0

Assume there is a set $S$ in $\mathbb{R}^2 $ that is a square with $x \in [-1 ,1]$ and $y\in [-1,1]$. I need to prove that this set is convex. Hence, I thought of the following:

Suppose $a, b \in S$ such that $a = (x_1, y_1)$, $b = (x_2, y_2)$ and $\forall \theta \in [0,1]$ the following must hold:

$$(x,y) = \theta a + (1-\theta) b \in S$$

So I need to prove that:

$$||(x,y)|| = ||(\theta x_1 + (1-\theta) x_2, \theta y_1 + (1-\theta ) y_2 )|| \leq \sqrt{2}$$

Is this correct and how can I continue?

mathguy69
  • 37
  • $a = (x_1, y_1)$ and $b = (x_2, y_2)$ right? – Falcon Nov 11 '20 at 14:26
  • Yes it is! I've added it. Thanks – mathguy69 Nov 11 '20 at 14:27
  • $u = (u_1, u_2) \in S$ if and only if $u_1 \in [-1, 1]$ and $u_2 \in [-1, 1]$. Therefore, you just have to prove that $(x, y)$ is such that $x \in [-1, 1]$ and $y \in [-1, 1]$ which seem quite obvious because of the definition of $\theta$. – Falcon Nov 11 '20 at 14:32
  • 2
    It's the intersection of the convex sets ${(x,y): x\le1}$, ${(x,y): x\ge-1}$,${(x,y): y\le1}$, ${(x,y): y\ge-1}$. –  Nov 11 '20 at 14:33

2 Answers2

2

It is not correct because the condition $||(x,y)|| \leq \sqrt{2}$ decribes a circular disk with center $0$ and radius $\sqrt 2$.

You know that $x_1,x_2,y_1,y_2 \in [-1,1]$. Hence $\theta x_1 + (1-\theta)x_2 \in [-1,1]$, $\theta y_1 + (1-\theta)y_2 \in [-1,1]$. Thus $\theta a + (1-\theta)b \in S$.

Paul Frost
  • 76,394
  • 12
  • 43
  • 125
0

As I am just a student take my answer carefully but I think that there is a simple answer using the definition that hasn't been yet written. I hope that it will help you or futur student with the same question.

My answer:
By def a square center at $(0;0)$ in $\mathbb{R^2}$ is describe by the following set $E_a=\left \{ (x;y) \in \mathbb{R}^2 | -a<x<a, -a<y<a \right \}=\left \{ (x;y) \in \mathbb{R}^2 | |x-y|+|x+y| <\frac{a}{2} \right \}$.
Now we want to verify $\forall \vec{x}=\binom{x_1}{x_2},\vec{y}=\binom{y_1}{y_2} \in E_a$ and that $ \forall 0 \leq t \leq 1$ we get $t \vec{x}+ (1-t) \vec{y}=\binom{tx_1+(1-t)y_1}{tx_2+(1-t)y_2}$ is still in $E_a$.
$|tx_1+(1-t)y_1-(tx_2+(1-t)y_2)|+|tx_1+(1-t)y_1+(tx_2+(1-t)y_2)|=|t(x_1-x_2)+(1-t)(y_1-y_2)|+|t(x_1+x_2)+(1-t)(y_1+y_2)|\leq t|x_1-x_2|+(1-t)|y_1-y_2|+t|x_1+x_2|+(1-t)|y_1+y_2|$
Now as $\vec{x} \in E_a$ by assumption we get that $|x_1-x_2|+|x_1+x_2|<\frac{a}{2} \Rightarrow t|x_1-x_2|+t|x_1+x_2|<t\frac{a}{2}$. Same argument for $\vec{y}$.
Hence we get $t|x_1-x_2|+ t|x_1+x_2| +(1-t)|y_1-y_2|+(1-t)|y_1+y_2|<t\frac{a}{2}+(1-t)\frac{a}{2}=\frac{a}{2}$

Rem: the generalisation for a square that is not center is zero is very easy (make a simple change of variable).

Q.E.D.

X0-user-0X
  • 795