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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.

Kyle Austin
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  • 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.) – GEdgar Dec 30 '18 at 13:51

1 Answers1

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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.

Tomasz Kania
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