The additive function is bounded

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  • #1
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Main Question or Discussion Point

Hi,
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],
 

Answers and Replies

  • #2
HallsofIvy
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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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