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 Bourbaki (1950).

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 ${\displaystyle \dim X>2}$ 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.

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 ${\displaystyle \mathbb {C} }$ (if considered as a complex vector space) or equal to ${\displaystyle \mathbb {R} ^{2}}$ (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 ${\displaystyle R(\theta )e^{i\theta }\in \mathbb {C} .}$ Let ${\textstyle S:=\bigcup _{\theta \in [0,2\pi )}S_{\theta }.}$ Then ${\displaystyle S}$ is always an absorbing subset of ${\displaystyle \mathbb {R} ^{2}}$ (a real vector space) but it is an absorbing subset of ${\displaystyle \mathbb {C} }$ (a complex vector space) if and only if it is a neighborhood of the origin. Moreover, ${\displaystyle S}$ is a balanced subset of ${\displaystyle \mathbb {R} ^{2}}$ if and only if ${\displaystyle R(\theta )=R(\pi +\theta )}$ for every ${\displaystyle 0\leq \theta <\pi }$ (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 ${\displaystyle \mathbb {C} }$ if and only it is an open or closed ball centered at the origin (of radius ${\displaystyle 0<r\leq \infty }$). In particular, barrels in ${\displaystyle \mathbb {C} }$ 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 ${\displaystyle \mathbb {R} ^{2}}$ but not absorbing in ${\displaystyle \mathbb {C} ,}$ 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 ${\displaystyle \mathbb {R} ^{2}}$ that is neither closed nor convex. To have ${\displaystyle S}$ be a balanced, absorbing, and closed subset of ${\displaystyle \mathbb {R} ^{2}}$ that is neither convex nor a neighborhood of the origin, define ${\displaystyle R}$ on ${\displaystyle [0,\pi )}$ as follows: for ${\displaystyle 0\leq \theta <\pi ,}$ let ${\displaystyle R(\theta ):=\pi -\theta }$ (alternatively, it can be any positive function on ${\displaystyle [0,\pi )}$ that is continuously differentiable, which guarantees that ${\textstyle \lim _{\theta \searrow 0}R(\theta )=R(0)>0}$ and that ${\displaystyle S}$ is closed, and that also satisfies ${\textstyle \lim _{\theta \nearrow \pi }R(\theta )=0,}$ 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 ${\displaystyle \mathbb {R} ^{2}.}$

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