Are these two functions norms on R²?

[gtfitzpatrick]

Homework Statement

Are the following norms defined on R ^2

1. $$\left\right\|\|$$+ defined by $$\left\right\|\|$$+ = $$\left|x1\right|$$ + $$\left|x2\right|$$

2. $$\left\right\|\|$$4 defined by $$\left\right\|\|$$4 = 4$$\left|x1\right|$$

Homework Equations

The Attempt at a Solution

 

[Fredrik]

You should post the definition of "norm", and your attempt to check if the conditions in the definition of "norm" are satisfied by the functions defined in the problem.

Also, the definitions of the two functions that you posted are incomplete. What do you mean by x1 and x2? Did you mean that $\|\ \|+$ is defined by $\|(x_1,x_2)\|+=|x_1|+|x_2|$?

 

[gtfitzpatrick]

for 1
$$\left|x1\right|$$ + $$\left|x2\right|$$ $$\geq$$ 0 for all x1,x2 $$\in$$ R so axiom 1 holds
for axiom 2 (= 0 ) iff x1 and x2 = 0 so axiom 2 holds
axiom 3 = $$\left|\alpha\right|$$ ($$\left|x1\right|$$ + $$\left|x2\right|$$) so axiom 3 holds
axiom 4 $$\left|x+y\right|$$ = $$\left|x1+y1\right|$$ + $$\left|x2+y2\right|$$ so axiom 4 holds right?

 

[gtfitzpatrick]

all my symbols are getting jumbled up when i save my msg, and some of them arent even there :(

 

[Fredrik]

You need to be more explicit. To verify that what you call "axiom 4" is satisfied, your calculation must start with

$$\|\vec x+\vec y\|_+=\|(x_1,x_2)+(y_1,y_2)\|_+=\|(x_1+y_1,x_2+y_2)\|=|x_1+y_1|+|x_2+y_2|$$

and end with

$$|x_1|+|x_2|+|y_1|+|y_2|=\|\vec x\|_+ +\|\vec y\|_+$$. (Edit: I made a mistake here that I didn't spot until I wrote post #10. I have corrected the mistake now).

and there must be a $\leq$ somwhere in the part of the calculation that connects the first line with the second. Can you show that

$$|x_1+y_1|+|x_2+y_2|\leq |x_1|+|x_2|+|y_1|+|y_2|$$?

all my symbols are getting jumbled up when i save my msg, and some of them arent even there :(

It's a known bug. You need to refresh and resend after each preview.

Edit: You can edit your posts for 12 hours. Looks like the LaTeX in the post below this one needs some work. You don't need that many tex tags. Click the quote button next to one of my posts, and you'll see how I'm doing it. Remember that you need to refresh and resend after each preview.

 

[gtfitzpatrick]

A normed linear space is a vector space together with a function( called the norm)(norm=ideal length of vector)

 

[gtfitzpatrick]

what the heck?

 

[Fredrik]

You can still edit the messed up post. See my comments at the end of post #5.

 

[gtfitzpatrick]

thanks for the help Fredrik. I was using firefox, don't think that was helping either.

$$\|\vec x+\vec y\|_+=\|(x_1,x_2)+(y_1,y_2)\|_+=\|(x_1+y_1,x_2+y_2 )\|=|x_1+y_1|+|x_2+y_2|$$


$$|x_1+y_1|\leq |x_1|+|y_1|$$

and

$$|x_2+y_2|\leq |x_2|+|y_2|$$

so implies


$$|x_1+y_1|+|x_2+y_2|\leq |x_1|+|y_1|+|x_2|+|y_2|$$

$$\leq |x_1|+|x_2|+|y_1|+|y_2|$$
but
$$|x_1|+|x_2|\geq |x_1+x_2|$$
and
$$|y_1|+|y_2|\geq |y_1+y_2|$$


so

$$|x_1+x_2|+|y_1+y_2|=\|\vec x\|_+ +\|\vec y\|_+$$

 

[Fredrik]

I was using firefox, don't think that was helping either.

I'm using Firefox too. It's the most popular browser here (acccording to this thread). The bug is the same in all browsers. I should also mention that if you use itex tags instead of tex tags, some of the larger characters might get their tops cut off.I see now that I made a mistake in #5. I used the definition of $\|\ \|_+$ incorrectly. I just edited #5 to correct it. Now that #5 is correct, I see that the only thing I didn't include in #5 is the triangle inequality for absolute values of real numbers, which you just used in #9.

$$\|\vec x+\vec y\|_+=\|(x_1,x_2)+(y_1,y_2)\|_+=\|(x_1+y_1,x_2+y_2)\|=|x_1+y_1|+|x_2+y_2|$$

$$\leq |x_1|+|y_1|+|x_2|+|y_2|=|x_1|+|x_2|+|y_1|+|y_2|=\|\vec x\|_+ +\|\vec y\|_+$$

 

[gtfitzpatrick]

Thanks fredrik for the help

it would be likewise if i started out with say
$$\|(x_1,x_2)\|+=|x_1|+3|x_2|$$

axiom 1 holds $$\|x_1|+3|x_2|> 0$$

axiom 2 holds $$\|x_1|+3|x_2|= 0$$ iff $$\vec x = \vec0$$

axiom 3 holds $$\|\alpha\vec x\|_+=|\alpha x_1|+ 3|\alpha x_2|=|\alpha| |x_1|+3|\alpha| |x_2|=|\alpha| (|x_1|+3|x_2|) = |\alpha| \|\vec x\|_+$$

axiom 4 holds $$\|\vec x+\vec y\|_+=\|(x_1,x_2)+(y_1,y_2)\|_+=\|(x_1+y_1,x_2+y_2 )\|=|x_1+y_1|+3|x_2+y_2|<br /> \leq |x_1|+|y_1|+3|x_2|+3|y_2|=|x_1|+3|x_2|+|y_1|+3|y_2|=\|\vec x\|_+ +\|\vec y\|_+$$

think this is ok? now i'll give part 2 a go

 

[gtfitzpatrick]

part 2

$$\|\vec x\|_3 = 3|x_1|$$

axiom 1 holds $$3|x_1|> 0$$ if $$x_1 > 0$$

axiom 2 doesnt hold $$3|x_1|= 0$$ if $$x_1 = 0$$ but $$x_2 > 0$$ so its not = 0 iff $$\|\vec x\|_3 = \vec 0$$

think this is ok? its not a norm

 

[Fredrik]

Looks pretty good.

To prove that a function isn't a norm, a single example of a vector that doesn't satisfy an axiom is sufficient. For example, (0,1) violates axiom 2:

$$\|(0,1)\|_3=0,\quad(0,1)\neq\vec 0$$

So there's no need to mention that you checked axiom 1. If there had been a need to show that you verified that axiom 1 holds, you should have said that for any $\vec x$, we have $\|\vec x\|_3=3|x_1|\geq 0$, because the absolute value of any real number is non-negative. The part where you say "if $x_1>0$" is strange and unnecessary.

 

[gtfitzpatrick]

Got it, thanks a million. i guess I've already said its a real number, and abolutle value must be positive.no need to restate it.
Thanks for all the help