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Thread: Additive functions

  1. #1
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    Additive functions

    Hi,
    I search for a proof that any nonlinear and additive function (f(x+y)=f(x)+f(y)),is unbounded on every interval.
    Thank's in advance.
    Linear means that f(x)=cx for some real constant c.
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  2. #2
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    Re: Additive functions

    What ideas do you have? I recommend generating an arbitrary basis for the reals over the rationals and going from there. I am not quite sure where it will lead, but it seems promising since $\displaystyle f\left( \sum_{k=1}^n a_k \right) = \sum_{k=1}^n f(a_k)$
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  3. #3
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    Re: Additive functions

    Quote Originally Posted by hedi View Post
    Hi,
    I search for a proof that any nonlinear and additive function (f(x+y)=f(x)+f(y)),is unbounded on every interval.
    Linear means that f(x)=cx for some real constant c.
    Do you really mean every interval?
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  4. #4
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    Re: Additive functions

    If we prove that the graf of f is dense in the plane then f must be unbounded on every interval.If f is nonlinear then there exist x1 and x2 in R such that f(x1)/x1is not equal to f(x2)/x2.Hence the two vectors (x1,f(x1)) and (x2,f(x2)) are linearly independent in R^2 as a vector space over the field R.So the span of these two vectors over Q (the rational field) is dense ii R^2.Now it is sufficient to show that tis span is contained in tne graf of f.Now show that for any two rational numbers a and b it holds that

    f(ax1+bx2)=af(x1)+bf(x2).Can someone help me proving this?
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  5. #5
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    Re: Additive functions

    Your last equation is almost trivial since for any rational q and real x, f(qx)=qf(x):

    1. For q a non-negative integer, an easy induction.
    2. If q=1/m with m positive, by 1, f(x)=f(m(x/m))=mf(x/m) or f(x/m)=(1/m)f(x)
    3. By 1 and 2, for m and n positive f((n/m)x)=nf((1/m)x)=(n/m)f(x)
    4. f(0)=0 -- obvious since f(0+0)=f(0)
    5. For any x f(-x)=-f(x) -- obvious from 4.
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  6. #6
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    Re: Additive functions

    yes I got this,thank's
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