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Jensen's inequality doubt

  1. Feb 25, 2013 #1
    1. The problem statement, all variables and given/known data

    Show that 2-norm is less equal to 1-norm

    But I've found this proof

    http://img825.imageshack.us/img825/5451/capturaklt.jpg [Broken]

    Which basically shows that if p=1 and q=2 then 1-norm is less equal than 2-norm, i.e. the opposite hypothesis

    2. Relevant equations
    None


    3. The attempt at a solution
    It's not homework, it's just a doubt
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Feb 26, 2013 #2

    jbunniii

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    In a general measure space, it need not be true that ##\|x\|_p \leq \|x\|_q## or ##\|x\|_q \leq \|x\|_p##. Indeed, in general it is not even true that ##L^p \subset L^q## or ##L^q \subset L^p##. In order to obtain this nesting, it is necessary to impose additional restrictions.

    If the space has finite measure (certainly the case for a probability measure space), then ##L^q \subset L^p## if ##1 \leq p \leq q \leq \infty##. The probability space is special because it has measure 1, which gives us ##||x||_p \leq ||x||_q##. In general, there would be a constant ##|x|_p \leq c \|x\|_q##. The intuitive explanation for why the containment goes this way is that a function in a finite measure space can have "thick singularities," but it can't have "thick tails."

    If the space uses counting measure (or in general, if it does not contain sets with arbitrarily small positive measure), then ##L^p \subset L^q##. For example, this is true for the sequence spaces ##\ell^p \subset \ell^q##. The intuitive reason the containment goes in this direction is that you can have "thick tails" but not "thick singularities."

    So it depends on what measure space you are working with. If we are in a finite-dimensional vector space, say ##\mathbb{R}^n##, then we have both ##\ell^p(\mathbb{R}^n) \subset \ell^q(\mathbb{R}^n)## and ##\ell^q(\mathbb{R}^n) \subset \ell^p(\mathbb{R}^n)##. In the special special case ##p = 1##, ##q = 2##, we will have ##\|x\|_2 \leq \|x\|_1## and ##\|x\|_1 \leq \sqrt{n} \|x\|_2##.
     
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