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Given $\epsilon>0$, can we always find a non-compact operator $T:l_\infty\to l_\infty$ of norm larger than $1$ such that the restriction of $T$ to $c_0$ is compact and has norm smaller than $\epsilon$?

Markus
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  • The norm thing is easy to establish by taking a Banach limes. Indeed, let $L \in (\ell_\infty)^*$, such that $L(x) = \lim_{n \to \infty} x_n$ for all convergent sequences. Then, the norm of $L$ is larger than one, but its restriction to $c_0$ is $0$. Thus, we can replace $T$ by $c , (T + x , L(\cdot))$ with some $c > 0$ and $x \in \ell_\infty \setminus {0}$ to get the desired norm properties. – gerw Jun 18 '16 at 20:10
  • @gerw, Banach limits certainly may have norm one. What is your $T$? – Tomasz Kania Jun 19 '16 at 22:03
  • Any $T$ which has the compactness property of the question. – gerw Jun 20 '16 at 09:03

2 Answers2

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There are two things that will help: First, whenever $c_0$ is isomorphic to a subspace of a separable Banach space $X$ then that subspace is complemented in $X$. Second, whenever $Y$ is a subspace of a normed space $X$ and $T:Y\to\ell_\infty$ is a continuous linear operator then there exists a continuous linear extension $\widetilde{T}:X\to\ell_\infty$ with $\|\widetilde{T}\|=\|T\|$.

Now we simply find a linear independent sequence $(y_n)_{n=1}^\infty\subseteq \ell_\infty\setminus c_0$. As $Z:=\overline{c_0+[y_n]_{n=1}^\infty}$ is a separable closed subspace of $\ell_\infty$ containing $c_0$, we can decompose it $Z=c_0\oplus Y$ for some infinite-dimensional subspace $Y$ of $\ell_\infty$. Write $I_{c_0}\in\mathcal{L}(c_0)$ andt $I_Y\in\mathcal{L}(Y)$ for the identity maps. Then for any $r\in\mathbb{R}$ we have $0I_{c_0}\oplus rI_Y\in\mathcal{L}(c_0\oplus Y)=\mathcal{L}(Z)$. Let $T$ denote its extension of $\ell_\infty$. Then $T|_{c_0}=0I_{c_0}$ is the zero mapping, trivially compact and with zero norm. On the other hand $\|T\|=r$, which we can make arbitrarily large.

Ben W
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One can even have $T|_{c_0}=0$. Indeed, first note that if $X\subseteq \ell_\infty$ is isomorphic to $\ell_2$, then $X\cap c_0$ is finite-dimensional so there is a subspace $X\subseteq \ell_\infty$ isometric to $\ell_2$ with $X\cap c_0=\{0\}$. (Actually, by simple repetition of the coordinates in the standard way of embedding separable spaces into $\ell_\infty$ one may guarantee from the start that the image has trivial intersection with $c_0$.) Secondly, $\ell_\infty / c_0$ has a quotient isomorphic to $\ell_2$ because it has (complemented) subspaces isomorphic to $\ell_\infty$ and the latter space has this property.

To finish the proof, take any linear surjection $Q\colon \ell_\infty/c_0\to \ell_2$ and let $S\colon \ell_2\to X$ be an isomorphic embedding. Let $\pi\colon \ell_\infty\to \ell_\infty / c_0$ be the quotient map. It is enough then to take $T=cSQ\pi$ for $c$ large enough.

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Tomasz Kania
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  • Thank you Ben and Tomek. Can you repeat the argument by taking instead of $l_\infty$ a separable this time dual space, and instead of $c_0$ a $w^*$-dense norm closed subspace? – Markus Jun 20 '16 at 00:30
  • Without the dual space criterion you could take $Z=c_0\oplus Y\subseteq\ell_\infty$ instead of the entire space $\ell_\infty$. As $c_0$ is weak star dense in $\ell_\infty$ (by Goldstine), it is weak-dense in $Z$. But of course $c_0$ is not a dual space and so probably $Z$ is not one either. But I can't immediately see how to find a weak-dense and norm-closed subspace of a separable dual space such that a compact operator between them can be extended to a noncompact operator--to say nothing of having arbitrarily large norm.Tomek could probably do it, as he is good at this kind of thing. – Ben W Jun 20 '16 at 02:20
  • @Markus, I don't think so. Take the the second dual of the James space. The original James space is then weak-dense in it, so the quotient map $\pi$ is already rank-one. In fact, $T\colon J^{}\to J^{*}$ is compact if and only if so is $T|_J$. – Tomasz Kania Jun 20 '16 at 08:09
  • @TomekKania Is that the case for any separable dual? Is the restriction to the weak-dense subspace compact only if the operator was compact, as in James? Or for some separable dual, as in the (non-separable) $l_\infty$ case,one can have a "big" operator with "small" restriction to a weak-dense subspace. – Markus Jun 20 '16 at 19:08