# There are compact operators that are not norm-limits of finite-rank operators

Given an example of a Banach space for which There are compact operators that are not norm-limits of finite-rank operators.

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 – user53153 Jan 28 at 16:52

A Banach space for which the finite rank operators are norm-dense in the compact operators is said to have the approximation property (AP). An explicit example of a Banach space without the AP is the space $B(H)$ of bounded linear operators on an infinite-dimensional Hilbert space by deep work of Szankowski.
 It is important that one considers operators to all other Banach spaces in the AP. It is still open whether it is sufficient to consider only operators from $X$ to itself (in the definition of the AP). – Martin Jan 28 at 18:33 thanks alot of you – Ali Qurbani Feb 7 at 15:21 @Martin While examples might not be easy to identify, if you look at subspaces, Szankowski showed that any $l_p$, with $p\neq 2$, has a subspace without the AP. In fact, even more general, if $X$ does not have type $2-\varepsilon$ and cotype $2+\varepsilon$ for any $\varepsilon$ (so in a sense it is not very close to a Hilbert space), then $X$ has a subspace without AP. Pretty great stuff. – Theo Feb 8 at 17:29 @Theo: Thanks for your comments. Yes, all this is fantastic stuff, no doubt! While many examples are known, my point is that it is still quite delicate to prove that there are spaces failing the AP in the first place. Most of the introductory texts on Banach space theory, let alone the general introductions to functional analysis skip the topic. – Martin Feb 8 at 21:01 @Martin Yes, as far as I know the only technique is still the Szankowsi's one, which in fact is a substantial simplification of Enflo's. At least it seemed substantial to me, Szankowski's proof was fairly easy to understand (a good presentation is in Lindenstrauss-Tzafriri vol. 2), while Enflo's original example gave me a lot of headaches :). – Theo Feb 9 at 0:19