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 $T$ is non-expansive.

Remark: We have that

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


Let $x,y\in c_0.$ WTS: there exists $0\leq k\leq 1,$ such that $\Vert T(x)-T(y) \Vert \leq k \Vert x-y \Vert . $

Now, \begin{align} \Vert T(x)-T(y) \Vert &=\Vert (1,x_1,x_2,\cdots)-(1,y_1,y_2,\cdots) \Vert\\&=\Vert ( 0,x_i-y_i)\Vert\\&= \sup\limits_{i\in \Bbb{N}} \{0,|x_i-y_i|\}\\&= \sup\limits_{i\in \Bbb{N}} \{|x_i-y_i|\}\\&=\Vert x-y \Vert \end{align} Taking $k=1,$ (based on mathworker21's advice and BigbearZzz's help), then

\begin{align} \Vert T(x)-T(y) \Vert \leq k \Vert x-y \Vert . \end{align} I'll be interested to see if you have a different kind of proof.

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    $\begingroup$ in your 'WTS', it's equivalent to take $k=1$. also do you know what the norm $||\cdot||$ is here? this problem is pretty trivial $\endgroup$ Dec 14, 2018 at 8:16
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    $\begingroup$ Yes, your proof is correct. $\endgroup$ Dec 14, 2018 at 8:21
  • $\begingroup$ @Kavi Rama Murthy: Thank you very much Sir. $\endgroup$ Dec 14, 2018 at 8:24

1 Answer 1


You've computed that $$\begin{align} ||Tx-Ty|| &= \sup \{0,|x_1-y_1|,|x_2-y_2|,\dots\}, \end{align}$$ whereas we know that $$ ||x-y|| = \sup \{|x_1-y_1|,|x_2-y_2|,\dots\}. $$ All you need to know to solve the problem is that norm for $c_0$ is the supremum norm.

  • $\begingroup$ Do we need to have $\sup?$, seeing that $c_0=\{\bar{x}=(x_1,x_2,\cdots) :x_n\to 0\;\text{as}\;n\to \infty\}.$ $\endgroup$ Dec 14, 2018 at 8:23
  • $\begingroup$ @Mike It's the standard norm on $c_0$ that makes it a Banach space. I think this should be written in your book/problem sheet. $\endgroup$
    – BigbearZzz
    Dec 14, 2018 at 8:24
  • $\begingroup$ The norm on $c_0$ is defined to be $||x|| = \sup_{i\in\Bbb N} |x_i|$. $\endgroup$
    – BigbearZzz
    Dec 14, 2018 at 8:25
  • $\begingroup$ Maybe I'll get top know that as time goes on! I'm teaching myself Functional analysis with the help of a book. Anyway, is my proof correct? $\endgroup$ Dec 14, 2018 at 8:26
  • $\begingroup$ Thanks about the norm info. That is an addition. Thanks! $\endgroup$ Dec 14, 2018 at 8:27

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