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Strange proof of Young's inequality

  1. Jan 16, 2012 #1
    I'm trying to absorb a perplexing proof of Young's inequality I've found. Young's inequality states that if [itex]A,B \geq 0[/itex] and [itex]0 \leq \theta \leq 1[/itex], then [itex]A^\theta B^{1-\theta} \leq \theta A + (1-\theta)B[/itex].

    The first step they take is the following: We can assume [itex]B \neq 0[/itex]. (I get that.) But then they say we can just replace [itex]A[/itex] with [itex]AB[/itex], in which case it suffices to prove [itex]A^\theta \leq \theta A - (1-\theta)[/itex].

    I understand why the replacement of A with AB in Young's inequality gets you to the new inequality in only A. But I don't see why you get to do that. Why is proving [itex]A^\theta \leq \theta A - (1-\theta)[/itex] logically equivalent to proving [itex]A^\theta B^{1-\theta} \leq \theta A + (1-\theta)B[/itex]?
     
  2. jcsd
  3. Jan 16, 2012 #2
    Wow. Don't I feel dumb. Suppose you know that for [itex]A \geq 0[/itex] and [itex]0 \leq \theta \leq 1[/itex], [itex]A^\theta \leq \theta A + (1-\theta)[/itex]. Then if [itex]B > 0[/itex], you know [itex]A/B \geq 0[/itex], so you have

    [tex]
    \left( \frac{A}{B} \right)^\theta \leq \theta \frac{A}{B} + (1-\theta).
    [/tex]

    Rewriting this and multiplying both sides by [itex]B[/itex], you obtain

    [tex]
    A^\theta B^{1-\theta} \leq \theta A + (1-\theta)B,
    [/tex]

    which is the desired result. So, I'm good to go. But I still think the book's way of explaining this bit of subterfuge ("replace A by AB") is poor.
     
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