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Let $ c_{0}:=\lbrace x:\mathbb{N}\rightarrow \mathbb{R} :\lim_{j\rightarrow\infty} x_{j}=0 \rbrace$ denote the usual Banach sequence spaces. Given Banach spaces $X,Y$ let $L(X,Y)$ denote the Banach space of bounded, linear operators from $X$ to $Y$ . Also let $ \ell_{1} :=\lbrace x:\mathbb{N}\rightarrow \mathbb{R} :\Sigma_{j=1}^{\infty}\vert x_{j}\vert <\infty \rbrace$.

is $ c_{0} $ isometrically embedding in $ L(c_{0},\ell_{1}) $ ?

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  • $\begingroup$ By Pitt's theorem $L(c_0,\ell^1)=K(c_0,\ell^1)=\ell^1 \tilde{\otimes}_\varepsilon \ell^1$. There is an article of Bombal, Fernandez-Unzueta, and Villanueva Local structure and copies of $c_0$ and $\ell^1$ in the tensor product of Banach spaces but, unfortunately, their main result theorem 2.1 apparently does not apply here. $\endgroup$
    – Jochen Wengenroth
    Commented May 3, 2016 at 13:45

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No. Kalton proved that if it contains a copy of $c_0$ it would also contain a copy of $\ell_{\infty}$, and this cannot happen since $L(c_0, \ell_1)=K(c_0, \ell_1)$.

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  • $\begingroup$ Could you please give a reference for Kalton's result and elaborate why $K(c_0,\ell_1)$ cannot contain $\ell_\infty$? $\endgroup$
    – Jochen Wengenroth
    Commented May 4, 2016 at 8:45
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    $\begingroup$ $K(c_0,\ell_1)$ is separable, Jochen; in fact, $e_j \otimes e_i$ is a Schauder basis for it. $\endgroup$
    – Bill Johnson
    Commented May 4, 2016 at 15:38
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    $\begingroup$ The relevant Kalton paper is Kalton, N. J. Spaces of compact operators. Math. Ann. 208 (1974), 267–278. It is arguably overkill to quote this paper because a standard gliding hump argument gives the non existence of an isomorphic copy of $c_0$ in $K(c_0,\ell_1)$. $\endgroup$
    – Bill Johnson
    Commented May 4, 2016 at 15:42

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