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Function with multiplicative property

  1. Jan 20, 2007 #1
    Here is the question:
    --------
    Let [itex]f: \mathbb{R} \rightarrow \mathbb{R}[/itex] satisfy [itex]f(x+y) = f(x)f(y) \ \forall x,y \in \mathbb{R}[/itex].


    Let a = f(1) > 0. Show that [itex]f(r)=a^r \ \forall r \in \mathbb{Q}[/itex].
    ---------

    This question actually has multiple parts to it. I have already proved the following:

    f(0)=1
    f(-x) = 1/f(x) for all x in R
    f(x)>0 for all x in R.
    [itex]f(n)=a^n \ \forall n \in \mathbb{N}[/itex]
    [itex]f(n)=a^z \ \forall z \in \mathbb{Z}[/itex]
    -----

    Let r=p/q p,q in Z, q not 0.

    To solve the problem I have been trying to show that [itex]f(1/q) = a^{1/q}[/itex] which I believe would lead to the solution, since we have the multiplicative property, and if not at least get a better idea of how to get the whole thing. However, this does not seem to be going anywhere, it feels like I need something else. I just had an idea of developing some type of composition property, and maybe that will get it. If you think this might work feel free to ignore this post, if not, any ideas are welcome. Not sure if that is good, basically I am thinking of trying to get some kind of identity involving f(xy) and then maybe something might happen? I suppose [itex]f(xy) = (f(x))^y[/itex].

    Thanks!
     
    Last edited: Jan 20, 2007
  2. jcsd
  3. Jan 20, 2007 #2

    StatusX

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    What is f(1/q+1/q+...+1/q), q times?
     
  4. Jan 20, 2007 #3
    Awesome!!! Fantastic Idea, thanks!
     
    Last edited: Jan 20, 2007
  5. Jan 20, 2007 #4

    HallsofIvy

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    Is there any condition that f be continuous on R? If not then f(x)= ax may not be true.
     
  6. Jan 20, 2007 #5
    Yes, that is the next part actually :smile: We then have to prove that f is continuous, which I did a while ago, and then use this and the f(r) = ar for r in Q to show that f(x) = ax for x in R.
     
  7. Jan 20, 2007 #6

    HallsofIvy

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    You are given only that f(x+ y)= f(x)f(y) and asked to prove that f is continuous?

    I am not familiar with f(x+ y)= f(x)f(y) but I know that the simpler equation
    f(x+ y)= f(x)+ f(y) has non-continuous solutions. IF f is continuous then the general solution is f(x)= Cx but non-continuous solutions are very complicated.
     
  8. Jan 20, 2007 #7
    Well, we are given that f is continuous at x = 0, and from there we conclude that f is continuous for every point in R.

    Here is my proof:

    Since f is continuous at x=0 we have that [itex]\lim_{x\rightarrow x_0}f(x) = f(0) = 1[/itex].

    From the multiplicative property of f and part (i) we get:

    [tex]\lim_{x\rightarrow x_0}\dfrac{f(x)}{f(x_0)} = \lim_{x\rightarrow x_0}f(x)f(-x_0) = \lim_{x\rightarrow x_0}f(x-x_0) = \lim_{(x-x_0) \rightarrow 0}f(x-x_0) = f(0) = 1[/tex]

    So, [tex]\lim_{x\rightarrow x_0}\dfrac{f(x)}{f(x_0)} = 1 \Rightarrow \lim_{x\rightarrow x_0}f(x) = f(x_0)[/tex]

    Therefore f is continuous at every point in [itex]\mathbb{R}[/itex].
     
    Last edited: Jan 20, 2007
  9. Jan 20, 2007 #8

    Hurkyl

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    You were given that. You forgot to give that to the rest of us. :wink:
     
  10. Jan 21, 2007 #9

    HallsofIvy

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    Yes, it is a really good idea not to withold information.

    If f(x+y)= f(x)+ f(y) (not YOUR equation!) and f is known to be continuous at x= 0, then you can prove that f is continuous at any point a:

    In [itex]\lim{x\rightarrow a} f(x)[/itex], let h= x- a so that [itex]\lim_{x\rightarrow a}f(x)= \lim_{h\rightarrow 0}f(a+ h)[/itex]
    [itex]= \lim_{h\rightarrow 0}(f(a)+ f(h))= f(a)+ \lim_{h\rightarrow 0}f(h)= f(a)[/itex]
    since you will already have proved f(0)= 0.

    You should be able to do something similar for f(x+ y)= f(x)f(y).
     
  11. Jan 21, 2007 #10
    Sorry for not posting that information :redface:
     
  12. Jan 22, 2007 #11
    Alright, I now have this question, continuation:

    Assume that f is continuous at x = 0, use the fact that f is then continuous for all x in R, and that f(r) = ar for all r in Q, to conclude that f(x) = ax for all x in R.
    ------------

    I think I understand why this is true (the rationals are dense in R, and we have the continuity to force the points to stay close together), but I just can't seem to get anywhere in the proof. It seems like I would like to grab some real number x, (and maybe another real y), then place some rationals as close as possible, and then somehow get that f(r) < f(x) < f(q), and then get that f(x) = ax (maybe with a limit or two and the continuity fact). However, I am not sure of how (or if what I said will work) to do this (and with proof). Any ideas for this one? Thanks!
     
  13. Jan 22, 2007 #12

    StatusX

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    It depends how you define a^x for real x. It's usually defined simply as the limit of a^r as r goes over a rational sequence approaching x, which makes the question trivial.
     
  14. Jan 22, 2007 #13
    Ahh yes, that other (equivalent) definition of continuity, that is perfect! Thanks!
     
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