Let $C \subset \mathbb{R}^n$ be a convex set. Additionally, $x_1, x_2,\dots, x_k \in C$ and $\theta_1,\theta_2,\dots,\theta_k \in \mathbb{R}, \theta_i \ge 0, \sum\theta_i = 1$. I have to proof that $\theta_1x_1+\theta_2x_2+\dots+\theta_kx_k \in C$. I decided to prove it by induction.
If I add a new term to the previous combination, I'll have $\theta_1x_1+\theta_2x_2+\dots+\theta_kx_k+\theta_{k+1}x_{k+1}$. Then, if I make $\theta_{k+1} = 0$, it is true that the latter combination belongs to $C$. By induction, and knowing that this is true for $k=2$, it will hold for any value of $k$.
I have always had trouble with proofs, so I am not sure when I arrive to such a trivial conclusion. Could anybody please help and tell me if this is a valid proof?
Thanks in advance.
This is not a valid proof, but you are on the right track. Induction will work, if you structure things right.
Here is a hint:
Fact 1: $\theta_1 + \cdots + \theta_k = 1 - \theta_{k+1} \geq 0,\;$ and
Fact 2: You can actually assume without loss of generality that $1 - \theta_{k+1} > 0\,$ (why?), and so,
$$ \theta_1 x_1 + \cdots + \theta_k x_k + \theta_{k+1} x_{k+1} = (1 - \theta_{k+1}) y + \theta_{k+1} x_{k+1} \> , $$ where $$ y = \frac{\theta_1 x_1 + \cdots + \theta_k x_k}{\theta_1 + \cdots + \theta_k} \>. $$
Under the right induction hypothesis, what do you know about $y$ and why is it true? Now, use this and what you know about convexity for two points to conclude.