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Definition. Let us define a Banach space $X$ to be co-Sobczyk if every linear bounded operator $T:Z\to c_0$ defined on a separable subspace $Z$ of $X$ extends to a bounded operator $\bar T:X\to c_0$.

By the classical Sobczyk Theorem, each separable Banach space is co-Sobczyk. But the class of co-Sobczyk spaces includes many non-separable Banach spaces. In particular, a Banach space $X$ is co-Sobczyk if each separable subspace of $X$ is contained in a complemented separable subspace. So, all classical Banach spaces $c_0(\Gamma)$ and $\ell_p(\Gamma)$ for $1\le p<\infty$, are co-Sobczyk for any set $\Gamma$.

I have a strong feeling that co-Sobczyk spaces have been studied in the theory of non-separable Banach spaces, so asking the MO commubnity for a proper reference and an existing terminology (I suspect that co-Sobczyk spaces are called differently).

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Nigel Kalton studied a similar but stronger notion: for $\lambda \geqslant 1$, he termed a Banach space $X$ to have the $(\lambda, \mathcal{C})$-extension property, when for any compact space $K$ you may find extensions of operators $T$ from subspaces of $X$ into $C(K)$ to operators from $X$ to $C(K)$ with norm at most $\lambda \|T\|$.

It is thus natural to term your spaces as having the $(\lambda, c_0)$-separable extension property if you care about the extension constant.

Update: Correa and Tausk call this separable $c_0$-extension property.

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  • $\begingroup$ Thank you for the answer. Has this term ``$(\lambda,c_0)$-separable extension property'' been used in any written paper? Because I thoughtalso about the name "$c_0$-coinjective". $\endgroup$ – Taras Banakh Jun 6 '19 at 10:26
  • $\begingroup$ @TarasBanakh, yes, up to a permutation. Here this is called separable $c_0$-extension property. sciencedirect.com/science/article/pii/S0022247X13002540 $\endgroup$ – Tomasz Kania Jun 6 '19 at 10:52

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