# Prove there's no differentiable f such that f(f(x)) = g(x) with g'(x) < 0 for every x

Suppose the real valued $$g$$ is defined on $$\mathbb{R}$$ and $$g'(x) < 0$$ for every real $$x$$. Prove there's no differentiable $$f: R \rightarrow R$$ such that $$f \circ f = g$$.

CompuChip
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My suggestion would be to look at (f o f)' (using the chain rule) and try to find some x for which this is necessarily > 0.

By the way, double posting here was unnecessary and is generally not appreciated here.

You want to prove that f is monotone. Then, you want to prove that f isn't monotone.

I don't see it. How do you show that f is bounded?

matt grime
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It doesn't matter whether f is bounded or not. It is clear that f'(x) and f'(f(x)) have opposite signs for all x. Pick any x. If f(x)=x, then this is a contradiction. If not, then by the intermediate value theorem, f' is zero at some point between x and f(x). That is essentially what the fixed point theorem says.

How do you know that f' is continuous?

whe only need to know that f is continuous to get f(x) = x, and by assuming that f is differential it is.

matt grime
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OK, let's assume that the function is once continuously differentiable if that makes you feel better. Off the top of my head I can't think of a function that is differentiable, but where the derivative is discontinuous....

The fixed point theorem requires f to be bounded. Otherwise, take f(x) = x + 1

See http://en.wikipedia.org/wiki/Smooth_function" [Broken] for a differentiable but not C^1 function.

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matt grime
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If you note, I didn't use the fixed point theorem, just the intermediate value theorem (I've no idea why anyone would invoke the f.p.t. - at best its proof is enlightening for this question). And I suspect that original poster's question was phrased in terms of smooth functions making the exceptional case of a differentiable but not C^1 function important as an aside.

my bad didn't see that the fix point theorem needed bounded. matt_grimes proof works, if you assume that f' is continuous.

$$A = \{x\in\mathbb{R}\;|\; f'(x)>0\}$$

$$B = \{x\in\mathbb{R}\;|\; f'(x)<0\}$$

Using continuity of $$f$$ and checking some preimages, don't you get a contradiction with the fact that the real line is connected?

edit: Ouch. I wasn't thinking this all the way to the end. Never mind if this is not working.

Anyway, my approach was to apply the mean value theorem to show that f is injective. This implies something contradictory about f'.