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A question involving selfcomposition. 
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#1
Jul2308, 11:29 PM

P: 244

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
Jul2408, 01:56 AM

Sci Advisor
P: 1,593

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
Jul2408, 11:09 AM

P: 244

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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