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The additive function is bounded

  1. Jan 4, 2012 #1
    If I have an additive function which is [itex]f(x+y)=f(x)+f(y)[/itex],
    the question is
    how can we prove that if this function has a limit at each real number then there is a number a greater than zero and M greater than zero
    such that
    [itex]|f(x)|\leq M[/itex], for all [itex]x\in[-a,a][/itex],
  2. jcsd
  3. Jan 4, 2012 #2


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    Science Advisor

    Can you show that, for any rational number, r, f(rx)= rf(x)? From that, you should be able to show that if the limit exists at any x, then f is continuous there and so is continuous for all x. From that it follows that f(x)= cf(x) where c is a real constant.
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