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Non archimedean norm

  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) \\
    Last edited: Dec 18, 2006
  2. jcsd
  3. Dec 18, 2006 #2


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    Does |ab| = |a||b| hold? Then for the first inequality, you can split the summand into the product of three norms. The first norm is that of an integer, hence is no greater than C. Then you have |xk||yn-k| which is no greater than max(|x|,|y|)n. The sum goes from k=0 to n, i.e. it has n+1 terms, accounting for the (n+1) factor on the right side of the inequality. Next, by establishing |x+y|n < X, where X is the rightmost expression on the first line, you get |x+y| < X1/n. n is arbitrary here, so this inequality holds for all n, and hence it holds for the limit as n goes to infinity. For an analogy, observe that 0 < 1/n for all n, hence 0 < [itex]\lim _{n\to\infty }1/n[/itex]. Finally, since [max(|x|,|y|)n]1/n = max(|x|,|y|), you can pull it out of the limit. Then I guess [(n+1)C]1/n goes to 1 as n goes to infinity.
  4. Dec 18, 2006 #3


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    I get it, thanks for helping.
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