# Rudin 8.6

## Homework Statement

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.

## 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).

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

For (a), would it not be easier to compare the power series of f(x) and ecx.
Perhaps, but I don't know how to prove that f is analytic.

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

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

lurflurf
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)

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)
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.

HallsofIvy
Homework Helper
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.

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.
I said I could do part a).

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