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Let $\Omega$ a convex set of $\mathbb{R}^n$ and let $x_1,x_2,..,x_n \in \Omega$ and $\alpha_1,\alpha_2,...,\alpha_n \in \mathbb{R}^+$ such that $\alpha_{1}+\alpha_{2}+...+\alpha_{n}=1$ proof that $\alpha_{1}x_{1}+\alpha_{2}x_{2}+...+\alpha_{n}x_{n} \in \Omega$

I'm trying to show this by induction, the case where n = 1 is trivial, now if we take n = 2 we get let $x_1,x_2\in \Omega$ and $\alpha_1, \alpha_2 \in \mathbb{R}^+$ such that $\alpha_1+\alpha_2=1$. We must proof that for all $t\in [0,1]$ $(1-t)(\alpha_1x_1+\alpha_2x_2)+t(\alpha_1x_1+\alpha_2x_2) \in \Omega$ But I have not been able to do it, and I think that with a similar argument the induction can proceed.

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    $\begingroup$ Think of it like this: if you have $\alpha_1 + \alpha_2 = 1$, then $\alpha_2 = 1-\alpha_1$, so $\alpha_1 x_1+ \alpha_2 x_2$ is going to be a convex combination of elements in $\Omega$. You can deduce this will be in the set. Now what would you do for n=3? $\endgroup$ – User203940 Sep 16 at 23:01
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Let's change the alphas to a for convenience.

For $n=3$ we want to show $a_1 x_1 + a_2 x_2 + a_3 x_3$ is in the set. Notice that from the $n=2$ case we have $$ a_1/(a_1+a_2)x_1 + a_2/(a_1+a_2)x_2 = y \in \Omega.$$ So $$ a_1x_1 + a_2 x_2 = (a_1+a_2)y.$$ Now we can use the $n=2$ case again to get that our original combination is in $\Omega.$

The induction argument will work like this.

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This is a sketch of a proof by induction on the number of terms in $(x_1,\ldots,x_n)\in \Omega^n$.

To show that for any $n\geq2$, $(x_1,\ldots,x_n)\in \Omega^n$ and $(a_1,\ldots,a_n)\subset\mathbb{R}^n_+$ with $\sum^n_{j=1}a_j=1$, $$\sum^n_{j=1}a_jx_j\in\Omega$$

  • For $n=2$ the statement follows by definition of convexity.

  • Suppose statement holds for an integer $n\geq2$. For $n+1$, let $(a_1,\ldots,a_{n+1})\in\mathbb{R}^{n+1}_+$ with $\sum^{n+1}_{j=1}a_j=1$ and $(x_1,\ldots,x_{n+1})\in\Omega^{n+1}$. Without loss of generality, suppose $0<a_{n+1}<1$. $$z:=a_1x_1+\ldots +a_nx_n+a_{n+1}x_{n+1}=(1-a_{n+1})\big(\frac{a_1x_1+\ldots +a_nx_n}{1-a_{n+1}}\big) +a_{n+1}x_{n+1}$$ Since $\sum^n_{j=1}\frac{a_j}{1-a_{n+1}}=1$, the induction hypothesis means that $y:=\big(\frac{a_1x_1+\ldots +a_nx_n}{1-a_{n+1}}\big)\in\Omega$. So, by definition of convexity, $$z=(1-a_{n+1})y+a_{n+1}x_{n+1}\in\Omega$$

This completes the proof by induction.

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  • $\begingroup$ why $\sum^n_{j=1}\frac{a_j}{1-a_{n+1}}=1$ ? $\endgroup$ – camilo Sep 17 at 6:16
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    $\begingroup$ Since $a_1 + \cdots + a_n + a_{n+1} = 1$, we have $a_1 + \cdots + a_n = 1 - a_{n+1}$. Now $\sum_{j=1}^n \frac{a_j}{1-a_{n+1}} = \frac{a_1 + \cdots + a_n}{1-a_{n+1}} = \frac{a_1 + \cdots + a_n}{a_1 + \cdots + a_n} = 1.$ $\endgroup$ – User203940 Sep 17 at 16:54
  • $\begingroup$ @OliverDiaz you took $ a_ {n + 1}> 0$ but eventually it may be the case that $ a_ {n + 1} = 1 $ and so $ 1-a_ {n + 1} = 0 $ in that case your reasoning would be wrong... I'm wrong? $\endgroup$ – camilo Sep 17 at 20:21
  • $\begingroup$ @camilo: I just made a small edit to my answer and set $0<a_{n+1}<1$. That this is enough follows by looking at the cases $a_{n+1}=0$ which is already taken cared of by the induction hypothesis, and $a_{n+1}=1$ which leads to the trivial case $z=x_{n+1}\in\Omega$ by the choice the the $x$'s. Let me know if that makes it clear for you now. $\endgroup$ – Oliver Diaz Sep 17 at 20:48
  • $\begingroup$ Yes, it's clear! $\endgroup$ – camilo Sep 17 at 20:50

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