In mathematics, specifically functional analysis, a Banach space is said to have the approximation property (AP), if every compact operator is a limit of finite-rank operators. The converse is always true.
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, much work in this area was done by Grothendieck (1955).
Later many other counterexamples were found. The space of bounded operators on an infinite-dimensional Hilbert space does not have the approximation property. [2] The spaces for and (see Sequence space) have closed subspaces that do not have the approximation property.
A locally convex topological vector space X 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. [3]
For a locally convex space X, the following are equivalent: [3]
where denotes the space of continuous linear operators from X to Y endowed with the topology of uniform convergence on pre-compact subsets of X.
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.
In mathematics, specifically functional analysis, a Banach space is said to have the approximation property (AP), if every compact operator is a limit of finite-rank operators. The converse is always true.
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, much work in this area was done by Grothendieck (1955).
Later many other counterexamples were found. The space of bounded operators on an infinite-dimensional Hilbert space does not have the approximation property. [2] The spaces for and (see Sequence space) have closed subspaces that do not have the approximation property.
A locally convex topological vector space X 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. [3]
For a locally convex space X, the following are equivalent: [3]
where denotes the space of continuous linear operators from X to Y endowed with the topology of uniform convergence on pre-compact subsets of X.
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.