Balanced set

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

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

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 ${\displaystyle X}$ be a vector space over the field ${\displaystyle \mathbb {K} }$ of real or complex numbers.

Notation

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

Balanced set

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

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

If ${\displaystyle S}$ is a convex set then this list may be extended to include:

9. ${\displaystyle aS\subseteq S}$ for all scalars ${\displaystyle a}$ satisfying ${\displaystyle |a|=1.}$

If ${\displaystyle \mathbb {K} =\mathbb {R} }$ then this list may be extended to include:

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

Examples

The empty set is a balanced set. As is any vector subspace of any (real or complex) vector space. In particular, ${\displaystyle \{0\}}$ 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 ${\displaystyle p}$ is a seminorm (or norm) on a vector space ${\displaystyle X}$ then for any constant ${\displaystyle c>0,}$ the set ${\displaystyle \{x\in X:p(x)\leq c\}}$ is balanced.

If ${\displaystyle S\subseteq X}$ is any subset and ${\displaystyle B_{1}:=\{a\in \mathbb {K}$ :|a|<1\}} then ${\displaystyle B_{1}S}$ is a balanced set. In particular, if ${\displaystyle U\subseteq X}$ is any balanced neighborhood of the origin in a topological vector space ${\displaystyle X}$ then $${\displaystyle \operatorname {Int} _{X}U~\subseteq ~B_{1}U~=~\bigcup _{0<|a|<1}aU~\subseteq ~U.}$$

Balanced sets in ${\displaystyle \mathbb {R} }$ and ${\displaystyle \mathbb {C} }$

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

The balanced subsets of ${\displaystyle X=\mathbb {K} }$ are exactly the following:

1. ${\displaystyle \varnothing }$
2. ${\displaystyle X}$
3. ${\displaystyle \{0\}}$
4. ${\displaystyle \{x\in X:|x|<r\}}$ for some real ${\displaystyle r>0}$
5. ${\displaystyle \{x\in X:|x|\leq r\}}$ for some real ${\displaystyle r>0.}$

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 ${\displaystyle \mathbb {C} }$ 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, ${\displaystyle \mathbb {C} }$ and ${\displaystyle \mathbb {R} ^{2}}$ are entirely different as far as scalar multiplication is concerned.

Balanced sets in ${\displaystyle \mathbb {R} ^{2}}$

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

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

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

For every ${\displaystyle 0\leq t\leq \pi ,}$ let ${\displaystyle r_{t}}$ be any positive real number and let ${\displaystyle B^{t}}$ be the (open or closed) line segment in ${\displaystyle X:=\mathbb {R} ^{2}}$ between the points ${\displaystyle (\cos t,\sin t)}$ and ${\displaystyle -(\cos t,\sin t).}$ Then the set ${\displaystyle B=\bigcup _{0\leq t<\pi }r_{t}B^{t}}$ 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 ${\displaystyle xy=1}$ in ${\displaystyle X=\mathbb {R} ^{2}.}$

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

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 ${\displaystyle B}$ is a balanced subset of ${\displaystyle X}$ then:

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

Properties of balanced hulls and balanced cores

For any collection ${\displaystyle {\mathcal {S}}}$ of subsets of ${\displaystyle X,}$ $${\displaystyle \operatorname {bal} \left(\bigcup _{S\in {\mathcal {S}}}S\right)=\bigcup _{S\in {\mathcal {S}}}\operatorname {bal} S\quad {\text{ and }}\quad \operatorname {balcore} \left(\bigcap _{S\in {\mathcal {S}}}S\right)=\bigcap _{S\in {\mathcal {S}}}\operatorname {balcore} S.}$$

In any topological vector space, the balanced hull of any open neighborhood of the origin is again open. If ${\displaystyle X}$ is a Hausdorff topological vector space and if ${\displaystyle K}$ is a compact subset of ${\displaystyle X}$ then the balanced hull of ${\displaystyle K}$ 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 ${\displaystyle S\subseteq X}$ and any scalar ${\displaystyle c,}$ ${\displaystyle \operatorname {bal} (c\,S)=c\operatorname {bal} S=|c|\operatorname {bal} S.}$

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

Related notions

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

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

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

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

See also

• Absolutely convex set – Convex and balanced 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