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L1 and l2 norm inequality.

  1. Oct 27, 2009 #1
    1. The problem statement, all variables and given/known data
    [tex]\|x\|_2\le\|x\|_1\le\sqrt{n}\|x\|_2[/tex]
    where |x|1 is the l1 norm and |x|2 is the l2 norm


    2. Relevant equations
    See above


    3. The attempt at a solution
    I have [tex]\|\mathbf{x}\|_1 := \sum_{i=1}^{n} |x_i|[/tex]
    and [tex]\|x\|_2 = \left(\sum_{i\in\mathbb N}|x_i|^2\right)^{\frac12}[/tex]
    I have tried to expand out the x 2 norm but i cant seem to figure out how to prove the inequality. Any suggestions?
     
    Last edited: Oct 27, 2009
  2. jcsd
  3. Oct 27, 2009 #2

    lanedance

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    for the first part of the inequality, you could try squaring both sides
     
  4. Oct 27, 2009 #3
    Yea that works for the first part. Thanks for the reply.

    Any idea on the second part (square root of n)?

    I am thinking it may have to do with the projection vector (such as (1,1,1,1,1,1)) in a scalar product or something like.
     
  5. Oct 27, 2009 #4

    lanedance

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    your idea should work with for the 2nd one with the use of Cauchy Schwarz
     
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