Show that $\Vert Tx-Tp \Vert \leq \Vert x-p \Vert $, where $p=0.$

Question

Let $X = l_\infty$ (the space of sequences of real numbers which are bounded). Let $K=\{x\in l_\infty:\Vert x \Vert_\infty\leq 1\}.$ Defined \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

1. $Tp=p$ if and only if $p=0;$
2. $\Vert Tx-Tp \Vert \leq \Vert x-p \Vert $, where $p=0.$

MY TRIAL

1.

\begin{align} Tp=p&\iff(0,p^2_1,p^2_2,p^2_3,\cdots)=(p_1,p_2,p_3,\cdots)\\ &\iff p_1=0,\;p_2=p^2_1,\;p_3=p^2_2,\;\cdots\\ &\iff p_n=0,\;\forall n\in \Bbb{N}\\ &\iff p=0\end{align}

2. Let $x,p\in K$ s.t. $p=0,$ then \begin{align} \Vert Tx-Tp \Vert=\Vert (0,x^2_1,x^2_2,x^2_3,\cdots)-(0,0,0,0,\cdots)\Vert\end{align} Honestly, I don't know what to do from here. Any help please?

Answer 1

The norm on $l^\infty$ is $||x||:=\sup_{n\in\Bbb N} |x_n|$.

Hint: If $|\lambda|\le 1$, then $|\lambda|^2\le |\lambda|$.

Answer 2

Your answer for 1 is maybe not very well written, but it looks correct to me.

For 2 notice that $Tp=p=0$, so you just need to show $\|Tx\|_\infty\leq\|x\|_\infty$. Maybe you can even find a closed form of $\|Tx\|_\infty$ in terms of $\|x\|_\infty$?