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 txt.

Barrels

A convex and balanced subset of a real or complex vector space is called a disk and it is said to be disked, absolutely convex, or convex balanced.

A ' or a Barrelled set' 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 \dim X \geq 2 and if S is any subset of X, then S is a convex, balanced, and absorbing set of X if and only if this is all true of S \cap Y in Y for every 2-dimensional vector subspace Y; thus if \dim X > 2 then the requirement that a barrel be a closed subset of X is the only defining property that does not depend solely on 2 (or lower)-dimensional vector subspaces of X.

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

Properties of barrels

  • In any topological vector space (TVS) X, every barrel in X absorbs every compact convex subset of X.[1]
  • In any locally convex Hausdorff TVS X, every barrel in X absorbs every convex bounded complete subset of X.[1]
  • If X is locally convex then a subset H of X^{\prime} is \sigma\left(X^{\prime}, X\right)-bounded if and only if there exists a barrel B in X such that H \subseteq B^{\circ}.[1]
  • Let (X, Y, b) be a pairing and let \nu be a locally convex topology on X consistent with duality. Then a subset B of X is a barrel in (X, \nu) if and only if B is the polar of some \sigma(Y, X, b)-bounded subset of Y.[1]
  • Suppose M is a vector subspace of finite codimension in a locally convex space X and B \subseteq M. If B is a barrel (resp. bornivorous barrel, bornivorous disk) in M then there exists a barrel (resp. bornivorous barrel, bornivorous disk) C in X such that B = C \cap M.[2]

Characterizations of barreled spaces

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

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

  1. X is barrelled.
  2. Definition: Every barrel in X 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 Y 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 some point of Y (not necessarily the origin).[2]
  3. For any Hausdorff TVS Y every pointwise bounded subset of L(X; Y) is equicontinuous.[3]
  4. For any F-space Y every pointwise bounded subset of L(X; Y) is equicontinuous.[3]
  5. Every closed linear operator from X into a complete metrizable TVS is continuous.[4]
    • A linear map F : X \to Y is called closed if its graph is a closed subset of X \times Y.
  6. Every Hausdorff TVS topology \nu on X that has a neighborhood basis of the origin consisting of \tau-closed set is coarser than \tau.[5]

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

  1. There exists a TVS Y not carrying the indiscrete topology (so in particular, Y \neq \{0\}) such that every pointwise bounded subset of L(X; Y) is equicontinuous.[2]
  2. For any locally convex TVS Y, every pointwise bounded subset of L(X; Y) is equicontinuous.[2]
    • 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 principle holds.
  3. Every \sigma\left(X^{\prime}, X\right)-bounded subset of the continuous dual space X is equicontinuous (this provides a partial converse to the Banach-Steinhaus theorem).[2][6]
  4. X carries the strong dual topology \beta\left(X, X^{\prime}\right).[2]
  5. Every lower semicontinuous seminorm on X is continuous.[2]
  6. Every linear map F : X \to Y into a locally convex space Y is almost continuous.[2]
    • A linear map F : X \to Y is called ' if for every neighborhood V of the origin in Y, the closure of F^{-1}(V) is a neighborhood of the origin in X.
  7. Every surjective linear map F : Y \to X from a locally convex space Y is almost open.[2]
    • This means that for every neighborhood V of 0 in Y, the closure of F(V) is a neighborhood of 0 in X.
  8. If \omega is a locally convex topology on X such that (X, \omega) has a neighborhood basis at the origin consisting of \tau-closed sets, then \omega is weaker than \tau.[2]

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

  1. Closed graph theorem: Every closed linear operator F : X \to Y into a Banach space Y is continuous.[7]
  2. For every subset A of the continuous dual space of X, the following properties are equivalent: A is[6]
    1. equicontinuous;
    2. relatively weakly compact;
    3. strongly bounded;
    4. weakly bounded.
  3. The 0-neighborhood bases in X and the fundamental families of bounded sets in X_{\beta}^{\prime} correspond to each other by polarity.[6]

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

  1. For any complete metrizable TVS Y every pointwise bounded sequence in L(X; Y) is equicontinuous.[3]

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

  1. (): The weak* topology on X^{\prime} is sequentially complete.[8]
  2. (): Every weak* bounded subset of X^{\prime} is \sigma\left(X^{\prime}, X\right)-relatively countably compact.[8]
  3. (): Every countable weak* bounded subset of X^{\prime} is equicontinuous.[8]
  4. (): X is not the union of an increase sequence of nowhere dense disks.[8]

Examples and sufficient conditions

Each of the following topological vector spaces is barreled:

  1. TVSs that are Baire space.
    • Consequently, every topological vector space that is of the second category in itself is barrelled.
  2. F-spaces, Fréchet spaces, Banach spaces, and Hilbert spaces.
    • However, there exist normed vector spaces that are not barrelled. For example, if the L^p-space L^2([0, 1]) is topologized as a subspace of L^1([0, 1]), then it is not barrelled.
  3. Complete pseudometrizable TVSs.[9]
    • Consequently, every finite-dimensional TVS is barrelled.
  4. Montel spaces.
  5. Strong dual spaces of Montel spaces (since they are necessarily Montel spaces).
  6. A locally convex quasi-barrelled space that is also a σ-barrelled space.[10]
  7. A sequentially complete quasibarrelled space.
  8. A quasi-complete Hausdorff locally convex infrabarrelled space.[2]
    • A TVS is called quasi-complete if every closed and bounded subset is complete.
  9. A TVS with a dense barrelled vector subspace.[2]
    • Thus the completion of a barreled space is barrelled.
  10. A Hausdorff locally convex TVS with a dense infrabarrelled vector subspace.[2]
    • Thus the completion of an infrabarrelled Hausdorff locally convex space is barrelled.[2]
  11. A vector subspace of a barrelled space that has countable codimensional.[2]
    • In particular, a finite codimensional vector subspace of a barrelled space is barreled.
  12. A locally convex ultrabarelled TVS.[11]
  13. A Hausdorff locally convex TVS X such that every weakly bounded subset of its continuous dual space is equicontinuous.[12]
  14. A locally convex TVS X such that for every Banach space B, a closed linear map of X into B is necessarily continuous.[13]
  15. A product of a family of barreled spaces.[14]
  16. A locally convex direct sum and the inductive limit of a family of barrelled spaces.[15]
  17. A quotient of a barrelled space.[16][15]
  18. A Hausdorff sequentially complete quasibarrelled boundedly summing TVS.[17]
  19. A locally convex Hausdorff reflexive space is barrelled.

Counterexamples

  • 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.[2]
  • A closed subspace of a barreled space is not necessarily countably quasi-barreled (and thus not necessarily barrelled).[18]
  • There exists a dense vector subspace of the Fréchet barrelled space \R^{\N} that is not barrelled.[2]
  • There exist complete locally convex TVSs that are not barrelled.[2]
  • 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).[2]

Properties of barreled spaces

Banach–Steinhaus generalization

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

The Banach-Steinhaus theorem is a corollary of the above result.[19] When the vector space Y consists of the complex numbers then the following generalization also holds.

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

Other properties

  • Every Hausdorff barrelled space is quasi-barrelled.[20]
  • A linear map from a barrelled space into a locally convex space is almost continuous.
  • A linear map from a locally convex space onto a barrelled space is almost open.
  • A separately continuous bilinear map from a product of barrelled spaces into a locally convex space is hypocontinuous.[21]
  • A linear map with a closed graph from a barreled TVS into a B_r-complete TVS is necessarily continuous.[13]

References

  1. ^ Narici & Beckenstein 2011, pp. 225–273.
  2. ^ Narici & Beckenstein 2011, pp. 371–423.
  3. ^ Adasch, Ernst & Keim 1978, p. 39.
  4. ^ Adasch, Ernst & Keim 1978, p. 43.
  5. ^ Adasch, Ernst & Keim 1978, p. 32.
  6. ^ Schaefer & Wolff 1999, pp. 127, 141Trèves 2006, p. 350.
  7. ^ Narici & Beckenstein 2011, p. 477.
  8. ^ Narici & Beckenstein 2011, p. 399.
  9. ^ Narici & Beckenstein 2011, p. 383.
  10. ^ Khaleelulla 1982, pp. 28–63.
  11. ^ Narici & Beckenstein 2011, pp. 418–419.
  12. ^ Trèves 2006, p. 350.
  13. ^ Schaefer & Wolff 1999, p. 166.
  14. ^ Schaefer & Wolff 1999, p. 138.
  15. ^ Schaefer & Wolff 1999, p. 61.
  16. ^ Trèves 2006, p. 346.
  17. ^ Adasch, Ernst & Keim 1978, p. 77.
  18. ^ Schaefer & Wolff 1999, pp. 103–110.
  19. ^ Trèves 2006, p. 348.
  20. ^ Adasch, Ernst & Keim 1978, pp. 70–73.
  21. ^ Trèves 2006, p. 424.

Bibliography