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Let $f:\mathbb{R} \to \mathbb{R}$ be a differentiable function. Suppose $|f'(x)|\le r<1,\forall x\in \mathbb{R}$ and for some $r\in \mathbb{R}$.Then by contraction mapping theorem f has a unique fixed point in $\mathbb{R}$. Now suppose the inequality changes as $f'(x)\le r<1, \forall x\in \mathbb{R}$ and for some $r\in \mathbb{R}$. Then is it true that $f$ has at least one fixed point. What about uniqueness if the existence is true? Conversely if there is unique fixed point for such a differentiable function, is it necessary that $f'(x)\le r<1, \forall x\in \mathbb{R}$ and for some $r\in \mathbb{R}$?

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Both existence and uniqueness still hold. You can easily see geometrically it by noticing that $f$ will always be increasing less than $id(x)=x$ and a fixed point is the same as an intersection of the graph of $f$ with the diagonal of $\mathbb{R}^2$ (which is the graph of $id$).

Formally, let $x\in\mathbb{R}$ and suppose $f(x)>x$. Let $k=\frac{f(x)-x}{1-r}$, which solves the equation $$f(x)+kr = x+k\ .$$ Then $$f(x+k) = f(x)+\int_x^{x+k}f'(t)dt\le f(x)+kr = x+k\ .$$ By the intermediate value theorem, it follows that $f$ has a fixed point. A similar proof gives a fixed point if $f(x)<x$, proving existence.

For uniqueness, suppose $x_0<x_1$ are two distinct fixed points. Then $$f(x_1) = f(x_0)+\int_{x_0}^{x_1}f'(t)dt \le x_0+(x_1-x_0)r<x_1\ ,$$ giving a contradiction. Thus the fixed point must be unique.

Notice that this works for $f:\mathbb{R}\to\mathbb{R}$, not for general functions in metric spaces.

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  • $\begingroup$ I cannot notice "f will always be increasing" $\endgroup$ – supremum Jul 16 '17 at 16:11
  • $\begingroup$ @supremum Read all the sentence, please, not just some part of it. $\endgroup$ – Daniel Robert-Nicoud Jul 16 '17 at 16:15
  • $\begingroup$ isn't f(x)=-2x satisfy those conditions still being decreasing and not comparable with Id(x) in general. $\endgroup$ – supremum Jul 16 '17 at 16:18
  • $\begingroup$ @supremum "increasing less than" does not preclude the function from being decreasing, it only says that it will never grow faster than it. $\endgroup$ – Daniel Robert-Nicoud Jul 16 '17 at 16:19
  • $\begingroup$ @supremum Also, you should perfectly understand what I mean from the formal proof below the intuitive statement. $\endgroup$ – Daniel Robert-Nicoud Jul 16 '17 at 16:21
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The function $g(x) := x- f(x)$ is differentiable and $$ g'(x) = 1 - f'(x) \geq 1 - r \qquad \forall x\in\mathbb{R}. $$ Since $1-r>0$, this proves that $g$ is strictly increasing (hence it can have at most one zero) and that $$ \lim_{x\to -\infty} g(x) = -\infty, \qquad \lim_{x\to +\infty} g(x) = +\infty, $$ so it has at least one zero (being continuous). The unique zero of $g$ is the unique fixed point of $f$.

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$f: x\mapsto \sin (x) $ has a unique fixed point $x=0$ but $f'(2\pi)=1$.

$g:x\mapsto 2x $ has one fixed point but $g'(x)=2$.

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