Show that $T:\,c_0\to c_0\;\;$, $x\mapsto T(x)=(1,x_1,x_2,\cdots),$ is non-expansive

Question

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.

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.