A $p-$ norm of $\mathbb{Q}_p$ is a function $||_p: \mathbb{Q}_p \to \mathbb{R}$
$$x \neq 0, x=p^{w_p(x)}u \mapsto p^{-w_p(x)}$$
$$\text{ For } x=0 \Leftrightarrow w_p(x)=\infty \\ p^{-\infty}:=0$$
Identity:
- $$|x+y|_p \leq \max \{ |x|_p, |y|_p\} \\ \leq |x|_p+|y|_p$$ If $|x|_p \neq |y|_p \Rightarrow |x+y|_p=\max \{ |x|_p, |y|_p\}$
In order to prove the identity, I tried the following:
$$x=p^{w_p(x)} u_1, u_1 \in \mathbb{Z}_p^*$$
$$y=p^{w_p(x)} u_2, u_2 \in \mathbb{Z}_p^*$$
$$|x|_p=|p^{w_p(x)} u_1|_p=p^{-w_p(x)}$$
$$|y|_p=|p^{w_p(y)} u_2|_p=p^{-w_p(y)}$$
$$x+y=p^{w_p(x)}u_1+p^{w_p(y)}u_2$$
We suppose that $w_p(x) > w_p(y)$
Then:
$$\max \{ |x|_p, |y|_p \}=\max \{p^{-w_p(x)}, p^{-w_p(y)} \}=p^{-w_p(y)}=|y|_p$$
$$x+y=p^{w_p(x)}u_1+p^{w_p(y)}u_2=p^{w_p(y)}(p^{w_p(x)-w_p(y)}u_1+u_2)$$
$$|x+y|_p=|p^{w_p(y)}(p^{w_p(x)-w_p(y)}u_1+u_2)|_p=p^{-w_p(y)}$$ since $p^{w_p(x)-w_p(y)}u_1+u_2 \in \mathbb{Z}_p^*$
So, when $|x|_p \neq |y|_p$, then $|x+y|_p= \max \{ |x|_p, |y|_p\}$.
If $|x|_p=|y|_p$, then $w_p(x)=w_p(y)$
$$x+y=p^{w_p(x)}u_1+p^{w_p(y)}u_2=p^{w_p(x)} (u_1+u_2)$$
$$|x+y|_p=|p^{w_p(x)} (u_1+u_2)|_p \leq p^{-w_p(x)}=\max \{ |x|_p, |y|_p \}$$
Could you tell me if it is right? And how can I show the other two identities?