Dual of a subspace

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.

• 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.) Dec 30, 2018 at 13:51

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.