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Homework Help: Holder's Inequality Help

  1. Jan 26, 2010 #1
    I'm currently reading a book and stuck on an exercise with no solutions.

    A proof would be great.
     

    Attached Files:

  2. jcsd
  3. Jan 26, 2010 #2

    sylas

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    I have a bit of an interest in Holders inequality, so I'm going to make the effort to convert your word docoument into something readable for the forums. In general, posting a word document is not adequate for readers. I have also converted an "a" to a "b" in the last formula.

    The problem is:

    Let [itex]1 < p < \infty[/itex] and 1/p + 1/q = 1

    Show that if [itex]a \in l_p[/itex] and [itex]b \in l_q[/itex] then the series
    [tex]S(a,b) = \sum_{n=1}^{\infty} a_n b_n[/tex]
    converges absolutely, and is bounded above by
    [tex]\left( \sum_{n=1}^\infty\left| a_n \right|^p \right)^{1/p}
    \left( \sum_{n=1}^\infty\left| b_n \right|^q \right)^{1/q}[/tex]

    My question. What does lp mean?
     
  4. Jan 27, 2010 #3
    Lp = {a = a(n) | Sigma |a(n)|^p < infinity}.

    Thanks in advance.
     
  5. Jan 27, 2010 #4

    sylas

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    Define
    [tex]\begin{align*}
    X & = \sum_{n=0}^\infty \left| a_n \right| ^p \\
    Y & = \sum_{n=0}^\infty \left| b_n \right| ^q \\
    x_n & = | a_n | / X^{1/p} \\
    y_n & = | b_n | / Y^{1/q} \\
    \intertext{Hence}
    \sum_{n=0}^\infty x_n^p & = \frac{\sum_{n=0}^\infty \left| a_n \right|^p}{X} & = 1 \\
    \sum_{n=0}^\infty y_n^q & = \frac{\sum_{n=0}^\infty \left| b_n \right|^q}{Y} & = 1
    \end{align*}[/tex]​

    I will also use Young's inequality. Given 1/p + 1/q = 1, for p,q > 0, a,b >= 0, we have
    [tex]ab \leq \frac{a^p}{p} + \frac{b^q}{q}[/tex]​

    Hence
    [tex]\begin{align*}
    \sum_{n=0}^\infty a_n b_n & \leq X^{1/p}Y^{1/q} \sum_{n=0}^\infty x_n y_n \\
    & \leq X^{1/p}Y^{1/q} \sum_{n=0}^\infty \left( \frac{x_n^p}{p} + \frac{y_n^q}{q} \right) \\
    & \leq \frac{X^{1/p}Y^{1/q}}{p} \sum_{n=0}^\infty x_n^p + \frac{X^{1/p}Y^{1/q}}{q} \sum_{n=0}^\infty y_n^q \\
    & = \frac{X^{1/p}Y^{1/q}}{p} + \frac{X^{1/p}Y^{1/q}}{q} \\
    & = X^{1/p}Y^{1/q} \left( \frac{1}{p} + \frac{1}{q} \right) \\
    & = X^{1/p}Y^{1/q}
    \end{align*}[/tex]​
    as required.

    Felicitations -- sylas
     
  6. Jan 27, 2010 #5
    Thanks Sylas. I can see from the proof that the series is bounded above. But where does the absolute convergence come in?

    Would you have to use the MCT?

    Also, do I not need to consider finite sums for Holder's Inequality?
     
    Last edited: Jan 27, 2010
  7. Jan 27, 2010 #6

    sylas

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    I am not sure what you mean. The given condition
    [tex]\sum_{n=0}^\infty |a_n|^p < \infty[/tex]​
    is simply saying that the sum is finite. So it has a finite value, and I give it a name, X.

    There are many forms of Holder's inequality. I gave the proof for the case you presented. A finite sum follows directly as a special case of the infinite sum, where all but an infinite number of the terms are zero.

    And by the way, I see you started all series at n=1, whereas I started at n=0. There's no difference; it is still just an infinite series.

    Cheers -- sylas
     
  8. Jan 28, 2010 #7
    Thanks.
     
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