any real α. In particular, the equality will hold if α is defined as

<math> \alpha:=\min_{1\leq j \leq k} \left\{ \tfrac{\lambda_j}{\mu_j}:\mu_j>0\right\}=\tfrac{\lambda_i}{\mu_i}.</math>

Note that α>0, and for every j between 1 and k,

<math>\lambda_j-\alpha\mu_j \geq 0.</math>

In particular, λi − αμi = 0 by definition of α. Therefore,

<math>\mathbf{x} = \sum_{j=1}^k (\lambda_j-\alpha\mu_j) \mathbf{x}_j</math>

where every <math>\lambda_j - \alpha \mu_j</math> is nonnegative, their sum is one , and furthermore, <math>\lambda_i-\alpha\mu_i=0</math>. 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. Q.E.D.

References

  • C. Caratheodory, Über den Variabilitätsbereich der Fourierschen Konstanten von positiven harmonischen Funktionen, Rend. Circ. Mat. Palermo, Vol. 32 (1911), 193-217.
  • E. Steinitz, Bedingt konvergente Reihen und konvexe Systeme, I-IV, J. Reine Angew. Math. Vol. 143 (1913), 128-175.

External links

View More Summaries on Carathéodory's theorem (convex hull)
 
Ask any question on Carathéodory's theorem (convex hull) and get it answered FAST!
Answer questions in BookRags Q&A and earn points toward
discounted or even FREE Study Guides and other BookRags products!
Learn more about BookRags Q&A
Copyrights
Carathéodory's theorem (convex hull) from Wíkipedia. ©2006 by Wíkipedia. Licensed under the GNU Free Documentation License. View a list of authors or edit this article.

Article Navigation
Join BookRagslearn moreJoin BookRags




About BookRags | Customer Service | Report an Error | Terms of Use | Privacy Policy