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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.

TonicWaterWithLime
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3 Answers3

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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.

Tom
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    Strictly speaking, this is not an answer but a counterexample. Placing as a comment would help the OP revise his question. – mlc May 03 '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.

Bargabbiati
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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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Will Jagy
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    Better than my example, since $|f'(x)|<1$, not just $f'(x)<1$. – Thomas Andrews May 03 '17 at 20:00
  • @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 $x_1$. – Tom May 03 '17 at 20:09
  • @szw1710 Let $f: [1,2] \to \Bbb{R}$ be defined by $f(x) = 0$ for all $x \in [1,2]$. Define the sequence $(x_n)$. – Tom May 03 '17 at 20:13
  • I have deleted my two comments because there was a mistake there. @Tom referred to them. – szw1710 May 03 '17 at 20:36