# Barrelled space

Short description: Type of topological vector 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 Bourbaki (1950).

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

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

### Properties of barrels

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

## Characterizations of barreled spaces

Denote by $\displaystyle{ L(X; Y) }$ the space of continuous linear maps from $\displaystyle{ X }$ into $\displaystyle{ Y. }$

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

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

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

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

If $\displaystyle{ X }$ is a Hausdorff locally convex space then this list may be extended by appending:

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

If $\displaystyle{ X }$ is metrizable topological vector space then this list may be extended by appending:

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

If $\displaystyle{ X }$ is a locally convex metrizable topological vector space then this list may be extended by appending:

1. (Property S): The weak* topology on $\displaystyle{ X^{\prime} }$ is sequentially complete.[8]
2. (Property C): Every weak* bounded subset of $\displaystyle{ X^{\prime} }$ is $\displaystyle{ \sigma\left(X^{\prime}, X\right) }$-relatively countably compact.[8]
3. (𝜎-barrelled): Every countable weak* bounded subset of $\displaystyle{ X^{\prime} }$ is equicontinuous.[8]
4. (Baire-like): $\displaystyle{ X }$ 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.
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 $\displaystyle{ X }$ such that every weakly bounded subset of its continuous dual space is equicontinuous.[12]
14. A locally convex TVS $\displaystyle{ X }$ such that for every Banach space $\displaystyle{ B, }$ a closed linear map of $\displaystyle{ X }$ into $\displaystyle{ B }$ 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.

### 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.[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 $\displaystyle{ \R^{\N} }$ 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.

Theorem[19] — Let $\displaystyle{ X }$ be a barrelled TVS and $\displaystyle{ Y }$ be a locally convex TVS. Let $\displaystyle{ H }$ be a subset of the space $\displaystyle{ L(X ;Y) }$ of continuous linear maps from $\displaystyle{ X }$ into $\displaystyle{ Y }$. The following are equivalent:

1. $\displaystyle{ H }$ is bounded for the topology of pointwise convergence;
2. $\displaystyle{ H }$ is bounded for the topology of bounded convergence;
3. $\displaystyle{ H }$ is equicontinuous.

The Banach-Steinhaus theorem is a corollary of the above result.[20] When the vector space $\displaystyle{ Y }$ consists of the complex numbers then the following generalization also holds.

Theorem[21] — If $\displaystyle{ X }$ is a barrelled TVS over the complex numbers and $\displaystyle{ H }$ is a subset of the continuous dual space of $\displaystyle{ X }$, then the following are equivalent:

1. $\displaystyle{ H }$ is weakly bounded;
2. $\displaystyle{ H }$ is strongly bounded;
3. $\displaystyle{ H }$ is equicontinuous;
4. $\displaystyle{ H }$ is relatively compact in the weak dual topology.

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

Closed Graph Theorem[22] — Every closed linear operator from a Hausdorff barrelled TVS into a complete metrizable TVS is continuous.

### Other properties

• Every Hausdorff barrelled space is quasi-barrelled.[23]
• 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.[24]
• A linear map with a closed graph from a barreled TVS into a $\displaystyle{ B_r }$-complete TVS is necessarily continuous.[13]