# $f:\mathbb R \to \mathbb R$ is a differentiable function such that $f'(x)\le r<1$ , does $f$ necessarily have a fixed point ? [duplicate]

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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 ?

## marked as duplicate by apnorton, Najib Idrissi, André Nicolas real-analysis StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Jan 5 '15 at 5:31

• @Woria That has a fixed point at $x=0$. – Matt Samuel Dec 29 '14 at 4:52
• If $f(x) = \frac{1}{2}x$, then $f(0) = 0$! – Robert Lewis Dec 29 '14 at 4:52
• @learnmore That doesn't satisfy $f'(x)<1$. – Matt Samuel Dec 29 '14 at 4:54
• @learnmore To answer your edit, $-x-3=x$ $\implies$ $-3=2x$ $\implies$ $x=-3/2$. – Matt Samuel Dec 29 '14 at 4:56
• @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 – user123733 Dec 29 '14 at 5:12
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 $x$.