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Holder's Inequality for Sums

  1. Jun 6, 2009 #1
    I am actually attempting the proof for Minkowski's inequality, but have not gotten that far yet. I am stuck on a step in Holder's inequality, and I have a feeling it's something very simple that I am just overlooking...

    I have easily been able to show [tex]ab \leq \frac{a^p}{p} + \frac{b^q}{q}[/tex]

    And if [tex]a,b[/tex] are normalized vectors then:

    [tex]\sum_k a_k b_k \leq \frac{1}{p}\sum_k {a_k}^p + \frac{1}{q}\sum_k {b_k}^q[/tex]

    And I am aware that through normalizing the vectors, I am supposed to be able to deduce the formula for Holder's inequality:

    [tex]\sum_k a_k b_k \leq (\sum_k {a_k}^p)^{\frac{1}{p}}(\sum_k {b_k}^q)^{\frac{1}{q}}[/tex]

    But I just cannot figure this step out for some reason! Please give me at least a hint... :confused:

    Thank you in advance!
     
  2. jcsd
  3. Jun 6, 2009 #2
    This inequality:
    [tex]\sum_k a_k b_k \leq \frac{1}{p}\sum_k {a_k}^p + \frac{1}{q}\sum_k {b_k}^q[/tex]

    holds for all vectors, not just those that are normalized.

    Whenever there are several things being added and you want to turn them into a single term, a good strategy is to do whatever it takes to turn the sum into a constant 1. Then after things simplify, bring back in the complexity that you threw away, but now the sum is gone. This happens all the time, and I think there is a deep meaning why it works so much, but I don't know what it is.

    In this case, if you restrict ||a||p = 1, and ||b||q = 1, then the equation becomes

    [tex]\sum_k a_k b_k \leq 1[/tex]

    for a and b normalized like so. Now bring back the complexity for arbitrary vectors x and y,

    [tex]\sum_k \frac{x_k}{||x||_p} \frac{y_k}{||y||_q} \leq 1[/tex]

    [tex]\sum_k x_k y_k \leq ||x||_p ||y||_q[/tex]
     
    Last edited: Jun 6, 2009
  4. Jun 6, 2009 #3
    Oh wow, okay. That's pretty nice and slick. Thank you!
     
  5. Jun 7, 2009 #4
    Also note that the same exact strategy will prove the integral form.
     
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