Every Hilbert space has this property. There are, however, Banach spaces which do not; Per Enflo published the first counterexample in a 1973 article. However, a lot of work in this area was done by Grothendieck (1955).
Later many other counterexamples were found. The space of bounded operators on does not have the approximation property (Szankowski). The spaces for and (see Sequence space) have closed subspaces that do not have the approximation property.
A locally convex topological vector space is said to have the approximation property, if the identity map can be approximated, uniformly on precompact sets, by continuous linear maps of finite rank. If X is a Banach space this requirement becomes that for every compact set and every , there is an operator of finite rank so that , for every .
Some other flavours of the AP are studied:
Let be a Banach space and let . We say that X has the -approximation property (-AP), if, for every compact set and every , there is an operator of finite rank so that , for every , and .
A Banach space is said to have bounded approximation property (BAP), if it has the -AP for some .
A Banach space is said to have metric approximation property (MAP), if it is 1-AP.
A Banach space is said to have compact approximation property (CAP), if in the definition of AP an operator of finite rank is replaced with a compact operator.
- Every projective limit of Hilbert spaces, as well as any subspace of such a projective limit, possesses the approximation property.
- Hence every nuclear space possesses the approximation property.
- Every subspace of an arbitrary product of Hilbert spaces possesses the approximation property.
- Every separable Frechet space that contains a Schauder basis possesses the approximation property.
- Every space with a Schauder basis has the AP (we can use the projections associated to the base as the 's in the definition), thus a lot of spaces with the AP can be found. For example, the spaces, or the symmetric Tsirelson space.
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