# Simultaneously extending the functionals of a subspace of a Banach space to the whole space

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.

• 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. – Marc Palm Mar 20 '14 at 14:08
• Be careful, there exists Banach spaces which do not admit a Schauder Basis. The first examples are due to Enflo. – Marc Palm Mar 20 '14 at 14:09
• 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 $X^*$. – Yemon Choi Mar 20 '14 at 14:35

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