# Show that $f$ has at most one fixed point

Let $f\colon\mathbb{R}\to\mathbb{R}$ be a differentiable function. $x\in\mathbb{R}$ is a fixed point of $f$ if $f(x)=x$. Show that if $f'(t)\neq 1\;\forall\;t\in\mathbb{R}$, then $f$ has at most one fixed point.

My biggest problem with this is that it doesn't seem to be true. For example, consider $f(x)=x^2$. Then certainly $f(0)=0$ and $f(1)=1 \Rightarrow 0$ and $1$ are fixed points. But $f'(x)=2x\neq 1 \;\forall\;x\in\mathbb{R}$. Is there some sort of formulation that makes this statement correct? Am I missing something obvious? This is a problem from an old exam, so I'm assuming that maybe there's some sort of typo or missing condition.

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If $f(x)=x^2$, then $f\,'(1/2)=2(1/2)=1$, so $f$ does not satisfy the hypotheses of the theorem. –  Brian M. Scott Oct 19 '12 at 18:33
Okay. I think I misunderstood something. I was assuming that the contrapositive is "if $f$ has more than one fixed point, then $f'(t)=1\;\forall\;t$" but in reality, the contrapositive is "if $f$ has more than one fixed point, then $\exists\;t$ such that $f'(t)=1$." So it was me missing something obvious. Thank you. –  chris Oct 19 '12 at 18:39

This problem is straight out of baby Rudin.

Assume by contradiction that $f$ has more than one fixed point. Select any two distinct fixed points, say, $x$ and $y$.

Then, $f(x) = x$ and $f(y) = y$. By the Mean Value Theorem, there exists some $\alpha \in (x,y)$ such that $f'(\alpha) = \frac{f(x)-f(y)}{x-y} = \frac{x-y}{x-y} = 1$, contradicting the hypothesis.

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Thank you. I was missing something obvious. I thought that the derivative needed to be $1$ for every $t$, not that there was some $t$ for which the derivative equaled one. –  chris Oct 19 '12 at 18:41
One good rule to remember is that when you have quantifiers, to negate the statement, you change "for alls" to "there exists" and reverse any inequalities/equalities. So your hypothesis is $\forall t (f'(t)\neq 1)$. Reversing this, you get $\exists t (f'(t)=1)$. So it is sufficient to show that there is a single $t$. :) –  Arkamis Oct 19 '12 at 18:43
Good point. Thanks again. –  chris Oct 19 '12 at 18:45

HINT: Let $g(t)=f(t)-t$. If $f\,'(t)\ne 1$ for each $t\in\Bbb R$, then $g'(t)\ne 0$ for all $t\in\Bbb R$. It follows that either $g'(t)>0$ for all $t\in\Bbb R$, or $g'(t)<0$ for all $t\in\Bbb R$; see Darboux’s theorem.

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Well, basically, if $f'(t)>1$ then, as this is the slope of $f$ at $x=t$, it means that $f$ grows faster than the $y=x$ function. Similarly for the case $f'(t)<1$, then it grows slower.

The basic thing, is that the range of the derivative, though not necessarily continuous, is still satisfies: if $f'(t_1)>1$ and $f'(t_2)<1$ then there will be a $t$ between with $f'(t)=1$.

So, suppopse, $x_0$ is a fixed point, then either $f$ grows faster, or slower..

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