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*I would only like a hint! Not a full proof.


If a function $f(x)$ is differentiable on $\mathbb{R}$ and $f'(x) < 1$ for all $x\in\mathbb{R}$, then $f(x)$ has at most one fixed point.

So far, I have applied the Mean Value Theorem and developed the following inequality, which may or may not be of any help:

$$f(b)-b < f(a) - a$$

For all $a,b\in\mathbb{R},\ a<b$. I have a feeling this is much easier than I am making it.

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Draw some graphs of functions with this property and use your mathematical intuition. – Ragnar Dec 21 '13 at 1:25
Suppose there are two fixed points and then apply the mean value theorem. – copper.hat Dec 21 '13 at 1:25
up vote 7 down vote accepted


Suppose $\;a<b\;$ are fixed points, then

$$\exists\,c\in (a,b)\;\;s.t.\;\; 1=\frac{f(b)-f(a)}{b-a}=f'(c)$$

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Ok guys, all of your answers have helped, so I wanted to post the proof I developed for critique:

Let $f$ be differentiable on $\mathbb{R}$, and $f'(x)<1$ for all $x\in\mathbb{R}$. Suppose $a,b\in\mathbb{R}$ are fixed points of the function $f$ such that $a<b$. Then, by the Mean Value Theorem, there exists a $c\in(a,b)$ such that: $$f'(c) = \frac{f(b)-f(a)}{b-a}.$$ Using the fact that $a$ and $b$ are fixed point gives: $$f'(c) = \frac{f(b)-f(a)}{b-a} = \frac{b-a}{b-a} = 1.$$ Which is a contradiction. So, $f$ cannot have more than 1 fixed point.

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That's fine...and exactly the very same proof as in my answer. :) – DonAntonio Dec 21 '13 at 1:44
Ha, yeah I figured I would at least type it up and make sure there aren't any gaps in my understanding. Thanks for your help Don! – chs21259 Dec 21 '13 at 1:50

Hint: Consider function $g(x)=f(x)-x$ and use property of monotonicity, especially on the zero point of $g(x)$.

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