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. ${\displaystyle S=B_{\leq 1}S.}$ ^[1]
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.^{[note 1]}

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 ${\displaystyle |a|=1.}$ ^[2]

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

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

Balanced hull

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

 

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]

Balanced core

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

 

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 ${\displaystyle 0\in S.}$ 

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:^[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

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 ${\displaystyle \mathbb {K} }$ ).

• 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 ${\displaystyle L:X\to Y}$  is a linear map and ${\displaystyle B\subseteq X}$  and ${\displaystyle C\subseteq Y}$  are balanced sets, then ${\displaystyle L(B)}$  and ${\displaystyle L^{-1}(C)}$  are balanced sets.

Balanced neighborhoods

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]

Proof

Let ${\displaystyle 0\in W\subseteq X}$  and define ${\displaystyle A=\bigcap _{|u|=1}uW}$  (where ${\displaystyle u}$  denotes elements of the field ${\displaystyle \mathbb {K} }$  of scalars). Taking ${\displaystyle u:=1}$  shows that ${\displaystyle A\subseteq W.}$  If ${\displaystyle W}$  is convex then so is ${\displaystyle A}$  (since an intersection of convex sets is convex) and thus so is ${\displaystyle A}$ 's interior. If ${\displaystyle |s|=1}$  then

$${\displaystyle sA=\bigcap _{|u|=1}suW\subseteq \bigcap _{|u|=1}uW=A}$$

 

and thus ${\displaystyle sA=A.}$  If ${\displaystyle W}$  is star shaped at the origin^{[note 2]} then so is every ${\displaystyle uW}$  (for ${\displaystyle |u|=1}$ ), which implies that for any ${\displaystyle 0\leq r\leq 1,}$ 

$${\displaystyle rA=\bigcap _{|u|=1}ruW\subseteq \bigcap _{|u|=1}uW=A}$$

 

thus proving that ${\displaystyle A}$  is balanced. If ${\displaystyle W}$  is convex and contains the origin then it is star shaped at the origin and so ${\displaystyle A}$  will be balanced.

Now suppose ${\displaystyle W}$  is a neighborhood of the origin in ${\displaystyle X.}$  Since scalar multiplication ${\displaystyle M:\mathbb {K} \times X\to X}$  (defined by ${\displaystyle M(a,x)=ax}$ ) is continuous at the origin ${\displaystyle (0,0)\in \mathbb {K} \times X}$  and ${\displaystyle M(0,0)=0\in W,}$  there exists some basic open neighborhood ${\displaystyle B_{r}\times V}$  (where ${\displaystyle r>0}$  and ${\displaystyle B_{r}:=\{c\in \mathbb {K} :|c|<r\}}$ ) of the origin in the product topology on ${\displaystyle \mathbb {K} \times X}$  such that ${\displaystyle M\left(B_{r}\times V\right)\subseteq W;}$  the set ${\displaystyle M\left(B_{r}\times V\right)=B_{r}V}$  is balanced and it is also open because it may be written as

$${\displaystyle B_{r}V=\bigcup _{|a|<r}aV=\bigcup _{0<|a|<r}aV\qquad {\text{ (since }}0\cdot V=\{0\}\subseteq aV{\text{ )}}}$$

 

where ${\displaystyle aV}$  is an open neighborhood of the origin whenever ${\displaystyle a\neq 0.}$  Finally,

$${\displaystyle A=\bigcap _{|u|=1}uW\supseteq \bigcap _{|u|=1}uB_{r}V=\bigcap _{|u|=1}B_{r}V=B_{r}V}$$

 

shows that ${\displaystyle A}$  is also a neighborhood of the origin. If ${\displaystyle A}$  is balanced then because its interior ${\displaystyle \operatorname {Int} _{X}A}$  contains the origin, ${\displaystyle \operatorname {Int} _{X}A}$  will also be balanced. If ${\displaystyle W}$  is convex then ${\displaystyle A}$  is convex and balanced and thus the same is true of ${\displaystyle \operatorname {Int} _{X}A.}$  ${\displaystyle \blacksquare }$ 

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

See also: Topological vector space § 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 ${\displaystyle x\in rB.}$ ^[2]
• 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.^[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 ${\displaystyle c.}$ 

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:^[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.

See also

• Absolutely convex set – convex and balanced setPages displaying wikidata descriptions as a fallback
• 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

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

1. Assuming that all vector spaces containing a set ${\displaystyle S}$  are over the same field, when describing the set as being "balanced", it is not necessary to mention a vector space containing ${\displaystyle S.}$  That is, "${\displaystyle S}$  is balanced" may be written in place of "${\displaystyle S}$  is a balanced subset of ${\displaystyle X}$ ".
2. ${\displaystyle W}$  being star shaped at the origin means that ${\displaystyle 0\in W}$  and ${\displaystyle rw\in W}$  for all ${\displaystyle 0\leq r\leq 1}$  and ${\displaystyle w\in W.}$ 

Proofs

1. Let ${\displaystyle B\subseteq X}$  be balanced. If its topological interior ${\displaystyle \operatorname {Int} _{X}B}$  is empty then it is balanced so assume otherwise and let ${\displaystyle |s|\leq 1}$  be a scalar. If ${\displaystyle s\neq 0}$  then the map ${\displaystyle X\to X}$  defined by ${\displaystyle x\mapsto sx}$  is a homeomorphism, which implies ${\displaystyle s\operatorname {Int} _{X}B=\operatorname {Int} _{X}(sB)\subseteq sB\subseteq B;}$  because ${\displaystyle s\operatorname {Int} _{X}B}$  is open, ${\displaystyle s\operatorname {Int} _{X}B\subseteq \operatorname {Int} _{X}B}$  so that it only remains to show that this is true for ${\displaystyle s=0.}$  However, ${\displaystyle 0\in \operatorname {Int} _{X}B}$  might not be true but when it is true then ${\displaystyle \operatorname {Int} _{X}B}$  will be balanced. ${\displaystyle \blacksquare }$ 

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