# 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.

-
–  user53153 Jan 28 '13 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 '13 at 18:33
@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 '13 at 17:29