In linear algebra and related areas of mathematics a balanced set, circled set or disk in a vector space (over a field with an absolute value function ) is a set such that for all scalars satisfying
The balanced hull or balanced envelope of a set is the smallest balanced set containing The balanced core of a set is the largest balanced set contained in
Balanced sets are ubiquitous in functional analysis because every neighborhood of the origin in every topological vector space (TVS) contains a balanced neighborhood of the origin and every convex neighborhood of the origin contains a balanced convex neighborhood of the origin (even if the TVS is not locally convex). This neighborhood can also be chosen to be an open set or, alternatively, a closed set.
Let be a vector space over the field of real or complex numbers.
Notation
If is a set, is a scalar, and then let and and for any let denote, respectively, the open ball and the closed ball of radius in the scalar field centered at where and Every balanced subset of the field is of the form or for some
Balanced set
A subset of is called a balanced set or balanced if it satisfies any of the following equivalent conditions:
If is a convex set then this list may be extended to include:
If then this list may be extended to include:
The balanced hull of a subset of denoted by is defined in any of the following equivalent ways:
The balanced core of a subset of denoted by is defined in any of the following equivalent ways:
The empty set is a balanced set. As is any vector subspace of any (real or complex) vector space. In particular, is always a balanced set.
Any non-empty set that does not contain the origin is not balanced and furthermore, the balanced core of such a set will equal the empty set.
Normed and topological vector spaces
The open and closed balls centered at the origin in a normed vector space are balanced sets. If is a seminorm (or norm) on a vector space then for any constant the set is balanced.
If is any subset and then is a balanced set. In particular, if is any balanced neighborhood of the origin in a topological vector space then
Balanced sets in and
Let be the field real numbers or complex numbers let denote the absolute value on and let denotes the vector space over So for example, if is the field of complex numbers then is a 1-dimensional complex vector space whereas if then is a 1-dimensional real vector space.
The balanced subsets of are exactly the following:^{[3]}
Consequently, both the balanced core and the balanced hull of every set of scalars is equal to one of the sets listed above.
The balanced sets are itself, the empty set and the open and closed discs centered at zero. Contrariwise, in the two dimensional Euclidean space there are many more balanced sets: any line segment with midpoint at the origin will do. As a result, and are entirely different as far as scalar multiplication is concerned.
Balanced sets in
Throughout, let (so is a vector space over ) and let is the closed unit ball in centered at the origin.
If is non-zero, and then the set is a closed, symmetric, and balanced neighborhood of the origin in More generally, if is any closed subset of such that then is a closed, symmetric, and balanced neighborhood of the origin in This example can be generalized to for any integer
Let be the union of the line segment between the points and and the line segment between and Then is balanced but not convex. Nor is is absorbing (despite the fact that is the entire vector space).
For every let be any positive real number and let be the (open or closed) line segment in between the points and Then the set is a balanced and absorbing set but it is not necessarily convex.
The balanced hull of a closed set need not be closed. Take for instance the graph of in
The next example shows that the balanced hull of a convex set may fail to be convex (however, the convex hull of a balanced set is always balanced). For an example, let the convex subset be which is a horizontal closed line segment lying above the axis in The balanced hull is a non-convex subset that is "hour glass shaped" and equal to the union of two closed and filled isosceles triangles and where and is the filled triangle whose vertices are the origin together with the endpoints of (said differently, is the convex hull of while is the convex hull of ).
A set is balanced if and only if it is equal to its balanced hull or to its balanced core in which case all three of these sets are equal:
The Cartesian product of a family of balanced sets is balanced in the product space of the corresponding vector spaces (over the same field ).
In any topological vector space, the closure of a balanced set is balanced.^{[5]} The union of the origin and the topological interior of a balanced set is balanced. Therefore, the topological interior of a balanced neighborhood of the origin is balanced.^{[5]}^{[proof 1]} However, is a balanced subset of that contains the origin but whose (nonempty) topological interior does not contain the origin and is therefore not a balanced set.^{[6]} Similarly for real vector spaces, if denotes the convex hull of and (a filled triangle whose vertices are these three points) then is an (hour glass shaped) balanced subset of whose non-empty topological interior does not contain the origin and so is not a balanced set (and although the set formed by adding the origin is balanced, it is neither an open set nor a neighborhood of the origin).
Every neighborhood (respectively, convex neighborhood) of the origin in a topological vector space contains a balanced (respectively, convex and balanced) open neighborhood of the origin. In fact, the following construction produces such balanced sets. Given the symmetric set will be convex (respectively, closed, balanced, bounded, a neighborhood of the origin, an absorbing subset of ) whenever this is true of It will be a balanced set if is a star shaped at the origin,^{[note 2]} which is true, for instance, when is convex and contains In particular, if is a convex neighborhood of the origin then will be a balanced convex neighborhood of the origin and so its topological interior will be a balanced convex open neighborhood of the origin.^{[5]}
Let and define (where denotes elements of the field of scalars). Taking shows that If is convex then so is (since an intersection of convex sets is convex) and thus so is 's interior. If then and thus If is star shaped at the origin^{[note 2]} then so is every (for ), which implies that for any thus proving that is balanced. If is convex and contains the origin then it is star shaped at the origin and so will be balanced.
Now suppose is a neighborhood of the origin in Since scalar multiplication (defined by ) is continuous at the origin and there exists some basic open neighborhood (where and ) of the origin in the product topology on such that the set is balanced and it is also open because it may be written as where is an open neighborhood of the origin whenever Finally, shows that is also a neighborhood of the origin. If is balanced then because its interior contains the origin, will also be balanced. If is convex then is convex and balanced and thus the same is true of
Suppose that is a convex and absorbing subset of Then will be convex balanced absorbing subset of which guarantees that the Minkowski functional of will be a seminorm on thereby making into a seminormed space that carries its canonical pseduometrizable topology. The set of scalar multiples as ranges over (or over any other set of non-zero scalars having as a limit point) forms a neighborhood basis of absorbing disks at the origin for this locally convex topology. If is a topological vector space and if this convex absorbing subset is also a bounded subset of then the same will be true of the absorbing disk if in addition does not contain any non-trivial vector subspace then will be a norm and will form what is known as an auxiliary normed space.^{[7]} If this normed space is a Banach space then is called a Banach disk.
Properties of balanced sets
A balanced set is not empty if and only if it contains the origin. By definition, a set is absolutely convex if and only if it is convex and balanced. Every balanced set is star-shaped (at 0) and a symmetric set. If is a balanced subset of then:
Properties of balanced hulls and balanced cores
For any collection of subsets of
In any topological vector space, the balanced hull of any open neighborhood of the origin is again open. If is a Hausdorff topological vector space and if is a compact subset of then the balanced hull of is compact.^{[8]}
If a set is closed (respectively, convex, absorbing, a neighborhood of the origin) then the same is true of its balanced core.
For any subset and any scalar
For any scalar This equality holds for if and only if Thus if or then for every scalar
A function on a real or complex vector space is said to be a balanced function if it satisfies any of the following equivalent conditions:^{[9]}
If is a balanced function then for every scalar and vector so in particular, for every unit length scalar (satisfying ) and every ^{[9]} Using shows that every balanced function is a symmetric function.
A real-valued function is a seminorm if and only if it is a balanced sublinear function.
Proofs