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Jan 13, 2016 at 22:32 vote accept Norbert
Jan 13, 2016 at 22:13 answer added Bill Johnson timeline score: 2
Jan 13, 2016 at 21:22 comment added Norbert @BillJohnson, why don't you post this as answer?
Jan 13, 2016 at 16:25 comment added Bill Johnson The $Y$ in Mikhail's answer has codimension one. Obviously $Y$ cannot be reflexive, but $Y$ can be of any non zero finite codimension or of infinite codimension. (Let $Z$ be any separable Banach space and let $Q$ be an operator from $L_1$ to $Z$ that maps the closed unit ball of $L_1$ onto the open unit ball of $Z$. The kernel $Y$ of such a quotient map $Q$ is antiproximinal. It is easy to built such an operator from $\ell_1$ onto $Z$; to get one from $L_1$ compose the operator from $\ell_1$ with a norm one projection from $L_1$ onto a subspace that is isometric to $\ell_1$.)
Jan 13, 2016 at 11:02 vote accept Norbert
Jan 13, 2016 at 22:32
Jan 13, 2016 at 5:25 answer added Mikhail Ostrovskii timeline score: 6
Jan 12, 2016 at 23:58 history asked Norbert CC BY-SA 3.0