Banach space preliminaries

The purpose of this section is to review some well-known concepts and results from separable Banach space theory. Our focus is the weak-*topology on the dual of a separable Banach space. A reference for most of this material is Chapter V of Dunford and Schwartz [11].

Let X be a Banach space. The weak topology on X is the weakest topology such that every bounded linear functional on X is continuous. The dual space of X is the space X^* of all bounded linear functionals on X. The norm of $x^*\in X^*$is defined by

$$\Vert x^*\Vert=\sup\bigl\{\vert x^*(x)\vert\bigm\vert\Vert x\Vert\le1\bigr\}\,.$$

The weak-* topology on X^* is the weakest topology such that for all $x\in X$ the functional $x^*\mapsto x^*(x)$ is continuous. The weak-* closure of a set $Z\subseteq X^*$ is denoted $\mbox{\rm cl}^*\,Z$. Note that about any point $x_0^*\in X^*$ there is a weak-* neighborhood basis consisting of all sets of the form

$$\bigl\{x^*\in X^*\bigm\vert \vert x^*(x_1)-x_0^*(x_1)\vert<1,\ldots, \vert x^*(x_n)-x_0^*(x_n)\vert<1\bigr\}$$

for some finite set $x_1,\ldots,x_n\in X$.

A key theorem concerning the weak-* topology is:

Theorem 2.1 (Banach-Alaoglu)   For any r>0, the closed ball

$$B_r(X^*)=\bigl\{x^*\in X^*\bigm\vert\Vert x^*\Vert\le r\bigr\}$$

is weak-*closed and weak-* compact. Furthermore, if X is separable then B_r(X^*) is weak-* metrizable.  

Proof. See [11], page 424.

In considering weak-* sequential convergence, an important fact to keep in mind is the following consequence of the Banach-Steinhaus theorem:

Theorem 2.2   Let $\{x_n^*\mid n\in{\mathbb{N} }\}$ be a countable set of points in X^*. If $\sup_n\vert x^*_n(x)\vert<\infty$ for all $x\in X$, then $\sup_n\Vert x_n^*\Vert<\infty$.

Proof. See [11], page 66.

Corollary 2.3   If $x_n^*\to x^*$ in the weak-* topology in X^*, then $\{x^*_n\mid n\in{\mathbb{N} }\}$ is bounded, i.e., $\{x^*_n\mid n\in{\mathbb{N} }\}\subseteq B_r(X^*)$ for some $0<r<\infty$.  

A set $Z\subseteq X^*$ is said to be weak-*sequentially closed if it is closed under weak-* limits of sequences, i.e., $x_n^*\to x^*$ weak-* and $x_n^*\in Z$ for all $n\in{\mathbb{N} }$ imply $x^*\in Z$.

Corollary 2.4   A set $Z\subseteq X^*$ is weak-* sequentially closed if and only if $Z\cap B_r(X^*)$ is weak-* sequentially closed for all r>0.  

For an arbitrary set $Z\subseteq X^*$, being weak-* sequentially closed is not in general equivalent to being weak-* closed. In particular, the weak-* topology is not in general metrizable, even when X is separable. This is shown by the following theorem.

Theorem 2.5   Let X be an infinite-dimensional separable Banach space. Then we can find a countable set $Z\subset X^*$ such that Z is weak-* sequentially closed yet weak-* dense in X^*.  

Proof. We first prove the following lemma.

Lemma 2.6   Let X be an infinite-dimensional Banach space. Then we can find a sequence of points $\langle{}x_k\mid k\in{\mathbb{N} }\rangle{}$ in Xsuch that $\lim_k\Vert x_k\Vert=\infty$ and $\Vert x_k\Vert>1$ for all $k\in{\mathbb{N} }$, yet 0 belongs to the weak closure of $\{x_k\mid k\in{\mathbb{N} }\}$.

Proof. For any finite set $F\subset X^*$, $\{x\in X\mid y^*(x)=0 \mbox{ for all }y^*\in F\}$ is a subspace of X of codimension at most the cardinality of F, and hence in particular it intersects $\{x\in X\mid\Vert x\Vert>n\}$ for each $n\in{\mathbb{N} }$. By the Banach-Alaoglu theorem, B_n(X^*) is weak-* compact, and so B_n(X^*)ⁿ is weak-* compact as well. Thus for each n we can find a finite set $G_n\subset\{x\in X\mid\Vert x\Vert>n\}$ such that, for each finite set $F\subset B_n(X^*)$ of cardinality n, $\{x\in X\mid\vert y^*(x)\vert<1\mbox{ for all }y^*\in F\}$intersects G_n. Letting $\{x_k\mid k\in{\mathbb{N} }\}$ be an enumeration without repetition of $\bigcup_{n=1}^\infty G_n$, we obtain the desired sequence.

Proof of Theorem 2.5. Since X is separable, it follows by Theorem 2.1 that X^* is weak-*separable, so let $\{x^*_n\mid n\in{\mathbb{N} }\}$ be a countable weak-* dense subset of X^*. By the preceding lemma, for each n we can find a sequence $\langle{}z^*_{nk}\mid k\in{\mathbb{N} }\rangle{}$ such that $\Vert z^*_{nk}\Vert>n$ for all k and $\lim_k\Vert z^*_{nk}\Vert=\infty$ and x^*_n belongs to the weak, and hence weak-*, closure of $\{z^*_{nk}\mid k\in{\mathbb{N} }\}$. Thus $Z=\{z^*_{nk}\mid n,k\in{\mathbb{N} }\}$ is a countable set which is weak-*dense in X^*. On the other hand, for each n, $Z\cap B_n(X^*)$ is finite, so by Corollary 2.4 Z is weak-* sequentially closed.

The previous theorem shows that a set in X^* can be weak-*sequentially closed yet far from weak-* closed, even when X is separable. Nevertheless, it turns out that for convex sets in X^*, being weak-* sequentially closed is equivalent to being weak-* closed, provided X is separable. We shall obtain this result as a consequence of the following well-known theorem:

Theorem 2.7 (Krein-Smulian)   Let X be a Banach space. A convex set in X^* is weak-* closed if and only if its intersection with B_r(X^*) is weak-* closed for every r>0.  

Proof. See [11], page 429.

The special case of the Krein-Smulian theorem for subspaces of X^* is originally due to Banach ([2] page 124):

Corollary 2.8 (Banach)   A subspace of X^* is weak-* closed if and only if its intersection with B₁(X^*) is weak-* closed.  

Proof. For a subspace Z of X^*, we have $Z\cap B_r(X^*)=r(Z\cap B_1(X^*))$. Hence $Z\cap B_r(X^*)$ is weak-* closed for all r if and only if $Z\cap B_1(X^*)$ is weak-* closed. The desired result follows immediately from the Krein-Smulian theorem.

A set Z in X^* is said to be bounded-weak-* closed if $Z\cap B_r(X^*)$ is weak-*-closed for all r>0. This defines yet another topology on X^*, the bounded-weak-* topology. By the Banach-Alaoglu theorem, weak-* closed sets are bounded-weak-*closed, but the converse does not hold in general. We can paraphrase the Krein-Smulian theorem by saying that a convex set in X^* is weak-* closed if and only if it is bounded-weak-* closed.

Lemma 2.9   Let X be a separable Banach space. A set $Z\subseteq X^*$ is bounded-weak-* closed if and only if it is weak-* sequentially closed.  

Proof. By definition, Z is bounded-weak-* closed if and only if $Z\cap B_r(X^*)$ is weak-* closed for all r. Since X is separable, we have by Theorem 2.1 that B_r(X^*) is weak-* compact and weak-* metrizable. Hence, for all r, $Z\cap B_r(X^*)$ is weak-*closed if and only if $Z\cap B_r(X^*)$ is weak-* sequentially closed. By Corollary 2.4 it now follows that $Z\cap B_r(X^*)$ is weak-* closed for all r if and only if Z is weak-* sequentially closed. This completes the proof.

We now obtain the desired result:

Theorem 2.10   Let X be a separable Banach space. A convex set $Z\subseteq X^*$ is weak-* closed if and only if it is weak-* sequentially closed.

Proof. Immediate from Theorem 2.7 plus Lemma 2.9.

Again, the special case when Z is a subspace of X^* is due to Banach ([2] page 124):

Corollary 2.11 (Banach)   Let X be a separable Banach space. A subspace Z of X^* is weak-* closed if and only if it is weak-* sequentially closed.  

We now turn to a discussion of weak-* sequential closure ordinals. For an arbitrary set $Z\subseteq X^*$, let Z' denote the set of weak-* limits of sequences from Z. Define a transfinite sequence of sets $Z^{(\alpha)}$, $\alpha$ an ordinal, by

$$\begin{eqnarray*}Z^{(0)} & = & Z\,,\\ Z^{(\alpha+1)} & = & (Z^{(\alpha)})'\,,\\... ...elta}Z^{(\alpha)} \quad\mbox{for }\delta\mbox{ a limit ordinal.} \end{eqnarray*}$$

Clearly $\alpha<\beta$ implies $Z^{(\alpha)}\subseteq Z^{(\beta)}$. Define $\mbox{\rm ord}(Z)$ to be the least ordinal $\alpha$ such that $Z^{(\alpha+1)}=Z^{(\alpha)}$. Thus $Z^{(\mbox{\rm ord}(Z))}$ is the weak-*sequential closure of Z. We call $\mbox{\rm ord}(Z)$ the closure ordinal of Z. Note that if Z is convex, then $Z^{(\alpha)}$ is convex for all $\alpha$; also, by the previous theorem $Z^{(\mbox{\rm ord}(Z))}=\mbox{\rm cl}^*\,Z$, the weak-* closure of Z.

In the remainder of this section and the next section, we shall prove some results which completely answer the question of which ordinals can arise as closure ordinals of subspaces of X^* where X is a separable Banach space. This question was first answered completely by McGehee [18] and Sarason [19,20,21].

Lemma 2.12   If X is a separable Banach space, then for any set $Z\subseteq X^*$ the closure ordinal $\mbox{\rm ord}(Z)$ is countable.

Proof. For r>0 and $\alpha$ an ordinal, let $C_r^\alpha$ be the weak-*closure of $Z^{(\alpha)}\cap B_r(X^*)$. For fixed r>0, the sets $C_r^\alpha$ form an increasing, transfinite sequence of compact subsets of a compact metric space, namely B_r(X^*) with the weak-*topology (Theorem 2.1). This transfinite sequence must therefore stabilize at some countable ordinal. Since $C_r^\alpha \subseteq Z^{(\alpha+1)}\cap B_r(X^*) \subseteq C_r^{\alpha+1}$ for all $\alpha$, it follows that the transfinite sequence $Z^{(\alpha)}\cap B_r(X^*)$ also stabilizes at a countable ordinal, call it $\alpha_r$. Put $\alpha=\sup\{\alpha_n\mid n\in{\mathbb{N} }\}$. Then $\alpha$ is a countable ordinal. We claim that $Z^{(\alpha)}$ is weak-* sequentially closed. To see this, suppose that x^* is the weak-* limit of a sequence $\langle{}x_k^*\mid k\in{\mathbb{N} }\rangle{}$ from $Z^{(\alpha)}$. By Corollary 2.3, there exists $n\in{\mathbb{N} }$such that $x^*_k\in B_n(X^*)$ for all $k\in{\mathbb{N} }$. Hence $x^*\in Z^{(\alpha+1)}\cap B_n(X^*)=Z^{(\alpha)}\cap B_n(X^*)$, so in particular $x^*\in Z^{(\alpha)}$ and our claim is proved. Thus $\mbox{\rm ord}(Z)\le\alpha<\omega_1$, i.e., $\mbox{\rm ord}(Z)$ is countable.

Specializing to subspaces of X^*, we can say more:

Theorem 2.13   Let X be a separable Banach space and let Z be a subspace of X^*. Then the closure ordinal $\mbox{\rm ord}(Z)$ is a countable successor ordinal, unless Z is already weak-* closed, in which case $\mbox{\rm ord}(Z)=0$.  

Proof. By the previous lemma, $\mbox{\rm ord}(Z)$ is countable. Suppose that $\mbox{\rm cl}^*\,Z=Z^{(\delta)}= \bigcup_{\alpha<\delta}Z^{(\alpha)}$, for some countable limit ordinal $\delta$. As $\mbox{\rm cl}^*\,Z$ is norm closed, this implies that $\mbox{\rm cl}^*\, Z=\bigcup_{\alpha<\delta}\mbox{\rm cl}\,{Z^{(\alpha)}}$, where $\mbox{\rm cl}\,Y$ denotes the norm closure of $Y\subseteq X^*$. Thus $\mbox{\rm cl}^*\,Z$ is a closed subset of a complete metric space, written as a countable union of closed sets. By the Baire category theorem, there must be an ordinal $\alpha<\delta$ such that $\mbox{\rm cl}\,{Z^{(\alpha)}}$ has non-void interior as a subset of $\mbox{\rm cl}^*\,Z$, i.e., $\mbox{\rm cl}\,{Z^{(\alpha)}}$ contains the intersection of an open (in norm) ball with $\mbox{\rm cl}^*\,Z$. But $\mbox{\rm cl}\,{Z^{(\alpha)}}$ is a subspace of $\mbox{\rm cl}^*\,Z$, so it must be all of $\mbox{\rm cl}^*\,Z$. Since $\mbox{\rm cl}\,{Z^{(\alpha)}}\subseteq Z^{(\alpha+1)}$, it follows that $Z^{(\alpha+1)}=\mbox{\rm cl}^*\,Z$, whence $\mbox{\rm ord}(Z)\ne\delta$. We have now shown that $\mbox{\rm ord}(Z)$ is not a limit ordinal. Thus $\mbox{\rm ord}(Z)$must be either 0 or a successor ordinal. If $\mbox{\rm ord}(Z)=0$ then Z is weak-* sequentially closed, and hence is weak-* closed by Corollary 2.11. This completes the proof. See also Kechris and Louveau [14], page 157.

It is known that the converse of the previous theorem also holds: For every countable successor ordinal $\alpha+1$, we can find a subspace Z of the dual X^* of a separable Banach space X such that $\mbox{\rm ord}(Z)=\alpha+1$. In the next section we present a proof of this result, with $Z\subseteq\ell_1=c_0^*$ and Z weak-*dense in $\ell _1$.

The study of closure ordinals of subspaces of X^* has an interesting history. Von Neumann ([24] page 380) exhibits a set $S\subset\ell_2$ such that 0 is in the weak closure of S yet no sequence from S converges weakly to 0. The first example of a subspace Z of X^* such that $\mbox{\rm ord}(Z)\ge2$ is due to Mazurkiewicz [17]. Banach ([2] pages 209-213) proves that for every $n\in{\mathbb{N} }$ there is a subspace Z of $\ell_1=c_0^*$ such that $\mbox{\rm ord}(Z)\ge n$ and states the analogous result for all countable ordinals. For the proof Banach refers to a ``forthcoming'' paper which seems never to have appeared, and this reference is omitted from the English translation [1]. Later, McGehee [18] proves the stronger result that for each countable ordinal $\alpha$ there exists a weak-* dense subspace of $\ell_1=c_0^*$ whose closure ordinal is exactly $\alpha+1$ (but note that McGehee's notation differs from ours). In the next section we reprove this result of McGehee. Sarason [19,20,21] proves a similar result for the spaces $H^\infty$ and $\ell_\infty$. In view of the theorem above, these results of McGehee and Sarason are in a sense best possible. While McGehee's proof uses sets of synthesis and uniqueness, our proof in the next section is much more elementary.