Proving Limit of f:R->R Exists at All Points

[Icebreaker]

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.

 

[AKG]

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

for all x, there exists L_x 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) - L_x| < 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

 

[benorin]

Hope this helps...

Try differentiating f(x), say

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)

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.

-Ben

 

[Icebreaker]

Note that y = (x-y) + x

Do you mean y=(y-x)+x?

a handy differential equation

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.

 

[AKG]

Do you mean y=(y-x)+x?

Yes. Can you see what to do with it?

 

[Icebreaker]

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?

 

[Icebreaker]

Nevermind, I got it. Thanks for the help.