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Balanced set - Wikipedia

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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} } 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.

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.}[1]
  5. For every {\displaystyle s\in S,} {\displaystyle S\cap \mathbb {K} s=B_{\leq 1}(S\cap \mathbb {K} s).}
  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).

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

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

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

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

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

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

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

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

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

  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)\}}).

Sufficient conditions

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

Balanced neighborhoods

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In any topological vector space, the closure of a balanced set is balanced.[5] The union of the origin {\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\}} is a balanced subset of {\displaystyle X=\mathbb {C} ^{2}} that contains the origin {\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} denotes the convex hull of {\displaystyle (0,0)} and {\displaystyle (\pm 1,1)} (a filled triangle whose vertices are these three points) then {\displaystyle B:=T\cup (-T)} is an (hour glass shaped) balanced subset of {\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} 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} 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,} the symmetric set {\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}) whenever this is true of {\displaystyle W.} It will be a balanced set if {\displaystyle W} is a star shaped at the origin,[note 2] which is true, for instance, when {\displaystyle W} is convex and contains {\displaystyle 0.} In particular, if {\displaystyle W} is a convex neighborhood of the origin then {\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} is a convex and absorbing subset of {\displaystyle X.} Then {\displaystyle D:=\bigcap _{|u|=1}uW} will be convex balanced absorbing subset of {\displaystyle X,} which guarantees that the Minkowski functional {\displaystyle p_{D}:X\to \mathbb {R} } of {\displaystyle D} will be a seminorm on {\displaystyle X,} thereby making {\displaystyle \left(X,p_{D}\right)} into a seminormed space that carries its canonical pseduometrizable topology. The set of scalar multiples {\displaystyle rD} as {\displaystyle r} ranges over {\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} as a limit point) forms a neighborhood basis of absorbing disks at the origin for this locally convex topology. If {\displaystyle X} is a topological vector space and if this convex absorbing subset {\displaystyle W} is also a bounded subset of {\displaystyle X,} then the same will be true of the absorbing disk {\displaystyle D:={\textstyle \bigcap \limits _{|u|=1}}uW;} if in addition {\displaystyle D} does not contain any non-trivial vector subspace then {\displaystyle p_{D}} will be a norm and {\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} 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} is a balanced subset of {\displaystyle X} then:

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.[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} 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.}

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:[9]

  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.}[9] 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.

  1. ^ a b Swartz 1992, pp. 4–8.
  2. ^ a b Narici & Beckenstein 2011, pp. 107–110.
  3. ^ Jarchow 1981, p. 34.
  4. ^ Narici & Beckenstein 2011, pp. 156–175.
  5. ^ a b c Rudin 1991, pp. 10–14.
  6. ^ Rudin 1991, p. 38.
  7. ^ Narici & Beckenstein 2011, pp. 115–154.
  8. ^ Trèves 2006, p. 56.
  9. ^ a b Schechter 1996, p. 313.

Proofs