Let $x_i, i=1\dots n$ be elements of a convex subset $K$ of a linear space $X$ over the reals. Then any linear combination $\sum\limits_{i=1}^n a_i x_i$ such that $a_i \geq 0$ and $\sum a_i = 1$ is also in the convex set.
My attempt involves first trying to prove it for the case $n = 3$. Let $x = a_1 x_1 + \dots a_3 x_3$, then one of the conditions gives $a_3 = 1 – a_2 – a_1$ so we have $a_1 x_1 + (1 – a_1)x_3 + a_2 (x_2 – x_3) = x$, the first two terms summing to an element of $K$ since $K$ is convex. I thought I had the next step on paper, but lost it and didn't take good notes. What's next?
Best Answer
I assume your definition of convexity is as follows: $K$ is said to be convex if whenever $x,y$ are in $K$, so too is $tx+(1-t)y$ for all $0\leq t\leq 1$.
If so, all you need to do is use induction.
Edit: Here's the answer:
Let $\sum_{i=1}^{n+1}t_{i}x_{i}$ be a convex combination. Then $\sum_{i=1}^{n+1}t_{i}x_{i}=\sum_{i=1}^{n}t_{i}x_{i}+t_{n+1}x_{n+1}$ where $t_{n+1}=1-\sum_{i=1}^{n}t_{i}$. We can write this as $$\left(\sum_{j=1}^{n}t_{j}\right)\left(\sum_{i=1}^{n}\frac{t_{i}}{\sum_{j=1}^{n}t_{j}}x_{i}\right)+\left(1-\sum_{i=1}^{n}t_{i}\right)x_{n+1},$$ which is a convex combination of two points in the set.