I can't for the life of me figure this one out all the way.(adsbygoogle = window.adsbygoogle || []).push({});

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.

**Physics Forums - The Fusion of Science and Community**

# A question involving self-composition.

Have something to add?

- Similar discussions for: A question involving self-composition.

Loading...

**Physics Forums - The Fusion of Science and Community**