Английская Википедия:Barrelled space

Шаблон:Short description In functional analysis and related areas of mathematics, a barrelled space (also written barreled space) is a topological vector space (TVS) for which every barrelled set in the space is a neighbourhood for the zero vector. A barrelled set or a barrel in a topological vector space is a set that is convex, balanced, absorbing, and closed. Barrelled spaces are studied because a form of the Banach–Steinhaus theorem still holds for them. Barrelled spaces were introduced by Шаблон:Harvs.

Barrels

A convex and balanced subset of a real or complex vector space is called a Шаблон:Em and it is said to be Шаблон:Em, Шаблон:Em, or Шаблон:Em.

A Шаблон:Em or a Шаблон:Em in a topological vector space (TVS) is a subset that is a closed absorbing disk; that is, a barrel is a convex, balanced, closed, and absorbing subset.

Every barrel must contain the origin. If <math>\dim X \geq 2</math> and if <math>S</math> is any subset of <math>X,</math> then <math>S</math> is a convex, balanced, and absorbing set of <math>X</math> if and only if this is all true of <math>S \cap Y</math> in <math>Y</math> for every <math>2</math>-dimensional vector subspace <math>Y;</math> thus if <math>\dim X > 2</math> then the requirement that a barrel be a closed subset of <math>X</math> is the only defining property that does not depend Шаблон:Em on <math>2</math> (or lower)-dimensional vector subspaces of <math>X.</math>

If <math>X</math> is any TVS then every closed convex and balanced neighborhood of the origin is necessarily a barrel in <math>X</math> (because every neighborhood of the origin is necessarily an absorbing subset). In fact, every locally convex topological vector space has a neighborhood basis at its origin consisting entirely of barrels. However, in general, there Шаблон:Em exist barrels that are not neighborhoods of the origin; "barrelled spaces" are exactly those TVSs in which every barrel is necessarily a neighborhood of the origin. Every finite dimensional topological vector space is a barrelled space so examples of barrels that are not neighborhoods of the origin can only be found in infinite dimensional spaces.

Examples of barrels and non-barrels

The closure of any convex, balanced, and absorbing subset is a barrel. This is because the closure of any convex (respectively, any balanced, any absorbing) subset has this same property.

A family of examples: Suppose that <math>X</math> is equal to <math>\Complex</math> (if considered as a complex vector space) or equal to <math>\R^2</math> (if considered as a real vector space). Regardless of whether <math>X</math> is a real or complex vector space, every barrel in <math>X</math> is necessarily a neighborhood of the origin (so <math>X</math> is an example of a barrelled space). Let <math>R : [0, 2\pi) \to (0, \infty]</math> be any function and for every angle <math>\theta \in [0, 2 \pi),</math> let <math>S_{\theta}</math> denote the closed line segment from the origin to the point <math>R(\theta) e^{i \theta} \in \Complex.</math> Let <math display="inline">S := \bigcup_{\theta \in [0, 2 \pi)} S_{\theta}.</math> Then <math>S</math> is always an absorbing subset of <math>\R^2</math> (a real vector space) but it is an absorbing subset of <math>\Complex</math> (a complex vector space) if and only if it is a neighborhood of the origin. Moreover, <math>S</math> is a balanced subset of <math>\R^2</math> if and only if <math>R(\theta) = R(\pi + \theta)</math> for every <math>0 \leq \theta < \pi</math> (if this is the case then <math>R</math> and <math>S</math> are completely determined by <math>R</math>'s values on <math>[0, \pi)</math>) but <math>S</math> is a balanced subset of <math>\Complex</math> if and only it is an open or closed ball centered at the origin (of radius <math>0 < r \leq \infty</math>). In particular, barrels in <math>\Complex</math> are exactly those closed balls centered at the origin with radius in <math>(0, \infty].</math> If <math>R(\theta) := 2 \pi - \theta</math> then <math>S</math> is a closed subset that is absorbing in <math>\R^2</math> but not absorbing in <math>\Complex,</math> and that is neither convex, balanced, nor a neighborhood of the origin in <math>X.</math> By an appropriate choice of the function <math>R,</math> it is also possible to have <math>S</math> be a balanced and absorbing subset of <math>\R^2</math> that is neither closed nor convex. To have <math>S</math> be a balanced, absorbing, and closed subset of <math>\R^2</math> that is Шаблон:Em convex nor a neighborhood of the origin, define <math>R</math> on <math>[0, \pi)</math> as follows: for <math>0 \leq \theta < \pi,</math> let <math>R(\theta) := \pi - \theta</math> (alternatively, it can be any positive function on <math>[0, \pi)</math> that is continuously differentiable, which guarantees that <math display="inline">\lim_{\theta \searrow 0} R(\theta) = R(0) > 0</math> and that <math>S</math> is closed, and that also satisfies <math display="inline">\lim_{\theta \nearrow \pi} R(\theta) = 0,</math> which prevents <math>S</math> from being a neighborhood of the origin) and then extend <math>R</math> to <math>[\pi, 2 \pi)</math> by defining <math>R(\theta) := R(\theta - \pi),</math> which guarantees that <math>S</math> is balanced in <math>\R^2.</math>

Properties of barrels

In any topological vector space (TVS) <math>X,</math> every barrel in <math>X</math> absorbs every compact convex subset of <math>X.</math>Шаблон:Sfn
In any locally convex Hausdorff TVS <math>X,</math> every barrel in <math>X</math> absorbs every convex bounded complete subset of <math>X.</math>Шаблон:Sfn
If <math>X</math> is locally convex then a subset <math>H</math> of <math>X^{\prime}</math> is <math>\sigma\left(X^{\prime}, X\right)</math>-bounded if and only if there exists a barrel <math>B</math> in <math>X</math> such that <math>H \subseteq B^{\circ}.</math>Шаблон:Sfn
Let <math>(X, Y, b)</math> be a pairing and let <math>\nu</math> be a locally convex topology on <math>X</math> consistent with duality. Then a subset <math>B</math> of <math>X</math> is a barrel in <math>(X, \nu)</math> if and only if <math>B</math> is the polar of some <math>\sigma(Y, X, b)</math>-bounded subset of <math>Y.</math>Шаблон:Sfn
Suppose <math>M</math> is a vector subspace of finite codimension in a locally convex space <math>X</math> and <math>B \subseteq M.</math> If <math>B</math> is a barrel (resp. bornivorous barrel, bornivorous disk) in <math>M</math> then there exists a barrel (resp. bornivorous barrel, bornivorous disk) <math>C</math> in <math>X</math> such that <math>B = C \cap M.</math>Шаблон:Sfn

Characterizations of barreled spaces

Denote by <math>L(X; Y)</math> the space of continuous linear maps from <math>X</math> into <math>Y.</math>

If <math>(X, \tau)</math> is a Hausdorff topological vector space (TVS) with continuous dual space <math>X^{\prime}</math> then the following are equivalent:

<math>X</math> is barrelled.
Шаблон:Em: Every barrel in <math>X</math> is a neighborhood of the origin.
- This definition is similar to a characterization of Baire TVSs proved by Saxon [1974], who proved that a TVS <math>Y</math> with a topology that is not the indiscrete topology is a Baire space if and only if every absorbing balanced subset is a neighborhood of Шаблон:Em point of <math>Y</math> (not necessarily the origin).Шаблон:Sfn
For any Hausdorff TVS <math>Y</math> every pointwise bounded subset of <math>L(X; Y)</math> is equicontinuous.Шаблон:Sfn
For any F-space <math>Y</math> every pointwise bounded subset of <math>L(X; Y)</math> is equicontinuous.Шаблон:Sfn
- An F-space is a complete metrizable TVS.
Every closed linear operator from <math>X</math> into a complete metrizable TVS is continuous.Шаблон:Sfn
- A linear map <math>F : X \to Y</math> is called closed if its graph is a closed subset of <math>X \times Y.</math>
Every Hausdorff TVS topology <math>\nu</math> on <math>X</math> that has a neighborhood basis of the origin consisting of <math>\tau</math>-closed set is course than <math>\tau.</math>Шаблон:Sfn

If <math>(X, \tau)</math> is locally convex space then this list may be extended by appending:

There exists a TVS <math>Y</math> not carrying the indiscrete topology (so in particular, <math>Y \neq \{0\}</math>) such that every pointwise bounded subset of <math>L(X; Y)</math> is equicontinuous.Шаблон:Sfn
For any locally convex TVS <math>Y,</math> every pointwise bounded subset of <math>L(X; Y)</math> is equicontinuous.Шаблон:Sfn
- It follows from the above two characterizations that in the class of locally convex TVS, barrelled spaces are exactly those for which the uniform boundedness principal holds.
Every <math>\sigma\left(X^{\prime}, X\right)</math>-bounded subset of the continuous dual space <math>X</math> is equicontinuous (this provides a partial converse to the Banach-Steinhaus theorem).Шаблон:Sfn^[1]
<math>X</math> carries the strong dual topology <math>\beta\left(X, X^{\prime}\right).</math>Шаблон:Sfn
Every lower semicontinuous seminorm on <math>X</math> is continuous.Шаблон:Sfn
Every linear map <math>F : X \to Y</math> into a locally convex space <math>Y</math> is almost continuous.Шаблон:Sfn
- A linear map <math>F : X \to Y</math> is called Шаблон:Em if for every neighborhood <math>V</math> of the origin in <math>Y,</math> the closure of <math>F^{-1}(V)</math> is a neighborhood of the origin in <math>X.</math>
Every surjective linear map <math>F : Y \to X</math> from a locally convex space <math>Y</math> is almost open.Шаблон:Sfn
- This means that for every neighborhood <math>V</math> of 0 in <math>Y,</math> the closure of <math>F(V)</math> is a neighborhood of 0 in <math>X.</math>
If <math>\omega</math> is a locally convex topology on <math>X</math> such that <math>(X, \omega)</math> has a neighborhood basis at the origin consisting of <math>\tau</math>-closed sets, then <math>\omega</math> is weaker than <math>\tau.</math>Шаблон:Sfn

If <math>X</math> is a Hausdorff locally convex space then this list may be extended by appending:

Closed graph theorem: Every closed linear operator <math>F : X \to Y</math> into a Banach space <math>Y</math> is continuous.Шаблон:Sfn
- The linear operator is called Шаблон:Em if its graph is a closed subset of <math>X \times Y.</math>
For every subset <math>A</math> of the continuous dual space of <math>X,</math> the following properties are equivalent: <math>A</math> is^[1]
1. equicontinuous;
2. relatively weakly compact;
3. strongly bounded;
4. weakly bounded.
The 0-neighborhood bases in <math>X</math> and the fundamental families of bounded sets in <math>X_{\beta}^{\prime}</math> correspond to each other by polarity.^[1]

If <math>X</math> is metrizable topological vector space then this list may be extended by appending:

For any complete metrizable TVS <math>Y</math> every pointwise bounded Шаблон:Em in <math>L(X; Y)</math> is equicontinuous.Шаблон:Sfn

If <math>X</math> is a locally convex metrizable topological vector space then this list may be extended by appending:

(Шаблон:Visible anchor): The weak* topology on <math>X^{\prime}</math> is sequentially complete.Шаблон:Sfn
(Шаблон:Visible anchor): Every weak* bounded subset of <math>X^{\prime}</math> is <math>\sigma\left(X^{\prime}, X\right)</math>-relatively countably compact.Шаблон:Sfn
(Шаблон:Visible anchor): Every countable weak* bounded subset of <math>X^{\prime}</math> is equicontinuous.Шаблон:Sfn
(Шаблон:Visible anchor): <math>X</math> is not the union of an increase sequence of nowhere dense disks.Шаблон:Sfn

Examples and sufficient conditions

Each of the following topological vector spaces is barreled:

TVSs that are Baire space.
- Consequently, every topological vector space that is of the second category in itself is barrelled.
F-spaces, Fréchet spaces, Banach spaces, and Hilbert spaces.
- However, there exist normed vector spaces that are Шаблон:Em barrelled. For example, if the <math>L^p</math>-space <math>L^2([0, 1])</math> is topologized as a subspace of <math>L^1([0, 1]),</math> then it is not barrelled.
Complete pseudometrizable TVSs.Шаблон:Sfn
- Consequently, every finite-dimensional TVS is barrelled.
Montel spaces.
Strong dual spaces of Montel spaces (since they are necessarily Montel spaces).
A locally convex quasi-barrelled space that is also a σ-barrelled space.Шаблон:Sfn
A sequentially complete quasibarrelled space.
A quasi-complete Hausdorff locally convex infrabarrelled space.Шаблон:Sfn
- A TVS is called quasi-complete if every closed and bounded subset is complete.
A TVS with a dense barrelled vector subspace.Шаблон:Sfn
- Thus the completion of a barreled space is barrelled.
A Hausdorff locally convex TVS with a dense infrabarrelled vector subspace.Шаблон:Sfn
- Thus the completion of an infrabarrelled Hausdorff locally convex space is barrelled.Шаблон:Sfn
A vector subspace of a barrelled space that has countable codimensional.Шаблон:Sfn
- In particular, a finite codimensional vector subspace of a barrelled space is barreled.
A locally convex ultrabarelled TVS.Шаблон:Sfn
A Hausdorff locally convex TVS <math>X</math> such that every weakly bounded subset of its continuous dual space is equicontinuous.Шаблон:Sfn
A locally convex TVS <math>X</math> such that for every Banach space <math>B,</math> a closed linear map of <math>X</math> into <math>B</math> is necessarily continuous.Шаблон:Sfn
A product of a family of barreled spaces.Шаблон:Sfn
A locally convex direct sum and the inductive limit of a family of barrelled spaces.Шаблон:Sfn
A quotient of a barrelled space.Шаблон:Sfn Шаблон:Sfn
A Hausdorff sequentially complete quasibarrelled boundedly summing TVS.Шаблон:Sfn
A locally convex Hausdorff reflexive space is barrelled.

Counter examples

A barrelled space need not be Montel, complete, metrizable, unordered Baire-like, nor the inductive limit of Banach spaces.
Not all normed spaces are barrelled. However, they are all infrabarrelled.Шаблон:Sfn
A closed subspace of a barreled space is not necessarily countably quasi-barreled (and thus not necessarily barrelled).Шаблон:Sfn
There exists a dense vector subspace of the Fréchet barrelled space <math>\R^{\N}</math> that is not barrelled.Шаблон:Sfn
There exist complete locally convex TVSs that are not barrelled.Шаблон:Sfn
The finest locally convex topology on an infinite-dimensional vector space is a Hausdorff barrelled space that is a meagre subset of itself (and thus not a Baire space).Шаблон:Sfn

Properties of barreled spaces

Banach–Steinhaus generalization

The importance of barrelled spaces is due mainly to the following results.

Шаблон:Math theorem

The Banach-Steinhaus theorem is a corollary of the above result.Шаблон:Sfn When the vector space <math>Y</math> consists of the complex numbers then the following generalization also holds.

Шаблон:Math theorem

Recall that a linear map <math>F : X \to Y</math> is called closed if its graph is a closed subset of <math>X \times Y.</math>

Шаблон:Math theorem

Other properties

Every Hausdorff barrelled space is quasi-barrelled.Шаблон:Sfn
A linear map from a barrelled space into a locally convex space is almost continuous.
A linear map from a locally convex space Шаблон:Emto a barrelled space is almost open.
A separately continuous bilinear map from a product of barrelled spaces into a locally convex space is hypocontinuous.Шаблон:Sfn
A linear map with a closed graph from a barreled TVS into a <math>B_r</math>-complete TVS is necessarily continuous.Шаблон:Sfn

References

Шаблон:Reflist Шаблон:Reflist

Bibliography

Шаблон:Refbegin

Шаблон:Refend

Шаблон:Functional analysis Шаблон:Boundedness and bornology Шаблон:Topological vector spaces

↑ ^1,0 ^1,1 ^1,2 Шаблон:Harvnb Шаблон:Harvnb.

[Schaefer_(1999)_p._127,_141,_Trèves_(2006)_p._350-1] 1,0 ^1,1 ^1,2 Шаблон:Harvnb Шаблон:Harvnb.

[1]

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Английская Википедия:Barrelled space

Содержание

Barrels

Examples of barrels and non-barrels

Properties of barrels

Characterizations of barreled spaces

Examples and sufficient conditions

Counter examples

Properties of barreled spaces

Banach–Steinhaus generalization

Other properties

See also

References

Bibliography

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