A Banach space E is said to have the approximation property if for every compact subset K of E and every e > 0 there exists a finite rank operator S: E —> E such that ||x- S x|| < e for every x in K.
Hi,
Every Banach space with Schauder basis has the approximation property (AP).
(See https://en.wikipedia.org/wiki/Approximation_property)
It is known that the usual disk algebra A(D) has a Schauder basis, so it also has AP.
(See https://en.wikipedia.org/wiki/Schauder_basis)
Best wishes,
Joel Feinstein
It is always a problem to find examples of spaces WITHOUT the approximation property (A.P. for short), and very easy to find spaces WITH this property. So the right question would be whether there is a uniform algebra WITHOUT the A.P. This is not an easy question. Say in the Lindenstrauss-Tsafriri book "Classical Banach spaces" the question whether the space H_\infty of all bounded analytic functions in the unit disc with the supremum norm has the A.P. is formulated as an open problem ( problem 1.e.10 ). Recall, that H_\infty is a uniform algebra. Since the book is old enough, maybe the problem is already solved, but for sure it is not something evident.
Concerning the previous communication by Luiz C. L. Botelho: a space with a basis must be separable, but a space with the A.P. can be non-separable. In particular, all L_p spaces and C(K) spaces (even non-separable ones) have the A.P.
Dear Luiz,
First of all about Schauder basis. If a Banach space E has a Schauder basis, then E is separable. In particular, if C(K) has a Schauder basis, then C(K) is separable. Separability of C(K) is equivalent to metrizability of the underlying compact K. If a compact K is metrizable, then K is separable, but there are many non-metrizable separable compact topological spaces.
Now concerning AP. You are completely right that the space Cb(X) of BOUNDED continuous functions on a regular topological space X is isomorphic to some C(K), so it has the AP.
All the best, Vladimir
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