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

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}

1. 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?
• If $p=0$ in 2.) then doesn't it mean that you merely want to prove $||Tx||\le ||x||$? – BigbearZzz Dec 14 '18 at 16:51
• @BigbearZzz: Yes, that's it! – Omojola Micheal Dec 14 '18 at 17:17

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

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

• Sorry, I don't quite get the hint. Can you please, break it down? – Omojola Micheal Dec 14 '18 at 17:18
• Your set $K$ consists of elements $x=(x_1,x_2,\dots)$ such that $|x_i|\le 1$ (see the definition of the norm on $l^\infty$). – BigbearZzz Dec 14 '18 at 17:22

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 $$\|x\|_\infty$$?

• Thanks for number one but how do I find the closed form of $\Vert Tx \Vert_\infty$ in terms of $\Vert x \Vert_\infty$? I'm new into Functional Analysis. – Omojola Micheal Dec 14 '18 at 17:13
• Maybe first try to find $\|Tx\|_\infty$ for $x=(t,0,0,...)$ in terms of $t$. – SmileyCraft Dec 14 '18 at 17:16