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Limits with respect to different norms

  1. Aug 6, 2005 #1
    A few weeks ago, I was rereading my multivariable calculus textbook, because I've recently studied linear algebra and hoped to more fully understand the subject. Anyway, I noticed that when discussing the epsilon-delta definition of limits, they defined it with respect to the standard L2 norm. Since one norm is as good as another, that got me thinking: can one use different norms to define different calculi? Or does it turn out that no matter what norm you pick, you end up with the same basic results?
     
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  3. Aug 7, 2005 #2

    HallsofIvy

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    There are 3 commonly used norms in multi-variable calculus:

    |ai+ bj+ ck|= [tex]\sqrt{a^2+ b^2+ c^2}[/tex]

    |ai+ bj+ ck|= |a|+ |b|+ |c|

    |ai+ bj+ ck|= max(|a|,|b|,|c|).

    If you look at the set of vectors that have norm equal to, say, 1, the first (standard) norm gives the surface of a sphere. The second norm gives the surface of "diamond" formed of two pyramids. The third norm gives the surface of cube. All "neighborhoods" of points using those norms are of those forms.
    You then define a set to be "open" if around every point in the set you can find a neighborhood of the point that is completely in the set. But it is easy to show that given any one of the different kinds of neighborhoods (different norms) you can find either of the other kinds of neighborhoods inside it. That means that if a set is open using one kind of norm, it is open using the other kinds also. Although "neighborhoods" are different, the open sets, and therefore all limit properties are exactly the same.

    HOWEVER, it is possible to choose unusual norms where that is not true. For example, the "discrete" norm is defined by : "the norm of the zero vector is 0, the norm of any non-zero vector is 1". In that case, every set, including singleton sets, is open with the result that no sequence converges. Of course, that norm is useful only for making up strange examples!

    Your use of "L2" made me think first of "square integrable functions": the norm there is [tex]||f(x)||= \sqrt{\int f(x)^2dx}[/tex] where the integration is over some given set.

    We also often use "L1" where [tex]||f(x)||= \int |f(x)|dx[/tex]. However, in those situations, the two sets of functions are not the same. functions for which the square is integrable are not in general functions whose absolute value is integrable. Limits may be quite different in those two sets.
    And, of course, there is the "uniform norm": If f(x) is a continuous function on a compact set, define the norm to be ||f(x)||= max |f(x)| where the max is taken over the entire set. That is called the "uniform" norm because convergence of a sequence of functions in that norm is uniform convergence.
     
  4. Aug 7, 2005 #3

    Hurkyl

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    You can construct some other (very ugly!) examples of norms by exploiting the fact that the reals are a vector space over the rationals, and that vector space has (uncountably) infinite dimension.
     
  5. Aug 7, 2005 #4

    matt grime

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    The answer is "yes and no" depending on what space you're talking about.

    Yes, there are different calculuses we can define. The reals are the clsure of the rationals in the euclidean norm. there are also "valuations" that give the p-adic numbers, one for each prime p. all of these are distinct.

    however, if we try to put a norm on any finite dimensional vector space then the resulting topological spaces are all equivalent, that is they define the same topology.
     
  6. Aug 7, 2005 #5
    Wow, what great replies (especially from you, HallsOfIvy)! This is why I like coming to PF: no half-assed, hand-wavy answers. Thank you all very much.
     
  7. Aug 7, 2005 #6
    If we were letting:

    ||a|| = <a,a> ^ (1/2)

    Then how could we define our inner product to get those other norms you mentioned?

    Is this always possible to do? I mean, can one always find an inner product for any given norm?
     
  8. Aug 7, 2005 #7

    mathwonk

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    halls' example of a discrete norm is actually not a norm, i.e. it does not satisfy the usual definition of a "norm" since it is not "homogeneous". as matt said, all norms define the same notion of differential calculus on all finite dimensional vector spaces, but on infinite dimensional banach spaces, then halls' first three norms define different notions of convergence.

    there is a notion of "equivalence" of norms, where two norms, ||x|| and |x|, are equivalent iff there exist positive numbers c,d such that for all x: |x| < c||x||, and ||x|| < d|x|. then equivalent norms define the same notions of convergence and of derivatives.

    if you compute these constans for R^n, and then let n go to infinjity, you can see why the three norms halls gave, become inequivalent in infinite dimensions.

    halls is thinking of a "discrete metric", which does not however derive from a norm.
     
  9. Aug 7, 2005 #8

    mathwonk

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    also not all norms derive from inner products, except in finite dimensions.
     
  10. Aug 8, 2005 #9

    matt grime

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    i meant to say "the closure of the rationals in each p-adic valuation is different for every p" rather than just implying the valuations are distinct

    the cirterion for getting an inner product from a norm is called the polarization identity. i can't recall it exactly off the top of my head, so try googling for something on the wolfram website.
     
  11. Aug 8, 2005 #10
    Wolfram doesn't have it, but wikipedia says
    [tex]
    \[
    \left\| {x + y} \right\|^2 = \left\| x \right\|^2 + \left\| y \right\|^2 + 2 < x,y >
    \]
    [/tex]
    is satisfied in a vector space whose norm is defined in terms of an inner product.

    Thats neat, I had discovered this identity for myself before and thought it was interesting since it was just like expanding a binomial.
    [tex]
    \[
    \left( {x + y} \right)^2 = x^2 + y^2 + 2xy
    \]
    [/tex]

    I had no idea it had a name though.

    (edit: fixed the typo matt grime mentioned)
     
    Last edited: Aug 8, 2005
  12. Aug 8, 2005 #11

    matt grime

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    that wasn't what i meant. that is trivial (though you made the same typo twice in ommiting the square on the x and y terms, and only true over real inner products, no i meant the identity that tells you when a norm defines an inner product, since an inner product always gives a norm. it must be called something else then, but if a norm satisfies some property then we can induce an inner product, perhaps it is called the parallelogram law or rule or something. can someone confirm i'm not going mad? had a quick serach but got bored of trying all the variations until i got a good hit. seems right though.
     
  13. Aug 8, 2005 #12

    shmoe

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    If you have a normed space and the norm satisfies the parallelogram law, then you can use the polarization identity (different for real or complex spaces) to define an inner product.
     
  14. Aug 8, 2005 #13
    The parallelogram law is:

    [tex]2\|x\|^2+2\|y\|^2=\|x+y\|^2+\|x-y\|^2[/tex]

    The special property you referenced is the fact that if you have a norm, then there exists an inner product such that [itex]\|x\|^2=\langle x,x\rangle[/itex] if and only if the parallelogram law is true.
     
  15. Aug 8, 2005 #14
    The parallelogram law is always true if the polarization identity is true...

    [tex]
    \[
    \left\| {x + y} \right\| = \left\| x \right\|^2 + \left\| y \right\|^2 + 2 < x,y >
    \]
    [/tex]
    implies:
    [tex]
    \[
    \left\| {x - y} \right\| = \left\| x \right\|^2 + \left\| y \right\|^2 - 2 < x,y >
    \]
    [/tex]
    So we can add the two equations and get the parallelogram law.

    So, the criteria master coda quoted could just as easily have been: there exists an inner product such that [itex]\|x\|^2=\langle x,x\rangle[/itex]
    if and only if the polarization identity is true.

    (sorry about the typo in the last post, I fixed it now)
     
    Last edited: Aug 8, 2005
  16. Aug 8, 2005 #15

    shmoe

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    You can't ask if the polarization identity is true if you don't know you have an inner product, so your reverse implication doesn't make much sense in the context of a normed space unless you were already assuming you had an inner product, related to your norm or otherwise.
     
  17. Aug 8, 2005 #16
    yeah, I see what you mean, thanks
     
  18. Aug 9, 2005 #17

    mathwonk

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    i meant "up to equivalence" of norms.
     
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