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Find my mistake

  1. Jun 5, 2008 #1

    Fredrik

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    I have to ask a dumb question. I seem to be doing something very wrong here, and it's probably trivial, but for some reason I don't see what it is. I decided to try to prove the Cauchy-Schwarz inequality without opening a book. I remember that a proof I read once started by noting that

    [tex]0\leq(x+ty,x+ty)=||x||^2+2 Re (t(x,y))+||y||^2[/tex]

    t is an arbitrary complex number. From here on it's supposed to be about choosing the right t. The inequality above is obviously satisfied when Re(t(x,y))>0, so let's choose Arg t so that it's negative instead.

    We get

    [tex]0\leq||x||^2-2|t||(x,y)|+||y||^2[/tex]

    or equivalently

    [tex]|(x,y)|\leq\frac{1}{2|t|}(||x||^2+||y||^2)[/tex]

    Now if we choose

    [tex]|t|=\frac{||x||^2+||y||^2}{2||x||||y||}[/tex]

    the inequality above turns into the Cauchy-Schwarz inequality.

    Here's my dumb question: By choosing |t| much larger, I can make |(x,y)| as small as I want, and this implies that (x,y)=0 for arbitrary x and y! This obviously can't be true, I must have done something wrong. What's my mistake?

    By the way, is there a better way to write the norm in LaTeX?
     
  2. jcsd
  3. Jun 5, 2008 #2

    Hurkyl

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    It's here!
     
  4. Jun 5, 2008 #3

    Fredrik

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    [tex](x+ty,x+ty)=(x,x)+t(x,y)+t^*(y,x)+(y,y)[/tex]

    [tex]=||x||^2+t(x,y)+(t(x,y))^*+||y||^2=||x||^2+2 Re (t(x,y))+||y||^2[/tex]
     
  5. Jun 5, 2008 #4

    Hurkyl

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    (x+ty,x+ty) = (x,x) + (x,ty) + (ty,x) + (ty,ty)
     
  6. Jun 5, 2008 #5

    Fredrik

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    D'oh...and LOL. Thanks. I don't know how I could stare at it for like an hour and not see it.
     
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