Let S be an arbitrary set in $mathbb{R}^n$.If $x in Coleft ( S right )$, then $x in Coleft ( x_1,x_2,….,x_n,x_{n+1} right )$.
Proof
Since $x in Coleft ( Sright )$, then $x$ is representated by a convex combination of a finite number of points in S, i.e.,
$x=displaystylesumlimits_{j=1}^k lambda_jx_j,displaystylesumlimits_{j=1}^k lambda_j=1, lambda_j geq 0$ and $x_j in S, forall j in left ( 1,k right )$
If $k leq n+1$, the result obtained is obviously true.
If $k geq n+1$, then $left ( x_2-x_1 right )left ( x_3-x_1 right ),….., left ( x_k-x_1 right )$ are linearly dependent.
$Rightarrow exists mu _j in mathbb{R}, 2leq jleq k$ (not all zero) such that $displaystylesumlimits_{j=2}^k mu _jleft ( x_j-x_1 right )=0$
Define $mu_1=-displaystylesumlimits_{j=2}^k mu _j$, then $displaystylesumlimits_{j=1}^k mu_j x_j=0, displaystylesumlimits_{j=1}^k mu_j=0$
where not all $mu_j”s$ are equal to zero. Since $displaystylesumlimits_{j=1}^k mu_j=0$, at least one of the $mu_j > 0,1 leq j leq k$
Then, $x=displaystylesumlimits_{1}^k lambda_j x_j+0$
$x=displaystylesumlimits_{1}^k lambda_j x_j- alpha displaystylesumlimits_{1}^k mu_j x_j$
$x=displaystylesumlimits_{1}^kleft ( lambda_j- alphamu_j right )x_j $
Choose $alpha$ such that $alpha=minleft { frac{lambda_j}{mu_j}, mu_jgeq 0 right }=frac{lambda_j}{mu _j},$ for some $i=1,2,…,k$
If $mu_jleq 0, lambda_j-alpha mu_jgeq 0$
If $mu_j> 0, then :frac{lambda _j}{mu_j}geq frac{lambda_i}{mu _i}=alpha Rightarrow lambda_j-alpha mu_jgeq 0, j=1,2,…k$
In particular, $lambda_i-alpha mu_i=0$, by definition of $alpha$
$x=displaystylesumlimits_{j=1}^k left ( lambda_j- alphamu_jright )x_j$,where
$lambda_j- alphamu_jgeq0$ and $displaystylesumlimits_{j=1}^kleft ( lambda_j- alphamu_jright )=1$ and $lambda_i- alphamu_i=0$
Thus, x can be representated as a convex combination of at most (k-1) points.
This reduction process can be repeated until x is representated as a convex combination of (n+1) elements.
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