SherbyNet - /page/wikibook/entry.mu

Micron Document

Balanced set

55.3 KB · ⇊ 33 parts · ⤓ Micron 55.3 KB · ⤓ HTML 400.9 KB

──────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────────

top

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

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

Contents

• Definition

• Balanced hull

• Balanced core

• Examples

• Sufficient conditions

• Balanced neighborhoods

• Properties

• Related notions

• See also

• References

• Sources

Definition

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

Notation

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

Balanced set

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

1. Definition: a s ∈ ∈ S {\displaystyle as\in S} for all s ∈ ∈ S {\displaystyle s\in S} and all scalars a {\displaystyle a} satisfying | a | ≤ ≤ 1. {\displaystyle |a|\leq 1.}

2. a S ⊆ ⊆ S {\displaystyle aS\subseteq S} for all scalars a {\displaystyle a} satisfying | a | ≤ ≤ 1. {\displaystyle |a|\leq 1.}

3. B ≤ ≤ 1 S ⊆ ⊆ S {\displaystyle B_{\leq 1}S\subseteq S} (where B ≤ ≤ 1 := { a ∈ ∈ K : | a | ≤ ≤ 1 } {\displaystyle B_{\leq 1}:=\{a\in \mathbb {K} :|a|\leq 1\}} ).

4. S = B ≤ ≤ 1 S . {\displaystyle S=B_{\leq 1}S.} cite-ref-footnoteswartz19924-8-1-0[1]

5. For every s ∈ ∈ S , {\displaystyle s\in S,} S ∩ ∩ K s = B ≤ ≤ 1 ( S ∩ ∩ K s ) . {\displaystyle S\cap \mathbb {K} s=B_{\leq 1}(S\cap \mathbb {K} s).}

• K s = span ⁡ ⁡ { s } {\displaystyle \mathbb {K} s=\operatorname {span} \{s\}} is a 0 {\displaystyle 0} (if s = 0 {\displaystyle s=0} ) or 1 {\displaystyle 1} (if s ≠ ≠ 0 {\displaystyle s\neq 0} ) dimensional vector subspace of X . {\displaystyle X.}

• If R := S ∩ ∩ K s {\displaystyle R:=S\cap \mathbb {K} s} then the above equality becomes R = B ≤ ≤ 1 R , {\displaystyle R=B_{\leq 1}R,} which is exactly the previous condition for a set to be balanced. Thus, S {\displaystyle S} is balanced if and only if for every s ∈ ∈ S , {\displaystyle s\in S,} S ∩ ∩ 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 Y {\displaystyle Y} of span ⁡ ⁡ S , {\displaystyle \operatorname {span} S,} S ∩ ∩ Y {\displaystyle S\cap Y} is a balanced set (according to any defining condition other than this one).

7. For every s ∈ ∈ S , {\displaystyle s\in S,} there exists some 0 ≤ ≤ r ≤ ≤ ∞ ∞ {\displaystyle 0\leq r\leq \infty } such that S ∩ ∩ K s = B r s {\displaystyle S\cap \mathbb {K} s=B_{r}s} or S ∩ ∩ K s = B ≤ ≤ r s . {\displaystyle S\cap \mathbb {K} s=B_{\leq r}s.}

8. S {\displaystyle S} is a balanced subset of span ⁡ ⁡ S {\displaystyle \operatorname {span} S} (according to any defining condition of "balanced" other than this one).

• Thus S {\displaystyle S} is a balanced subset of X {\displaystyle X} if and only if it is balanced subset of every (equivalently, of some) vector space over the field K {\displaystyle \mathbb {K} } that contains S . {\displaystyle S.} So assuming that the field K {\displaystyle \mathbb {K} } is clear from context, this justifies writing " S {\displaystyle S} is balanced" without mentioning any vector space.cite-ref-2[note 1]

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

1. a S ⊆ ⊆ S {\displaystyle aS\subseteq S} for all scalars a {\displaystyle a} satisfying | a | = 1. {\displaystyle |a|=1.} cite-ref-footnotenaricibeckenstein2011107-110-3-0[2]

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

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

Balanced hull

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

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

1. Definition: bal ⁡ ⁡ S {\displaystyle \operatorname {bal} S} is the smallest (with respect to ⊆ ⊆ {\displaystyle \,\subseteq \,} ) balanced subset of X {\displaystyle X} containing S . {\displaystyle S.}

2. bal ⁡ ⁡ S {\displaystyle \operatorname {bal} S} is the intersection of all balanced sets containing S . {\displaystyle S.}

3. bal ⁡ ⁡ S = ⋃ ⋃ | a | ≤ ≤ 1 ( a S ) . {\displaystyle \operatorname {bal} S=\bigcup _{|a|\leq 1}(aS).}

4. bal ⁡ ⁡ S = B ≤ ≤ 1 S . {\displaystyle \operatorname {bal} S=B_{\leq 1}S.} cite-ref-footnoteswartz19924-8-1-1[1]

Balanced core

balcore ⁡ ⁡ S = { ⋂ ⋂ | a | ≥ ≥ 1 a S if 0 ∈ ∈ S ∅ ∅ if 0 ∉ S {\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 S {\displaystyle S} of X , {\displaystyle X,} denoted by balcore ⁡ ⁡ S , {\displaystyle \operatorname {balcore} S,} is defined in any of the following equivalent ways:

1. Definition: balcore ⁡ ⁡ S {\displaystyle \operatorname {balcore} S} is the largest (with respect to ⊆ ⊆ {\displaystyle \,\subseteq \,} ) balanced subset of S . {\displaystyle S.}

2. balcore ⁡ ⁡ S {\displaystyle \operatorname {balcore} S} is the union of all balanced subsets of S . {\displaystyle S.}

3. balcore ⁡ ⁡ S = ∅ ∅ {\displaystyle \operatorname {balcore} S=\varnothing } if 0 ∉ S {\displaystyle 0\not \in S} while balcore ⁡ ⁡ S = ⋂ ⋂ | a | ≥ ≥ 1 ( a S ) {\displaystyle \operatorname {balcore} S=\bigcap _{|a|\geq 1}(aS)} if 0 ∈ ∈ S . {\displaystyle 0\in S.}

Examples

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

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

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

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

The balanced subsets of X = K {\displaystyle X=\mathbb {K} } are exactly the following:cite-ref-footnotejarchow198134-4-0[3]

1. ∅ ∅ {\displaystyle \varnothing }

2. X {\displaystyle X}

3. { 0 } {\displaystyle \{0\}}

4. { x ∈ ∈ X : | x | < r } {\displaystyle \{x\in X:|x|<r\}} for some real r > 0 {\displaystyle r>0}

5. { x ∈ ∈ X : | x | ≤ ≤ r } {\displaystyle \{x\in X:|x|\leq r\}} for some real 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 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, C {\displaystyle \mathbb {C} } and R 2 {\displaystyle \mathbb {R} ^{2}} are entirely different as far as scalar multiplication is concerned.

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

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

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

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

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

Sufficient conditions

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

• The balanced hull of a compact (respectively, totally bounded, bounded) set has the same property.cite-ref-footnotenaricibeckenstein2011156-175-5-0[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 L : X → → Y {\displaystyle L:X\to Y} is a linear map and B ⊆ ⊆ X {\displaystyle B\subseteq X} and C ⊆ ⊆ Y {\displaystyle C\subseteq Y} are balanced sets, then L ( B ) {\displaystyle L(B)} and L − − 1 ( C ) {\displaystyle L^{-1}(C)} are balanced sets.

Balanced neighborhoods

In any topological vector space, the closure of a balanced set is balanced.cite-ref-footnoterudin199110-14-6-0[5] The union of the origin { 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.cite-ref-footnoterudin199110-14-6-1[5]cite-ref-7[proof 1] However, { ( z , w ) ∈ ∈ C 2 : | z | ≤ ≤ | w | } {\displaystyle \left\{(z,w)\in \mathbb {C} ^{2}:|z|\leq |w|\right\}} is a balanced subset of X = C 2 {\displaystyle X=\mathbb {C} ^{2}} that contains the origin ( 0 , 0 ) ∈ ∈ X {\displaystyle (0,0)\in X} but whose (nonempty) topological interior does not contain the origin and is therefore not a balanced set.cite-ref-footnoterudin199138-8-0[6] Similarly for real vector spaces, if T {\displaystyle T} denotes the convex hull of ( 0 , 0 ) {\displaystyle (0,0)} and ( ± ± 1 , 1 ) {\displaystyle (\pm 1,1)} (a filled triangle whose vertices are these three points) then B := T ∪ ∪ ( − − T ) {\displaystyle B:=T\cup (-T)} is an (hour glass shaped) balanced subset of X := 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 { ( 0 , 0 ) } ∪ ∪ 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 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 W ⊆ ⊆ X , {\displaystyle W\subseteq X,} the symmetric set ⋂ ⋂ | u | = 1 u W ⊆ ⊆ W {\displaystyle \bigcap _{|u|=1}uW\subseteq W} will be convex (respectively, closed, balanced, bounded, a neighborhood of the origin, an absorbing subset of X {\displaystyle X} ) whenever this is true of W . {\displaystyle W.} It will be a balanced set if W {\displaystyle W} is a star shaped at the origin,cite-ref-defstarshapedatorigin-9-0[note 2] which is true, for instance, when W {\displaystyle W} is convex and contains 0. {\displaystyle 0.} In particular, if W {\displaystyle W} is a convex neighborhood of the origin then ⋂ ⋂ | u | = 1 u W {\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.cite-ref-footnoterudin199110-14-6-2[5]

Proof

Let 0 ∈ ∈ W ⊆ ⊆ X {\displaystyle 0\in W\subseteq X} and define A = ⋂ ⋂ | u | = 1 u W {\displaystyle A=\bigcap _{|u|=1}uW} (where u {\displaystyle u} denotes elements of the field K {\displaystyle \mathbb {K} } of scalars). Taking u := 1 {\displaystyle u:=1} shows that A ⊆ ⊆ W . {\displaystyle A\subseteq W.} If W {\displaystyle W} is convex then so is A {\displaystyle A} (since an intersection of convex sets is convex) and thus so is A {\displaystyle A} 's interior. If | s | = 1 {\displaystyle |s|=1} then s A = ⋂ ⋂ | u | = 1 s u W ⊆ ⊆ ⋂ ⋂ | u | = 1 u W = A {\displaystyle sA=\bigcap _{|u|=1}suW\subseteq \bigcap _{|u|=1}uW=A} and thus s A = A . {\displaystyle sA=A.} If W {\displaystyle W} is star shaped at the origincite-ref-defstarshapedatorigin-9-1[note 2] then so is every u W {\displaystyle uW} (for | u | = 1 {\displaystyle |u|=1} ), which implies that for any 0 ≤ ≤ r ≤ ≤ 1 , {\displaystyle 0\leq r\leq 1,} r A = ⋂ ⋂ | u | = 1 r u W ⊆ ⊆ ⋂ ⋂ | u | = 1 u W = A {\displaystyle rA=\bigcap _{|u|=1}ruW\subseteq \bigcap _{|u|=1}uW=A} thus proving that A {\displaystyle A} is balanced. If W {\displaystyle W} is convex and contains the origin then it is star shaped at the origin and so A {\displaystyle A} will be balanced.

Now suppose W {\displaystyle W} is a neighborhood of the origin in X . {\displaystyle X.} Since scalar multiplication M : K × × X → → X {\displaystyle M:\mathbb {K} \times X\to X} (defined by M ( a , x ) = a x {\displaystyle M(a,x)=ax} ) is continuous at the origin ( 0 , 0 ) ∈ ∈ K × × X {\displaystyle (0,0)\in \mathbb {K} \times X} and M ( 0 , 0 ) = 0 ∈ ∈ W , {\displaystyle M(0,0)=0\in W,} there exists some basic open neighborhood B r × × V {\displaystyle B_{r}\times V} (where r > 0 {\displaystyle r>0} and B r := { c ∈ ∈ K : | c | < r } {\displaystyle B_{r}:=\{c\in \mathbb {K} :|c|<r\}} ) of the origin in the product topology on K × × X {\displaystyle \mathbb {K} \times X} such that M ( B r × × V ) ⊆ ⊆ W ; {\displaystyle M\left(B_{r}\times V\right)\subseteq W;} the set M ( B r × × V ) = B r V {\displaystyle M\left(B_{r}\times V\right)=B_{r}V} is balanced and it is also open because it may be written as B r V = ⋃ ⋃ | a | < r a V = ⋃ ⋃ 0 < | a | < r a V (since 0 ⋅ ⋅ V = { 0 } ⊆ ⊆ a V ) {\displaystyle B_{r}V=\bigcup _{|a|<r}aV=\bigcup _{0<|a|<r}aV\qquad {\text{ (since }}0\cdot V=\{0\}\subseteq aV{\text{ )}}} where a V {\displaystyle aV} is an open neighborhood of the origin whenever a ≠ ≠ 0. {\displaystyle a\neq 0.} Finally, A = ⋂ ⋂ | u | = 1 u W ⊇ ⊇ ⋂ ⋂ | u | = 1 u B r V = ⋂ ⋂ | u | = 1 B r V = B r V {\displaystyle A=\bigcap _{|u|=1}uW\supseteq \bigcap _{|u|=1}uB_{r}V=\bigcap _{|u|=1}B_{r}V=B_{r}V} shows that A {\displaystyle A} is also a neighborhood of the origin. If A {\displaystyle A} is balanced then because its interior Int X ⁡ ⁡ A {\displaystyle \operatorname {Int} _{X}A} contains the origin, Int X ⁡ ⁡ A {\displaystyle \operatorname {Int} _{X}A} will also be balanced. If W {\displaystyle W} is convex then A {\displaystyle A} is convex and balanced and thus the same is true of Int X ⁡ ⁡ A . {\displaystyle \operatorname {Int} _{X}A.} ◼ {\displaystyle \blacksquare }

Suppose that W {\displaystyle W} is a convex and absorbing subset of X . {\displaystyle X.} Then D := ⋂ ⋂ | u | = 1 u W {\displaystyle D:=\bigcap _{|u|=1}uW} will be convex balanced absorbing subset of X , {\displaystyle X,} which guarantees that the Minkowski functional p D : X → → R {\displaystyle p_{D}:X\to \mathbb {R} } of D {\displaystyle D} will be a seminorm on X , {\displaystyle X,} thereby making ( X , p D ) {\displaystyle \left(X,p_{D}\right)} into a seminormed space that carries its canonical pseduometrizable topology. The set of scalar multiples r D {\displaystyle rD} as r {\displaystyle r} ranges over { 1 2 , 1 3 , 1 4 , … … } {\displaystyle \left\{{\tfrac {1}{2}},{\tfrac {1}{3}},{\tfrac {1}{4}},\ldots \right\}} (or over any other set of non-zero scalars having 0 {\displaystyle 0} as a limit point) forms a neighborhood basis of absorbing disks at the origin for this locally convex topology. If X {\displaystyle X} is a topological vector space and if this convex absorbing subset W {\displaystyle W} is also a bounded subset of X , {\displaystyle X,} then the same will be true of the absorbing disk D := ⋂ ⋂ | u | = 1 u W ; {\displaystyle D:={\textstyle \bigcap \limits _{|u|=1}}uW;} if in addition D {\displaystyle D} does not contain any non-trivial vector subspace then p D {\displaystyle p_{D}} will be a norm and ( X , p D ) {\displaystyle \left(X,p_{D}\right)} will form what is known as an auxiliary normed space.cite-ref-footnotenaricibeckenstein2011115-154-10-0[7] If this normed space is a Banach space then D {\displaystyle D} 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 B {\displaystyle B} is a balanced subset of X {\displaystyle X} then:

• for any scalars c {\displaystyle c} and d , {\displaystyle d,} if | c | ≤ ≤ | d | {\displaystyle |c|\leq |d|} then c B ⊆ ⊆ d B {\displaystyle cB\subseteq dB} and c B = | c | B . {\displaystyle cB=|c|B.} Thus if c {\displaystyle c} and d {\displaystyle d} are any scalars then ( c B ) ∩ ∩ ( d B ) = min { | c | , | d | } B . {\displaystyle (cB)\cap (dB)=\min _{}\{|c|,|d|\}B.}

• B {\displaystyle B} is absorbing in X {\displaystyle X} if and only if for all x ∈ ∈ X , {\displaystyle x\in X,} there exists r > 0 {\displaystyle r>0} such that x ∈ ∈ r B . {\displaystyle x\in rB.} cite-ref-footnotenaricibeckenstein2011107-110-3-1[2]

• for any 1-dimensional vector subspace Y {\displaystyle Y} of X , {\displaystyle X,} the set B ∩ ∩ Y {\displaystyle B\cap Y} is convex and balanced. If B {\displaystyle B} is not empty and if Y {\displaystyle Y} is a 1-dimensional vector subspace of span ⁡ ⁡ B {\displaystyle \operatorname {span} B} then B ∩ ∩ Y {\displaystyle B\cap Y} is either { 0 } {\displaystyle \{0\}} or else it is absorbing in Y . {\displaystyle Y.}

• for any x ∈ ∈ X , {\displaystyle x\in X,} if B ∩ ∩ span ⁡ ⁡ x {\displaystyle B\cap \operatorname {span} x} contains more than one point then it is a convex and balanced neighborhood of 0 {\displaystyle 0} in the 1-dimensional vector space span ⁡ ⁡ x {\displaystyle \operatorname {span} x} when this space is endowed with the Hausdorff Euclidean topology; and the set B ∩ ∩ R x {\displaystyle B\cap \mathbb {R} x} is a convex balanced subset of the real vector space R x {\displaystyle \mathbb {R} x} that contains the origin.

Properties of balanced hulls and balanced cores

For any collection S {\displaystyle {\mathcal {S}}} of subsets of X , {\displaystyle X,} bal ⁡ ⁡ ( ⋃ ⋃ S ∈ ∈ S S ) = ⋃ ⋃ S ∈ ∈ S bal ⁡ ⁡ S and balcore ⁡ ⁡ ( ⋂ ⋂ S ∈ ∈ S S ) = ⋂ ⋂ S ∈ ∈ S 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 X {\displaystyle X} is a Hausdorff topological vector space and if K {\displaystyle K} is a compact subset of X {\displaystyle X} then the balanced hull of K {\displaystyle K} is compact.cite-ref-footnotetr-ves200656-11-0[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 S ⊆ ⊆ X {\displaystyle S\subseteq X} and any scalar c , {\displaystyle c,} bal ⁡ ⁡ ( c S ) = c bal ⁡ ⁡ S = | c | bal ⁡ ⁡ S . {\displaystyle \operatorname {bal} (c\,S)=c\operatorname {bal} S=|c|\operatorname {bal} S.}

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

Related notions

A function p : X → → [ 0 , ∞ ∞ ) {\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:cite-ref-footnoteschechter1996313-12-0[9]

1. p ( a x ) ≤ ≤ p ( x ) {\displaystyle p(ax)\leq p(x)} whenever a {\displaystyle a} is a scalar satisfying | a | ≤ ≤ 1 {\displaystyle |a|\leq 1} and x ∈ ∈ X . {\displaystyle x\in X.}

2. p ( a x ) ≤ ≤ p ( b x ) {\displaystyle p(ax)\leq p(bx)} whenever a {\displaystyle a} and b {\displaystyle b} are scalars satisfying | a | ≤ ≤ | b | {\displaystyle |a|\leq |b|} and x ∈ ∈ X . {\displaystyle x\in X.}

3. { x ∈ ∈ X : p ( x ) ≤ ≤ t } {\displaystyle \{x\in X:p(x)\leq t\}} is a balanced set for every non-negative real t ≥ ≥ 0. {\displaystyle t\geq 0.}

If p {\displaystyle p} is a balanced function then p ( a x ) = p ( | a | x ) {\displaystyle p(ax)=p(|a|x)} for every scalar a {\displaystyle a} and vector x ∈ ∈ X ; {\displaystyle x\in X;} so in particular, p ( u x ) = p ( x ) {\displaystyle p(ux)=p(x)} for every unit length scalar u {\displaystyle u} (satisfying | u | = 1 {\displaystyle |u|=1} ) and every x ∈ ∈ X . {\displaystyle x\in X.} cite-ref-footnoteschechter1996313-12-1[9] Using u := − − 1 {\displaystyle u:=-1} shows that every balanced function is a symmetric function.

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