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$x \in \mathbb{R}^{n}$ is a convex combination $C$ if there $p=p(x)\in \mathbb{N}$, $\lbrace \lambda_i\rbrace_{i=1}^{p} \subseteq [0,1]$ y $\lbrace x_i\rbrace_{i=1}^{p} \subseteq C$ such that $$ x=\sum_{i=1}^{p}\lambda_ix_i \ , \ \ \sum_{i=1}^{p}\lambda_i=1$$

For a triangle in $\mathbb{R}^{2}$, with vertices a, b, c. if x is a convex combination of {a,b,c} then $\lambda_1=\dfrac{\Vert x-a\Vert}{\Vert x-a\Vert +\Vert x-b\Vert +\Vert x-c\Vert }$? and so similarly for $\lambda_2,\lambda_3$??

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Otherwise you may give me a counterexample, please – helmonio Dec 11 '12 at 19:11
When $x=a$ we want $\lambda_1 = 1$ but that is not true for your formula. – GEdgar Dec 11 '12 at 19:23
Sure, but if $ C $ has dimension greater than one? – helmonio Dec 11 '12 at 19:26
I was considering the question about a triangle in $\mathbb R^2$. – GEdgar Dec 11 '12 at 19:29
up vote 3 down vote accepted

I am assuming that $a,b,c$ are not collinear.

Your answer cannot be correct. If $x=a$, your formula gives $\lambda_1 = 0$, when it should be $\lambda_1 = 1$. It is impossible for the formula to have any $\lambda_i = 1$.

However, it is straightforward to compute the multipliers:

You can show show that $A = \begin{bmatrix} 1 & 1 & 1 \\ a & b & c\end{bmatrix}$ in invertible, and that the multipliers $\lambda$ satisfy: $A \lambda = \begin{bmatrix} 1 \\ x\end{bmatrix}$, hence $\lambda = A^{-1} \begin{bmatrix} 1 \\ x\end{bmatrix} = A^{-1} \begin{bmatrix} 1 \\ 0\end{bmatrix}+A^{-1} \begin{bmatrix} 0 \\ x\end{bmatrix}$ . In particular, the multipliers are affine functions (ie, linear plus a constant) of $x$.

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Thank you very much for the reply, you can generalize? – helmonio Dec 11 '12 at 19:55
You are very welcome. – copper.hat Dec 11 '12 at 19:55
Generalize in what way? If the points $a_k$ are affinely independent (ie, $\binom{1}{a_k}$ are linearly independent), then the same technique works for more points... – copper.hat Dec 11 '12 at 20:26

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