A simple way to obtain certain examples of locally bounded spaces E of the following kind is described: E is nonlocally convex but contains a locally convex subspace K such that E/K is locally convex.The purpose of this note is to give a quite simple construction which proves the following:Theorem 1 (Independently proved by Kalton [5]).77iere is a separated locally bounded space E with an uncomplemented one-dimensional subspace L such that E/L is isomorphic to lx.Theorem 2. Indeed, there is an uncountable class of such spaces E of which no one can be mapped into another by a continuous linear mapping which does not annihilate the uncomplemented line.Theorem 3A.There is a locally bounded space E with a subspace K such that K is isomorphic to l and E/K to /', and such that for any infinite-dimensional subspace Ex c E, the subspace Ex + K fails to be locally convex.Theorem 3B.Let A be a set of the cardinality of the continuum.There is a locally bounded space E with a subspace K such that K is isomorphic to /(A) and E/K to /', and such that for any infinite-dimensional subspace Ex c E, the subspace Ex + K fails to be separated by its dual.Kalton [5] also proves that in Theorem 1, /' cannot be replaced by lp for any p ^ 1 (0 < p < oo).For some further recent related results, see Dierolf
No takes yet. Share an insight, caveat, or question.
M. Ribe (1979) studied this question.