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Consider $$ T\colon\ell^1\to\ell^1, (s_n)\mapsto\left(\frac{s_{n+1}}{n}\right). $$ Calculate the norm of $T$ and show that $T$ is compact.

1.) Operator norm of $T$

What I have is the following: $$ \lVert T\rVert=\sup\limits_{\lVert x\rVert_{\ell^1}=1}\lVert Tx\rVert_{\ell^1}=\sup\limits_{\lVert x\rVert_{\ell^1}=1}\sum\limits_{i=1}^{\infty}\frac{\lvert s_{i+1}\rvert}{i}\leq\sup\limits_{\lVert x\rVert_{\ell^1}=1}\sum\limits_{i=1}^{\infty}\lvert s_i\rvert=1 $$ Now I do not know how to continue.

2.) Compactness

The operator $T$ can be written as $$ T=D\circ L, $$ where $L$ is the Leftshift and $D\colon\ell^1\to\ell^1, (s_n)\mapsto\left(\frac{s_n}{n}\right)$.

The operator $L$ is bounded. Therefore I have to show that $D$ is compact, because then $T$ is (as a convolution of a bounded and a compact operator) compact, too.

How can I I show the compactness of $D$?

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    $\begingroup$ 1) First show that $\Vert T\Vert\le 1$ (you just need to estimate the right most expression in your string of equalities). Then compute $T(0,1,0,\ldots)$. 2) See this post. $\endgroup$ – David Mitra Feb 16 '13 at 18:11
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    $\begingroup$ For intuition with compact operators, I think of them as finite rank operators. For the most part this simplification (and my addled mind needs simplifications) works well. $\endgroup$ – copper.hat Feb 16 '13 at 18:25
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1) Note that $$ \sum_{i\geq 1}\frac{|s_{i+1}|}{i}\leq \sum_{i\geq 1}|s_i| $$ so $\|T\| \leq 1$.

Then for $s=(0,1,0,\ldots)$, $Ts=(1,0,\ldots)$ so $\|Ts\|=\|s\|=1$ and so $\|T\|\geq 1$.

Therefore $\|T\|=1$.

2) It is not a convolution, but a composition. But you are right, this amounts to showing that $D$ is compact.

You can easily show that $D$ is a limit of finite rank operators, so it is compact.

Hint: consider the finite rank operators $$ D_n:s\longmapsto(s_1,\frac{s_2}{2},\ldots,\frac{s_n}{n},0,\ldots) $$ and observe that $\|D-D_n\|\leq 1/n$.

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  • $\begingroup$ @copper.hat Thanks for paying attention. I did a typo. I was actually approximating the $D$ of the OP, not $T$. $\endgroup$ – Julien Feb 16 '13 at 18:28
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    $\begingroup$ Oops, missed that! $\endgroup$ – copper.hat Feb 16 '13 at 18:30
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$\|Tx\| = \sum_n |[Tx]_n| = \sum_n \frac{|x_{n+1}|}{n} \leq \sum_n |x_{n+1}| \leq \|x\|$, so $\|T\| \leq 1$. Since $T e_2 = e_1$, it follows that $\|T \| = 1$.

Let $T_k x = \sum_{n=1}^k \frac{x_{n+1}}{n} e_n$, and note that a similar computation to above shows that $\|T-T_k\| \leq \frac{1}{k+1}$. Since $T_k$ has finite rank it is compact, and since $T$ is the limit of compact operators, it follows that $T$ is compact.

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  • $\begingroup$ Did I miss something? $\endgroup$ – copper.hat Feb 16 '13 at 20:59
  • $\begingroup$ D has to be approximated, right? $\endgroup$ – math12 Feb 16 '13 at 22:04
  • $\begingroup$ There is no need to bother with $D$. The above approximates $T$ in a natural, finite rank way. $\endgroup$ – copper.hat Feb 16 '13 at 22:28
  • $\begingroup$ But it is not false to do so? $\endgroup$ – math12 Feb 16 '13 at 23:06
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    $\begingroup$ Both ways are fine. Showing $D_n$ approximates $D$ involves exactly the same work as showing $T_n$ approximates $T$. $\endgroup$ – copper.hat Feb 16 '13 at 23:23

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