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Equivalent norms

  1. Feb 20, 2010 #1
    1. The problem statement, all variables and given/known data

    Show that the Euclidean and supremum norms are equivalent norms on R2

    3. The attempt at a solution

    The Euclidean Norm is
    [tex]\left|\left|[/tex]X[tex]\left|\left|[/tex]1 = [tex]\sqrt{x1^2 + x2^2}[/tex]

    The Supremum norm is
    [tex]\left|\left|[/tex]X[tex]\left|\left|[/tex][tex]\infty[/tex] = max ([tex]\left|[/tex]x1[tex]\left|[/tex],[tex]\left|[/tex]x2[tex]\left|[/tex])

    so for them to be equivalent:

    a([tex]\sqrt{x1^2 + x2^2}[/tex])[tex]\leq[/tex] max ([tex]\left|[/tex]x1[tex]\left|[/tex],[tex]\left|[/tex]x2[tex]\left|[/tex]) [tex]\leq[/tex] b([tex]\sqrt{x1^2 + x2^2}[/tex])

    I think i'm going right with this?
    im not sure how to work with the supremum norm in the middle?
     
  2. jcsd
  3. Feb 20, 2010 #2

    Hurkyl

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    That sounds like a correct algebraic translation of the problem. In more detail, you want to prove:



    There exist positive real numbers a and b such that
    For any real numbers x1 and x2
    [tex]a \sqrt{x_1^2 + x_2^2} \leq \max \{ |x_1|, |x_2| \} \leq b \sqrt{x_1^2 + x_2^2}[/tex]​



    The supremum in the middle is not essential -- with a little effort you can rearrange that inequality so that there are suprema is on the outside and one Euclidean norm in the middle. Really, for the purposes of solving for a and b, you probably just want to look at it as two separate inequalities.
     
  4. Feb 20, 2010 #3

    LCKurtz

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    Yes, what you have to prove is you can find the a and b. You might start by stating that without loss of generality, you may assume [itex]|x_1| \le |x_2|[/itex], so that:

    max{|x1|,|x2|} = |x2|

    and work with assumption.
     
  5. Feb 20, 2010 #4
    thanks for the help lads

    [tex]
    a \sqrt{x_1^2 + x_2^2} \leq b \sqrt{x_1^2 + x_2^2} \geq \max \{ |x_1|, |x_2| \} }
    [/tex]

    [tex]\Rightarrow[/tex]

    [tex]
    a \sqrt{x_1^2 + x_2^2} \leq b \sqrt{x_1^2 + x_2^2} \geq |x_2| }
    [/tex]

    [tex]\Rightarrow[/tex]

    [tex]
    a \leq b \geq |x_2|/\sqrt{x_1^2 + x_2^2} }
    [/tex]

    a constant?
     
  6. Feb 20, 2010 #5

    Dick

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    Just draw a right triangle. Make the lengths of the legs x1 and x2. sqrt(x1^2+x2^2)=h is the length of the hypotenuse. In terms of h, what's the longest the longest leg can be? What's the shortest the longest leg can be?
     
  7. Feb 20, 2010 #6
    the shortest the longest leg can be is x1=x2 so then x2 = [tex]\sqrt{(h^2)/2}[/tex] ?
     
  8. Feb 20, 2010 #7

    Dick

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    Exactly. h/sqrt(2). How about the longest the longest leg can be? So the max norm is bounded by constants times the euclidean norm, right?
     
  9. Feb 20, 2010 #8
    the longest the longest leg can be is x1 = 0 so then x2 = h?
     
  10. Feb 20, 2010 #9

    Dick

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    I would have said 'the longest the longest can be is h'. But I think you've got the idea. If h is the hypotenuse, the longest leg is between h and h/sqrt(2).
     
  11. Feb 20, 2010 #10
    ok, so the longest leg is between h and h/sqrt(2), i get that with the triangle but im not sure how it shows the norms are equivalent?sorry and thanks for the help!
     
  12. Feb 20, 2010 #11

    Dick

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    The hypotenuse is the euclidean norm of the point (x1,x2) in R^2. The longest leg is the max norm of (x1,x2).
     
  13. Feb 20, 2010 #12
    oh god,yes-thanks a million...that actually really,really explains it to me. Damn thanks a million.Thanks dick. Im actually happy now :biggrin:
     
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