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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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2 Answers 2

up vote -3 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 ${f(x)-f(x')\over x-x'}=f'(c)<1$, $c\in [x,x']$ –  El Angel Exterminador 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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