Show that $T: K\to K,$ defined by $Tx=(0,x^2_1,x^2_2,x^2_3,\cdots)$ is not non-expansive

Question

Let $X = l_\infty$ (the space of sequences of real numbers which are bounded). Let $K=\{x\in l_\infty:\Vert x \Vert_\infty\leq 1\}.$ Define \begin{align} T:& K\to K \\&x\mapsto Tx=(0,x^2_1,x^2_2,x^2_3,\cdots)\end{align}

I want to show that $T$ is not non-expansive.

MY TRIAL

$T$ is not non-expansive if for all $k\in [0,1]$, $\Vert Tx-Ty \Vert \geq k\Vert x-y \Vert $. (Note: I am not sure of this definition). Now, take $x=\left(\frac{3}{4},\frac{3}{4},\frac{3}{4},\cdots \right)$ and $y=\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\cdots\right).$ Then,

\begin{align} \Vert Tx-Ty \Vert &=\Vert T\left(\frac{3}{4},\frac{3}{4},\frac{3}{4},\cdots \right)-T\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\cdots\right)\Vert\\&=\Vert \left(0,\frac{9}{16},\frac{9}{16},\frac{9}{16},\cdots\right)-\left(0,\frac{1}{4},\frac{1}{4},\frac{1}{4},\cdots\right)\Vert \\&=\sup \{ 0,\left|\frac{9}{16}-\frac{1}{4}\right|,\left|\frac{9}{16}-\frac{1}{4}\right|\cdots\}\\&=\frac{5}{4}\sup \{ 0,\left|\frac{3}{4}-\frac{1}{2}\right|,\left|\frac{3}{4}-\frac{1}{2}\right|\cdots\}\\&\geq\sup \{ \left|\frac{3}{4}-\frac{1}{2}\right|,\left|\frac{3}{4}-\frac{1}{2}\right|\cdots\}\\&=\Vert x-y \Vert.\end{align}

I am not sure of my work. Kindly check please? If I'm wrong, then kindly provide a way-out. Thanks!

Answer 1

The only problem I see is the part where you write $$ \frac{5}{4}\sup \{ 0,\left|\frac{3}{4}-\frac{1}{2}\right|,\left|\frac{3}{4}-\frac{1}{2}\right|\cdots\} \geq\sup \{ \left|\frac{3}{4}-\frac{1}{2}\right|,\left|\frac{3}{4}-\frac{1}{2}\right|\cdots\}, $$ the sign $\ge$ should be $>$ instead because in order to show that $T$ is not nonexpansive we need to show that $\exists x,y\in K$ such that $$ ||Tx-Ty||>||x-y||. $$