How can I prove that $\theta_p$ is a p-Norm?

[evinda]

Hello! (Wave)

Let $c \in \mathbb{R}, 0<c<1$ and $p \in \mathbb{P}$. We consider the function $\theta_p:\mathbb{Q}\rightarrow \mathbb{R}$
$x=p^{w(x)}u \mapsto c^{w(x)}$.
Show that $\theta_p$ is a p-Norm.

How could I show this? (Thinking)

 

[Euge]

Hello! (Wave)

Let $c \in \mathbb{R}, 0<c<1$ and $p \in \mathbb{P}$. We consider the function $\theta_p:\mathbb{Q}\rightarrow \mathbb{R}$
$x=p^{w(x)}u \mapsto c^{w(x)}$.
Show that $\theta_p$ is a p-Norm.

How could I show this? (Thinking)

Hi evinda,

What do you mean by a $p$-Norm? Are you claiming that $\theta_p$ satisfies the strong triangle inequality $\theta_p(x + y) \le \max\{\theta_p(x), \theta_p(y)\}$?

 

[evinda]

Hi evinda,

What do you mean by a $p$-Norm? Are you claiming that $\theta_p$ satisfies the strong triangle inequality $\theta_p(x + y) \le \max\{\theta_p(x), \theta_p(y)\}$?

According to my notes:

A $p-$ norm of $\mathbb{Q}_p$ is a function $||_p: \mathbb{Q}_p \to \mathbb{R}$

$$x \neq 0, x=p^{w_p(x)}u \mapsto p^{-w_p(x)}$$

$$\text{ For } x=0 \Leftrightarrow w_p(x)=\infty \\ p^{-\infty}:=0$$

Identities:

1. $$|x|_p \geq 0 \text{ and } |x|_p=0 \Leftrightarrow x=0$$
2. $$|xy|_p=|x|_p |y|_p$$
3. $$|x+y|_p \leq \max \{ |x|_p, |y|_p\} \\ \leq |x|_p+|y|_p$$
   If $|x|_p \neq |y|_p \Rightarrow |x+y|_p=\max \{ |x|_p, |y|_p\}$
4. $$\mathbb{Z}_p=\{ x \in \mathbb{Q}_p | |x|_p \leq 1 \}$$
5. $$\mathbb{Z}_{p}^*=\{ x \in \mathbb{Z}_p | |x|_p=1 \}$$

 

[mathbalarka]

I don't even understand the question. If you are defining $p$-norm by the usual $p$-adic norm, how does it make sense to say that some other function $x \mapsto c^{\nu_p(x)}$ is a $p$-norm?

In any case, it is quite straightforward that $\theta_p : x \mapsto c^{\nu_p(x)}$ satisfies the strong triangle ineq : we know that $\nu_p(x + y) \geq \text{min}(\nu_p(x), \nu_p(y))$. Thus $\theta_p(x + y) = c^{\nu_p(x + y)} \leq c^{\min(\nu_p(x), \nu_p(y))} = \text{min}(c^{\nu_p(x)}, c^{\nu_p(y)}) = \text{min}(\theta_p(x), \theta_p(y))$ $\blacksquare$

 

[evinda]

I don't even understand the question. If you are defining $p$-norm by the usual $p$-adic norm, how does it make sense to say that some other function $x \mapsto c^{\nu_p(x)}$ is a $p$-norm?

In any case, it is quite straightforward that $\theta_p : x \mapsto c^{\nu_p(x)}$ satisfies the strong triangle ineq : we know that $\nu_p(x + y) \geq \text{min}(\nu_p(x), \nu_p(y))$. Thus $\theta_p(x + y) = c^{\nu_p(x + y)} \leq c^{\min(\nu_p(x), \nu_p(y))} = \text{min}(c^{\nu_p(x)}, c^{\nu_p(y)}) = \text{min}(\theta_p(x), \theta_p(y))$ $\blacksquare$

So, do I not have to show that $\theta_p$ satisfies all of the above $5$ identities? (Thinking)

 

[mathbalarka]

So, do I not have to show that $\theta_p$ satisfies all of the above $5$ identities? (Thinking)

I don't know. As I said, I don't understand your question. $\theta_p$ is certainly not a $p$-adic norm. So clarify what a $p$-norm is first.

 

[evinda]

I don't know. As I said, I don't understand your question. $\theta_p$ is certainly not a $p$-adic norm. So clarify what a $p$-norm is first.

The exercise asks to show that $\theta_p$ is a $p-$ norm.

The definition of $p-$norm, that is given in my notes, is the one I wrote in post #3.

That's why I thought, that I should show that $\theta_p$ satisfies the $5$ identities.

Do, $\theta_p$ satisfy the identities, indeed? (Thinking)

 

[evinda]

That's what I have tried to prove that $\theta_p$ satisfies the first three identities:

• 
  For $$x \neq 0, x=p^{w(x)}u \mapsto c^{w(x)}>0$$, since $$c>0$$.
  What can I say for $$x=0$$, in order to show that $$|x|_{\theta_p}=0$$?
  
• $$x=p^{w(x)}u_1, y=p^{w(y)}u_2, u_1, u_2 \in \mathbb{Z}_p^*$$
  
  $$|x|_{\theta_p}=c^{w(x)}$$
  
  $$|y|_{\theta_p}=c^{w(y)}$$
  
  $$|xy|_{\theta_p}=|p^{w(x)+w(y)} u_1u_2|_{\theta_p}=c^{w(x)+w(y)}=c^{w(x)} \cdot c^{w(y)}=|x|_{\theta_p} \cdot |y|_{\theta_p}$$
  
• $$x+y=p^{w(x)}u_1+p^{w(y)}u_2$$
  
  Let $$w(x)>w(y) \Rightarrow |x|_{\theta_p} \neq |y|_{\theta_p}$$
  
  Then, $$x+y=p^{w(x)}[p^{w(x)-w(y)}u_1+u_2]$$
  
  Therefore, $$|x+y|_{\theta_p}=c^{w(y)}=|y|_p=\max \{ |x|_p, |y|_p\}$$
  
  If $$w(x)=w(y) \Rightarrow |x|_{\theta_p}=|y|_{\theta_p}$$.
  
  Then,
  
  $$x+y=p^{w(y)}[u_1+u_2]$$
  
  $$|x+y|_{\theta_p}=|p^{w(y)}(u_1+u_2)|_{\theta_p} \leq c^{w(y)}=\max \{ |x|_{\theta_p}, |y|_{\theta} \}$$

What can I do, to show that $$|x|_{\theta_p}=0$$, when $$x=0$$?
Also, is that what I have tried right or have I done something wrong?