0
$\begingroup$

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!

3 Answers 3

1

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

  • 0
    The question is asking: "What must the norm function be, to make the unit ball look like this?" Now do you get it?2012-01-17
0

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.

0

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.