Prove that for every $x\in M$ the sequence $\{T^nx\}$ converges to a fixed point of $T$.

Question

Let $(M, d)$ be a complete metric space, let $T:M\to M$ be a continuous map and let $\varphi:M\to\mathbb{R}$ be a function which is bounded below. Assume that together they satisfy $$d(x,Tx)\leq\varphi(x)-\varphi(Tx)$$ Prove that for every $x\in M$ the sequence $\{T^nx\}$ converges to a fixed point of $T$.

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My idea is to use banach fixed point theorem, that is, we need to show $d(Tx,Ty)\leq cd(x,y)$ where $c\in[0,1)$. $$d(Tx,Ty)\leq d(Tx,x)+d(x,y)+d(y,Ty)\leq\cdots$$ I don't follow the condition of $\varphi$. And that confuses me how $T$ and $\varphi$ satisfy the inequality.

Hints and good observations are welcome! Thanks in advance!

Answer 1

Hints: $d(T^{n}x,T^{n+1}x) \leq \phi (T^{n}x)-\phi (T^{n+1}x)$. From this and the fact that $\phi$ is bounded below show that $\sum d(T^{n}x,T^{n+1}x) <\infty$ [You will have telescopic sum on the right hand side]. From this and triangle inequality show that $(T^{n}(x))$ is Cauchy sequence. If $T^{n}x \to y$ prove that $Ty=y$.