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Let $(X,d)$ be a complete metric space, $a \in X$ and $r>0$. Let $f:O(a,r)→X$ be a contraction with a contraction constant $L$, $d(f(x),f(y))<Ld(x,y)$, for all $x,y \in O(a,r)$ and $d(f(a),a) \le \alpha r$. If $\alpha = 1-L$, can we prove that the function $f$ has a unique fixed point in $O(a,r)$, or is there a counterexample?

It is easy to prove that when $0<\alpha < 1-L$, the function $f$ has a unique fixed point in $O(a,r)$. But $\alpha = 1-L$ seems difficult for me.

I find a similar question here, but it is a little different.

shwsq
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  • Did you check in the book Fixed Point Theory by Granas and Dugundji? – Jakobian Sep 19 '24 at 03:08
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    A little less general, but corollary 1.2 of that book answers that $f$ must have a fixed point if $d(f(a), a)\leq \alpha r$ is replaced by $d(f(a), a) < \alpha r$. – Jakobian Sep 19 '24 at 03:14
  • Oh yes I have solved it! Thank you very much! – shwsq Sep 19 '24 at 03:33
  • I have come up with another question. If $d(f(x),f(y))<Ld(x,y)$ is changed into $d(f(x),f(y)) \le Ld(x,y)$, is it still true? – shwsq Sep 19 '24 at 03:39

1 Answers1

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That's not true when $d(f(x),f(y))<Ld(x,y)$ is changed into $d(f(x),f(y))\leq Ld(x,y)$.

For a counterexample, let $r=1, L=\alpha=\frac{1}{2}, X =[0, 1],a=0$ where $X$ is equipped with Euclidean metric.

Define $f:O(0, 1)\to X$ by $f(x) = \frac{1}{2}+\frac{1}{2}x$, then $f$ satisfies the conditions, but $f$ doesn't have a fixed point in $O(0,1) = [0, 1)$.

Jakobian
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Isllier
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