Do I have to check all the identities?

Hello!!!

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? :confused:

Reply 1

tombayes

12

Original post by evinda

Hello!!!

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? :confused:

What have you done so far?

Remember: what properties must the p-norm satisfy

(edited 9 years ago)

Reply 2

evinda

OP

1

Original post by tombayes

Remember: what properties must the p-norm satisfy

So, do I have to show that

\theta_p

satisfies the following properties?

•

$|x|_p \geq 0 \text{ and } |x|_p=0 \Leftrightarrow x=0$

•

$|xy|_p=|x|_p |y|_p$

•

$|x+y|_p \leq \max \{ |x|_p, |y|_p\} \\ \leq |x|_p+|y|_p$

If

|x|_p \neq |y|_p \Rightarrow |x+y|_p=\max \{ |x|_p, |y|_p\}

•

$\mathbb{Z}_p=\{ x \in \mathbb{Q}_p | |x|_p \leq 1 \}$

•

$\mathbb{Z}_{p}^*=\{ x \in \mathbb{Z}_p | |x|_p=1 \}$

If so, that's what I have tried to prove the first two identities:

1)For

x>0

:

|x|_{\theta_p}=|p^{w(x)}u|_p=c^{w_{x}}

For

x=0

:

|x|_{\theta_p}=|p^{\infty} u|=c^{\infty}=0

, since

0<c<1

2)

x=p^{w_p(x)}u_1, u_1 \in \mathbb{Z}_p^*

y=p^{w_p(y)}u_2, u_2 \in \mathbb{Z}_p^*

|x|_{\theta_p}=c^{w_p(x)}

|y|_{\theta_p}=c^{w_p(y)}

|xy|_{\theta_p}=|p^{w_p(x)+w_p(y)}u_1u_2|_{\theta_p}=c^{w_p(x)+w_p(y)}=c^{w_p(x)} \cdot c^{w_p(y)}=|x|_{\theta_p} |y|_{\theta_p}

Could you tell me if it is right? And how can I prove the other three identities? :confused:

Reply 3

tombayes

12

Original post by evinda

So, do I have to show that

\theta_p

satisfies the following properties?

•

$|x|_p \geq 0 \text{ and } |x|_p=0 \Leftrightarrow x=0$

•

$|xy|_p=|x|_p |y|_p$

•

$|x+y|_p \leq \max \{ |x|_p, |y|_p\} \\ \leq |x|_p+|y|_p$

If

|x|_p \neq |y|_p \Rightarrow |x+y|_p=\max \{ |x|_p, |y|_p\}

•

$\mathbb{Z}_p=\{ x \in \mathbb{Q}_p | |x|_p \leq 1 \}$

•

$\mathbb{Z}_{p}^*=\{ x \in \mathbb{Z}_p | |x|_p=1 \}$

If so, that's what I have tried to prove the first two identities:

1)For

x>0

:

|x|_{\theta_p}=|p^{w(x)}u|_p=c^{w_{x}}

For

x=0

:

|x|_{\theta_p}=|p^{\infty} u|=c^{\infty}=0

, since

0<c<1

2)

x=p^{w_p(x)}u_1, u_1 \in \mathbb{Z}_p^*

y=p^{w_p(y)}u_2, u_2 \in \mathbb{Z}_p^*

|x|_{\theta_p}=c^{w_p(x)}

|y|_{\theta_p}=c^{w_p(y)}

|xy|_{\theta_p}=|p^{w_p(x)+w_p(y)}u_1u_2|_{\theta_p}=c^{w_p(x)+w_p(y)}=c^{w_p(x)} \cdot c^{w_p(y)}=|x|_{\theta_p} |y|_{\theta_p}

Could you tell me if it is right? And how can I prove the other three identities? :confused:

looks right i think :biggrin:

. But is

p\ge1

because you can have a p-norm with

0<p<1

using a different definition. Also, I thought there were only three properties:
1. the zero vector one you proved
2. the multiplication one you proved
3. the triangle inequality

Are you using a stronger definition?

Reply 4

evinda

OP

1

Original post by tombayes

looks right i think :biggrin:

. But is

p\ge1

because you can have a p-norm with

0<p<1

using a different definition. Also, I thought there were only three properties:
1. the zero vector one you proved
2. the multiplication one you proved
3. the triangle inequality

Are you using a stronger definition?

I think that

p

is a prime, so it must be

p \geq 2

.
In my notes, there are the five identities of the p-norm that I wrote in post #3.

So, to show that

\theta_p

is a p-norm, do I just have to show the three identities you mentioned? :confused:

Reply 5

tombayes

12

Original post by evinda

I think that

p

is a prime, so it must be

p \geq 2

.
In my notes, there are the five identities of the p-norm that I wrote in post #3.

So, to show that

\theta_p

is a p-norm, do I just have to show the three identities you mentioned? :confused:

hang on why is p prime? I suspect we may be talking about different (but similar things) especially if your notes give you more properties.

I was talking about

L_p

space. That said I am no analyst so perhaps I am not the best person to ask.

Do I have to check all the identities?

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you