Dual to compact operators

Question

Let $X=\mathcal{K}(c_0,\ell_1)$ be the space of compact operators. What is the dual space $X^*$? Actually do we have $X^*=\mathcal{N}(\ell_1,\ell_{\infty})$?

Answer 1

As $\ell_1$ has the approximation property, $$\mathcal{K}(c_0, \ell_1)\cong c_0^* \otimes_\varepsilon \ell_1\cong \ell_1 \otimes_\varepsilon \ell_1,$$ where $\otimes_\varepsilon$ is the injective tensor product. See Corollary 4.13 in Ryan's Introduction to Tensor Products of Banach Spaces for more details. Proposition 3.22 therein confirms your guess that the dual space is $\mathcal{N}(\ell_1, \ell_\infty)$.

By the way, note that $X$ contains a complemented copy of $\ell_1$ so it has non-separable dual, so in particular, this dual space couldn't be $\mathcal{N}(\ell_1, c_0)$ which is separable.