Is being isomorphic to dual hereditary?

Question

Let $V$ be a vector space, and assume that $V$ is isomorphic to its dual, i.e., $V \simeq V^*$. Is every linear subspace $U$ of $V$ also isomorphic to its dual, i.e., $U \simeq U^*$?

This is certainly true in finite dimensions, and I believe also for Hilbert spaces, so assume $V$ is infinite-dimensional and not a Hilbert space. I believe that, if there is any chance of the above being true, we also need to assume that $V$ is a topological vector space, $V^*$ is the continuous/topological dual space (rather than algebraic dual space), and the isomorphism $V \simeq V^*$ is continuous linear (with continuous linear inverse).

Infinite dimensional spaces are not isomorphic to their duel. You run into cardinality considerations right away that prevent it. — CyclotomicField, Jan 17 '23 at 15:19
@CyclotomicField The question is about continuous duals. But admittedly, the OP thinks wrong about Hilbert spaces. — Anne Bauval, Jan 17 '23 at 15:20
So to clarify: if $V$ is a topological vector space isomorphic to its continuous dual $V'$ and $U$ is a closed subspace of $V$, then $U$ is isomorphic to its continuous dual $U'$? Do you have a proof in mind? — Jon Warneke, Jan 17 '23 at 20:19
@JonWarneke I don't think $U$ being closed is enough, I think we actually need $U$ to be complemented — Ben Grossmann, Jan 17 '23 at 22:21
@JonWarneke Also, it is notable that the isomorphism between $V$ and $V^$ doesn't necessarily "induce" an invertible map from $U$ to $U^$. For example, taking $\phi : \Bbb C^2 \to [\Bbb C^2]^*$ to be the map $$ \phi([x_1,x_2])([y_1,y_2]) = x_1y_1 - x_2 y_2, $$ we find that if $U$ is the span of $(1,1)$, then for $u \in U$, the restriction of $\phi(u)$ to $U$ is simply the zero map, so we don't recover the full dual space of $U$ in this fashion. — Ben Grossmann, Jan 17 '23 at 22:28

score 0 · Answer 1 · answered Jan 17 '23 at 18:38

I assume that you're referring to the continuous dual of a normed (or at least topological) vector-space.

If we don't specify that $U$ is a closed subspace, then the answer is no, not even for Hilbert spaces. As an example, take $V = \ell^2$ and $U = c_{00} \subset V$. $c_{00}$ is separable, but $c_{00}^* \cong \ell_\infty$ is not.

Is being isomorphic to dual hereditary?

1 Answers1