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As a follow-up to my previous question Show that $T:\,c_0\to c_0\;\;$, $x\mapsto T(x)=(1,x_1,x_2,\cdots),$ is non-expansive. Let $X$ be a normed linear space and $X=c_0$ (the space of sequences of real numbers which converge to $0$). Define \begin{align} T:\,&c_0\to c_0 \\ &x\mapsto T(x)=(1,x_1,x_2,\cdots), \end{align} for arbitrary $x=(1,x_1,x_2,\cdots)\in c_0.$ I want to show that $T$ has no fixed points.

Remark: We have that

$$c_0=\{\bar{x}=(x_1,x_2,\cdots) :x_n\to 0\;\text{as}\;n\to \infty\}.$$

MY TRIAL

Suppose for contradiction that $T$ has fixed points in $X$, then there exists $u\in c_0$ s.t. $T(u)=u.$ That is, \begin{align} T(u_1,u_2,\cdots)=(u_1,u_2,\cdots), \end{align} Please, how do I draw out a contradiction from this?

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  • $\begingroup$ which book of this question can you suggest me..please $\endgroup$ – Inverse Problem Mar 24 at 4:38
  • $\begingroup$ @Inverse Problem: I got this from my the book "C.E. Chidume, Applicable Functional Analysis, Ibadan University Press Publishing House, 2014, University of Ibadan, Ibadan, Nigeria. ISBN: 978-978-8456-31-5". There are several other books too! $\endgroup$ – Omojola Micheal Mar 25 at 21:56
  • $\begingroup$ ........can you tell me...math.stackexchange.com/questions/3159556/… math.stackexchange.com/questions/3160256/… which have this questions? do you have idea $\endgroup$ – Inverse Problem Mar 26 at 15:30
  • $\begingroup$ Applicable Functional Analysis u have this book? $\endgroup$ – Inverse Problem Mar 26 at 15:37
  • $\begingroup$ @Inverse Problem: Yes, I do! $\endgroup$ – Omojola Micheal Mar 27 at 18:12
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$(1,u_1,u_2,...)=(u_1,u_2,..)$ implies $1=u_1,u_1=u_2$ etc so $u_n=1$ for all $n$. But then $(u_n) \notin c_0$.

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  • $\begingroup$ Sorry Sir, how's $u_n\notin c_0$? $\endgroup$ – Omojola Micheal Dec 14 '18 at 8:58
  • $\begingroup$ Let me guess, it's because $u_n$ does not go to zero and $n\to\infty.$ Therefore, $u_n\notin c_o. $ I'm I right? $\endgroup$ – Omojola Micheal Dec 14 '18 at 9:01
  • $\begingroup$ Yes, $u_n \to 1$. $\endgroup$ – Kavi Rama Murthy Dec 14 '18 at 9:02

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