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Mar 18 at 18:13 comment added PPB Indeed, I am talking about the convergence of $\{x_n\}$ here. You are right, sir. It may need some other criteria to be fulfilled for the convergence of $\{x_n\}$.
Mar 18 at 13:09 comment added Willie Wong Also, any particular reason you write "$\{x_n\}$ is Cauchy" and not that it converges? You are working in a Banach space after all.
Mar 18 at 13:08 comment added Willie Wong 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).
Mar 18 at 8:17 history edited PPB CC BY-SA 4.0
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Mar 18 at 8:15 comment added PPB Also, I have made some changes to clarify my question. Thank you.
Mar 18 at 8:12 history edited PPB CC BY-SA 4.0
added 210 characters in body
Mar 18 at 8:09 comment added PPB @Willie Wong, my question is: Under what assumptions on T or any other hypothesis does the sequence $\{x_n \}$ become Cauchy?
Mar 18 at 6:57 comment added Pietro Majer Maybe assumptions on K are in order
Mar 18 at 6:12 comment added Willie Wong 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.
Mar 18 at 3:32 history asked PPB CC BY-SA 4.0