# A question involving self-composition.

1. Jul 23, 2008

### zpconn

I can't for the life of me figure this one out all the way.

Suppose f : R -> R is differentiable, and consider g(x) = f(f(x)). Show that if g is monotone decreasing, then g must be constant.

Here's what I've done so far (I'd hesitate to call it "progress"):

By the chain rule, g'(x) = f'(f(x)) f'(x). Suppose that g is strictly decreasing so that f'(f(x)) f'(x) < 0. One of the factors is positive and the other is negative. Since, by Darboux's theorem, f'(x) has the intermediate value property, there exists a q between x and f(x) such that f'(q) = 0. Then g'(q) = f'(f(q)) f'(q) = 0, a contradiction. Therefore g is not strictly decreasing.

I've investigated the fixed points of f and found lots of interesting facts, but none of them seems to lead anywhere on this problem.

Any ideas or suggestions? Thanks a lot.

2. Jul 24, 2008

### Ben Niehoff

You have your proof already. "Strictly decreasing" and "monotone decreasing" mean two different things. If g is monotone decreasing but not strictly decreasing, then g must be constant.

3. Jul 24, 2008

### zpconn

Why is that?

If g is monotone decreasing and not strictly decreasing, than means nothing more than that there's at least one u at which g'(u) = 0, i.e. g has at least one inflection point. However, if g were constant, it'd have g'(x) = 0 everywhere. Is there some standard result about monotone but not strictly decreasing functions that I'm unaware of?

Last edited: Jul 24, 2008