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Proof of an inner product space

  1. Oct 3, 2008 #1
    1. The problem statement, all variables and given/known data
    Prove that the normed linear space [tex]l_{\infty}^{2}[/tex] is not an inner product space.


    2. Relevant equations
    parallelogram law;
    [tex]\left\|x+y\right\|^2+\left\|x-y\right\|^2=2\left\|x\right\|^2+2\left\|y\right\|^2[/tex]


    3. The attempt at a solution
    Well, I tried to apply parallelogram law to the [tex]l_{\infty}^{2}[/tex] space where

    [tex]x=(\alpha^1,\alpha^2)[/tex] and [tex]y=(\beta^1,\beta^2)\in l_{\infty}^{2} [/tex] .

    [tex]\left\|x\right\|=max\left\{\left|\alpha^1\right|,\left|\alpha^2\right|\right\} and

    \left\|y\right\|=max\left\{\left|\beta^1\right|,\left|\beta^2\right|\right\} [/tex]

    If one puts these norms into the parallelogram law equation, one gets a fuzzy expression on both sides of the equation, therefore it is important to put out expressions inside the max{} function which I could not achieve to do.

    Is there another method to solve this problem or am I misapplying the law to [tex]l_{\infty}^{2}[/tex] space?
     
    Last edited: Oct 3, 2008
  2. jcsd
  3. Oct 3, 2008 #2

    Dick

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    You only need a counterexample to show it's not an inner product space. Start putting some actual numbers in for the alphas and betas.
     
  4. Oct 3, 2008 #3
    Is it a mathematically correct method?
     
  5. Oct 3, 2008 #4

    Dick

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    What would not be 'mathematical' or 'correct' about it? If I claim all primes are odd, and somebody points out 2 is even, that's enough to prove me wrong. Just do it.
     
  6. Oct 3, 2008 #5

    HallsofIvy

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    Yes, as Dick said, proving that a general statement is NOT true by counterexample is a perfectly correct method.
     
  7. Oct 3, 2008 #6
    thanks for the post!
     
  8. Oct 3, 2008 #7
    well,
    I tried the parallelogram law for x=(0,3) and y=(2,5) and it perfectly worked on both sides of the equation.
    Should I choose the complex field for that space?

    What's wrong with that?
     
    Last edited: Oct 3, 2008
  9. Oct 3, 2008 #8

    Dick

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    You must have put some effort in to find a case where it works. Almost all other cases don't work. Like x=(1,0) and y=(0,1). It has to work for all cases or your norm doesn't come from an inner product.
     
  10. Oct 3, 2008 #9
    Finally, It worked :)
    I thought any ordered pair would work as a counterexample.

    Thanks a lot!
     
  11. Oct 5, 2008 #10

    HallsofIvy

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    No, and ordered pair won't work because R2 DOES make a inner product space. This question was about [itex]l_\infty^2[/itex]. Your counterexample must be from that space. What are the vectors in that space?
     
  12. Oct 5, 2008 #11

    Dick

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    I took it to be R^2 with the max norm. You think it's bounded functions on R^2, right? It's still pretty easy to find a counterexample with bounded functions.
     
  13. Oct 5, 2008 #12
    I have no clue whether one must use a bounded function or not in order to prove that.
     
  14. Oct 5, 2008 #13

    Dick

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    The question is, what is the definition of the space l^2_infinity? Can you tell us what it is?
     
  15. Oct 12, 2008 #14
    Sorry for so late reply:

    The definition of [tex]l_{\infty}^{n} [/tex] :
    On the linear space [tex]V_{n}(F) [/tex] with the infinity norm defined by
    [tex]\left\|x\right\|_p=\left[\sum^{\infty}_{i=1}\left|\alpha^{i}\right|^p\right]^{1/p}[/tex]

    where [tex] x=(\alpha^i) [/tex].
    The corresponding linear space to this norm is denoted by [tex]l_{\infty}^{n} [/tex].
     
    Last edited: Oct 12, 2008
  16. Oct 12, 2008 #15

    HallsofIvy

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    No, I was assuming infinite sequences {an} such that {a2n} was summable.
     
  17. Oct 12, 2008 #16
    so [tex]l_{\infty}^{2} [/tex] defines that norm which is basically the square of root total sum of square of each element of 2 vectors defined over R^2 field.

    Eventually , I think that you're example is correct, but besides I have no idea about whether functions defined in [tex]l_{\infty}^{2} [/tex] is bounded or not (since I didn't take any real analysis course during undergrad.)
     
  18. Oct 12, 2008 #17

    Dick

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    I don't think that's what the problem is supposed to be about.
     
  19. Oct 12, 2008 #18

    Dick

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    I think what you just described is l^2_2. The 'infinity' usually designates a max norm (supremum) rather than a power root norm.
     
  20. Oct 12, 2008 #19
    oh, ****!

    you're right , It was supposed to be max norm! I miswrote the definition.
     
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