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F. Pittnauer’s Fixed Point Theorem. Let $(X,d)$ be a complete metric space and $T:X\rightarrow X$ continuous. Assume that there exists an integer $n$ and $0\leq k<1$ such that $$d(Tx,Ty)\leq k[d(x,T^nz)+d(y,T^nz)]$$ for all $x,y,z\in X$. Then $T$ has a unique fixed point.

I have been struggling proving this problem. I tried to apply the triangle inequality on the right hand side of the given inequality to simplify the terms, but it is getting messy by doing so. I am lost, so I literally need help.

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The uniqueness is easy.

For the existence, define $x_p=T^p(x_0)$ for any $x_0$ and show that for $p \geq n$, $d(Tx_p,Tx_{p+1}) \leq kd(x_p,T^nx_{p-n})+kd(x_{p+1},T^nx_{p-n})$.

As a consequence, $\sum_p{d(x_p,x_{p+1})}$ converges, thus $(T^p(x_0))_p$ converges and the rest is standard.

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  • $\begingroup$ I am just curious why do you consider $x_{p-n}$ in the inequality? what happens if $p<n$? Why do way that $\sum_p d(x_p,x_{p+1})<\infty$. $\endgroup$ – Nothingone Jul 30 at 23:42
  • $\begingroup$ I noticed that the given inequality became a contraction inequality if $x$ or $y$ was in the range of $T_n$. If $p < n$ we don’t know enough to apply the inequality interestingly. $\endgroup$ – Mindlack Jul 30 at 23:50

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