Topological Vector Spaces Part 2: Bounded Sets

Continuing the previous post.

Bounded Sets

Definition. Let $X$ be a topological vector space. A set $E \sbe X$ is topologically bounded or von Neumann bounded if it is absorbed by all neighborhoods of the origin. In terms of seminorms, $E$ is bounded if $p(E) < \infty$ for all seminorms generating the topology.

While a neighborhood of the origin only needs to absorb some element in the local base, a bounded set $E$ needs to be absorbed by all elements of the local base. We can think of bounded sets as a collection of points uniformly close to the origin. Continuity of multiplication implies that singletons are absorbed by all neighborhoods of the origin, hence bounded.

Fact. Cauchy sequences, convergent sequences, and precompact sets are bounded.

A normed space is very special, whose local base is a single element, the unit ball $B(0, 1)$, hence bounded means within a scalar factor of $B(0,1)$, and the unit ball itself is bounded. Conversely, if any neighborhood of zero is bounded, then that neighborhood of zero alone forms a local base, and the space must be normed.

Generically we should think of the local base as having many different open sets, each of which may be fatter in certain directions and thinner in others. For a set to be bounded, it must be so thin that it collapses to something much smaller.

Example. We show that if $E \sbe \mathcal{s}$ is bounded, then there exists some positive $y_n \in \mathcal{s}$ such that $\vt x_n \vt \leq y_n$ for all $x_n \in E$. We say that $E$ is dominated by $y_n$.

Proof. From the definition of bounded, for all $k$, there is some $f(k)$ such that

\[ p_k(x_n) := \sup_{n \in \N} \vt n^k x_n \vt \leq f(k) \]

for all $x_n \in E$. Therefore, $\vt x_n \vt \leq n^{-k} f(k) $ for all $k$ and all $x_n$. Define $y_n = \inf_{k \in \N} n^{-k} f(k)$, so $E$ is dominated by $y_n$. To check $y_n \in \mathcal{s}$, for any $m \in \N$,

$$ p_m(y_n) = \sup_{n \in \N} \inf_{k \in \N} n^{m-k} f(k) \leq \sup_{n \in \N} f(m) = f(m) < \infty. $$

This concludes the proof.

Exercise. Check the following:

• If $E \sbe \R^\N$ is bounded, there must be a positive sequence $y_n \in \R^\N$ such that $E$ is dominated by $y_n$.
• If $E \sbe c_{0,0}$ is bounded, there must be a positive sequence $y_n \in c_{0,0}$ such that $E$ is dominated by $y_n$.
• Verify that in all three cases, no bounded sets are neighborhoods of the origin.

The case $c_{0,0}$ is special, the topology is so strong, all bounded sets must be finite dimensional. More generally, we have the following:

Fact. Let $X = \bigcup_{i=1}^\infty$ be an LF space. If $E \sbe X$ is bounded, then there must exist some $i$ where $E \sbe X_i$, and $E$ is bounded in $X_i$.

In particular, since Cauchy sequences are bounded, a Cauchy sequence in an LF space must eventually stabilize in some $X_i$, and since $X_i$ is complete, it follows that the full $X$ is complete.

Bornologies

Definition. We say that a linear map between topological vector spaces is bounded if it maps bounded sets to bounded sets.

Important Exercise. Show that any continuous linear map is bounded.

While open sets pull back under homomorphisms or are contravariant, bounded sets push forward or are covariant, so in this sense they are dual to each other. Roughly speaking, bounded sets are the largest reasonable covariant objects. Other important covariant objects include finite sets, convergent sequences, and precompact sets. Remarkably, all are bounded.

Idea. A set $E \sbe X$ is bounded if, when viewed as a subset of $X \simeq L(\R,X)$, $E$ is equicontinuous.

This is a more abstract criteria for boundedness that doesn’t explicitly use absorbing, instead only relying on the primitive topological notion of equicontinuity and the bijection $X \simeq L(\R,X)$ given by scalar multiplication. ¹

Given a topology, we can define von Neumann boundedness. Conversely, given a collection of sets $\beta$ but no topology on $X$, we can define a topology on $X$ to be the strongest locally convex topology such that all $E \in \beta$ are equicontinuous. An absolutely convex set $V$ is declared to be a neighborhood of the origin if it absorbs all $E \in \beta$.

We make no loss of generality by assuming $\beta$ is closed under subsets and finite unions, this will not change equicontinuity. As long as $\beta$ contains all points, the topology on $X$ will be singleton-absorbing, hence a valid topology. We call a translation-invariant collection $\beta$ that satisfies the previous assumptions a vector space bornology (borné meaning bounded in French). Vector space bornologies form its own category, where morphisms are bounded (rather than continuous) linear maps, and worth studying in their own right. However, we will focus on bornologies exclusively as they related to topologies.

• From the bornology of finite sets, the induced topology is the strongest locally convex topology on $X$ that makes scalar multiplication continuous.
• From the bornology of von Neumann bounded sets, the resulting topology $X$ is the bornologicification of the original topology, which is generally finer than the original topology.

Given a topology on $X$, if all absolutely convex, bounded-set-absorbing sets are neighborhoods of the origin, then $X$ is its bornologification, the data contained in bounded sets and the topology are identical. We say that $X$ is a bornological space.

Exercise. Show that F-spaces are bornological. Hint: ²

Exercise. Suppose that $X, Y$ are locally convex, $X$ is bornological. Show that any bounded linear map from $X$ to $Y$ is continuous.

Many nice enough and natural topological vector spaces are bornological. ³ This certainly does not require the space to be normed - all of our motivating examples are bornological. From our motivating examples, because we are working with sequence spaces, we can characterize boundedness as being dominated by a single element. If we didn’t know the appropriate topology beforehand, but we managed to pick this bornology, then we could have derived the matching topology!

Space of Continuous Linear Maps

Given two topological vector spaces $X, Y$, it is most natural to consider the continuous linear maps $L(X, Y)$ as its own topological vector space. However, we run into an immediate issue, there is no topology we can place on $L(X,Y)$ generally using only the topology on $X$ and $Y$.

One thing we can do is define the equicontinuous bornology. Given $H \sbe L(X, Y)$, we say that $H$ is bounded if $H$ is equicontinuous, or for any $V \sbe Y$ neighborhood of the origin, there is $U$ neighborhood of the origin such that $T(U) \sbe V$ for all $T \in H$. This is a valid bornology, as it contains singletons, all $T \in L(X,Y)$ is equicontinuous as a singleton.

From the most natural equicontinuous bornology, we can obtain a topology on $L(X,Y)$, which is the strongest topology we have on $L(X,Y)$, but too indirect to work with. Instead, we define a topology on $L(X,Y)$ directly while using a bornology on $X$. In any case, we cannot avoid a bornology somewhere.

Definition. Given a bornology $\beta$ on $X$, we place on $L(X, Y)$ the topology of uniform convergence on $\beta$. ⁴ A local base of $L(X,Y)$ is given by

$$ U_{E,V} = \{ T \in L(X,Y) : T(E) \sbe V \} $$

over all $E \in \beta$ and $V$ in a local base. If $X$ has a topology, then $\beta$ defaults to the von Neumann bornology.

If $V$ is absolutely convex, then so is $U_{E,V}$. For the topology to be valid, it is necessary that $U_{E,V}$ is absorbing for all $E, V$. Thus, for any $T \in L(X,Y)$, there must is a scalar $t > 0$ such that $tT \in U_{E,V}$. Since $E$ is bounded, $T(E)$ is bounded, hence there is $tT(E) \sbe V$, as needed. This would not work if in $U_{E,V}$ we took $E$ to be neighborhoods.

In the case $Y = \R$, we denote the dual space $X' = L(X,\R)$. It suffices to take $V = [-1,1]$, and we call

the polar of $E$, denoted $E^\circ$. Hence, our construction of topologies on $X'$ are all polar topologies.

• When $\beta$ are finite subset of $X$, the topology on $X'$ is called the weak topology, denoted $X'_\sigma$.
• When $\beta$ are compact subset of $X$, the topology on $X'$ is simply called the topology of uniform convergence on compact sets.
• When $\beta$ are bounded subset of $X$, the topology on $X'$ is called the strong topology, denoted $X'_b$. Unless otherwise indicated, $X'_b$ is the default topology on $X'$.

Back to Examples

We end this post by looking at strong duals of our motivating examples. Because we are working with spaces of sequences, elements of the dual also can be represented as sequences. Any $z_n \in X'$ acts on $x_n \in X$ via $z_n(x_n) = \sum_{n=1}^\infty x_n z_n$.

Example. The dual of the space of rapidly decaying sequences $\mathcal{s}$ is the space of slowly increasing sequences $\mathcal{s}'$.

Proof. If $z_n$ is continuous, there must be a $k$ and associated semi-norm $p_k$ such that $p_k(x_n) \leq r$ implies $\vt z_n(x_n) \vt \leq 1$. By testing all sequences $x_n$ that is equal to $n^k$ for a single index and zero otherwise, we have that $\vt z_n n^k \vt \leq r^{-1}$ for all $n$, hence $z_n \in O(n^k)$. Conversely, if $z_n \in O(n^k)$, then $p_k$ witnesses continuity of $z_n$. Therefore, $\mathcal{s}' = \bigcup_{k=0}^\infty Z_k$ where $Z_k$ is the space of sequences that has asymptotics $O(n^k)$.

You may want to skip the next part of the proof on the first reading, and come back after attempting the exercises. I’m showing the derivation for $\mathcal{s}'$ since it is the hardest.

Next we show $\mathcal{s}'$ has the LB topology with respect to the increasing subspaces $Z_k$, where each $Z_k$ has the obvious norm $p_{-k}(z_n) := \sup_{n \in \N} \vt z_n n^{-k} \vt$. Recall that any bounded set $E \sbe \mathcal{s}$ is dominated by a single positive $x_n \in \mathcal{s}$. Then, a local base of $\mathcal{s}$ are the polars

\[ U_{x_n} = \{ x_n \}^\circ = \{ z_n \in \mathcal{s}' : \vt \sum_{n=1}^\infty x_n z_n \vt \leq 1 \}. \]

On the other hand, a local base of the LB topology is $U = \bigcup_{k=0}^\infty B_k(0,r_k)$ where $B_k(0,r_k) = \{ z_n \in Z_k : p_{-k}(z_n) \leq r_k \}$, and $r_k$ may decrease arbitrairly fast. First we show any $U_{x_n}$ contains some $U$. For any $k$,

\[ \vt \sum_{n=1}^\infty x_n z_n \vt = \vt \sum_{n=1}^\infty x_n n^{k+2} z_n n^{-k} n^{-2} \vt \leq p_{k+2} (x_n) p_{-k} (z_n) \sum_{n=1}^\infty n^{-2}. \]

Hence, taking $r_k^{-1} \geq p_{k+2} (x_n) \zeta(2)$ is enough to construct the desired $U$. ⁵

Before doing the other direction, we observe that for any $z_n \in \mathcal{s}$, the tail sequence of $z_n$ converges to zero in the LB topology. Indeed, if $z_n \in Z_\ell$, then for large enough $m$,

\[ p_{\ell+1}(\delta_{n \geq m} z_n) = \sup_{n \geq m} \vt z_n n^{-\ell-1} \vt \leq m^{-1} p_{-\ell} (z_n), \]

so the tail actually converges in $Z_{\ell+1}$. For the other direction, we want to show any $U$ defined by $r_k$ contains some $U_{x_n}$. Pick sufficiently fast growing $m_1, m_2, \ldots$ such that $\prod_{i=1}^k m_i \geq r_k^{-1}$, and define $x_n \in \mathcal{s}$ with

\[ x_n = \left( \prod_{i=1}^k m_i \right) n^{-k} \text{ if } n \in [m_k, m_{k+1}]. \]

Thus, $x_n$ is piecewise $n^{-k}$ while continuous across intervals. Moreover, $x_n n^k \geq r_k^{-1}$ if $n \in [1, m_{k+1}]$, note that all previous intervals are included. Now take $z_n \in U_{x_n}$. By splitting $z_n$ into a finite head sequence and a tail sequence at $m$ so that the tail $\in U$, suppose that $m \in [m_k, m_{k+1}]$. Then, $\vt x_n z_n \vt \leq 1$ for $n \leq m$ implies the head will be in $B(0, r_k)$, and finally $z_n \in 2U$.

Exercise. Show the following:

• The strong dual of $\ell^\infty$ and $c_0$ are both the space of absolutely summable sequences $\ell^1$ with the $\ell^1$-norm topology.
• The strong dual of $c_{0,0}$ is $\R^\N$. Hint: ⁶
• The strong dual of $\R^\N$ is $c_{0,0}$.

The dual of a normed space is normed, since the local base of the dual has a single element. On the other hand, it is a fact that the dual of a Frechet space is never Frechet, unless the space was actually normed.

Generally, the strong dual of an Frechet space is a so-called DF-space, and vice versa. LB-spaces (including $c_{0,0}$ and $\mathcal{s}'$) are particular instances of DF-spaces, though LF-spaces are generally not DF-spaces. A priori, it is a coincidence that both $\mathcal{s}$ and $\R^\N$ have LB-space duals, it means that both $\R^\N$ and $\mathcal{s}$ are quite special even among Frechet spaces.

We are lucky to have a few examples where two spaces are strong duals of each other, though not always, as $\ell^\infty$ and $c_0$ both have $\ell^1$ as the dual.

When we instead take the spaces $C_b(\Omega), C_0(\Omega), C_c(\Omega)$, then the duals of these spaces are generally signed measures. For convenience assume $\Omega \sbe \R^n$ is a domain, though the result can be generalized to Polish spaces and suitable topological spaces.

• $C_b(\Omega)'$ is the space of finitely additive (rather than countably additive) finite Borel measures with the total variation norm.
• $C_0(\Omega)'$ is the space of finite Borel measures with the total variation norm.
• $C_c(\Omega)'$ is best described as the space of measures which act like finite Borel measures on compact sets. ⁷

Finally, the ultimate example is $C_c^\infty(\Omega)'$, the space of distributions. A distribution $T$ is a continuous functional that is continuous on each of $C_c^\infty(K_i)$. On each $C_c^\infty(K_i)$, there is some order $m$ that witnesses continuity of $T$. We say that $T$ has order $m$ if it has one universal order over all compacts $K$. There are distributions that do not have finite order.