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This question already has an answer here:

Let $f:\mathbb R \to \mathbb R$ be a differentiable function . If $\exists r \in \mathbb R $ such that $|f'(x)|\le r<1 , \forall x \in \mathbb R$ then using Lagrange's theorem one can show $f$ is a Lipscitz contraction and then use Banach contraction principle to conclude $f$ has a unique fixed-point. My question is what happens if $f'(x)\le r<1 , \forall x \in \mathbb R$ ? Then does $f$ even have a fixed point ?

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marked as duplicate by apnorton, Najib Idrissi, André Nicolas real-analysis Jan 5 '15 at 5:31

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ @Woria That has a fixed point at $x=0$. $\endgroup$ – Matt Samuel Dec 29 '14 at 4:52
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    $\begingroup$ If $f(x) = \frac{1}{2}x$, then $f(0) = 0$! $\endgroup$ – Robert Lewis Dec 29 '14 at 4:52
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    $\begingroup$ @learnmore That doesn't satisfy $f'(x)<1$. $\endgroup$ – Matt Samuel Dec 29 '14 at 4:54
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    $\begingroup$ @learnmore To answer your edit, $-x-3=x$ $\implies$ $-3=2x$ $\implies$ $x=-3/2$. $\endgroup$ – Matt Samuel Dec 29 '14 at 4:56
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    $\begingroup$ @user21820: But none of the answers $f(x)=\sqrt {x^2+1}$ ; $f(x)=\log(1+e^x)$ satisfies the $r$ condition ; for example if there existed such an $r$ for say the second function then $e^x/(1+e^x) \le r , \forall x \implies \lim_{x \to \infty} e^x/(1+e^x)=1 \le r <1$ contradiction! I don't know how that answer got accepted $\endgroup$ – user123733 Dec 29 '14 at 5:12
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If $f$ has no fixed point then $f(x)<x$ for all $x$ or $f(x)>x$ for all $x$. Assume that $f(x)>x$ for all $x$, then $f(x)-f(0)>x-f(0)$ for all $x>0$. Take $x=f(0)t$ for $t>0$, then $f(f(0)t)-f(0)>f(0)(t-1)$. Dividing $f(0)t$ both sides then $$\frac{f(f(0)t)-f(0)}{f(0)t}\ge 1-\frac{1}{t}.$$ By mean value theorem then we can find for some $c\in (0,f(0)t)$ such that $f'(c)\ge 1-1/t$, so $f'$ cannot be bounded by $r<1$.

Similarly, you can check that $f'$ cannot be bounded by $r<1$ in the case $f(x)<x$ for all $x$.

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