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Proof regarding property of continuity

  1. Oct 12, 2008 #1
    1. The problem statement, all variables and given/known data

    Prove that if f(x) satisfies the functional equation f(x+y) = f(x) + f(y) and if f is continous then f(x) = cx for some constant c.

    2. Relevant equations


    3. The attempt at a solution

    Assume |f(a)| > |ca| for some a in the domain of f. Since f is continuous at every point then for every e > 0 we can find a d > 0 such that |f(x) - f(x+a)| < e whenever |x - (x+a)| < d.
    We have |f(x) - (f(x) + f(a)| = |-f(a)| = |f(a)| < e whenever |a| < d. Since |ca| < f(a)| < e then |ca| < e, implying that |a| < e/|c|. Thus we can take d = e/|c|, implying that
    |c|d = e. Now, since d would be points on the x axis and e points on the y axis, then we have y = |c|x, or f(x) = cx for all x in the domain of f. However, we said that |f(a)| > |ca|. So this is a contradiction.

    If we assume |f(a)| < |ca| for some a in the domain of f, then we have again |f(a)| < e whenever |a| < d. However, we don't know if |f(a)| < |ca| < e or |f(a)| < e < |ca|. If it's |f(a)| < |ca| < e then we can continuous with our proof using the same argument as before. If it's |f(a)| < e < |ca| then we can't. However, the function g(x) = cx is also continuous for all x and so we have |f(a)| < |ca| < e'. Then |a| < e'/|c| and we can choose d = e'|c|, implying that |c|d = e', implying that |c|x = y = f(x). This is again a contradiction, so therefore |f(x)| = |cx|. So if f(x) > 0 then f(x) = cx. And if f(x) < 0 then -f(x) = -(cx), implying that f(x) = cx. Therefore f(x) = cx. QED

    Is this a valid proof? I found the question difficult so I want to make sure that the proof is correct.
  2. jcsd
  3. Oct 13, 2008 #2
    The reason why I think my proof is not valid is because the assumption is that for some x = a, with a < d, we have f(a) < ca and f(a) > ca. Since a is less then d, then it's okay if |c|d = e. But on the other hand, the equation |c|d = e must hold for every positive d, and if d is negative, we have |c|(-d) = -e, also, it is easy to prove that f(0) = 0 for this function, so shouldn't the whole function also be f(x) = |c|x?
  4. Oct 13, 2008 #3


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    I haven't really read your proof, because it looks like it's overly complicated. This problem is simpler than that. You said you noticed that f(0)=0. Try proving that in fact f(nx)=nf(x) for every integer n (positive and negative). Then use this to prove that f(rx)=rf(x) for every rational r. Finally, use the density of the rationals.

    Of course all of this is assuming f is defined from R into R.
  5. Oct 13, 2008 #4
    Assuming f is defined in the appropriate domain:
    If f(a) is not zero for some a (otherwise we are done).
    What is [tex]\lim_{n \to \infty}\frac{f(a/n)}{a/n}[/tex]?

    What does that tell us about [tex]\lim_{h \to 0}\frac{f(h)}{h}[/tex]?

    Finally, can we determine if f'(x) exists and what it is?
  6. Oct 13, 2008 #5
    It seems that the continuity of the function is not needed in such proof? I'm not sure if I miss something...
  7. Oct 13, 2008 #6


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    You need the continuity to go from f(rx)=rf(x) for all rational r to all real r.
  8. Oct 13, 2008 #7
    yea... I got it now. it needs the continuity to have [tex]f(x)=lim_{rational r->x}f(r)=lim_{r->x}rf(1)=xf(1).[/tex]
  9. Oct 13, 2008 #8
    Morphism, thanks for the help. I managed to prove f(nx) = nf(x), now I'm working on it for f(rx) = rf(x). I'll post up my proof tomorrow as tonight is thanks giving (over here in Canada at least) and I'll be busy all night.
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