# Proof of a norm

Prove that:
$\|\cdot\|_1:\mathbb{R}^n \rightarrow \mathbb{R};\vec{x} \mapsto \sum_{j=1}^n |x_j|$ is a norm defined on the vector space $\mathbb{R}^n$.

1) Zero vector:
$\sum_{j=1}^n |x_j| = |x|_1 + |x|_2+...+|x|_n$ for $x= 0 =>\|0\|_1=\sum_{j=1}^n |0|=0_1+0_2+...0_n <=> \sum_{j=1}^n |0|=\vec{0}=\|0\|_1$

2)positive homogeneity:
$\forall a \in \mathbb{R};$
$\|a*x\|_1=\sum_{j=1}^n |a*x_j|=|a|*\sum_{j=1}^n |x_j|=|a|\|x\|_1$

3)triangle inequality:
$\forall u,v \in \mathbb{R};$
$\|u+v\|=\sum_{j=1}^n |u_j+v_j|\leq\sum_{j=1}^n |u_j|+\sum_{j=1}^n |v_j|=\|u\|_1+\|v\|_1$

I don't really know if that prove is correct, it would be great if somebody could help me.

-

Your proofs of 2) and 3) look good. For 1), you have certainly shown that $\|0\|_1=0$; however for the converse you need to show that if $\|x\|_1=0$, we must have $x=0$. You haven't done this yet. I will edit to include this proof if you can't figure it out.
if $\|x\|_1=0 \leftrightarrow \sum_{j=1}^n |x_j|=0 \leftrightarrow |x|_1 + |x|_2+...+|x|_n =|0|_1 + |0|_2+...+|0|_n=0 => x=0$. Is that right? – Phil Jan 25 '13 at 0:47
@Phil: you might mean this, but stop before the second equivalence; then remark "as each |$x_i$| $\geq 0$, this implies $x_i = 0$ for all $i$, and so $x =0$". – gnometorule Jan 25 '13 at 0:58
@Phil: as I tried to convey, you might mean the same thing. It's a bit clearer. You don't explicitly say why all $x_i = 0$, even though you probably meant this. – gnometorule Jan 25 '13 at 1:04