If we let $X$ be a complete metric space, and let $S:X\to X$ be a map, such that $S^m$ is a contraction. We now want to show, that $S$ has a unique fixed point

This is what I've thought so far:

Due to Banachs fixed point theorem is it enough to show, that $S$ is a contraction. Due to $S^m$ being a contraction, we know this about $S^m$(the definition of being a contraction): $$\exists\beta, 0\le\beta\lt1:d(S^mx,S^my)\le\beta d(x,y),\forall x,y\in X$$

I'm not really sure how to show that S is a contraction.. Any ideas ad to how to approach this?

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    $\begingroup$ You can prove this result without $S$ being a contraction. Anyway, I dont think that $S$ is a contraction. $\endgroup$ – Tomás Feb 27 '13 at 22:48
  • $\begingroup$ My book says that a fixed point is a $x\in X$ such that $Sx=x$. I thought this was only possible when S was a contraction? $\endgroup$ – MBrown Feb 27 '13 at 22:53
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    $\begingroup$ No, a map can have a fixed point without being an contraction. $\endgroup$ – Tomás Feb 27 '13 at 22:59
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    $\begingroup$ $f(x) = 10 x$. $f(x)=x$ iff $x=0$. $\endgroup$ – copper.hat Feb 27 '13 at 23:00
  • $\begingroup$ Ah, of course, that makes sense. Thank you for clarifying that! $\endgroup$ – MBrown Feb 27 '13 at 23:02

If $x,y$ are fixed points of $S$, then $d(S^n x, S^n y) = d(x,y)\leq \beta d(x,y)$, from which we conclude that $d(x,y) = 0$. Hence there is at most one fixed point.

Since $S^n$ is a contraction, it has a unique fixed point $x_0$. Then we have $x_k = S^k x_0$. Since $x_{k+n} = x_k$, we see that each $x_0,...,x_{n-1}$ is a fixed point of $S^n$, from which it follows that $x_k = x_0$ for all $k$, and hence $S x_0 = x_0$.

  • $\begingroup$ This shows that $S$ has at most one fixed point. $\endgroup$ – Zev Chonoles Feb 27 '13 at 22:52
  • $\begingroup$ $S(S^{m-1}(x))=x$ for at least one x because $S^m$ is a contraction $\endgroup$ – Sean Ballentine Feb 27 '13 at 22:53
  • $\begingroup$ @Sean: I agree that $x$ (the unique fixed point of $S^m$) is in the image of $S$; I'm still not seeing why $x$ must be fixed by $S$. $\endgroup$ – Zev Chonoles Feb 27 '13 at 22:55
  • $\begingroup$ Me neither, ignore me $\endgroup$ – Sean Ballentine Feb 27 '13 at 22:55
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    $\begingroup$ Well, $x_0$ is a fixed point of $S^n$. I am defining $x_k = S^k x_0$. So the sequence $x_0, x_1=S x_0, x_2 = S^2 x_0, .... x_{n-1} = S^{n-1} x_0$ must repeat every $n$ (since $S^n x_0 = x_0$). And since $x_0$ repeats, then so do all the subsequent points. $\endgroup$ – copper.hat Feb 27 '13 at 23:58

Here's another way around, which is from an exercise in Fixed point theory and applications Agarwal et al (2001).

In general, it's not necessarily true $\operatorname{S}$ is a contraction with respect to $d$. But it can be, provided you find a new metric $\sigma$, which is defined as: $$\sigma(x,y): = d(x,y)+\frac{1}{\beta}d(\operatorname{S}x, \operatorname{S}y)+ \ldots+\frac{1}{\beta^{m-1}}d(\operatorname{S}^{m-1}x, \operatorname{S}^{m-1}y)$$

It's not difficult to show it's indeed qualified as a metric.And we have: $$\sigma(\operatorname{S}x, \operatorname{S}y) = \beta \sigma(x,y) -\beta d(x,y) + \frac{1}{\beta^{m-1}}d(\operatorname{S}^{m}x, \operatorname{S}^{m}y)$$ Hence $\sigma(\operatorname{S}x, \operatorname{S}y) \leq \beta \sigma(x,y)$.


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