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Limit of f

  1. Nov 5, 2005 #1
    Let f:R->R satisfy f(x+y)=f(x)f(y) for all x,y in R. Suppose f has a limit at 0, prove that f has a limit at all points.

    f could be either an exponential function which base is non-zero, or identically 0 or 1. Of course, this can't serve to prove anything unless I can prove that f cannot be anything else.

    f(x)=f(0+x)=f(0)f(x). If f isn't identically 0, then f(0)=1. However I fail to see how the limit of f at 0 can help me.

    Any pointers will be appreciated.
  2. jcsd
  3. Nov 5, 2005 #2


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    What's an obvious choice for what the limit should be at x? I'll denote it as Lx, although you should be able to figure out what it should be. What you need to prove is:

    for all x, there exists Lx such that for all e > 0, there exists d > 0 such that for all y satisfying 0 < |x-y| < d, the following holds:

    |f(y) - Lx| < e

    Note that y = (x-y) + x

    Also note that you know that:

    There exists L0 such that for all e > 0, there exists d > 0 such that for all h satisfying 0 < |h| < d, the following holds:

    |f(h) - L0| < e
  4. Nov 5, 2005 #3


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    Hope this helps...

    Try differentiating f(x), say

    [tex]f^{\prime}(x)=\lim_{h\rightarrow 0}\frac{f(x+h) -f(x)}{h}=\lim_{h\rightarrow 0}\frac{f(x)f(h) -f(x)}{h}=f(x)\lim_{h\rightarrow 0}\frac{f(h) -1}{h}=f(x)\lim_{h\rightarrow 0}\frac{f(h) -f(0)}{h}=f(x)f^{\prime}(0)[/tex]

    so that if f is differentiable at 0, then you have earned your self a handy differential equation from which to prove that f must be an exponential.
    But I am tired... hope I helped.

  5. Nov 5, 2005 #4
    Do you mean y=(y-x)+x?

    I haven't done differential equations in analysis. The last time I used something ahead in my proof, I got the comment "Please prove _______ if you use it at this level". Not sure I want to do that again.
    Last edited by a moderator: Nov 5, 2005
  6. Nov 5, 2005 #5


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    Yes. Can you see what to do with it?
  7. Nov 6, 2005 #6
    f(y) = f(y-x)f(x).

    Let h = y-x. |f(h)f(x)-Lx|<e whenever 0<|h|<d. But isn't the existence of Lx the very question I have to solve?
  8. Nov 6, 2005 #7
    Nevermind, I got it. Thanks for the help.
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