Suppose $\,x = \left(x_1, x_2\right)$, then $\,l^2\,$ norm ball is $\,\left\lbrace x\;\big\vert\;\, \sqrt{\left\lvert x_1 \right\rvert^2 + \left\lvert x_2 \right\rvert^2} \leq 1\right\rbrace$

Easily we can see that $\left\lvert x_2 \right\rvert = \pm\sqrt{1 - \left\lvert x_1 \right\rvert^2}$

And the set of $\,\left(x_1,x_2\right)\,$ forms a circle in $\, \mathbb R^2$.

Now suppose $\,x = \left(x_1, x_2\right)$, then $\,l^\infty\,$ norm ball is $\,\big\lbrace x\, \mid\,\left\|x\right\|_\infty \leq 1\big\rbrace$

where $\,\left\|x\right\|_\infty \leq 1 = \max\big\lbrace\left\lvert x_1 \right\rvert, \left\lvert x_2 \right\rvert\big\rbrace \leq 1,\,$ but in this case we cannot use the same technique as in $\,l^2\,$ case to write $\,x_2\,$ as a function of $\,x_1$.

Can anyone tell me how they came up with the following diagram?

enter image description here

  • $\begingroup$ The $\ell^{\infty}$ norm of a point in $\mathbb{R}^2$ is just the maximum of the two coordinates, and the square encloses all points where neither of the two coordinates exceeds 1, right? $\endgroup$ – gogurt Sep 14 '15 at 17:18
  • $\begingroup$ It's the limit $||x||_\infty = \lim \limits_{n \to \infty} ||x||_n$! Draw pictures as $n$ grows. $\endgroup$ – Anthony Peter Sep 14 '15 at 17:22
  • $\begingroup$ @gogurt How do you justify the line on the negative $x_1$ $x_2$ axis?, The vertical line on the left is clearly $|x_1| = -1$ and the horizontal line at the bottom is $|x_2| = -1$ which is both impossible $\endgroup$ – Carlos - the Mongoose - Danger Sep 14 '15 at 17:22
  • $\begingroup$ Think of this as $\left \{ (x_{1},x_{2}) \in \mathbb{R}|\left | x_{1} \right | \leq \left | x_{2} \right | \leq 1) \right \} \bigcup \left \{ (x_{1},x_{2}) \in \mathbb{R}|\left | x_{2} \right | < \left | x_{1} \right | \leq 1) \right \} $ $\endgroup$ – rtybase Sep 14 '15 at 17:23
  • $\begingroup$ @IllegalImmigrant how are they impossible? $\endgroup$ – gogurt Sep 14 '15 at 17:24

Think of this as $$\left \{ (x_{1},x_{2}) \in \mathbb{R}|\left | x_{1} \right | \leq \left | x_{2} \right | \leq 1) \right \} \bigcup \left \{ (x_{1},x_{2}) \in \mathbb{R}|\left | x_{2} \right | < \left | x_{1} \right | \leq 1) \right \} $$

Which is enter image description here

  • 1
    $\begingroup$ Okay, so the trick is that $|x_1| \leq 1 \Leftrightarrow -|x_1| \geq -1$ Thanks that really was my confusion as to how you got the negative areas $\endgroup$ – Carlos - the Mongoose - Danger Sep 14 '15 at 20:53

They looked for all points $(x,y)$ in the plane such that $|x|\leq 1$ and also $|y|\leq 1$.


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