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Given the operator $T:l^1\to c_0$ with $T(x=(x_k))=\left(\sum\limits_{k=1}^\infty x_k,\sum\limits_{k=2}^\infty x_k,...,\sum\limits_{k=n}^\infty x_k,...\right)$.

I have to decide if this is a compact operator. I think it is, and to prove it I have the sequence of operators:

$T_n(x=(x_k))=\left(\sum\limits_{k=1}^\infty x_k,\sum\limits_{k=2}^\infty x_k,...,\sum\limits_{k=n}^\infty x_k,0,0,...\right)$

whose image is finite dimensional. So if I prove that $T_n\to T$, I have that $T$ is compact.

I get $||T_n(x)-T(x)||\leq \sum\limits_{k=n+1}^\infty|x_k|$, but I want to find $||T_n(x)-T(x)||\leq \varepsilon_n||x||$ where $\varepsilon_n\to0$. If I prove that, I have that $||T_n-T||\leq\varepsilon_n\to 0$ and I conclude my proof.

Can somebody help me to find that $\varepsilon_n$? Thank you

JN_2605
  • 479

2 Answers2

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It doesn't exist, and this operator is not compact.

Consider elements $x \in \ell^1$ of the form $(0,0,\dots, 0,-1,1,0,0,\dots)$. They are all in the ball of radius 2. Look at what the corresponding elements $T(x)$ are, and show that this is an infinite set with no limit points.

Nate Eldredge
  • 97,710
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In fact it's pretty clear that $||T_n-T||=1$ (for example because the standard isomorphism of $\ell_1^*$ and $\ell_\infty$ is an isometry.) That doesn't prove that $T$ is not compact, but it's not; for example $||Te_n-Te_m||=1$ for $n\ne m$, so the sequence $(Te_n)$ has no convergent subsequence.

David C. Ullrich
  • 89,985