Supremum norm in $\mathbb{R}^{2}$

Question

Let be $\textbf{u}$ and $\textbf{v}$ two vectors in $\mathbb{R}^{2}$:

$\textbf{u} = (u_{1}, u_{2})$ , $\textbf{v} = (v_{1},v_{2})$

I'm very close to show that $\lVert \textbf{v}\rVert = max (|v_{1}| , |v_{2}|)$ is a norm (i think is called the Supremum norm), but in order to do it i'm stuck because i need to show that this claim is true:

$max (|u_{1}| + |v_{1}| , |u_{2}| + |v_{2}|) \leq max (|u_{1}|,|u_{2}|) + max(|v_{1}|,|v_{2}|)$

How can i prove this?

At least, is this intuitive? Why? I cannot see it! Thank's!

Answer 1

It is intuitive, because for the same target function $f(x,y)=|x|+|y|$, the lhs. is the maximum of one image of that function, but the rhs. takes two maximum of different preimages. The intuition is that two independent maxima are better than one in a same set.

Hint: To prove that expression, you can just write 12 cases, some of them are trivial. Actually, by symmetry you can reduce to 6 cases.

Answer 2

Note that $|u_1|\le\max(|u_1|, |u_2|)$ and $|v_1|\le\max(|v_1|, |v_2|)$.

So $$|u_1|+|v_1|\le\max(|u_1|, |u_2|)+\max(|v_1|, |v_2|)$$ Analogously $$|u_2|+|v_2|\le\max(|u_1|, |u_2|)+\max(|v_1|, |v_2|)$$ and therefore

$$\max(|u_1|+|v_1|, |u_2|+|v_2|)\le\max(|u_1|, |u_2|)+\max(|v_1|, |v_2|)$$