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Rudin 8.6

  1. May 8, 2008 #1
    1. The problem statement, all variables and given/known data
    Suppose f(x)f(y)=f(x+y) for all real x and y,
    (a) Assuming that f is differentiable and not zero, prove that

    [tex]f(x) = e^{cx}[/tex]

    where c is a constant.

    (b) Prove the same thing, assuming only that f is continuous.


    2. Relevant equations



    3. The attempt at a solution
    (a) The given information implies that f(x)^2 = f(2 x) and furthermore that f(x)^n = f(n x) for any natural number n. We can differentiate that to obtain f(x)^(n-1) f'(x) = f'(n x) for any natural number n.

    It is also not hard to show that f'(x) f(y) = f'(y) f(x) for all real x and y.

    We can take y = 0 to find that f(0) = 1 (I have used the fact that f is nonzero here).
     
  2. jcsd
  3. May 8, 2008 #2
    For (a), would it not be easier to compare the power series of f(x) and ecx.
     
  4. May 8, 2008 #3
    Perhaps, but I don't know how to prove that f is analytic.
     
  5. May 8, 2008 #4
    You're told f is differentiable which is just another way of saying that it's analytic.
     
  6. May 8, 2008 #5
    differentiability implies analyticity only in complex analysis. Rudin even gives an example of a function that is real-differentiable but not analytic in Exercise 8.1
     
  7. May 8, 2008 #6

    lurflurf

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    Homework Helper

    show that a and b each imply
    lim_{h->0}[f(h)-1]/h=c
    and then
    f(x)f(y)=f(x+y) for all real x and y
    with
    lim_{h->0}[f(h)-1]/h=c
    imply
    f(x)=exp(cx)
     
  8. May 8, 2008 #7
    I can prove part a). I can see how it would be enough to show that continuity implies lim_{h->0}[f(h)-f(0)]/h exists but I don't know how to do that.
     
  9. May 9, 2008 #8

    HallsofIvy

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    Staff Emeritus
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    You are given that f is differentiable.

    f(x+h)- f(x)= f(x)f(h)- f(x)= f(x)[f(h)- 1].

    [tex]f'(x)= \lim_{\stack{h\rightarrow 0}}\frac{f(x+h)- f(x)}{h}= f(x) \lim_{\stack{h\rightarrow 0}}\frac{f(h)- 1}{h}[/itex]
    and because f is differentiable, that limit exists.
     
  10. May 9, 2008 #9
    I said I could do part a).

    But lurflurf said that (b) also implied that the limit exists.
     
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