# Thread: is this true for all norm on Rn

1. ## is this true for all norm on Rn

Consider a norm on Rn.

there are two vectors x,y in Rn
x=(x1,x2,x3,...,xn)
y=(y1,y2,...,yn)

if, y1>x1,
y2>x2,
y3>x3,
.
.
.
yn>xn
.
is it true that norm : ||y||>= ||x|| for any norm on Rn?

2. Originally Posted by szpengchao
Consider a norm on Rn.

there are two vectors x,y in Rn
x=(x1,x2,x3,...,xn)
y=(y1,y2,...,yn)

if, y1>x1,
y2>x2,
y3>x3,
.
.
.
yn>xn
.
is it true that norm : ||y||>= ||x|| for any norm on Rn?
consider $\| y \|^2$ how does it compare to $\| x \|^2$? what does that tell you?

3. ## not defined

||y|| is not defined as \sqrt( y1^2+y2^2...+yn^2)

so how can u compare ||y||,||x||

4. Originally Posted by szpengchao
||y|| is not defined as \sqrt( y1^2+y2^2...+yn^2)

so how can u compare ||y||,||x||
yes, i know what the norm is defined as. i asked you to compare $\| y \|^2$ with $\| x \|^2$ (we can do this since we were given comparisons component wise).

anyway, just make a counter example, it shouldn't be hard. consider vectors in $\mathbb{R}^2$ given by

$\vec y = \left< 1,1 \right>$ and $\vec x = \left< -5, -5 \right>$

do those vectors satisfy the conditions given? is $\| y \| \ge \| x \|$ ?

5. ## .

i m sorry. forgot one condition:

y1,y2,...,>0
x1,x2,...,>0

is there any counterexample ?

6. Originally Posted by szpengchao
i m sorry. forgot one condition:

y1,y2,...,>0
x1,x2,...,>0

is there any counterexample ?
that's a big condition to forget

in that case, note that $y_i > x_i \implies y_i^2 > x_i^2$ for $1 \le i \le n$

now go back and do what i told you to do before

7. ## no idea

no idea. i said the norm is not given as:

\sqrt( y1^2+y2^2+...+yn^2)

8. Originally Posted by szpengchao
no idea. i said the norm is not given as:

\sqrt( y1^2+y2^2+...+yn^2)
and if you square that, what do you get?

9. ## i dont know

i dont know

10. Originally Posted by szpengchao
i dont know
you don't know what, say, $(\sqrt{a})^2$ is? what if i asked you to find $(\sqrt{3})^2$, you wouldn't be able to tell me?

11. ## ...

i know what you mean, but i m not talking about euclidean norm.

12. Originally Posted by szpengchao
i know what you mean, but i m not talking about euclidean norm.
we are dealing with the Euclidean n-space, why do you think this is not the Euclidean norm. and even if not, how does that matter. we have a definition with a square root in it and i asked you to square it. how does it matter what kind of definition it is?

13. ## The question is

The question is about a norm on Rn, but that norm needs not to be Euclidean. That argument is obviously true if it is euclidean norm.
But there are other norms on Rn, i just want to find a counterexample.

14. Originally Posted by szpengchao
The question is about a norm on Rn, but that norm needs not to be Euclidean. That argument is obviously true if it is euclidean norm.
But there are other norms on Rn, i just want to find a counterexample.
you told me the definition of norm you are using. it is the default. we take that definition unless another one is specified

15. $\begin{array}{ccl} \|y \|^2 & = & \left( \sqrt{y_1^2 + y_2^2 + \cdots + y_n^2} \right)^2 \\
& & \\
& = & y_1^2 + y_2^2 + \cdots + y_n^2 \\
& & \\
& \ge & x_1^2 + x_2^2 + \cdots + x_n^2 \\
& & \\
& = & \| x \|^2 \end{array}$

the result follows by taking the square roots of both sides