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I need to find the norm of $x \in \mathbb{R}^2$ if the unit ball is defined by this inequality:

$B=(\begin{pmatrix} x_1\\ x_2 \end{pmatrix}: -a_1\leq x_1\leq a_1, -a_2\leq x_2\leq a_2 ) $.

What exactly I am asked to do? Clearly $|x_1| \leq a_1$ and $|x_2| \leq a_2$ so any norm is smaller or equal to $\sqrt{a_1^2+a_2^2}$.

Thanks a lot!

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Hint: the unit ball $B$ consists of all points $(x_1,x_2)\in \mathbb{R}^2$ such that $$\max \left(\frac{|x_1|}{a_1},\frac{|x_2|}{a_2}\right) \le 1$$

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  • $\begingroup$ Thank you. I don't really get it, can you please extend your answer? $\endgroup$ – Jozef Jan 17 '12 at 8:08
  • $\begingroup$ The question is asking: "What must the norm function be, to make the unit ball look like this?" Now do you get it? $\endgroup$ – TonyK Jan 17 '12 at 8:17
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Hint: You want a function $p:\mathbb{R}^2 \to \mathbb{R}$ which sends any point on the boundary of $B$ to $1$, and where $p(k\boldsymbol{v}) = |k|p(\boldsymbol{v})$ for real $k$.

When you have found it you can, if you wish, check the definitions of a norm are satisfied.

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Hint: If $(x,y)$ is non-zero and not on the boundary of your unit ball, then it lies in the subspace of an (essentially) unique vector on the boundary of the unit ball.

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