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I have the following question related to $\ell^\infty(\mathbb{N}).$ How can I construct a bounded, linear operator from $\ell^\infty(\mathbb{N})$ into $c_0(\mathbb{N})$ which is non-compact?

It is clear that $\ell^\infty$ is a Grothendieck space with Dunford-Pettis property, hence any operator from $\ell^\infty$ into a separable Banach space must be strictly singular. But I do not know any example above which is non-compact.

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From problem VII.3 in Diestel's Sequences and Series in Banach Spaces: The bounded linear operators from $\ell_\infty$ to $c_0$ correspond to the weak$^*$ null sequences in $\ell_\infty^*$ and a bounded operator $T$ from $\ell_\infty$ to $c_0$ is compact iff the sequence $(T^*e_i^*)$ is norm null in $\ell_\infty^*$. – David Mitra Jun 6 '12 at 14:51
Thank you very much, I will check it! – leka Jun 6 '12 at 16:54
up vote 4 down vote accepted

A bounded operator $T:\ell_\infty\rightarrow c_0$ has the form $Tx=(x_n^*(x))$ for some weak$^*$ null sequence $(x_n^*)$ in $\ell_\infty^*$. A set $K\subset c_0$ is relatively compact if and only if there is a $x\in c_0$ such that $|k_n|\le |x_n|$ for all $k\in K$ and all $n\ge1$. From these two facts, it follows that $T(B({\ell_\infty}))$ is relatively compact if and only if the representing sequence $(x_n^*)$ is norm-null.

So, you need only find a sequence in $\ell_\infty^*$ that is weak$^*$ null, but not norm null. Such a sequence exists in $\ell_\infty^*$ since: 1) weak$^*$ convergent sequences in $\ell_\infty^*$ are weakly convergent ($\ell_\infty^*$ has the Grothendieck property), and 2) $\ell_\infty^*$ does not have the Schur property (weakly convergent sequences are norm convergent).

(There may be a less roundabout way of showing the the result of the preceding paragraph.)

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What is a null sequence? – Nate Eldredge Jun 6 '12 at 18:21
@NateEldredge Sorry, a sequence that converges to the zero vector. – David Mitra Jun 6 '12 at 18:24
I expected a direct example but now I understand why this is not immediate. Thank you! – leka Jun 7 '12 at 16:41

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