Composition of Functions, Derivatives, Hard Problem

[JG89]

Homework Statement

Does there exist a continuous function f: R -> R such that $$f'(f(x)) = x$$?

Homework Equations

The Attempt at a Solution

I was trying to find an example, but wasn't able to, and if I had to take a guess whether such a function exists or not, I'd say no. Here is my "proof" (quotation marks because I think the proof is flawed):

Assume $$f'(f(x)) = x$$. Then integration gives us $$f(f(x)) = \frac{x^2}{2}$$. Now, $$(f(f(x))' = f'(f(x))*f'(x) = xf'(x)$$. But since $$f(f(x)) = \frac{x^2}{2}$$ then $$(f(f(x))' = (x^2/2)' = x$$. So $$xf'(x) = x$$, implying f'(x) = 1. But this contradicts the assumption that f'(f(x)) = x.


I think the flaw is where I did the integration. I don't think I integrated properly.

 

[lanedance]

try differentiating as a check
$$\frac{d}{dx}f(f(x)) = f'(f(x))*f'(x)$$