Barrelled space

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.

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

A ' or a ' 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 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 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 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 $$X.$$ In any locally convex Hausdorff TVS $$X,$$ every barrel in $$X$$ absorbs every convex bounded complete subset of $$X.$$ 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 $$H \subseteq B^{\circ}.$$ 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 $$Y.$$ 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 $$B = C \cap M.$$ 

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:

 $$X$$ is barrelled. : Every barrel in $$X$$ is a neighborhood of the origin. <li>For any Hausdorff TVS $$Y$$ every pointwise bounded subset of $$L(X; Y)$$ is equicontinuous.</li> <li>For any F-space $$Y$$ every pointwise bounded subset of $$L(X; Y)$$ is equicontinuous. <li>Every closed linear operator from $$X$$ into a complete metrizable TVS is continuous. <li>Every Hausdorff TVS topology $$\nu$$ on $$X$$ that has a neighborhood basis of the origin consisting of $$\tau$$-closed set is course than $$\tau.$$</li> </ol>
 * 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 point of $$Y$$ (not necessarily the origin).</li>
 * An F-space is a complete metrizable TVS.</li>
 * A linear map $$F : X \to Y$$ is called closed if its graph is a closed subset of $$X \times Y.$$</li>

If $$(X, \tau)$$ is locally convex space then this list may be extended by appending: <li>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.</li> <li>For any locally convex TVS $$Y,$$ every pointwise bounded subset of $$L(X; Y)$$ is equicontinuous. <li>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). </li> <li>$$X$$ carries the strong dual topology $$\beta\left(X, X^{\prime}\right).$$</li> <li>Every lower semicontinuous seminorm on $$X$$ is continuous.</li> <li>Every linear map $$F : X \to Y$$ into a locally convex space $$Y$$ is almost continuous. <li>Every surjective linear map $$F : Y \to X$$ from a locally convex space $$Y$$ is almost open. <li>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 $$\tau.$$</li> </ol>
 * 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.</li>
 * 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 $$X.$$</li>
 * This means that for every neighborhood $$V$$ of 0 in $$Y,$$ the closure of $$F(V)$$ is a neighborhood of 0 in $$X.$$</li>

If $$X$$ is a Hausdorff locally convex space then this list may be extended by appending: <li>Closed graph theorem: Every closed linear operator $$F : X \to Y$$ into a Banach space $$Y$$ is continuous. <li>For every subset $$A$$ of the continuous dual space of $$X,$$ the following properties are equivalent: $$A$$ is <ol style="list-style-type: lower-roman;"> <li>equicontinuous;</li> <li>relatively weakly compact;</li> <li>strongly bounded;</li> <li>weakly bounded.</li> </ol></li> <li>The 0-neighborhood bases in $$X$$ and the fundamental families of bounded sets in $$X_{\beta}^{\prime}$$ correspond to each other by polarity. </li> </ol>
 * The linear operator is called if its graph is a closed subset of $$X \times Y.$$</li>

If $$X$$ is metrizable topological vector space then this list may be extended by appending: <li>For any complete metrizable TVS $$Y$$ every pointwise bounded in $$L(X; Y)$$ is equicontinuous.</li> </ol>

If $$X$$ is a locally convex metrizable topological vector space then this list may be extended by appending: <li>: The weak* topology on $$X^{\prime}$$ is sequentially complete.</li> <li>: Every weak* bounded subset of $$X^{\prime}$$ is $$\sigma\left(X^{\prime}, X\right)$$-relatively countably compact.</li> <li>: Every countable weak* bounded subset of $$X^{\prime}$$ is equicontinuous.</li> <li>: $$X$$ is not the union of an increase sequence of nowhere dense disks.</li> </ol>

Examples and sufficient conditions
Each of the following topological vector spaces is barreled: <ol> <li>TVSs that are Baire space. <li>F-spaces, Fréchet spaces, Banach spaces, and Hilbert spaces. <li>Complete pseudometrizable TVSs. <li>Montel spaces.</li> <li>Strong dual spaces of Montel spaces (since they are necessarily Montel spaces).</li> <li>A locally convex quasi-barrelled space that is also a σ-barrelled space.</li> <li>A sequentially complete quasibarrelled space.</li> <li>A quasi-complete Hausdorff locally convex infrabarrelled space. <li>A TVS with a dense barrelled vector subspace. <li>A Hausdorff locally convex TVS with a dense infrabarrelled vector subspace. <li>A vector subspace of a barrelled space that has countable codimensional. <li>A locally convex ultrabarelled TVS.</li> <li>A Hausdorff locally convex TVS $$X$$ such that every weakly bounded subset of its continuous dual space is equicontinuous.</li> <li>A locally convex TVS $$X$$ such that for every Banach space $$B,$$ a closed linear map of $$X$$ into $$B$$ is necessarily continuous.</li> <li>A product of a family of barreled spaces.</li> <li>A locally convex direct sum and the inductive limit of a family of barrelled spaces.</li> <li>A quotient of a barrelled space.</li> <li>A Hausdorff sequentially complete quasibarrelled boundedly summing TVS.</li> <li>A locally convex Hausdorff reflexive space is barrelled.</li> </ol>
 * Consequently, every topological vector space that is of the second category in itself is barrelled.</li>
 * However, there exist normed vector spaces that are 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.</li>
 * Consequently, every finite-dimensional TVS is barrelled.</li>
 * A TVS is called quasi-complete if every closed and bounded subset is complete.</li>
 * Thus the completion of a barreled space is barrelled.</li>
 * Thus the completion of an infrabarrelled Hausdorff locally convex space is barrelled.</li>
 * In particular, a finite codimensional vector subspace of a barrelled space is barreled.</li>

Counter examples
<ul> <li>A barrelled space need not be Montel, complete, metrizable, unordered Baire-like, nor the inductive limit of Banach spaces.</li> <li>Not all normed spaces are barrelled. However, they are all infrabarrelled.</li> <li>A closed subspace of a barreled space is not necessarily countably quasi-barreled (and thus not necessarily barrelled).</li> <li>There exists a dense vector subspace of the Fréchet barrelled space $$\R^{\N}$$ that is not barrelled.</li> <li>There exist complete locally convex TVSs that are not barrelled.</li> <li>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).</li> </ul>

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