Topological Vector Spaces Part 3: Duality

Continuing the last two posts. Additional references for this time:

• Topological Vector Spaces, Distributions and Kernels, Treves. The reference on topological vector spaces.
• A Course on Topological Vector Spaces, Voigt. I like this source on duality-related topics.

Hahn-Banach

Theorem (Hahn-Banach). Let $X$ be a topological vector space, $M \sbe X$ a subspace and a linear functional $f$ defined on $M$, if $U$ is an absolutely convex neighborhood of the origin with $\vt f(U \cap M) \vt \leq 1$, then there is $x' \in X$ such that $\vt f(U) \vt \leq 1$ and $x' = f$ on $M$.

Taken literally, Hahn-Banach says that we can extend a partially defined linear functional $f$ while keeping its continuity, as witnessed by $U$. Absolutely convex is used to convert $U$ to a seminorm, and the proof of just Zorn’s lemma plus algebra, which I find not too interesting.

An alternative explanation of Hahn-Banach is that it is the key reason we work with locally convex spaces rather than arbitrary topological vector spaces, If $X$ is locally convex, we can always restrict $U$ to be absolutely convex. Hahn-Banach has three important applications. 1. Showing that $X'$ separates points on $X$. 2. Deriving the bipolar theorem. 3. Constructing dual elements.

For example, given finitely many linearly independent $x_1,\ldots,x_n$, let $f(x_j) = 1$ if $j = 1$ and zero otherwise. Since $f$ is finite-dimensional, it is trivially continuous, though we want to extend it to a continuous functional on all of $X$. Hahn-Banach says that we can do this, and any neighborhood of the origin $U$ can be used, it absorbs finite sets. If we do this $n$ times, we get $x_i'$ such that $x_i'(x_j) = \delta_{ij}$, so $x_i'$ is a collection of linear functionals which, restricted to the finite dimension subspace, forms a dual basis. In particular, it separates the points $x_1,\ldots,x_n$. When $X$ is normed, if $\Vt x_j \Vt = 1$ and $U$ is the unit ball, $\Vt x'_i \Vt = 1$.

Topologies on $X$ and $X'$

We first remind ourselves of the initial topology and final topolgoy construction. Let $X$ be a set, and $\{f_\alpha\}_{\alpha \in \Lambda}$ a collection of functions $f_\alpha : X \rightarrow Y_\alpha$ map out of $X$ into topological spaces $Y_\alpha$. The initial topology on $X$ defined to be the weakeast topology such that all $f_\alpha$ are continuous, weakeast since any stronger topology would automatically have $f_\alpha$ continuous. Concretely, the topology on $X$ is generated by finite intersections and arbitrary unions of sets of form $f_\alpha^{-1} (U)$ where $U \sbe Y_\alpha$ is open. If $\Lambda$ is a singleton the initial topology would just be the pull-back of the topology on $Y_\alpha$. If $Z$ is another topological space and a map $g : Z \rightarrow X$ has composition

continuous for all $\alpha$, then $g$ must be continuous, since the push-forward of the topology of $Z$ onto $X$ must be stronger than the initial topology on $X$.

An important example of an initial topology is the product topology, which is the initial topology with respect to projections $p_\alpha$. Specifically, sets of the form $\bigcap_{\alpha \in I} p_\alpha^{-1}(U_\alpha)$ for finite $I$ are called cylindrical sets, and form a base of the product topology.

Conversely, if functions $f_\alpha : Y_\alpha \rightarrow X$ map into $X$, than the final topology is the strongest topology such that all $f_\alpha$ are continuous. If $Z$ is another topological space and a map $g : X \rightarrow Z$ has composition

continuous for all $\alpha$, then $g$ must be continuous, The LB and LF topology are examples of a final topology, and another similar example is the topology on CW complexes. ¹

Generally, given a vector space $X$ with no topology, there are a few things we can do, which we examined in the previous post.

• Form the strongest or final locally convex topology with resepct to the collection over $x \in X$ of scalar multiplication maps $t \mapsto tx$. A linear map from $X$ to another topological vector space $Y$ is automatically continuous, since it is continuous when composed with scalar multiplication.
• Given a bornology $\beta$, form the strongest topology where all $\beta$ are topologically bounded.

Given $X$ with locally convex topology, the dual space $X'$ has various topologies. We will study these topologies more in this and coming posts.

• The weak or weak star topology on $X'$, denoted $X'_\sigma$, is the topology of pointwise convergence on $X$.
• The strong topology on $X'$, denoted $X'_b$, is the topology of uniform convergence on von Neumann bounded sets of $X$.
• The equicontinuous bornology on $X'$ is the bornology of equicontinuous sets.

Given $X$ with an original locally convex topology $\tau_0$, consider the space of all continuous linear functionals $X' := (X,\tau_0)'$. Now, fix $X'$, but vary $\tau_0$. These topologies will be the focus of this post.

• The weak topology on $X$, denoted $X_\sigma$, is the topology of pointwise convergence on points of $X'$.
• The Mackey topology on $X$, denoted $X_\tau$, is the topology of uniform convergence on equicontinuous sets of $X'$.

Exercise. Verify that all topologies above are locally convex.

Weak Topologies

The weak and weak star topologies on $X$ and $X'$ are similar so it does not hurt to call both weak topologies. The weak topology on $X$ is the weakest or initial topology with respect to the collection of functions $\{ x \mapsto x'(x) : x' \in X'\}$. In terms of seminorms, the topology is generated by the family $p_{x'} (x) = \vt x'(x) \vt$. Since $X$ is locally convex, for any $x$, Hahn-Banach implies that there is some $x'$ where $x'(x) \neq 0$, hence the weak topology is Hausdorff.

The weak star topology is the initial topology with respect to $\{ x' \mapsto x'(x) : x \in X \}$, and with seminorms, the family $p_x (x') = \vt x'(x) \vt$. The fact that the weak star topology is Hausdorff follows by definition of $x' \neq 0$. Often we denote $\langle x', x \rangle = x'(x)$. The weak topology is the weakest topology such that $\langle \cdot , \cdot \rangle : X' \times X \rightarrow X$ is separately continuous.

Fact. Let $X$ be a locally convex space with an original topology. Let $X_\sigma'$ be the dual of $X$ equipped with the weak topology. Then, $(X_\sigma')' \simeq X$.

Proof. Given any $x \in X$, $x$ acts on $X'$ by $x(x') = x'(x)$, thus there is an inclusion $j : X \hookrightarrow (X_\sigma')'$. Since $X'$ separates points on $X$, $j$ is injective. Conversely, given $\phi \in (X_\sigma')'$, by continuity of $\phi$, for any $\epsilon > 0$, there exists an open set of $X_\sigma'$ such that $\vt \phi(U) \vt < \epsilon$. The set $U$ has form

\[ U = \{ x' \in X' : \vt x'(x_i) \vt < \delta \text{ for } 1 \leq i \leq n \} \]

for some finite set of points $x_1,\ldots,x_n$.

Since $\phi$ is linear, we may scale $\delta, \epsilon$ together, and after passing $\delta, \epsilon \rightarrow 0$, we have $\phi(x') = 0$ if $x'(x_i) = 0$ for $1 \leq i \leq n$. This implies that $\phi$ has finite co-dimension, so $\phi$ must be represented as a finite linear combination in $x_i$. ²

This proof I hope is illustrative, it shows that the weak topology only has control on finite dimension subspaces.

Compactness

The weak topology is generally difficult to work with, it is generated by many seminorms, and certainly is not normed even if $X$ is normed. Given a function defined on a space $X$, continuity with respect to the weak topology on $X$ is harder to show than continuity with respect to other topologies. However, the weak topology also has one key utility for existence results in analysis.

Idea. Since the weak topologies only has control over finite dimensions, it allows for compactness in infinite dimension spaces.

We have two main results, both of which involve embedding the space of linear continuous functions into the space of continuous functions and space of all functions, all endowed with the topology of pointwise convergence, and invoking results from general topology. ³ There is little use of the vector space structure, other than the fact that the space of linear functions is closed.

Theorem (Tychonoff). Let $X$ be an arbitrary set. If $E \sbe \R^X$ is uniformly bounded, then it is precompact in the pointwise topology.

This statement is a special case of the more general Tychonoff theorem, though it has the advantage of being closer to what we need.

Recall that in an arbitrary topological space, the notions of compact and sequentially compact differ. We also note that a compact set is always closed in Hausdorff spaces, but sequentially compact sets need not be closed. There is a third notion, countably compact, every sequence has a limit point, which is generally weaker and implied by both compact and sequentially compact.

Theorem (Eberlein-Grothendieck). Let $K$ be a compact topological space, $E \sbe C(K) \sbe \R^K$ be such that any sequence in $E$ has a limit point in $C(K)$. Then, the closure of $E$ in $\R^K$ is compact, sequentially compact, and lies in $C(K)$.

We may now conclude the two following facts.

Fact (Banach-Alaoglu). Let $X$ be locally convex, $E \sbe X'$ equicontinuous. Then, $E$ is weak-star precompact.

Proof. Since $E$ is equicontinuous, there is some $U \sbe X$ neighborhood of zero such that $\vt x'(U) \vt \leq 1$ for all $x' \in E$. Regard $E \sbe \R^U$, and invoke Tychonoff, $E$ is precompact in the pointwise topology on $\R^U$.

However, the pointwise topology on $\R^U$ and $\R^X$, when restricted to linear functions, are identical. A local base of open sets on $L(X)$ is generated by $\{x' \in L(X) : \vt x'(x) \vt < 1 \}$ for some $x \in X$, but since $U$ is absorbing, if $tx \in U$, then $\{ x' \in L(U) : \vt x'(tx) \vt < t \}$ is the same open set, and part of the local base on $L(U)$.

Often in practice $X$ is separable, and a simple construction shows that any weak star compact set in $X'$ is metrizable, hence we also get sequential compactness. Generally there is less we can say on $X$. We will only derive the following result regarding normed spaces, though generalizations do exist.

Exercise (Eberlein). Let $X$ be normed, $E \sbe X$ be weakly countably compact. Then, $E$ is weakly compact and weakly sequentially compact. Hint: ⁴

Duality

Definition. A dual pair $\langle Y, X \rangle$ is a pair $X, Y$ of vector spaces and a bilinear pairing $\langle \cdot, \cdot \rangle : Y \times X \rightarrow \R$ that separates points.

The theory of dual pairs generalizes our existing weak topology setup. If we have an original locally convex $X$ and its dual $X'$, we may forget about the topology on both $X$ and $X'$, and only remember $\langle \cdot, \cdot \rangle$ to get a dual pair. The weak topology of a dual pair is the weakest topologies on $X, X'$ such that $\langle \cdot, \cdot \rangle$ is separately continuous. With the dual pair perspective, $X_\sigma$ and $X'_\sigma$ are treated equally. ⁵

Definition. Given a dual pair $\langle Y, X \rangle$, we say the topology $\tau$ on $X$ is compatible with the dual pair if $(X,\tau)' = Y$. A dual pair with topologies on both $X, Y$ compatible is a compatible dual pair.

Aside from the weak topology, another very practical dual pair is $X$ and its strong dual $X_b'$, and the key engine in many existence results in analysis. However, in order for the dual pair to be compatible, we need $(X_b')' = X$. More on this later.

Sets on two spaces in a dual pair talk to each other via the following definition.

Definition. Let $\langle Y, X \rangle $ be a dual pair. Given $E \sbe X$, define the polar of $E$ as

\[ E^\circ = \{ y \in Y : \vt \langle y, E \rangle \vt \leq 1 \}. \]

Clearly, the polar reverses set-inclusion, $A \sbe B$ implies $B^\circ \sbe A^\circ$. We have seen the polar of a neighborhood of the origin appear a few times in the previous section. A set $F \sbe X'$ is equicontinuous precisely when it is contained in the polar of a neighborhood of the origin, hence the polar of neighborhoods of the origin in $X$ generate the equicontinuous bornology on $X$.

Theorem (Bipolar). Let $\langle Y, X \rangle$ be a dual pair, the topology on $X$ compatible. Let $C \sbe X$ be symmetric about the origin. Then, $C^{\circ \circ}$ is the closed convex hull of $C$.

I find the bipolar theorem to be the most important result in this post. It converts the abstract Hahn-Banach theorem into a very useful observation that closed convex sets are determined by a collection of linear functionals, and this result is foundational in convex analysis.

A more familiar phrasing of the bipolar theorem is the separating hyperplane theorem in Euclidean spaces, which says that given a set of points $C$, the polar $C^\circ$ is all halfspaces containing $C$, then the intersection of all halfspaces, which is the bipolar $C^{\circ \circ}$, is precisely the closed convex hull of $C$. ⁶

Proof. Since $C^\circ \sbe Y$, the bipolar $C^{\circ \circ}$ lives in $X$. Given any $x \in C$, $\vt \langle y, x \rangle \vt \leq 1$ for all $y \in C^\circ$, so $x \in C^{\circ \circ}$. Conversely, if $x \notin C$, let $U$ be absolutely convex neighborhood of origin that separates $x$ and $C$, specifically such that $(x + U) \cap (C + U) = \emptyset$. Now, $D = C + U$ is an absolutely convex and containing $C$. By Hahn-Banach there is a $y \in X' = Y$ where $\langle y, x \rangle = p_D(x) \geq 1$ and $\langle y, C \rangle \leq 1$. Then, $y \in C^\circ$, so $x \notin C^{\circ \circ}$.

Note that we assumed nothing of the topology on $X$ other than $X' = Y$. As a corollary of the bipolar theorem, any topology on $X$ which is compatible with $Y$ has the same closed convex hulls. In particular, a convex set that is weakly closed is actually originally closed.

Definition. The Mackey topology of $X$ in a dual pair $\langle Y, X \rangle$ is the strongest topology on $X$ such that $X' = Y$.

Theorem (Mackey-Arens). The Mackey topology $\tau(X, Y)$ on $X$ is generated by polars of weakly compact sets on $Y$.

Proof. Let $K \sbe Y$ be the collection of weakly compact, absolutely convex sets, and define local base $K^\circ \sbe X$ for the Mackey topology $\tau$. The sets $K^\circ$ are absorbing, since for any $x \in X$, $\langle K, x \rangle$ must be compact in $\R$.

To verify that $(X, \tau)'$ is not larger than $Y$, let $X^\#$ be the algebraic dual space of $X$, where both $Y \sbe X^\#$ and $(X,\tau)' \sbe X^\#$. The bipolar theorem holds for subsets of $X$ on the pair $\langle X, X^\# \rangle$. ⁷ Denote the polars in this dual pair by $\bullet$. Then $K^{\circ \bullet} = K^{\bullet \bullet} = K$. Suppose $f \in (X,\tau)'$, that is, $f \in X^\#$ is continuous with respect to $\tau$, and let $K^\circ$ be the witness, then $f \in K^{\circ \bullet} = K \sbe Y$.

Finally we observe $\tau$ is the strongest. If another topology $\tau_0$ on $X$ has local base $U_\beta$ with $(X, \tau_0)' = Y$, the polars $U_\beta^\circ$ are equicontinuous, hence weakly compact by Banach-Alaoglu, and the collection $(U_\beta^\circ)^\circ = U_\beta$ is a subset of the $\tau$ topology.

We can think of Mackey-Arens as a converse to Banach-Alaoglu, and I would say that we really have one combined result.

We do have some unanswered questions. The first is, given $X$ and its dual $X'$ as a dual pair, when is the Mackey topology on $X$ the original topology. This turns out to be true for nice enough spaces like Frechet spaces. Another question is, when is $(X_b')' = X$. We know that this is not true even for some Banach spaces like $c_0$, though we may still hope to find a nice criteria. This is a good time to stop, since to proceed, we need to introduce a new definition.