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Can someone give me an idea how to generalize Banach's fixed-point theorem for complete metric spaces such that the constant contraction coefficient $c$ (as in $d(Tx,Ty)\leq c \ d(x,y)$ ) may be replaced by a sequence $(r_n)_{n \in \mathbb{N}}$ so that $T:X\rightarrow X$ has a unique fixed point if $d(Tx,Ty)\leq r_n d(x,y)$ holds for all $x,y \in X$ and $\sum r_n$ converges?

I tried modifying the original proof, but of $\sum r_n$ only converges (not necessarily with a limit in the interval $[0,1)$ I can't use the original idea...

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How should we interpret inequality $d(Tx,Ty)\leq r_n d(x,y)$. Does it holds for all $n\in\mathbb{N}$ or some $n\in\mathbb{N}$? –  no identity Dec 10 '11 at 20:42
    
I think that the statement should be "for every $x,y$ there exists $r_n$ with the property" ... –  Beni Bogosel Dec 10 '11 at 21:00
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Then this definition is useless. Because we have $d(Tx,Ty)\leq(\inf\limits_{n\in\mathbb{N}}r_n) d(x,y)$. Moreover, since $\sum\limits_{n\in\mathbb{N}}r_n$ converges we have $\inf\limits_{n\in\mathbb{N}}r_n=0$ –  no identity Dec 10 '11 at 21:22
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Can you please state the result that you want to prove in Theorem form? –  leo Dec 11 '11 at 0:18

1 Answer 1

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There is a simple result, if the space is compact, all you need is that whenever $x \neq y,$ we have $d(Tx,Ty) < d(x,y).$ The trick is to define $f(x) = d(x,Tx).$ As the space is compact, the minimum value of $f$ is achieved at some point, call it $x_0.$ If $f(x_0) \neq 0,$ that is $Tx_0 \neq x_0,$ we have a contradiction from $$ f(Tx_0) = d(Tx_0, T(Tx_0)) < d(x_0, Tx_0) = f(x_0).$$

I'm afraid I cannot make any sense of your version with the $r_n.$ Meanwhile, if you actually want a vector space over the reals, that is not going to be compact. That comes to mind because of your use of the notation $Tx$ for your function, which suggests an interest in linear transformations. I'm not sure.

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