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Showing a Norm is not an Inner Product

  1. Mar 27, 2012 #1
    Show the taxicab norm is not an IP.

    taxicab norm is v=(x[itex]_{1}[/itex].....x[itex]_{n}[/itex])
    then ||V||= |x[itex]_{1}[/itex]|+.....+|x[itex]_{n}[/itex]|)

    I was thinking about using the parallelogram law

    but I would get this nasty thing


    (|x[itex]_{1}[/itex]+w[itex]_{1}[/itex]|+...........+|x[itex]_{n}[/itex]+w[itex]_{n}[/itex]|)[itex]^{2}[/itex]+(|x[itex]_{1}[/itex]-w[itex]_{1}[/itex]}+...........+|x[itex]_{n}[/itex]-w[itex]_{n}[/itex]|[itex])^{2}[/itex]=2(|x[itex]_{1}[/itex]|+.....+|x[itex]_{n}[/itex]|)[itex]^{2}[/itex]+(|w[itex]_{1}[/itex]|+.....+|w[itex]_{n}[/itex]|)[itex]^{2}[/itex]


    Im not sure how to work with this. Am I going about this wrong?

    Also there might be some typos with the absolute value signs the latex get messy
     
  2. jcsd
  3. Mar 27, 2012 #2
    Hint: the norm doesn't satisfy linearity (it does satisfy the triangle inequality).
     
  4. Mar 27, 2012 #3
    Do you mean that ||v+w||is not an inner product? This is easy to show! Just go through the definition of inner product -- one of the axioms is clearly not satisfied in this case.
     
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