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A question involving self-composition.

by zpconn
Tags: involving, selfcomposition
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zpconn
#1
Jul23-08, 11:29 PM
P: 243
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.
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Ben Niehoff
#2
Jul24-08, 01:56 AM
Sci Advisor
P: 1,588
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.
zpconn
#3
Jul24-08, 11:09 AM
P: 243
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?


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