How to prove $(1-x) y \ge 0$ is a convex set?

Question

$x \epsilon [0,1], y> 0 $

Let $(1-\underline{x}) \underline{y} \geq 0 $ and $(1-\bar{x}) \bar{y} \geq 0 $

Let $t \epsilon [0,1]$

$[1- (t\underline{x}+ (1-t)\bar{x})] (t\underline{y}+ (1-t)\bar{y})$

$= (t\underline{y}+ (1-t)\bar{y})-(t\underline{x}+ (1-t)\bar{x})(t\underline{y}+ (1-t)\bar{y}) $

$=(t\underline{y}+ (1-t)\bar{y})-(t\underline{x}t\underline{y}+(1-t)\bar{x}(1-t)\bar{y}+t\underline{x}(1-t)\bar{y}+(1-t)\bar{x}t\underline{y}$

$=[t\underline{y}-t^2\underline{x}\underline{y}]+[(1-t)\bar{y}-(1-t)^2\bar{x}\bar{y}]- t\underline{x}(1-t)\bar{y}-(1-t)\bar{x}t\underline{y}$

All the terms in brackets are greater than $0$. How about the rest? I assume the region is convex since when I plot it, it looks like it.

Also please see question $1$ part c. http://mjo.osborne.economics.utoronto.ca/index.php/tutorial/index/1/cvn/x

Thank you.

Answer 1

The set $A = \{ (x,y) \in \mathbb{R}^2 \ \vert \ (1-x) y \geq 0 \}$ is not a convex set. To see this, observe that $(1, -1), (0, 0) \in A$ but $$\frac{1}{2} \cdot (1, -1) + \frac{1}{2} \cdot \left(0, 0\right) = \left(\frac{1}{2}, -\frac{1}{2}\right) \notin A$$

However, let's work through the other example that you suggested. The set in question is $B = \{ (x,y) \in \mathbb{R}^2 \ \vert \ x y \geq 1, x > 0, y > 0 \}$. Let's show that $B$ satistfies the definition of convexity; that is, for any $(x_1, y_1), (x_2, y_2) \in B$ and $\lambda \in [0,1]$ we have that $\lambda (x_1, y_1) + (1-\lambda) (x_2, y_2) \in B$. First, note that

\begin{align*} \lambda (x_1, y_1) + (1-\lambda) (x_2, y_2) &= \left(\lambda x_1 + (1-\lambda) x_2, \lambda y_1 + (1-\lambda) y_2 \right) \\ &:= (x_3, y_3) \\ \end{align*}

Because $(x_1, y_1), (x_2, y_2) \in B$ we have that $x_1, x_2, y_1, y_2 > 0$. This, together with $\lambda \in [0,1]$ gives us that $x_3, y_3 > 0$. Finally, let us check that

\begin{align*} x_3 \cdot y_3 &= \left[ \lambda x_1 + (1-\lambda) x_2 \right] \cdot \left[ \lambda y_1 + (1-\lambda) y_2 \right] \\ &= \lambda^2 x_1 y_1 + \lambda (1-\lambda) \left(x_1 y_2 + x_2 y_1\right) + (1-\lambda)^2 x_2 y_2 \\ &> \lambda^2 \cdot 0 + \lambda (1-\lambda) \left(x_1 y_2 + x_2 y_1\right) + (1-\lambda)^2 \cdot 0 \\ &= \lambda (1-\lambda) \left(x_1 y_2 + x_2 y_1\right) \\ &> 0 \end{align*} where the both inequalities come from the fact that $(x_1, y_1), (x_2, y_2) \in B$. In general, this is how you'll want to prove that a set in convex. Please let me know if you have any questions or other examples you want to try to work out.