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For a differentiable function $f:\mathbb R\to\mathbb R$ I need to choose the correct statement(s):

  1. $f'(x)\le r<1~\forall~x\in\mathbb R\implies f$ has at least one fixed point.

  2. $f'(x)\le r<1~\forall~x\in\mathbb R\implies f$ has a unique fixed point.

My attempt:

  • $g:\mathbb R\to\mathbb R:x\mapsto f(x)-x$ is differentiable on $\mathbb R.$ The search is for the number of roots of $g.$ Now $f'(x)\le r<1~\forall~x\in\mathbb R\implies g'(x)\leq r-1<0~\forall~x\in\mathbb R\implies g$ is decreasing everywhere on $\mathbb R.$ Can I say something from this observation$?$
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$g$ is strictly decreasing, so it has at most one zero. Thus $f$ has at most one fixed point. Now, since $g'(x) \leqslant r-1$ for all $x$, we have $g(x) \leqslant g(0) + x\cdot (r-1)$ for $x > 0$, hence $\lim\limits_{x\to+\infty} g(x) = -\infty$, and for $x < 0$ we have $g(x) > g(0) + x\cdot (r-1) = g(0) + \lvert x\rvert\cdot \lvert r-1\rvert$, hence $\lim\limits_{x\to -\infty} g(x) = +\infty$. By the intermediate value theorem, $g$ has at least one zero, so $f$ has at least one fixed point. Together it follows that such an $f$ has a unique fixed point. – Daniel Fischer Feb 21 '15 at 15:32
up vote -2 down vote accepted

MVT will imply that $|f(x)-f(x')|<|x-x'|, x\neq x'$, all three functions satisfies your conditions but none has fixed points, You need compact metric space to have fixed points other than $\infty$,$f(x)=x+\frac{1}{x},x\in[1,\infty)$, $f(x)=\sqrt{x^2+1},x\in\mathbb{R}$, $f(x)=\log(1+e^x),x\in\mathbb{R}$

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excuse me can you explain what are you getting from MVT? – Topology Jun 6 '14 at 11:11
MVT on $[x,x']$ allows me ${f(x)-f(x')\over x-x'}=f'(c)<1$, $c\in [x,x']$ – Un Chien Andalou Jun 6 '14 at 15:27
Wait your examples don't satisfy the condition if $r$ is fixed. – user21820 Dec 29 '14 at 5:10

I can show that if it has fixed point than it has to be unique. Suppose you have two fixed points $x_1<x_2$. Than by MVT there is $\xi$ that $x_2-x_1=f(x_2)-f(x_1) = f'(\xi)(x_2 - x_1 )=r(x_2-x_1)$. So $r=1$ and that is contradiction.

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I don't see why your last step should be true. We do not require $f'\equiv r$, only $f'\leq r$. But this will still yield the desired contradiction. – PhoemueX Dec 29 '14 at 5:22

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