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Let me consider the iteration $x_{n+1}=Tx_n$ and $T$ is a self-map from a non-empty subset $K$ of a smooth Banach space $X$ to itself, satisfying $W(Tx, Ty) \leq W(x, y)$, where $W(x, y)=\Vert x \Vert^2-2 \langle x, Jy \rangle + \Vert y \Vert^2$. Here, $J$ is the normalized duality mapping $J:X \to X^{'}$ given by $$ J(x)=\{ Jx \in X^{'} : \langle x, Jx \rangle = \Vert x \Vert^2=\Vert Jx \Vert^2 \}. $$ It should be noted that $J$ is single-valued when $X$ is smooth. I have come to an inequality that $W(x_{n+1}, x_n) \leq W(x_n, x_{n-1})$. Then, using the above criteria, we will further reduce to the inequality $W(x_{n+1}, x_n) \leq W(x_1, x_0)$. Then we will get $$h(\Vert x_{n+1} - x_n \Vert) \leq W(x_{n+1}, x_n) \leq W(x_1, x_0);$$ where $h:[0, \infty) \to [0, \infty)$ is a continuous, strictly increasing, convex function with $h(0)=0$, does $(x_n)$ behaves like a Cauchy sequence here. If not, what extra condition is required to show that $\{x_n\}$ is Cauchy? Here $X^{'}$ is the dual of $X$. Kindly help me out. Thank you.

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    $\begingroup$ As stated, no. You can take $X = \ell_2(\mathbb{N})$. Then $W(x,y) = \|x - y\|^2$. Let $T$ be the shift map, and choose $x_1 = (1,0,0,\ldots)$. Then $x_n$ is not Cauchy. // But I wonder whether the question you typed above is actually the question you want to ask. Please double check and revise. $\endgroup$ Commented Mar 18 at 6:12
  • $\begingroup$ Maybe assumptions on K are in order $\endgroup$ Commented Mar 18 at 6:57
  • $\begingroup$ @Willie Wong, my question is: Under what assumptions on T or any other hypothesis does the sequence $\{x_n \}$ become Cauchy? $\endgroup$
    – PPB
    Commented Mar 18 at 8:09
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    $\begingroup$ I feel that this question is too broad. Certainly you have sufficient conditions like contraction mapping. But since your conclusion only concerns the values of $T$ on a countable subset, it certainly cannot be used to constrain $T$ in the general situation (so at the level of generalities I don't see any hope for a necessary condition). $\endgroup$ Commented Mar 18 at 13:08
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    $\begingroup$ Also, any particular reason you write "$\{x_n\}$ is Cauchy" and not that it converges? You are working in a Banach space after all. $\endgroup$ Commented Mar 18 at 13:09

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