Norm Verification for Vectors in R^2

[bugatti79]

Homework Statement

Consider the vector space R^2, Decide whether or not the following are norms defined on R^2. If they are, verify all axioms of a norm, if not, demonstrate by counter example some axioms which fails. For x=(x_1,x_2) in R^2

Homework Equations

(i) || ||_#: R^2 defined by ||x||_#=|x_1|+2|x_2|

(ii) || ||_3:R^2 defined by ||x||=3|x_1|

The Attempt at a Solution

How do I decide whether or not the above are norms on R^2?

 

[CompuChip]

What is the definition of a norm?

 

[bugatti79]

What is the definition of a norm?

Definition of a norm is a norm on a vector space V is a function || ||: V m to R satisfying for x,y in V and for all scalar alpha the following

1) ||x||>=0
2)||x||=0 iff x=0
3)||aX||=|a|||x||
4) The triangle in equality

But what do I need to do to decide whether or not they are norms?

 

[Mark44]

Does each of your two norms satisfy all four requirements in the definition above? If so, it's a norm. If not, it isn't.

 

[bugatti79]

Does each of your two norms satisfy all four requirements in the definition above? If so, it's a norm. If not, it isn't.

but how does one check this?

I mean we don't know the value of x1 and x2, they could both be easily less than 1 and hence it wouldn't satisfy axiom 1?
The value of x1 and x2 could both be 0 and then axiom 2 is satisfied.

I don't know how to argue axiom 3 and 4...?

If it is a norm, how does one 'verify' it?

thanks

 

[bugatti79]

Any tips on this guys? Thanks

 

[Fredrik]

but how does one check this?

What do you mean? You just do. Say something like "let x be an arbitrary member of ℝ²" and then use the definition of the function that may or may not be a norm to see if it satisfies the first condition. If it's satisfied, then you move on to the second condition. If it looks like it's not, you try to find an example that proves that it's not satisfied.

I mean we don't know the value of x1 and x2, they could both be easily less than 1 and hence it wouldn't satisfy axiom 1?

I don't know what you're talking about here. Both of the functions you mentioned in your post very obviously satisfy axiom 1.

The value of x1 and x2 could both be 0 and then axiom 2 is satisfied.

Axiom 2 is a "for all" statement, so you can't test it with one specific value of x, unless you're trying to prove that it's not satisfied. Note that all of the axioms are "for all" statements.

 

[HallsofIvy]

Why would being "less than 1" contradict "greater than or equal to 0"?

 

[bugatti79]

but how does one check this?

I mean we don't know the value of x1 and x2, they could both be easily less than 0 and hence it wouldn't satisfy axiom 1?
The value of x1 and x2 could both be 0 and then axiom 2 is satisfied.


thanks

Why would being "less than 1" contradict "greater than or equal to 0"?

Sorry, this is a typo, I meant the value of x could be less than 0 not 1...and then it would contradict axoim 1..?

 

[Fredrik]

Sorry, this is a typo, I meant the value of x could be less than 0 not 1...and then it would contradict axoim 1..?

Assuming that you made another mistake here, and meant x_1 and x_2 when you said "x", then no, it wouldn't, because of the absolute value signs in the definitions. If you had been working with another definition, one that implies that there's an x such that $\|x\|<0$, then the axiom is violated, but that doesn't make it harder to check if the axioms are satisfied. It makes it easier, because then all you have to do is to find that x, show that $\|x\|<0$, and you're done.

 

[bugatti79]

Homework Statement

Consider the vector space R^2, Decide whether or not the following are norms defined on R^2. If they are, verify all axioms of a norm, if not, demonstrate by counter example some axioms which fails. For x=(x_1,x_2) in R^2

Homework Equations

(i) || ||_#: R^2 defined by ||x||_#=|x_1|+2|x_2|

(ii) || ||_3:R^2 defined by ||x||=3|x_1|

The Attempt at a Solution

How do I decide whether or not the above are norms on R^2?

Definition of a norm is a norm on a vector space V is a function || ||: V m to R satisfying for x,y in V and for all scalar alpha the following

1) ||x||>=0
2)||x||=0 iff x=0
3)||aX||=|a|||x||
4) The triangle in equality

But what do I need to do to decide whether or not they are norms?

Ok here's my attempt,

1) since x=(x1,x2) is in R^2 and we know |x1|>=0 and |x2|>=0 implies ||x||>=0 for all x in R^2 hence N1 holds

2) ||x||=0 iff both x1=0 and x2=0, ie x=(0,0)=0 the zero vector implies N2 holds

3) ||a x || = |a| |x|
=|a|(|x1|+2|x2|)
= |a| |x1| + 2 |a| |x2| (all I am doing is throwing a on both side of equation so of course it holds, right?)

4) The triangle inequality is ||x+y|| <= ||x||+ ||y||
I don't know how to check this one...

Thanks

 

[Fredrik]

1. You can't say "Since x is in R^2" when you haven't even mentioned x before. You need to start with something like "Let x be an arbitrary member of ℝ²". You also need to make it absolutely clear what definitions you're using. I don't see anything that indicates which of the functions you're trying to prove something about.

2. What is x in the iff statement? Why does that statement hold. Which one(s) of the functions are you talking about?

3. This doesn't make any sense. The first equality is what you're trying to prove, except that the right-hand side should say $|a|\,\|x\|$, not |a| |x|. You're obviously not allowed to use what you're trying to prove. The calculation you did after that adds nothing at all. You need to start with $\|ax\|=\dots$, then use the definition, and end it with $\dots=|a|\,\|x\|$.

You must always make it clear if your variables are part of "for all" or "there exists" statements, or if they are defined to represent some specific member of the set you're dealing with. You must always make it absolutely clear how you're using the definitions of the terms or symbols that you're trying to prove something about.

 

[bugatti79]

Homework Statement

Consider the vector space R^2, Decide whether or not the following are norms defined on R^2. If they are, verify all axioms of a norm, if not, demonstrate by counter example some axioms which fails. For x=(x_1,x_2) in R^2


(i) || ||_#: R^2 defined by ||x||_#=|x_1|+2|x_2|

1. You can't say "Since x is in R^2" when you haven't even mentioned x before. You need to start with something like "Let x be an arbitrary member of ℝ²". You also need to make it absolutely clear what definitions you're using. I don't see anything that indicates which of the functions you're trying to prove something about.

2. What is x in the iff statement? Why does that statement hold. Which one(s) of the functions are you talking about?

3. This doesn't make any sense. The first equality is what you're trying to prove, except that the right-hand side should say $|a|\,\|x\|$, not |a| |x|. You're obviously not allowed to use what you're trying to prove. The calculation you did after that adds nothing at all. You need to start with $\|ax\|=\dots$, then use the definition, and end it with $\dots=|a|\,\|x\|$.

You must always make it clear if your variables are part of "for all" or "there exists" statements, or if they are defined to represent some specific member of the set you're dealing with. You must always make it absolutely clear how you're using the definitions of the terms or symbols that you're trying to prove something about.

I am referring to part (i). I am not sure what the subscript 'hatch' or 'star' means

1) Since we are given that x=(x1,x2) is in R^2 and we have that ||x||_#=|x1|+2|x2|, then ||x||>=0 because we have that |x1| and |x2| are both >=0 implies ||x||>=0 for all x in R

2)||x||_#=0? We have ||x||=|x1|+2|x2|=0 iff |x1|=0 and |x2|=0 for all x in R

3) ||ax||_#=...? I didn't know one was not allowed use what we are trying to prove. Then what else can we do?

4) ?

Thanks

 

[Fredrik]

I am not sure what the subscript 'hatch' or 'star' means

It's just a notation for that specific function.

1) Since we are given that x=(x1,x2) is in R^2 and we have that ||x||_#=|x1|+2|x2|, then ||x||>=0 because we have that |x1| and |x2| are both >=0 implies ||x||>=0 for all x in R

This is OK, but don't say that you're given that x is in $R^2$. You're not given an x, or any variable at all. You're just supposed to prove a statement of the form "For all v in $R^2$, we have P(v)", where P(v) is a statement that involves v. In this case, P(v) is the statement "$\|v\|\geq 0$". This is how I'd do it:

Let $x\in\mathbb R^2$ be arbitrary. By definition of $\|\ \|_{\#}$, we have
$$\|x\|_{\#}=\underbrace{|x_1|}_{\geq 0}+2\underbrace{|x_2|}_{\geq 0}\geq 0.$$

2)||x||_#=0? We have ||x||=|x1|+2|x2|=0 iff |x1|=0 and |x2|=0 for all x in R

This is OK too. It's sufficient to say that since $\|x\|_{\#}=|x_1|+2|x_2|$, it's obvious that $\|x\|_{\#}=0$ if and only if $x=0$.

3) ||ax||_#=...? I didn't know one was not allowed use what we are trying to prove. Then what else can we do?

I answered that in my previous post, and I'm doing it again below.

4) ?

Same idea as in all of these problems. Start by saying "Let $x\in\mathbb R^2$ be arbitrary". Then write down the left-hand side of the equality or inequality you want to prove. Then use the definition of $\|\ \|_{\#}$ and things you know about $\mathbb R^2$ until you end up with the right-hand side of the result you want to prove.

 

[bugatti79]

It's just a notation for that specific function. This is OK, but don't say that you're given that x is in $R^2$. You're not given an x, or any variable at all. You're just supposed to prove a statement of the form "For all v in $R^2$, we have P(v)", where P(v) is a statement that involves v. In this case, P(v) is the statement "$\|v\|\geq 0$". This is how I'd do it:

Let $x\in\mathbb R^2$ be arbitrary. By definition of $\|\ \|_{\#}$, we have
$$\|x\|_{\#}=\underbrace{|x_1|}_{\geq 0}+2\underbrace{|x_2|}_{\geq 0}\geq 0.$$

This is OK too. It's sufficient to say that since $\|x\|_{\#}=|x_1|+2|x_2|$, it's obvious that $\|x\|_{\#}=0$ if and only if $x=0$. I answered that in my previous post, and I'm doing it again below.Same idea as in all of these problems. Start by saying "Let $x\in\mathbb R^2$ be arbitrary". Then write down the left-hand side of the equality or inequality you want to prove. Then use the definition of $\|\ \|_{\#}$ and things you know about $\mathbb R^2$ until you end up with the right-hand side of the result you want to prove.

3) Let $x \in \mathbb R^2.$ By defintion we have $\|x \|_{\#} = \|x_1\|+2\|x_2\|$

$\|a x \|_{\#}=\|ax_1\| +2 \|ax_2\|=|a| ( |x_1|+2|x_2|)=|a| \| x \|_{\#} \forall a, x \in \mathbb R^2$ N3 holds

4) Let $y \in \mathbb R^2$ where $\|y\|_{\#}=|y_1|+2|y_2|$

Therefore $\|x+y\|= |x_1|+2|x_2| + |y_1||+2|y_2|$

By defintion of the inequality $\|x+y\| \le \|x\|+\|y\|$ we have

$\|x+y\|_{\#} \le |x_1|+2|x_2|+|y_1|+2|y_2|$
$$ \le \|x\|_{\#}+\|y\|_{\#} \forall x,y \in \mathbb R^2$$ N4 holds.....?

Tackle part 2 in same way?

Thanks

 

[Fredrik]

3) Let $x \in \mathbb R^2.$ By defintion we have $\|x \|_{\#} = \|x_1\|+2\|x_2\|$

The first sentence should say "Let $x\in\mathbb R^2$ be arbitrary". The definition of $\|\ \|_{\#}$ tells us that $\|x\|_{\#}=|x_1|+2|x_2|$.

$\|a x \|_{\#}=\|ax_1\| +2 \|ax_2\|=|a| ( |x_1|+2|x_2|)=|a| \| x \|_{\#} \forall a, x \in \mathbb R^2$ N3 holds

This is mostly OK. The norm symbols after the first equality sign should of course be absolute value signs. a is not a member of $\mathbb R^2$. This is what you should have said: For all $a\in\mathbb R$ and all $x\in\mathbb R^2$, we have
$$\|ax\|_{\#}=\|(ax_1,ax_2)\|_{\#}=|ax_1|+2|ax_2|=|a|(|x_1|+2|x_2|)=|a|\|x\|_{\#}.$$ Note that the "for all" statement makes it unnecessary to also say "Let ... be arbitrary". The point of "let ... be arbitrary" statements is just that they enable us to break up a long sentence into several short sentences.

4) Let $y \in \mathbb R^2$ where $\|y\|_{\#}=|y_1|+2|y_2|$

Therefore $\|x+y\|= |x_1|+2|x_2| + |y_1||+2|y_2|$

This is not what the definition of $\|\ \|_{\#}$ says. Try again. The proof should start like this: For all $x,y\in\mathbb R^2$, we have
$$\|x+y\|_{\#}=\dots$$ Then you just use the definitions of x+y and $\|\ \|_{\#}$, and whatever else you need.

 

[bugatti79]

The first sentence should say "Let $x\in\mathbb R^2$ be arbitrary". The definition of $\|\ \|_{\#}$ tells us that $\|x\|_{\#}=|x_1|+2|x_2|$.


This is mostly OK. The norm symbols after the first equality sign should of course be absolute value signs. a is not a member of $\mathbb R^2$. This is what you should have said: For all $a\in\mathbb R$ and all $x\in\mathbb R^2$, we have
$$\|ax\|_{\#}=\|(ax_1,ax_2)\|_{\#}=|ax_1|+2|ax_2|=|a|(|x_1|+2|x_2|)=|a|\|x\|.$$ Note that the "for all" statement makes it unnecessary to also say "Let ... be arbitrary". The point of "let ... be arbitrary" statements is just that they enable us to break up a long sentence into several short sentences.


This is not what the definition of $\|\ \|_{\#}$ says. Try again. The proof should start like this: For all $x,y\in\mathbb R^2$, we have
$$\|x+y\|_{\#}=\dots$$ Then you just use the definitions of x+y and $\|\ \|_{\#}$, and whatever else you need.

Ok, thanks for the tips.

4) $\forall x,y \in \mathbb R^2$ we have $x+y = x_1+2x_2 +y_1 +2y_2$ therefore

$\|x+y|\_{\#}= |x_1|+2|x_2| +|y_1|+2|y_2|$ but by defintion of triangle inequality we have

$\|x+y\|_{\#} \le |x_1|+2|x_2|+|y_1|+2|y_2| \le \|x\|_{\#}+\|y\|_{\#}$...?

 

[Fredrik]

4) $\forall x,y \in \mathbb R^2$ we have $x+y = x_1+2x_2 +y_1 +2y_2$

Surely, you must have meant to put something else on the right-hand side. The left-hand side is a member of ℝ², and the right-hand side a member of ℝ.

$\|x+y|\_{\#}= |x_1|+2|x_2| +|y_1|+2|y_2|$

This is wrong. The right-hand side is equal to $\|x\|_{\#}+\|y\|_{\#}$, but it's certainly not true that $\|x+y\|_{\#}=\|x\|_{\#}+\|y\|_{\#}$ for all x,y in ℝ². For example, when x=(0,1) and y=(0,-1), we have $\|x+y\|_{\#}=0$ and $\|x\|_{\#}+\|y\|_{\#}=4$.

 

[bugatti79]

Surely, you must have meant to put something else on the right-hand side. The left-hand side is a member of ℝ², and the right-hand side a member of ℝ.This is wrong. The right-hand side is equal to $\|x\|_{\#}+\|y\|_{\#}$, but it's certainly not true that $\|x+y\|_{\#}=\|x\|_{\#}+\|y\|_{\#}$ for all x,y in ℝ². For example, when x=(0,1) and y=(0,-1), we have $\|x+y\|_{\#}=0$ and $\|x\|_{\#}+\|y\|_{\#}=4$.

4) $\forall x,y \in \mathbb R^2$ and $x_1, y_1 \in \mathbb R$

we have $x+y = x_1+2x_2 +y_1 +2y_2$ ...?

Ok,so basically this question is not a norm in R^2 because axiom 4 does not hold. You have demonstrated that by counter example. but how do you know that axoim 4 does not hold without having to play around with actual values of x and y to test it?

 

[micromass]

Ok,so basically this question is not a norm in R^2 because axiom 4 does not hold. You have demonstrated that by counter example. but how do you know that axoim 4 does not hold without having to play around with actual values of x and y to test it?

No, Fredrik gave a counterexample to the equality. That doesn't mean that axiom 4 doesn't hold.

 

[micromass]

we have $x+y = x_1+2x_2 +y_1 +2y_2$ ...?

Are you just guessing now??

 

[Fredrik]

we have $x+y = x_1+2x_2 +y_1 +2y_2$ ...?

No, we don't. The left-hand side isn't even in the same set as the right-hand side (as I have already explained).

Ok,so basically this question is not a norm in R^2 because axiom 4 does not hold. You have demonstrated that by counter example.

As micromass said, I only proved that it's not the case that $\|x+y\|_{\#}=\|x\|_{\#}+\|y\|_{\#}$ for all $x,y\in\mathbb R^2$. You still have to determine if $\|x+y\|_{\#}\leq\|x\|_{\#}+\|y\|_{\#}$ for all $x,y\in\mathbb R^2$. (Since 0≤4, my choice of x and y doesn't contradict this inequality).

 

[HallsofIvy]

4) $\forall x,y \in \mathbb R^2$ and $x_1, y_1 \in \mathbb R$

we have $x+y = x_1+2x_2 +y_1 +2y_2$ ...?

If you mean that x= &lt;x_1, x_2&gt; and y= &lt;y_1, y_2&gt; then you should say that. Also, the result would be x+ y= &lt;x_1+ 2y_1, x_2+ 2y_2&gt;.

Ok,so basically this question is not a norm in R^2 because axiom 4 does not hold. You have demonstrated that by counter example. but how do you know that axoim 4 does not hold without having to play around with actual values of x and y to test it?

 

[jgens]

The first sentence should say "Let $x\in\mathbb R^2$ be arbitrary".

The first sentence was fine as it was.

 

[bugatti79]

Are you just guessing now??

No, I was just building up the equations bit by bit and then convert it into a norm (putting it inside the vertical brackets as shown on the following line of that thread) and then test it. Obviously I am wrong.

No, we don't. The left-hand side isn't even in the same set as the right-hand side (as I have already explained).
As micromass said, I only proved that it's not the case that $\|x+y\|_{\#}=\|x\|_{\#}+\|y\|_{\#}$ for all $x,y\in\mathbb R^2$. You still have to determine if $\|x+y\|_{\#}\leq\|x\|_{\#}+\|y\|_{\#}$ for all $x,y\in\mathbb R^2$. (Since 0≤4, my choice of x and y doesn't contradict this inequality).

"we have $x+y = x_1+2x_2 +y_1 +2y_2$ ...?"
I don't understand why the LHS would not be = the RHS when I specify for the LHS that x,y are in R^2 and for the RHS x1 and y1 are in R.
I have an example where the triangle inequality is tested for a taxi cab norm ie || ||_1 and I understand it but I don't know how to do it for a || ||_#...?

 

[Mark44]

"we have $x+y = x_1+2x_2 +y_1 +2y_2$ ...?"
I don't understand why the LHS would not be = the RHS when I specify for the LHS that x,y are in R^2 and for the RHS x1 and y1 are in R.

Because you can't compare two things that aren't in the same space. A vector in R² can't possibly be equal to a scalar. The two things aren't comparable.

I have an example where the triangle inequality is tested for a taxi cab norm ie || ||_1 and I understand it but I don't know how to do it for a || ||_#...?

 

[Fredrik]

The first sentence was fine as it was.

Yes, it's OK to say "Let $x\in\mathbb R^2$" instead of the more precise "Let $x\in\mathbb R^2$ be arbitrary" or the even more precise "Let x be an arbitrary member of $\mathbb R^2$". When the proof is written by someone who understands proofs, and is read by someone who understands proofs, the statement "Let $x\in\mathbb R^2$" will be interpreted as what it really is: A way to break up a long "for all" statement into several sentences. If you had started a proof with "Let $x\in\mathbb R^2$" in a thread where you're discussing something with me, I wouldn't have said anything.

However, a person who doesn't understand these things, can, in my opinion, never put too much effort into improving the clarity of the proofs he's writing. I suggested that he should change the opening sentence because I strongly doubted that he understood the significance of what he wrote. In my experience, one of the most common mistakes that people make when they're just getting started with proofs is that that they fail to make it clear if their variables are part of "for all" or "there exists" statements.

 

[Fredrik]

"we have $x+y = x_1+2x_2 +y_1 +2y_2$ ...?"
I don't understand why the LHS would not be = the RHS when I specify for the LHS that x,y are in R^2 and for the RHS x1 and y1 are in R.

An equality sign means that what you have on the left is the same thing as what you have on the right, not just that they're related in some unspecified way.

Maybe an example can make it even clearer. When x=(1,1), y=(1,0). We have x+y=(2,1) and $x_1+2x_2+y_1+2y_2=1+2·1+1+2·0=4$. So in this case, your equality says (2,1)=4. This is obviously incorrect. Even if x and y have values such that the equality says (0,0)=0, where each 0 denotes the additive identity (the "zero") of ℝ, the equality is still incorrect, because even though (0,0) and 0 are both the additive identities of the vector spaces they are members of (ℝ² and ℝ) respectively, they are members of different sets. (0,0) is an ordered pair of real numbers, while 0 is a real number, so you can never put an equality sign between them. (It is however OK to write (0,0)=0 if it's clear from the context that the 0 on the left denotes the additive identity of ℝ and the 0 on the right denotes the additive identity of ℝ²).

I don't know how to do it for a || ||_#...?

I have already told you:

The proof should start like this: For all $x,y\in\mathbb R^2$, we have
$$\|x+y\|_{\#}=\dots$$ Then you just use the definitions of x+y and $\|\ \|_{\#}$, and whatever else you need.

If you keep asking yourself "how is this thing defined" at each step, you shouldn't have any problems with this. You can't just write things down that have nothing to do with the definitions. When you prove something, you always have to look for ways to use the definitions. A "proof" that doesn't use the definitions of the terms and symbols that are part of the statement is never correct.

 

[bugatti79]

I have already told you:
If you keep asking yourself "how is this thing defined" at each step, you shouldn't have any problems with this. You can't just write things down that have nothing to do with the definitions. When you prove something, you always have to look for ways to use the definitions. A "proof" that doesn't use the definitions of the terms and symbols that are part of the statement is never correct.

I have mulled over all the threads and I'm afraid I don't see any light. I don't know how to proceed further. Its simple I know but I don't see it...

 

[Fredrik]

What is the first step that you're not 100% sure about? Do you know the definition of x+y? If yes, then why not use it? Then use the definition of $\|\ \|_{\#}$. After that you're almost done.

 

[bugatti79]

What is the first step that you're not 100% sure about? Do you know the definition of x+y? If yes, then why not use it? Then use the definition of $\|\ \|_{\#}$. After that you're almost done.

I don't know how to write the definition of x+y apart from what I gave originally..thanks for your patience.

 

[Mark44]

4) $\forall x,y \in \mathbb R^2$ and $x_1, y_1 \in \mathbb R$
we have $x+y = x_1+2x_2 +y_1 +2y_2$ ...?

So, x and y are elements of R², meaning that you can write them as
x = <x₁, x₂>
y = <y₁, y₂>

Then what is x + y? It is NOT as you wrote above.

I don't know how to write the definition of x+y apart from what I gave originally..thanks for your patience.

 

[Fredrik]

I don't know how to write the definition of x+y apart from what I gave originally..thanks for your patience.

I don't recall seeing a definition of x+y in this thread.

Edit: If you meant the equality $x+y = x_1+2x_2 +y_1 +2y_2$ (which is extremely incorrect, since the left-hand side is a member of a different set than the right-hand side), then no, this looks nothing at all like the definition of x+y. It looks like it was inspired by the definition of $\|\ \|_{\#}$, but I see nothing in it that reminds me of the definition of x+y.

I'm going to bed early tonight, but I'm sure Mark44 or someone else can answer your next post.

 

[bugatti79]

I don't recall seeing a definition of x+y in this thread.

Edit: If you meant the equality $x+y = x_1+2x_2 +y_1 +2y_2$ (which is extremely incorrect, since the left-hand side is a member of a different set than the right-hand side), then no, this looks nothing at all like the definition of x+y. It looks like it was inspired by the definition of $\|\ \|_{\#}$, but I see nothing in it that reminds me of the definition of x+y.

I'm going to bed early tonight, but I'm sure Mark44 or someone else can answer your next post.

Yes, you are correct. Ok, thanks.

It is $\|x+y\|_\#=...$ that I should be focusing on...

 

[Fredrik]

Yes, but before you focus on $\|x+y\|_{\#}$, you should focus on x+y. Do you understand what the notation $\mathbb R^2$ means? What sort of objects are members of $\mathbb R^2$. How do you add one of those objects to another? Do you at least understand that the notation x+y is completely meaningless until it has been defined? What is the standard definition of x+y for $x,y\in\mathbb R^2$?

 

[Mark44]

To elaborate on what Fredrik said, how is addition usually defined for vectors in R²? That's basically what he's asking about for the expression x + y. This has nothing to do with norms.

 

[bugatti79]

To elaborate on what Fredrik said, how is addition usually defined for vectors in R²? That's basically what he's asking about for the expression x + y. This has nothing to do with norms.

Yes, but before you focus on $\|x+y\|_{\#}$, you should focus on x+y. Do you understand what the notation $\mathbb R^2$ means? What sort of objects are members of $\mathbb R^2$. How do you add one of those objects to another? Do you at least understand that the notation x+y is completely meaningless until it has been defined? What is the standard definition of x+y for $x,y\in\mathbb R^2$?

$\|x+y\| \le \|x_1+y_1\|+ \|2x_2 +2y_2\|$

$\le \|x_1+y_1\|+2\|x_2+y_2\|$

If the value of y is 0, isn't it still possible that the inequality holds? ie, the LHS can still be less than the RHS ie

$\|x\| \le \|x_1\|+ \|2x_2 \|$

So I am saying that axiom 4 holds. Is this how one examines it? Thanks

 

[micromass]

$\|x+y\| \le \|x_1+y_1\|+ \|2x_2 +2y_2\|$

$\le \|x_1+y_1\|+2\|x_2+y_2\|$

If the value of y is 0, isn't it still possible that the inequality holds? ie, the LHS can still be less than the RHS ie

$\|x\| \le \|x_1\|+ \|2x_2 \|$

So I am saying that axiom 4 holds. Is this how one examines it? Thanks

No. What you wrote makes no sense.

We asked you several times, how is x+y defined??

Let x=(x_1,y_1) and y=(y_1,y_2). Then how did we define

(x_1,x_2)+(y_1,y_2)

?

 

[Fredrik]

$\|x+y\| \le \|x_1+y_1\|+ \|2x_2 +2y_2\|$

$\le \|x_1+y_1\|+2\|x_2+y_2\|$

Try doing just one thing at a time. For all $x,y\in\mathbb R^2$, we have
$$\|x+y\|_{\#}=...$$ The first step is to use the definition of + to rewrite x+y as an ordered pair of real numbers. The second step is to use the definition of $\|\ \|_{\#}$. Then there's a third step. Don't try to do them all at once. Do them one at a time.

You also need to think about when to use the notation |something| and when to use the notation $\|\text{something}\|_{\#}$. Also, in each step, use an equality sign if what you have on the left is the same thing as what you have on the right. Use ≤ only if you have reason to think that the thing on the right may be greater than the thing on the left.

 

[Mark44]

In addition, in this problem ||x + y|| is completely irrelevant. The norm you appear to be working with is ||x + y||_#. Use the definition of this norm to first say what this expression equals, and then work on showing that the triangle inequality holds.

 

[bugatti79]

In addition, in this problem ||x + y|| is completely irrelevant. The norm you appear to be working with is ||x + y||_#. Use the definition of this norm to first say what this expression equals, and then work on showing that the triangle inequality holds.

Try doing just one thing at a time. For all $x,y\in\mathbb R^2$, we have
$$\|x+y\|_{\#}=...$$ The first step is to use the definition of + to rewrite x+y as an ordered pair of real numbers. The second step is to use the definition of $\|\ \|_{\#}$. Then there's a third step. Don't try to do them all at once. Do them one at a time.

You also need to think about when to use the notation |something| and when to use the notation $\|\text{something}\|_{\#}$. Also, in each step, use an equality sign if what you have on the left is the same thing as what you have on the right. Use ≤ only if you have reason to think that the thing on the right may be greater than the thing on the left.

$x+y=(x_1+y_1,2x_2+2y_2)$?

 

[Mark44]

$x+y=(x_1+y_1,2x_2+2y_2)$?

NO! Do you not understand how to add two vectors together? You should not be attempting to work problems about norms if you don't understand the basics of vector operations.

 

[Fredrik]

$x+y=(x_1+y_1,2x_2+2y_2)$?

No, this is wrong. Note that we're just doing addition here. Those functions that may or may not be norms have nothing to do with it.

 

[micromass]

Bugatti, I think we could help you better if you would explain us your situation. What course are you taking now? What courses did you already take??
I'm asking this because I feel you miss some preliminary knowledge. We can help you rectify it by suggesting things you should look at.

 

[bugatti79]

NO! Do you not understand how to add two vectors together? You should not be attempting to work problems about norms if you don't understand the basics of vector operations.

If v=(v1,v2) and w=(w1,w2) then v+w=(v1+w1,v2+w2)? Thats all I have done above?

No, this is wrong. Note that we're just doing addition here. Those functions that may or may not be norms have nothing to do with it.

Bugatti, I think we could help you better if you would explain us your situation. What course are you taking now? What courses did you already take??
I'm asking this because I feel you miss some preliminary knowledge. We can help you rectify it by suggesting things you should look at.

I am studying topics in analysis part time so I may be over tired from work. I have done PDE's and Calculus 1,2 last year.

 

[Mark44]

If v=(v1,v2) and w=(w1,w2) then v+w=(v1+w1,v2+w2)? Thats all I have done above?

No, it isn't. You were mixing vector addition with the || ||_# norm, producing a meaningless mush. Here is what I'm referring to.

x+y=x₁ +2x₂ +y₁ +2y₂

 

[bugatti79]

No, it isn't. You were mixing vector addition with the || ||_# norm, producing a meaningless mush. Here is what I'm referring to.

So for x,y in R^2 where x=(x1,x2) and y=(y1,y2)

$x+ y= <x_1+ 2y_1, x_2+ 2y_2>$?

 

[micromass]

So for x,y in R^2 where x=(x1,x2) and y=(y1,y2)

$x+ y= <x_1+ 2y_1, x_2+ 2y_2>$?

No! You wrote it correctly in in your post 45. Why do you keep insisting on the 2?

 

[bugatti79]

No! You wrote it correctly in in your post 45. Why do you keep insisting on the 2?

$x+y=(x_1+y_1, x_2+y_2)$

$\| x+y\|_{\#}= \| |x1|+|y1|, 2|x2|+2|y2| \|_{\#}$...?

 

[Fredrik]

$x+y=(x_1+y_1, x_2+y_2)$

$\| x+y\|_{\#}= \| |x1|+|y1|, 2|x2|+2|y2| \|_{\#}$...?

Here you're adding x and y correctly, but why don't you just insert the result you've found for x+y into $\|x+y\|_{\#}$, and then use the definition of $\|\ \|_{\#}$? What you wrote doesn't make any sense. The function $\|\ \|_{\#}$ takes a member of $\mathbb R^2$ as input. Here you used a real number as input. That doesn't make sense.

Edit: Why is there a comma in there? Did you mean $\|(|x_1|+|y_1|,2|x_2|+2|y_2|)\|_{\#}$? That doesn't make much sense either. I mean, it it's not a nonsense expression, but I have no idea why you would consider an ordered pair with those components.