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Modulus and inequalities

  1. Mar 31, 2010 #1

    I've got

    [tex] |a|^{2} = |a - b + b|^{2} [/tex]

    What can I do with this guy? Usually when the square isn't there I use the triangle inequality and things fall out pretty quick, for example,

    [tex] |a| = |a - b + b| \leq |a-b| + |b| [/tex]

    Is there something like that I can do with the orginial guy?
  2. jcsd
  3. Mar 31, 2010 #2


    Staff: Mentor

    Well, |a - b + b|2 = |a|2
  4. Mar 31, 2010 #3


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    What are you looking for?
  5. Mar 31, 2010 #4
    I know a property concerning |a-b|2 (it's less than some epsilon) and I know what |b|2 is, so, if I can break this down into an inequality, I can say something about |a|2. Does that make sense? this is part of a bigger problem concerning metric spaces.
  6. Mar 31, 2010 #5
    Are you looking for something like this?


    Now, if [itex]\left|a^2-b^2\right|\leq\epsilon[/itex] and [itex]\left|b^2\right|\leq\epsilon[/itex], then [itex]\left|a\right|^2\leq 2\epsilon[/itex]
  7. Mar 31, 2010 #6
    No, my a is a function in C[0,1] so that statement may not be true. I'm looking for a way to massage out that to which I responded to mathman.
  8. Mar 31, 2010 #7
    Well, my guess is that you are working with the sup norm, and want to prove something related to uniform convergence, but I think you must give more details. After all, you stated that you had information about |a2-b2|, |b2| and needed to relate these to |a2|. Given only this, I can't add anything else.
  9. Mar 31, 2010 #8
    I have

    [tex]|f_{n}(t)-f(t)|^{2} \leq\epsilon [/tex]

    and [tex] |f_{n}(t)|^{2} \leq 1 [/tex]

    I wish to show

    [tex] |f(t)|^{2} \leq 1 [/tex]

    as well.

    As much as I'd like to, I cannot say

    [tex] |f(t)|^{2} = |f(t) - f_{n} (t) + f_{n} (t)|^{2} \leq |f_{n}(t) - f(t)|^{2} + |f_{n}(t)|^{2} [/tex]
  10. Mar 31, 2010 #9
    No, but you can say:

    |f(t)|^{2} = |f(t) - f_{n} (t) + f_{n} (t)|^{2} \leq |f_{n}(t) - f(t)|^{2} +2|f_{n} (t)||f_{n}(t) - f(t)|+ |f_{n}(t)|^{2}

    And you know how to estimate all terms on the right; just keep in mind that [itex]\epsilon[/itex] is arbitrarily small (or at least I think it is).
  11. Mar 31, 2010 #10
    I see you just expanded it...I didn't know you could do that inside absolute value.
  12. Mar 31, 2010 #11
    It's just one of those Analysis tricks. After doing a fair number of these proofs, you start to spot them more rapidly.:smile:
  13. Mar 31, 2010 #12
    Well thank you so much! And yes, you are right, it did begin with the sup norm.
  14. Mar 31, 2010 #13
    Glad to help.:wink:
  15. Apr 1, 2010 #14


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    When you expand the square of the sum or difference of two terms, you get two squares and a product. Taking absolute values for each term can only make the sum of the three terms bigger.
  16. Apr 1, 2010 #15

    [tex] |f_n(t) - f(t)|^2 \leq \epsilon \implies [/tex]
    [tex] |f_n(t) - f(t)| \leq \epsilon' \implies [/tex] by reverse triangle inequality
    [tex] |f(t)| - |f_n(t)| \leq \epsilon' \implies [/tex]
    [tex] |f(t)| \leq 1 + \epsilon' \implies [/tex]
    [tex] |f(t)|^2 \leq 1 [/tex]

    edit: this is true for one n if the OP's statements still hold. as in if [itex] |f_n(t) - f(t) | \leq \epsilon [/itex] for all epsilon without incrementing n.
    Last edited: Apr 2, 2010
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