general Convex set definition

Question

A set is $C$ is convex if $tC+(1-t)C\subset C$ where $0\leq t \leq 1$. how do we manage to extend it for n vectors?

I manage to get $t^2 C + t(1-t)C + t(1-t)C + (1-t)^2 C \subset C$ but induction through this process only gives convexity definition for 2,4,8.. vectors only.

There is an alternate derivation for the following definition: $C$ is convex $\iff$ $\sum_{i=1}^{n} a_i C \subset C$ with $\sum_{i=1}^{n}a_i=1$ and $0 \leq a_i \leq 1$ . can somebody how the two defintion are equivalent and can be derived from each other?

Answer 1

Suppose $a_i\in[0,1]$ ($i=1,2,3$) with $a_1+a_2+a_3=1$, and $x_1,x_2,x_3\in C$. At least one of the $a_i$s is less than $1$; let's assume it's $a_3$. Then $$ a_1x_1+a_2x_2+a_3x_3 = (1-a_3)\left[{a_1\over 1-a_3}x_1+{a_2\over 1-a_3}x_2\right]+a_3x_3. $$ Now $x^*:={a_1\over 1-a_3}x_1+{a_2\over 1-a_3}x_2\in C$ because ${a_1\over 1-a_3}+{a_2\over 1-a_3}=1$. Therefore $$ a_1x_1+a_2x_2+a_3x_3 = (1-a_3)x^*+a_3x_3\in C $$ as well. You should be able to base a full induction argument on this observation.

Answer 2

Fix $x_1,\ldots, x_n,x_{n+1}$ and positive numbers $w_1,\ldots, w_n,w_{n+1}$ with $\sum_{i=1}^{n+1}w_i=1$. Then $$\sum_{i=1}^{n+1} w_i = (1-w_{n+1})x+w_{n+1}x_{n+1},$$ where $$x=\sum_{i=1}^n \frac{w_i}{1-w_{n+1}}x_i.$$