2.1 Let $C \subseteq \mathbf{R}^n$ be a convex set, with $x_1, \ldots, x_k \in C$, and let $\theta_1, \ldots, \theta_k \in \mathbf{R}$ satisfy $\theta_i \geq 0$, $\theta_1+\cdots+\theta_k=1$. Show that $\theta_1 x_1+\cdots+\theta_k x_k \in C$. (The definition of convexity is that this holds for $k=2$; you must show it for arbitrary $k$.) Hint. Use induction on $k$.

Solution. This is readily shown by induction from the definition of convex set. We illustrate the idea for $k=3$, leaving the general case to the reader. Suppose that $x_1, x_2, x_3 \in C$, and $\theta_1+\theta_2+\theta_3=1$ with $\theta_1, \theta_2, \theta_3 \geq 0$. We will show that $y=\theta_1 x_1+\theta_2 x_2+\theta_3 x_3 \in C$. At least one of the $\theta_i$ is not equal to one; without loss of generality we can assume that $\theta_1 \neq 1$. Then we can write
$$y=\theta_1 x_1+\left(1-\theta_1\right)\left(\mu_2 x_2+\mu_3 x_3\right)$$
where $\mu_2=\theta_2 /\left(1-\theta_1\right)$ and $\mu_2=\theta_3 /\left(1-\theta_1\right)$. Note that $\mu_2, \mu_3 \geq 0$ and
$$\mu_1+\mu_2=\frac{\theta_2+\theta_3}{1-\theta_1}=\frac{1-\theta_1}{1-\theta_1}=1 .$$
Since $C$ is convex and $x_2, x_3 \in C$, we conclude that $\mu_2 x_2+\mu_3 x_3 \in C$. Since this point and $x_1$ are in $C, y \in C$.

2.2 Show that a set is convex if and only if its intersection with any line is convex. Show that a set is affine if and only if its intersection with any line is affine.

Solution. We prove the first part. The intersection of two convex sets is convex. Therefore if $S$ is a convex set, the intersection of $S$ with a line is convex.
Conversely, suppose the intersection of $S$ with any line is convex. Take any two distinct points $x_1$ and $x_2 \in S$. The intersection of $S$ with the line through $x_1$ and $x_2$ is convex. Therefore convex combinations of $x_1$ and $x_2$ belong to the intersection, hence also to $S$.

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