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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?
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  • $\begingroup$ If $p=0$ in 2.) then doesn't it mean that you merely want to prove $||Tx||\le ||x||$? $\endgroup$ – BigbearZzz Dec 14 '18 at 16:51
  • $\begingroup$ @BigbearZzz: Yes, that's it! $\endgroup$ – Omojola Micheal Dec 14 '18 at 17:17
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The norm on $l^\infty$ is $||x||:=\sup_{n\in\Bbb N} |x_n|$.

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

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  • $\begingroup$ Sorry, I don't quite get the hint. Can you please, break it down? $\endgroup$ – Omojola Micheal Dec 14 '18 at 17:18
  • $\begingroup$ 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 $l^\infty$). $\endgroup$ – BigbearZzz Dec 14 '18 at 17:22
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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 $\|x\|_\infty$?

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  • $\begingroup$ 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. $\endgroup$ – Omojola Micheal Dec 14 '18 at 17:13
  • $\begingroup$ Maybe first try to find $\|Tx\|_\infty$ for $x=(t,0,0,...)$ in terms of $t$. $\endgroup$ – SmileyCraft Dec 14 '18 at 17:16

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