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
editSuppose is a topological vector space (TVS) over a field
A subset of is called von Neumann bounded or just bounded in if any of the following equivalent conditions are satisfied:
- Definition: For every neighborhood of the origin there exists a real such that [note 1] for all scalars satisfying [1]
- This was the definition introduced by John von Neumann in 1935.[1]
- is absorbed by every neighborhood of the origin.[2]
- For every neighborhood of the origin there exists a scalar such that
- For every neighborhood of the origin there exists a real such that for all scalars satisfying [1]
- For every neighborhood of the origin there exists a real such that for all real [3]
- 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: is bounded if and only if is absorbed by every balanced neighborhood of the origin.[1]
- If is locally convex then the adjective "convex" may be also be added to any of these 5 replacements.
- For every sequence of scalars that converges to and every sequence in the sequence converges to in [1]
- 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.[1]
- For every sequence in the sequence converges to in [4]
- Every countable subset of is bounded (according to any defining condition other than this one).[1]
If is a neighborhood basis for at the origin then this list may be extended to include:
- Any one of statements (1) through (5) above but with the neighborhoods limited to those belonging to
- e.g. Statement (3) may become: For every there exists a scalar such that
If is a locally convex space whose topology is defined by a family of continuous seminorms, then this list may be extended to include:
- is bounded for all [1]
- There exists a sequence of non-zero scalars such that for every sequence in the sequence is bounded in (according to any defining condition other than this one).[1]
- For all is bounded (according to any defining condition other than this one) in the semi normed space
- B is weakly bounded, i.e. every continuous linear functional is bounded on B[5]
If is a normed space with norm (or more generally, if it is a seminormed space and is merely a seminorm),[note 2] then this list may be extended to include:
- is a norm bounded subset of By definition, this means that there exists a real number such that for all [1]
-
- Thus, if is a linear map between two normed (or seminormed) spaces and if is the closed (alternatively, open) unit ball in centered at the origin, then is a bounded linear operator (which recall means that its operator norm is finite) if and only if the image of this ball under is a norm bounded subset of
- is a subset of some (open or closed) ball.[note 3]
- This ball need not be centered at the origin, but its radius must (as usual) be positive and finite.
If is a vector subspace of the TVS then this list may be extended to include:
- is contained in the closure of [1]
- In other words, a vector subspace of is bounded if and only if it is a subset of (the vector space)
- Recall that is a Hausdorff space if and only if is closed in So the only bounded vector subspace of a Hausdorff TVS is
A subset that is not bounded is called unbounded.
Bornology and fundamental systems of bounded sets
editThe collection of all bounded sets on a topological vector space is called the von Neumann bornology or the (canonical) bornology of
A base or fundamental system of bounded sets of is a set of bounded subsets of such that every bounded subset of is a subset of some [1] The set of all bounded subsets of trivially forms a fundamental system of bounded sets of
Examples
editIn any locally convex TVS, the set of closed and bounded disks are a base of bounded set.[1]
Examples and sufficient conditions
editUnless indicated otherwise, a topological vector space (TVS) need not be Hausdorff nor locally convex.
- Finite sets are bounded.[1]
- Every totally bounded subset of a TVS is bounded.[1]
- 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 is the unique largest (with respect to set inclusion ) bounded vector subspace of In particular, if is a bounded subset of then so is
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 having a bounded subset and also a dense vector subspace such that is not contained in the closure (in ) of any bounded subset of [6]
Stability properties
edit- In any TVS, finite unions, finite Minkowski sums, scalar multiples, translations, subsets, closures, interiors, and balanced hulls of bounded sets are again bounded.[1]
- In any locally convex TVS, the convex hull (also called the convex envelope) of a bounded set is again bounded.[7] However, this may be false if the space is not locally convex, as the (non-locally convex) Lp space spaces for have no nontrivial open convex subsets.[7]
- The image of a bounded set under a continuous linear map is a bounded subset of the codomain.[1]
- A subset of an arbitrary (Cartesian) product of TVSs is bounded if and only if its image under every coordinate projections is bounded.
- If and is a topological vector subspace of then is bounded in if and only if is bounded in [1]
- In other words, a subset is bounded in if and only if it is bounded in every (or equivalently, in some) topological vector superspace of
Properties
editA locally convex topological vector space has a bounded neighborhood of zero if and only if its topology can be defined by a single seminorm.
The polar of a bounded set is an absolutely convex and absorbing set.
Mackey's countability condition[8] — If is a countable sequence of bounded subsets of a metrizable locally convex topological vector space then there exists a bounded subset of and a sequence of positive real numbers such that for all (or equivalently, such that ).
Using the definition of uniformly bounded sets given below, Mackey's countability condition can be restated as: If are bounded subsets of a metrizable locally convex space then there exists a sequence of positive real numbers such that 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
editUniformly bounded sets
editA family of sets of subsets of a topological vector space is said to be uniformly bounded in if there exists some bounded subset of such that which happens if and only if its union is a bounded subset of In the case of a normed (or seminormed) space, a family is uniformly bounded if and only if its union is norm bounded, meaning that there exists some real such that for every or equivalently, if and only if
A set of maps from to is said to be uniformly bounded on a given set if the family is uniformly bounded in which by definition means that there exists some bounded subset of such that or equivalently, if and only if is a bounded subset of A set of linear maps between two normed (or seminormed) spaces and is uniformly bounded on some (or equivalently, every) open ball (and/or non-degenerate closed ball) in if and only if their operator norms are uniformly bounded; that is, if and only if
Proposition[9] — Let be a set of continuous linear operators between two topological vector spaces and and let be any bounded subset of Then is uniformly bounded on (that is, the family is uniformly bounded in ) if any of the following conditions are satisfied:
- is equicontinuous.
- is a convex compact Hausdorff subspace of and for every the orbit is a bounded subset of
Proof of part (1)[9]
|
---|
Assume is equicontinuous and let be a neighborhood of the origin in Since is equicontinuous, there exists a neighborhood of the origin in such that for every Because is bounded in there exists some real such that if then So for every and every which implies that Thus is bounded in Q.E.D. |
Proof of part (2)[10]
|
---|
Let be a balanced neighborhood of the origin in and let be a closed balanced neighborhood of the origin in such that Define which is a closed subset of (since is closed while every is continuous) that satisfies for every Note that for every non-zero scalar the set is closed in (since scalar multiplication by is a homeomorphism) and so every is closed in It will now be shown that from which follows. If then being bounded guarantees the existence of some positive integer such that where the linearity of every now implies thus and hence as desired. Thus expresses as a countable union of closed (in ) sets. Since 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 such that has non-empty interior in Let be any point belonging to this open subset of Let be any balanced open neighborhood of the origin in such that The sets form an increasing (meaning implies ) cover of the compact space so there exists some such that (and thus ). It will be shown that for every thus demonstrating that is uniformly bounded in and completing the proof. So fix and Let The convexity of guarantees and moreover, since Thus which is a subset of Since is balanced and we have which combined with gives Finally, and imply as desired. Q.E.D. |
Since every singleton subset of is also a bounded subset, it follows that if is an equicontinuous set of continuous linear operators between two topological vector spaces and (not necessarily Hausdorff or locally convex), then the orbit of every is a bounded subset of
Bounded subsets of topological modules
editThe definition of bounded sets can be generalized to topological modules. A subset of a topological module over a topological ring is bounded if for any neighborhood of there exists a neighborhood of such that
See also
edit- Bornological space – Space where bounded operators are continuous
- Bornivorous set – A set that can absorb any bounded subset
- Bounded function – A mathematical function the set of whose values is bounded
- Bounded operator – Linear transformation between topological vector spaces
- Bounding point – Mathematical concept related to subsets of vector spaces
- Compact space – Type of mathematical space
- Kolmogorov's normability criterion – Characterization of normable spaces
- Local boundedness
- Totally bounded space – Generalization of compactness
References
edit- ^ a b c d e f g h i j k l m n o p q r Narici & Beckenstein 2011, pp. 156–175.
- ^ Schaefer 1970, p. 25.
- ^ Rudin 1991, p. 8.
- ^ Wilansky 2013, p. 47.
- ^ Narici Beckenstein (2011). Topological Vector Spaces (2nd ed.). pp. 253, Theorem 8.8.7. ISBN 978-1-58488-866-6.
- ^ Wilansky 2013, p. 57.
- ^ a b Narici & Beckenstein 2011, p. 162.
- ^ Narici & Beckenstein 2011, p. 174.
- ^ a b Rudin 1991, pp. 42−47.
- ^ Rudin 1991, pp. 46−47.
Notes
- ^ For any set and scalar the notation denotes the set
- ^ This means that the topology on is equal to the topology induced on it by 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.
- ^ If is a normed space or a seminormed space, then the open and closed balls of radius (where is a real number) centered at a point are, respectively, the sets and Any such set is called a (non-degenerate) ball.
Bibliography
edit- Adasch, Norbert; Ernst, Bruno; Keim, Dieter (1978). Topological Vector Spaces: The Theory Without Convexity Conditions. Lecture Notes in Mathematics. Vol. 639. Berlin New York: Springer-Verlag. ISBN 978-3-540-08662-8. OCLC 297140003.
- Berberian, Sterling K. (1974). Lectures in Functional Analysis and Operator Theory. Graduate Texts in Mathematics. Vol. 15. New York: Springer. ISBN 978-0-387-90081-0. OCLC 878109401.
- Bourbaki, Nicolas (1987) [1981]. Topological Vector Spaces: Chapters 1–5. Éléments de mathématique. Translated by Eggleston, H.G.; Madan, S. Berlin New York: Springer-Verlag. ISBN 3-540-13627-4. OCLC 17499190.
- Conway, John (1990). A course in functional analysis. Graduate Texts in Mathematics. Vol. 96 (2nd ed.). New York: Springer-Verlag. ISBN 978-0-387-97245-9. OCLC 21195908.
- Edwards, Robert E. (1995). Functional Analysis: Theory and Applications. New York: Dover Publications. ISBN 978-0-486-68143-6. OCLC 30593138.
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- Robertson, Alex P.; Robertson, Wendy J. (1980). Topological Vector Spaces. Cambridge Tracts in Mathematics. Vol. 53. Cambridge England: Cambridge University Press. ISBN 978-0-521-29882-7. OCLC 589250.
- Schaefer, H.H. (1970). Topological Vector Spaces. GTM. Vol. 3. Springer-Verlag. pp. 25–26. ISBN 0-387-05380-8.
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