Bounded set (topological vector space)

In functional analysis and related areas of mathematics, a set in a topological vector space is called bounded or von Neumann bounded, if every neighborhood of the zero vector can be ''inflated'' to include the set.
A set that is not bounded is called unbounded.

Bounded sets are a natural way to define locally convex polar topologies on the vector spaces in a dual pair, as the polar set of a bounded set is an absolutely convex and absorbing set.
The concept was first introduced by John von Neumann and Andrey Kolmogorov in 1935.

Definition

Suppose $X$ is a topological vector space (TVS) over a field $\mathbb.$

A subset $B$ of $X$ is called or just in $X$ if any of the following equivalent conditions are satisfied:

1. : For every neighborhood $V$ of the origin there exists a real $r > 0$ such that $B \subseteq s V$For any set $A$ and scalar $s,$ the notation $s A$ is denotes the set $s A := \.$ for all scalars $s$ satisfying $, s, \geq r.$
   * This was the definition introduced by John von Neumann in 1935.


2. $B$ is absorbed by every neighborhood of the origin.


3. For every neighborhood $V$ of the origin there exists a scalar $s$ such that $B \subseteq s V.$


4. For every neighborhood $V$ of the origin there exists a real $r > 0$ such that $s B \subseteq V$ for all scalars $s$ satisfying $, s, \leq r.$


5. For every neighborhood $V$ of the origin there exists a real $r > 0$ such that $t B \subseteq V$ for all real $0 < t \leq r.$


6. Any one of statements (1) through (5) above but with the word "neighborhood" replaced by any of the following: " balanced neighborhood," "open balanced neighborhood," "closed balanced neighborhood," "open neighborhood," "closed neighborhood".
   * e.g. Statement (2) may become: $B$ is bounded if and only if $B$ is absorbed by every balanced neighborhood of the origin.
   * If $X$ is locally convex then the adjective "convex" may be also be added to any of these 5 replacements.


7. For every sequence of scalars $s_1, s_2, s_3, \ldots$ that converges to $0$ and every sequence $b_1, b_2, b_3, \ldots$ in $B,$ the sequence $s_1 b_1, s_2 b_2, s_3 b_3, \ldots$ converges to $0$ in $X.$
   * This was the definition of "bounded" that Andrey Kolmogorov used in 1934, which is the same as the definition introduced by Stanisław Mazur and Władysław Orlicz in 1933 for metrizable TVS. Kolmogorov used this definition to prove that a TVS is seminormable if and only if it has a bounded convex neighborhood of the origin.


8. For every sequence $b_1, b_2, b_3, \ldots$ in $B,$ the sequence $\left(\tfrac b_i\right)_^$ converges to $0$ in $X.$


9. Every countable subset of $B$ is bounded (according to any defining condition other than this one).

If $\mathcal$ is a neighborhood basis for $X$ at the origin then this list may be extended to include:

10. Any one of statements (1) through (5) above but with the neighborhoods limited to those belonging to $\mathcal.$
    * e.g. Statement (3) may become: For every $V \in \mathcal$ there exists a scalar $s$ such that $B \subseteq s V.$

If $X$ is a locally convex space whose topology is defined by a family $\mathcal$ of continuous seminorms, then this list may be extended to include:

11. $p(B)$ is bounded for all $p \in \mathcal.$


12. There exists a sequence of non-zero scalars $s_1, s_2, s_3, \ldots$ such that for every sequence $b_1, b_2, b_3, \ldots$ in $B,$ the sequence $b_1 s_1, b_2 s_2, b_3 s_3, \ldots$ is bounded in $X$ (according to any defining condition other than this one).


13. For all $p \in \mathcal,$ $B$ is bounded (according to any defining condition other than this one) in the semi normed space $(X, p).$

If $X$ is a normed space with norm $\, \cdot\,$ (or more generally, if it is a seminormed space and $\, \cdot\,$ is merely a seminorm),This means that the topology on $X$ is equal to the topology induced on it by $\, \cdot\, .$ Note that every normed space is a seminormed space and every norm is a seminorm. The definition of the topology induced by a seminorm is identical to the definition of the topology induced by a norm. then this list may be extended to include:

14. $B$ is a ''norm bounded'' subset of $(X, \, \cdot\, ).$ By definition, this means that there exists a real number $r > 0$ such that $\, b\, \leq r$ for all $b \in B.$


15. $\sup_ \, b\, < \infty.$
    * Thus, if $L : (X, \, \cdot\, ) \to (Y, \, \cdot\, )$ is a linear map between two normed (or seminormed) spaces and if $B$ is the closed (alternatively, open) unit ball in $(X, \, \cdot\, )$ centered at the origin, then $L$ is a bounded linear operator (which recall means that its operator norm $\, L\, := \sup_ \, L(b)\, < \infty$ is finite) if and only if the image $L(B)$ of this ball under $L$ is a norm bounded subset of $(Y, \, \cdot\, ).$


16. $B$ is a subset of some (open or closed) ball.If $(X, \, \cdot\, )$ is a normed space or a seminormed space, then the open and closed balls of radius $r > 0$ (where $r \neq \infty$ is a real number) centered at a point $x \in X$ are, respectively, the sets $B_(x) := \$ and $B_(x) := \.$ Any such set is call a (non-degenerate) .
    * This ball need not be centered at the origin, but its radius must (as usual) be positive and finite.

If $B$ is a vector subspace of the TVS $X$ then this list may be extended to include:

17. $B$ is contained in the closure of $\.$
    * In other words, a vector subspace of $X$ is bounded if and only if it is a subset of (the vector space) $\operatorname_X \.$
    * Recall that $X$ is a Hausdorff space if and only if $\$ is closed in $X.$ So the only bounded vector subspace of a Hausdorff TVS is $\.$

A subset that is not bounded is called .

Bornology and fundamental systems of bounded sets

The collection of all bounded sets on a topological vector space $X$ is called the or the ()

A or of $X$ is a set $\mathcal$ of bounded subsets of $X$ such that every bounded subset of $X$ is a subset of some $B \in \mathcal.$
The set of all bounded subsets of $X$ trivially forms a fundamental system of bounded sets of $X.$

Examples

In any locally convex TVS, the set of closed and bounded disks are a base of bounded set.

Examples and sufficient conditions

Unless indicated otherwise, a topological vector space (TVS) need not be Hausdorff nor locally convex.

• Finite sets are bounded.


• Every totally bounded subset of a TVS is bounded.


• Every relatively compact set in a topological vector space is bounded. If the space is equipped with the weak topology the converse is also true.


• The set of points of a Cauchy sequence is bounded, the set of points of a Cauchy net need not be bounded.


• The closure of the origin (referring to the closure of the set $\$) is always a bounded closed vector subspace. This set $\operatorname_X \$ is the unique largest (with respect to set inclusion $\,\subseteq\,$) bounded vector subspace of $X.$ In particular, if $B \subseteq X$ is a bounded subset of $X$ then so is $B + \operatorname_X \.$

Unbounded sets

A set that is not bounded is said to be ''unbounded''.

Any vector subspace of a TVS that is not a contained in the closure of $\$ is unbounded

There exists a Fréchet space $X$ having a bounded subset $B$ and also a dense vector subspace $M$ such that $B$ is contained in the closure (in $X$) of any bounded subset of $M.$

Stability properties

• In any TVS, finite unions, finite Minkowski sums, scalar multiples, translations, subsets, closures, interiors, and balanced hulls of bounded sets are again bounded.


• In any locally convex TVS, the convex hull (also called the convex envelope) of a bounded set is again bounded. However, this may be false if the space is not locally convex, as the (non-locally convex) Lp space $L^p$ spaces for $0 < p < 1$ have no nontrivial open convex subsets.


• The image of a bounded set under a continuous linear map is a bounded subset of the codomain.


• A subset of an arbitrary (Cartesian) product of TVSs is bounded if and only if its image under every coordinate projections is bounded.


• If $S \subseteq X \subseteq Y$ and $X$ is a topological vector subspace of $Y,$ then $S$ is bounded in $X$ if and only if $S$ is bounded in $Y.$
  * In other words, a subset $S \subseteq X$ is bounded in $X$ if and only if it is bounded in every (or equivalently, in some) topological vector superspace of $X.$

Properties

A locally convex topological vector space has a bounded neighborhood of zero if and only if its topology can be defined by a seminorm.

The polar of a bounded set is an absolutely convex and absorbing set.

Using the definition of uniformly bounded sets given below, Mackey's countability condition can be restated as: If $B_1, B_2, B_3, \ldots$ are bounded subsets of a metrizable locally convex space then there exists a sequence $t_1, t_2, t_3, \ldots$ of positive real numbers such that $t_1 B_1, \, t_2 B_2, \, t_3 B_3, \ldots$ are uniformly bounded.
In words, given any countable family of bounded sets in a metrizable locally convex space, it is possible to scale each set by its own positive real so that they become uniformly bounded.

Generalizations

Uniformly bounded sets

A family of sets $\mathcal$ of subsets of a topological vector space $Y$ is said to be in $Y,$ if there exists some bounded subset $D$ of $Y$ such that
$B \subseteq D \quad \text B \in \mathcal,$
which happens if and only if its union
$\cup \mathcal ~:=~ \bigcup_ B$
is a bounded subset of $Y.$
In the case of a normed (or seminormed) space, a family $\mathcal$ is uniformly bounded if and only if its union $\cup \mathcal$ is ''norm bounded'', meaning that there exists some real $M \geq 0$ such that $\, b\, \leq M$ for every $b \in \cup \mathcal,$ or equivalently, if and only if $\sup_ \, b\, < \infty.$

A set $H$ of maps from $X$ to $Y$ is said to be $C \subseteq X$ if the family $H(C) := \$ is uniformly bounded in $Y,$ which by definition means that there exists some bounded subset $D$ of $Y$ such that $h(C) \subseteq D \text h \in H,$ or equivalently, if and only if $\cup H(C) := \bigcup_ h(C)$ is a bounded subset of $Y.$
A set $H$ of linear maps between two normed (or seminormed) spaces $X$ and $Y$ is uniformly bounded on some (or equivalently, every) open ball (and/or non-degenerate closed ball) in $X$ if and only if their operator norms are uniformly bounded; that is, if and only if $\sup_ \, h\, < \infty.$

Assume $H$ is equicontinuous and let $W$ be a neighborhood of the origin in $Y.$
Since $H$ is equicontinuous, there exists a neighborhood $U$ of the origin in $X$ such that $h(U) \subseteq W$ for every $h \in H.$
Because $C$ is bounded in $X,$ there exists some real $r > 0$ such that if $t \geq r$ then $C \subseteq t U.$
So for every $h \in H$ and every $t \geq r,$ $h(C) \subseteq h(t U) = t h(U) \subseteq t W,$ which implies that $\bigcup_ h(C) \subseteq t W.$ Thus $\bigcup_ h(C)$ is bounded in $Y.$ Q.E.D.

Let $W$ be a balanced neighborhood of the origin in $Y$ and let $V$ be a closed balanced neighborhood of the origin in $Y$ such that $V + V \subseteq W.$
Define
$E ~:=~ \bigcap_ h^(V),$
which is a closed subset of $X$ (since $V$ is closed while every $h : X \to Y$ is continuous) that satisfies $h(E) \subseteq V$ for every $h \in H.$
Note that for every non-zero scalar $n \neq 0,$ the set $n E$ is closed in $X$ (since scalar multiplication by $n \neq 0$ is a homeomorphism) and so every $C \cap n E$ is closed in $C.$

It will now be shown that $C \subseteq \bigcup_ n E,$ from which $C = \bigcup_ (C \cap n E)$ follows.
If $c \in C$ then $H(c)$ being bounded guarantees the existence of some positive integer $n = n_c \in \N$ such that $H(c) \subseteq n_c V,$ where the linearity of every $h \in H$ now implies $\tfrac c \in h^(V);$ thus $\tfrac c \in \bigcap_ h^(V) = E$ and hence $C \subseteq \bigcup_ n E,$ as desired.

Thus
$C = (C \cap 1 E) \cup (C \cap 2 E) \cup (C \cap 3 E) \cup \cdots$
expresses $C$ as a countable union of closed (in $C$) sets.
Since $C$ is a nonmeager subset of itself (as it is a Baire space by the Baire category theorem), this is only possible if there is some integer $n \in \N$ such that $C \cap n E$ has non-empty interior in $C.$
Let $k \in \operatorname_C (C \cap n E)$ be any point belonging to this open subset of $C.$
Let $U$ be any balanced open neighborhood of the origin in $X$ such that
$C \cap (k + U) ~\subseteq~ \operatorname_C (C \cap n E).$

The sets $\$ form an increasing (meaning $p \leq q$ implies $k + p U \subseteq k + q U$) cover of the compact space $C,$ so there exists some $p > 1$ such that $C \subseteq k + p U$ (and thus $\tfrac(C - k) \subseteq U$).
It will be shown that $h(C) \subseteq p n W$ for every $h \in H,$ thus demonstrating that $\$ is uniformly bounded in $Y$ and completing the proof.
So fix $h \in H$ and $c \in C.$
Let
$z ~:=~ \tfrac k + \tfrac c.$

The convexity of $C$ guarantees $z \in C$ and moreover, $z \in k + U$ since
$z - k = \tfrac k + \tfrac c = \tfrac (c - k) \in \tfrac(C - k) \subseteq U.$
Thus $z \in C \cap (k + U),$ which is a subset of $\operatorname_C (C \cap n E).$
Since $n V$ is balanced and $, 1 - p, = p - 1 < p,$ we have $(1 - p) n V \subseteq p n V,$ which combined with $h(E) \subseteq V$ gives
$p n h(E) + (1 - p) n h(E) ~\subseteq~ p n V + (1 - p) n V ~\subseteq~ p n V + p n V ~\subseteq~ p n (V + V) ~\subseteq~ p n W.$
Finally, $c = p z + (1 - p) k$ and $k, z \in n E$ imply
$h(c) ~=~ p h(z) + (1 - p) h(k) ~\in~ p n h(E) + (1 - p) n h(E) ~\subseteq~ p n W,$
as desired. Q.E.D.

Since every singleton subset of $X$ is also a bounded subset, it follows that if $H \subseteq L(X, Y)$ is an equicontinuous set of continuous linear operators between two topological vector spaces $X$ and $Y$ (not necessarily Hausdorff or locally convex), then the orbit $H(x) := \$ of every $x \in X$ is a bounded subset of $Y.$

Bounded subsets of topological modules

The definition of bounded sets can be generalized to topological modules.
A subset $A$ of a topological module $M$ over a topological ring $R$ is bounded if for any neighborhood $N$ of $0_M$ there exists a neighborhood $w$ of $0_R$ such that $w A \subseteq B.$

See also

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References

Notes

Bibliography

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{{Topological vector spaces

Topological vector spaces