How can a non-reflexive space have a reflexive subspace?

Question

I know that this can happen (take any one-dimensional subspace, for instance), but I had the following thought while reading Kadison and Ringrose (Fundamentals of the Theory of Operator Algebras, Vol I):

If $V_0$ is a subspace of $V$, then $V^*$ can be thought as a subspace of $V_0^*$. Now, for each element $\rho \in V^{**}$, Theorem 1.6.1. of this book,

Theorem 1.6.1.: If $X_0$ is a subspace of a normed space $X$ and $\rho_0$ is a bounded linear functional on $X_0$, there is a bounded linear functional $\rho$ on $X$ such that $\Vert \rho \Vert = \Vert \rho_0 \Vert$ and $\rho(x) = \rho_0(x)$ when $x \in X_0$.

implies that we can extend $\rho$ to a functional over $V_0^*$, that is, $\rho \in V_0^{**}$.

Thus, if $V_0$ is reflexive, then there exists $x_0 \in V_0$ such that $\rho(\eta) = \eta(x_0)$, for all $\eta \in V_0^*$. Consequently, $x_0 \in V$ and $\rho(\eta) = \eta(x_0)$, for all $\eta \in V^*$, since $V^* \subset V_0^*$.

Where is the flaw in this argument?

Answer 1

Let's examine your proof closely, in light of your comment.

We start with a $\rho \in V^{**}$ (good), i.e. a linear functional on $V^*$. You then want to extend $\rho$ to a linear functional on $V_0^*$. As per Qiaochu Yuan's comment that you were replying to, "extend" is really the wrong verb here. But, even so, you seem to be searching for a surjective map, and there doesn't seem to be a canonical surjection from $V^{**}$ to $V_0^{**}$. This would have to be clarified in the proof.

Even if you do manage to construct one such surjection, remember that it's not enough to simply construct some kind of surjective map from $V^{**}$ to $V$, it must be the canonical map! For example, James' Space is a non-reflexive space that is isometrically isomorphic to its second dual. So, while it's possible to stumble through and find a suitable surjection (indeed, surjective linear isometry), it is the wrong surjection! So, you should be careful not just about finding linear maps, but making sure they are the right linear maps.

Answer 2

I believe I understand the issue more clearly and I thank for this those who answered! The problem does lie in my assumption that I can identify the dual space $V^*$ of $V$ as a subspace of $V_0^*$ and so we cannot simply extend a $\Omega \in V^{**}$ so that it acts over $V_0*$ (I changed here the notation for elements in the double dual so that we can differentiate more easily between elements of $V^*$ and $V^{**}$).

If anyone is interested in the long answer, here it is:

Starting with a $\Omega \in V^{**}$, we could in principle construct a $\Pi \in V_0^{**}$, but as @TheoBendit explained, this is not a proper extension and so it would not yield a canonical mapping between the second dual $V^{**}$ and the space $V$ itself.

We cannot view $V^*$ as a linear subspace of $V_0^*$, as was pointed out by @QiaochuYuan, since the "embedding" given by the restriction of a linear functional $\rho \in V^*$ to $V_0$ is not injective (although it is always surjective, fact demonstrated by the Theorem I presented).

We can, however, identify the whole $V_0^*$ with the quotient space $V^*/V_0^{\perp}$, where $V_0^{\perp}$ is the linear subspace of functionals acting on $V$ which vanish at $V_0$. That is, we can construct a (isometric) isomorphism

\begin{split} T : V^*/V_0^{\perp} & \rightarrow V_0^* \\ \rho + V_0^{\perp} & \rightarrow \rho\vert_{V_0} \end{split}

Therefore, if we pick, for each $\rho_0 \in V_0^*$ (again, this is a surjective mapping), one pre-image $\rho \in V^*$ in the equivalence class $\{ \sigma \in V^* \; ; \; T(\sigma) = \rho_0 \}$ and call this $T^{-1} (\rho_0)$, then we can form a $\Pi \in V_0^{**}$, defined by the equation

$$ \Pi(\rho_0) = \Omega (T^{-1} (\rho_0)). $$

From $V_0$ reflexive, there exists a $x_0 \in V_0$ such that $\Pi = \hat{x_0}$, that is, $\Pi(\eta) = \hat{x_0}(\eta) = \eta(x_0)$, for all $\eta \in V_0^*$.

Here comes the flaw in this argument: For some $\rho \in V^*$, we will not be able to retrieve, from $\Pi(\rho_0)$, the application of $\Omega$. Take, for example, a $\rho$ which is an element of the equivalence class of $T^{-1}(\rho_0)$ (which essentially means that $\rho\vert_{V_0} = \rho_0$), but different from it, we have

$$ \Omega(\rho) \neq \Omega(T^{-1}(\rho_0)) = \Pi (\rho_0) = \rho_0(x_0) = \rho(x_0). $$

(Of course the first inequality could be satisfied for some $\rho$ in the equivalence class of $T^{-1}(\rho_0)$, but this is not true in general and thus the result cannot be valid for all $\Omega$).