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Is this a norm?

  1. Sep 2, 2010 #1
    1. The problem statement, all variables and given/known data
    check whether [tex]||.|| : \Re^{2} -> \Re_{+}[/tex] given by

    [tex](x,y) = |x| + |y|^{2}[/tex]

    is a norm on R2


    2. Relevant equations
    For all a in F and all u and v in V,

    1. p(a v) = |a| p(v), (positive homogeneity or positive scalability)
    2. p(u + v) ≤ p(u) + p(v) (triangle inequality or subadditivity).
    3. p(v) = 0 if and only if v is the zero vector (positive definiteness).

    A simple consequence of the first two axioms, positive homogeneity and the triangle inequality, is p(0) = 0 and thus

    p(v) ≥ 0 (positivity).


    3. The attempt at a solution

    I'm going to say it isnt a norm because 1 above does not hold(assume a > 0)
    [tex]||a(x,y)|| = ||(ax,ay)|| = |ax| + |ay|^{2} = a|x| + a^{2}y^{2} \neq a||(x,y)|| = a|x| + a|y|^{2}[/tex]

    is this correct?
     
  2. jcsd
  3. Sep 2, 2010 #2

    Hurkyl

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    Have you proven that [itex]a|x| + a^2 y^2 = a |x| + a |y|^2[/itex] is not an identity?

    (Incidentally, [itex]a|x| + a^2 y^2 \neq a |x| + a |y|^2[/itex] is not an identity either)

    My apologies if this really is obvious to you -- but it is common for people to think two expressions are unequal just because they look different after simplification.
     
  4. Sep 2, 2010 #3
    Well I should say first that I don't exactly get what you are saying (maybe I should say I don't know why I would have to prove that...)

    Would you have asked the same question if I would have wrote this instead(notice absolute value in left hand of expression)
    [tex]a|x| + a^{2}|y|^{2} \neq a|x| + a|y|^{2}[/tex]

    is that statement obvious and clear?
     
  5. Sep 2, 2010 #4
    Okay, I see what I missed now

    That statement can be true if we consider the trivial case (a = 0) and a = 1
     
  6. Sep 2, 2010 #5

    Dick

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    You could just give a counterexample of an (x,y) and a where it's not true. That's enough.
     
  7. Sep 2, 2010 #6
    but it's fair enough to do my route correct?

    say it works for a = 0 and 1 but for no other a which implies it isnt a norm
     
  8. Sep 2, 2010 #7

    Dick

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    If you can prove it only works for a=0 and a=1 independent of |x| and |y|. I think it's easier to just pick x and y to begin with.
     
  9. Sep 3, 2010 #8

    Hurkyl

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    The difference between your work being acceptable and unacceptable is entirely up to how much your professor is willing to assume you understand.

    I wouldn't have given it a second thought if one of my colleagues wrote it. But coming from a student, I would be suspicious they don't get the idea of disproving an alleged identity. But maybe I'm more skeptical than your professor? I don't know.


    (Incidentally, the equation holds for y=0 too)
     
  10. Sep 3, 2010 #9

    ehild

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    Hurkyl, an identity has to hold for all values of the symbols in it. Why has somebody prove that a|x|+a2y2=a|x|+ay2 is not an identity? The definition of norm has to hold for any vectors, that is, ordered (x,y) pairs and for any scalar a. It was a pity to confuse the OP who did a good job. :frown:


    ehild
     
  11. Sep 3, 2010 #10

    Hurkyl

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    That's just the thing -- I don't know that he did. Ignoring the sloppiness with the ordering of the chained equations/inequations, I cannot infer from his posts whether he omitted a proof that [itex]a|x|+a^2 y^2[/itex] and [itex]a|x| + a|y|[/itex] are inequivalent because he know how and thinks it's so obvious it's not worth writing down, or because he really doesn't know there's something he hasn't written down.

    Yes, I agree that his work is good. But I want to know if he really understands it! There is a common mistake (assuming two expressions are inequivalent just because thay look different) and a common misunderstanding here (not knowing proof by counterexample), both of which could be lurking here.


    Because, given the reduction the opening poster correctly performed, "that equation is an identity" is the same thing as "the first part of the definition holds for any vector and any scalar".
     
    Last edited: Sep 3, 2010
  12. Sep 3, 2010 #11

    ehild

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    I agree that those chain of equations/inequation was not a professional thing. But [itex]a|x|+a^2 y^2[/itex] and [itex]a|x| + a|y|[/itex] are inequivalent is should be really obvious in a Calculus class.
    As I am not familiar how Maths derivations are required in the English-speaking countries could you please show how would you solve this problem?


    ehild
     
  13. Sep 3, 2010 #12
    Thanks for your input Hurkyl and ehild.

    I know you are not berating me and I see your point - the reason I ask is because I am unfamiliar with writing proofs!

    Now I know it is sufficient to provide a counterexample (for some silly reason in my mind I find that not rigorous enough... but who am I to judge that? heh).

    Instead of the long chain of equations(which you two agree is non-professional), what would you have me do instead (and I don't necessarily mean in regards to this problem)?
     
  14. Sep 3, 2010 #13

    Hurkyl

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    There's a difference between something being intuitively obvious, and knowing how to actually demonstrate it!

    And intuition fails people -- we frequently get people with questions like
    I did an integral and got log(3x)+C, but the answer book said the answer was log(x)+C. What did I do wrong? Help!​
    because the poster was fooled by seeing two solutions that were "obviously" different when they were, in fact, the same.


    The reason I make a big deal because the easiest way to show that [itex]a|x|+a^2 y^2[/itex] and [itex]a|x| + a|y|[/itex] are inequivalent (provide a counter-example) is also the easiest way to show that the first condition of being a norm fails too, without having to bother with the algebraic simplification.

    This is why I responded the way I did -- to distinguish between the two cases:
    • The opening poster already knows about using counter-examples to disprove identities, but decided upon this proof instead because he already worked through it and it is a little more informative than a straight counter-example
    • The opening poster doesn't yet instinctively consider plugging in examples as a means of testing alleged identities
     
  15. Sep 4, 2010 #14

    ehild

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    Well, I also prefer your derivation instead of using a simple counter-example, but Hurkyl is right, it would have been enough. Anyway, you presented correctly that the given procedure contradicts the first requirement of being a norm. You could have proceeded a little further after that last two expressions

    a|x|+a^2 y^2

    and

    a|x| + ay^2

    saying that two linear combinations of independent quantities can be equivalent only if all coefficients agree in both expressions.

    Instead of using a long chain when you prove something, just write the equations under each other, and use some explanatory text, refer to the definition, theorems, laws if needed. Do so if you write a thesis, or paper, or you explain it at a lecture in front of students :smile: For me, here on the Forum, that chain was OK, I do the same chains when I derive something for myself. I like big sheets of paper with enough place for chains. And I also forget about counter-examples and use it only when I can not prove something in a "nicer" way. But I am not a Mathematician.

    And it was useful that you did this derivation. You can proceed on the same way if you need to prove that something is a norm. For that, it is not enough to show one example.


    ehild
     
    Last edited: Sep 4, 2010
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