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Let $X$ be a Banach space and $Y$ a closed subspace of $X$. If $\varphi\in Y^*$, then Hahn-Banach allows us to extend $\varphi$ to a $\tilde\varphi\in X^*$, such that $\|\tilde\varphi\|=\|\varphi\|$. This extension can happen in infinitely many ways, in general.

My question is the following: Can we guarantee the existence of a bounded linear transformation $$ F : Y^*\to X^*, $$ such that $F(\varphi)$ is an extension of $\varphi\in Y^*$ as a bounded linear functional on $X$? (Preferably with $\|F\|=1$.)

First unsuccessful attempt: Define $F$ on a (Hamel) basis of $Y^*$, and then linearly extend to the whole of $Y^*$. But this $F$ is not necessarily bounded.

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  • $\begingroup$ There are results available, when the extension is unique, e.g. in a Hilbert space or more general results can be found here jstor.org/discover/10.2307/…. Of course, Hamel basis are not to be chosen in topological vector space, e.g. for Banach spaces one works with a Schauder basis. $\endgroup$
    – Marc Palm
    Commented Mar 20, 2014 at 14:08
  • $\begingroup$ Be careful, there exists Banach spaces which do not admit a Schauder Basis. The first examples are due to Enflo. $\endgroup$
    – Marc Palm
    Commented Mar 20, 2014 at 14:09
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    $\begingroup$ Having a Schauder basis still might not save you; see Bill Johnson's answer. Note that every Banach space $Y$ embeds isometrically into some $X=\ell^\infty(\Gamma)$, but it is relatively rare that $Y^\perp$ will be complemented in $X^*$. $\endgroup$
    – Yemon Choi
    Commented Mar 20, 2014 at 14:35

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No. Note that your condition is equivalent to having the dual to the short exact sequence

$Y\to X \to X/Y$

split, which is equivalent to having $Y^\perp$ complemented in $X^*$.

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