# Pittnauer’s Fixed Point Theorem

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.

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.
• I am just curious why do you consider $x_{p-n}$ in the inequality? what happens if $p<n$? Why do way that $\sum_p d(x_p,x_{p+1})<\infty$. – Nothingone Jul 30 '19 at 23:42
• 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. – Mindlack Jul 30 '19 at 23:50