# Preannihilators of subspaces of separable duals

If $Y\subset X^*$ is a closed subspace (where $X$ is a separable Banach space), the preannihilator of $Y$ in $X$ is $Y_{\perp}:=\{x\in X : y^*(x)=0, \forall y^*\in Y \}$. If $Y$ is a proper subspace of $X^*$, and $X$ is reflexive then it can be proved that $Y_{\perp}\neq\{0\}$. On the other hand if $X=l_1$ and $Y=c_0\subset l_{\infty}$, then $Y_{\perp}=\{0\}$.

• 1) If $X^*$ is separable, is it still possible that $Y_{\perp}=\{0\}$?
• 2) Can $c_0$ be a subspace of a separable dual? This question is probably unrelated, and most likely well known, but I realized I do not know an answer to it.

For (1), take $X = c_0$ so that $X^* = \ell^1$, which is separable. Take $Y = \{y \in \ell^1 : \sum_i y(i) = 0\}$ which is a proper closed subspace. Since $Y$ contains all the elements of the form $e_i - e_{i+1}$, where $e_i$ are the usual basis functions, you can see that any $x \in Y_\perp$ satisfies $x(1) = x(2) = x(3) = \dots$. The only such element of $c_0$ is $0$.

The intuitive idea is that by Hahn-Banach, $Y$ will certainly have a nontrivial annihilator living in $X^{**}$. In this case it is the constants in $X^{**} = \ell^\infty$ which are not in $X$.

More generally, whenever $X$ is non-reflexive, you should be able to choose any $f \in X^{**} \setminus X$ and take $Y$ to be its kernel. Then if $x \in Y_\perp$, it is also in the annihilator $Y^\perp$ when considered as an element $f_x$ of $X^{**}$. The kernel of $f_x$ contains the kernel of $f$, so $f_x$ is a scalar multiple of $f$. Since $f_x \in X$, this is only possible if $f_x = 0$, i.e. $x=0$.

The answer to Q2 is negative. One way to see this (which might not be the first, and might not be the easiest) is to combine the following results.

1. A separable dual (Banach) space has the Radon-Nikodym property.

2. The RNP is inherited by closed subspaces

3. $c_0$ does not have the RNP.

(The statement that $c_0$ does not embed as a closed subspace of a separable dual space can be found as Theorem 6.3.7 in the book of Albiac-Kalton; the argument above is sketched by the same authors after Corollary 6.3.9.)

A negative answer to the second question can be obtained by a norming technique. Assume $0<a\leq b<\infty$ and $(x^*_n)_{n=1}^\infty\subset X^*$ are such that $$1/a\leq \inf \{\|\sum_{i=1}^n a_ix_i^*\|: n\in\mathbb{N}, \max_i |a_i|=1\} \leqslant \sup \{\|\sum_{i=1}^n a_ix_i^*\|: n\in\mathbb{N}, \max_i |a_i|=1\}\leq b.$$ In particular, $(x^*_n)_{n=1}^\infty$ is a basic sequence equivalent to the $c_0$ basis (and it must therefore be weakly null). Now let $(\phi_n)_{n=1}^\infty$ be the coordinate functionals to the $(x^*_n)_{n=1}^\infty$ and let $x^{**}_n$ be a Hahn-Banach extension of $\phi_n$. Then $x^{**}_n(x_m)=1$ if $m=n$ and $0$ otherwise.

Our assumptions imply that $\|x^{**}_n\|\leq a$ for all $n\in \mathbb{N}$. By Helly's Lemma (or Goldstine's theorem + perturbation, or local reflexivity if you don't mind overkill), we may find a sequence $(x_n)_{n=1}^\infty\subset 2a B_X$ such that $x^*_n(x_n)=1$ and $x^*_m(x_n)=0$ for $1\leq m<n$ (note that we do not control $x^*_m(x_n)$ for $m>n$).

Now fix a sequence $(\epsilon_n)_{n=1}^\infty$ of positive numbers such that $\sum_{n=1}^\infty \epsilon_n=1/2$. Since the sequence $(x^*_n)_{n=1}^\infty$ is weakly null, we may recursively select $n_1<n_2<\ldots$ such that, with $y^*_i=x^*_{n_i}$ and $y_i=x_{n_i}$, $|y^*_i(y_j)|<\epsilon_i$ for $i>j$. Note that these sequences retain the properties that $y^*_i(y_i)=1$ and $y^*_i(y_j)=0$ for $i<j$. Thus we have extracted subsequences $(y_i)_{i=1}^\infty$ and $(y^*_i)_{i=1}^\infty$ which are almost biorthogonal.'' Then the $c_0$ behavior of the $(y^*_i)_{i=1}^\infty$ sequence allows us to show $\ell_1$ behavior of $(y_i)_{i=1}^\infty$, which we do next.

Now fix $n\in \mathbb{N}$ and scalars $(b_j)_{j=1}^n$. For each $1\leq j\leq n$, fix $a_j$ with $|a_j|=1$ and $a_jb_j=|b_j|$. Let $x=\sum_{j=1}^n b_j y_j$ and $x^*=\sum_{j=1}^n a_j y^*_j$. Note that $\|x^*\|\leq b$, so \begin{align*} b\|x\| & \geq |x^*(x)| = \sum_{j=1}^n a_jb_jy^*_j(y_j)-\sum_{j=1}^n |b_j| \sum_{i=j+1}^n |a_jy^*_i(y_j)| \\ & =\sum_{j=1}^n |b_j| - \sum_{j=1}^n |b_j|\frac{1}{2} = \frac{1}{2}\sum_{j=1}^n |b_j|. \end{align*}

By the triangle inequality, $\|x\|\leq 2a\sum_{j=1}^n |b_j|$, so $(y_j)_{j=1}^\infty$ is a basis for a closed subspace $Y$ of $X$ isomorphic to $\ell_1$. This means $X^*$ has a quotient isomorphic to $\ell_\infty$, and $X^*$ is not separable.

By working slightly harder, one may also choose the $(y_j)_{j=1}^\infty$ to have $Y$ complemented in $X$, so $X^*$ actually has not just a quotient, but a complemented subspace isomorphic to $\ell_\infty$. Using results of James and sharper constants above, for $\epsilon>0$, we may have $Y$ $1+\epsilon$-isomorphic to $\ell_1$ and the range of a projection of norm at most $1+\epsilon$.

• This is much more informative than the answer I found by "combining black boxes" :) – Yemon Choi Aug 16 '18 at 23:12
• As my PhD advisor likes to point out to me repeatedly, I have an attitude of "Why say less when you can go on and on and on..." – user114263 Aug 17 '18 at 21:25