Prove that half plane is convex

Question

I'm having problems to prove this statement mathematically:

Prove that the set $\{(x,y) \in \mathbb{R}^2| y\ge 0\}$ is convex.

Answer 1

For any convex combination $(x,y)=t(x_1,y_1)+(1-t)(x_2,y_2)$ with $y_1\ge 0$. $y_2\ge 0$, $0\le t\le 1$, we have $$y=\underbrace{ty_1}_{\ge 0}+\underbrace{(1-t)y_2}_{\ge 0}\ge 0. $$

Answer 2

I prefer to think in terms of functions:

If $f$ is a linear functional, then the set $C= \{ x \in \mathbb{R}^n | f(x) \le 0 \}$ is convex.

If $x_1,x_2 \in C$ and $\lambda \in [0,1]$, then $f(\lambda x_1 + (1-\lambda)x_1) = \lambda f(x_1)+ (1-\lambda) f(x_2) \le 0$ (since $\lambda \ge 0 , 1-\lambda \ge 0$).

In the above case, take $f(x) = - x_2$.

The same analysis holds (with $=$ replaced by $\le$) for any convex functional $f$. This is why I prefer the $\le$ rather than $\ge$, but this is a purely personal preference.