Balanced set

Short description: Construct in functional analysis

In linear algebra and related areas of mathematics a balanced set, circled set or disk in a vector space (over a field [math]\displaystyle{ \mathbb{K} }[/math] with an absolute value function [math]\displaystyle{ |\cdot | }[/math]) is a set [math]\displaystyle{ S }[/math] such that [math]\displaystyle{ a S \subseteq S }[/math] for all scalars [math]\displaystyle{ a }[/math] satisfying [math]\displaystyle{ |a| \leq 1. }[/math]

The balanced hull or balanced envelope of a set [math]\displaystyle{ S }[/math] is the smallest balanced set containing [math]\displaystyle{ S. }[/math] The balanced core of a set [math]\displaystyle{ S }[/math] is the largest balanced set contained in [math]\displaystyle{ S. }[/math]

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.

Definition

Let [math]\displaystyle{ X }[/math] be a vector space over the field [math]\displaystyle{ \mathbb{K} }[/math] of real or complex numbers.

Notation

If [math]\displaystyle{ S }[/math] is a set, [math]\displaystyle{ a }[/math] is a scalar, and [math]\displaystyle{ B \subseteq \mathbb{K} }[/math] then let [math]\displaystyle{ a S = \{a s : s \in S\} }[/math] and [math]\displaystyle{ B S = \{b s : b \in B, s \in S\} }[/math] and for any [math]\displaystyle{ 0 \leq r \leq \infty, }[/math] let [math]\displaystyle{ B_r = \{a \in \mathbb{K} : |a| \lt r\} \qquad \text{ and } \qquad B_{\leq r} = \{ a \in \mathbb{K} : |a| \leq r\}. }[/math] denote, respectively, the open ball and the closed ball of radius [math]\displaystyle{ r }[/math] in the scalar field [math]\displaystyle{ \mathbb{K} }[/math] centered at [math]\displaystyle{ 0 }[/math] where [math]\displaystyle{ B_0 = \varnothing, B_{\leq 0} = \{0\}, }[/math] and [math]\displaystyle{ B_{\infty} = B_{\leq \infty} = \mathbb{K}. }[/math] Every balanced subset of the field [math]\displaystyle{ \mathbb{K} }[/math] is of the form [math]\displaystyle{ B_{\leq r} }[/math] or [math]\displaystyle{ B_r }[/math] for some [math]\displaystyle{ 0 \leq r \leq \infty. }[/math]

Balanced set

A subset [math]\displaystyle{ S }[/math] of [math]\displaystyle{ X }[/math] is called a balanced set or balanced if it satisfies any of the following equivalent conditions:

1. Definition: [math]\displaystyle{ a s \in S }[/math] for all [math]\displaystyle{ s \in S }[/math] and all scalars [math]\displaystyle{ a }[/math] satisfying [math]\displaystyle{ |a| \leq 1. }[/math]
2. [math]\displaystyle{ a S \subseteq S }[/math] for all scalars [math]\displaystyle{ a }[/math] satisfying [math]\displaystyle{ |a| \leq 1. }[/math]
3. [math]\displaystyle{ B_{\leq 1} S \subseteq S }[/math] (where [math]\displaystyle{ B_{\leq 1} := \{a \in \mathbb{K} : |a| \leq 1\} }[/math]).
4. [math]\displaystyle{ S = B_{\leq 1} S. }[/math]
5. For every [math]\displaystyle{ s \in S, }[/math] [math]\displaystyle{ S \cap \mathbb{K} s = B_{\leq 1} (S \cap \mathbb{K} s). }[/math]
   • [math]\displaystyle{ \mathbb{K} s = \operatorname{span} \{s\} }[/math] is a [math]\displaystyle{ 0 }[/math] (if [math]\displaystyle{ s = 0 }[/math]) or [math]\displaystyle{ 1 }[/math] (if [math]\displaystyle{ s \neq 0 }[/math]) dimensional vector subspace of [math]\displaystyle{ X. }[/math]
   • If [math]\displaystyle{ R := S \cap \mathbb{K} s }[/math] then the above equality becomes [math]\displaystyle{ R = B_{\leq 1} R, }[/math] which is exactly the previous condition for a set to be balanced. Thus, [math]\displaystyle{ S }[/math] is balanced if and only if for every [math]\displaystyle{ s \in S, }[/math] [math]\displaystyle{ S \cap \mathbb{K} s }[/math] is a balanced set (according to any of the previous defining conditions).
6. For every 1-dimensional vector subspace [math]\displaystyle{ Y }[/math] of [math]\displaystyle{ \operatorname{span} S, }[/math] [math]\displaystyle{ S \cap Y }[/math] is a balanced set (according to any defining condition other than this one).
7. For every [math]\displaystyle{ s \in S, }[/math] there exists some [math]\displaystyle{ 0 \leq r \leq \infty }[/math] such that [math]\displaystyle{ S \cap \mathbb{K} s = B_r s }[/math] or [math]\displaystyle{ S \cap \mathbb{K} s = B_{\leq r} s. }[/math]
8. [math]\displaystyle{ S }[/math] is a balanced subset of [math]\displaystyle{ \operatorname{span} S }[/math] (according to any defining condition of "balanced" other than this one).
   • Thus [math]\displaystyle{ S }[/math] is a balanced subset of [math]\displaystyle{ X }[/math] if and only if it is balanced subset of every (equivalently, of some) vector space over the field [math]\displaystyle{ \mathbb{K} }[/math] that contains [math]\displaystyle{ S. }[/math] So assuming that the field [math]\displaystyle{ \mathbb{K} }[/math] is clear from context, this justifies writing "[math]\displaystyle{ S }[/math] is balanced" without mentioning any vector space.

If [math]\displaystyle{ S }[/math] is a convex set then this list may be extended to include:

9. [math]\displaystyle{ a S \subseteq S }[/math] for all scalars [math]\displaystyle{ a }[/math] satisfying [math]\displaystyle{ |a| = 1. }[/math]

If [math]\displaystyle{ \mathbb{K} = \R }[/math] then this list may be extended to include:

10. [math]\displaystyle{ S }[/math] is symmetric (meaning [math]\displaystyle{ - S = S }[/math]) and [math]\displaystyle{ [0, 1) S \subseteq S. }[/math]

Balanced hull

[math]\displaystyle{ \operatorname{bal} S ~=~ \bigcup_{|a| \leq 1} a S = B_{\leq 1} S }[/math]

The balanced hull of a subset [math]\displaystyle{ S }[/math] of [math]\displaystyle{ X, }[/math] denoted by [math]\displaystyle{ \operatorname{bal} S, }[/math] is defined in any of the following equivalent ways:

1. Definition: [math]\displaystyle{ \operatorname{bal} S }[/math] is the smallest (with respect to [math]\displaystyle{ \,\subseteq\, }[/math]) balanced subset of [math]\displaystyle{ X }[/math] containing [math]\displaystyle{ S. }[/math]
2. [math]\displaystyle{ \operatorname{bal} S }[/math] is the intersection of all balanced sets containing [math]\displaystyle{ S. }[/math]
3. [math]\displaystyle{ \operatorname{bal} S = \bigcup_{|a| \leq 1} (a S). }[/math]
4. [math]\displaystyle{ \operatorname{bal} S = B_{\leq 1} S. }[/math]

Balanced core

[math]\displaystyle{ \operatorname{balcore} S ~=~ \begin{cases} \displaystyle\bigcap_{|a| \geq 1} a S & \text{ if } 0 \in S \\ \varnothing & \text{ if } 0 \not\in S \\ \end{cases} }[/math]

The balanced core of a subset [math]\displaystyle{ S }[/math] of [math]\displaystyle{ X, }[/math] denoted by [math]\displaystyle{ \operatorname{balcore} S, }[/math] is defined in any of the following equivalent ways:

1. Definition: [math]\displaystyle{ \operatorname{balcore} S }[/math] is the largest (with respect to [math]\displaystyle{ \,\subseteq\, }[/math]) balanced subset of [math]\displaystyle{ S. }[/math]
2. [math]\displaystyle{ \operatorname{balcore} S }[/math] is the union of all balanced subsets of [math]\displaystyle{ S. }[/math]
3. [math]\displaystyle{ \operatorname{balcore} S = \varnothing }[/math] if [math]\displaystyle{ 0 \not\in S }[/math] while [math]\displaystyle{ \operatorname{balcore} S = \bigcap_{|a| \geq 1} (a S) }[/math] if [math]\displaystyle{ 0 \in S. }[/math]

Examples

The empty set is a balanced set. As is any vector subspace of any (real or complex) vector space. In particular, [math]\displaystyle{ \{0\} }[/math] 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 vectors spaces

The open and closed balls centered at the origin in a normed vector space are balanced sets. If [math]\displaystyle{ p }[/math] is a seminorm (or norm) on a vector space [math]\displaystyle{ X }[/math] then for any constant [math]\displaystyle{ c \gt 0, }[/math] the set [math]\displaystyle{ \{x \in X : p(x) \leq c\} }[/math] is balanced.

If [math]\displaystyle{ S \subseteq X }[/math] is any subset and [math]\displaystyle{ B_1 := \{a \in \mathbb{K} : |a| \lt 1\} }[/math] then [math]\displaystyle{ B_1 S }[/math] is a balanced set. In particular, if [math]\displaystyle{ U \subseteq X }[/math] is any balanced neighborhood of the origin in a topological vector space [math]\displaystyle{ X }[/math] then [math]\displaystyle{ \operatorname{Int}_X U ~\subseteq~ B_1 U ~=~ \bigcup_{0 \lt |a| \lt 1} a U ~\subseteq~ U. }[/math]

Balanced sets in [math]\displaystyle{ \R }[/math] and [math]\displaystyle{ \Complex }[/math]

Let [math]\displaystyle{ \mathbb{K} }[/math] be the field real numbers [math]\displaystyle{ \R }[/math] or complex numbers [math]\displaystyle{ \Complex, }[/math] let [math]\displaystyle{ |\cdot| }[/math] denote the absolute value on [math]\displaystyle{ \mathbb{K}, }[/math] and let [math]\displaystyle{ X := \mathbb{K} }[/math] denotes the vector space over [math]\displaystyle{ \mathbb{K}. }[/math] So for example, if [math]\displaystyle{ \mathbb{K} := \Complex }[/math] is the field of complex numbers then [math]\displaystyle{ X = \mathbb{K} = \Complex }[/math] is a 1-dimensional complex vector space whereas if [math]\displaystyle{ \mathbb{K} := \R }[/math] then [math]\displaystyle{ X = \mathbb{K} = \R }[/math] is a 1-dimensional real vector space.

The balanced subsets of [math]\displaystyle{ X = \mathbb{K} }[/math] are exactly the following:

1. [math]\displaystyle{ \varnothing }[/math]
2. [math]\displaystyle{ X }[/math]
3. [math]\displaystyle{ \{0\} }[/math]
4. [math]\displaystyle{ \{x \in X : |x| \lt r\} }[/math] for some real [math]\displaystyle{ r \gt 0 }[/math]
5. [math]\displaystyle{ \{x \in X : |x| \leq r\} }[/math] for some real [math]\displaystyle{ r \gt 0. }[/math]

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 [math]\displaystyle{ \Complex }[/math] 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, [math]\displaystyle{ \Complex }[/math] and [math]\displaystyle{ \R^2 }[/math] are entirely different as far as scalar multiplication is concerned.

Balanced sets in [math]\displaystyle{ \R^2 }[/math]

Throughout, let [math]\displaystyle{ X = \R^2 }[/math] (so [math]\displaystyle{ X }[/math] is a vector space over [math]\displaystyle{ \R }[/math]) and let [math]\displaystyle{ B_{\leq 1} }[/math] is the closed unit ball in [math]\displaystyle{ X }[/math] centered at the origin.

If [math]\displaystyle{ x_0 \in X = \R^2 }[/math] is non-zero, and [math]\displaystyle{ L := \R x_0, }[/math] then the set [math]\displaystyle{ R := B_{\leq 1} \cup L }[/math] is a closed, symmetric, and balanced neighborhood of the origin in [math]\displaystyle{ X. }[/math] More generally, if [math]\displaystyle{ C }[/math] is any closed subset of [math]\displaystyle{ X }[/math] such that [math]\displaystyle{ (0, 1) C \subseteq C, }[/math] then [math]\displaystyle{ S := B_{\leq 1} \cup C \cup (-C) }[/math] is a closed, symmetric, and balanced neighborhood of the origin in [math]\displaystyle{ X. }[/math] This example can be generalized to [math]\displaystyle{ \R^n }[/math] for any integer [math]\displaystyle{ n \geq 1. }[/math]

Let [math]\displaystyle{ B \subseteq \R^2 }[/math] be the union of the line segment between the points [math]\displaystyle{ (-1, 0) }[/math] and [math]\displaystyle{ (1, 0) }[/math] and the line segment between [math]\displaystyle{ (0, -1) }[/math] and [math]\displaystyle{ (0, 1). }[/math] Then [math]\displaystyle{ B }[/math] is balanced but not convex. Nor is [math]\displaystyle{ B }[/math] is absorbing (despite the fact that [math]\displaystyle{ \operatorname{span} B = \R^2 }[/math] is the entire vector space).

For every [math]\displaystyle{ 0 \leq t \leq \pi, }[/math] let [math]\displaystyle{ r_t }[/math] be any positive real number and let [math]\displaystyle{ B^t }[/math] be the (open or closed) line segment in [math]\displaystyle{ X := \R^2 }[/math] between the points [math]\displaystyle{ (\cos t, \sin t) }[/math] and [math]\displaystyle{ - (\cos t, \sin t). }[/math] Then the set [math]\displaystyle{ B = \bigcup_{0 \leq t \lt \pi} r_t B^t }[/math] 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 [math]\displaystyle{ x y = 1 }[/math] in [math]\displaystyle{ X = \R^2. }[/math]

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 [math]\displaystyle{ S := [-1, 1] \times \{1\}, }[/math] which is a horizontal closed line segment lying above the [math]\displaystyle{ x- }[/math]axis in [math]\displaystyle{ X := \R^2. }[/math] The balanced hull [math]\displaystyle{ \operatorname{bal} S }[/math] is a non-convex subset that is "hour glass shaped" and equal to the union of two closed and filled isosceles triangles [math]\displaystyle{ T_1 }[/math] and [math]\displaystyle{ T_2, }[/math] where [math]\displaystyle{ T_2 = - T_1 }[/math] and [math]\displaystyle{ T_1 }[/math] is the filled triangle whose vertices are the origin together with the endpoints of [math]\displaystyle{ S }[/math] (said differently, [math]\displaystyle{ T_1 }[/math] is the convex hull of [math]\displaystyle{ S \cup \{(0,0)\} }[/math] while [math]\displaystyle{ T_2 }[/math] is the convex hull of [math]\displaystyle{ (-S) \cup \{(0,0)\} }[/math]).

Sufficient conditions

A set [math]\displaystyle{ T }[/math] is balanced if and only if it is equal to its balanced hull [math]\displaystyle{ \operatorname{bal} T }[/math] or to its balanced core [math]\displaystyle{ \operatorname{balcore} T, }[/math] in which case all three of these sets are equal: [math]\displaystyle{ T = \operatorname{bal} T = \operatorname{balcore} T. }[/math]

The Cartesian product of a family of balanced sets is balanced in the product space of the corresponding vector spaces (over the same field [math]\displaystyle{ \mathbb{K} }[/math]).

• The balanced hull of a compact (respectively, totally bounded, bounded) set has the same property.
• The convex hull of a balanced set is convex and balanced (that is, it is absolutely convex). However, the balanced hull of a convex set may fail to be convex (a counter-example is given above).
• Arbitrary unions of balanced sets are balanced, and the same is true of arbitrary intersections of balanced sets.
• Scalar multiples and (finite) Minkowski sums of balanced sets are again balanced.
• Images and preimages of balanced sets under linear maps are again balanced. Explicitly, if [math]\displaystyle{ L : X \to Y }[/math] is a linear map and [math]\displaystyle{ B \subseteq X }[/math] and [math]\displaystyle{ C \subseteq Y }[/math] are balanced sets, then [math]\displaystyle{ L(B) }[/math] and [math]\displaystyle{ L^{-1}(C) }[/math] are balanced sets.

Balanced neighborhoods

In any topological vector space, the closure of a balanced set is balanced. The union of the origin [math]\displaystyle{ \{0\} }[/math] and the topological interior of a balanced set is balanced. Therefore, the topological interior of a balanced neighborhood of the origin is balanced. However, [math]\displaystyle{ \left\{(z, w) \in \Complex^2 : |z| \leq |w|\right\} }[/math] is a balanced subset of [math]\displaystyle{ X = \Complex^2 }[/math] that contains the origin [math]\displaystyle{ (0, 0) \in X }[/math] but whose (nonempty) topological interior does not contain the origin and is therefore not a balanced set. Similarly for real vector spaces, if [math]\displaystyle{ T }[/math] denotes the convex hull of [math]\displaystyle{ (0, 0) }[/math] and [math]\displaystyle{ (\pm 1, 1) }[/math] (a filled triangle whose vertices are these three points) then [math]\displaystyle{ B := T \cup (-T) }[/math] is an (hour glass shaped) balanced subset of [math]\displaystyle{ X := \Reals^2 }[/math] whose non-empty topological interior does not contain the origin and so is not a balanced set (and although the set [math]\displaystyle{ \{(0, 0)\} \cup \operatorname{Int}_X B }[/math] 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 [math]\displaystyle{ X }[/math] contains a balanced (respectively, convex and balanced) open neighborhood of the origin. In fact, the following construction produces such balanced sets. Given [math]\displaystyle{ W \subseteq X, }[/math] the symmetric set [math]\displaystyle{ \bigcap_{|u|=1} u W \subseteq W }[/math] will be convex (respectively, closed, balanced, bounded, a neighborhood of the origin, an absorbing subset of [math]\displaystyle{ X }[/math]) whenever this is true of [math]\displaystyle{ W. }[/math] It will be a balanced set if [math]\displaystyle{ W }[/math] is a star shaped at the origin, which is true, for instance, when [math]\displaystyle{ W }[/math] is convex and contains [math]\displaystyle{ 0. }[/math] In particular, if [math]\displaystyle{ W }[/math] is a convex neighborhood of the origin then [math]\displaystyle{ \bigcap_{|u|=1} u W }[/math] will be a balanced convex neighborhood of the origin and so its topological interior will be a balanced convex open neighborhood of the origin.

Proof

Let [math]\displaystyle{ 0 \in W \subseteq X }[/math] and define [math]\displaystyle{ A = \bigcap_{|u|=1} u W }[/math] (where [math]\displaystyle{ u }[/math] denotes elements of the field [math]\displaystyle{ \mathbb{K} }[/math] of scalars). Taking [math]\displaystyle{ u := 1 }[/math] shows that [math]\displaystyle{ A \subseteq W. }[/math] If [math]\displaystyle{ W }[/math] is convex then so is [math]\displaystyle{ A }[/math] (since an intersection of convex sets is convex) and thus so is [math]\displaystyle{ A }[/math]'s interior. If [math]\displaystyle{ |s| = 1 }[/math] then [math]\displaystyle{ s A = \bigcap_{|u|=1} s u W \subseteq \bigcap_{|u|=1} u W = A }[/math] and thus [math]\displaystyle{ s A = A. }[/math] If [math]\displaystyle{ W }[/math] is star shaped at the origin then so is every [math]\displaystyle{ u W }[/math] (for [math]\displaystyle{ |u| = 1 }[/math]), which implies that for any [math]\displaystyle{ 0 \leq r \leq 1, }[/math] [math]\displaystyle{ r A = \bigcap_{|u|=1} r u W \subseteq \bigcap_{|u|=1} u W = A }[/math] thus proving that [math]\displaystyle{ A }[/math] is balanced. If [math]\displaystyle{ W }[/math] is convex and contains the origin then it is star shaped at the origin and so [math]\displaystyle{ A }[/math] will be balanced.

Now suppose [math]\displaystyle{ W }[/math] is a neighborhood of the origin in [math]\displaystyle{ X. }[/math] Since scalar multiplication [math]\displaystyle{ M : \mathbb{K} \times X \to X }[/math] (defined by [math]\displaystyle{ M(a, x) = a x }[/math]) is continuous at the origin [math]\displaystyle{ (0, 0) \in \mathbb{K} \times X }[/math] and [math]\displaystyle{ M(0, 0) = 0 \in W, }[/math] there exists some basic open neighborhood [math]\displaystyle{ B_r \times V }[/math] (where [math]\displaystyle{ r \gt 0 }[/math] and [math]\displaystyle{ B_r := \{c \in \mathbb{K} : |c| \lt r\} }[/math]) of the origin in the product topology on [math]\displaystyle{ \mathbb{K} \times X }[/math] such that [math]\displaystyle{ M\left(B_r \times V\right) \subseteq W; }[/math] the set [math]\displaystyle{ M\left(B_r \times V\right) = B_r V }[/math] is balanced and it is also open because it may be written as [math]\displaystyle{ B_r V = \bigcup_{|a| \lt r} a V = \bigcup_{0 \lt |a| \lt r} a V \qquad \text{ (since } 0 \cdot V = \{0\} \subseteq a V \text{ )} }[/math] where [math]\displaystyle{ a V }[/math] is an open neighborhood of the origin whenever [math]\displaystyle{ a \neq 0. }[/math] Finally, [math]\displaystyle{ A = \bigcap_{|u|=1} u W \supseteq \bigcap_{|u|=1} u B_r V = \bigcap_{|u|=1} B_r V = B_r V }[/math] shows that [math]\displaystyle{ A }[/math] is also a neighborhood of the origin. If [math]\displaystyle{ A }[/math] is balanced then because its interior [math]\displaystyle{ \operatorname{Int}_X A }[/math] contains the origin, [math]\displaystyle{ \operatorname{Int}_X A }[/math] will also be balanced. If [math]\displaystyle{ W }[/math] is convex then [math]\displaystyle{ A }[/math] is convex and balanced and thus the same is true of [math]\displaystyle{ \operatorname{Int}_X A. }[/math] [math]\displaystyle{ \blacksquare }[/math]

Suppose that [math]\displaystyle{ W }[/math] is a convex and absorbing subset of [math]\displaystyle{ X. }[/math] Then [math]\displaystyle{ D := \bigcap_{|u|=1} u W }[/math] will be convex balanced absorbing subset of [math]\displaystyle{ X, }[/math] which guarantees that the Minkowski functional [math]\displaystyle{ p_D : X \to \R }[/math] of [math]\displaystyle{ D }[/math] will be a seminorm on [math]\displaystyle{ X, }[/math] thereby making [math]\displaystyle{ \left(X, p_D\right) }[/math] into a seminormed space that carries its canonical pseduometrizable topology. The set of scalar multiples [math]\displaystyle{ r D }[/math] as [math]\displaystyle{ r }[/math] ranges over [math]\displaystyle{ \left\{\tfrac{1}{2}, \tfrac{1}{3}, \tfrac{1}{4}, \ldots\right\} }[/math] (or over any other set of non-zero scalars having [math]\displaystyle{ 0 }[/math] as a limit point) forms a neighborhood basis of absorbing disks at the origin for this locally convex topology. If [math]\displaystyle{ X }[/math] is a topological vector space and if this convex absorbing subset [math]\displaystyle{ W }[/math] is also a bounded subset of [math]\displaystyle{ X, }[/math] then the same will be true of the absorbing disk [math]\displaystyle{ D := {\textstyle\bigcap\limits_{|u|=1}} u W; }[/math] if in addition [math]\displaystyle{ D }[/math] does not contain any non-trivial vector subspace then [math]\displaystyle{ p_D }[/math] will be a norm and [math]\displaystyle{ \left(X, p_D\right) }[/math] will form what is known as an auxiliary normed space. If this normed space is a Banach space then [math]\displaystyle{ D }[/math] is called a Banach disk.

Properties

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 [math]\displaystyle{ B }[/math] is a balanced subset of [math]\displaystyle{ X }[/math] then:

• for any scalars [math]\displaystyle{ c }[/math] and [math]\displaystyle{ d, }[/math] if [math]\displaystyle{ |c| \leq |d| }[/math] then [math]\displaystyle{ c B \subseteq d B }[/math] and [math]\displaystyle{ c B = |c| B. }[/math] Thus if [math]\displaystyle{ c }[/math] and [math]\displaystyle{ d }[/math] are any scalars then [math]\displaystyle{ (c B) \cap (d B) = \min_{} \{|c|, |d|\} B. }[/math]
• [math]\displaystyle{ B }[/math] is absorbing in [math]\displaystyle{ X }[/math] if and only if for all [math]\displaystyle{ x \in X, }[/math] there exists [math]\displaystyle{ r \gt 0 }[/math] such that [math]\displaystyle{ x \in r B. }[/math]
• for any 1-dimensional vector subspace [math]\displaystyle{ Y }[/math] of [math]\displaystyle{ X, }[/math] the set [math]\displaystyle{ B \cap Y }[/math] is convex and balanced. If [math]\displaystyle{ B }[/math] is not empty and if [math]\displaystyle{ Y }[/math] is a 1-dimensional vector subspace of [math]\displaystyle{ \operatorname{span} B }[/math] then [math]\displaystyle{ B \cap Y }[/math] is either [math]\displaystyle{ \{0\} }[/math] or else it is absorbing in [math]\displaystyle{ Y. }[/math]
• for any [math]\displaystyle{ x \in X, }[/math] if [math]\displaystyle{ B \cap \operatorname{span} x }[/math] contains more than one point then it is a convex and balanced neighborhood of [math]\displaystyle{ 0 }[/math] in the 1-dimensional vector space [math]\displaystyle{ \operatorname{span} x }[/math] when this space is endowed with the Hausdorff Euclidean topology; and the set [math]\displaystyle{ B \cap \R x }[/math] is a convex balanced subset of the real vector space [math]\displaystyle{ \R x }[/math] that contains the origin.

Properties of balanced hulls and balanced cores

For any collection [math]\displaystyle{ \mathcal{S} }[/math] of subsets of [math]\displaystyle{ X, }[/math] [math]\displaystyle{ \operatorname{bal} \left(\bigcup_{S \in \mathcal{S}} S\right) = \bigcup_{S \in \mathcal{S}} \operatorname{bal} S \quad \text{ and } \quad \operatorname{balcore} \left(\bigcap_{S \in \mathcal{S}} S\right) = \bigcap_{S \in \mathcal{S}} \operatorname{balcore} S. }[/math]

In any topological vector space, the balanced hull of any open neighborhood of the origin is again open. If [math]\displaystyle{ X }[/math] is a Hausdorff topological vector space and if [math]\displaystyle{ K }[/math] is a compact subset of [math]\displaystyle{ X }[/math] then the balanced hull of [math]\displaystyle{ K }[/math] is compact.

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 [math]\displaystyle{ S \subseteq X }[/math] and any scalar [math]\displaystyle{ c, }[/math] [math]\displaystyle{ \operatorname{bal} (c \, S) = c \operatorname{bal} S = |c| \operatorname{bal} S. }[/math]

For any scalar [math]\displaystyle{ c \neq 0, }[/math] [math]\displaystyle{ \operatorname{balcore} (c \, S) = c \operatorname{balcore} S = |c| \operatorname{balcore} S. }[/math] This equality holds for [math]\displaystyle{ c = 0 }[/math] if and only if [math]\displaystyle{ S \subseteq \{0\}. }[/math] Thus if [math]\displaystyle{ 0 \in S }[/math] or [math]\displaystyle{ S = \varnothing }[/math] then [math]\displaystyle{ \operatorname{balcore} (c \, S) = c \operatorname{balcore} S = |c| \operatorname{balcore} S }[/math] for every scalar [math]\displaystyle{ c. }[/math]

Related notions

A function [math]\displaystyle{ p : X \to [0, \infty) }[/math] on a real or complex vector space is said to be a balanced function if it satisfies any of the following equivalent conditions:

1. [math]\displaystyle{ p(a x) \leq p(x) }[/math] whenever [math]\displaystyle{ a }[/math] is a scalar satisfying [math]\displaystyle{ |a| \leq 1 }[/math] and [math]\displaystyle{ x \in X. }[/math]
2. [math]\displaystyle{ p(a x) \leq p(b x) }[/math] whenever [math]\displaystyle{ a }[/math] and [math]\displaystyle{ b }[/math] are scalars satisfying [math]\displaystyle{ |a| \leq |b| }[/math] and [math]\displaystyle{ x \in X. }[/math]
3. [math]\displaystyle{ \{x \in X : p(x) \leq t\} }[/math] is a balanced set for every non-negative real [math]\displaystyle{ t \geq 0. }[/math]

If [math]\displaystyle{ p }[/math] is a balanced function then [math]\displaystyle{ p(a x) = p(|a| x) }[/math] for every scalar [math]\displaystyle{ a }[/math] and vector [math]\displaystyle{ x \in X; }[/math] so in particular, [math]\displaystyle{ p(u x) = p(x) }[/math] for every unit length scalar [math]\displaystyle{ u }[/math] (satisfying [math]\displaystyle{ |u| = 1 }[/math]) and every [math]\displaystyle{ x \in X. }[/math] Using [math]\displaystyle{ u := -1 }[/math] shows that every balanced function is a symmetric function.

A real-valued function [math]\displaystyle{ p : X \to \R }[/math] is a seminorm if and only if it is a balanced sublinear function.

See also

• Absolutely convex set
• Absorbing set – Set that can be "inflated" to reach any point
• Bounded set (topological vector space) – Generalization of boundedness
• Convex set – In geometry, set whose intersection with every line is a single line segment
• Star domain – Property of point sets in Euclidean spaces
• Symmetric set – Property of group subsets (mathematics)
• Topological vector space – Vector space with a notion of nearness

References

Proofs

Sources

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• Template:Dunford Schwartz Linear Operators Part 1 General Theory
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