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I'm trying to use the mean value theorem to show that for differentiable $f:\mathbb{R} \to \mathbb{R}$, if $f'(x) < 1 \forall x\in\mathbb{R}$, then $f$ has exactly one fixed point. I know how to show that if $f'(x)\neq 1$ then $f$ can have at most one fixed point. I think I need to use that $f'(x) \leq 1$ to see that $f$ is decreasing, but I'm not sure how to use that.

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  • $\begingroup$ If $f'\le 1$, $f$ need not to be decreasing. See $f(x)=\frac{1}{2}x$. $\endgroup$ – szw1710 May 3 '17 at 19:36
  • $\begingroup$ You need a strict inequality, $f \colon x \mapsto x$ satisfies $f'(x) \leqslant 1$ for all $x$ but has more than one fixed point. $f \colon x \mapsto x+ 1$ has no fixed point. $\endgroup$ – Daniel Fischer May 3 '17 at 19:38
  • $\begingroup$ I realized I had a typo. It should have ben that $f'(x) < 1$ not that $f'(x) \leq 1$. $\endgroup$ – TonicWaterWithLime May 3 '17 at 19:41
  • $\begingroup$ This doesn't yet imply that $f$ has a fixed point. Consider $f(x) = x - e^x$. Even $\lvert f'(x)\rvert < 1$ for all $x$ doesn't imply the existence of a fixed point. $\endgroup$ – Daniel Fischer May 3 '17 at 19:48
  • $\begingroup$ I think if $f(0) > 0$ is specified then this can be proved. $\endgroup$ – Χpẘ May 3 '17 at 19:54
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Let $f(x) = x+1$ for $x \in \mathbb{R}$. Then $f'(x) = 1$ for all $x$ and $f$ has no fixed points.

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    $\begingroup$ Strictly speaking, this is not an answer but a counterexample. Placing as a comment would help the OP revise his question. $\endgroup$ – mlc May 3 '17 at 19:40
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It must be $f'(x)<1$. Then let $a,b$ be two fixed points. Then you have $|f(a)-f(b)|=f'(x)|a-b|=f'(x)|f(a)-f(b)|$, so $f'(x)=1$, that is not possible. The first equality is due to Lagrange Theorem.

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The thing you are trying to prove is false, even with strict inequality.

$$ f(x) = \frac{x + \sqrt {1+x^2}}{2} $$ has derivative below $1$ but no fixed point: we always have $f(x) > x$

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    $\begingroup$ Better than my example, since $|f'(x)|<1$, not just $f'(x)<1$. $\endgroup$ – Thomas Andrews May 3 '17 at 20:00
  • $\begingroup$ @szw1710 How can you define the sequence $(x_n)_{n=1}^{\infty}$ such that $f(x_n) = nx_n$? For one, you'd already have a fixed point with $x_1$. $\endgroup$ – Tom May 3 '17 at 20:09
  • $\begingroup$ @szw1710 Let $f: [1,2] \to \Bbb{R}$ be defined by $f(x) = 0$ for all $x \in [1,2]$. Define the sequence $(x_n)$. $\endgroup$ – Tom May 3 '17 at 20:13
  • $\begingroup$ I have deleted my two comments because there was a mistake there. @Tom referred to them. $\endgroup$ – szw1710 May 3 '17 at 20:36

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