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If $ \ T:X \longrightarrow X \ $ is contraction, then using Banach fixed-point theorem we know that the fixed point exists and all other points converge to that point. But what happens if $T$ is not contraction?

Let $\quad X = (0, \infty) \quad $ and $ \quad Tx = \sqrt{x} \quad $ for all $x \in X$.

Distance between any arbitrary points $ x,y \in X$ is always (strictly! (if $x \neq y$)) decreasing when we transform they using T, but because Lipschitz constant is not $ < 1 $, but equal $1$, we don't have contraction. But still, the fixed-point exists (it is equal $1$) and all other points converge to $1$, even if we don't have Banach fixed-point theorem.

Are there some theories, or generalizations of the Banach's theorem, that allows us to say something about transformations like mentioned one, while it does not meet conditions of normal Banach's theorem?

P.S. Sorry for mz half-baked english, feel free to edit if you spot some irregularities.

Edit: Like Eric Towers said it is not strictly decreasing, my mistake, but still, all $(T^{n}x)$ sequences are Cauchy sequences.

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  • $\begingroup$ Distance between arbitrary points does not strictly decrease. The length $T[1/9,1/4]$ is $1/6$, but the length of $[1/9,1/4]$ is $5/36 < 1/6$. $\endgroup$ – Eric Towers Mar 20 '14 at 22:35
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    $\begingroup$ Yes, there are many generalizations. So many that there is a journal devoted to publishing them. $\endgroup$ – user127096 Mar 24 '14 at 16:10
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I think that the phenomenon you describe is best explained by splitting your space into two parts. In each part there is a unique fixed point (which is, obviously, $1$ in both cases) which is obtained by iteration but for different reasons.

  1. on $[1,\infty[$ the mapping is a contraction---so you can use the BFPT.

  2. on $]0,1]$ because the function is increasing, each starting value produces a monotone increasing sequence which must converge to a fixed point and so to the only FP, i.e. $1$.

The importance of these abstract considerations for a simple equation that can be solved by inspection is, of course, that it applies to more general functions for which the FP equation might not be explicitly solvable by elementary methods.

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  • $\begingroup$ Thank you, I see it now :) $\endgroup$ – Kusavil Mar 26 '14 at 21:56
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There is the Brouwer fixed point theorem that says any continuous map on a closed convex subset of $\mathbb R^n$ has a fixed point. But simple 2D examples show that you won't find this via iteration.

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  • $\begingroup$ Thanks. Can you give an example of such a simple 2D example? By not being able to find this, do you mean that we can only prove the existence, but we cannot construct the fixed point itself? $\endgroup$ – Kusavil Mar 26 '14 at 21:53
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    $\begingroup$ Something like the function rotates by an angle $\theta$ on the unit disk. It has $0$ as its fixed point, but it is clear that iterating any point other than $0$ will not get this fixed point. $\endgroup$ – Stephen Montgomery-Smith Mar 26 '14 at 22:11
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it seems that you can use its dual space $X^{'}$

$\vert$$T_{n}^{'}(x)$$\vert$$\le$$\frac{\vert{T}\vert}{\sqrt{x}}$$\le$$\frac{1}{\sqrt{x}}$$\Vert{x}\Vert$$_{\infty}$

then, by the definition of dual space, you will find the fixed point !

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There is a Generalization of Banach's Fixed Point Theorem in which we may relax to the contraction requirement to that of a weak contraction, i.e. for a function $f:X \to X$ and $x,y \in X$

$$ d(x,y) > d(f(x),f(y)). $$

Notice that No Lipschitz coefficient is required under this definition. Now, if $f:X \to X$ is a such a weak contraction, and $X$ is compact, then we still have a unique fixed point.

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