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Let $X$ be a Banach Space and let $Y$ be a closed subspace of $X^{**}$ such that $X\bigcap Y=0$. Let $P$ be the quotient map from $X^{**}$ onto $X^{**}/ Y$. I need to prove or refute that $P\left|_{X}\right.$ has a closed range (or equivalently is bicontinuous), or is at least a semiembedding (meaning that closed balls are mapped into closed sets).

The case when $X$ is a Frechet Space (not Banach) is also of interest.

I feel I may be overlooking something obvious, and I am sorry if this is the case.

Thank you.

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I think that $P|_X$ does not have to be a semi-embedding and this can be done as follows. Let $X$ be hereditary $c_0$, $M$ be a total nonnorming subspace in $X$, and $Y=M^{\perp}\subset X^{**}$. Let $P:X^{**}\to X^{**}/Y$ be the quotient map. Since $X^{**}/Y$ can be identified with $M^*$ we get that $||Px||=\sup\{|f(x)|:~ f\in M, ||f||=1\}$. Since $M$ is nonnorming, this map is not an isomorphism, and since $X$ is hereditary $c_0$, the restriction $P|_X$ is not a semi-embedding by L.Drewnowski, [Proc. Edinburgh Math. Soc. (2) 26 (1983), no. 2, 163–167].

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  • $\begingroup$ Sorry, this is not my area, so a stupid question: why can we always find a total but nonnorming subspace of the dual? $\endgroup$
    – erz
    Commented Apr 10, 2015 at 5:25
  • $\begingroup$ Ok, the question apparently is not so stupid, but it is proven by Davis and Lindenstrauss [Proc. Amer. Math. Soc. 31 1972 109–111.] that a Banach Space is not quasi-reflexive if and only if it has a total nonnorming subspace in its dual. $C(K)$ satisfies this property and is hereditary $c_0$. Thank you! $\endgroup$
    – erz
    Commented Apr 10, 2015 at 5:42
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    $\begingroup$ It should be mentioned that $C(K)$ is hereditary $c_0$ only for special kinds of compacta (e.g. countable). The simplest example of a space satisfying the conditions is $c_0$ itself. $\endgroup$
    – Mikhail Ostrovskii
    Commented Apr 10, 2015 at 13:31

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