Balanced set

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

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

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

Notation

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

Balanced set

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

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

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

9. ${\displaystyle aS\subseteq S}${\displaystyle aS\subseteq S} for all scalars ${\displaystyle a}${\displaystyle a} satisfying ${\displaystyle |a|=1.}${\displaystyle |a|=1.}^[2]

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

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

$${\displaystyle \operatorname {bal} S~=~\bigcup _{|a|\leq 1}aS=B_{\leq 1}S}$${\displaystyle \operatorname {bal} S~=~\bigcup _{|a|\leq 1}aS=B_{\leq 1}S}

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

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

$${\displaystyle \operatorname {balcore} S~=~{\begin{cases}\displaystyle \bigcap _{|a|\geq 1}aS&{\text{ if }}0\in S\\\varnothing &{\text{ if }}0\not \in S\\\end{cases}}}$${\displaystyle \operatorname {balcore} S~=~{\begin{cases}\displaystyle \bigcap _{|a|\geq 1}aS&{\text{ if }}0\in S\\\varnothing &{\text{ if }}0\not \in S\\\end{cases}}}

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

1. Definition: ${\displaystyle \operatorname {balcore} S}${\displaystyle \operatorname {balcore} S} is the largest (with respect to ${\displaystyle ,円\subseteq ,円}${\displaystyle ,円\subseteq ,円}) balanced subset of ${\displaystyle S.}${\displaystyle S.}
2. ${\displaystyle \operatorname {balcore} S}${\displaystyle \operatorname {balcore} S} is the union of all balanced subsets of ${\displaystyle S.}${\displaystyle S.}
3. ${\displaystyle \operatorname {balcore} S=\varnothing }${\displaystyle \operatorname {balcore} S=\varnothing } if ${\displaystyle 0\not \in S}${\displaystyle 0\not \in S} while ${\displaystyle \operatorname {balcore} S=\bigcap _{|a|\geq 1}(aS)}${\displaystyle \operatorname {balcore} S=\bigcap _{|a|\geq 1}(aS)} if ${\displaystyle 0\in S.}${\displaystyle 0\in S.}

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

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

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

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

The balanced subsets of ${\displaystyle X=\mathbb {K} }${\displaystyle X=\mathbb {K} } are exactly the following:^[3]

1. ${\displaystyle \varnothing }${\displaystyle \varnothing }
2. ${\displaystyle X}${\displaystyle X}
3. ${\displaystyle \{0\}}${\displaystyle \{0\}}
4. ${\displaystyle \{x\in X:|x|<r\}}${\displaystyle \{x\in X:|x|<r\}} for some real ${\displaystyle r>0}${\displaystyle r>0}
5. ${\displaystyle \{x\in X:|x|\leq r\}}${\displaystyle \{x\in X:|x|\leq r\}} for some real ${\displaystyle r>0.}${\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 ${\displaystyle \mathbb {C} }${\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, ${\displaystyle \mathbb {C} }${\displaystyle \mathbb {C} } and ${\displaystyle \mathbb {R} ^{2}}${\displaystyle \mathbb {R} ^{2}} are entirely different as far as scalar multiplication is concerned.

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

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

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

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

For every ${\displaystyle 0\leq t\leq \pi ,}${\displaystyle 0\leq t\leq \pi ,} let ${\displaystyle r_{t}}${\displaystyle r_{t}} be any positive real number and let ${\displaystyle B^{t}}${\displaystyle B^{t}} be the (open or closed) line segment in ${\displaystyle X:=\mathbb {R} ^{2}}${\displaystyle X:=\mathbb {R} ^{2}} between the points ${\displaystyle (\cos t,\sin t)}${\displaystyle (\cos t,\sin t)} and ${\displaystyle -(\cos t,\sin t).}${\displaystyle -(\cos t,\sin t).} Then the set ${\displaystyle B=\bigcup _{0\leq t<\pi }r_{t}B^{t}}${\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 ${\displaystyle xy=1}${\displaystyle xy=1} in ${\displaystyle X=\mathbb {R} ^{2}.}${\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 ${\displaystyle S:=[-1,1]\times \{1\},}${\displaystyle S:=[-1,1]\times \{1\},} which is a horizontal closed line segment lying above the ${\displaystyle x-}${\displaystyle x-}axis in ${\displaystyle X:=\mathbb {R} ^{2}.}${\displaystyle X:=\mathbb {R} ^{2}.} The balanced hull ${\displaystyle \operatorname {bal} S}${\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 ${\displaystyle T_{1}}${\displaystyle T_{1}} and ${\displaystyle T_{2},}${\displaystyle T_{2},} where ${\displaystyle T_{2}=-T_{1}}${\displaystyle T_{2}=-T_{1}} and ${\displaystyle T_{1}}${\displaystyle T_{1}} is the filled triangle whose vertices are the origin together with the endpoints of ${\displaystyle S}${\displaystyle S} (said differently, ${\displaystyle T_{1}}${\displaystyle T_{1}} is the convex hull of ${\displaystyle S\cup \{(0,0)\}}${\displaystyle S\cup \{(0,0)\}} while ${\displaystyle T_{2}}${\displaystyle T_{2}} is the convex hull of ${\displaystyle (-S)\cup \{(0,0)\}}${\displaystyle (-S)\cup \{(0,0)\}}).

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

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

• The balanced hull of a compact (respectively, totally bounded, bounded) set has the same property.^[4]
• 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 ${\displaystyle L:X\to Y}${\displaystyle L:X\to Y} is a linear map and ${\displaystyle B\subseteq X}${\displaystyle B\subseteq X} and ${\displaystyle C\subseteq Y}${\displaystyle C\subseteq Y} are balanced sets, then ${\displaystyle L(B)}${\displaystyle L(B)} and ${\displaystyle L^{-1}(C)}${\displaystyle L^{-1}(C)} are balanced sets.

In any topological vector space, the closure of a balanced set is balanced.^[5] The union of the origin ${\displaystyle \{0\}}${\displaystyle \{0\}} 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, ${\displaystyle \left\{(z,w)\in \mathbb {C} ^{2}:|z|\leq |w|\right\}}${\displaystyle \left\{(z,w)\in \mathbb {C} ^{2}:|z|\leq |w|\right\}} is a balanced subset of ${\displaystyle X=\mathbb {C} ^{2}}${\displaystyle X=\mathbb {C} ^{2}} that contains the origin ${\displaystyle (0,0)\in X}${\displaystyle (0,0)\in X} but whose (nonempty) topological interior does not contain the origin and is therefore not a balanced set.^[6] Similarly for real vector spaces, if ${\displaystyle T}${\displaystyle T} denotes the convex hull of ${\displaystyle (0,0)}${\displaystyle (0,0)} and ${\displaystyle (\pm 1,1)}${\displaystyle (\pm 1,1)} (a filled triangle whose vertices are these three points) then ${\displaystyle B:=T\cup (-T)}${\displaystyle B:=T\cup (-T)} is an (hour glass shaped) balanced subset of ${\displaystyle X:=\mathbb {R} ^{2}}${\displaystyle X:=\mathbb {R} ^{2}} whose non-empty topological interior does not contain the origin and so is not a balanced set (and although the set ${\displaystyle \{(0,0)\}\cup \operatorname {Int} _{X}B}${\displaystyle \{(0,0)\}\cup \operatorname {Int} _{X}B} 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 ${\displaystyle X}${\displaystyle X} contains a balanced (respectively, convex and balanced) open neighborhood of the origin. In fact, the following construction produces such balanced sets. Given ${\displaystyle W\subseteq X,}${\displaystyle W\subseteq X,} the symmetric set ${\displaystyle \bigcap _{|u|=1}uW\subseteq W}${\displaystyle \bigcap _{|u|=1}uW\subseteq W} will be convex (respectively, closed, balanced, bounded, a neighborhood of the origin, an absorbing subset of ${\displaystyle X}${\displaystyle X}) whenever this is true of ${\displaystyle W.}${\displaystyle W.} It will be a balanced set if ${\displaystyle W}${\displaystyle W} is a star shaped at the origin,^{[note 2]} which is true, for instance, when ${\displaystyle W}${\displaystyle W} is convex and contains ${\displaystyle 0.}${\displaystyle 0.} In particular, if ${\displaystyle W}${\displaystyle W} is a convex neighborhood of the origin then ${\displaystyle \bigcap _{|u|=1}uW}${\displaystyle \bigcap _{|u|=1}uW} 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]

Suppose that ${\displaystyle W}${\displaystyle W} is a convex and absorbing subset of ${\displaystyle X.}${\displaystyle X.} Then ${\displaystyle D:=\bigcap _{|u|=1}uW}${\displaystyle D:=\bigcap _{|u|=1}uW} will be convex balanced absorbing subset of ${\displaystyle X,}${\displaystyle X,} which guarantees that the Minkowski functional ${\displaystyle p_{D}:X\to \mathbb {R} }${\displaystyle p_{D}:X\to \mathbb {R} } of ${\displaystyle D}${\displaystyle D} will be a seminorm on ${\displaystyle X,}${\displaystyle X,} thereby making ${\displaystyle \left(X,p_{D}\right)}${\displaystyle \left(X,p_{D}\right)} into a seminormed space that carries its canonical pseduometrizable topology. The set of scalar multiples ${\displaystyle rD}${\displaystyle rD} as ${\displaystyle r}${\displaystyle r} ranges over ${\displaystyle \left\{{\tfrac {1}{2}},{\tfrac {1}{3}},{\tfrac {1}{4}},\ldots \right\}}${\displaystyle \left\{{\tfrac {1}{2}},{\tfrac {1}{3}},{\tfrac {1}{4}},\ldots \right\}} (or over any other set of non-zero scalars having ${\displaystyle 0}${\displaystyle 0} as a limit point) forms a neighborhood basis of absorbing disks at the origin for this locally convex topology. If ${\displaystyle X}${\displaystyle X} is a topological vector space and if this convex absorbing subset ${\displaystyle W}${\displaystyle W} is also a bounded subset of ${\displaystyle X,}${\displaystyle X,} then the same will be true of the absorbing disk ${\displaystyle D:={\textstyle \bigcap \limits _{|u|=1}}uW;}${\displaystyle D:={\textstyle \bigcap \limits _{|u|=1}}uW;} if in addition ${\displaystyle D}${\displaystyle D} does not contain any non-trivial vector subspace then ${\displaystyle p_{D}}${\displaystyle p_{D}} will be a norm and ${\displaystyle \left(X,p_{D}\right)}${\displaystyle \left(X,p_{D}\right)} will form what is known as an auxiliary normed space.^[7] If this normed space is a Banach space then ${\displaystyle D}${\displaystyle D} 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 ${\displaystyle B}${\displaystyle B} is a balanced subset of ${\displaystyle X}${\displaystyle X} then:

• for any scalars ${\displaystyle c}${\displaystyle c} and ${\displaystyle d,}${\displaystyle d,} if ${\displaystyle |c|\leq |d|}${\displaystyle |c|\leq |d|} then ${\displaystyle cB\subseteq dB}${\displaystyle cB\subseteq dB} and ${\displaystyle cB=|c|B.}${\displaystyle cB=|c|B.} Thus if ${\displaystyle c}${\displaystyle c} and ${\displaystyle d}${\displaystyle d} are any scalars then ${\displaystyle (cB)\cap (dB)=\min _{}\{|c|,|d|\}B.}${\displaystyle (cB)\cap (dB)=\min _{}\{|c|,|d|\}B.}
• ${\displaystyle B}${\displaystyle B} is absorbing in ${\displaystyle X}${\displaystyle X} if and only if for all ${\displaystyle x\in X,}${\displaystyle x\in X,} there exists ${\displaystyle r>0}${\displaystyle r>0} such that ${\displaystyle x\in rB.}${\displaystyle x\in rB.}^[2]
• for any 1-dimensional vector subspace ${\displaystyle Y}${\displaystyle Y} of ${\displaystyle X,}${\displaystyle X,} the set ${\displaystyle B\cap Y}${\displaystyle B\cap Y} is convex and balanced. If ${\displaystyle B}${\displaystyle B} is not empty and if ${\displaystyle Y}${\displaystyle Y} is a 1-dimensional vector subspace of ${\displaystyle \operatorname {span} B}${\displaystyle \operatorname {span} B} then ${\displaystyle B\cap Y}${\displaystyle B\cap Y} is either ${\displaystyle \{0\}}${\displaystyle \{0\}} or else it is absorbing in ${\displaystyle Y.}${\displaystyle Y.}
• for any ${\displaystyle x\in X,}${\displaystyle x\in X,} if ${\displaystyle B\cap \operatorname {span} x}${\displaystyle B\cap \operatorname {span} x} contains more than one point then it is a convex and balanced neighborhood of ${\displaystyle 0}${\displaystyle 0} in the 1-dimensional vector space ${\displaystyle \operatorname {span} x}${\displaystyle \operatorname {span} x} when this space is endowed with the Hausdorff Euclidean topology; and the set ${\displaystyle B\cap \mathbb {R} x}${\displaystyle B\cap \mathbb {R} x} is a convex balanced subset of the real vector space ${\displaystyle \mathbb {R} x}${\displaystyle \mathbb {R} x} that contains the origin.

Properties of balanced hulls and balanced cores

For any collection ${\displaystyle {\mathcal {S}}}${\displaystyle {\mathcal {S}}} of subsets of ${\displaystyle X,}${\displaystyle X,} $${\displaystyle \operatorname {bal} \left(\bigcup _{S\in {\mathcal {S}}}S\right)=\bigcup _{S\in {\mathcal {S}}}\operatorname {bal} S\quad {\text{ and }}\quad \operatorname {balcore} \left(\bigcap _{S\in {\mathcal {S}}}S\right)=\bigcap _{S\in {\mathcal {S}}}\operatorname {balcore} S.}$${\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 ${\displaystyle X}${\displaystyle X} is a Hausdorff topological vector space and if ${\displaystyle K}${\displaystyle K} is a compact subset of ${\displaystyle X}${\displaystyle X} then the balanced hull of ${\displaystyle K}${\displaystyle K} 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 ${\displaystyle S\subseteq X}${\displaystyle S\subseteq X} and any scalar ${\displaystyle c,}${\displaystyle c,} ${\displaystyle \operatorname {bal} (c,円S)=c\operatorname {bal} S=|c|\operatorname {bal} S.}${\displaystyle \operatorname {bal} (c,円S)=c\operatorname {bal} S=|c|\operatorname {bal} S.}

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

A function ${\displaystyle p:X\to [0,\infty )}${\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:^[9]

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

If ${\displaystyle p}${\displaystyle p} is a balanced function then ${\displaystyle p(ax)=p(|a|x)}${\displaystyle p(ax)=p(|a|x)} for every scalar ${\displaystyle a}${\displaystyle a} and vector ${\displaystyle x\in X;}${\displaystyle x\in X;} so in particular, ${\displaystyle p(ux)=p(x)}${\displaystyle p(ux)=p(x)} for every unit length scalar ${\displaystyle u}${\displaystyle u} (satisfying ${\displaystyle |u|=1}${\displaystyle |u|=1}) and every ${\displaystyle x\in X.}${\displaystyle x\in X.}^[9] Using ${\displaystyle u:=-1}${\displaystyle u:=-1} shows that every balanced function is a symmetric function.

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

• 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

Proofs

• Bourbaki, Nicolas (1987) [1981]. Topological Vector Spaces: Chapters 1–5. Éléments de mathématique. Translated by Eggleston, H.G.; Madan, S. Berlin New York: Springer-Verlag. ISBN 3-540-13627-4. OCLC 17499190.
• Conway, John (1990). A course in functional analysis. Graduate Texts in Mathematics. Vol. 96 (2nd ed.). New York: Springer-Verlag. ISBN 978-0-387-97245-9. OCLC 21195908.
• Dunford, Nelson; Schwartz, Jacob T. (1988). Linear Operators . Pure and applied mathematics. Vol. 1. New York: Wiley-Interscience. ISBN 978-0-471-60848-6. OCLC 18412261.
• Edwards, Robert E. (1995). Functional Analysis: Theory and Applications. New York: Dover Publications. ISBN 978-0-486-68143-6. OCLC 30593138.
• Jarchow, Hans (1981). Locally convex spaces. Stuttgart: B.G. Teubner. ISBN 978-3-519-02224-4. OCLC 8210342.
• Köthe, Gottfried (1983) [1969]. Topological Vector Spaces I. Grundlehren der mathematischen Wissenschaften. Vol. 159. Translated by Garling, D.J.H. New York: Springer Science & Business Media. ISBN 978-3-642-64988-2. MR 0248498. OCLC 840293704.
• Köthe, Gottfried (1979). Topological Vector Spaces II. Grundlehren der mathematischen Wissenschaften. Vol. 237. New York: Springer Science & Business Media. ISBN 978-0-387-90400-9. OCLC 180577972.
• Narici, Lawrence; Beckenstein, Edward (2011). Topological Vector Spaces. Pure and applied mathematics (Second ed.). Boca Raton, FL: CRC Press. ISBN 978-1584888666. OCLC 144216834.
• Robertson, Alex P.; Robertson, Wendy J. (1980). Topological Vector Spaces. Cambridge Tracts in Mathematics. Vol. 53. Cambridge England: Cambridge University Press. ISBN 978-0-521-29882-7. OCLC 589250.
• Rudin, Walter (1991). Functional Analysis. International Series in Pure and Applied Mathematics. Vol. 8 (Second ed.). New York, NY: McGraw-Hill Science/Engineering/Math. ISBN 978-0-07-054236-5. OCLC 21163277.
• Schaefer, Helmut H.; Wolff, Manfred P. (1999). Topological Vector Spaces. GTM. Vol. 8 (Second ed.). New York, NY: Springer New York Imprint Springer. ISBN 978-1-4612-7155-0. OCLC 840278135.
• Schechter, Eric (October 24, 1996). Handbook of Analysis and Its Foundations. Academic Press. ISBN 978-0-08-053299-8.
• Swartz, Charles (1992). An introduction to Functional Analysis. New York: M. Dekker. ISBN 978-0-8247-8643-4. OCLC 24909067.
• Trèves, François (2006) [1967]. Topological Vector Spaces, Distributions and Kernels. Mineola, N.Y.: Dover Publications. ISBN 978-0-486-45352-1. OCLC 853623322.
• Wilansky, Albert (2013). Modern Methods in Topological Vector Spaces. Mineola, New York: Dover Publications, Inc. ISBN 978-0-486-49353-4. OCLC 849801114.

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