Convex Optimization Exercise 2.1

Problem:

Solution:

The case for $k = 1$ is trivial. The case for $k = 2$ is given as definition. Using induction, we can assume $\theta_1 x_1 + \cdots + \theta_{k-1} x_{k-1} \in C$ work for any set of $\theta$ that sums to 1.

Now consider

$\begin{eqnarray*} & & \theta_1 x_1 + \cdots + \theta_k x_k \\ &=& \theta_1 x_1 + \cdots + \theta_{k-1} x_{k-1} + \theta_k x_k \\ &=& \frac{1 - \theta_k}{1 - \theta_k}(\theta_1 x_1 + \cdots + \theta_{k-1} x_{k-1}) + \theta_k x_k \\ &=& (1 - \theta_k)(\frac{\theta_1}{1 - \theta_k} x_1 + \cdots + \frac{\theta_{k-1}}{1 - \theta_k} x_{k-1}) + \theta_k x_k \end{eqnarray*}$

Now we know the sum inside the bracket is in $C$ by the induction hypothesis, and therefore the point we wanted to prove is also in $C$ because of the definition.