most recent 30 from http://mathoverflow.net 2013-05-23T03:15:05Z http://mathoverflow.net/feeds/question/17801 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/17801/approximation-property Robb Fry 2010-03-11T02:03:37Z 2010-03-13T22:01:04Z

<p>It seems to be a folk result that l_infinity has the approximation property, even the bounded approximation property, and also, I think, even the so-called propery pi (approximation property) of Lindenstrauss. </p> <p>This is alluded to in a few texts, but I cannot seem to find the proof, which is presumably obvious. </p> <p>Does anyone have a reference or an easy solution?</p> <p>Cheers,</p> <p>R. Fry</p>

http://mathoverflow.net/questions/17801/approximation-property/17804#17804 Bill Johnson 2010-03-11T02:20:20Z 2010-03-13T22:01:04Z

<p>Partition the measure space into finitely many sets and consider the span of their indicator functions. This space is isometrically isomorphic to $\ell_\infty^n$ for appropriate $n$ and hence is the range of a norm one projection. Index the partitions by refinement to get a net of norm one finite rank projections that converge strongly to the identity.</p>