# 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 $$T$$ is non-expansive.

Remark: We have that

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

MY TRIAL

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.

• 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 Dec 14, 2018 at 8:16
• Yes, your proof is correct. Dec 14, 2018 at 8:21
• @Kavi Rama Murthy: Thank you very much Sir. Dec 14, 2018 at 8:24

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.
• Do we need to have $\sup?$, seeing that $c_0=\{\bar{x}=(x_1,x_2,\cdots) :x_n\to 0\;\text{as}\;n\to \infty\}.$ Dec 14, 2018 at 8:23
• @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. Dec 14, 2018 at 8:24
• The norm on $c_0$ is defined to be $||x|| = \sup_{i\in\Bbb N} |x_i|$. Dec 14, 2018 at 8:25