In question Can a subspace have a larger dual?, they discussed the following type of situtation:

Suppose that $Y\subset X$ is a proper embedding of Banach spaces (i.e. the inclusion of a proper closed subspace). By Hahn-Banach, we may view $Y^*\subset X^*$, but this seems to me to be only at the level of topology. What I mean by that is that this doesnt seem to be an an embedding of Banach spaces anymore. For example, it may happen that the extensions you choose for $x,y\in Y^*$ may have the property that their sum is not the extension you chose for $x+y$. Can one realize $Y^*\subset X^*$ as a linear subspace? Or better still, as a closed linear subspace?

There are examples of this happening that I can think of; for example, if you have a surjection $\phi:X\to Y$, the pullback map gives an embedding $\ell^1(Y)\subset \ell^1(X)$ and we do have that $(\ell^1(Y))^*\subset (\ell^1(X))^*$ (as this is just the pullback embedding of $\ell^\infty(Y)\subset \ell^\infty(X)$). I dont see how to make this work at the level of choosing extensions using Hahn-Banach in a systematic way.

  • $\begingroup$ In general, if $Y$ is a subspace of $X$, then $Y^*$ is a quotient of $X^*$. It need not be a subspace. (Your Hahn-Banach thing may not even be topological.) $\endgroup$
    – GEdgar
    Dec 30, 2018 at 13:51

1 Answer 1


It is not quite correct to say that Hahn–Banach implies that $Y^*\subset X^*$. It only says that $$\{f|_Y\colon f\in X^*\}=Y^*.$$

As for concrete examples where this may fail, let us take $X=C[0,1]$. By the Banach–Mazur theorem, $X$ contains an isometric copy of every separable Banach space so let $Y$ be a subspace of $X$ isometric to $\ell_1$. The dual space of $X^*$ is weakly sequentially complete and this property passes to closed subspaces. On the other hand $Y^*$ is isometric to $\ell_\infty$, so it is not weakly sequentially complete.


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.