How to show that this set is convex?

Question

Prove that the following set is convex.

$$X = \{ x \in \mathbb{R}^2 \mid x_1^2 \leq x_2, x_1 \geq 0, x_2 \geq 0 \}$$

Answer 1

Let $(x_1, y_1), (x_2, y_2) \in X$ and consider the line segment \begin{align} (x, y)= t(x_1, y_1)+(1-t)(x_2, y_2) \ \ \text{ for } \ \ 0\leq t \leq 1. \end{align} Since $x_1^2 \leq y_1$ and $x_2^2 \leq y_2$ then \begin{align} x^2= (tx_1+(1-t)x_2)^2 =&\ t^2x_1^2+(1-t)^2x_2^2+2t(1-t)x_1x_2 \\ \leq&\ t^2x_1^2+(1-t)^2x_2^2+t(1-t)(x_1^2+x_2^2)\\ \leq&\ t^2y_1+(1-t)^2y_2 + t(1-t)(y_1+y_2)\\ =&\ ty_1+(1-t)y_2 = y. \end{align}

Answer 2

The set is defined by the following linear matrix inequality (LMI)

$$\begin{bmatrix} 1 & x_1 & 0\\ x_1 & x_2 & 0\\ 0 & 0 & x_1\end{bmatrix} \succeq \mathrm O_3$$

and, thus, it is convex.

Answer 3

For any convex function f $X=\{(x_1,x_2)\in \mathbb R^{2}: f(x_1) \leq x_2, x_1 \geq 0, x_2 \geq 0\}$ is a convex set as you can verify immediately from the definition. The function $f(x)=x^{2}$ is convex.