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Prove 1-norm is => 2-norm for vectors

  1. Feb 10, 2016 #1
    1. The problem statement, all variables and given/known data
    Show that ##||x||_1\geq ||x||_2##


    2. Relevant equations

    ##||x||_1 = \sum_{i=1}^n |x_i|##
    ##||x||_2 = (\sum_{i=1}^n |x_i|^2)^.5##

    3. The attempt at a solution
    I am having a hard time with this, because the question just seems so trivial, that I don't even know how to prove it. By looking at the relevant equations we can easily see that the 1-norm is the sum of the absolute value of each element of x-vector. We also see that the 2-norm is the square root of the sum of squares of each absolute value of each element in x. Just by looking at the definition you can easily see that the 1-norm is always greater or equal to the 2-norm.

    is there an actual formal way to do this?
    It really just seems trivial to me.
     
  2. jcsd
  3. Feb 10, 2016 #2

    Ray Vickson

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    Compare ##(||x||_1)^2## and ##(||x||_2)^2##.
     
  4. Feb 10, 2016 #3
    Thanks. But isn't that just a different way of saying what I said above?
     
  5. Feb 10, 2016 #4

    Ray Vickson

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    No. It constitutes a proof, rather than just a claim.
     
  6. Feb 10, 2016 #5
    Hmmm. I might be missing something here:
    So if I square both sides I obtain:
    ##||x||_1= (\sum |x_i|)^2##
    and
    ##||x||_2=(\sum |x_i|^2)##

    Since, the square of sums of absolute values is always greater or equal to the sum of squares, we have proved that ##||x||_1 \geq ||x||_2##.

    Is what I just stated considered a proof or claim?
     
  7. Feb 10, 2016 #6

    Ray Vickson

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    Now it would be a proof if you explained why "square of sums of absolute values is always greater or equal to the sum of squares".

    It has to do, of course, with the presence of (non-negative) cross-product terms of the form ##2 |x_1| \cdot |x_j|## in one expansion but not the other.
     
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