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Given Banach spaces $X$, $Y$ and a bounded operator $T:X\to Y$ with non-closed range, a perturbation argument shows that there exists an infinite-dimensional closed subspace $M$ of $X$ such that the restriction of $T$ to $M$ is compact.

Let $J:L_\infty(0,1)\to L_1(0,1)$ denote the natural inclusion. Is it posible to find a non-separable closed subspace $N$ of $L_\infty(0,1)$ such that the restriction of $J$ to $N$ is compact?

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Take the functions which are constant on the intervals $[1/(n+1),1/n)$ for all positive integers $n$.

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There is even a non separable subspace $X$ of $L_\infty$ such that if $T: L_\infty \to Y$ is any weakly compact operator from $L_\infty$, then the restriction of $T$ to $X$ is compact. Indeed, if $Z$ is a subspace of $L_\infty$ and $T: L_\infty \to Y$ is a weakly compact operator for which $T_{|Z}$ is not compact, then since $L_\infty$ has the Dunford-Pettis property, $\ell_1$ must embed into $Z$. In 2006, Argyros, Arvanakis, and Tolias constructed a separable space whose dual $X$ is non separable and is hereditarily indecomposable and hence $\ell_1$ does not embed into $X$. The space $X$ embeds into $L_\infty$ because its predual, being separable, is a quotient of $L_1$.

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Every operator $\ell_\infty\to\text{separable}$ is weakly compact, thus, its restriction to an isomorphic predual of $\ell_1(I)$ is compact. And such nonseparable things exist in $\ell_\infty$: say, the span of characteristic functions of $\{n\}$ and an uncountable almost disjoint family of subsets of $\mathbb N$.

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  • $\begingroup$ Better than my example and a space that I know well. $\endgroup$ – Bill Johnson Nov 9 '18 at 14:08

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