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Let $(X,d)$ a metric space, $\alpha >0$ (fixed ) and $T: X \rightarrow X$ a map such that exist $n \in N$, where :

$$ d(T^n x , T^n y) \leq \alpha^n d(x,y), \forall x,y \in X$$

Define $h(x,y) = [d^2 (x,y) + \frac{1}{\alpha^2} d(Tx,ty)+...+ \frac{1}{\alpha^{2(n-1)}} d(T^{n-1}x,T^{n-1}y)]^{1/2}$. I want to prove that $h $ is a metric on $X$. My problem is the triangle inequality ... I am getting anywhere .. Someone could give me a help?

Thanks in advance

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  • $\begingroup$ When you say 'for a fixed $n\in N$...' is that we only know there is some $n$, or it works for all $n$? I just want to be clear on the quantifier. $\endgroup$ – Dom Jan 6 '15 at 5:06
  • $\begingroup$ "there is some n" is the right interpretation. thanks for note this. thanks for your commentary . I edited the question $\endgroup$ – math student Jan 6 '15 at 5:18
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First, I suppose that your $h(x,y)$ is $$h(x,y)=[d^2 (x,y) + \frac{1}{\alpha^2} d^2(Tx,Ty)+...+ \frac{1}{\alpha^{2(n-1)}} d^2(T^{n-1}x,T^{n-1}y)]^{1/2}$$ Please edit if this is true. Now put $$u(x,y)=(d (x,y), \frac{1}{\alpha} d(Tx,Ty),\cdots, \frac{1}{\alpha^{n-1}} d(T^{n-1}x,T^{n-1}y))\in \mathbb{R}^n$$, and $u(x,y)=(u_k(x,y), 0\leq k\leq n-1)$. Let $N$ be the Euclidian norm on $\mathbb{R}^n$. Then you have $h(x,y)=N(u(x,y))$.

Let now be $x,y,z$ fixed. For each $k$ we have $$u_k(x,y)\leq u_k(x,z)+u_k(z,y)$$ Hence $$h(x,y)\leq N(u(x,z)+u(z,y))\leq N(u(x,z))+N(u(z,y)=h(x,z)+h(z,y)$$ and we are done

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