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Could someone give a reference or construct an example of closed subspace of $Y\subset L_1[0,1]$ such that $\operatorname{dist}(x,Y)$ is not attained of for any $x\notin Y$.

I read somewhere that $Y$ is necessarily of infinite dimension and codimension.

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  • $\begingroup$ 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 $\ell_1$.) $\endgroup$ Jan 13, 2016 at 16:25
  • $\begingroup$ @BillJohnson, why don't you post this as answer? $\endgroup$
    – Norbert
    Jan 13, 2016 at 21:22

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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 $\ell_1$.)

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Consider any functional $f$ which does not attain its norm on $L_1[0,1]$ (such $f$ exists by James's theorem, but in this case one can find it without, as an $L_\infty$-function with essential supremum equal to $1$, which is not attained on a set of non-zero measure) and let $Y$ be the kernel of $f$. Let $x\notin Y$, if $\hbox{dist}(x,Y)$ is attained at $y\in Y$, then $f$ would attain its norm on $(x-y)/||x-y||$.

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  • $\begingroup$ I was always sure that this trick works only for some $x\notin Y$, not for all of them. Thank you. $\endgroup$
    – Norbert
    Jan 13, 2016 at 11:02

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