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Math Help - final proof involving real functional equation

  1. #1
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    final proof involving real functional equation

    Suppose that f:R->R is a continuous function such that f(x+y) = f(x) +f(y) for all x, y element of R. Prove that there exists k element of R such that f(x) = kx, for every x element of R
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    A beautiful beautiful problem. My very first question that I asked on this forum*. The necessary and sufficient conditions is for f to be continous at 0. Since f is continous everywhere it must be continous at 0. Follow the elegant solution hpe gave.










    *)Actually it is find all such functions.
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  3. #3
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by luckyc1423 View Post
    Suppose that f:R->R is a continuous function such that f(x+y) = f(x) +f(y) for all x, y element of R. Prove that there exists k element of R such that f(x) = kx, for every x element of R
    I believe this can be done through induction (if not, then someone correct me please).

    First: check to see that it works for the first term: x = 1
    f(1) = k

    Second: assume that it works for some value: x = n
    f(n) = kn

    Third: show that it works for the subsequent term: x = n + 1
    f(n + 1) = k(n + 1)

    Since f(x + y) = f(x) + f(y), we have that
    f(n + 1) = f(n) + f(1) = kn + k = k(n + 1)
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  4. #4
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    Quote Originally Posted by ecMathGeek View Post
    I believe this can be done through induction (if not, then someone correct me please).
    Exactly, if you follow the link, first you do it with integers, like you did it. Extended it rationals. Then use a continuity arguement that it works for reals as well.
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  5. #5
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    wow, that is a wierd proof, not sure that i follow his method through all the steps
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    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by ecMathGeek View Post
    I believe this can be done through induction (if not, then someone correct me please).

    First: check to see that it works for the first term: x = 1
    f(1) = k

    Second: assume that it works for some value: x = n
    f(n) = kn

    Third: show that it works for the subsequent term: x = n + 1
    f(n + 1) = k(n + 1)

    Since f(x + y) = f(x) + f(y), we have that
    f(n + 1) = f(n) + f(1) = kn + k = k(n + 1)
    I only showed this works for whole numbers. I like how hpe did it.
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    Quote Originally Posted by luckyc1423 View Post
    wow, that is a wierd proof, not sure that i follow his method through all the steps
    Ignore the thing in the end with the "Hamel Basis" and the Axiom of Choice. If that is what is confusing you. That is just showing that there exists a non-linear function (without actually knowing what it is ).
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  8. #8
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    Exactly, if you follow the link, first you do it with integers, like you did it. Extended it rationals. Then use a continuity arguement that it works for reals as well.
    THP, I have a hard time reading the jumbled up math-program language this site used to use. I understand most of the proof but I get lost trying to read anything surrounded by  ...
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  9. #9
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    ditto
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  10. #10
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    For ecMathGeek I copied it. (I hope he got no copyright).


    Quote Originally Posted by lucky
    ditto
    What does that mean?
    Attached Thumbnails Attached Thumbnails final proof involving real functional equation-picture15.gif  
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  11. #11
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    ditto just means I feal the same way
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  12. #12
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    This functional equation happens to be very old.

    Cauchy Functional Equation
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