# Rudin 8.6

1. May 8, 2008

### ehrenfest

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

$$f(x) = e^{cx}$$

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. May 8, 2008

### e(ho0n3

For (a), would it not be easier to compare the power series of f(x) and ecx.

3. May 8, 2008

### ehrenfest

Perhaps, but I don't know how to prove that f is analytic.

4. May 8, 2008

### e(ho0n3

You're told f is differentiable which is just another way of saying that it's analytic.

5. May 8, 2008

### ehrenfest

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

6. May 8, 2008

### lurflurf

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)

7. May 8, 2008

### ehrenfest

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.

8. May 9, 2008

### HallsofIvy

Staff Emeritus
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.

9. May 9, 2008

### ehrenfest

I said I could do part a).

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