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Additive Montone Functions Proof

  1. Nov 30, 2004 #1
    So if f is a monotone function which takes elements of the Reals to the Reals. If f is additive, how do I show that f is continuous?
     
  2. jcsd
  3. Nov 30, 2004 #2

    Hurkyl

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    Probably by assuming it's not continuous. You know what f(0) is, right? That's often a key place to analyze. Making a conjecture for a formula for f might help too.
     
    Last edited: Nov 30, 2004
  4. Nov 30, 2004 #3
    Nope, I don't know what f(0) is.
     
  5. Nov 30, 2004 #4

    Hurkyl

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    You can figure it out from the definition of additive, though!

    You can probably figure out a lot of values of f in terms of f(1), too!
     
  6. Nov 30, 2004 #5
    I see that f(0)=0 since f(0)=f(0)+f(0) when x=y=0

    And I also see that f(2)=2f(1) and f(3)=3f(1) and f(n)=nf(n) but this is only true when n is an integer. And n is not strictly limited to being an integer, it can be any rational number I believe.

    And I am not sure what to do with this information.
     
  7. Dec 1, 2004 #6

    Hurkyl

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    So, you know that for n rational, f(n) = n f(1).
    You also know that f is monotone -- does that help you with figuring out values of f at irrational numbers?
     
  8. Dec 1, 2004 #7
    Perhaps it would help someone smarter, but not me. I really do appreciate your help very much. I hope I am not annoying you.

    1)I can't find a logical connection between the rationals and irrationals here. I am not even convinced that f(n)=nf(1) for all rationals. How do I know f(1/2)=1/2f(1) for instance?

    2)I am not sure how to use the fact that the function is monotone to help me with this proof. I know what monotone means, but how do I know that no jump discontinuities exist? I know that a montone function on the Reals can only have countably many jump discontinuities, and the irrationals aren't countable so f must be continuous at some irrational number. How do I jump to the conclusion that f is continuous for all of them?
     
    Last edited: Dec 1, 2004
  9. Dec 1, 2004 #8

    arildno

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    Here's how to show that f(a)=af(1) for a rational:
    Let p,n be integers:
    [tex]pf(1)=f(p)=f(\frac{p}{n}+++\frac{p}{n})=nf(\frac{p}{n})[/tex]

    The "+++" means we have n terms in our argument.
     
  10. Dec 1, 2004 #9

    Hurkyl

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    Suppose one does. What does that say about the values at the rational points?



    or



    How can you identify irrational points using only rational points?
     
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