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Let $X$ be a Banach space and let $A$ be a closed, convex and balanced subset of $B_{X^{*}}$ (where $B_{X^{*}}$ denotes the closed unit ball of the dual $X^{*}$). Is there a closed subspace $M$ of $X$ such that $Q^{*}_{M}$ maps $B_{(X/M)^{*}}$ onto $A$, where $Q_{M}:X\rightarrow X/M$ is the quotient map?

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    $\begingroup$ What happens when $X=\mathbb{R}^n$? Suppose $A$ is any closed convex balanced set with nonempty interior, other than the ball itself. If $M \ne 0$ then $Q_M^*$ has rank less than $n$ and so its image cannot cover $A$, and if $M=0$ then $Q_M^*$ is the identity map and it maps the ball to itself. $\endgroup$ – Nate Eldredge Feb 15 at 13:33
  • $\begingroup$ Thanks, Nate. What happens if $X$ is infinite-dimensional? $\endgroup$ – Dongyang Chen Feb 15 at 14:36
  • $\begingroup$ I was just thinking about that. More generally, the image of $Q_M^*$ will always equal the annihilator of $M$, right? If $M \ne 0$ this is a proper closed subspace. So take any $A$ which is not contained in a proper closed subspace (e.g. any $A$ with nonempty interior) and is not the ball, and I think that is a counterexample. $\endgroup$ – Nate Eldredge Feb 15 at 14:39
  • $\begingroup$ Indeed, $Q^{*}_{M}B_{(X/M)^{*}}$ is equal to the closed unit ball of the annilator of $M$. If we take $A$ with nonempty interior, then $M=0$. $\endgroup$ – Dongyang Chen Feb 15 at 14:57

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