Show that $C$ is convex

Question

Let $C=\{x_2\geq-x_1; x_2\geq x_1; x_2 \leq 3\}$ determine if the set $C$ is convex.

By deffinition a a set $C$ is convex if for all $x_1,x_2\in C$ and $t\in[0,1]$, $$tx_1 + (1-t)x_2 \in C$$

I drew the set and I can tell that is convex. In order to prove that $C$ is convex I just have to follow the definition but I find it difficult to involuve all three equations at the same time. Any suggestions would be great!

Answer 1

I take $v$ and $w$ two points in $C$ then they satisfy the inequilities of $C$ by separeted. Now take $t\in (0,1)$ and consider $z=tv+(1-t)w$.

So $z_{1}=tv_{1}+(1-t)w_{1}$ and $z_{2}=tv_{2}+(1-t)w_{2}$.

The firts question is: $z_{2}\geq-z_{1}$?

The answer is yes, if and only if : $tv_{2}+(1-t)w_{2}\geq -tv_{1}-(1-t)w_{1}$, but we know that $v_{1}\geq -v_{2}$ and as $t>0$ this implies $tv_{1}\geq -tv_{2}$. Similar argument applies to the second addend. Then the first inequality holds.

The other two are similar.

Answer 2

Hint: Use that intersections of convex sets are convex. If needed, prove that first.

Answer 3

Suppose $(x_1,x_2), (y_1,y_2) \in C$ and $0\leq t \leq 1$. Then $x_2 \geq -x_1$ and $y_2 \geq -y_1$. Multiply the first by $t$ the second by $(1-t)$ and add. You get $tx_2+(1-t)y_2 \geq -[tx_1+(1-t)y_1]$. Similarly you get two other inequalities. This proves that $C$ is convex.