Barreled 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 $\dim X\geq 2$ 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 $\mathbb {C}$ (if considered as a complex vector space) or equal to $\mathbb {R} ^{2}$ (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 $R(\theta )e^{i\theta }\in \mathbb {C} .$ Let ${\textstyle S:=\bigcup _{\theta \in [0,2\pi )}S_{\theta }.}$ Then $S$ is always an absorbing subset of $\mathbb {R} ^{2}$ (a real vector space) but it is an absorbing subset of $\mathbb {C}$ (a complex vector space) if and only if it is a neighborhood of the origin. Moreover, $S$ is a balanced subset of $\mathbb {R} ^{2}$ 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 $\mathbb {C}$ if and only it is an open or closed ball centered at the origin (of radius $0<r\leq \infty$ ). In particular, barrels in $\mathbb {C}$ 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 $\mathbb {R} ^{2}$ but not absorbing in $\mathbb {C} ,$ 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 $\mathbb {R} ^{2}$ that is neither closed nor convex. To have $S$ be a balanced, absorbing, and closed subset of $\mathbb {R} ^{2}$ 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 ${\textstyle \lim _{\theta \searrow 0}R(\theta )=R(0)>0}$ and that $S$ is closed, and that also satisfies ${\textstyle \lim _{\theta \nearrow \pi }R(\theta )=0,}$ 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 $\mathbb {R} ^{2}.$

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:

$X$ is barrelled.
Definition: Every barrel in X {\displaystyle X} $X$ 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]
For any Hausdorff TVS $Y$ every pointwise bounded subset of $L(X;Y)$ is equicontinuous.^[3]
For any F-space Y {\displaystyle Y} $Y$ every pointwise bounded subset of L ( X ; Y ) {\displaystyle L(X;Y)} $L(X;Y)$ is equicontinuous.^[3]
- An F-space is a complete metrizable TVS.
Every closed linear operator from X {\displaystyle X} $X$ 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 $X\times Y.$
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:

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]
For any locally convex TVS Y , {\displaystyle Y,} $Y,$ every pointwise bounded subset of L ( X ; Y ) {\displaystyle L(X;Y)} $L(X;Y)$ 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.
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]
$X$ carries the strong dual topology $\beta \left(X,X^{\prime }\right).$ ^[2]
Every lower semicontinuous seminorm on $X$ is continuous.^[2]
Every linear map F : X → Y {\displaystyle F:X\to Y} $F:X\to Y$ into a locally convex space Y {\displaystyle Y} $Y$ is almost continuous.^[2]
- A linear map $F:X\to Y$ is called almost continuous if for every neighborhood $V$ of the origin in $Y,$ the closure of $F^{-1}(V)$ is a neighborhood of the origin in $X.$
Every surjective linear map F : Y → X {\displaystyle F:Y\to X} $F:Y\to X$ from a locally convex space Y {\displaystyle Y} $Y$ 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 $X.$
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:

Closed graph theorem: Every closed linear operator F : X → Y {\displaystyle F:X\to Y} $F:X\to Y$ into a Banach space Y {\displaystyle Y} $Y$ is continuous.^[7]
- The linear operator is called closed if its graph is a closed subset of $X\times Y.$
For every subset A {\displaystyle A} $A$ of the continuous dual space of X , {\displaystyle X,} $X,$ the following properties are equivalent: A {\displaystyle A} $A$ is^[6]
1. equicontinuous;
2. relatively weakly compact;
3. strongly bounded;
4. weakly bounded.
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:

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:

(Property S): The weak* topology on $X^{\prime }$ is sequentially complete.^[8]
(Property C): Every weak* bounded subset of $X^{\prime }$ is $\sigma \left(X^{\prime },X\right)$ -relatively countably compact.^[8]
(𝜎-barrelled): Every countable weak* bounded subset of $X^{\prime }$ is equicontinuous.^[8]
(Baire-like): $X$ 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:

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 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.
Complete pseudometrizable TVSs.^[9]
- 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.^[10]
A sequentially complete quasibarrelled space.
A quasi-complete Hausdorff locally convex infrabarrelled space.^[2]
- A TVS is called quasi-complete if every closed and bounded subset is complete.
A TVS with a dense barrelled vector subspace.^[2]
- Thus the completion of a barreled space is barrelled.
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]
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.
A locally convex ultrabarelled TVS.^[11]
A Hausdorff locally convex TVS $X$ such that every weakly bounded subset of its continuous dual space is equicontinuous.^[12]
A locally convex TVS $X$ such that for every Banach space $B,$ a closed linear map of $X$ into $B$ is necessarily continuous.^[13]
A product of a family of barreled spaces.^[14]
A locally convex direct sum and the inductive limit of a family of barrelled spaces.^[15]
A quotient of a barrelled space.^[16]^[15]
A Hausdorff sequentially complete quasibarrelled boundedly summing TVS.^[17]
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 $\mathbb {R} ^{\mathbb {N} }$ 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 $X$ be a barrelled TVS and $Y$ be a locally convex TVS. Let $H$ be a subset of the space $L(X;Y)$ of continuous linear maps from $X$ into $Y$ . The following are equivalent:

$H$ is bounded for the topology of pointwise convergence;
$H$ is bounded for the topology of bounded convergence;
$H$ is equicontinuous.

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

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

$H$ is weakly bounded;
$H$ is strongly bounded;
$H$ is equicontinuous;
$H$ is relatively compact in the weak dual topology.

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

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 $B_{r}$ -complete TVS is necessarily continuous.^[13]

References

Narici & Beckenstein 2011, pp. 225–273.
Narici & Beckenstein 2011, pp. 371–423.
Adasch, Ernst & Keim 1978, p. 39.
Adasch, Ernst & Keim 1978, p. 43.
Adasch, Ernst & Keim 1978, p. 32.
Schaefer & Wolff 1999, pp. 127, 141Trèves 2006, p. 350.
Narici & Beckenstein 2011, p. 477.
Narici & Beckenstein 2011, p. 399.
Narici & Beckenstein 2011, p. 383.
Khaleelulla 1982, pp. 28–63.
Narici & Beckenstein 2011, pp. 418–419.
Trèves 2006, p. 350.
Schaefer & Wolff 1999, p. 166.
Schaefer & Wolff 1999, p. 138.
Schaefer & Wolff 1999, p. 61.
Trèves 2006, p. 346.
Adasch, Ernst & Keim 1978, p. 77.
Schaefer & Wolff 1999, pp. 103–110.
Trèves 2006, p. 347.
Trèves 2006, p. 348.
Trèves 2006, p. 349.
Adasch, Ernst & Keim 1978, p. 41.
Adasch, Ernst & Keim 1978, pp. 70–73.
Trèves 2006, p. 424.

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