2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 Mar 20 '14 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 Mar 20 '14 at 14:09
  • 1
    $\begingroup$ 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 Mar 20 '14 at 14:35
8
$\begingroup$

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 $X^*$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.