Convex set in $\mathbb{R}^2$

Question

I want to show that the set $A=\{(x, y)\in\mathbb{R}^2: x^2\leq y\} $ is convex.

Answer 1

Definition: a convex function is a function $f:\Bbb{R}\to\Bbb{R}$ such that $A_f = \{(x,y):y\geq f(x)\}$ is convex.

Proposition: If for any $x_1, x_2\in \Bbb{R}$ and $t\in [0,1]$ we have $f(tx_1+(1-t)x_2)\leq tf(x_1)+(1-t)f(x_2)$, then $f$ is convex.

Proof: Suppose the hypothesis holds. Pick $(u_1,u_2), (v_1,v_2)\in A_f$, so $u_2\geq f(u_1)$ and $v_2\geq f(v_1)$ *. Pick $t\in [0,1]$. We wish to show that $t\mathbf{u}+ (1-t)\mathbf{v}\in A$, i.e. $tu_2+(1-t)v_2\geq f(tu_1+(1-t)v_1)$. By hypothesis, we have $$f(tu_1+(1-t)v_1) \leq tf(u_1)+(1-t)f(v_1) \leq tu_2 + (1-t)v_2$$ Where the last inequality is by *. This completes the proof.

Now we show that the proposition holds for $f(x) = x^2$. Let $x,y\in \Bbb{R}$, $t\in [0,1]$. Then $$tf(x)+(1-t)f(y) - f(tx+(1-t)y) = $$ $$tx^2 + (1-t)y^2 - t^2x^2 - (1-t)^2y^2-2t(1-t)xy = $$ $$t(1-t)x^2+t(1-t)y^2-2t(1-t)xy = $$ $$t(1-t)(x-y)^2 \geq 0$$

Verifying that $f$ is convex. The proposition here is very useful for proving functions are convex; it's actually an "if and only if", but I omitted the other half as it wasn't necessary here.