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I have a problem with the following exercise:

We have the operator $T: l^1 \to l^1$ given by

$$T(x_1,x_2,x_3,\dots)=\left(\left(1-\frac11\right)x_1, \left(1-\frac12\right)x_2, \dots\right)$$ for $(x_1,x_2,x_3,\dots)$ in $l^1$. Showing that this operator is bounded is easy, but I am really desperate with showing that the norm $\|T\| = 1$.

I know that for bounded operators the norm is defined as $\|T\|=\sup{\left\{\|T(x)\|: \|x\| \le 1\right\}}$.

I am also wondering if there exists a x in $ l^1$ such that $\|x\|=1 $

and $\|T(x)\|= \|T\|$

Thank you! :)

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3 Answers 3

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First observe that $\|T\|\le 1$.

Fix a large $n$ and let $x_n=(\dots,0,1,0,\dots)$ with $1$ exactly at the $n$th position and $0$ everywhere else. We have $Tx_n = (1-\frac1n) x_n$, $\|x_n\|=1$ and $\|Tx_n\| = 1-\frac1n.$ This shows $\|T\|\ge 1- \frac1n$. Now let $n\to\infty$ to conclude $\|T\|\ge 1$.

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  • $\begingroup$ Thank you a lot, but why does this show that ||T|| >= 1-1/n ?? $\endgroup$
    – Yuhe
    Oct 25, 2016 at 10:42
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    $\begingroup$ Because $\|T\|=\sup\{ \|Tx\| \,:\,\|x\|=1\} \ge \|T x_n\| = 1-\frac1n$. $\endgroup$
    – J.R.
    Oct 25, 2016 at 14:36
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$$\left|\left| T((x_n)_{n\in\mathbb{N}})\right|\right|_{\ell_1}=\left|\left| \left(\left(1-\frac{1}{n}\right)x_n\right)_{n\in\mathbb{N}}\right|\right|_{\ell_1}=\sum_j \left|\left(1-\frac{1}{j}\right)x_j\right|\leqslant \sum_j |x_j |=||(x_n )_{n\in\mathbb{N}} ||_{\ell_1}$$

hence

$$||T||\leqslant 1$$

but $$||T||\geqslant \sup_j ||Te_j || =\sup_j \left(1-\frac{1}{j}\right) =1$$

Thus $$||T||=1.$$

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  • $\begingroup$ Thank you, I edited the question, could you help me with part 2? $\endgroup$
    – Yuhe
    Oct 25, 2016 at 10:50
  • $\begingroup$ No there does not exist a $x$ such that $||x||=1$ and $||Tx||=1$ $\endgroup$
    – user235708
    Oct 25, 2016 at 10:54
  • $\begingroup$ Thats what I was assuming too, but could you give me small proof for that? $\endgroup$
    – Yuhe
    Oct 25, 2016 at 10:56
  • $\begingroup$ $\sum (1 - \frac{1}{n}) |x_n| < \sum |x_n| \leq 1$ therefore $||Tx|| <1$ $\endgroup$
    – Hermès
    Oct 25, 2016 at 12:37
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Hint: Define $X^n = (\underbrace{\frac{1}{n}, \dots ,\frac{1}{n}}_{n}, 0, 0, ...)$. Show that for every $\epsilon >0$, there exists $N$ such that $||X^N||_{l^1} > 1-\epsilon$, thus proving the statement.

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  • $\begingroup$ Thank you, could you help me with part 2 as well? $\endgroup$
    – Yuhe
    Oct 25, 2016 at 12:12

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