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Homework Help: Sequences in lp spaces (Functional Analysis)

  1. Mar 2, 2008 #1
    [SOLVED] Sequences in lp spaces... (Functional Analysis)

    1. The problem statement, all variables and given/known data
    Find a sequence which converges to zero but is not in any lp space where 1<=p<infinity.


    2. Relevant equations
    N/A


    3. The attempt at a solution
    I strongly suspect 1/ln(n+1) is a solution.

    Since ln(n+1) -> infinity as n -> infinity, we have 1/ln(n+1) -> 0 as n-> infinity.

    I attempted using the integration test to show that partial sums for the sequence (1/ln(n+1))^p do not converge, but integrating 1/ln(n) became a bit problematic!

    Attempt two was to show that each successive term in the sequence {(1/ln(n+1))^p} is larger than each successive term in some tail of the harmonic sequence.

    i.e. There exists m such that
    (1/ln(2))^p >= 1/(m+0) or m >= (ln(2))^p
    (1/ln(3))^p >= 1/(m+1) or m >= (ln(3))^p - 1
    (1/ln(4))^p >= 1/(m+2) or m >= (ln(4))^p - 2
    .
    .
    .
    In general for such an m to exist, m >= (ln(x+2))^p - x

    And therefore I need to show that this right hand side is bounded above. I suspect this is true, and with a lot of hand waving can convince myself that the "ln" part of the RHS is "stronger" than the "^p" part and thus (ln(x+2))^p will eventually "slow down enough" so as to be less than x for large enough values of x.

    But my log work leaves a little to be desired and I can't even prove there is a solution to the equation (ln(x+2))^p = x.

    Any help much appreciated!

    Best regards,
    Justin
     
    Last edited: Mar 2, 2008
  2. jcsd
  3. Mar 2, 2008 #2

    quasar987

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    Never mind. :x
     
  4. Mar 2, 2008 #3

    quasar987

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    There is a sweet test when it comes to series with logarithm in the argument. It says if a sequence a_n is decreasing and non-negative, then the series of a_n converges if and only if the series of 2^n*a_{2^n} does.

    So here,

    [tex]a_n=\frac{1}{\ln(n)^p}[/tex]

    and so

    [tex]2^na_{2^n} = \frac{2^n}{n^p\ln(2)^p}[/tex]

    It's called the Cauchy condensation test: http://en.wikipedia.org/wiki/Cauchy_condensation_test
     
    Last edited: Mar 2, 2008
  5. Mar 2, 2008 #4
    That IS sweet!

    A gazillion thanks!

    J
     
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