Operators from $\ell^\infty$ into $c_0$ 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. 

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.) 
