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If we are given a complete metric space X and $f$ a contraction from into itself, then the Banach fixed Point theorem assures the existance of a unique fixed Point. How about the reverse of this conclusion, i.e. given a fixed point of a function defined on a complete metric space, can we suggest that it will be a contraction close to the fixed point ? If not, under which conditions it will be a contraction ?

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  • $\begingroup$ No. Take a linear (or "affine") function $f:\mathbb{R}\rightarrow\mathbb{R}$ and play around with the slope. $\endgroup$ – Michael Oct 16 '17 at 22:16
  • $\begingroup$ A more interesting question would be the following: Given we have a unique fixed point $x_0$ of $f$ in $X$, is there a bijective mapping $g : X\to X$ such that $g^{-1}\circ f\circ g$ is a contraction and $g(x_0) = x_0$? $\endgroup$ – amsmath Oct 16 '17 at 22:59
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Let $f:\Bbb R\to\Bbb R$, $f(x)=2x$. The unique fixed point is $0$, while $f$ is celarly not a contraction, rather expanding map.

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  • $\begingroup$ Thanks. In the case it is not differentiable at the fixed point, i want to figure out under which conditions $f$ will be a contraction ? $\endgroup$ – user249018 Oct 16 '17 at 22:37
  • $\begingroup$ A contraction in $\mathbb R$ is a.e. differentiable since it is Lipschitz continuous and therefore absolutely continuous and therefore of bounded variation. $\endgroup$ – amsmath Oct 16 '17 at 22:47
  • $\begingroup$ A.e. differentiable means that it must not be differentiable at the fixed point. Do you agree ? $\endgroup$ – user249018 Oct 16 '17 at 22:59
  • $\begingroup$ @user249018 Completely wrong. There are many differentiable functions with fixed points. A differentiable function is (in particular) almost differentiable. $\endgroup$ – szw1710 Oct 17 '17 at 6:25

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