# proof that $\alpha_{1}x_{1}+\alpha_{2}x_{2}+…+\alpha_{n}x_{n} \in \Omega$

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.

• 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? – User203940 Sep 16 at 23:01

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.

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 $$0. $$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.

• why $\sum^n_{j=1}\frac{a_j}{1-a_{n+1}}=1$ ? – camilo Sep 17 at 6:16
• Since $a_1 + \cdots + a_n + a_{n+1} = 1$, we have $a_1 + \cdots + a_n = 1 - a_{n+1}$. Now $\sum_{j=1}^n \frac{a_j}{1-a_{n+1}} = \frac{a_1 + \cdots + a_n}{1-a_{n+1}} = \frac{a_1 + \cdots + a_n}{a_1 + \cdots + a_n} = 1.$ – User203940 Sep 17 at 16:54
• @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? – camilo Sep 17 at 20:21
• @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. – Oliver Diaz Sep 17 at 20:48
• Yes, it's clear! – camilo Sep 17 at 20:50