Notation: $X$ is a banach space, $X'$ is the dual space to $X$. When $V \subset X'$, we write $\ker V = \cap_{l \in V} \ker l$, and when $W \subset X$, we write $ann \; W = \{l \in X' \mid l(w) = 0 \text{ for all } w \in W \}$.

When $V$ is a finite-dimensional subspace of $X'$, it is well-known (at least to math.stackexchange, where it's been asked about 3,000 times) that $$ V = ann \; (\ker V) $$

My question: is this true when $V$ is a closed, finite CO-dimensional subspace? For me, this means that there is a complement $W \subset X'$ of finite dimension to $V$ in $W$, such that the splitting $X' = W \oplus V$ is topological.


This is false.

Let $X$ be the Banach space $c^0$ of real sequences tending to zero. Let $\ell^1$ be the space of absolutely summable real sequences. We can identify $\ell^1$ with $X'$ via the isomorphism $x \to (y \to \sum_{i=1}^\infty x_iy_i)$.

Now let $\tau : \ell^1 \to \mathbb{R}$ be given by $\tau(x) = \sum_{i=1}^\infty x_i$. This is a continuous linear functional, so $V := \ker \tau$ is a closed subspace of $\ell^1$ of codimension $1$.

Let $e_i \in \ell^1$ be such that $(e_i)_j = \delta_{ij}$. Then $e_i - e_j \in V$ whenever $i \neq j$.

Suppose that $c \in \ker V$. Then $(e_i - e_j)(c) = c_i - c_j = 0$ whenever $i \neq j$, so $c$ must be a constant sequence. But $c_i \to 0$ as $i \to \infty$ so $c = 0$. Hence $\ker V = 0$ and $ann(\ker V) = X'$ properly contains $V$ because $e_1 \in X' \backslash V$.

You can show that the Banach space $X$ satisfies your condition precisely when it is reflexive, i.e. the canonical map $X \to (X')'$ is an isomorphism.

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  • $\begingroup$ @DanielFischer I made an error; this answer addresses my question on the nose. $\endgroup$ – A Blumenthal Jan 9 '14 at 22:26
  • $\begingroup$ @ABlumenthal Just saw the edit. $\endgroup$ – Daniel Fischer Jan 9 '14 at 22:26
  • $\begingroup$ @KonstantinArdakov When $X$ is reflexive, does the desired formula hold for any closed subspace? $\endgroup$ – A Blumenthal Jan 10 '14 at 21:53
  • $\begingroup$ @ABlumenthal Yes. You can use the Hahn-Banach theorem to find some $\theta \in X''$ such that $\theta(V)=0$ but $\theta$ is not zero on $ann \ker V$, and derive a contradiction if $X$ is reflexive. $\endgroup$ – Konstantin Ardakov Jan 10 '14 at 23:39

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