Convexity of certain set

Question

I need to show that $S = \{(x,y) \in \mathbb{R}^2 : y \geqslant x^2\}$ is a convex set, but I'm having a bit of trouble. Simply applying the definition of convexity has got me nowhere.

Answer 1

If you are allowed to use the fact that $f(x)=x^2$ is a convex function, the proof would be utterly simple: for any $(x_1,y_1), (x_2,y_2)\in S$ and $t\in[0,1]$, we have \begin{align} [tx_1 + (1-t)x_2]^2 &= f(tx_1 + (1-t)x_2)\\ &\le tf(x_1) + (1-t)f(x_2)\\ &\le ty_1 + (1-t)y_2. \end{align} Therefore $\left(tx_1 + (1-t)x_2,\ ty_1 + (1-t)y_2\right)$ also lies in $S$ and hence $S$ is convex.

The above argument actually shows that if $\varphi(x)$ is a convex function defined on $\mathbb{R}$, $\{(x,y)\in\mathbb{R}^2: y\ge\varphi(x)\}$ is automatically a convex set, but I guess the point of the exercise is to show that the square function is convex ... :-D

Answer 2

Let $(a,b),(c,d)\in S$, we need to show that for all $\lambda\in (0,1)$ $(\lambda a+(1-\lambda)c,\lambda b+(1-\lambda)d)\in S$.

Proof: Firstly, we note that $a^2\leq b$ and $c^2 \leq d$

Since $ac\leq \sqrt {bd}\leq \frac {b+d}{2}$, therefore $2\lambda (1-\lambda) ac\leq (\lambda-\lambda^2)b+(\lambda-\lambda^2)d$. Since $\lambda^2a^2\leq\lambda^2b$ and $\lambda^2c^2\leq\lambda^2d$. By adding the last three inequlities we get: $$(\lambda a+(1-\lambda)c)^2=\lambda^2a^2+2\lambda (1-\lambda) ac+\lambda^2c^2\leq \lambda b+(1-\lambda d) $$