Proving No Differentiable f Such That f Circ f = g

[mathwizarddud]

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]

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.

 

[Sqens]

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

 

[mrandersdk]

think you can do it by using CompuChip's idea and then use

http://mathworld.wolfram.com/FixedPointTheorem.html

 

[Sqens]

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

 

[matt grime]

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.

 

[Sqens]

How do you know that f' is continuous?

 

[mrandersdk]

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]

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...

 

[Sqens]

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

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

 

[matt grime]

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.

 

[mrandersdk]

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

 

[jostpuur]

<br /> A = \{x\in\mathbb{R}\;|\; f&#039;(x)&gt;0\}<br />

<br /> B = \{x\in\mathbb{R}\;|\; f&#039;(x)&lt;0\}<br />

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.

 

[Sqens]

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