Prove a function is a norm?

Question

I want to show that $${||x||}_{\infty} = \max{\{|x_1|,|x_1|,...,|x_n|\}}$$ is a norm.

Where the properties of a norm are

1. ${||x||} \ge 0$ for all $x$ in${\mathbb R}^n$
2. ${||x||} = 0$ when $x=0$
3. ${||ax||} = |a|{||x||}$ for all $a$ in ${\mathbb R}$
4. $||x+y|| \le ||x||+||y||$

I think I intuitively understand that 1-3 hold, but I'm not certain how to actually prove this. I'm also not sure how to think about 4. What is $y$ in this case?

Answer 1

For 1. $\max = |x_i| \geq 0$ by the definition of absolute value, where $i$ is corresponds to the max component. For 2. let $||x||=0 \rightarrow 0 = \max \geq |x_i| \geq 0 \forall x_i \rightarrow x=0$. Now let $x=0 \rightarrow \forall x_i, x_i = 0 \rightarrow \max = 0 \rightarrow ||x||=0$. For 3. $||ax|| = \max |ax| = |a||x_i| = |a|\max |x| = |a|||x||$, where $|x_i|$ is the max element. For 4. $||x+y|| = |x_i+y_i| \leq |x_i| + |y_i| \leq \max x + \max y = ||x|| + ||y||$, where $|x_i + y_i|$ is the max element of $|x+y|$ (abs applied componentwise).

Answer 2

To prove 4., notice that

$$ ||x+y||_{\infty} = |x_l+y_l | $$

for some $l$ by the definition of max. Now apply the usual triangle inequality: $|x+y| \leq |x| + |y| $ and since for example $|x_l| \leq \max |x_i| = ||x|| $, the result follows