Prove that $C \subset \mathbb{R}^n$ is convex iff $\forall m \in \mathbb{N}$ and every set of $m$ points $\{x_1,...,x_m\} \subset C$ we have that $\sum_{i=1}^m \lambda_i x_i \in C$
Where $\sum_{i=1}^m \lambda_i = 1$ and $\lambda_i \geq 0 $ for all $i=1,...,m$.
My attempt at this deviously straightforward question:
Assume that $\sum_{i=1}^m \lambda_i x_i \in C$ then for any two points $x_a,x_b \in C$ $$\lambda_a x_a + \lambda_b x_b \in C$$
Since $\lambda_i \geq 0$ then for convexity $\{\lambda x_a + (1-\lambda)x_b\} 0 \leq \lambda \leq 1\}$ is satisfied (have I shown sufficient proof of this?)
Conversely assume $\sum_{i=1}^m \lambda_i x_i \notin C$
Then there exists two points $x_\alpha , x_\beta$ such that $\lambda_\alpha x_\alpha + \lambda_\beta x_\beta \notin C$
Hence there exists some $\lambda$ such that $\lambda x_\alpha + (1-\lambda)x_\beta \notin C$ for $0 \leq \lambda \leq 1$ and therefore $C$ is not convex.
Can anyone give me pointers on how to improve my attempt (if it needs any) or where I've made any logical fallacies (if I have).
Regards, ~e
Assume $\;C\;$ is convex and let $\;x_1,...,x_n\in C\;$ . Proceed inductively: for $\;n=2\;$ it is exactly the definition of convexity, so assume for $\;n-1\;$ and prove for $\;n\;$ : let $\;\lambda_i\ge 0\;$ be such that $\;\sum\limits_{k=1}^n\lambda_k=1\;$ .The induction hypothesis gives us that
$$\sum_{k=2}^n\left(\frac{\lambda_k}{\sum\limits_{i=2}^n\lambda_i}\right)x_k= c\in C$$
Check the above carefully! Get convinced that the sum of those coefficients indeed is $\;1\;$ ...and check the following slowly:
$$\sum_{k=1}^n\lambda_kx_k=\lambda_1x_1+\sum_{k=2}^n\lambda_k\left[\sum_{k=2}^n\left(\frac{\lambda_k}{\sum\limits_{i=2}^n\lambda_i}\right)x_k\right]=\lambda_1x_1+c\sum_{k=2}^n\lambda_k=$$
$$=\lambda_1x_1+(1-\lambda_1)c\in C$$
and we're done.
The other direction, particularly once the above is properly understood, is almost trivial.