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Mar 20, 2014 at 14:35 comment added Yemon Choi Having a Schauder basis still might not save you; see Bill Johnson's answer. Note that every Banach space $Y$ embeds isometrically into some $X=\ell^\infty(\Gamma)$, but it is relatively rare that $Y^\perp$ will be complemented in $X^*$.
Mar 20, 2014 at 14:18 history edited smyrlis CC BY-SA 3.0
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Mar 20, 2014 at 14:11 answer added Bill Johnson timeline score: 8
Mar 20, 2014 at 14:09 comment added Marc Palm Be careful, there exists Banach spaces which do not admit a Schauder Basis. The first examples are due to Enflo.
Mar 20, 2014 at 14:08 comment added Marc Palm There are results available, when the extension is unique, e.g. in a Hilbert space or more general results can be found here jstor.org/discover/10.2307/…. Of course, Hamel basis are not to be chosen in topological vector space, e.g. for Banach spaces one works with a Schauder basis.
Mar 20, 2014 at 13:53 history asked smyrlis CC BY-SA 3.0