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$$S = \{(a,b)\in \mathbb{R}^{2}:ax_1+bx_2\leq 1, \forall (x_1,x_2)\in \mathbb{R}^{2}\text{ such that }|x_1|+|x_2|\leq 1\}$$

I want to sketch the solutions of this set in $\mathbb{R}^{2},$ but I am not sure what it looks like. I tried taking specific vectors $(x_1,x_2)$ but this does not help either. Any ideas will be much appreciated.

Edit:

If I take the vector $(\pm 1,0)$ then $|a|\leq 1$ and similarily $(0,\pm1)$ implies that $|b|\leq 1.$ Can I still conclude that $S$ is a square such that $|a|\leq 1$ and $|b|\leq 1.$

Student
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1 Answers1

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$$|x_1|+|x_2|\leq 1\implies \begin{cases}x_1+x_2\leq1 & x_1,x_2\geq0\\-x_1+x_2\leq1 &x_1<0,\:x_2\geq0\\x_1-x_2\leq1 &x_1\geq0,\:x_2<0\\-x_1-x_2\leq1 &x_1,x_2<0\end{cases}$$

Thus $$ax_1+bx_2\leq 1\: \text{such that}\: |x_1|+|x_2|\leq 1\implies \begin{cases}ax_1+bx_2\leq1 & x_1,x_2\geq0\\-ax_1+bx_2\leq1 &x_1<0,\:x_2\geq0\\ax_1-bx_2\leq1 &x_1\geq0,\:x_2<0\\-ax_1-bx_2\leq1 &x_1,x_2<0\end{cases}$$

(*You shall end up with some form of a rectangle in the plane $\mathbb{R}^2$).

Yadati Kiran
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  • Hey could you give feedback on the edit I made? – Student Nov 21 '18 at 00:19
  • $S$ will be a square only if $a: &: b$ are equal. Imagine your set had to satisfy $\frac{1}{2}x_1+x_2\leq1 $ subject to the same above constraints. Will the $S$ still be a square? – Yadati Kiran Nov 21 '18 at 00:22