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Question about vector space l^p

  1. Apr 3, 2005 #1
    So if we let x>0, For which 0<p<=infinity is {1/n^x} an element of l^p?

    If x=1, then 1/n^x is clearly an element of l^p for p>=2, since for all these vector spaces, the series of 1/n will converge?

    But if x<1, then in it seems that only for p=infinity, will {1/n^x} be an element of l^p. Is this correct?
     
  2. jcsd
  3. Apr 4, 2005 #2

    matt grime

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    Why don't you work out the length of 1/n^x in the l_p norm.
     
  4. Apr 5, 2005 #3
    The series 1 +1/2+1/3+...+1/n doesnt converge. Now if we have the summation of 1/n^x, and x>1, will 1 +1/2^x + 1/3^x + ... + 1/n^x converge? If this does always converge, then as long as p>x the series of {1/n^x} will be an element of l^p,right? That is what I meant to ask before.
     
  5. Apr 5, 2005 #4

    matt grime

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    the sum of 1/n^r converges if and only if r>1. HOpefully you can use that with the l^p norm: what is the l^p norm of the series 1/n^x?
     
  6. Apr 5, 2005 #5
    [ (1/1)^p + (1/2^x)^p +...+ (1/n^x)^p]^(1/p)

    And as I stated in my previous post, this should only converge where p>x unless I am missing something.


    One more question. If p<q, then why must l^p be a subset of l^q? I know that there are two cases to prove here. One in which q is finite, and one in which q is infinity. But I am not sure how I can show that this is true.
     
  7. Apr 5, 2005 #6

    matt grime

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    If you think all the series are finite, then yes, you are missing something.

    I think you missing something else too.

    (a_1,a_2,....) is in l^p iff the sum of (a_i)^p converges, right?

    so a_r = 1/r^x is in l^p iff the sum of (1/r^x)^p converges, which is iff px>1.

    Right? Or are we talking at cross purposes? It should be clear now why the containment you mention is true.
     
    Last edited: Apr 5, 2005
  8. Apr 5, 2005 #7
    It isn't clear to me where p = infinity
     
  9. Apr 6, 2005 #8

    matt grime

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    What isn't clear to you? (a_1,a_2,...) is in L^{infinity} if and only if its terms are bounded, so 1/n^x is in it iff x>0 (which agrees with the notion of px>1 as it happens, as p tends to infinity.

    Or do you mean the containment?

    If a sum converges its terms tend to zero, and in particular are bounded so its l infinity.
     
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