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Proof, strong triangle inequality

  1. Dec 18, 2006 #1


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    can someone explain this proof please, I added a star to the inequalities I don't see/understand.

    if | | is a norm on a field K and if there is a C > 0 so that for all integers n |n.1| is smaller than or equal to C, the norm is non archimedean (ie the strong triangle inequality is true)

    proof: if x and y in K

    |x + y|^n \le \sum\limits_{k = 0}^n {|\frac{{n!}}{{k!(n - k)!}}} x^k y^{n - k} | \le *(n + 1).C.\max \left( {|x|,|y|} \right)^n \\
    |x + y| \le *\mathop {\lim }\limits_{n \to \infty } \left[ {(n + 1)C.\max \left( {|x|,|y|} \right)^n } \right]^{1/n} * = \max \left( {|x|,|y|} \right) \\

    I understand everything except the parts I marked with a *
    Last edited: Dec 18, 2006
  2. jcsd
  3. Dec 18, 2006 #2

    matt grime

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    The first * seems to assume that the binomial coefficient is less than C in norm, though I don't claim to be able to explain that right now, and is the only thing to do with non-rchmideanness. The second * is just taking roots in the first line. And the third * is just taking the limit. The last is standard - the n'th root of a constant tends to 1 as n tends to infinity, as does the n'th root of n+1.
  4. Dec 18, 2006 #3


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    thank you very much
    Last edited: Dec 18, 2006
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