# How does sketching norms show that they are equivalent?

I have the following statement in my notes:

"You might want to check by drawing the sets of all $x\in\mathbb R^2$ such that $\|x\|_1=1$,$\|x\|_2=1$,$\|x\|_\infty=1$ that indeed these norms are equivalent."

Let $x:=(x,y)$. I have managed to sketch the sets using wolfram alpha, with each drawing corresponding the respective norm as follows:

$\|(x,y)\|_1=|x|+|y|=1$

$\|(x,y)\|_2=\sqrt{x^2+y^2}=1$

$\|(x,y)\|_\infty=\sup\{|x|,|y|\}=1$

My question now is how is it exactly that these drawings show that the three norms are equivalent? What can we take from them shows this fact? Has it to do with the fact that they are all within $|x|,|y|=1$?

## 1 Answer

Remember the definition of equivalence is that there exists numbers $c,d$ such that $c\Vert x\Vert_2 \leq \Vert x\Vert_1 \leq d\Vert x\Vert_2$. What the c and d do in the picture is to stretch or shrink the shapes you've drawn. What the inequality represents is one shape fitting inside another. What equivalence means is that I could shrink the circle for $\Vert \Vert_2$ and fit it inside the diamond for $\Vert \Vert_1$, and I could also stretch the circle so that the diamond fits inside of it.

In this case, $c=1$ and $d = \sqrt{2}$ works, and it looks like this: